Question: If two integers are chosen at random out of first 5 positive integers, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?

A. 2/5

B. 3/5

C. 7/10

D. 4/5

E. 9/10

Solution: This might look like a probability question but isn’t. Questions like these are the reason we ask you to go through basics of every topic including probability. If you do not know probability at all, you may skip this question even though it needs very basic knowledge of probability.

Probability will tell you that

Required probability = Favorable cases/Total cases

Total cases are very easy to find: 5C2 = 10 or 5*4/2 = 10 whatever you prefer. This is the number of ways in which you select any 2 distinct numbers out of the given 5 distinct numbers.

Number of favorable cases is the challenge here. That is why it is a number properties question and not so much a probability question. Let’s focus on the main part of the question:

First five positive integers: 1, 2, 3, 4, 5

We need to select two integers such that their product is of the form a^2 – b^2. What does a^2 – b^2 remind you of? It reminds me of (a + b)(a – b). So the product needs to be of the form (a + b)(a – b). So is it necessary that of the two numbers we pick, one must be of the form (a + b) and the other must be (a – b)? No. Note that we should be able to write the product in this form. It is not necessary that the numbers must be of this form only.

But first let’s focus on numbers which are already of the form (a + b) and (a – b).

Say you pick two numbers, 2 and 5. Are they of the form (a + b) and (a – b) such that a and b are integers? No.

5 = 3.5 + 1.5

2 = 3.5 – 1.5

So a = 3.5, b = 1.5.

a and b are not integers.

What about numbers such as 3 and 5? Are they of the form (a + b) and (a – b) such that a and b are integers? Yes.

5 = 4 + 1

3 = 4 – 1

Note that whenever the average of the numbers will be an integer, we will be able to write them as a+b and a – b because one number will be some number more than the average and the other will be the same number less than average. So a will be the average and the amount more or less will be b.

When will the average of two numbers (Number1 + Number2)/2 be an integer? When the sum of the two numbers is even! When is the sum of two numbers even? It is when both the numbers are even or when both are odd. So then does the question boil down to “favorable cases are when we select both numbers even or both numbers odd?” Yes and No. When we select both even numbers or both odd numbers, the product can be written as a^2 – b^2. But are those the only cases when the product can be written as a^2 – b^2?

The question is not so much as whether both the numbers are even or both are odd as whether the product of the numbers can be written as product of two even numbers or two odd numbers. We need to be able to write the product (whatever we obtain) as product of two even or two odd numbers.

To explain this, let’s say we pick two numbers 4 and 5

4*5 = 20

Can we write 20 as product of two even numbers? Yes 2*10.

So even though, 4 is even and 5 is odd, their product can be written as product of two even numbers. So in which all cases will this happen?

- Whenever you have at least 4 in the product, you can write it as product of two even numbers: give one 2 to one number and the other 2 to the other number to make both even.

If the product is even but not a multiple of 4, it cannot be written as product of two even numbers or product of two odd numbers. It can only be written as product of one even and one odd number.

If the product is odd, it can always be written as product of two odd numbers.

Let’s go back to our question:

We have 5 numbers: 1, 2, 3, 4, 5

Our favorable cases constitute those in which either both numbers are odd or the product has 4 as a factor.

3 Odd numbers: 1, 3, 5

2 Even numbers: 2, 4

Number of cases when both numbers are odd = 3C2 = 3 (select 2 of the 3 odd numbers)

Number of cases when 4 is a factor of the product = Number of cases such that we select 4 and any other number = 1*4C1 = 4

Total number of favorable cases = 3 + 4 = 7

Note that this includes the case where we take both even numbers. Had there been more even numbers such as 6, we would have included more cases where we pick both even numbers such as 2 and 6 since their product would have 4 as a factor.

Required Probability = 7/10

Answer (C)

Takeaway:

When can we write a number as difference of squares?

- When the number is odd

or

- When the number has 4 as a factor

*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 two items are sold at the same selling price, one at a profit of x% and the other at a loss of x%, there is an overall loss. The loss% = (x^2/100)%

We will see how this formula is derived but the algebra involved is tedious. You can skip it if you wish.

Say two items are sold at $S each. On one, a profit of x% is made and on the other a loss of x% is made.

Say, cost price of the article on which profit was made = Ct

Ct (1 + x/100) = S

Ct = S/(1 + x/100)

Cost Price of the article on which loss was made = Cs

Cs (1 – x/100) = S

Cs = S/(1 – x/100)

Total Cost Price of both articles together = Ct + Cs = S/(1 + x/100) + S/(1 – x/100)

Ct + Cs = S[1/(1 + x/100) + 1/(1 - x/100)]

Ct + Cs = 2S/(1 – (x/100)^2)

Total Selling Price of both articles together = 2S

Overall Profit/Loss = 2S – (Ct + Cs)

Overall Profit/Loss % = [2S – (Ct + Cs)]/[Ct + Cs] * 100

= [2S/(Ct + Cs) – 1] * 100

= [2S/[2S/(1 – (x/100)^2)] – 1] * 100

= (x/100)^2 * 100

= x^2/100

Overall there is a loss of (x^2/100)%.

Let’ see how this formula works on a GMAT Prep question.

**Question**: John bought 2 shares and sold them for $96 each. If he had a profit of 20% on the sale of one of the shares but a loss of 20% on the sale of the other share, then on the sale of both shares John had

(A) a profit of $10

(B) a profit of $8

(C) a loss of $8

(D) a loss of $10

(E) neither a profit nor a loss

**Solution**:

Note that the question would have been straight forward had the COST price been the same, say $100. A 20% profit would mean a gain of $20 and a 20% loss would mean a loss of $20. Overall, there would have been no profit no loss.

Here the two shares are sold at the same SALE price. One at a profit of 20% on cost price which must be lower than the sale price (to get a profit) and the other at a loss of 20% on cost price which must be higher than the sale price (to get a loss). 20% of a lower amount will be less in dollar terms and hence overall, there will be a loss.

The loss % = (20)^2/100 % = 4%.

But we need the amount of loss, not the percentage of loss.

Total Sale price of the two shares = 2*96 = $192

Since there is a loss of 4%, the 96% of the total cost price must be the total sale price

(96/100)*Cost Price = Sale Price

Cost Price = $200

Loss = $200 – $192 = $8

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!*

There are some triplets that you should know out cold: (3, 4, 5), (5, 12, 13) and (8, 15, 17). Usually you will find one of these three or their multiples on GMAT. Given a right triangle and two sides, say the two legs, of length 20 and 48, we need to immediately bring them down to the lowest form 20:48 = 5:12. So we know that we are talking about the 5, 12, 13 triplet and the hypotenuse will be 13*4 = 52. These little things help us save a lot of time. Why is it that some people get done with the Quant section in less than an hour while others fall short on time? It is these little things that an adept test taker has mastered which make all the difference.

Anyway, let us go on to the question we have in mind.

Question: In the figure given below, the length of PQ is 12 and the length of PR is 15. The area of right triangle STU is equal to the area of the shaded region. If the ratio of the length of ST to the length of TU is equal to the ratio of the length of PQ to the length of QR, what is the length of TU?

(A) (9√2)/4

(B) 9/2

(C) (9√2)/2

(D) 6√2

(E) 12

Solution: The information given in the question seems to overwhelm us but let’s take it a bit at a time.

“length of PQ is 12 and the length of PR is 15”

PQR is a right triangle such that PQ = 12 and PR = 15. So PQ:PR = 4:5. Recall the 3-4-5 triplet. A multiple triplet of 3-4-5 is 9-12-15. This means QR = 9.

“ratio of the length of ST to the length of TU is equal to the ratio of the length of PQ to the length of QR”

ST/TU = PQ/QR

The ratio of two sides of PQR is equal to the ratio of two sides of STU and the included angle between the sides is same ( = 90). Using SAS, triangles PQR and STU are similar.

“The area of right triangle STU is equal to the area of the shaded region”

Area of triangle PQR = Area of triangle STU + Area of Shaded Region

Since area of triangle STU = area of shaded region, (area of triangle PQR) = 2*(area of triangle STU)

In similar triangles, if the sides are in the ratio k, the areas of the triangles are in the ratio k^2. If the ratio of the areas is given as 2 (i.e. k^2 is 2), the sides must be in the ratio √2 (i.e. k must be √2).

Since QR = 9, TU must be 9/√2. But there is no 9/√2 in the options – in the options the denominators are rationalized. TU = 9/√2 = (9*√2)/(√2*√2) = (9*√2)/2.

Answer (C)

The question could take a long time if we do not remember the Pythagorean triplets and the area of similar triangles property.

Takeaways:

- Pythagorean triplets you should know: (3, 4, 5), (5, 12, 13) and (8, 15, 17) and their multiples.
- In similar triangles, if the sides are in the ratio k, the areas of the triangles are in the ratio k^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!*

We are also assuming that you are comfortable with the figures that beg you to think about similar triangles such as

Try to figure out the similar triangles and the reason they are similar in each one of these cases. (Angles that look 90 are 90). Most of the figures have right angles/parallel lines.

This topic was also discussed by David Newland in a rather engaging post last week. You must check it out for its content as well as its context!

What we would like to discuss today are situations where most people do not think about similar triangles but if they do, it would make the question very easy for them. But before we do that, we would like to discuss a concept related to similar triangles which is very useful but not discussed often.

We already know that sides of similar triangles are in the same ratio. Say two triangles have sides a, b, c and A, B, C respectively. Then, a/A = b/B = c/C = k

Note that the altitudes of the two triangles will also be in the same ratio, ‘k’, since all lengths have the ratio ‘k’.

Then what is the relation between the areas of the two triangles? Since the ratio of the bases is k and the ratio of the altitudes is also k, the ratio of the areas will be k*k = k^2.

So if there are two similar triangles such that their sides are in the ratio 1:2, their areas will be in the ratio 1:4.

Now we are all ready to tackle the question we have in mind.

Question: In the given figure, ABCD is a parallelogram and E, F, G and H are midpoints of its respective sides. What is the ratio of the shaded area to that of the un-shaded area?

(A) 3:8

(B) 3:5

(C) 5:8

(D) 8:5

(E) 5:3

Solution: There are many ways to do this question but we will look at the method using similar triangles (obviously!).

Assume the area of the parallelogram is 8P. In a parallelogram, the lengths of opposite sides are the same. The two triangles formed by the diagonal and two sides are similar by SSS and the ratio of their sides is 1. So they will have equal areas of 4P each (look at the figures in second row below)

Now look at the original figure.

HE is formed by joining the mid-points of AD and AB. So AH/AD = AE/AB = 1/2 and included angle A is common. Hence by SAS rule, triangle AHE is similar to triangle ADB. If the ratio of sides is 1/2, ratio of areas will be 1/4.

Since area of triangle ADB is 4P, area of AHE is P. We have 3 such triangles, AHE, DHG and CGF which are not shaded so the area of these three triangles together will be 3P.

The total area of parallelogram is 8P and the unshaded region is 3P. So the shaded region must be 5P.

Hence, area of shaded region : Area of unshaded region = 5:3

Answer (E)

Try to think of other ways in which you can solve this question.

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

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It actually is a little harder than your standard GMAT questions but the point is that it can be easily solved using all concepts relevant to GMAT. Hence it certainly makes sense to understand how to solve it.

Question: What is the remainder when 3^(7^11) is divided by 5? (here, 3 is raised to the power (7^11))

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

Solution: As we said last week, this question can easily be solved using cyclicity and negative remainders. What is the remainder when a number is divided by 5? Say, what is the remainder when 2387646 is divided by 5? Are you going to do this division to find the remainder? No! Note that every number ending in 5 or 0 is divisible by 5.

2387646 = 2387645 + 1

i.e. the given number is 1 more than a multiple of 5. Obviously then, when the number is divided by 5, the remainder will be 1. Hence the last digit of a number decides what the remainder is when the number is divided by 5.

On the same lines,

What is the remainder when 36793 is divided by 5? It is 3 (since it is 3 more than 36790 – a multiple of 5).

What is the remainder when 46^8 is divided by 5? It is 1. Why? Because 46 to any power will always end with 6 so it will always be 1 more than a multiple of 5.

On the same lines, if we can find the last digit of 3^(7^11), we will be able to find the remainder when it is divided by 5.

Recall from the discussion in your books, 3 has a cyclicity of 4 i.e. the last digit of 3 to any power takes one of 4 values in succession.

3^1 = 3

3^2 = 9

3^3 = 27

3^4 = 81

3^5 = 243

3^6 = 729

and so on… The last digits of powers of 3 are 3, 9, 7, 1, 3, 9, 7, 1 … Every time the power is a multiple of 4, the last digit is 1. If it is 1 more than a multiple of 4, the last digit is 3. If it is 2 more than a multiple of 4, the last digit is 9 and if it 3 more than a multiple of 4, the last digit is 7.

What about the power here 7^(11)? Is it a multiple of 4, 1 more than a multiple of 4, 2 more than a multiple of 4 or 3 more than a multiple of 4? We need to find the remainder when 7^(11) is divided by 4 to know that.

Do you remember the binomial theorem concept we discussed many weeks back? If no, check it out here.

7^(11) = (8 – 1)^(11)

When this is divided by 4, the remainder will be the last term of this expansion which will be (-1)^11. A remainder of -1 means a positive remainder of 3 (if you are not sure why this is so, check last week’s post here). Mind you, you are not to mark the answer as (D) here and move on! The solution is not complete yet. 3 is just the remainder when 7^(11) is divided by 4.

So 7^(11) is 3 more than a multiple of 4.

Review what we just discussed above: If the power of 3 is 3 more than a multiple of 4, the last digit of 3^(power) will be 7.

So the last digit of 3^(7^11) is 7.

If the last digit of a number is 7, when it is divided by 5, the remainder will be 2. Now we got the answer!

Answer (C)

Interesting question, isn’t it?

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

Since we did miss it, we will discuss it in detail today but you must review the link given above before we proceed.

Consider this: When n is divided by 3, it leaves a remainder 1.

This means that when we divide n balls in groups of 3 balls each, we are left with 1 ball.

This also means that n is 1 MORE than a multiple of 3. Or, it also means that n is 2 less than the next multiple of 3, doesn’t it?

Say, n is 16. When you split 16 balls into groups of 3 balls each, you get 5 groups of 3 balls each and there is one ball leftover. n is 1 more than a multiple of 3 (the multiple being 15). But we can also say that it is 2 LESS than the next multiple of 3 (which is 18). Hence, the negative remainder in this case is -2 which is equivalent to a positive remainder of 1.

Generally speaking, if n is divided by m and it leaves a remainder r, the negative remainder in this case is -(m – r).

When n is divided by 7, it leaves a remainder of 4. This is equivalent to a remainder of -3.

n is 3 more than a multiple of m. It is also 2 less than the next multiple of m. This means m is 5.

This concept is very useful to us sometimes, especially when the divisor and the remainder are big numbers.

Let’s take a question to see how.

Question 1: What is the remainder when 1555 * 1557 * 1559 is divided by 13?

(A) 0

(B) 2

(C) 4

(D) 9

(E) 11

Solution: Since it is a GMAT question (a question for which we will have no calculator), multiplying the 3 numbers and then dividing by 13 is absolutely out of question! There has to be another method.

Say n = 1555 * 1557 * 1559

When we divide 1555 by 13, we get a quotient of 119 (irrelevant to our question) and remainder of 8. So the remainder when we divide 1557 by 13 will be 8+2 = 10 (since 1557 is 2 more than 1555) and when we divide 1559 by 13, the remainder will be 10+2 = 12 (since 1559 is 2 more than 1557).

So n = (13*119 + 8)*(13*119 + 10)*(13*119 + 12) (you can choose to ignore the quotient and just write it as ‘a’ since it is irrelevant to our discussion)

So we need to find the remainder when n is divided by 13.

Note that when we multiply these factors, all terms we obtain will have 13 in them except the last term which is obtained by multiplying the constants together i.e. 8*10*12.

Since all other terms are multiples of 13, we can say that n is 8*10*12 (= 960) more than a multiple of 13. There are many more groups of 13 balls that we can form out of 960.

960 divided by 13 gives a remainder of 11.

Hence n is actually 11 more than a multiple of 13.

We did not use the negative remainders concept here. Let’s see how using negative remainders makes our calculations easier here. The remainder of 8, 10 and 12 imply that the negative remainders are -5, -3 and -1 respectively.

Now n = (13a – 5) * (13a – 3) * (13a – 1)

The last term in this case is -5*-3*-1 = -15

This means that n is 15 less than a multiple of 13 i.e. actually 2 less than a multiple of 13 because when you go back 13 steps, you get another multiple of 13. This gives us a negative remainder of -2 which means the positive remainder in this case will be 11.

Here we avoided some big calculations.

I will leave you now with a question which you should try to solve using negative remainders.

Question 2: What is the remainder when 3^(7^11) is divided by 5? (here, 3 is raised to the power (7^11))

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

Hint: I solved this question orally in a few secs using cyclicity and negative remainders. Don’t get lost in calculations!

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

First let me give you the question stem of that question:

Which of the following, if true, supplies the best reason to suspect that the proposed new course will increase interest in the metropolitan cooking academy?

So do we have to find a reason which indicates that the new course WILL increase interest in the academy or do we have to find a reason that indicates that the new course WILL NOT increase interest in the academy? That is, is this a type of ‘strengthen’ question or the opposite – a ‘weaken’ question?

I would have expected most people to tag it as a strengthen question i.e. we are looking for a reason which indicates that the new course WILL increase interest in the academy but that is not the case. Many people get waylaid by the word ‘suspect’ and incorrectly tag it as a weaken question. Yes, suspect could mean ‘doubtful’ but just because the question stem has it, it doesn’t mean you have a weaken question at hand. Similar to the situation where the word ‘support’ in the question stem doesn’t necessarily imply that you have been given a strengthen question to deal with.

Let’s discuss various meanings of the word ‘suspect’ and how they are used (using merriam-webster.com):

Suspect can be used as a verb, noun or adjective. In our question stem, it is used as a verb and that’s what we will focus most on but let’s take up the other two briefly first.

SUSPECT (- NOUN) – one that is suspected; especially a person suspected of a crime

- One suspect has been arrested.
- She is a possible suspect in connection with the kidnapping.

SUSPECT (- ADJECTIVE) regarded or deserving to be regarded with suspicion

doubtful, questionable

- The room had a suspect odor.
- Since she was carrying no cash or credit cards, her claim to the store’s detectives that she had intended to pay for the items was suspect.

SUSPECT ( – VERB) – The verb suspect can be used in 3 different ways:

- to imagine (one) to be guilty or culpable on slight evidence or without proof

For Example: He’s suspected in four burglaries.

**2.** to have doubts of**:** distrust

For Example: The fire chief suspects arson. I suspect his intentions.

**3. **to imagine to exist or be true, likely, or probable

For Example: I suspect it will rain.

Given a construction “I suspect A will happen”, which meaning will it have? In this case, it has the meaning of ‘imagine to be true’ or ‘think to be true’. There is absolutely no ambiguity here. When I ask “Which option supplies the best reason to suspect that the new course will increase interest in the academy?” you are definitely looking for the option that indicates that the new course WILL increase interest. Let me give you the whole question now. I am sure you will be able to solve it easily.

Question: The metropolitan cooking academy surveyed prospective students and found that students wanted a curriculum that focused on today’s healthy dining trends. In order to reverse the trend of declining interest in the school’s programs, administrators propose a series of new courses focused on cooking exotic species of fish, alternative grains such as quinoa, and organically produced vegetables.

Which of the following, if true, supplies the best reason to suspect that the proposed new course will increase interest in the metropolitan cooking academy?

(A) Cooking fish, grains, and vegetables relies on same culinary fundamentals as does the preparation of other ingredients.

(B) In the food and beverage industry, many employers no longer have time to train apprentices and therefore demand basic culinary skills from their new hires.

(C) Local producers in the area near the Metropolitan cooking academy are excellent sources of exotic fish and organic vegetables.

(D) Many other cooking schools have found a decline in the level of interest in their program.

(E) Many advocates of healthy dining stress the importance of including fish, grains and organically produced vegetables in one’s diet.

Solution: Let’s break down the argument:

Premises:

A survey found that students wanted a curriculum that focused on today’s healthy dining trends.

Administrators propose a series of new courses focused on cooking fish, alternative grains and organically produced vegetables

What will indicate that the new course will increase interest in the academy? If fish, grains and organic vegetables are considered ‘today’s healthy dining trends’, then probably the course will become popular. That is what option (E) says. Hence the answer is (E).

I suspect that the word suspect will not waylay any reader of this article anymore.

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

Knowing these basic properties can help us quickly analyze the question and arrive at the answer without getting stuck in analyzing different ranges, a cumbersome procedure.

First we will look at a GMAT Prep question.

**Question 1:** Is |m – n| > |m| – |n|?

Statement 1: n < m

Statement 2: mn < 0

**Solution 1:**

Recall the property number 2

**Property 2: For all real x and y, |x – y| >= |x| – |y|**

**|x – y| = |x| – |y| when (1) x and y have the same sign and x has greater (or equal) absolute value than y (2) y is 0**

**|x – y| > |x| – |y| in all other cases**

So if m and n have the same sign with |m| >= |n|, equality will hold.

Also, if n is 0, equality will hold.

If we can prove that both these conditions are not met, then we can say that |m – n| is definitely greater than |m| – |n|.

Statement 1: n < m

We have no idea about the signs of m and n. Are they same? Are they opposite? We don’t know. Also n may or may not be 0. Hence we don’t know whether the equality will hold or the inequality. Statement 1 alone is not sufficient to answer the question.

Statement 2: mn < 0

Since mn is negative, it means one of m and n is positive and the other is negative. This also implies that n is definitely not 0. So we know that m and n do not have the same sign and n is not 0. So under no condition will the equality hold. Hence |m – n| is definitely greater than |m| – |n|. Statement 2 alone is sufficient to answer the question.

Answer (B)

Let’s look at one more question now.

**Question 2:** If xyz ≠ 0, is x(y + z) >= 0?

Statement 1: |y + z| = |y| + |z|

Statement 2: |x + y| = |x| + |y|

**Solution 2:** xyz ≠ 0 implies that all, x, y and z, are non zero numbers.

Question: Is x(y + z) >= 0?

If we can prove that x(y + z) is not negative that is x and (y+z) do not have opposite signs, we can say that x(y + z) is positive or 0.

Looking at the statements given, let’s review our property number 1:

**Property 1: For all real x and y, |x + y| <= |x| + |y|**

**|x + y| = |x| + |y| when (1) x and y have the same sign (2) at least one of x and y is 0.**

**|x + y| < |x| + |y| when (1) x and y have opposite signs**

The two statements give us equalities which means that the relevant part of the property is this:

|x + y| = |x| + |y| when (1) x and y have the same sign (2) at least one of x and y is 0.

We are also given in the question stem that x, y and z are not 0. Hence, given |x + y| = |x| + |y|, we can infer that x and y have the same sign.

Statement 1: |y + z| = |y| + |z|

This implies that y and z have the same signs. But we have no information about the sign of x hence this statement alone is not sufficient.

Statement 2: |x + y| = |x| + |y|

This implies that x and y have the same signs. But we have no information about the sign of z hence this statement alone is not sufficient.

Using both statements together, we know that x, y and z have the same sign. Whatever is the sign of y and z, the same will be the sign of (y+z). Hence x and (y+z) have the same sign. This implies that x(y + z) cannot be negative.

Hence we can answer our question with a definite ‘yes’.

Answer (C).

Mind you, both these questions can get time consuming (even though they aren’t really tough) if you don’t understand these properties well. You can certainly start your thinking from the scratch, arrive at the properties and then proceed or resort to more desperate measures such as number plugging but that is best avoided in DS questions.

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

(III) |x – y| = 0 implies x = y

x and y could be positive/negative integer/fraction; if the absolute value of their difference is 0, it means x = y. They cannot have opposite signs while having the same absolute value. They must be equal. This also means that if and only if x = y, the absolute value of their difference will be 0.

Mind you, this is different from ‘difference of their absolute values’

|x| – |y| = 0 implies that the absolute value of x is equal to the absolute value of y. So x and y could be equal or they could have opposite signs while having the same absolute value.

Let’s now take up the question we were talking about.

Question: Is |x + y| < |x| + |y|?

Statement 1: | x | ≠ | y |

Statement 2: | x – y | > | x + y |

Solution: One of the properties we discussed last week was

*“For all real x and y, |x + y| <= |x| + |y|*

*|x + y| = |x| + |y| when (1) x and y have the same sign (2) at least one of x and y is 0.*

*|x + y| < |x| + |y| when (1) x and y have opposite signs”*

We discussed in detail the reason absolute values behave this way.

So our question “Is |x + y| < |x| + |y|?” now becomes:

Question: Do x and y have opposite signs?

We do not care which one is greater – the one with the positive sign or the one with the negative sign. All we want to know is whether they have opposite signs (opposite sign also implies that neither one of x and y can be 0)? If we can answer this question definitively with a ‘Yes’ or a ‘No’, the statement will be sufficient to answer the question. Let’s go on to the statements now.

Statement 1: | x | ≠ | y |

This statement tells us that absolute value of x is not equal to absolute value of y. It doesn’t tell us anything about the signs of x and y and whether they are same or opposite. So this statement alone is not sufficient.

Statement 2:| x – y | > | x + y |

Let’s think along the same lines as last week – when will | x – y | be greater than | x + y |? When will the absolute value of subtraction of two numbers be greater than the absolute value of their addition? This will happen only when x and y have opposite signs. In that case, while subtracting, we would actually be adding the absolute values of the two and while adding, we would actually be subtracting the absolute values of the two. That is when the absolute value of the subtraction will be more than the absolute value of the addition.

For Example: x = 3, y = -2

| x – y | = |3 – (-2)| = 5

| x + y | = |3 – 2| = 1

or

x = -3, y = 2

| x – y | = |-3 – 2| = 5

| x + y | = |-3 + 2| = 1

If instead, x and y have the same sign, | x + y | will be greater than| x – y |.

If at least one of x and y is 0, | x + y | will be equal to| x – y |.

Since this statement tells us that | x – y | > | x + y |, it implies that x and y have opposite signs. So this statement alone is sufficient to answer the question with a ‘Yes’.

Answer (B)

Takeaway from this question:

If x and y have the same signs, | x + y | >| x – y |.

If x and y have opposite signs, | x + y | <| x – y |.

If at least one of x and y is 0, | x + y | =| x – y |.

You don’t need to ‘learn this up’. Understand the logic here. You can easily recreate it in the exam if need be.

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

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(I) For all real x and y, |x + y| <= |x| + |y|

(II) For all real x and y, |x – y| >= |x| – |y|

We don’t need to learn them of course and there is no need to look at how to prove them either. All we need to do is understand them – why do they hold, when is the equality sign applicable and when can they be useful. Let’s look at both the properties one by one.

**(I) For all real x and y, |x + y| <= |x| + |y|**

The result of both the left hand side and the right hand side will be positive or zero. On the right hand side, the absolute values of x and y will always get added irrespective of the signs of x and y. On the left hand side, the absolute values of x and y might get added or subtracted depending on whether they have the same sign or different signs. Hence the result of the left hand side might be smaller than or equal to that of the right hand side.

For which values of x and y will the equality hold and for which values will the inequality hold? Let’s think logically about it.

The absolute values of x and y get added on the right hand side. We want the absolute values of x and y to get added on the left hand side too for the equality to hold. This will happen when x and y have the same sign. So the equality should hold when they have the same signs.

For example, x = 4, y = 8:

|4 + 8| = |4| + |8| = 12

OR x = -3, y = -4:

|-3 -4| = |-3| + |-4| = 7

Also, when at least one of x and y is 0, the equality will hold.

For example, x = 0, y = 8:

|0 + 8| = |0| + |8| = 8

OR x = -3, y = 0:

|-3 + 0| = |-3| + |0| = 3

What happens when x and y have opposite signs? On the left hand side, the absolute values of x and y get subtracted hence the left hand side will be smaller than the right hand side (where they still get added). That is when the inequality holds i.e. |x + y| < |x| + |y|

For example, x = -4, y = 8:

|-4 + 8| < |-4| + |8|

4 < 12

OR x = 3, y = -4:

|3 -4| < |3| + |-4|

1 < 7

Let’s look at our second property now:

**(II) For all real x and y, |x – y| >= |x| – |y|**

Thinking on similar lines as above, we see that the right hand side of the inequality will always lead to subtraction of the absolute values of x and y whereas the left hand side could lead to addition or subtraction depending on the signs of x and y. The left hand side will always be positive whereas the right hand side could be negative too. So in any case, the left hand side will be either greater than or equal to the right hand side.

When will the equality hold?

When x and y have the same sign and x has greater (or equal) absolute value than y, both sides will yield a positive result which will be the difference between their absolute values

For example, x = 9, y = 2;

|9 – 2| = |9| – |2| = 7

OR x = -7, y = -3

|-7 – (-3)| = |-7| – |-3| = 4

Also when y is 0, the equality will hold.

For example, x = 8, y = 0:

|8 – 0| = |8| – |0| = 8

OR x = -3, y = 0:

|-3 – 0| = |-3| – |0| = 3

What happens when x and y have the same sign but absolute value of y is greater than that of x?

It is easy to see that in that case both sides have the same absolute value but the right hand side becomes negative.

For example, x = -4, y = -9

|x – y| = |-4 – (-9)| = 5

|x| – |y| = |-4| – |-9| = -5

So even though the absolute values will be the same since we will get the difference of the absolute values of x and y on both sides, the right hand side will be negative. If we were to take further absolute value of the right hand side, the two will become equal i.e. the right hand side will become |(|x| – |y|)| = |-5| = 5 in our example above. In that case, the equality will hold again.

Similarly, what happens when only x = 0? The right hand side becomes negative again so taking further absolute value will make both sides equal.

For example, x = 0, y = -5

|x – y| = |0 – (-5)| = 5

|x| – |y| = |0| – |5| = -5

Taking further absolute value, |(|x| – |y|)| = |-5| = 5

So when we take further absolute value of the right hand side, this property becomes similar to property 1 above: |x – y| = |(|x| – |y|)| when x and y have the same sign or at least one of x and y is 0.

Now let’s look at the inequality part of property 2.

Whenever x and y have opposite signs, |x – y| > |x| – |y|

On the left hand side, the absolute values will get added while on the right hand side, the absolute values will get subtracted. So the absolute value of the left hand side will always be greater than the absolute value of the right hand side. The left hand side will always be positive while the right hand side could be negative too. Hence even if we take the further absolute value of the right hand side, the inequality will hold: |x – y| > |(|x| – |y|)| when x and y have opposite signs

For example, x = -4, y = 8:

|-4 – 8| > |-4| – |8|

12 > -4

Taking further absolute value of the right hand side, we get |(|x| – |y|)| = |-4| = 4

Still, 12 > 4 i.e. |x – y| > |(|x| – |y|)|

OR x = 3, y = -4:

|3 –(-4)| > |3| – |-4|

7 > -1

Taking further absolute value of the right hand side, we get |(|x| – |y|)| = |-1| = 1

Still, 7 > 1 i.e. |x – y| > |(|x| – |y|)|

Note that the inequality of the original property 2 also holds when x and y have the same sign but absolute value of y is greater than the absolute value of x since the right hand side becomes negative. It also holds when x is 0 but y is not.

**To sum it all neatly,**

(I) For all real x and y, |x + y| <= |x| + |y|

|x + y| = |x| + |y| when (1) x and y have the same sign (2) at least one of x and y is 0.

|x + y| < |x| + |y| when (1) x and y have opposite signs

(II) For all real x and y, |x – y| >= |x| – |y|

|x – y| = |x| – |y|when (1) x and y have the same sign and x has greater (or equal) absolute value than y (2) y is 0

|x – y| > |x| – |y| in all other cases

(III) For all real x and y, |x – y| >= |(|x| – |y|)|

|x – y| = |(|x| – |y|)| when (1) x and y have the same sign (2) at least one of x and y is 0.

|x – y| > |(|x| – |y|)| when (1) x and y have opposite signs

Note that property (III) matches property (I).

There is another property we would like to discuss but let’s take it up next week along with some GMAT questions where we put these properties to use.

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