Surds & Indices CAT Previous Year Questions with Answer PDF

​​​​2024

Q1: The sum of all real values of k for which  is
(a) 2/3
(b) 4/3
(c) -4/3
(d) -2/3

Ans: d

Sol: 

Q2: (a + b√3)² = 52 + 30√3, where a and b are natural numbers, then a + b equals 
(a) 9
(b) 7
(c) 8
(d) 10

Ans: c

Sol: 

2023

Q1: Let a, b, m and n be natural numbers such that a > 1 and b > 1. If a^mbⁿ = 144¹⁴⁵, then the largest possible value of n mis  [2023]
​​(a) 580
(b) 290
(c) 289
(d) 579

Ans: d

Sol: 
​It is given that a^m . bⁿ = 144¹⁴⁵, where a > 1 and b > 1.

144 can be written as 144 = 2⁴ x 3²

Hence, a^m . bⁿ = 144¹⁴⁵ can be written as a^m . bⁿ = (2⁴ x 3²)¹⁴⁵ = 2⁵⁸⁰ x 3²⁹⁰

We know that 3²⁹⁰ is a natural number, which implies it can be written as a^l, where a > 1

Hence, the least possible value of m is 1. Similarly, the largest value of n is 580.

Hence, the largest value of (n-m) is (580-1) = 579

The correct option is D

Q2: Let n be any natural number such that 5^n-1 < 3ⁿ⁺¹. Then, the least integer value of m that satisfies 3ⁿ⁺¹ < 2^n+m  for each such n, is  [2023]

​Ans: 5

Sol: It is given that 5^n-1 < 3ⁿ⁺¹, where n is a natural number. By inspection, we can say that the inequality holds when n = 1, 2, 3 4, and 5. 

Now, we need to find the least integer value of m that satisfies 3ⁿ⁺¹ < 2^n+m 

For, n =1, the least integer value of m is 2.

For, n = 2, the least integer value of m is 3

For, n = 3, the least integer value of m is 4.

For, n = 4, the least integer value of m is 4.

For, n= 5, the least integer value of m is 5.

Hence, the least integer value of m such that for all the values of n, the equation holds is 5.

2022

Q1: Let A be the largest positive integer that divides all the numbers of the form 3^k + 4^k + 5^k, and B be the largest positive integer that divides all the numbers of the form 4^k + 3(4^k) + 4^k+2, where k is any positive integer. Then (A + B) equals  [2022]

Ans: 82

Sol: A is the HCF of 3^k + 4^k + 5^k for different values of k.

For k = 1, value is 12

For k = 2, value is 50

For k = 3, value is 216

HCF is 2. Therefore, A = 2

4^k + 3 (4^k) + 4^k+2 = 4^k (1 + 3) + 4^k+2 = 4^k+1 + 4^k+2 = 4^k+1 (1 + 4) = 5 · 4^k+1 

HCF of the values is when k = 1, i.e. 5*16 = 80

Therefore, B = 80

A + B = 82

2020

Q1: If x = ( 4096)^7+4√3 , then which of the following equals 64?   [2020]

(a) 
(b) 
(c) 
(d) 

Ans: d

Sol: x = ( 4096)^7+4√3

On rationalizing 7+4√3, we get

∴
∴
=

Q2: If a, b, c are non-zero and 14^a = 36^b = 84^c, then  is equal to  [2020]

Ans: 3

Sol: Let 14^a = 36^b = 84^c = k
⇒
Similarly,

Required answer,

=
= 3