+91 9600 121 800

Plans

Dashboard

Daily & Speed

Quant

Verbal

DILR

Compete

Free Stuff

calendarBack
Quant

/

Numbers

/

Surds & Indices

Surds And Indices

MODULES

Basics of Surds
Comparison of Surds
Root of Surds
Indices Rules
Comparing Indices
Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Surds & Indices 1
-/10

PRACTICE

Surds & Indices : Level 1
Surds & Indices : Level 2
Surds & Indices : Level 3
ALL MODULES

CAT 2025 Lesson : Surds & Indices - Past Questions

bookmarked

3. Past Questions

Question 1

Which among 21/22^{1/2}21/2, 31/33^{1/3}31/3, 41/44^{1/4}41/4, 61/66^{1/6}61/6, 121/1212^{1/12}121/12 is the largest?
[CAT 2006]

21/22^{1/2}21/2
31/33^{1/3}31/3
41/44^{1/4}41/4
61/66^{1/6}61/6
121/1212^{1/12}121/12

Observation/Strategy
1) The powers are in fractions and the denominators are the same.
2) We can multiply each of the powers by the LCM of the denominators and compare.

LCM(2,3,4,6,12)=12(2, 3, 4, 6, 12) = 12(2,3,4,6,12)=12

2112×12=26=642^{\frac{1}{12} \times 12} = 2^6 = 642121​×12=26=64

313×12=34=813^{\frac{1}{3} \times 12} = 3^4 = 81331​×12=34=81

414×12=43=644^{\frac{1}{4} \times 12} = 4^3 = 64441​×12=43=64

616×12=62=366^{\frac{1}{6} \times 12} = 6^2 = 36661​×12=62=36

12112×12=1212^{\frac{1}{12} \times 12} = 1212121​×12=12

The largest number here is
818181.

∴
3133^{\frac{1}{3}}331​ is the largest.

Answer: (2)
31/33^{1/3}31/3

Question 2

ppp and qqq are positive numbers such that pq=qpp^{q} = q^{p}pq=qp and q=9pq = 9pq=9p. The value of ppp is
[XAT 2013]

9\sqrt{9}9​
96\sqrt[6]{9}69​
99\sqrt[9]{9}99​
98\sqrt[8]{9}89​
93\sqrt[3]{9}39​

Observation/Strategy
1) There are 2 equations with 2 variables. So, we need to substitute one in the other.

Substituting q=9p⇒p9p=(9p)pq = 9p ⇒ p^{9p} = (9p)^{p}q=9p⇒p9p=(9p)p

⇒
p8p×pp=9p×ppp^{8p} \times p^{p} = 9^{p} \times p^{p}p8p×pp=9p×pp

⇒
(p8)p=9p(p^{8})^{p} = 9^{p}(p8)p=9p

⇒
p8=9p^{8} = 9p8=9

⇒
p=918=98p = 9^{\frac{1}{8}} = \sqrt[8]{9}p=981​=89​

Answer: (4)
98\sqrt[8]{9}89​

Question 3

The simplest value of the expression (4p+14×2×2p2×2−p)1p\left( \dfrac{4^{p + \frac{1}{4}} \times \sqrt{2 \times 2^{p}}}{2 \times \sqrt{2^{-p}}} \right)^{\frac{1}{p}}(2×2−p​4p+41​×2×2p​​)p1​ is:
[IIFT 2015]

444
888
4p4^p4p
8p8^p8p

(4p+14×2×2p2×2−p)1p=(22×(p+14)×(2p+1)122×2−p2)1p\left( \dfrac{4^{p + \frac{1}{4}} \times \sqrt{2 \times 2^{p}}}{2 \times \sqrt{2^{-p}}} \right)^{\frac{1}{p}} = \left( \dfrac{2^{2 \times (p + \frac{1}{4})} \times (2^{p + 1})^{\frac{1}{2}}}{2 \times 2^{\frac{-p}{2}}} \right)^{\frac{1}{p}}(2×2−p​4p+41​×2×2p​​)p1​=(2×22−p​22×(p+41​)×(2p+1)21​​)p1​

=(22p+12×2p+12×2p2−1)1p= \left( 2^{2p + \frac{1}{2}} \times 2^{\frac{p + 1}{2}} \times 2^{\frac{p}{2} - 1} \right)^{\frac{1}{p}}=(22p+21​×22p+1​×22p​−1)p1​

=(23p)1p=23=8= (2^{3p})^{\frac{1}{p}} = 2^3 = 8=(23p)p1​=23=8

Answer: (2)
888

Want to read the full content

Unlock this content & enjoy all the features of the platform

Subscribe Now arrow-right
videovideo-lock