# 25 days to CAT 2015 | Quick quant

We are back with another set of CAT questions. Try to solve this set in five minutes.

Q.1 The remainder when (15^23 + 23^23) is divided by 19 is (CAT 2004)
(A) 4 (B) 15 (C) 0 (D) 18

Q.2 If x = (16^3 + 17^3 + 18^3 + 19^3), then x divided by 70 leaves a remainder of (CAT 2004)
(A) 0 (B) 1 (C) 69 (D) 35

Q.3 Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is a perfect square whose square root equals the total of the three original integers. Which of the following best describes the minimum, say m, of these three integers? (CAT 2008)
(1) 1 ≤ m ≤ 3 (2) 4 ≤ m ≤ 6 (3) 7 ≤ m ≤ 9 (4) 10 ≤ m ≤ 12 (5) 13 ≤ m ≤ 15

Q.4 In a certain examination paper, there are n questions. For j = 1, 2, 3… n, there are 2^(n-j) students who answered j or more questions wrongly. If the total number of wrong answers is 4095, then the value of n is (CAT 2003)
(1) 12 (2) 11 (3) 10 (4) 9

Q.5 The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the same progression. Then which element of the series should necessarily be equal to zero? (CAT 2003)
(1) 1st (2) 9th (3) 12th (4) none of these

Solutions

Q.1 The remainder when (15^23 + 23^23) is divided by 19 is (CAT 2004)

Solution:

A number of the form a+b raised to n where n is odd will have a+b as a factor. Hence, 38 is a factor in the expression. Thus, the remainder is 0.
Hence Option C

Q.2 If x = (16^3 + 17^3 + 18^3 + 19^3), then x divided by 70 leaves a remainder of (CAT 2004)

Solution:

Similar to the above question. Number of the above form will be divisible by 70. Hence option A.

Q.3 Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is a perfect square whose square root equals the total of the three original integers. Which of the following best describes the minimum, say m, of these three integers? (CAT 2008)

Solution:

Let these three numbers be (a-1), a, (a+1). The advantage of taking these numbers is that (a-1) and (a+1) cancel the plus one and minus effect.
(a-1) + a^2 + (a+1)^3
a – 1 + a^2 + a^3 + 3a^2 + 3a + 1
Root of a^3 + 4a^2 + 4a = 3a
a^3 + 4a^2 + 4a = 9a^2
a^3 + 4a = 5a^2
a^2 + 4 = 5a
a^2 – 5a + 4 = 0
(a – 1)(a – 4) = 0
A = 1 or 4.

Hence, either three numbers are 0, 1, 2 or 3, 4, 5. Answer will be 1 ≤ m ≤ 3

Q.4 In a certain examination paper, there are n questions. For j = 1, 2, 3… n, there are 2^(n-j) students who answered j or more questions wrongly. If the total number of wrong answers is 4095, then the value of n is (CAT 2003)

Solution:

Lets say j = 1. Hence, there are 2^(n-1) students who answered 1 or more questions wrongly.
If n = 1, 2^(n-1) = 1
If n = 2, 2^(n-1) = 2; 2^(n-2) = 1; 2 + 1 = 3
If n = 3, 2^(n-1) = 4; 2^(n-2) = 2; 2^(n-3) = 1; 4 + 2 + 1 = 7
This can be generalized to 2^n-1.
If 4095 = 2^n-1.
N = 12

Tip: Note numbers like 31, 63, 1023. Some classic 2^n-1 have been formed with these numbers.

Q.5 The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the same progression. Then which element of the series should necessarily be equal to zero? (CAT 2003)

Solution:

Using standard approach, let’s consider first term as a & difference as d. Using this logic:
(a + 2d) + (a + 14d) = (a + 5d) + (a + 10d) + (a + 12d)
2a + 16d = 3a + 27d
This gives us a + 11d = 0. Hence, 12th term is zero

Alternate approach: Draw a line. Every number indicates nth term.
1 – 2 – 3 – 4 – 5 …. – 13 – 14 – 15
If first term is zero, we basically add d 2 and 14 = 16 times to get the sum of 3rd and 15th term. Whereas we add difference 5 + 10 + 12 = 27 times to get three terms. Not matching
If the 9th term is zero, 6 + 6 = 12 on LHS and 2 + 4 + (-3) = 3 on RHS. Not matching
If the 12th term is zero, 3 + (-9) = -6 on LHS and (-6) + 1 – 1 = -6 on RHS. Matching and hence the answer is 12th.

Read our 75 days to CAT series. You will find everything here: from A game of sticks to Cauchy Schwarz Inequality; from Critical Reasoning to Dangling & Misplaced Modifiers. Keep reading!