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QUESTION: 1

Set P consists of the first n positive multiples of 3 and set Q consists of the first m positive multiples of 5. The sum of all the numbers in set P is equal to R and the sum of all the numbers in set Q is equal to S. If m and n are positive integers, is the difference between R and S odd?

(1) m is odd and n is even

(2) m can be expressed in the form of 4x +3 and n can be expressed in the form of 2x, where x is a positive integer

Solution:

__Steps 1 & 2: Understand Question and Draw Inferences__

- Difference of two integers will be odd, only if one of the numbers is odd.
- Let’s see when R and S will be odd:
- The even-odd nature of R will depend on the values of n and n + 1. Since n and n+ 1 are consecutive integers, only one of them can be even.
- Now, in the expression of R as n(n+1) is divided by 2, for R to be even, either of n or n + 1 should be a multiple of 4, i.e. they should be of the form 4k, where k is a positive integer. Hence, following cases arise:
- n is of the form 4k, i.e. n = 4k OR
- n+1 is of the form 4k, i.e. n + 1 = 4k
- So, n = 4k – 1 = 4(k -1) + 4 -3 = 4(k-1) + 3

- So, we can say that R will be even if n is of the form 4k or 4k + 3. For all the other cases, R will be odd

- Since the expression of S is similar to R, we can say that S will be even if m is of the form 4k or 4k + 3
- So, we need to look for values of m and n in the statements.

__Step 3: Analyze Statement 1 independently__

(1) m is odd and n is even

- As we do not know if m or n is of the form of 4k or 4k + 3, we cannot say, if R and S are odd or even.
- Insufficient to answer.

__Step 4: Analyze Statement 2 independently__

(2) m can be expressed in the form of 4x +3 and n can be expressed in the form of 2x, where x is a positive integer

- As m can be expressed in the form of 4x + 3, S will be even.
- As n can be expressed in the form of 2x, n may be expressed in the form of 4x or may not be expressed in the form of 4x.
- So, R may be odd or R may be even.
- If R is odd, R – S = odd
- If R is even, R – S = even.

- So, R may be odd or R may be even.
- Hence, we cannot say for sure if R – S is odd.
- Insufficient to answer.

__Step 5: Analyze Both Statements Together (if needed)__

(1) From statement-1, we know that is odd and n is even

(2) From statement-2, we know that R may be even or odd and S is even

- Combining both the statements does not tell us anything about the even-odd nature of R, hence we cannot say for sure if R – S = odd.
- Insufficient to answer.

__Answer: E__

QUESTION: 2

If *a*,*b* and *c* are three positive integers such that at least one of them is odd, which of the following statements must be true?

Solution:

__Given__

- a, b, c are integers > 0
- At least one of a, b, c is odd. So possible cases can be:
- Only one of (a, b, c) is odd
- Two of (a, b, c) are odd
- All (a, b, c, ) are odd

- At least one of a, b, c is odd. So possible cases can be:

__To Find__: the option that must be true (for all possible values of a,b and c)

__Approach__

- As the options involve the combination of sum of a, b and c, we will need to evaluate the options to see if they are always true for the possible combinations of the numbers (a, b, c) being odd or even

__Working Out__

- a+b+c is odd
- Sum of 3 integers will be odd only if:
- One of the integers is odd
- OR all three integers are odd

- However, they may be a case where two of integers are odd and one is even. In such a case, a + b + c = even.
- So, option-I is not always true.

- Sum of 3 integers will be odd only if:
- a+b+c and abc have the same even-odd nature
- Consider the case, where abc is even. This is possible if:
- 1 of a, b, c is even→ In this case, a+ b + c = even
- 2 of a, b, c, is even→ In this case, a+ b + c = odd
- All of a, b, c, are even→ In this case a + b + c = even

- As we can see from above that the even-odd nature of a+b+c and abc are not always the same, this option is not always true

- Consider the case, where abc is even. This is possible if:
- 100a + 10b + c is odd
- The even-odd nature of 100a +10b + c will depend on even-odd nature of c, because 100a and 10b are always even.
- If c is odd, 100a +10b +c is odd
- If c is even, 100a +10b + c is even

- So, option-III is not always true

- The even-odd nature of 100a +10b + c will depend on even-odd nature of c, because 100a and 10b are always even.
- If b/2 and c/2 are integers, is even
- Since b/2 and c/2 are integers, b and c must be even
- As 2a must be even, b/2+c/2 should be even for to be even.
- Sum of two integers can be even, only if both are even or both are odd. Following cases can arise:
- If b and c are multiples of 4→b/2 and c/2 will be even
- Hence will be even

- If either of b and c are not multiples of 4→ In this case only one of b/2 or c/2 will be even and the other will be odd.

- If b and c are multiples of 4→b/2 and c/2 will be even

- Sum of two integers can be even, only if both are even or both are odd. Following cases can arise:

Hence b/2+c/2 will be odd.

- Hence will be odd

So, option-IV is not always true.

As none of the options are true, the correct answer will be option E

__Answer: E__

QUESTION: 3

If n is an integer, the number of integers between, but not including, n and 3n cannot be

I. 13

II. 14

III. 15

Solution:

__Given:__

- Integer n

** To find:** Which of the following CANNOT be the number of integers between, but not including, n and 3n: {13, 14, 15}

__Approach:__

- To answer this question, we’ll work out a constraint on the number of integers between, but not including, n and 3n
- Then, we’ll use this constraint to check which of the 3 values is not possible.

__Working Out:__

- Number of integers between 3n and n = 3n – n = 2n
- But this count includes 3n itself.
- For example, say n = 1. Then, if you do 3 – 1, you get 2. That is, 2 integers lie between 1 and 3. This count includes 3.

- So, the Number of integers between, but not including, 3n and n = 2n – 1

- But this count includes 3n itself.

- Thus, the number of integers between and not including n and 3n can only be an odd number.
- So, this number cannot be 14.

Looking at the answer choices, we see that the correct answer is Option B

QUESTION: 4

If z is an integer greater than 1, is z even?

(1) 2z is not a factor of 8

(2) 3z/4 is a factor of 6

Solution:

__Steps 1 & 2: Understand Question and Draw Inferences__

**Given:** Integer z > 1

**To find:** Is z even?** **

__Step 3: Analyze Statement 1 independently__

- 2z is not a factor of 8
- Factors of 8: {1, 2, 4, 8}

- This means, 2z is not equal to 1, 2, 4, 8
- z is not equal to ½ or 1(anyways not possible since z is an integer greater than 1)
- And, z is not 2 or 4

- But z can be equal to 8 or 10 or 12 etc.
- That is, z can still be an even number.

Thus, Statement 1 is not sufficient to answer the question

__Step 4: Analyze Statement 2 independently__

- 3z/4 is a factor of 6
- Factors of 6: {1, 2, 3, 6}
- This means 3z/4 is an integer and is equal to one of these 4 numbers.
- Since 3z/4 is an integer, z must be divisible by 4.

- Therefore, z is indeed even.

Statement 2 is sufficient to answer the question

__Step 5: Analyze Both Statements Together (if needed)__

Since we’ve already arrived at a unique answer in Step 4, this step is not required

Answer: Option B

QUESTION: 5

Set A consists of a set of n consecutive integers. Is the sum of all the integers in set A even?

(1) If -5 is added to set A, the set would become symmetric about 0

(2) If the largest integer of set A is removed, the sum of the remaining integers is even

Solution:

__Steps 1 & 2: Understand Question and Draw Inferences__

- Let the smallest integer be in set A be a. So, the other integers in set A will be a+1, a+2….a+n-1
- Sum of all integers in set

__Step 3: Analyze Statement 1 independently__

(1) If -5 is added to set A, the set would become symmetric about 0

- As the set becomes symmetric about 0, that would mean that there will be (upon the inclusion of -5 in set A) an equal number of terms on either side of 0 with the same magnitude (but opposite signs).
- Let’s call this new set as A’
- So, the mean of (Set A’) = 0
- So, the sum of (All the integers in set A’) = 0

- Hence, the sum of all the integers in set A is odd.
- Sufficient to answer.

__Step 4: Analyze Statement 2 independently__

(2) If the largest integer of set A is removed, the sum of the remaining integers is even.

- The largest integer of Set A = a +n -1.
- So, (Sum of all the terms in set A – largest integer of set A) = even
- Let’s think of the possible ways in which difference of two integers is even:
- Case-I: even – even = even
- So, if the largest integer removed is even, the sum of all the integers in set A would be even.
- Hence will be even
- Case-II: odd – odd = even
- So, if the largest integer removed is odd, the sum of all the integers in set A would be odd.
- Hence will be odd
- As we do not have a unique answer, the statement is insufficient to answer.

- Case-I: even – even = even

__Step 5: Analyze Both Statements Together (if needed)__

- As we have a unique answer from step-3, this step is not required.

__Answer: A__

QUESTION: 6

If A and B are integers, is the product AB even?

Solution:

__Steps 1 & 2: Understand Question and Draw Inferences__

**Given: **Integers A and B

**To find:** Is the product AB even?

- The answer will be YES if at least one of A and B is even

__Step 3: Analyze Statement 1 independently__

- In the above equation, 2AB and 10A will always be even. So, we can write:

**Case 1: B is even**

- Then, B
^{2}and 3B will both be even - The difference of B
^{2}– 3B will be Even as well - A
^{2 }+ Even = Even

- This is possible only if A is even
- Since both A and B are even, the product AB will be even

**Case 2: B is odd**

- Then, B
^{2}and 3B will both be odd - The difference of B
^{2}– 3B will be Even - A
^{2 }+ Even = Even

- This is possible only if A is even
- Since A is even, the product AB will be even

This means, the product AB is definitely even.

Statement 1 is sufficient to answer the question

__Step 4: Analyze Statement 2 independently__

- The left hand side of this equation is the difference of 2 even terms. So, the left hand side is Even
- This means, the Right hand side of the equation must be Even as well. So, we can write:

13A^{2 }- AB =Even

**Case 1: B is even**- Then, AB is even
- Since the above equation holds true, 13A
^{2}must be Even as well. - This means, A is even as well (since 13A
^{2}has the same even-odd nature as A) - Since both A and B are even, the product AB is Even.

**Case 2: B is odd**- Then, AB has the same even-odd nature as A
- Also, 13A
^{2}will have the same even-odd nature as A - Thus, the left hand side of the equation has the difference of two terms which are either both odd or both even.
- We know that Odd – Odd = Even and Even- Even = Even
- Thus, the above equation will be satisfied whether A is even or odd
- If A is even, then the product AB will be even
- But if A is odd, then the product AB will be odd

- Thus, we see that the product AB may be either even or odd. Statement 2 is not sufficient to determine a definite answer about the even-odd nature of AB.

__Step 5: Analyze Both Statements Together (if needed)__

- Since we’ve already arrived at a unique answer in Step 3, this step is not required
- Answer: Option A

QUESTION: 7

If A and B are positive integers, is A – B even?

1. The product of A and B is even

2. A + B is odd

Solution:

__Steps 1 & 2: Understand Question and Draw Inferences__

**Given: **Integers A, B > 0

**To find: **Is A – B even?

- The answer will be YES if A and B have the same even-odd nature, i.e. A and B are both even or both odd

__Step 3: Analyze Statement 1 independently__

Statement 1 says that ‘The product of A and B is even’

- This means, that one number out of A and B is definitely even
- The other number may be even or odd
- So, A – B may be:
- Either Even – Odd = Odd
- Or Even – Even = Even

- Thus, Statement 1 is not sufficient to give a unique Yes/ No answer to the question

__Step 4: Analyze Statement 2 independently__

Statement 2 says that ‘A + B is odd’

- For A + B to be odd, one of A or B has to be even and the other odd.
- So, Even – Odd = Odd OR Odd – Even = Odd

- Hence A – B will be odd.
- Therefore, Statement 2 is sufficient to answer the question

__Step 5: Analyze Both Statements Together (if needed)__

Since we’ve already arrived at a unique answer in Step 4, this step is not required

Answer: **Option B**

QUESTION: 8

If P and Q are positive integers, then is (P+2)(Q-1) an even number?

(1) p/3Q is an even integer

(2) is a positive odd integer

Solution:

__Steps 1 & 2: Understand Question and Draw Inferences__

**Given: **Integers P, Q > 0

**To find: ** Is (P+2)(Q-1) even?

- The answer is YES if
- At least one out of P + 2 and Q – 1 is even
- P + 2 is even if P is even
- Q – 1 is even if Q is odd

- At least one out of P + 2 and Q – 1 is even
- So, the answer is YES if
- Either P is Even
- Or Q is odd, or both

__Step 3: Analyze Statement 1 independently__

- P/3Q is an even integer.

- This means, P = (3Q)(2n) where n is a positive integer

- So, P is an even number
- Therefore, the answer is YES

Thus, Statement 1 is sufficient to answer the question

__Step 4: Analyze Statement 2 independently__

- is an odd integer

- So, we can write: Q= ( ) (2m+1), where m is an integer
- Since Q and 2m + 1 are both integers,( must be an integer as well

- Therefore,√P is an integer
- Implies, that P is a perfect square

- √P has the same even-odd nature as P

**Case 1: P is even**- We’ve already determined in Steps 1 and 2 that if P is even, the answer to the question is YES

**Case 2: P is odd**- Q = (Odd - Odd)(Odd)
- Q = Even * Odd
- Q = Even

- Thus, Statement 2 doesn’t lead us to a unique answer to the question.

Therefore, Statement 2 is not sufficient.

__Step 5: Analyze Both Statements Together (if needed)__

Since we’ve already arrived at a unique answer in Step 3, this step is not required

Answer: Option A

QUESTION: 9

There are N students in a class. When the students are distributed into groups that contain 4A number of students each, 3 students are left without a group. When the students are distributed into groups that contain A/3 number of students each, no students are left without a group. Which of the following statements is correct?

I. If the students are distributed into groups that contain A+ 1 students each, the number of students that are left without a group can be 2

II. If the students are distributed into groups that contain 3 students each, no students are left without a group

III. If the students are distributed into groups that contain 12 students each, 9 students are left without a group

Solution:

__Given:__

- N = (4A)x + 3, where quotient x is an integer
- N=(A/3)k

- , where k is an integer
- Since N denotes the number of students, N must be a positive integer

** To find:** Which of the 3 statements is/are correct?

__Approach:__

- To answer the question, we first need to get an idea about the values of N and A
- Then, we’ll evaluate the 3 statements one by one. A statement will be a must be true statement only if it holds true for
*all*possible values of N and A

__Working Out:__

**Drawing inferences about N and A**- We’re given that N is of the form (4A)x + 3
- Now, the term (4A)x will be even for all values of A and x
- So, N is of the form Even + Odd
- So, N is odd

- We’re given that N is of the form (4A)x + 3
- Also, in the second piece of given information, we’re told that each student group contains A/3 students
- Since number of students cannot be in fractions, this means A must be a multiple of 3

- Also, N=(A/3)k
- We’ve inferred above that N is odd
- So, the product of integers (A/3)
- and k is odd
- This is possible only if (A/3)
- and k themselves are odd
- So, (A/3) is odd

- Thus, A = (Odd number 3)*(Another odd integer)
- So, A is odd as well
- Thus, combining our above inferences about A, we can say that A is an odd multiple of 3.
- So, we can write: A = 3(2B+1) = 6B + 3, where B is an integer

**Evaluating Statement I**- If the students are distributed into groups that contain A + 1 students each, the number of students that are left without a group can be 2
- A + 1 = (6B+3) + 1 = 6B + 4 = Even number
- Let the number of groups that contain A + 1 students be G.
- (Odd Number N) = (Even Number A+1)*(G) + Remainder
- So, (Odd Integer) = Even Integer + Remainder
- So, Remainder = Odd Integer minus Even Integer

- So, the remainder when N is divided by A +1 will be an odd integer.
- Therefore, the remainder cannot be 2

- So, Statement I is not correct

**Evaluating Statement II**- If the students are distributed into groups that contain 3 students each, no students are left without a group
- To answer this question, we need to evaluate if N is divisible by 3 or not
- If it is, then Statement II is indeed a must be true statement.

- Now, we’ve analyzed above that:
- N = (4A)x + 3
- And, A = 6B+3

- So, we can write: N = 4(6B+3)x + 3
- = 4*3(2B+1)x + 3
- =3{4(2B+1)x + 1}

- Thus, we see that N is completely divisible by 3
- So, Statement II is correct

**Evaluating Statement III**- If the students are distributed into groups that contain 12 students each, 9 students are left without a group
- To evaluate this statement, we need to determine if the remainder when N is divided by 12 is 9
- If it is, then this is a must be true statement.

- In our analysis of Statement II above, we’ve inferred that:
- N = 3{4(2B+1)x + 1}
- Simplifying this: N = 12(2B+1)x + 3
- So, when N is divided by 12, the first term in the above expression: 12(2B+1)x will be completely divisible by 12
- So, the remainder when N is divided by 12 will be 3

- Therefore, Statement III is incorrect

**Getting to the answer**- From our analysis above, we’ve concluded that only Statement II is always correct.

Looking at the answer choices, we see that the correct answer is Option B

QUESTION: 10

Which one of the following is even?

Solution:

**Step 1:**

5 × 7 = 35

1 × 3 = 3

9 × 5 = 45

**Step 2:**

Only 3 × 4 = 12 is even. Rest are odd.

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