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!*

Writing Factors of an Ugly Number

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!*

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!*

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.

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

– 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.

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

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.**

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

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:

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.

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

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)

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**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.

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

x – y = 55 – 10 = 45

Answer (C)

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

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?

(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.

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.

*GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*