How to Compare Effectively During the GMAT

Quarter Wit, Quarter WisdomA lot of GMAT test takers complain about insufficient time. This is understandable as far as the Verbal section is concerned. We all have different reading speeds and that itself accounts for a lot of time issues in the Verbal section. Obviously then there are other factors – your comfort with the language, your comprehension skills, your conceptual understanding of the Verbal question types, etc.

However, timing issues should not arise in the Quant section. Your reading speed has very little effect on the overall timing scheme because most of the time during the Quant section is spent in solving the question. So if you are falling short on time, it means the methods you are using are not appropriate. We have said it before and will say it again – most GMAT Quant questions can be done in under one minute if you just look for the right thing.

For example, of the four listed numbers below, which number is the greatest and which is the least?

2/3

2^2/3^2

2^3/3^3

Sqrt(2)/Sqrt(3)

Now, how much time you take to solve this depends on how you approach this problem. If you get into ugly calculations, you will end up wasting a ton of time.

2/3 = .667

2^2/3^2 = 4/9 = .444

2^3/3^3 = 8/27 = .296

Sqrt(2)/Sqrt(3) = 1.414/1.732 = .816

So we know that the greatest is Sqrt(2)/Sqrt(3) and the least is 2^3/3^3. We got the answer but we wasted at least 2-3 mins in getting it.

We can do the same thing very quickly. We know that the squares/cubes/roots etc of numbers vary according to where the numbers lie on the number line.

2/3 lies in between 0 and 1, as does 1/4.

The Sqrt(1/4) = 1/2, which is greater than 1/4, so we know that the Sqrt(2/3) will be greater than 2/3 as well.

Also, the square and cube of 1/4 is less than 1/4, so the square and cube of 2/3 will also be less than 2/3. So the comparison will look like this:

(2/3)^3 < (2/3)^2 < 2/3 < Sqrt(2/3)

That is all you need to do! We arrived at the same answer using less than 30 secs.

Using this technique, let’s solve a question:

Which of the following represents the greatest value?

(A) Sqrt(3)/Sqrt(5) + Sqrt(5)/Sqrt(7) + Sqrt(7)/Sqrt(9)

(B) 3/5 + 5/7 + 7/9

(C) 3^2/5^2 + 5^2/7^2 + 7^2/9^2

(D) 3^3/5^3 + 5^3/7^3 + 7^3/9^3

(E) 3/5 + 1 – 5/7 + 7/9

Such a question can baffle someone who believes in calculating everything. We know better than that!

Note that the base values in all the options are 3/5, 5/7 and 7/9. This should hint that we need to compare term to term and not the entire expressions. Also, all values lie between 0 and 1 so they will behave the same way.

Sqrt(3)/Sqrt(5) is the same as Sqrt(3/5). The square root of a number between 0 and 1 is greater than the number itself.

3^2/5^2 is the same as (3/5)^2. The square (and cube) of a number between 0 and 1 is less than the number itself.

So, the comparison will look like this:

(3/5)^3 < (3/5)^2 < 3/5 < Sqrt(3/5)

(5/7)^3 < (5/7)^2 < 5/7 < Sqrt(5/7)

(7/9)^3 < (7/9)^2 < 7/9 < Sqrt(7/9)

This means that out of (A), (B), (C) and (D), the greatest one is (A).

Now we just need to analyse (E) and compare it with (B).

The first term is the same, 3/5.

The last term is the same, 7/9.

The only difference is that (B) has 5/7 in the middle and (E) has 1 – 5/7 = 2/7 in the middle. So (E) is certainly less than (B).

We already know that (A) is greater than (B), so we can say that (A) must be the greatest value.

A quick recap of important number properties:

Case 1: N > 1

N^2, N^3, etc. will be greater than N.

The Sqrt(N) and the CubeRoot(N) will be less than N.

The relation will look like this:

… CubeRoot(N) < Sqrt(N) < N < N^2 < N^3 …

Case II: 0 < N < 1

N^2, N^3 etc will be less than N.

The Sqrt(N) and the CubeRoot(N) will be greater than N.

The relation will look like this:

… N^3 < N^2 < N < Sqrt(N) < CubeRoot(N)  …

Case III: -1 < N < 0

Even powers will be greater than N and positive; Odd powers will be greater than N but negative.

The square root will not be defined, and the cube root of N will be less than N.

CubeRoot(N) < N < N^3 < 0 < N^2

Case IV: N < -1

Even powers will be greater than N and positive; Odd powers will be less than N.

The square root will not be defined, and the cube root of N will be greater than N.

N^3 < N < CubeRoot(N) < 0 < N^2

Note that you don’t need to actually remember these relations, just take a value in each range and you will know how all the numbers in that range behave.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Finding the Product of Factors on GMAT Questions

Quarter Wit, Quarter WisdomWe have discussed how to find the factors of a number and their properties in these two posts:

Writing Factors of an Ugly Number

Factors of Perfect Squares

Today let’s discuss the concept of ‘product of the factors of a number’.

From the two posts above, we know that the factors equidistant from the centre multiply to give the number. We also know that the behaviour is a little different for perfect squares. Let’s take two examples to understand this.

Example 1: Say N = 6

Factors of 6 are 1, 2, 3, 6

1*6 = 6 (first factor * last factor)

2*3 = 6 (second factor and second last factor)

Product of the four factors of 6 is given by 1*6 * 2*3 = 6*6 = 6^2 = [Sqrt(N)]^4

Example 2: Say N = 25 (a perfect square)

Factors of 25 are 1, 5, 25

1*25 = 25 (first factor * last factor)

5*5 = 25 (middle factor multiplied by itself)

Product of the three factors of 25 is given by 1*25 * 5 = 5^3 = [Sqrt(N)]^3

If a number, N, can be expressed as: 2^a * 3^b * 5^c *…

The total number of factors f = (a+1)*(b+1)*(c+1)…

The product of all factors of N is given by [Sqrt(N)]^f i.e. N^(f/2)

Let’s look at a couple of questions based on this principle:

Question 1: If the product of all the factors of a positive integer, N, is

2^(18) * 3^(12), how many values can N take?

(A) None

(B) 1

(C) 2

(D) 3

(E) 4

Solution: Since the product of all factors of N has only 2 and 3 as prime factors, N must have two prime factors only: 2 and 3.

Let N = 2^a * 3^b

Given that N^(f/2) = 2^(18) * 3^(12)

(2^a * 3^b)^[(a+1)(b+1)/2] = 2^(18) * 3^(12)

a*(a+1)*(b+1)/2 = 18

b*(a+1)*(b+1)/2 = 12

Dividing the two equations, we get a/b = 3/2

Smallest values: a = 3, b = 2. It satisfies our two equations.

Can we have more values for a and b? Can a = 6 and b = 4? No. Then the product a*(a+1)*(b+1)/2 would be much larger than 18.

So N = 2^3 * 3^2

There is only one such value of N.

Answer (B)

Question 2: If the product of all the factors of a positive integer, N, is 2^9 * 3^9, how many values can N take?

(A) None

(B) 1

(C) 2

(D) 3

(E) 4

Solution: Since the product of all factors of N has only 2 and 3 as prime factors, N must have two prime factors only: 2 and 3.

Let N = 2^a * 3^b

Given that N^(f/2) = 2^(9) * 3^(9)

(2^a * 3^b)^[(a+1)(b+1)/2] = 2^(9) * 3^(9)

a*(a+1)*(b+1)/2 = 9

b*(a+1)*(b+1)/2 = 9

Dividing the two equations, we get a/b = 1/1

Smallest values: a = 1, b = 1 – Does not satisfy our equation

Next set of values: a = 2, b = 2 – Satisfies our equations

All larger values will not satisfy our equations.

Answer (B)

Note that we can easily use hit and trial in these questions without actually working through the equations.

This is how we will do it:

N^(f/2) = 2^(18) * 3^(12)

Case 1: Assume values of f/2 from common factors of 18 and 12 – say 2

[2^9 * 3^6]^2

Can f/2 = 2 i.e. can f = 4?

If N = 2^9 * 3^6, total number of factors f = (9+1)*(6+1) = 70

This doesn’t work.

Case 2: Assume f/2 is 6

[2^3 * 3^2]^6

Can f/2 = 6 i.e. can f = 12?

If N = 2^3 * 3^2, total number of factors f = (3+1)*(2+1) = 12

This works.

The reason hit and trial isn’t a bad idea is that there will be only one such set of values. If we can quickly find it, we are done.

Why should we then bother to find it at all. Shouldn’t we just answer with option ‘B’ in both cases? Think of a case in which the product of all factors is given as 2^(16) * 3^(14). Will there be any value of N in such a case?

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure tofind us on Facebook and Google+, and follow us on Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

How to Solve Relative Rate of Work Questions on the GMAT

Quarter Wit, Quarter WisdomToday, we look at the relative rate concept  of work, rate and time – the parallel of relative speed of distance, speed and time.

But before we do that, we will first look at one fundamental principle of work, rate and time (which has a parallel in distance, speed and time).

Say, there is a straight long track with a red flag at one end. Mr A is standing on the track 100 feet away from the flag and Mr B is standing on the track at a distance 700 feet away from the flag. So they have a distance of 600 feet between them. They start walking towards each other. Where will they meet? Is it necessary that they will meet at 400 feet from the red flag – the mid point of the distance between them? Think about it – say Mr A walks very slowly and Mr B is super fast. Of the 600 feet between them, Mr A will cover very little distance and Mr B will cover most of the distance. So where they meet depends on their rate of walking. They will not necessarily meet at the mid point. When do they meet at the mid point? When their rate of walking is the same. When they both cover equal distance.

Now imagine that you have two pools of water. Pool A has 100 gallons of water in it and the Pool B has 700 gallons. Say, water is being pumped into pool A and water is being pumped out of pool B. When will the two pools have equal water level? Is it necessary that they both have to hit the 400 gallons mark to have equal amount of water? Again, it depends on the rate of work on the two pools. If water is being pumped into pool A very slowly but water is being pumped out of pool B very fast, at some point, they both might have 200 gallons of water in them. They will both have 400 gallons at the same time only when their rate of pumping is the same. This case is exactly like the case above.

Now let’s go on to the question from the GMAT Club tests which tests this understanding and the concept of  relative rate of work:

Question: Tanks X and Y contain 500 and 200 gallons of water respectively. If water is being pumped out of tank X at a rate of K gallons per minute and water is being added to tank Y at a rate of M gallons per minute, how many hours will elapse before the two tanks contain equal amounts of water?

(A) 5/(M+K) hours

(B) 6/(M+K) hours

(C) 300/(M+K) hours

(D) 300/(M−K) hours

(E) 60/(M−K) hours

Solution: There are two tanks with different water levels. Note that the rate of pumping is given as K gallons per min and M gallons per min i.e. they are different. So we cannot say that they both will have equal amount of water when they have 350 gallons. They could very well have equal amount of water at 300 gallons or 400 gallons etc. So when one expects that water in both tanks will be at 350 gallon level, one is making a mistake. The two tanks are working for the same time to get their level equal but their rates are different. So the work done is different. Note here that equal level does not imply equal work done. The equal level could be achieved at 300 gallons when work done would be different – 200 gallons removed from tank X and 100 gallons added to tank Y. The equal level could be achieved at 400 gallons when work done would be different again – 100 gallons removed from tank X and 200 gallons added to tank Y.

To achieve the ‘equal level,’ tank Y needs to gain water and tank X needs to lose water. Total 300 gallons (500 gallons – 200 gallons) of work needs to be done. Which tank will do how much depends on their respective rates.

Work to be done together = 300 gallons

Relative rate of work = (K + M) gallons/minute

The rates get added because they are working in opposite directions – one is removing water and the other is adding water. So we get relative rate (which is same as relative speed) by adding the individual rates.

Note here that rate is given in gallons per minute. But the options have hours so we must convert the rate to gallons per hour.

Relative rate of work = (K + M) gallons/minute = (K + M) gallons/(1/60) hour = 60*(K + M)  gallons/hour

Time taken to complete the work = 300/60(K+M) hours = 5/(K+M) hours

Answer (A)

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Identifying the Paradox on GMAT Critical Reasoning Questions

Quarter Wit, Quarter WisdomLet’s take a look at a very tricky GMAT Prep critical reasoning problem today. Problems such as these make CR more attractive than RC and SC to people who have a Quantitative bent of mind. It’s one of the “explain the paradox” problems, which usually tend to be easy if you know exactly how to tackle them, but the issue here is that it is hard to put your finger on the paradox.

Once you do, then the problem is quite easy.

 

Question: Technological improvements and reduced equipment costs have made converting solar energy directly into electricity far more cost-efficient in the last decade. However, the threshold of economic viability for solar power (that is, the price per barrel to which oil would have to rise in order for new solar power plants to be more economical than new oil-fired power plants) is unchanged at thirty-five dollars.

Which of the following, if true, does most to help explain why the increased cost-efficiency of solar power has not decreased its threshold of economic viability?

(A) The cost of oil has fallen dramatically.

(B) The reduction in the cost of solar-power equipment has occurred despite increased raw material costs for that equipment.

(C) Technological changes have increased the efficiency of oil-fired power plants.

(D) Most electricity is generated by coal-fired or nuclear, rather than oil-fired, power plants.

(E) When the price of oil increases, reserves of oil not previously worth exploiting become economically viable.

Solution: We really need to understand this $35 figure that is given. The argument calls it “the threshold of economic viability for solar plant.” It is further explained as price per barrel to which oil would have to rise in order for new solar power plants to be more economical than new oil-fired power plants.

Note the exact meaning of this “threshold of economic viability”. It is the price TO WHICH oil would have to rise to make solar power more economical i.e. the price to which oil would have to rise to make electricity generated out of oil power plants more expensive than electricity generated out of solar power plants. So this is a hypothetical price of oil. It is not the price BY WHICH oil would have to rise. So this number 35 has nothing to do with the actual price of oil right now – it could be $10 or $15. The threshold of economic viability will remain 35.

So what the argument tells us is that tech improvements have made solar power cheaper but the price to which oil should rise has stayed the same. If you are not sure where the paradox is, let’s take some numbers to understand:

Previous Situation:

– Sunlight is free. Infrastructure needed to convert it to electricity is expensive. Say for every one unit of electricity, you need to spend $50 in a solar power plant.

– Oil is expensive. Infrastructure needed to convert it to electricity, not so much. Say for every one unit of electricity, the oil needed costs $25 and cost of infrastructure to produce a unit of electricity is $15. So total you spend $40 for a unit of electricity in an oil fired plant.

Oil based electricity is cheaper. If the cost of oil rises by $10 and becomes $35 from $25 assumed above, solar power will become viable. Electricity produced from both sources will cost the same.

Again, note properly what the $35 implies.

Raw material cost in solar plant + Infrastructure cost in solar plant = Raw material cost in oil plant + Infrastructure cost in oil plant

0 + 50 = Hypothetical cost of oil + 15

Hypothetical cost of oil = 50 – 15

That is, this $35 = Infra price per unit in solar plant – Infra price per unit in oil plant

This threshold of economic viability for solar power is the hypothetical price per barrel to which oil would have to rise (mind you, this isn’t the actual price of oil) to make solar power viable.

What happens if you need to spend only $45 in a solar power plant for a unit of electricity? Now, for solar viability, ‘cost of oil + cost of infrastructure in oil power plant’ should be only $45. If ‘cost of infrastructure in oil power plant’ = 15, we need the oil to go up to $30 only. That will make solar power plants viable. So the threshold of economic viability will be expected to decrease.

Now here lies the paradox – The argument tells you that even though the cost of production in solar power plant has come down, the threshold of economic viability for solar power is still $35! It doesn’t decrease. How can this be possible? How can you resolve it?

One way of doing it is by saying that ‘Cost of infrastructure in oil power plant’ has also gone down by $5.

Raw material cost in solar plant + Infrastructure cost in solar plant = Raw material cost in oil plant + Infrastructure cost in oil plant

0 + $45 = $35 + Infrastructure cost in oil plant

Infrastructure cost in oil plant = $10

Current Situation:

– Sunlight is free. Infrastructure needed to convert it to electricity is expensive. For every one unit of electricity, you need to spend $45 in a solar power plant.

– Oil is expensive. Infrastructure needed to convert it to electricity, not so much. For every one unit of electricity, you need to spend $25 + $10 = $35 in an oil fired power plant.

You still need the oil price to go up to $35 so that cost of electricity generation in oil power plant is $45.

So you explained the paradox by saying that “Technological changes have increased the efficiency of oil-fired power plants.” i.e. price of infrastructure in oil power plant has also decreased.

Hence, option (C) is correct.

The other option which seems viable to many people is (A). But think about it, the actual price of the oil has nothing to do with ‘the threshold of economic viability for solar power’. This threshold is $35 so you need the oil to go up to $35. Whether the actual price of oil is $10 or $15 or $20, it doesn’t matter. It still needs to go up to $35 for solar viability. So option (A) is irrelevant.

We hope the paradox and its solution make sense.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

3 Important Concepts for Statistics Questions on the GMAT

Quarter Wit, Quarter WisdomWe have discussed these three concepts of statistics in detail:

– Arithmetic mean is the number that can represent/replace all the numbers of the sequence. It lies somewhere in between the smallest and the largest values.

– Median is the middle number (in case the total number of numbers is odd) or the average of two middle numbers (in case the total number of numbers is even).

– Standard deviation is a measure of the dispersion of the values around the mean.

A conceptual question is how these three measures change when all the numbers of the set are varied is a similar fashion.

For example, how does the mean of a set change when all the numbers are increased by say, 10? How does the median change? And what about the standard deviation? What happens when you multiply each element of a set by the same number?

Let’s discuss all these cases in detail but before we start, we would like to point out that the discussion will be conceptual. We will not get into formulas though you can arrive at the answer by manipulating the respective formulas.

When you talk about mean or median or standard deviation of a list of numbers, imagine the numbers lying on the number line. They would be spread on the number line in a certain way. For example,

——0—a———b—c———————d———e————————f—g———————

Case I:

When you add the same positive number (say x) to all the elements, the entire bunch of numbers moves ahead together on the number line. The new numbers a’, b’, c’, d’, e’, f’ and g’ would look like this

——0——————a’———b’—c’———————d’———e’————————f’—g’——————

The relative placement of the numbers does not change. They are still at the same distance from each other. Note that the numbers have moved further to the right of 0 now to show that they have moved ahead on the number line.

The mean lies somewhere in the middle of the bunch and will move forward by the added number. Say, if the mean was d, the new mean will be d’ = d + x.

So when you add the same number to each element of a list, 

New mean = Old mean + Added number.

On similar lines, the median is the middle number (d in this case) and will move ahead by the added number. The new median will be d’ = d + x

So when you add the same number to each element of a list, 

New median = Old median + Added number

Standard deviation is a measure of dispersion of the numbers around the mean and this dispersion does not change when the whole bunch moves ahead as it is. Standard deviation does not depend on where the numbers lie on the number line. It depends on how far the numbers are from the mean. So standard deviation of 3, 5, 7 and 9 is the same as the standard deviation of 13, 15, 17 and 19. The relative placement of the numbers in both the cases will be the same. Hence, if you add the same number to each element of a list, the standard deviation will stay the same.

Case II:

Let’s now move on to the discussion of multiplying each element by the same positive number.

The original placing of the numbers on the number line looked like this:

——0—a———b—c———————d———e————————f—g———————

The new placing of the numbers on the  number line will look something like this:

——0———a’——————b’———c’————————————d’—————————e—- etc

The numbers spread out. To understand this, take an example. Say, the initial numbers were 10, 20 and 30. If you multiply each number by 2, the new numbers are 20, 40 and 60. The difference between them has increased from 10 to 20.

If you multiply each number by x, the mean also gets multiplied by x. So, if d was the mean initially, d’ will be the new mean which is x*d.

New mean = Old mean * Multiplied number

Similarly, the median will also get multiplied by x.

New median = Old median * Multiplied number

What happens to standard deviation in this case? It changes! Since the numbers are now further apart from the mean, their dispersion increases and hence the standard deviation also increases. The new standard deviation will be x times the old standard deviation. You can also establish this using the standard deviation formula.

New standard deviation = Old standard deviation * Multiplied number

The same concept is applicable when you increase each number by the same percentage. It is akin to multiplying each element by the same number. Say, if you increase each number by 20%, you are, in effect, multiplying each number by 1.2. So our case II applies here.

Now, think about what happens when you subtract/divide each element by the same number.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

What to Do When Math Fails You on the GMAT!

Quarter Wit, Quarter WisdomThey say Mathematics is a perfect Science. There is a debate over this among scientists but we can definitely say that Mathematical methods are not perfect so we cannot use them blindly. We could very well use the standard method for some given numbers and get stranded with “no solution.” The issue is what do we do when that happens?

For example, review this post on averages.

Here we saw that:

Average Speed = 2ab/(a + b)

Applicable when one travels at speed a for half the distance and speed b for other half of the distance. In this case, average speed is the harmonic mean of the two speeds.

So now, say if we have a question which looks like this:

Question: In the morning, Chris drives from Toronto to Oakville and in the evening he drives back from Oakville to Toronto on the same road. Was his average speed for the entire round trip less than 100 miles per hour?

Statement 1: In the morning, Chris drove at an average speed of at least 10 miles per hour while travelling from Toronto to Oakville.

Statement 2: In the evening, Chris drove at an average speed which was no more than 50 miles per hour while travelling from Oakville to Toronto.

Solution: We know that the question involves average speed. The case involves travelling at a particular average speed for one half of the journey and at another average speed for the other half of the journey.

So average speed of the entire trip will be given by 2ab/(a+b)

But the first problem is that we are given a range of speeds. How do we handle ‘at least 10’ and ‘no more than 50’ in equation form? We have learnt that we should focus on the extremities so let’s analyse the problem by taking the numbers are the extremities:10 and 50

Statement 1: In the morning, Chris drove at an average speed of at least 10 miles per hour while travelling from Toronto to Oakville.

What if Chris drives at an average speed of 10 mph in the morning and averages 100 mph for the entire journey? What will be his average speed in the evening? Perhaps around 200, right? Let’s see.

100 = 2*10*b/(10 + b)

1000 + 100b = 20b

1000 = -80b

b = – 1000/80

How can speed be negative?

Let’s hold on here and try the same calculation for statement 2 too.

Statement 2: In the evening, Chris drove at an average speed which no more than 50 miles per hour while travelling from Oakville to Toronto.

If Chris drives at an average speed of 50 mph in the evening, and averages 100 mph, let’s find his average speed in the morning.

100 = 2a*50/(a + 50)

100a + 5000 = 100a

5000 = 0

This doesn’t make any sense either!

What is going wrong? Look at it conceptually:

Say, Toronto is 100 miles away from Oakville. If Chris wants his average speed to be 100 mph over the entire trip, he should cover 100+100 = 200 miles in 2 hrs.

What happens when he travels at 10 mph in the morning? He takes 100/10 = 10 hrs to reach Oakville in the morning. He has already taken more time than what he had allotted for the entire round trip. Now, no matter what his speed in the evening, his average speed cannot be 100mph. Even if he reaches Oakville to Toronto in the blink of an eye, he would have taken 10 hours and then some time to cover the total 200 miles distance. So his average speed cannot be equal to or more than 200/10 = 20 mph.

Similarly, if he travels at 50 mph in the evening, he takes 2 full hours to travel 100 miles (one side distance). In the morning, he would have taken some time to travel 100 miles from Toronto to Oakville. Even if that time is just a few seconds, his average speed cannot be 100 mph under any circumstances.

But statement 1 says that his speed in morning was at least 10 mph which means that he could have traveled at 10 mph in the morning or at 100 mph. In one case, his average speed for the round trip cannot be 100 mph and in the other case, it can very well be. Hence statement 1 alone is not sufficient.

On the other hand, statement 2 says that his speed in the evening was 50 mph or less. This means he would have taken AT LEAST 2 hours in the morning. So his average speed for the round trip cannot be 100 mph under any circumstances. So statement 2 alone is sufficient to answer this question with ‘No’.

Answer (B)

Takeaway: If your average speed is s for a certain trip, your average speed for half the distance must be more than s/2.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Expression vs Equation on GMAT

Quarter Wit, Quarter WisdomToday, we want to take up a conceptual discussion on expressions and equations and the differences between them. The concept is quite simple but a discussion on these is warranted because of the similarity between the two.

An expression contains numbers, variables and operators.

For example

x + 4

2x – 4x^2

5x^2 + 4x -18

and so on…

These are all expressions. We CANNOT equate these expressions to 0 by default. We cannot solve for x in these cases. As the value of x changes, the value of the expression changes.

For example, given x + 4, if x is 1, value of the expression is 5. If x is 2, value of the expression is 6. If value of the expression is given to be 10, x is 6 and so on.

We cannot say, “Solve x + 4.”

If we set an algebraic expression equal to something, with an “=“ sign, we have an equation.

So here are some ways of converting the above expressions into equations:

I. x + 4 = -3

II. 2x – 4x^2 = 0

III. 5x^2 + 4x -18 = 3x

Now the equation can be solved. Note that the right hand side of the equation needn’t always be 0. It might be something other than 0 and you might need to make it 0 by bringing whatever is on the right hand side to the left hand side or by segregating the variable if possible:

I. x + 4 = -3

x + 7 = 0

x = -7

II. 2x – 4x^2 = 0

2x(1 – 2x) = 0

x = 0 or 1/2

III. 5x^2 + 4x -18 – 3x = 0

5x^2 + x – 18 = 0

5x^2 + 10x – 9x – 18 = 0

5x(x + 2) -9(x + 2) = 0

(x + 2)(5x – 9) = 0

x = -2, 9/5

In each of these cases, we get only a few values for x because we were given equations.

Think about what you mean by “solving an equation”. Let’s take a particular type of equation – a quadratic.

This is how you usually depict a quadratic:

f(x) = ax^2 + bx + c

or

y = ax^2 + bx + c

This is a parabola – upward facing if a is positive and downward facing if a is negative.

When we solve ax^2 + bx + c = 0 for x, it means, when y = 0, what is the value of x? So you are looking for x intercepts.

When we solve ax^2 + bx + c = d for x, it means, when y = d, what is the value of x? Depending on the values of a, b, c and d, you may or may not get values for x.

Let’s take an example:

x^2 – 2x – 3 = 0

(x + 1)(x – 3) = 0

x = -1 or 3

This is what it looks like:

images

When y is 0, x can take two values: -1 and 3.

So what do we do when we have x^2 – 2x -3 = -3?

We solve it in the same way:

x^2 – 2x -3 + 3 = 0

x(x – 2) = 0

x = 0 or 2

So when y is -3, x is 0 or 2. It has 2 values for y = -3 as is apparent from the graph too.

Similarly, you can solve for it when y = 5 and get two values for x.

What happens when you put y = -5? x will have no value for y = -5 so the equation x^2 – 2x – 3 = – 5 has no real solutions (so ‘no solutions’ as far as we are concerned).

We hope you understand the difference between an expression and an equation now and also that you cannot equate any given expression to 0 and solve it.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

When Not to Use Number Plugging on the GMAT

Quarter Wit, Quarter WisdomA few weeks back we discussed the kind of questions which beg you to think of the process of elimination – a strategy probably next only to number plugging in popularity.

Today we discuss the kind of questions which beg you to stay away from number plugging (but somehow, people still insist on using it because they see variables).

Not every question with variables is suitable for number plugging. If there are too many variables, it can be confusing and error prone. Then there are some other cases where number plugging is not suitable. Today we discuss an official question where you face two of these problems.

Question: If m, p , s and v are positive, and m/p < s/v, which of the following must be between m/p and s/v?

I. (m+s)/(p+v)

II. ms/pv

III. s/v – m/p

(A) None

(B) I only

(C) II only

(D) III only

(E) I and II both

Solution: The moment people see m, p, s and v variables, they jump to m = 1, p = 2 etc.

But two things should put you off number plugging here:

– There are four variables – just too many to plug in and manage.

– The question is a “must be true” question. Plugging in numbers is not the best strategy for ‘must be true’ questions. If you know that say, statement 1 holds for some particular values of m, p, s and v (say, 1, 2, 3 and 4), that’s fine but how do you know that it will be true for every set of valid values of m, p, s and v? You cannot try every set because the variables can take an infinite variety of values. If you find a set of values for the variables such that statement 1 does not hold, then you know for sure that it may not be true. In this case, number plugging does have some use but it may be a while before you can arrive at values which do not satisfy the conditions. In such questions, it is far better to take the conceptual approach.

We can solve this question using some number line and averaging concepts.

We are given that m/p < s/v

This means, this is how they look on the number line:

…………. 0 ……………….. m/p …………………… s/v ……………..

(since m, p, s and v are all positive (not necessarily integers though) so m/p and s/v are to the right of 0)

Let’s look at statement II and III first since they look relatively easy.

II. ms/pv

Think of the case when m/p and s/v are both less than 1. When you multiply them, they will become even smaller. Say .2*.3 = .06. So the product ms/pv may not lie between m/p and s/v.

Tip: When working with number properties, you should imagine the number line split into four parts:

  • less than -1
  • between -1 and 0
  • between 0 and 1
  • greater than 1

Numbers lying in these different parts behave differently. You should have a good idea about how they behave.

III. s/v – m/p

Think of a case such as this:

…………. 0 ………………………… m/p … s/v ……….

s/v – m/p will be much smaller than both m/p and s/v and will lie somewhere “here”:

…………. 0 ……… here ………………… m/p … s/v ……….

So the difference between them needn’t actually lie between them on the number line.

Hence s/v – m/p may not be between m/p and s/v.

I. (m+s)/(p+v)

This is a little tricky. Think of the four numbers as N1, N2, D1, D2 for ease and given fractions as N1/D1 and N2/D2.

(m+s)/(p+v)

= [(m+s)/2]/[(p+v)/2]

= (Average of N1 and N2)/(Average of D1 and D2)

Now average of the numerators will lie between N1 and N2 and average of the denominators will lie between D1 and D2. So (Average of N1 and N2)/(Average of D1 and D2) will lie between N1/D1 and N2/D2. Try to think this through.

We will try to explain this but you must take some examples to ensure that you understand it fully. When is one fraction smaller than another fraction?

When N1/D1 < N2/D2, one of these five cases will hold:

  • N1 < N2 and D1 = D2 . For example: 2/9 and 4/9

Average of numerators/Average of denominators = 3/9 (between N1/D1 and N2/D2)

  • N1 < N2 and D1 > D2. For example: 2/11 and 4/9

Average of numerators/Average of denominators = 3/10 (between N1/D1 and N2/D2)

  • N1 << N2 and D1 < D2. For example: 2/9 and 20/19 i.e. N1 is much smaller than N2 as compared with D1 to D2.

Average of numerators/Average of denominators = 11/14 (between N1/D1 and N2/D2)

  • N1 = N2 but D1 > D2. For example: 2/9 and 2/7

Average of numerators/Average of denominators = 2/8 (between N1/D1 and N2/D2)

  • N1 > N2 but D1 >> D2. For example: 4/9 and 2/1

Average of numerators/Average of denominators = 3/5 (between N1/D1 and N2/D2)

In each of these cases, (average of N1 and N2)/(average of D1 and D2) will be greater than N1/D1 but smaller than N2/D2. Take some more numbers to understand why this makes sense. Note that you are not expected to conduct this analysis during the test. The following should be your takeaway from this question:

Takeaway: (Average of N1 and N2)/(Average of D1 and D2) will lie somewhere in between N1/D1 and N2/D2 (provided N1. N2, D1 and D2 are positive)

(m+s)/(p+v) must lie between m/p and s/v.

Answer (B)

Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Solving Inference-Based 700+ Level Official GMAT Questions

Quarter Wit, Quarter WisdomSometimes, to solve some tough questions, we need to make inferences. Those inferences may not be apparent at first but once you practice, they do become intuitive. Today we will discuss one such inference based high level question of an official GMAT practice test.

Question: In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x – y is:

(A) 65

(B) 55

(C) 45

(D) 35

(E) 25

Solution: We need to find the value of x – y

What is x? It is the greatest possible number of households that have all three devices

What is y? It is the lowest possible number of households that have all three devices

Say there are 100 households and we have three sets:

Set DVD including 75 households

Set Cell including 80 households

Set MP3 including 55 households

We need to find the values of x and y to get x – y.

We need to maximise the overlap of all three sets to get the value of x and we need to minimise the overlap of all three sets to get the value of y.

Maximum number of households that have all three devices:

We want to bring the circles to overlap as much as possible.

The smallest set is the MP3 set which has 55 households. Let’s make it overlap with both DVD set and Cell set. These 55 households are the maximum that can have all 3 things. The rest of the 45 households will definitely not have an MP3 player. Hence the value of x must be 55.

Sets Min Max Inference Im1

Note here that the number of households having no device may or may not be 0 (it doesn’t concern us anyway but confuses people sometimes). There are 75 – 55 = 20 households that have DVD but no MP3 player. There are 80 – 55 = 25 households that have Cell phone but no MP3 player. So they could make up the rest of the 45 households (20 + 25) such that these 45 households have exactly one device or there could be an overlap in them and hence there may be some households with no device. In the figure we show the case where none = 0.

Now, let’s focus on the value of y i.e. minimum number of households with all three devices:

How will we do that? Before we delve into it, let us consider a simpler example:

Say you have 3 siblings (A,B and C) and 5 chocolates which you want to distribute among them in any way you wish. Now you want to minimise the number of your siblings who get 3 chocolates. No one gets more than 3. What do you do?

Will you leave out one sibling without any chocolates (even if he did rat you out to your folks!)? No. Because if one sibling gets no chocolates, the other siblings get more chocolates and then more of them will get 3 chocolates. So instead you give 1 to each and then give the leftover 2 to 2 of them (one each). This way, no sibling gets 3 chocolates and you have successfully minimised the number of siblings who get 3 chocolates. Basically, you spread out the goodies to ensure that minimum people get too many of them.

This is the same concept.

When you want to minimise the overlap, you basically want to spread the goodies around. You want minimum people to have all three. So you give at least one to all of them. Here there will be no household which has no device. Every household will have at least one device.

So you have 80 households which have cell phone. The rest of the 20 households say, have a DVD player so the leftover 55 households (75 – 20) with DVD player will have both a cell phone and a DVD player. There are 55 households who already have two devices and 45 households with just one device.

Now how will you distribute the MP3 players such that the overlap between all three is minimum? Give the MP3 players to the households which have just one device so 45 MP3 player households are accounted for. But we still need to distribute 10 more MP3 players. These 10 will fall on the 55 overlap of the previous two sets. Hence there are a minimum of 10 households which will have all three devices. This means y = 10

Sets Min Max Inference Im2

x – y = 55 – 10 = 45

Answer (C)

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

When to Make Assumptions on GMAT Problem Solving Questions

Quarter Wit, Quarter WisdomToday we will discuss the flip side of “do not assume anything in Data Sufficiency” i.e. we will discuss “go ahead and assume in Problem Solving!”

Problem solving questions have five definite options, that is, “cannot be determined” and “data not sufficient” are not given as options. So this means that in all cases, data is sufficient for us to answer the question. So as long as the data we assume conforms to all the data given in the question, we are free to assume and make the problem simpler for ourselves. The concept is not new – you have been already doing it all along – every time you assume the total to be 100 in percentage questions or the value of n to be 0 or 1, you are assuming that as long as your assumed data conforms to the data given, the relation should hold for every value of the unknown. So the relation should be the same when n is 0 and also the same when n is 1.

Now all you have to do is go a step further and, using the same concept, assume that the given figure is more symmetrical than may seem. The reason is that say, you want to find the value of x. Since in problem solving questions, you are required to find a single unique value of x, the value will stay the same even if you make the figure more symmetrical – provided it conforms to the given data.

Let us give an example from Official Guide 13th edition to show you what we mean:

Question: In the figure shown, what is the value of v+x+y+z+w?

Star

(A) 45

(B) 90

(C) 180

(D) 270

(E) 360

We see that the leg with the angle w seems a bit narrower – i.e. the star does not look symmetrical. But the good news is that we can assume it to be symmetrical because we are not given that angle w is smaller than the other angles.  We can do this because the value of v+x+y+z+w would be unique. So whether w is much smaller than the other angles or almost the same, it doesn’t matter to us. The total sum will remain the same. Whatever is the total sum when w is very close to the other angles, will also be the sum when w is much smaller. So for our convenience, we can assume that all the angles are the same.

Now it is very simple to solve. Imagine that the star is inscribed in a circle.

Star in a circle

Now, arc MN subtends the angle w at the circumference of the circle; this angle w will be half of the central angle subtended by MN (by the central angle theorem discussed in your book).

Arc NP subtends angle v at the circumference of the circle; this angle v will be half of the central angle subtended by NP and so on for all the arcs which form the full circle i.e. PQ, QR and RM.

All the central angles combined measure 360 degrees so all the subtended angles w + v + x + y + z will add up to half of it i.e. 360/2 = 180.

Answer (C)

There are many other ways of solving this question including long winded algebraic methods but this is the best method, in my opinion.

This was possible because we assumed that the figure is symmetrical, which we can in problem solving questions!

But beware of question prompts which look like this:

– Which of the following cannot be the value of x?

– Which of the following must be true?

You cannot assume anything here since we are not looking for a unique value that exists. If a bunch of values are possible for x, then x will take different values in different circumstances.

If we know that the unknown has a unique value, then we are free to assume as long as we are working under the constraints of the question. Finally, we would like to mention here that this is a relatively advanced technique. Use it only if you understand fully when and what you can assume.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Advanced Applications of Common Factors on the GMAT – Part II

Quarter Wit, Quarter WisdomThere is something about factors and divisibility that people find hard to wrap their heads around. Every advanced application of a basic concept knocks people out of their seats! Needless to say, that the topic is quite important so we are trying to cover the ground for you. Here is another post on the topic discussing another important concept.

In a previous post, we saw that

“Two consecutive integers can have only 1 common factor and that is 1.”

This implies that N and N+1 have no common factor other than 1. (N is an integer)

Similarly,

N + 5 and N + 6 have no common factor other than 1. (N is an integer)

N – 3 and N – 2 have no common factor other than 1. (N is an integer)

2N and 2N + 1 have no common factor other than 1. (N is an integer)

We are sure you have no problem up until now.

How about:

N and 2N+1 have no common factor other than 1. (N is an integer)

It is a simple application of the same concept but makes for a 700 level question!

2N and 2N+1 have no common factor other than 1 – we know

The factors of N will be a subset of the factors of 2N. It will not have any factors which are not there in the list of factors of 2N. So if 2N and another number have no common factors other than 1, N and the same other number can certainly not have any common factor other than 1.

Taking an example, say N = 6

Factors of 2N (which is 12) are 1, 2, 3, 4, 6, 12.

Factors of 2N + 1 (which is 13) are 1, 13.

2N and 2N + 1 can have no common factors.

Now think, what are the factors of N? They are 1, 2, 3, 6 (a subset of the factors of 2N)

They will obviously not have any factor in common with 2N+1 (except 1) since these are the same factors as those of 2N except that these are fewer.

So we can deduce the following (N and M are integers):

M and NM +1 will have no common factor other than 1.

8 and 8M + 1 will have no common factor other than 1.

M and NM – 1 will have no common factor other than 1.

and so on…

Here is the 700 level official question of this concept:

Question: If x and y are positive integers such that x = 8y + 12, what is the greatest common divisor of x and y?

Statement 1: x = 12u, where u is an integer.

Statement 2: y = 12z, where z is an integer.

Solution:

x = 8y + 12

We need to find the greatest common divisor of x and y. We have 8y in the equation. A couple of immediate deductions:

The factors of y will be a subset of the factors of 8y.

The difference between x and 8y is 12 so the greatest common divisor of x and 8y will be a factor of 12 (discussed in this post a few weeks back).

This implies that the greatest factor that x and y can have must be a factor of 12.

Looking at the statements now:

Statement 1: x = 12u, where u is an integer.

Now we know that x has 12 as a factor. The problem is that we don’t know whether y has 12 as a factor.

y could be 3 —> x = 8*3 + 12 = 36 (a multiple of 12). Here greatest common divisor of x and y will be 3.

or y could be 12 —> x = 8*12 + 12 = 108 (a multiple 12). Here greatest common divisor of x and y will be 12.

So this statement alone is not sufficient.

Statement 2: y = 12z, where z is an integer.

This statement tells us that y also has 12 as a factor. So now do we just mark (C) as the answer and move on? Well no! It seems like an easy (C) now, doesn’t it? We must analyse this statement alone.

Substituting y = 12z in the given equation:

x = 8*12z + 12

x = 12*(8z + 1)

So this already gives us that x has 12 as a factor. We don’t really need statement 1.

Since both x and y have 12 as a factor and the highest common factor they can have is 12, greatest common divisor of x and y must be 12.

This statement alone is sufficient to find the greatest common divisor of x and y.

Answer (B)

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Important Caveat on Joint Variation GMAT Questions

Quarter Wit, Quarter WisdomBefore we start today’s discussion, recall a previous post on joint variation. A question arose some days back on the applicability of this concept. This official question was the case in point:

Question: In a certain business, production index p is directly proportional to efficiency index e, which is in turn directly proportional to investment i. What is p if i = 70?

Statement I: e = 0.5 whenever i = 60

Statement II: p = 2.0 whenever i = 50

This was the issue that was raised:

If one were to follow the method given in the post on joint variation, one would arrive at this solution:

p/e = k (a constant)

e/i = m (another constant)

Hence, p*i/e = n is the joint variation expression

(where k, m and n are constants)

So we get that p is inversely proportional to i, that is, p*i = Constant 

Statement II gives us the values of p and i which can help us get the value of the Constant.

2*50 = Constant 

The question asks us the value of p given the value of i = 70. If Constant = 100, 

p = 100/70.

But actually, this is wrong and the value that you get for p in this question is different.

The question is “why is it wrong?”

Valid question, right? It certainly seems like a joint variation scenario – relation between three variables. Then why does’t it work in this case?

The takeaway from this question is very important and before you proceed, we would like you to think about it on your own for a while and then proceed to the the rest of the discussion.

Here is how this question is actually done:

Taking one statement at a time:

“production index p is directly proportional to efficiency index e,”

implies p = ke (k is the constant of proportionality)

“e is in turn directly proportional to investment i”

implies e = mi (m is the constant of proportionality. Note here that we haven’t taken the constant of proportionality as k since the constant above and this constant could be different)

Then, p = kmi (km is the constant of proportionality here. It doesn’t matter that we depict it using two variables. It is still just a number)

Here, p seems to be directly proportional to i!

So if you have i and need p, you either need this constant directly (as you can find from statement II) or you need both k and m (statement I only gives you m).

So the issue now is that is p inversely proportional to i or is it directly proportional to i?

Review the joint variation post – In it we discussed that joint variation gives you the relation between 2 quantities keeping the third (or more) constant.

p will vary inversely with i if and only if e is kept constant.

Think of it this way: if p increases, e increases. But we need to keep e constant, we will have to decrease i to decrease e back to original value. So an increase in p leads to a decrease in i to keep e constant.

But if we don’t have to keep e constant, an increase in p will lead to an increase in e which will increase i.

It is all about the sequence of increases/decreases

Here, we are not given that e needs to be kept constant. So we will not use the joint variation approach.

Note how the independent question is framed in the joint variation post:

The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical M present and inversely proportional to the concentration of chemical N present. If the concentration of chemical N is increased by 100 percent, which of the following is closest to the percent change in the concentration of chemical M required to keep the reaction rate unchanged?

You need relation between N and M when reaction rate is constant.

You are given no such constraint here. So an increase in p leads to an increase in e which in turn, increases i.

So let’s complete the solution to our original question:

p = ke

e = mi

p = kmi

Statement I: e = 0.5 whenever i = 60

0.5 = m * 60

m = 0.5/60

We do not know k so we cannot find p given i and m.

This statement alone is not sufficient.

Statement II: p = 2.0 whenever i = 50

2 = km * 50

km = 1/25

If i = 70, p = (1/25)*70 = 14/5

This statement alone is sufficient.

Answer (B)

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

When Not to Use Parallelism on the GMAT

Quarter Wit, Quarter WisdomWe know that we are often tested on parallelism on the GMAT. The logically parallel entities should be grammatically parallel. But today, we need to talk about circumstances where you might be tempted to employ parallelism but it would be incorrect to do so.

For example, look at this sentence:

A New York City ordinance of 1897 regulated the use of bicycles, mandated a maximum speed of eight miles an hour, required cyclists to keep feet on pedals and hands on handlebars at all times, and granted pedestrians right-of-way.

Is everything ok here? Well, it certainly seems so. We have four elements in parallel:

regulated …

mandated …

required …

granted …

But actually, there is a problem in this sentence:

‘regulated…’ will not be parallel to the rest of the three elements. The rest of the three elements will be in parallel.

Before we explain why, let’s take a simpler example:

The girl sitting next to me wears blue everyday, eats only waffles, and listens to music in office.

The sentence will not be ‘The girl sits next to me…’ because ‘sit’ is not parallel to other verbs. “sit” modifies the girl and is not used as a verb here. It is a present participle modifier modifying ‘girl’. It specifies the girl about whom we are talking.

Similarly, in the original sentence, ‘regulate’ is modifying ‘ordinance of 1897’. It is telling you which ordinance of 1897.

The other verbs ‘mandated’, ‘required’ and ‘granted’ are used as verbs and are parallel. They are assimilated under ‘regulate’. They tell you how the ordinance regulated.

How did it regulate?

mandated …

required …

granted …

Hence, you cannot use ‘regulated’ here. You must use ‘regulating’  – the present participle modifier to modify the ordinance. So you have to think logically – are the items in the given list actually parallel? Are they equal elements? If yes, then they need to be grammatically parallel too; else not.

Here is the complete official question:

Question: A New York City ordinance of 1897 regulated the use of bicycles, mandated a maximum speed of eight miles an hour, required of cyclists to keep feet on pedals and hands on handlebars at all times, and it granted pedestrians right-of-way.

(A) regulated the use of bicycles, mandated a maximum speed of eight miles an

hour, required of cyclists to keep feet on pedals and hands on handlebars at all

times, and it granted

(B) regulated the use of bicycles, mandated a maximum speed of eight miles an

hour, required cyclists to keep feet on pedals and hands on handlebars at all

times, granting

(C) regulating the use of bicycles mandated a maximum speed of eight miles an

hour, required cyclists that they keep feet on pedals and hands on handlebars

at all times, and it granted

(D) regulating the use of bicycles, mandating a maximum speed of eight miles an

hour, requiring of cyclists that they keep feet on pedals and hands on

handlebars at all times, and granted

(E) regulating the use of bicycles mandated a maximum speed of eight miles an

hour, required cyclists to keep feet on pedals and hands on handlebars at all

times, and granted

Solution:

From our above discussion, we know that we have choose one of (C), and (E).

(A), (B) and (D) put regulate parallel to the other verbs.

Still, let’s point out all the errors of these options:

(A) regulated the use of bicycles, mandated a maximum speed of eight miles an

hour, required of cyclists to keep feet on pedals and hands on handlebars at all

times, and it granted

Parallelism problem – regulated cannot be parallel to mandated and other verbs. Also, ‘mandated’ is not parallel to ‘it granted’. Besides, ‘required of X to do Y’ is unidiomatic.

(B) regulated the use of bicycles, mandated a maximum speed of eight miles an

hour, required cyclists to keep feet on pedals and hands on handlebars at all

times, granting

Parallelism problem – ‘regulated’ is parallel to ‘mandated’ though it should not be.

‘granting’ is not parallel to ‘mandated’ and ‘required’ though it needs to be parallel.

You also need an ‘and’ before the last element of the list ‘and granted …’

(D) regulating the use of bicycles, mandating a maximum speed of eight miles an

hour, requiring of cyclists that they keep feet on pedals and hands on

handlebars at all times, and granted

This is not a valid sentence because the main clause does not have a verb. ‘regulating…’, ‘mandating…’ and ‘requiring…’ are the present participle modifiers.

‘granted…’ is not parallel to the other elements. Besides, ‘requiring of X that they do Y’ is unidiomatic.

Now let’s look at the leftover options:

(C) regulating the use of bicycles mandated a maximum speed of eight miles an

hour, required cyclists that they keep feet on pedals and hands on handlebars

at all times, and it granted

‘it granted’ is not parallel to the other verbs. Besides, ‘required X that they do Y’ is unidiomatic.

(E) regulating the use of bicycles mandated a maximum speed of eight miles an

hour, required cyclists to keep feet on pedals and hands on handlebars at all

times, and granted

Perfect! All issues sorted out!

Answer (E)

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

2 Possible Ways to Solve this GMAT Quant Question

Quarter Wit, Quarter WisdomProcess of elimination is only next to number plugging in popularity as a strategy for solving Quant questions on the GMAT. I am not a fan of either method. Yes, they are useful sometimes, and even necessary in some questions but for most questions, I like to use logic/reasoning.

That said, there is a set of questions in which we should think of these strategies. Number plugging is very useful when you have one or two variables in the options. Algebra can be time consuming in these cases because of equation manipulation required.

Similarly, some questions beg you to use the process of elimination. Their question stem goes something like ”which of the following options can be the value of x?”, “which of  the following options cannot be the sum of a and b?” etc. These questions are framed like this because often they have multiple solutions. x could possibly take many different values but the options would have only one of them. So it makes sense to check which values x can take from the options. Let’s look at one such instance of a tricky question where process of elimination can be very useful.

Question:

A list of numbers has six positive integers. Three of those integers are known – 4, 5 and 24 and three of those are unknown – x, y and z. The three unknowns are known to be distinct. It is also known that the mean of the list is 10 and the median lies between 7 and 8 (exclusive).

Which of the following CANNOT be the value of any one of the unknowns?

(A) 13

(B) 12

(C) 11

(D) 10

(E) 5

Solution: The question gives us concrete information about mean – it is 10 – but not about median – it is between 7 and 8 (exclusive). What can we say about median from this? That it cannot be 7 or 8 but anything in between. But we know that the list has all integers. When we have even number of integers, we know that the median is the average of the middle two numbers – when all are placed in increasing order. So can the average of the two middle numbers be, say, 7.1? Which two positive integers can average to give 7.1? None! Note that if the average of two integers is a decimal, the decimal must be (some number).5 such as 7.5 or 9.5 or 22.5 etc. This happens in case one number is odd and the other is even. In all other cases, the average would be an integer.

Since the median is given to be between 7 and 8, the median of the list of the six positive integers must be 7.5 only.

Now we know that the mean = 10 and median = 7.5

Method 1: Algebra/Logic

Let’s try to solve the question algebraically/logically first.

There are 6 elements in the list. The average of the list is 10 which means the sum of all 6 elements = 6*10 = 60

4 + 5 + 24 + x + y + z = 60

x + y + z = 27

Median of the list = 7.5

So sum of third and fourth elements must be 7.5 * 2 = 15

There are two cases possible:

Case 1: Two of the three integers x, y and z could be the third and the fourth numbers. In that case, since already 4 and 5 are less than 7.5, one of the unknown number would be less than 7.5 (the third number) and the other two would be greater than 7.5.

The sum of the third and fourth elements of the list is 15 so

15 + z = 27

z = 12

So, two numbers whose sum is 15 such that one is less than 7.5 and the other greater than 7.5 could be

5 and 10

6 and 9

7 and 8

x, y and z could take values 5, 6, 7, 8, 9, 10 and 12.

Case 2: The known 5 could be the third number in which case one of the unknown numbers is less than 5 and two of the unknown numbers would be more than 7.5.

If the third number is 5, the fourth number has to be 10 to get a median of 7.5. Hence, 10 must be one of the unknown numbers.

The sum of the other two unknown numbers would be 27 – 10 = 17.

One of them must be less than 5 and the other greater than 10. So possible options are

4 and 13

3 and 14

2 and 15

1 and 16

x, y and z could take various values but none of them could be 11

Answer (C)

Method 2: Process of Elimination

Let’s now try to look at the process of elimination here and see if we can find an easier way.

The three unknowns need to add up to 10*6 – 4 – 5 – 24 = 27.

Two of the given options are 5 and 10. They have a median of 7.5 so lets assume that two of the unknown numbers are 5 and 10 (5 can be one of the unknowns since we are not given that all six integers need to be distinct). If two unknowns make up third and fourth numbers in the list and have a median of 7.5, their sum would be 15 and the third unknown will be 12 (to get the mean of 10). This case (5, 10, 12) satisfies all conditions so options (B), (D) and (E) are out of play.

Now we are left with two options 13 and 11. Check any one of them and you will know which one is not possible. Let’s check 13.

From the given options, any number greater than 7.5 must be either the fourth number or the fifth number. 13 cannot be the fourth number since the third number would need to be 2 in that case to get median 7.5. But we have 4 and 5 more than 2 so it cannot be the third number. So 13 must be the fifth number of the list. We saw in the case above that if two unknowns are third and fourth numbers then the fifth number HAS TO BE 12. So the already present 5 must be the third number and the fourth number must be 10. In that case, the leftover unknown would be 4 (to get a sum of 27). So the three unknowns would be 4, 10 and 13. This satisfies all conditions and is possible. Hence answer must be (C). 11 will not be possible.

Let’s see what would have happened had you picked 11 to try out. If 11 were the fourth number, to get a median of 7.5, we would need 4 as the third number. That is not possible since we already have a 5 given. So 11 must have been the fifth number. This would mean that the already present 5 and one unknown 10 would make the median of 7.5. So the third unknown in this case would be 6 (to get a sum of 27). But 6 would be the third number and the median in this case would be (6 + 10)/2 = 8. So one of the numbers cannot be 11.

Answer (C)

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

The Symmetry Puzzle on the GMAT

Quarter Wit, Quarter WisdomA few days back, a student of ours asked me this question – in which cases is symmetry useful to us? Honestly, I don’t think I can create an exhaustive list of the topics where it could be useful. The first thing that comes to mind is of course, Geometry. Circles/equilateral triangles/squares/cubes are symmetrical figures. Symmetry helps us simplify questions which are based on these figures. We have also seen the uses of symmetry in dice throwing. In arrangements too, symmetry helped decrease our work substantially.

Today, let’s look at a puzzle where symmetry helps.

Question: A spider is sitting on one corner of a cube. It wants to get to the most distant corner but it can crawl only along the edges of the cube and cannot revisit a place where it has already been. In how many different ways can the spider go to the most distant corner?

(A) 6

(B) 12

(C) 18

(D) 24

(E) 30

Solution: The question is a puzzle type combinatorics questions. It seems like we will have to painstakingly calculate the various paths that the spider can take. But notice that the figure we have is a cube – a symmetrical figure. Let’s draw the figure to see what the question is asking.

 

Ques3

Now, assume that the spider is at A. In that case, he has to go to F – the farthest vertex from A. Every vertex has only one vertex farthest to it. C, E and G are equidistant from A but they are in the same plane. F is further off than C, E and G. So it needs to go from A to F:

Step 1:

It can crawl only along the edges so from A, it can take three different paths – AB or AH or AD. As far as F is concerned, all the points D, H and B  are similar.

Step 2:

Now, from each of these 3 points, the spider has two path options. If it is at D, it can crawl on DE or DC. The third path from D leads back to A but the spider is not allowed to revisit a place. So there are only two forward options for it – DE or DC.

Similarly, if it is at H, it can crawl on HE or on HG. If it is at B, it can crawl on BG or BC. So the total number of paths that we have found till now are  3*2.

Till now, we hope you did not face any problems.

Step 3:

Now comes the tricky part. The spider is at one of three vertices – E, C or G. Assume it came the AD – DE route and is now at E. There are multiple ways in which it can reach F. The obvious one is directly from E to F. But it can also go to F via H because it has not visited a number of other vertices (H, G, B, C)

There are three ways in which it can reach F now:

  • directly E to F
  • a three path EH – HG – GF
  • a five path EH – HG – GB – BC – CF

This takes care of all the ways in which it can reach F from E.

Since we found 3 different paths from E, it is obvious that we will find 3 different paths from C and from G too. It is a symmetrical figure and hence we don’t need to calculate the number of paths from each point. In any case, we have 3 ways to reach F now.

So total different paths to reach the farthest vertex = 3*2*3 = 18

Answer (C)

Hope you see how symmetry helps us reduce our work substantially.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Easy Logic to a Difficult Combinatorics GMAT Question!

Quarter Wit, Quarter WisdomSometimes, you come across some seriously interesting questions in Combinatorics. For example, this question I came across seemed like any other Combinatorics question, though it was a little cumbersome. But when I saw the answer, it got me thinking – it couldn’t have been a coincidence. There had to be a simpler logic to it and there was! I just wish I had thought of it before going the long route. So I must share it with you; you never know what might come in handy on test day!

But before I tell you what that question was, let’s solve a couple of questions which are similar to some questions you might have seen before  (for the sake of brevity,  let’s ignore the options):

Question 1: In how many ways can 10 different paintings be distributed between two collectors – Dave and Mona – if Dave should get either three paintings or five paintings? (All paintings should be given away).

Solution 1:

There are two ways of distributing the paintings in this case:

Dave gets 3 paintings and Mona gets the rest: You select 3 of the 10 paintings and give them to Dave. This can be done in 10C3 = 120 ways

Dave gets 5 paintings and Mona gets the rest: You select 5 of the 10 paintings and give them to Dave. This can be done in 10C5 = 252 ways

Total number of ways in which you can distribute the paintings = 120 + 252 = 372 ways

Simple enough, right? Let’s take a  look at another simple similar question.

Question 2: In how many ways can 10 different paintings be distributed between two collectors – Dave and Mona – if Dave should get at least two paintings? (All paintings should be given away.)

Solution 2:

Dave should get at least two paintings so it means he can get 2 or 3 or 4 or more up to 10 paintings. Calculating all those cases would be tedious so this is a perfect opportunity to use ‘Total – Opposite’ method.

Total ways in which you can distribute 10 paintings between two people without any constraints: Each painting can be given away in two ways – either to Dave or to Mona. So the paintings can be distributed in 2*2*2*…*2 = 2^10 = 1024 ways

Number of ways in which Dave gets 0 paintings or 1 painting: 1 + 10C1 = 11 ways

So number of ways in which Dave gets 2 or 3 or 4 … upto 10 (i.e. at least 2 paintings) = 1024 – 11 = 1013 ways

Another ‘seen before, know how to solve it’ kind of question. Now let’s come to the question of the day which doesn’t look much different but actually is.

Question 3: In how many ways can 10 different paintings be distributed between two collectors – Dave and Mona – if both collectors should get an even number of paintings? (All paintings should be given away.)

Solution 3:

Paintings can be distributed in the following ways:

0, 10 – One person gets 0 paintings and the other gets 10

2, 8 – One person gets 2 paintings and the other gets 8

4, 6 – One person gets 4 paintings and the other gets 6

You will need to calculate each one of these ways and then add them. Note that the ‘Total – Opposite’ method does not work here because finding the number of ways in which each person gets odd number of paintings is equally daunting.

Case 1: 0, 10

One person gets 0 paintings and the other gets 10. This can be done in 2 ways – either Dave gets all the paintings or Mona gets them.

Case 2: 2, 8

One person gets 2 paintings and the other gets 8. Select 2 paintings out of 10 for Dave in 10C2 = 45 ways. Mona could also get the 2 selected paintings so total number of ways = 45*2 = 90 ways

Case 3: 4, 6

One person gets 4 paintings and the other gets 6. Select 4 paintings out of 10 for Dave in 10C4 = 210 ways. Mona could also get the 4 selected paintings so total number of ways = 210*2 = 420

Total number of ways such that each person gets even number of paintings = 2 + 90 + 420 = 512 ways

But 512 is 2^9 – in form, suspiciously close to 2^10 we used in question 2 above. Is there some logic which leads to the answer 2^(n-1)? There is!

You have 10 different paintings. Each painting can be given to one of the 2 people in 2 ways. You do that with 9 paintings in 2*2*2…  = 2^9 ways. When you distribute 9 paintings, one person will have odd number of paintings and one will have even number of paintings (0 + 9 or 1 + 8 or 2 + 7 or 3 + 6 or 4 + 5).

The tenth painting needs to be given to the person who has the odd number of paintings so you give the tenth painting in only one way. This accounts for all cases in which both get even number of paintings.

Total ways = 2^9 * 1 = 512

On the same lines, now think about this:

Question 4: In how many ways can 10 different paintings be distributed between two collectors – Dave and Mona – if both collectors should get an odd number of paintings? (All paintings should be given away.)

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Advanced Applications of Common Factors on the GMAT

Quarter Wit, Quarter WisdomToday we will discuss the logic behind common factors (other than 1) of two numbers.

Without actually finding all the factors of two numbers, how do we know whether they have any common factors (ignoring 1)?

Let’s take some examples:

  • If the integers are even, we know that they must have at least one common factor – 2. Let’s say we have two numbers 476 and 478. How many common factors can they have? We know that 2 is a factor common to them. Can they have any other common factor? Note that the difference between them is 2. So if 4 were a factor of 476, could it be a factor of 478? No. If 4 were a factor of 476, it would be a factor of 480 next (4 away from 476). Similarly, if 7 were a factor of 476, it would not be a factor of 478, definitely. It would be a factor of 483 (7 away from 476). In fact, since the difference between the two numbers is 2, the only factor they can have in common is 2.
  • Now consider that the two numbers are 476 and 484. They have a difference of 8 between them. The common factors they can have are 2, 4 and 8 (the factors of 8). If any of these factors is a factor of 476, it will be a factor of 484 too. Obviously, 476 and 484 will have many other factors but they will not have any other common factor. 7 is a factor of 476. The next multiple of 7 will be 483 and the next will be 490. 7 cannot be a factor of 484.
  • What happens when both integers are odd? Say 525 and 531. The difference between them is 6. The factors of 6 are 2 and 3. Both 525 and 531 are odd numbers so 2 cannot be their factor. If 3 is a factor of 525, it will be a factor of 531 too else it will not be a factor of both the numbers. Any other number can be a factor of one of them, but not both.

This is what we can deduce:

The only factors that CAN be common (it’s not necessary that they will be common) between two numbers are the factors of the difference between them.

If any factor of the difference between them is a factor of one of the numbers, it will be a factor of the other number too. If it is not a factor of one number, it will not be a factor of the other number.

Take a look at a question based on these concepts:

Question: Given that x is a positive integer, what is the greatest common divisor (GCD) of the two positive integers, (x+m) and (x-m)?

Statement 1: m^2 – 10m + 16 = 0

Statement 2: x + 26 is a prime number.

Solution:

The two given positive integers are (x + m) and (x – m). x is a positive integer so m must be an integer too. Whether m is positive or negative, we don’t know.

To know the GCD of two numbers, we need to know their common divisors. As of now, we have no idea about their common divisors, but we know that the difference between the two numbers is 2m. Their common factors must be factors of 2m.

Let’s look at the two statements:

Statement 1: m^2 – 10m + 16 = 0

We know that the quadratic will give us two values for m so we will not be able to find a unique value for m. But let’s solve it in case we get some other clues from it.

m^2 – 10m +16 = 0

m^2 – 2m – 8m + 16 = 0

m (m – 2) – 8 (m – 2) = 0

(m – 2)*(m – 8) = 0

m is either 2 or 8. So 2m is either 4 or 16.

The factors of 2m will be 1, 2 and 4 and additionally, 8 and 16 (if 2m is 16). We have no idea whether x+m and x-m will have these factors so this statement alone is not sufficient.

Statement 2: x + 26 is a prime number.

What does it tell us about x? Other than 2, all prime numbers are odd numbers. Since x is a positive integer,  x+26 cannot be 2. It must be a prime number greater than 2 and hence, must be odd. But 26 is even. So x must be an odd integer (Odd + Even = Odd). But we have no information about m so this statement alone is not sufficient.

Using both statements together, since x is an odd integer and m is definitely even (either 2 or 8), both the numbers (x + m) and (x – m) are odd integers. Odd integers will not have any of these factors: 2, 4, 8, 16.

So (x + m) and (x – m) must have 1 as the only common factor. Hence their greatest common divisor must be 1.

Together, the two statements are sufficient to answer the question.

Answer (C)

To recap: Any common factor of two numbers has to be a factor of the difference between them. This also implies that the GCD of two numbers has to be a factor of the difference between them.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Is this GMAT Question an Alphametic or Simple Number Properties Question?

Quarter Wit, Quarter WisdomAs noticed in the first post of Alphametics, a data sufficiency alphametic is far more complicated than a problem solving alphametic. An alphametic can have multiple solutions and establishing that it does not, is time consuming. Hence, it is less likely that you will see a DS alphametic in the actual exam.

In fact, what may look like an alphametic problem, might actually be a number properties problem only.

We will look at an example below:

Question:

AlphameticPost3

In the correctly-worked multiplication problem above, each symbol represents a different nonzero digit. What is the value of C?

Statement 1: D is prime.

Statement 2: B is not prime.

Solution: We multiply two two-digit integers and get 1995. The good thing is that we know the result of the multiplication will be 1995. Usually, multiplication alphametics are harder since they involve multiple levels, but here the multiplication is actually a blessing. There are many many ways in which you can ADD two integers to give 1995 but there are only a few ways in which you can multiply two integers to give you 1995.

Let’s prime factorize 1995:

1995 = 3*5*7*19

We can probably count on our fingers the number of ways in which we can select AB and CD.

19 needs to be multiplied with one other factor to give us a two digit number since 5*3*7 = 105 (a three digit number) so AB and CD cannot be 19 and 105.

19*3 = 57, 5*7 = 35 – This is not possible since two of the four digits are same here – 5.

19*5 = 95, 3*7 = 21 – This is one option for AB and CD.

19*7 = 133 – Three digit number not possible.

Hence AB and CD can only take values out of 21 and 95.

As of now, C can be 2 or 9. We need to find whether the given statements give us a unique value of C.

Statement 1: D is prime

D is the units digit of CD. So D can be 1 or 5.

1 is not prime so CD cannot be 21. Hence, CD must be 95 and AB must be 21.

Hence, C must be 9.

This statement alone is sufficient.

Statement 2: B is not prime

If B is not prime then AB cannot be 95. Hence AB must be 21.

This means CD will be 95 and C will be 9.

This statement alone is sufficient.

Answer (D)

Note that the entire question was just about number properties – prime factors, prime numbers etc. Actually it required no iterative steps and no hit and trial. Rest assured that if it is a GMAT question, it will be reasoning based and will not require painful calculations.

Are you studying for the GMAT? We have free online GMAT seminars running all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

How to Solve Alphametic Multiplication Questions on the GMAT

Quarter Wit, Quarter WisdomLast week, we looked at alphametics involving addition and subtraction. The logic becomes a little more involved when the alphametic involves multiplication. When a two digit number is multiplied by another two digit number, the process of finding the result is composed of multiple levels. Today, let’s see how to handle those multiple levels.  The question involves quite a few steps and observations using number properties. Hence, you are unlikely to see such a question in actual GMAT but you might see a simpler version so it’s good to be prepared.

Question: The following alphametic shows multiplication of two numbers, IF and DR. The product you obtain is DORF.

AlphameticPost2Fig1

What is the value of D + O + R + F?
(A) 17
(B) 20
(C) 22
(D) 23
(E) 30

Solution: The good thing is that we know D + O + R + F has a single value. This means there will be a logic to obtain the value of each of D, O, R and F.

As discussed last week, we first focus on the big picture, but we will have to go one level at a time.

(i) IF * R = OFF

(ii) IF * D = IF

(iii) OF + IF = DOR

A few interesting points to note from the above:

– From (ii), IF * D = IF

When you multiply IF by D, you get IF itself. This means that D must be 1. D can take no other value.

D = 1

– From (iii), F + F has unit’s digit of R.

Also O + I gives O as unit’s digit and 1 as tens digit (D of DORF obtained from above). How can this happen? Say, if O = 4, 4 + I = 14. This is possible only when I = 9 and there is a 1 carry over from the previous addition of F + F. This means that F must be 5 or greater to have a carryover of 1. It cannot be 5 because 5+5 will give you 10 making R = 0. This would mean that F*R would end in R (0). But in (i), F * R has unit’s digit of F, not R. So F cannot be 5.

D = 1, I  = 9

– Another interesting point: From (i), F * R has unit’s digit of F. This is possible only when F = 0 or F = 5 or R = 6 (Think of multiplication tables of numbers to convince yourself why this is so)

Since F has to be greater than 5 (as seen above), R must be 6.
If R = 6, then from (iii), F + F has unit’s digit of 6 and a carryover of 1 so F = 8. When you add 8 + 8, you will get 16 (units digit of 6 and a carryover)

D = 1, I = 9, R = 6, F = 8

– From (i), when we multiply IF by R, we get OFF. That is, when we multiply 98 by 6, we get 588. So O must be 5.

This gives us: D + O + R + F = 1 + 5 + 6 + 8 = 20

Answer (B)

This question uses your understanding of numbers and how they are added and multiplied. It certainly takes time to get to the answer. Such questions can help you get a feel for numbers and their behavior.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

An Introduction to Solving Alphametic Questions on the GMAT

Quarter Wit, Quarter WisdomToday, let’s learn how to solve alphametics. An alphametic is a mathematical puzzle where every letter stands for a digit from 0 – 9. The mapping of letters to numbers is one-to-one; that is, the same letter always stands for the same digit, and the same digit is always represented by the same letter.

First focus on the big picture of the alphametic – such as, a two number is added to another two digit number to give a three digit number etc. Then look at the nitty gritty – for which digit can each letter stand?

Question 1: With # and & each representing different digits in the problem below, the difference between #&& and ## is 667. What is the value of &?

Alphametic0

(A) 3
(B) 4
(C) 5
(D) 8
(E) 9

Solution: The big picture: A two digit number is subtracted from a three digit number to give 667. So the three digit number must be a bit larger than 667. This means that the hundreds digit of #&& must be either 6 or 7. It cannot be 8 because you cannot obtain 800+ by adding a two digit number to 667.

Let’s look at both cases:

# is 6: If you subtract 66 from 6&&, you will not get 667 – the largest value you can get is 699 – 66 = 633. So # cannot be 6.

# must be 7.

Now the question is very simple
7&& – 77 = 667
7&& = 667 + 77 = 744

Answer (B)

There are many other ways in which you can solve this question including plugging in the answer choices. We should now take a look at a DS question on alphametics.

Question 2:

Alphametic

In the correctly worked addition problem above, M, N, P, R, S, T and V are distinct digits. Is R > 3?

Statement 1: M, N and P are positive even integers.

Statement 2: S = 2

Solution: This is certainly harder than the PS question but our process will remain the same.

First, let’s see what information we are given in the question – the units digits of all three numbers are the same. The two-digit numbers add up to give a three digit number. The hundreds digit, S, is either 1 or 2. Three two-digit numbers cannot add up to give a number 300 or more since 99 + 99 + 99 = 297. We have no information on what the value of R can be. All we know is that R cannot be 0 because 0+0+0 = 0 but V needs to be different from R.

Let’s look at the statements now.

Statement 1: M, N and P are positive even integers.

At first, it may seem that this has nothing to do with the value of R but we must analyze what is given to be sure.

M, N and P must take distinct values out of 2, 4, 6 and 8 and add up to give the units digit of T (again, distinct)

Every time you add three even numbers, you will get an even number. Let’s see which combinations we can get:

2 + 4 + 6 = 12

2 + 4 + 8 = 14

2 + 6 + 8 = 16

4 + 6 + 8 = 18

Note that in all four cases, the units digit is one of the numbers but T must be distinct. This means that there must have been a carryover from the previous addition. So when we added the three Rs, we must have got a carryover. Had R been 3 or less, we would not have got a carryover since 1+1+1 = 3, 2+2+2 = 6 and 3+3+3 = 9. So R must be greater than 3.

One such case would be

Alphametic1

This statement alone is sufficient.

Statement 2: S = 2

The result of addition gives us a number which is more than 200. In statement 1 we saw a case in which S is 2 and R is greater than 3. Now all we have to do is find a case in which S is 2 and R is less than 3. One of these cases is

Alphametic2

So this statement alone is not sufficient.

Answer (A)

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

How to Solve Advanced Compound Interest Questions on the GMAT

Quarter Wit, Quarter WisdomWe have discussed simple and compound interest in a previous post.

We saw that simple and compound interest (compounded annually) in the first year is the same. In the second year, the only difference is that in compound interest, you earn interest on previous year’s interest too. Hence, the total two year interest in compound interest exceeds the two year interest in case of simple interest by an amount which is interest on year 1 interest.

So a question such as this one is very simple to solve:

Question 1: Bob invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest (compounded annually) for the same 2 years at the same rate of interest and received $605 as interest. What was the annual rate of interest?

(A) 5%
(B) 10%
(C) 12%
(D) 15%
(E) 20%

Solution:

Simple Interest for two years = $550

So simple interest per year = 550/2 = 275

But in case of compound interest, you earn an extra 605 – 550 = $55

This $55 is interest earned on year 1 interest i.e. if rate of interest is R, it is

55 = R% of 275

R = 20

Answer (E)

The question is – what happens in case you have 3 years here, instead of 2? How do you solve it then? Here is a small table of the difference between simple and compound interest to help you.

Say the Principal is P and the rate of interest if R

Compound Interest

It gets a bit more complicated though not very hard to solve. All you need to do is solve a quadratic, which, if the values are well thought out, is fairly simple to solve. Let’s look at the same question adjusted for three years.

Question 2: Bob invested one half of his savings in a bond that paid simple interest for 3 years and received $825 as interest. He invested the remaining in a bond that paid compound interest (compounded annually) for the same 3 years at the same rate of interest and received $1001 as interest. What was the annual rate of interest?

(A) 5%
(B) 10%
(C) 12%
(D) 15%
(E) 20%

Simple Interest for three years = $825

So simple interest per year = 825/3 = $275

But in case of compound interest, you earn an extra $1001 – $825 = $176

What all is included in this extra $176? This is the extra interest earned by compounding.

This is R% of interest of Year1 + R% of total interest accumulated in Year2

This is R% of 275 + R% of (275 + 275 + R% of 275) = 176

(R/100) *[825 + (R/100)*275] = 176

Assuming R/100 = x to make the equation easier,

275x^2 + 825x – 176 = 0

25x^2 + 75x – 16 = 0

25x^2 + 80x – 5x – 16 = 0

5x(5x + 16) – 1(5x + 16) = 0

x = 1/5 or -16/5

Ignore the negative value to get R/100 = 1/5 or R = 20

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Taking the Best Approach When Solving Data Sufficiency GMAT Questions

Quarter Wit, Quarter WisdomSometimes, we get questions which we cannot neatly bracket as Arithmetic/Algebra/Geometry etc. In fact, the higher level questions usually focus on more than one subject area.  The trick in these questions is to assimilate all your knowledge from various areas and then think how best to solve.

Today we have such a question for you – you could get really lost in it or could solve it in a few seconds if you take the right track. The trick is starting on the right track and that is why you have 2 mins per question available to you else 40 secs per question would have been sufficient!

Question: This game season, five divisions are going to play. Out of all the teams in each division, 6, 9, 12, 13 and 14 teams have qualified from the respective divisions. Each division will hold its own tournament – where a team is eliminated from the tournament upon losing two games – in order to determine its champion. The five division champions will then play in a knock-off tournament – a team is eliminated as soon as it loses a game – in order to determine the overall champion. Assuming that there were no ties and no forfeits, what is the maximum number of games that could have been played in order to determine the overall league champion?

(A) 89

(B) 100

(C) 102

(D) 107

(E) 112

Solution: Is it a max-min problem? Perhaps, but which guiding principle of max-min will we use to solve this problem? First think on your own how you will solve this problem.

Will you focus on the method or the result i.e. will you worry about who plays against whom or just focus on each result which gives one loss and one win? If you don’t worry about the method and just focus on the result, you can use a concept of mixtures here. In mixture questions, we focus on one component and how it changes. Here, we need to keep track of losses. Let’s focus on those and forget about the wins. As given, there were no ties so every loss has a win on the other side.

Every time a game is played, someone loses. We can give at most 2 losses to a team since after that it is out of the tournament. Don’t worry about against who it plays those two games. Whenever a team loses 2 games, it is out. The team could have won many games but we are not counting the wins and hence, are not concerned about its wins. As discussed, we are counting the losses so each win of that team will be counted on the other side i.e. as a loss for the other team.

Consider the division which has 6 teams – what happens when 12 games are played? There are 12 losses and each team gets 2 losses (we can’t give more than 2 to a team since the team gets kicked out after 2 losses), so all are out of the tournament. But we need a winner so we play only 11 games so that the winning team gets only 1 loss. We want to maximize the losses (and hence the number of games), therefore the winning team must be given a loss too.

So maximum number of games that can be played by the district in its own tournament = 2*6 – 1 = 11

Similarly, the division with 9 teams can play at most 2*9 – 1 = 17 games.

The division with 12 teams can play at most 2*12 – 1 = 23 games.

The division with 13 teams can play at most 2*13 – 1 = 25 games.

The division with 14 teams can play at most 2*14 – 1 = 27 games.

This totals up to 11 + 17 + 23 + 25 + 27 = 103 games

Now we come to the games between the district champions.

We have 5 teams. 1 loss gets a team kicked out. If the teams play 4 games, there are 4 losses and 4 teams get kicked out. We have a final winner!
Hence the total number of games = 103 + 4 = 107

There are a lot of variations you can consider for this question.

Say, if we need to minimize the number of games, how many total games would have been played?

Notice that the only games you can avoid are the ones in which the 5 district champions lost. You do still need 2 losses for each team to get the district champion and one loss each for four district champions to get the winner. Hence, at least 107 – 5 = 102 games need to be played.

Look at it in another way:

To kick out a team, it needs to have 2 losses so if the district had 6 teams, there would be 5*2 = 10 games played.

Similarly, the division with 9 teams will play at least 2*8 = 16 games.

The division with 12 teams will play at least 2*11 = 22 games.

The division with 13 teams will play at least 2*12 = 24 games.

The division with 14 teams will play at least 2*13 = 26 games.

This totals up to 10 + 16 + 22 + 24 + 26 = 98 games

Now we have 5 champions and they will need to play at least 4 games to pick a winner.

Therefore, at least 98 + 4 = 102 games need to be played.

You can try other similar variations – what happens when a team is kicked out after it loses 3 games instead of 2? What happens if you don’t have the knock-off tournament and instead need each district champion to lose 2 games to get knocked out?

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Identifying and Correcting Run-On Sentences on GMAT Verbal Questions

Quarter Wit, Quarter WisdomOn the GMAT, most sentence correction questions involve compound/complex sentences with multiple phrases, clauses and modifiers. Hence it is very likely that you will see some run-on sentences on your test. In the complicated sentences that we get on the GMAT, it is very easy to overlook that we are dealing with run-on sentences.

A run-on sentence has at least two independent clauses which are not connected properly. There are various ways in which a sentence may be run-on. Here are some of the most common circumstances:

  1. When an independent clause gives a suggestion/advice/command based on what was said in the prior independent clause:

GMAT is a very tricky test, you should work hard.

Here, we should either split the two clauses into two sentences by putting in a full stop or we should put a semi colon between the two clauses.

  1. When two independent clauses are connected by a conjunctive adverb such as however, moreover, nevertheless.

My grandmother is supposed to travel tomorrow, however, she is not feeling well.

Here, we should either split the two clauses into two sentences by putting in a full stop or we should put a semi colon between the two clauses

To read more on conjunctive adverbs, check out this post.

  1. When the second of two independent clauses contains a pronoun that connects it to the first independent clause.

Marcy is thrilled, she got permission to go to the school dance.

Although these two clauses are quite brief, and the ideas are closely related, this is a run-on sentence. We need to put a full stop or a semi colon in place of the comma.

Now that we have an idea of what run-on sentences are, let’s look at a GMAT Prep question where this concept is tested extensively.

Question: The Anasazi settlements at Chaco Canyon were built on a spectacular scale with more than 75 carefully engineered structures, of up to 600 rooms each, were connected by a complex regional system of roads.

(A) with more than 75 carefully engineered structures, of up to 600 rooms each, were

(B) with more than 75 carefully engineered structures, of up to 600 rooms each,

(C) of more than 75 carefully engineered structures of up to 600 rooms, each that had been

(D) of more than 75 carefully engineered structures of up to 600 rooms and with each

(E) of more than 75 carefully engineered structures of up to 600 rooms each had been

Solution:

Consider option (A): Remove all the unnecessary elements and get the skeleton of the sentence (primarily the subject and the verbs):

The settlements were built on a spectacular scale…, were connected …

The action verb “were connected” has no subject here. If it were to have the same subject as the first clause “the Anasazi settlements”, then there should have been a conjunction joining the two clauses together. This is a run-on sentence.

Consider option (B): The problem of run-on sentence has been rectified here by using past participle instead.
The settlements were built on a spectacular scale with more than 75 carefully engineered structures, of up to 600 rooms each, connected by a complex regional system of roads.

Remove the non essential modifier “of up to 600 rooms each” and you see that the 75 carefully engineered structures were the ones connected by a complex system of roads. Now it all makes sense.

To read more about participles, check this post.

Let’s look at the other options too.

Option (C): You cannot say “built on a spectacular scale of more than 75 structures”. You need “with” instead of “of”. The same problem exists with options (D) and (E) too.

Also, in option (C), the use of past perfect “had been” is not justified.

Option (D): As discussed above, “scale of” is incorrect in option (D).
It is also illogical “and with each connected” doesn’t clarify what of each is connected by roads.

Option (E): As discussed above, “scale of” is incorrect in option (E).

Also, it is a run-on sentence.

The settlements were built on a spectacular scale of more than 75 carefully engineered structures of up to 600 rooms each had been connected by a complex regional system of roads.

The two different clauses do not even have a comma in between here. Also, the use of past perfect “had been” is not justified.

Hope you understand run-on sentences a little better now.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Playing the Devil’s Advocate on the GMAT

Quarter Wit, Quarter WisdomConfess it – while watching Harvey Specter and Mike Ross on ‘Suits’, many of you have wondered how ‘cool’ it would be to be a lawyer. It’s surprising how they question every assumption, every reason and come up with an innovative solution which looks as if the magician just pulled a rabbit out of a hat.

Well, in high level GMAT questions, you have a chance to play the Devil’s Advocate. If your best thought out logic says that answer has to be 2, still think why it cannot be 1. The higher level questions are quite tricky and if you play at 700+ level, you will need to be extra careful – if it seems too easy, it probably is! To illustrate, we have quite a brilliant little question from GMAT Prep.

Question 1: If 5x^2 has two different prime factors, at most how many different prime factors does x have?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Solution: So here is the logic with which most of us would come up – 5x^2 has two different prime factors – one would be 5 since it is already there so x must have one more prime factor. x^2  has only those prime factors that x has so 5x^2 will have two prime factors – one 5 and the other from x. Sounds perfectly reasonable and the answer should be 1 – x has 1 prime factor. In fact, it must have at least one prime factor and it cannot have more than one prime factor.

But then, and here we have a hint in the question to play the devil’s advocate – why does the question ask “at most how many different prime factors”. If there were a single value for different prime factors of x, the question would have probably said “how many different prime factors…”. There would have been no need for those words ‘at most’.

Then look at the other options. Is it possible that x has 2 prime factors? It certainly cannot have more than 2 distinct prime factors since then, 5x^2 will have more than two distinct prime factors. Actually, x can have two prime factors! x can have 5 as a factor too. Sneaky – eh? We already have a 5 in 5x^2 but that doesn’t mean that we cannot have a 5 in x^2 too. It will still count as a single prime factor. x can have another prime factor such as 2 or 3 or 7 or 11 etc. In that case, x can have two distinct prime factors.

So x can be 15 (two different prime factors) and x can be 25 (one prime factor)

Answer (B)

Note that this question has no calculations and no time consuming equations but still, this little trick makes this question quite hard. If most people get it wrong because of missing this trick, the question will be termed hard.

Now here is a trickier version of this question:

Question 2: How many prime factors does positive integer n have?

Statement 1: n/7 has only one prime factor.
Statement 2: 3*n^2 has two different prime factors.

Solution: Let’s keep in mind our learning from above while trying to solve this question.

Statement 1: n/7 has only one prime factor.

n/7 has a factor so obviously, it is an integer. Hence n must have a 7 as a factor. So we might jump to the conclusion that n has two prime factors –7 and another one which is left when n is divided off by 7. So n would be something like 7*3 so that n/7 = 7*3/7 = 3 (only one prime factor).

But what we wouldn’t have considered in this case is that n may have multiple 7s so that when a 7 is cancelled in n/7, you would still be left with 7 i.e. if n is 7*7, then n/7 = 7*7/7 = 7. In this case, n has only one distinct prime factor.

So n can have either one or two prime factors. This statement alone is not sufficient.

Statement 2: 3*n^2 has two different prime factors.

This is the same as our previous question. 3n^2 has two different prime factors but n itself can have either one or two prime factors (one of which will be 3). For example, n can be 7 or n can be 3*7. This statement alone is not sufficient.

Using both statements, n could have one or two prime factors i.e. n could be 49 (only one prime factor – 7) or n could be 21 (two prime factors).

Hence, even using both the statements, we cannot say how many prime factors n has.

Answer (E)

Now this question might not have seemed very complicated since we discussed the logic in the first question above. Remember to play the devil’s advocate in high level questions.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

A Closer Look at GMAT Function Questions

Quarter Wit, Quarter WisdomLast week, we looked at the basics of how to handle function questions. Today, let’s look at a couple of questions. We will start with an easier one and then go on to a slightly tougher one.

Question 1: If f(x) = 343/x^3, what is the value of f(7x)* f(x/7) in terms of f(x)?

(A)  f(x^2)

(B) (f(x))^2

(C) f(x^3)

(D) (f(x))^3

(E) f(343x)

Solution: We discussed that to get f(a) given f(x), all you need to do is substitute x with a.

f(x) = 343/x^3

f(7x) = 343/(7x)^3 = 1/x^3

f(x/7) = 343/(x/7)^3 = 343*343/x^3

So we get f(7x) * f(x/7) = (1/x^3) * (343*343/x^3) = (343/x^3)^2

But we know that 343/x^3 = f(x)

So, f(7x) * f(x/7) = (f(x))^2

Answer(B)

There are other ways of solving this too:

Say x = 1, then f(1) = 343

f(7x)* f(x/7) = f(7)*f(1/7) = (343/7^3) * (343/(1/7)^3 = (343)^2

So f(7x)* f(x/7) = (f(1))^2

We hope you see that the question was not difficult to solve. Once you get over your fear of symbols, it is quite straight forward.

Now, let’s take a question similar to an official question from the GMAT paper tests:

Question 2: The function f is defined for each positive three-digit integer n by f(n) = 2^x * 3^y * 5^z, where x, y and z are the hundreds, tens, and units digits of n, respectively. If m and v are three-digit positive integers such that f(m) = 25f(v), then m-v=?

(A) 2
(B) 9
(C) 18
(D) 20
(E) 100

Solution: The question may seem a bit difficult to understand first so let’s take one sentence at a time:

f is defined for each positive three-digit integer n by f(n) = 2^x * 3^y * 5^z, where x, y and z are the hundreds, tens, and units digits of n

Let’s take an example to see how to make sense of it: say 146 is a three digit positive integer. So f(146) = 2^1 * 3^4 * 5^6

In the same way, f(283) = 2^2 * 3^8 * 5^3

If m and v are three-digit positive integers such that f(m) = 25f(v)

So f(m) = 5^2 * f(v)

If m is represented as abc and v as def, then (2^a * 3^b * 5^c) = 5^2 * (2^d * 3^e * 5^f)

Note that for the left hand side to be equal to right hand side, a = d, b = e and c = 2 + f.

So the units digit of m is 2 more than the units digit of v but their tens and hundreds digits are the same.

So m – v = 2.

Answer (A)

If you are still not sure how we arrived at this, take an example.

f(m) = f(266) = 2^2 * 3^6 * 5^6

f(v) = f(264) = 2^2 * 3^6 * 5^4

The difference between f(m) and f(v) will be of 5^2 since their units digits are 2 away from each other.

Hope next time you see a functions question, you will not get spooked and will, instead, take it in your stride!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Functions on GMAT

Quarter Wit, Quarter WisdomLet’s discuss how to handle functions today. People usually perceive functions as an advanced topic mainly because of the notation. But actually, the function questions are very simplistic and can be solved with a simple process. If we ask you the value of 5x^3 where x = 3, would you be worried about what to do? We assume you won’t be. Then there should be no problem with “given f(x) = 5x^3, what is the value of f(3)?”

Just keep a few simple things in mind:

– f(x) = …. will be followed by an expression in x. This is the core of your calculations. You can turn a blind eye to f(x) – just focus on the expression. For example: f(x) = (x^2+1)/5x. Keep your eye on (x^2+1)/5x.

– When faced with “what is f(a)?” all you have to do is recall the expression given and put x = a in that. It doesn’t matter what a is – wherever you have x, just put ‘a’ there. For example: if f(x) = (x^2+1)/5x, what is f(5x^3)? Don’t get confused. Here, a = 5x^3. Look for x in the expression and put 5x^3 in its place.

f(5x^3) = ((5x^3)^2+1)/5(5x^3)

If you seem to be getting lost in too many exponents, terms etc, in place of x, put 5x^3 and put brackets around it as done above. Then simplify by opening the brackets.

f(5x^3) = (25x^6+1)/25x^3

– When we are given that f(a) = B, put x = a in the expression and equate the whole expression to B. For example: f(x) = (x^2+1)/5x, given that f(a) = 2/5, what is the value of a?

We know how to find f(a). It is simply (a^2+1)/5a. We are given that this is 2/5.

(a^2+1)/5a = 2/5

a^2 + 1 = 2a

a^2 – 2a + 1 = 0

(a – 1)^2 = 0

a = 1

That is pretty much all you need. Let’s look at a GMAT Prep question on functions.

Question: For which of the following functions f is f(x) = f(1-x) for all x?

(A) f(x) = 1 – x

(B) f(x) = 1 – x^2

(C) f(x) = x^2 – (1 – x)^2

(D) f(x) = x^2*(1 – x)^2

(E) f(x) = x/(1 – x)

Solution: What does this mean: f(x) = f(1-x)? It means that given a certain expression in x called f(x), for which function will that be the same as f(1-x) i.e. when you substitute x by (1-x), which expression will stay the same? Let’s look at each option:

(A) f(x) = 1 – x

Substitute (1 – x) in place of x to see what f(1 – x) looks like.

f(1 – x) = 1 – (1 – x)

f(1 – x) = x

f(x) is not the same as f(1-x) here. Ignore this option.

(B) f(x) = 1 – x^2

Substitute (1 – x) in place of x to see what f(1 – x) looks like.

f(1 – x) = 1 – (1 – x)^2

f(1 – x) = 2x – x^2

f(x) is not the same as f(1-x) here. Ignore this option.

(C) f(x) = x^2 – (1 – x)^2

Substitute (1 – x) in place of x to see what f(1 – x) looks like.

f(1 – x) = (1 – x)^2 – (1 – (1-x))^2

f(1 – x) = (1 – x)^2 – x^2

f(1 – x) = -x^2 + (1 – x)^2

f(x) is not the same as f(1-x) here. Ignore this option.

(D) f(x) = x^2*(1 – x)^2

Substitute (1 – x) in place of x to see what f(1 – x) looks like.

f(1 – x) = (1 – x)^2 * (1 – (1 – x))^2

f(1 – x) = (1 – x)^2 * x^2

f(1 – x) = x^2 * (1 – x)^2

Note that here, f(x) = f(1 – x), so this must be our answer. Still, let’s take a look at (E) as well for practice.

(E) f(x) = x/(1 – x)

Substitute (1 – x) in place of x to see what f(1 – x) looks like.

f(1 – x) = (1 – x)/(1 – (1-x))

f(1 – x) = (1 – x)/x

f(x) is not the same as f(1-x). Ignore this option.

Answer (D)

A cursory look back at the solution might make you feel that it involves some complicated manipulations but we hope you do see that it is anything but complicated. Now there are some other ways of handling this question too. If you are comfortable with the above, continue with the rest of the post.

Method 2: Number Plugging

We want the expression for which f(x) = f(1 – x) for ALL values of x. So no matter what value we give x, f(x) should be same as f(1 – x).

Say, if x = 0, for which function is f(x) = f(1 –x )? i.e. for which function is f(0) = f(1)

(A) f(x) = 1 – x

f(0) = 1 and f(1) = 0. Not equal.

(B) f(x) = 1 – x^2

f(0) = 1 and f(1) = 0. Not equal.

(C) f(x) = x^2 – (1 – x)^2

f(0) = -1 and f(1) = 1. Not equal.

(D) f(x) = x^2*(1 – x)^2

f(0) = 0 and f(1) = 0. Equal. But when using number plugging, you need to check all options because multiple options could give you equal values. In that case, you would need to try for another value of x.

(E) f(x) = x/(1 – x)

f(0) = 0 and f(1) is not defined. Just to be sure, say x = -1.

f(-1) = -1/2 and f(2) = -2. Not equal.

Answer (D)

Method 3: Intuitive Approach

Try to first focus on the options where terms are added/multiplied rather than subtracted/divided. They are more symmetrical and a substitution may not change the expression. Intuitively, we should check (D) first since it involves multiplication of the terms.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Past Perfect without Past Tense on GMAT Sentence Correction Questions

Quarter Wit, Quarter WisdomRecall the golden rule of past perfect tense –

The Past Perfect expresses the idea that something occurred before another action in the past. It can also show that something happened before a specific time in the past.

We often ignore the “something happened before a specific time in the past” part of the tense.

For example, look at this sentence: Robin had never cooked pasta before last night.

Here, we use past perfect “had cooked” without another verb in the past tense – why? Because we use past perfect to show that something happened before a specific time in the past i.e. before last night.

Similarly, sometimes in GMAT too, you may see past perfect where it seems reasonable but you may not find a verb in past tense. It could be because an action happened before a specific time in the past or there is an implied action in the past. There is a reason why we brought up this point – check out the sentence given below:

According to some economists, the gains in the stock market reflect growing confidence that the economy will avoid the recession that many had feared earlier in the year and instead come in for a ‘soft landing’.

The sentence is similar to a correct sentence given in Official Guide. Note the use of “had feared” – many people question the use of past perfect here. The reason past perfect is correct here is this:

“According to some economists” implies an action in the past – something similar to “Some economists said” or in other words, it implies a specific time in the past – the time when the economists expressed their opinion. In the sentence, “earlier in the year” is a time before the economists expressed their opinion and hence it makes sense to use past perfect. In such cases, our use of common sense is more important than the mere retention of grammar rules. Another thing that helps in such situations is that all other options would have a major fault.

Let’s show you the actual OG question:

Question: According to some analysts, the gains in the stock market reflect growing confidence that the economy will avoid the recession that many had feared earlier in the year and instead come in for a “soft landing,” followed by a gradual increase in business activity.

(A) that the economy will avoid the recession that many had feared earlier in the year and instead come
(B) in the economy to avoid the recession, what many feared earlier in the year, rather to come
(C) in the economy’s ability to avoid the recession, something earlier in the year many had feared,
and instead to come
(D) in the economy to avoid the recession many were fearing earlier in the year, and rather to come
(E) that the economy will avoid the recession that was feared earlier this year by many, with it instead coming

Let’s look at the errors in the other options:

(B) in the economy to avoid the recession, what many feared earlier in the year, rather to come

You cannot use “what” in place of “which”. Also, the use of “confidence in A to avoid” is not correct. It should be “confidence that A will avoid”.

(C) in the economy’s ability to avoid the recession, something earlier in the year many had feared,
and instead to come

The placement of “earlier in the year” is incorrect here. It should come after “had feared”.

(D) in the economy to avoid the recession many were fearing earlier in the year, and rather to come

Again, the use of “confidence in A to avoid” is not correct. It should be “confidence that A will avoid”.

(E) that the economy will avoid the recession that was feared earlier this year by many, with it instead coming

“With it instead coming” doesn’t make any sense so this option isn’t correct either.

So we see that all other options have fatal flaws. Hence, in this case, option (A) is our best bet even though the use of past perfect isn’t the way we usually see it.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

4 Average Speed Formulas You Need to Know for the GMAT

Quarter Wit, Quarter WisdomMany people have asked us to clear the confusion surrounding the various formulas of average speed. We will start with the bottom line – There is a single versatile formula for ALL average speed questions and that is

Average Speed = Total Distance/Total Time

No matter which formula you choose to use, it will always boil down to this one. Keeping this in mind, let’s discuss the various formulas we come across:

1. Average Speed = (a + b)/2

Applicable when one travels at speed a for half the time and speed b for other half of the time. In this case, average speed is the arithmetic mean of the two speeds.

2. Average Speed = 2ab/(a + b)

Applicable when one travels at speed a for half the distance and speed b for other half of the distance. In this case, average speed is the harmonic mean of the two speeds. On similar lines, you can modify this formula for one-third distance.

3. Average Speed = 3abc/(ab + bc + ca)

Applicable when one travels at speed a for one-third of the distance, at speed b for another one-third of the distance and speed c for rest of the one-third of the distance.

Note that the generic Harmonic mean formula for n numbers is

Harmonic Mean = n/(1/a + 1/b + 1/c + …)

4. You can also use weighted averages. Note that in case of average speed, the weight is always ‘time’. So in case you are given the average speed, you can find the ratio of time as

t1/t2 = (a – Avg)/(Avg – b)

As you already know, this is just our weighted average formula.

Now, let’s look at some simple questions where you can use these formulas.

Question 1: Myra drove at an average speed of 30 miles per hour for T hours and then at an average speed of 60 miles/hr for the next T hours. If she made no stops during the trip and reached her destination in 2T hours, what was her average speed in miles per hour for the entire trip?

(A)   40

(B)   45

(C)   48

(D)   50

(E)    55

Solution: Here, time for which Myra traveled at the two speeds is same.

Average Speed = (a + b)/2 = (30 + 60)/2 = 45 miles per hour

Answer (B)

Question 2: Myra drove at an average speed of 30 miles per hour for the first 30 miles of a trip & then at an average speed of 60 miles/hr for the remaining 30 miles of the trip. If she made no stops during the trip what was her average speed in miles/hr for the entire trip?

(A) 35
(B) 40
(C) 45
(D) 50
(E) 55

Solution: Here, distance for which Myra traveled at the two speeds is same.

Average Speed = 2ab/(a+b) = 2*30*60/(30 + 60) = 40 mph

Answer (B)

Question 3: Myra drove at an average speed of 30 miles per hour for the first 30 miles of a trip, at an average speed of 60 miles per hour for the next 30 miles and at a average speed of 90 miles/hr for the remaining 30 miles of the trip. If she made no stops during the trip, Myra’s average speed in miles/hr for the entire trip was closest to

(A) 35
(B) 40
(C) 45
(D) 50
(E) 55

Solution: Here, Myra traveled at three speeds for one-third distance each.

Average Speed = 3abc/(ab + bc + ca) = 3*30*60*90/(30*60 + 60*90 + 30*90)

Average Speed = 3*2*90/(2 + 6 + 3) = 540/11

This is a bit less than 50 so answer (D).

Question 4: Myra drove at an average speed of 30 miles per hour for some time and then at an average speed of 60 miles/hr for the rest of the journey. If she made no stops during the trip and her average speed for the entire journey was 50 miles per hour, for what fraction of the total time did she drive at 30 miles/hour?

(A) 1/5
(B) 1/3
(C) 2/5
(D) 2/3
(E) 3/5

Solution: We know the average speed and must find the fraction of time taken at a particular speed.

t1/t2 = (A2 – Aavg)/(Aavg – A1)

t1/t2 = (60 – 50)/(50 – 30) = 1/2

So out of a total of 3 parts of the journey time, she drove at 30 mph for 1 part and at 60 mph for 2 parts of the time. Fraction of the total time for which she drove at 30 mph is 1/3.

Answer (B)

Hope this sorts out some of your average speed formula confusion.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Using Symmetry in Probability on the GMAT

Quarter Wit, Quarter WisdomWe know that Combinatorics and Probability are tricky topics. It is easy to misinterpret questions of these topics and get the incorrect answer – which, unfortunately, we often find in the options, giving us a false sense of accomplishment.

In many questions, we need to account for different cases one by one but we don’t really see such questions on the GMAT since we have limited time. Also, we don’t tire of repeating this again and again – GMAT questions are more reasoning based than calculation intensive. Usually, there will be an intellectual method to solve every GMAT question – a method that will help you solve it in seconds.

We have discussed using symmetry in Combinatorics before. It can be used in many questions though most people don’t realize that. In our ongoing endeavor to expose you to intellectual methods, here we present how most people tackle a question and how you can tackle it instead to be in the top 1%ile.

Question: Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for which the first term is not 2. A permutation is chosen randomly from S. The probability that the second term is 5 is given by a/b (in lowest terms). What is a+b?

(A) 5
(B) 6
(C) 11
(D) 16
(E) 19

Solution:

Most Common Solution:

What are the permutations of sequence S? They are the different ways in which we can arrange the elements of S. For example, 3, 2, 4, 5, 6 or 4, 2, 3, 6, 5 or 6, 3, 4, 5, 2 etc

In how many different ways can we make the sequence? The first element can be chosen in 4 ways – one of 3, 4, 5 and 6. (You are given that 2 cannot be the first element).

The second element can be chosen in 4 ways (2 and the leftover 3 numbers).

The third element can be chosen in 3 ways.

The fourth element can be chosen in 2 ways.

And finally there will be only 1 element left for the last spot.

Number of ways of making set S = 4*4*3*2*1 = 96

In how many of these sets will 5 be in the second spot?

If 5 is reserved for the second spot, there are only 3 ways of filling the first spot (3 or 4 or 6).

The second spot has to be taken by 5.

The third element will be chosen in 3 ways (ignoring 5 and the first spot)

The fourth element can be chosen in 2 ways.

And finally there will be only 1 element left for the last spot.

Number of favorable cases = 3*1*3*2*1 = 18

Required Probability = Favorable Cases/Total Cases = 18/96 = 3/16 = a/b

a+b = 3 + 16 = 19

Answer (E)

Intellectual Approach:

Use a bit of logic of symmetry to solve this question without any calculations.

Set S would include all such sequences as 3, 2, 4, 5, 6 or 4, 2, 3, 6, 5 or 6, 3, 4, 5, 2 etc – starting with 3, with 4, with 5 or with 6 with equal probability.
By symmetry, note that 1/4th of them will start with 5 – which we need to ignore – so we are left with the rest of the 3/4th sequences.

Now, in these 3/4th sequences which start with either 3 or 4 or 6, 5 could occupy any one of the 4 positions – second, third, fourth or fifth with equal probability. So we need 1/4th of these sequences i.e. only those sequences in which 5 is in the second spot.

Probability that 5 is the second element of the sequence = (3/4)*(1/4) = 3/16

Therefore, a+b = 3+16 = 19

Answer (E)

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Of Opinions and Facts in GMAT Critical Reasoning Questions

Quarter Wit, Quarter WisdomToday, we would like to discuss with you one of our most debated critical reasoning questions. It is an absolutely brilliant question – not just because the correct option fits in beautifully but because the other four options are also very well thought out. It is easy to write the incorrect four options such that the student community will be split between 2 options – the correct one and one of the four incorrect ones but when the jury is split between 4 or all 5 options, that’s when we know that we have come up with an absolute masterpiece. Of course, in such questions, a lot of effort is needed to convince everyone of the correct answer but it is well worth it.

This question brings an important point to the fore – the correct option in strengthen/weaken question is the one that supplies new information but in most cases, the new information has to be a fact, not an opinion. Let’s explain this in detail with the help of this question.

Question: According to recent research, a blindfolded person whose nostrils have been pinched so that smelling is impossible will have great difficulty in differentiating a bite of an apple from a bite of a raw potato. This clearly demonstrates that taste buds are not the only sense organs involved in determining the taste of a piece of food.

Which of the following premises, is an assumption required by the argument?

(A) All people agree that an apple and a potato differ in taste.

(B) There are no other senses involved in tasting other than taste, smell, and sight.

(C) The word “taste” can be used to describe an experience that involves sight or smell or both.

(D) The research was based on experiments that were conducted on a broad spectrum of the general population.

(E) People who have been blindfolded and whose nostrils are pinched can differentiate a bite of an apple from a bite of an onion more easily than they can differentiate a bite of an apple from a bite of a raw potato.

Solution:

Argument:

– If you remove sight and smell, people will have great difficulty in differentiating a bite of an apple from a bite of a raw potato.

Conclusion: Taste buds are not the only sense organs involved in determining the taste of a piece of food.

We will look at the options one by one:

(A) All people agree that an apple and a potato differ in taste.

Note that usually, people’s opinion will not count for much. Facts are the ones which are important. The only opinion we care about is the author’s. We cannot strengthen/weaken the author’s opinion by giving similar/dissimilar opinions of other people.

Say, the conclusion of an argument is:

Daniel Day-Lewis is the greatest actor of the 21st century.

The premises would perhaps list his great performances, talk about his acting prowess, his Oscars and so on.

Can you strengthen the conclusion by saying that “My friend also believes that he is the greatest actor.”? No. You cannot strengthen your opinion by giving the opinion of other people. You need to give facts to strengthen your view.

So this option is already suspect. It is giving you the opinion of people “All people agree that an apple and a potato differ in taste.” So it doesn’t seem to be the right choice.

Anyway, let’s try to negate (A) just to be sure since this is an assumption question.

Negation: Not all people agree that an apple and a potato differ in taste.
This means there is at least one person who does not agree that an apple and a potato differ in taste. Perhaps he feels that the experience of eating an apple – the smell, the look, the sweetness etc is the same as the experience of eating a potato. It is still possible that taste buds are not the only sense organs involved in determining the taste of a piece of food. Even after we negate (A), the conclusion is possible so (A) is not an assumption.

Think of it in another way: During the research, blindfolded people with pinched noses found it very hard to differentiate the taste. One person comes up and says that he himself cannot differentiate between the two while looking and smelling. Does it mean that senses other than taste buds are not involved? No. There could be many other people who feel that they can easily differentiate between an apple and a potato taste. So other senses could be involved and (A) is not your answer.

(B) There are no other senses involved in tasting other than taste, smell, and sight.

This is not an assumption. All we are saying is that taste buds are not the only sense organs involved in determining the taste of a piece of food. Any other organs could be involved including smell and sight.

(C) The word “taste” can be used to describe an experience that involves sight or smell or both.

This option highlights a very basic thing that needs to be true for our conclusion to hold. When we conclude: taste buds are not the only sense organs involved in determining the taste of a piece of food, how do we define “taste”? Taste buds, we know, tell us whether the food is salty/sweet/sour etc. But how do we say that “taste” is not defined by only these features? We are assuming that taste is defined by not just how the food sits on our tongue but by other features such as sight/smell too. If this option were not true, then we would have needed only taste buds to find the taste of food. Hence, our conclusion would fall apart. Hence, (C) is the correct answer.

(D) The research was based on experiments that were conducted on a broad spectrum of the general population.

The conclusion does not say that for people of most classes/regions, taste buds are not the only sense organs involved in determining the taste of a piece of food. It is acceptable if the research was conducted on a few people and it was determined that other senses are involved. Even if some people found it difficult to differentiate between the two things, we can say that other senses are involved.

(E) People who have been blindfolded and whose nostrils are pinched can differentiate a bite of an apple from a bite of an onion more easily than they can differentiate a bite of an apple from a bite of a raw potato.

This option tells us that apples and onions are more different on the tongue than apples and potatoes. This is out of scope and is certainly not an assumption.

Answer (C)

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Advanced Number Properties on the GMAT – Part V

Quarter Wit, Quarter WisdomToday, let’s look in detail at a relation between arithmetic mean and geometric mean of two numbers. It is one of those properties which make sense the moment someone explains to us but are very hard to arrive on our own.

When two positive numbers are equal, their Arithmetic Mean = Geometric Mean = The number itself

Say, the two numbers are m and n (and are equal). Their arithmetic mean = (m+n)/2 = 2m/2 = m

Their geometric mean = sqrt(m*n) = sqrt(m^2) = m (the numbers are positive so |m| = m)

We also know that Arithmetic Mean >= Geometric Mean

So when arithmetic mean is equal to geometric mean, it means the arithmetic mean is taking its minimum value. So when (m+n)/2 is minimum, it implies (m+n) is minimum. Therefore, sum of numbers takes its minimum value when the numbers are equal.

When geometric mean is equal to arithmetic mean, it means the geometric mean is taking its maximum value. So when sqrt(m*n) is maximum, it means m*n is maximum. Therefore, product of numbers takes its maximum value when the numbers are equal.

Let’s see how to solve a difficult question using this concept.

Question: If x and y are positive, is x^2 + y^2 > 100?

Statement 1: 2xy < 100

Statement 2: (x + y)^2 > 200

Solution:

We need to find whether x^2 + y^2 must be greater than 100.

Statement 1: 2xy < 100

Plug in some easy values to see that this is not sufficient alone.

If x = 0 and y = 0, 2xy < 100 and x^2 + y^2 < 100

If x = 40 and y = 1, 2xy < 100 but x^2 + y^2 > 100

So x^2 + y^2 may be less than or greater than 100.

Statement 2: (x + y)^2 > 200

There are two ways to deal with this statement. One is the algebra way which is easier to understand but far less intuitive. Another is using the concept we discussed above. Let’s look at both:

Algebra solution:

We know that (x – y)^2 >= 0 because a square is never negative.

So x^2 + y^2 – 2xy >= 0

x^2 + y^2 >= 2xy

This will be true for all values of x and y.

Now, statement 2 gives us x^2 + y^2 + 2xy > 200. The left hand side is greater than 200. If on the left we substitute 2xy with (x^2 + y^2), the left hand side will either become greater than or same as before. So in any case, the left hand side will remain greater than 200.

x^2 + y^2 + (x^2 + y^2) > 200

2(x^2 + y^2) > 200

x^2 + y^2 > 100

This statement alone is sufficient to say that x^2 + y^2 will be greater than 100. But, we agree that the first step where we start with (x – y)^2 is not intuitive. It may not hit you at all. Hence, here is another way to analyze this statement.

Logical solution:

Let’s try to find the minimum value of x^2 + y^2. It will take minimum value when x^2 = y^2 i.e. when  x = y (x and y are both positive)

We are given that (x+y)^2 > 200

(x+x)^2 > 200

x > sqrt(50)

So x^2 + y^2 will take a value greater than [sqrt(50)]^2 + [sqrt(50)]^2 = 100.

So in any case, x^2 + y^2 will be greater than 100. This statement alone is sufficient to answer the question.

Answer (B)

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Pre-thinking in Quant GMAT Questions

Quarter Wit, Quarter WisdomWe all know about the role of pre-thinking in Critical Reasoning and how anticipating the answer can be supremely beneficial in not just the physical aspect of saving time in analyzing options but also the psychological aspect of promoting our self-confidence – we were thinking that the answer should look like this and that is exactly what we found! Pre-thinking puts us in the driver’s seat and we feel energized without consuming any red bull!

The exciting thing is that pre-thinking is useful in Quant too. If you take a step back to review what the question asks and think about what you are going to do and what you expect to get, it is highly likely that you will not get distracted mid-way during your solution. Let’s show you with the help of an example:

Question: Superfast train A leaves Newcastle for Birmingham at 3 PM and travels at the constant speed of 100 km/hr. An hour later, it passes superfast train B, which is making the trip from Birmingham to Newcastle on the same route at a constant speed. If train B left Birmingham at 3:50 PM and if the sum of the total travel time of the two trains is 2 hours, at what time did train B arrive at Newcastle?

Statement I: Train B arrived at Newcastle before train A arrived at Birmingham.
Statement II: The distance between Newcastle and Birmingham is greater than 140 km.

Following are the things that would ideally constitute pre-thinking on this question:

– Quite a bit of data is given in the question stem with some speed and time taken.

– Distance traveled by both the trains is the same since they travel along the same route.

– We could possibly make an equation by equating the two distances and come up with multiple answers for the time at which train B arrived at Newcastle.

– The statements do not provide any concrete data. We cannot make any equation using them but they might help us choose one of the answers we get from the equation of the question stem.

Mind you, the thinking about the statements helping us to arrive at the answer is just speculation. The answer may well be (E). But all we wanted to do at this point was find a direction.

The diagram given above incorporates the data given in the question stem. Train A starts from Newcastle toward Birmingham at 3:00 and meets train B at 4:00. Train B starts from Birmingham toward Newcastle at 3:50 and meets train A at 4:00. Let x be the distance from Birmingham to the meeting point.

Speed of train A = 100 km/hr
Speed of train B = Distance/Time = x/(10 min) = x/(1/6) km/hr = 6x km/hr (converted min to hour)

If we get the value of x, we get the value of speed of train B and  that tells us the time it takes to travel from the meeting point to Newcastle (a distance of 100 km). So all we need to figure out is whether the statements can give us a unique value of x.

By 4:00, train A has already travelled for 1 hour and train B has already travelled for 10 mins i.e. 1/6 hour. Total time taken by both is 2 hrs. The remaining (5/6) hrs is the time needed by both together to reach their respective destinations.

Time taken by train A to reach Birmingham + Time taken by train B to reach Newcastle = 5/6

Distance(x)/Speed of train A + 100/Speed of train B = 5/6

x/100 + 100/6x = 5/6

3x^2 – 250x + 5000 = 0

3x^2 – 150x – 100x + 5000 = 0

3x(x – 50) – 100(x – 50) = 0

(3x – 100)(x – 50) = 0

x = 100/3 or 50

So speed of train B = 6x = 200 km/hr or 300 km/hr

Statement 1: Train B arrived at Newcastle before Train A arrived at Birmingham.
If x = 50, time taken by train A to reach Birmingham = 50/100 = 1/2 hour and time taken by train B to reach Newcastle = 100/300 = 1/3 hour. Train B takes lesser time so it arrives first.
If x = 33.33, time taken by train A to reach Birmingham = (100/3)/100 = 1/3 hour and time taken by train B to reach Newcastle = 100/200 = 1/2 hour. Here, train A takes lesser time so it arrives first at its destination.
Since train B arrived first, x must be 50 and train B must have taken 1/3 hour i.e. 20 mins to arrive at Newcastle. So train B must have arrived at 4:20.

This statement is sufficient alone.

Statement 2: The distance between Newcastle and Birmingham is greater than 140 km.

Total distance between Newcastle and Birmingham = (100 + x) km. x must be 50 to make total distance more than 140.

Time taken by train B must be 1/3 hr (as calculated above) and it must have arrived at 4:20.

This statement is sufficient alone.

Answer (D)

So our speculation was right. Each of the statements provided us relevant information to choose one of the two values that the quadratic gave us.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Intelligent Guessing on GMAT

Quarter Wit, Quarter WisdomWe often tell you that if you are short on time, you can guess intelligently on a few questions and move on. Today we will discuss what we mean by “intelligent guessing”. There are many techniques – most of them involving your reasoning skills to eliminate some options and hence generating a higher probability of an accurate guess. Let’s look at one such method to get values in the ballpark.

A few months back, we had discussed a 700 level ‘Races’ question.

Question 1: A and B run a race of 2000 m. First, A gives B a head start of 200 m and beats him by 30 seconds. Next, A gives B a head start of 3 mins and is beaten by 1000 m. Find the time in minutes in which A and B can run the race separately.

(A)   8, 10

(B)    4, 5

(C)   5, 9

(D)   6, 9

(E)    7, 10

Check out its complete solution here.

Now, what if we had only 30 seconds to guess on it and move on?  Then we could have easily guessed (B) here and moved on. Actually, the question implies that the only possible options are those in which the time taken by B is somewhere between 3 mins and 6 mins (excluding) – we would guess 4 mins or 5 mins. Since only option (B) has time taken by (B) as 5 mins, that must be the answer – no chances of error here – perfect! Had there been 2 options with 4 mins/5 mins, we would have increased the probability of getting the correct answer to 50% from a mere 20% within 30 seconds.

Now you are probably curious as to how we got the 3 min to 6 min range. Here is the logic:

Read one sentence of the question at a time –

A and B run a race of 2000 m. First, A gives B a head start of 200 m and beats him by 30 seconds.

So first, A gives B a head start of 1/10th of the race but still beats him. This means B is certainly quite a bit slower than A. This should run through your mind on reading this sentence.

Next, A gives B a head start of 3 mins and is beaten by 1000 m.

Next, A gives B a head start of 3 mins and B beats him by 1000 m i.e. half of the race. What does this imply? It implies that B ran more than half the race in 3 mins. To understand this, say B covers x meters in 3 mins. Once A, who is faster, starts running, he starts reducing the distance between them since he is covering more distance than B every second. At the end, the distance between them is still 1000 m. This means the initial distance that B created between them by running for 3 mins was certainly more than 1000 m (This was intuitively shown in the diagram in this post). Since B covered more than 1000 m in 3 mins, he would have taken less than 6 mins to cover the length of the race i.e. 2000 m. A must be even faster and hence would take even lesser time.

Only option (B) has time taken by B as 5 mins (less than 6 mins) and hence satisfies our range! So the answer has to be (B).

Let’s try the same technique on another question.

Question 2: If 12 men and 16 women can do a piece of work in 5 days and 13 men and 24 women can do it in 4 days, how long will 7 men and 10 women take to do it?

(A)   4.2 days

(B)   6.8 days

(C)   8.3 days

(D)   9.8 days

(E)    10.2 days

Solution: If we try to use algebra here, the calculations involved will be quite complicated. The options are not very close together so we can try to get a ballpark value and move forward. Let’s take each sentence at a time:

If 12 men and 16 women can do a piece of work in 5 days

Say rate of work of each man is M and that of each woman is W. This statement gives us that

12M + 16W = 1/5  (Combined rate done per day)

In lowest terms, it is 3M + 4W = 1/20

13 men and 24 women can do it in 4 days,”

This gives us 13M + 24W = 1/4

how long will 7 men and 10 women take to do it?”

Required: 7M + 10W = ?

Solving the two equations above will be tedious so let’s estimate:

(3M + 4W = 1/20) * 6 gives 6M + 8W = 1/10

So 6 men and 8 women working together will take 10 days. Hence, 7 men and 10 women will certainly take fewer than 10 days.

(13M + 24W = 1/4) / 2 gives 6.5M + 12W = 1/8

So 7 men and 10 women might take about 8 or perhaps a little bit more than 8 days to complete the work. There is only 0.5 additional man (hypothetically) but 2 fewer women to complete it. So we would guess that the number of days would lie between 8 to 10 and closer to 8 days.

Answer (C) fits.

Note that it seems like there are many equations here but all you have actually done is made two equations. Once you write them down, you don’t even need to actually multiply them with some integer to get them close to the required equation. Just looking at the first one, you can say that 6 men and 8 women will take 10 days. It takes but a couple of seconds to arrive at these conclusions.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

The Speed and Accuracy Trade Off on the GMAT

Quarter Wit, Quarter WisdomWe know that speed is important in GMAT. We have about 2 mins per question and we always have questions in which we get stuck, waste 3-4 mins and probably still answer incorrectly. So we are always trying to go faster, rush, complete the easy ones in less time! In our bid to save time, sometimes we sacrifice accuracy. We should know that accuracy is most important. No point running through questions and completing all of them before time if at the end of it all, most of our answers are incorrect – there are no bonus points for completing the test before time, after all!

In your haste to complete the test on time, don’t overlook the important details. Getting too many easy questions wrong is certainly disastrous. Take a step back and ensure that what they asked is what you have found and that your logic is solid. To illustrate the problem, let’s give you a question – people gloss over it, consider it an easy remainders problem, answer it incorrectly and move on. But guess what, it isn’t as easy as it looks!

Question: If m and n are positive integers such that m > n, what is the remainder when m^2 – n^2 is divided by 21?

Statement 1: The remainder when (m + n) is divided by 7 is 1.

Statement 2: The remainder when (m – n) is divided by 3 is 1.

First let’s give you the incorrect solution provided by many.

Question: What is the remainder when (m^2 – n^2) is divided by 21?

Statement 1: The remainder when (m + n) is divided by 7 is 1.

(m + n) = 7a + 1

Statement 2: The remainder when (m – n) is divided by 3 is 1.

(m – n) = 3b + 1

Therefore, remainder of product (m^2 – n^2) = (m + n)*(m – n) = (7a + 1)(3b + 1) when it is divided by 21 is 1.

Answer (C)

This would have been correct had the statements been:

Statement 1: The remainder when (m + n) is divided by 21 is 1.
Statement 2: The remainder when (m – n) is divided by 21 is 1.

Statement 1: (m + n) = 21a + 1
Statement 2: (m – n) = 21b + 1
(m^2 – n^2) = (m + n)*(m – n) = (21a + 1)*(21b + 1) = 21*21ab + 21a + 21b + 1

Here, every term is divisible by 21 except the last term 1. So when we divide (m^2 – n^2) by 21, the remainder will be 1.

But let’s go back to our original question. If you solved it the way given above and got the answer as (C), you are not the only one who jumped the gun. Many people end up doing just that. But here is the correct solution:

The statements given are:

Statement 1: The remainder when (m + n) is divided by 7 is 1.

(m + n) = 7a + 1

Statement 2: The remainder when (m – n) is divided by 3 is 1.

(m – n) = 3b + 1

This gives us (m^2 – n^2) = (m + n)*(m – n) = (7a + 1)(3b + 1) = 21ab + 7a + 3b + 1

Here only the first term is divisible by 21. We have no clue about the other terms. We cannot say that 7a is divisible by 21. It may or may not be depending on the value of a. Similarly, 3b may or may not be divisible by 21 depending on the value of b. So how can we say here that the remainder must be 1? We cannot. We do not know what the remainder will be in this case even with both statements together.

Say, if a = 1 and b = 1,

m^2 – n^2 = 21*1*1 + 7*1 + 3*1 + 1 = 21 + 11

The remainder when you divide m^2 – n^2 by 21 will be 11.

Say, if a = 2 and b = 1,

m^2 – n^2 = 21*2*1 + 7*2 + 3*1 + 1 = 21*2 + 18

The remainder when you divide m^2 – n^2 by 21 will be 18.

Hence, both statements together are not sufficient to answer the question.

Answer (E)

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Bringing Back the Lazy Genius to Solve GMAT Questions!

Quarter Wit, Quarter WisdomThose of you who have seen the previous version of our curriculum would know that we had tips and tricks under the heading of ‘Lazy Genius’. These used to discuss innovative shortcuts for various questions – the way very smart people would solve the question – without putting in too much effort!

Today, let’s bring back the beloved lazy genius through a question. Try to solve it lazily i.e. try to do minimum work on paper. This means making equations and solving them is a big no-no and doing too many calculations is cumbersome.

Question: A tank has two water pumps Alpha and Beta and one drain Gamma. Pump Alpha alone can fill the whole tank in x hours, and pump Beta alone can fill the whole tank in y hours. The drain can empty the whole tank in z hours such that z>x. When the tank was empty, pumps Alpha and Beta started pumping water in the tank and the drain Gamma simultaneously was draining water from the tank. When finally the tank is full, which of the following represents the amount of water in terms of the fraction of the tank which pump Alpha pumped into the tank?

(A) yz/(x+y+z)

(B) yz/(yz + xz – xy)

(C) yz/(yz + xz + xy)

(D) xyz/(yz + xz – xy)

(E) (yz + xz – xy)/yz

Note that you have variables in the question and the options. Since we are looking for a lazy solution, making equations out of the variables is not acceptable. So then, should we plug in numbers? With three variables to take care of, that might involve a lot of calculations too. Then what else?

Here is our minimum-work-solution to this problem; try to think one of your own and don’t forget to share it with us.

Plugging in numbers for the variables can be troublesome but you can give some very convenient values to the variables so that the effect of a pump and a drain will cancel off.
There are no constraints on the values of x, y and z except z > x (drain Gamma empties slower than pipe Alpha fills)

Let’s say, x = 2 hrs, y = 4 hrs, z = 4 hrs
What did we do here? We made the rate of Beta same as the rate of Gamma i.e. 1/4 of the tank each. This means, whenever both of them are working together, drain Gamma cancels out the work of pump Beta. Every hour, pump Beta fills 1/4th of the tank and every hour drain Gamma empties 1/4th of the tank. So the entire tank will be filled by pump Alpha alone. Hence, if y = z, pump Alpha fills the entire tank i.e. the amount of water in terms of fraction of the tank pumped by Alpha will be 1.
In the options, put y = z and see which option gives you 1. Note that you don’t have to put in the values of 2, 4 and 4. We gave those values only for illustration purpose.

If y = z, xy = xz.

So in option (B), xz cancels xy in the denominator giving yz/yz = 1

Again, in option (E), xz cancels xy in the numerator giving yz/yz = 1

The other options will not simplify to 1 even though when we put y = z, the answer should be 1 irrespective of the value of x, y and z. The other options will depend on the values of x and/or y. Hence the only possible options are (B) and (E). But we still need to pick one out of these two.

Now let’s say, x = 4, y = 2, z = 4.00001 ( z should be greater than x but let’s assume it is infinitesimally greater than x such that we can approximate it to 4 only)
Rate of work of Gamma (1/4th of the tank per hour) is half the rate of work of Beta (1/2 the tank per hour). Rate of work of Gamma is same as rate of work of Alpha. Half the work done by pump Beta is removed by drain Gamma. So if pump Beta fills the tank, drain Gamma empties half of it in that time – the other half would be filled by pump Alpha i.e. the amount of water in terms of fraction of the tank pumped by Alpha will be 1/2.
Put x = z in the options (B) and (E). The one that gives you 1/2 with these values should be the answer. Again, you don’t need to plug in the actual values till the end.

If x = z, yx = yz

(B) yz/(yz + xz – xy)

yz cancels xy in the denominator giving us yz/xz = y/x = 2/4 = 1/2

(E) (yz + xz – xy)/yz

yz cancels xy in the numerator giving us xz/yz = x/y = 4/2 = 2

Only option (B) gives 1/2. Answer (B)

Even if you end up feeling that this method is complicated, try and wrap your head around it. If you do, you are on your way to becoming a lazy genius yourself!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Finding the Last Two Digits on GMAT Quant Questions – Part III

Quarter Wit, Quarter WisdomAs promised last week, we will look at another question which involves finding the last two digits of the product of some random numbers. In this question, along with the concepts discussed last week, we will assimilate the concept of negative remainders too discussed some weeks ago.

Let’s recap the concepts before we see the question:

I. When you divide a number by 100, the remainder is formed by the last two digits of the number.

II. When you divide (px + a)(qx + b)*…*(tx + e) by x, the remainder will be the remainder obtained by dividing a*b*…*e by x.

III. We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again.

These three were discussed with examples last week.

IV. When m is divided by n and a negative remainder (–r) is obtained, we can find the actual remainder simply as (n – r).

This is discussed with examples in this post.

Now, let’s solve a question involving all these concepts.

Question: What are the last two digits of (301*402*503*604*646*547*448*349)^2

(A) 96

(B) 76

(C) 56

(D) 36

(E) 16

Solution: We need to find the last two digits of the product. It means we need to find the remainder when we divide the product by 100.

Find remainder of (301*402*503*604*646*547*448*349)^2/100

Note that 301 = 300 + 1 which gives us a small remainder 1 to work with but 349 = 300 + 49, a large remainder with which calculations will become cumbersome. But note that 349 is close to 350. All the numbers in the product are quite close to a multiple of 50, if not to a multiple of 100.

We need to find the remainder of:

(301*402*503*604*646*547*448*349)*(301*402*503*604*646*547*448*349)/100

This implies we need to find the remainder of:

(301*201*503*604*646*547*448*349)*(301*402*503*604*646*547*448*349)/50

We cancel off a 2 (of 402) from the numerator with a 2 of the denominator to make the divisor 50. We will multiply the remainder we obtain by 2 back at the end.

We need the remainder of:

(300 + 1)*(200+1)*(500+3)*(600+4)*(650-4)*(550-3)*(450-2)*(350-1)* (300+ 1)*(400+2)*(500+3)*(600+4)*(650-4)*(550-3)*(450-2)*(350-1)/50

Note that in this product, all terms obtained will have a multiple of 50 except the last term obtained by multiplying the remainders together:

We need:

the remainder of 1*1*3*4*(-4)*(-3)*(-2)*(-1)*1*2*3*4*(-4)*(-3)*(-2)*(-1) / 50

the remainder of (12)*(24)*(24)*(24) / 50

Notice now that the remainders are far too great but they are close to 25. So let’s bring the divisor down to 25 by canceling off another 2.

We need:

the remainder of (6)*(24)*(24)*(24) / 25

the remainder of (6)*(25-1)*(25-1)*(25-1) / 25

Again, the only product we need to worry about is the last one obtained by multiplying the remainders together:

the remainder of 6*(-1)*(-1)*(-1) / 25

The remainder is -6 which is negative. To get the positive remainder, 25 – 6 = 19. But remember that we had divided the divisor twice by 2 so to get the actual remainder, we must multiply the remainder obtained back by 4: the actual remainder is 4*19 = 76

Answer (B)

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Finding the Last Two Digits on GMAT Quant Questions – Part II

Quarter Wit, Quarter WisdomLet’s continue the discussion of last two digits we started last week. We discussed the concept of pattern recognition and how it can help us determine the last two digits in case of numbers raised to some powers. Today we look at what happens when there is no pattern to determine! What if we are asked to determine the last two digits of the product of a bunch of numbers. We know that getting the last digit in this case is very easy – just multiply the last digits of the numbers together. But last TWO digits would seem much more complicated.

Actually, we can find the last two digits quite easily in most such cases by using the concepts of remainders.

There are two concepts you need to understand before we go on to see how to solve such questions:

I. When you divide a number by 100, the remainder is formed by the last two digits of the number.  Say, you divide 138 by 100, the remainder will be 38 (last two digits). Take another example – divide 1275 by 100, the remainder will be 75 and so on.

II. When you divide (px + a)(qx + b)*…*(tx + e) by x, the remainder will be the remainder obtained by dividing a*b*…*e by x. This should remind you of the binomial theorem we discussed many weeks ago.  When we multiply all these terms together (px + a), (qx + b) etc, each term obtained will have at least one x except the last term which is obtained by multiplying the remainders together. To get a better idea, let’s take some numbers:

Let’s say we need to find the remainder when we divide 12*23*52*81 by 10.

K = 12*23*52*81 = (10 + 2)*(20 + 3)*(50 +2)*(80 + 1)

When you multiply these four terms together, you will get many terms such as 10*20*50*80, 10*20*50*1, 10*20*2*80 etc. All these will have a multiple of 10 except the last one. The last one will be 2*3*2*1 = 12. That doesn’t have a multiple of 10. Now divide 12 by 10 to get the remainder 2. So when you divide K by 10, the remainder will be 2.

Now, let’s look at a question:

Question 1: What are the last two digits of 63*35*37*82*71*41?

(A) 10

(B) 30

(C) 40

(D) 70

(E) 80

Solution: Using concept 1, we know that to find the last two digits, we need to find the remainder we get when we divide the product by 100.

Remainder of (63*35*37*82*71*41)/ 100

Note that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note:

Note: We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again.
So say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct.
Take another example to reinforce this – what is the remainder when 85 is divided by 20? It is 5.
We might rephrase it as – what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct.

So keeping this very important point in mind, let’s go ahead and cancel the common 5 and 2.

We need the

Remainder of (63*7*37*41*71*41*5*2)/10*5*2

Remainder of (63*7*37*41*71*41)/10

Now using concept 2, let’s write the numbers in form of multiples of 10

Remainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10

Remainder of 3*7*7*1*1*1/10

Remainder of 147/10 = 7

Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70.

When 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 70.

Answer (D)

Next week, we will see some more complicated questions using these and other fundamentals.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Finding the Last Two Digits on GMAT Quant Questions – Part I

Quarter Wit, Quarter WisdomWe all know how to find the last digit using cyclicity when we are given a number raised to a power. Last digit of a number depends only on the last digit of the base.  You must be quite familiar with something like this –

Last Digit of Base:

0 – Last digit of expression with any power will be 0.

1 – Last digit of expression with any power will be 1.

2 – 2, 4, 8, 6, 2, 4, 8, 6… Cyclicity is 4.

3 – 3, 9, 7, 1, 3, 9, 7, 1… Cyclicity is 4.

4 – 4, 6, 4, 6, 4, 6, 4, 6… Cyclicity is 2.

5 –  Last digit of expression with any power will be 5.

6 – Last digit of expression with any power will be 6.

7 – 7, 9, 3, 1, 7, 9, 3, 1… Cyclicity is 4.

8 – 8, 4, 2, 6, 8, 4, 2, 6… Cyclicity is 4.

9 – 9, 1, 9, 1, 9, 1, 9, 1… Cyclicity is 2.

Cyclicity is nothing but pattern recognition. You see that when you multiply 2 by  itself, there is a pattern of last digit which goes 2, 4, 8, 6, 2, 4, 8, 6 and so on. We can use the same principle for when a question asks us for the last two digits of the expression. Let me remind you first that here at QWQW, we sometimes flirt with the lines that define GMAT scope. Obviously, we do point out whenever we are indulging and that’s exactly what we are going to do in this post. We are carrying on for the love of Math and the Q51 score.

The last two digits of the base decide the last two digits of the expression. For example,

Example 1: Let’s look at powers of 11.

11^1 = 11

11^2 = 121

11^3 = 1331

11^4 = …41 (we should just multiply the last two digits together and ignore the rest)

11^5 = …51

11^6 = …61

11^7 = …71

Note that the last two digits are displaying a pattern depending on the power. So we expect the cyclicity here to be 10.

11^8 = …81

11^9 = …91

11^(10) = …01

11^(11) = …11

11^(12) = …21

and so on. So the last two digits should go from 11, 21 to 91, 01 and then go to 11 again. The cycle of 10 starts from power of 1, 11, 21 etc. This means that 11^(46) should have last two digits as 61, 11^(92) should have last two digits as 21 and 11^(168) should have last two digits as 81.

Let’s look at some other numbers now:

Example 2: Say, we need the last two digits of 6^{58}

6^1 = 6 (No second last digit)

6^2 = 36

6^3 = 216

6^4 = …96 (Just multiply the last two digits)

6^5 = …76

6^6 = …56

6^7 = …36

and hence starts the cycle again:

3, 1, 9, 7, 5, 3, 1, 9, 7, 5 and so on.

The new cycle with tens digit of 3 begins at the powers of 2, 7, 12, 17, 22, 27 etc. So the new cycle will also begin at power of 57 and 6^58 will have 1 as the tens digit.

Example 3: How about the last two digits of 7^102?

7^1 = 7 (No second last digit)

7^2 = 49

7^3 = 343

7^4 = …01

7^5 = …07

7^6 = …49

7^7 = …43

We see a cyclicity of 4 here: 49, 43, 01, 07, 49, 43, 01, 07 … and so on. The new cycle begins at 2, 6, 10, 14 i.e. even powers which are not multiples of 4. So a new cycle will begin at 102 too. So the last two digits of 7^(102) will be 49.

Now there can be many variations in the questions asking us to find the last two digits. We will use different concepts for different question types. Today we saw how to use pattern recognition. We will look at some other methods next week.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Figuring Out the Topic of Discussion on the GMAT

Quarter Wit, Quarter WisdomYou must have come across questions which you thought tested one concept but later found out could be easily dealt with using another concept.  Often, crafty little mixture problems belong to this category. For example:

Mark is playing poker at a casino. Mark starts playing with 140 chips, 20% of which are $100 chips and 80% of which are $20 chips. For his first bet, Mark places chips, 10% of which are $100 chips, in the center of the table. If 70% of Mark’s remaining chips are $20 chips, how much money did Mark bet?

You can view this as a word problem where you assume the number of chips and then go splitting them up or you can view this as a mixtures problem even though it doesn’t use words such as ‘mixture’, ‘solution’, ‘combined’ etc. As we have seen enough number of times, our mixture problems are solved in seconds using the weighted average concept.

The question discussed here also belongs to the same category – looks super tricky but can be easily solved with weighted averages formula. But we have seen plenty and more of such questions in our blog posts. Today we will take a look at a different type of sinister question and I suggest you to think about the concept being tested in that before trying to solve it.

Question: Mark owns four low quality watches. Watch1 loses 15 minutes every hour. Watch2 gains 15 minutes every hour relative to watch1 (that is, as watch1 moves from 12:00 to 1:00, watch2 moves from 12:00 to 1:15). Watch3 loses 20 minutes every hour relative to watch2. Finally, watch4 gains 20 minutes every hour relative to watch3. If Mark resets all four watches to the correct time at 12 noon, what time will watch4 show at 12 midnight that day?

(A)10:00
(B)10:34
(C)11:02
(D)11:48
(E)12:20

Before we look at the solution, think about the concept being tested here – clocks? Circular motion?

Neither!

Solution: Note that when giving data about watch1, you are told how it varies with the actual time. Data about all other watches tells us about the time they show relative to the incorrect watches. The concept being tested here is Relative Speed!

What do we mean by “gains 15 mins” or “loses 20 mins” etc? When a watch gains 15 mins every hour, it means that even though it should show that one hour has passed, it shows that 1 hr 15 mins have passed. So the watch runs faster than it should. Hence the speed of the watch is more than the speed of a correct watch. Now the question is how much more? The minute hand of the correct watch travels one full circle in one hour. The minute hand of the incorrect watch travels one full circle and then a quarter circle in one hour (to show that 1 hour 15 mins have passed even when only an hour has passed). So it is 5/4 times the speed of a correct watch. On the same lines, let’s analyze each watch.

Say the speed of a correct watch is s.

– “Watch1 loses 15 minutes every hour. “

Watch1 covers only three quarters of the circle in an hour.

Speed of watch1 = (3/4)*s

– “Watch2 gains 15 minutes every hour relative to watch1 (that is, as watch1 moves from 12:00 to 1:00, watch2 moves from 12:00 to 1:15).”

Now we have the speed of watch2 relative to speed of watch1. Speed of watch2 is (5/4) times the speed of watch1.

Speed of watch2 = (5/4)*(3/4)s = (15/16)*s

– “Watch3 loses 20 minutes every hour relative to watch2.”

Watch3 loses 20 mins every hour means its speed is (2/3)rd the speed of watch2

Speed of watch3 = (2/3)*(15/16)*s = (5/8)*s

– “Finally, watch4 gains 20 minutes every hour relative to watch3.”

Speed of watch4 = (4/3)*Speed of watch3 = (4/3)*(5/8)*s = (5/6)*s

So the speed of watch4 is (5/6)th the speed of a correct watch. So if a correct watch shows that 6 hours have passed, watch4 will show that 5 hours have passed. If a correct watch shows that 12 hours have passed, watch4 will show that 10 hours have passed. From 12 noon to 12 midnight, a correct watch would have covered 12 hours. Watch4 will cover 10 hours and will show the time as 10:00.

Answer (A)

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

What are the Weights in Weighted Averages?

Quarter Wit, Quarter WisdomWe have discussed weighted averages in detail here but one thing we are yet to talk about is how you decide what the weights will be in weighted average problems. It is not always straight forward to identify the weights. For example, in a question such as this one,

While traveling from Detroit to Novi, a car averaged 10 miles per gallon, and while traveling from Novi to Lapeer, it averaged 18 miles per gallon. If the distance between Detroit and Novi is half the distance between Novi and Lapeer, what is the average miles per gallon for the entire journey?

We have two figures for mileage given here – 10 miles per gallon and 18 miles per gallon. We need to find the average mileage. So we can use the weighted average formula but what will the weights be? Will they be 1:2 since the distance between the two cities is given to be in the ratio 1:2? If you think that taking the distance to be the weights in this problem is correct, then you fell for the trap in this question.

To explain the concept, let us use a simpler example first:

When talking about average speed, what are the weights? We know that the weight given to each speed is the time for which that speed was maintained, right? Yes! But why?

Let’s review our weighted average formula:

Cavg = (C1*w1 + C2*w2)/(w1 + w2)
Average Speed = (Speed1*Time1 + Speed2*Time2)/(Time1 + Time2)
Average Speed = (Distance1 + Distance2)/(Time 1 + Time2)
Average Speed = Total Distance/Total Time

This is an accurate representation of average speed.

Now see what happens when you use distance as the weights.

Cavg = (C1*w1 + C2*w2)/(w1 + w2)
Average Speed = (Speed1*Distance1 + Speed2*Distance2)/(Distance1 + Distance2)

Speed*Distance doesn’t represent any physical quantity. So this doesn’t make sense. The units of the quantities will help you see the relation clearly.

Cavg = (C1*w1 + C2*w2)/(w1 + w2)
Average Speed = (Speed1*Time1 + Speed2*Time2)/(Time1 + Time2)

Average Speed = (miles/hour * hour + miles/hour * hour)/(hour + hour)

Average Speed = (miles + miles)/(hour + hour)

Average Speed = Total miles/Total hours

What happens when you take distance as the weights?

Cavg = (C1*w1 + C2*w2)/(w1 + w2)
Average Speed = (Speed1*Distance1 + Speed2*Distance2)/(Distance1 + Distance2)

Average Speed = (miles/hour * miles + miles/hour * miles)/(miles + miles)

miles^2/hour doesn’t represent a physical quantity and hence doesn’t make sense here. Therefore, whenever you are confused what the weights should be, look at the units.

Let’s go back to the original question now. Average required is miles per gallon. So you are trying to find the weighted average of two quantities whose units must be miles/gallon.

Cavg = (C1*w1 + C2*w2)/(w1 + w2)

The unit of Cavg, C1 and C2 is miles/gallon so w1 and w2 should be in gallons to get

miles/gallon = (miles/gallon * gallon + miles/gallon * gallon)/(gallon + gallon)

miles/gallon = Total miles/Total gallons

So how will we actually solve this question?

Question: While traveling from Detroit to Novi, a car averaged 10 miles per gallon while traveling from Novi to Lapeer, it averaged 18 miles per gallon. If the distance between Detroit and Novi is half the distance between Novi and Lapeer, what is the average miles per gallon for the entire journey?

Solution:

Let the distance between Detroit and Novi be D. So the distance between Novi and Lapeer must be 2D.

Amount of fuel used to cover distance D = D/10

Amount of fuel used to cover distance 2D = 2D/18 = D/9

So the two weights used must be D/10 and D/9

Average miles/gallon = (10*D/10 + 18*D/9)/(D/10 + D/9) = 3D*90/19D = 270/19 = 14.2 miles/gallon

Or simply, Average miles/gallon = Total miles/Total gallons = 3D/(D/10 + D/9) = 14.2 miles/gallon

Food for thought: Which one of the following can you solve?

– If a vendor sold 10 apples at a profit of 10% and 15 oranges at a profit of 20%, what was his overall profit%?

– If a vendor sold apples at a profit of 10% and oranges at a profit of 20%, what was his overall profit% if cost price of each apple was $0.20 and the cost price of each orange was $.06?

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!