Description

This mock test of Test: Ratio And Proportion, Indices, Logarithms - 2 for CA Foundation helps you for every CA Foundation entrance exam.
This contains 40 Multiple Choice Questions for CA Foundation Test: Ratio And Proportion, Indices, Logarithms - 2 (mcq) to study with solutions a complete question bank.
The solved questions answers in this Test: Ratio And Proportion, Indices, Logarithms - 2 quiz give you a good mix of easy questions and tough questions. CA Foundation
students definitely take this Test: Ratio And Proportion, Indices, Logarithms - 2 exercise for a better result in the exam. You can find other Test: Ratio And Proportion, Indices, Logarithms - 2 extra questions,
long questions & short questions for CA Foundation on EduRev as well by searching above.

QUESTION: 1

Division of Rs. 750 into 3 parts in the ratio 4 : 5 : 6 is

Solution:

QUESTION: 2

The sum of the ages of 3 persons is 150 years. 10 years ago their ages were in the ratio 7 : 8 : 9. Their present ages are

Solution:

QUESTION: 3

The numbers 14, 16, 35, 42 are not in proportion. The fourth term for which they will be in proportion is

Solution:

QUESTION: 4

If x/y = z/w, implies y/x = w/z, then the process is called

Solution:

QUESTION: 5

If p/q = r/s = p–r/q–s, the process is called

Solution:

QUESTION: 6

If a/b = c/d, implies (a+b)/(a–b) = (c+d)/(c–d), the process is called

Solution:

QUESTION: 7

If u/v = w/p, then (u–v)/(u+v) = (w–p)/(w+p). The process is called

Solution:

QUESTION: 8

12, 16, *, 20 are in proportion. Then * is

Solution:

QUESTION: 9

4, *, 9, 13½ are in proportion. Then * is

Solution:

QUESTION: 10

The mean proportional between 1.4 gms and 5.6 gms is

Solution:

Let 1.4 gms, x gms and 5.6 gms be in continuous proportion.

(Mean proportional terms)^2 = Product of extremes.

Therefore x^2 = 1.4 x 5.6

therefore x^2 = 7.84

therefore x= 2.8 gms.

QUESTION: 11

Solution:

QUESTION: 12

Two numbers are in the ratio 3 : 4; if 6 be added to each terms of the ratio, then the new ratio will be 4 : 5, then the numbers are

Solution:

QUESTION: 13

Solution:

QUESTION: 14

Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is:

Solution:

QUESTION: 15

if then (b-c)x + (c-a)y+(a-b)z is

Solution:

Let X=(b+c-a);y=(c+a-b);z=(a+b-c)

=(b-c)(b+c-a)+(c-a)(c+a-b)+(a-b)(a+b-c)

=b^2-c^2-ab+ac+c^2-a^2-bc+ab+a^2-b^2-ac+bc

=0

QUESTION: 16

4x^{–1/4} is expressed as

Solution:

QUESTION: 17

The value of 8^{1/3} is

Solution:

QUESTION: 18

The value of 2 × (32) ^{1/5} is

Solution:

QUESTION: 19

The value of 4/(32)^{1/5} is

Solution:

QUESTION: 20

The value of (8/27)^{1/3} is

Solution:

QUESTION: 21

2^{½} .4^{¾} is equal to

Solution:

QUESTION: 22

has simplified value equal to

Solution:

QUESTION: 23

x^{a–b} × x^{b–c} × x^{c–a} is equal to

Solution:

QUESTION: 24

The value of is equal to

Solution:

QUESTION: 25

{(3^{3})^{2} × (4^{2})^{3} × (5^{3})^{2}} / {(3^{2})^{3} × (4^{3})^{2} × (5^{2})^{3}} is

Solution:

QUESTION: 26

Which is True ?

Solution:

QUESTION: 27

If x^{1/p} = y^{1/q} = z^{1/r} and xyz = 1, then the value of p+q+r is

Solution:

QUESTION: 28

The value of y^{a–b} × y^{b–c} × y^{c–a} × y–^{a–b} is

Solution:

QUESTION: 29

The True option is

Solution:

QUESTION: 30

The simplified value of 16x^{–3}y^{2 }× 8^{–1}x^{3}y^{–2} is

Solution:

QUESTION: 31

The value of (8/27)^{–1/3} × (32/243)^{–1/5} is

Solution:

QUESTION: 32

The value of {(x+y)^{2/3} (x–y)^{3/2}/√x+y × √ (x–y)^{3}}^{6} is

Solution:

QUESTION: 33

Simplified value of (125)^{2/3} × √25 × ^{3}√5^{3 }× 5^{1/2} is

Solution:

QUESTION: 34

[{(2)^{1/2} . (4)^{3/4} . (8)^{5/6} . (16)^{7/8} . (32)^{9/10}}4]^{3/25} is

Solution:

QUESTION: 35

[1–{1–(1–x^{2})^{–1}}^{–1}]^{–1/2} is equal to

Solution:

**ANSWER :- a**

**Solution :- [1–{1–(1–x ^{2})^{–1}}^{–1}]^{–1/2}**

**=[1 - { 1- x ^{2} -1/(1-x^{2}) }^{-1}]^{-1/2} here we used a^{-n} = 1/a^{n}**

**= [1 - {(-x^2)/(1-x^2)}^-1]^-1/2**

**= [ 1 - { x^2 /( x^2 -1) } ^{-1}]^{-1/2}**

**= [ 1- (x ^{2}-1)/ x^{2}]^{-1/2}**

**= [ (x ^{2} -x^{2}+1)/ x^{2}]^{-1/2}**

**= [1/x ^{2}]^{-1/2}**

**= [x ^{2}]^{1/2}**

**= x ^{2*1/2}**

**= x**

QUESTION: 36

Ajay and Raj together have Rs. 1050. On taking Rs. 150 from Ajay, Ajay will have same amount as what Raj had earlier. Find the ratio of amounts with Ajay and Raj initially.

Solution:

QUESTION: 37

If a^{3}–b^{3 }= (a–b) (a^{2} + ab + b^{2}), then the simplified form of

Solution:

QUESTION: 38

Using (a–b)^{3} = a^{3}–b^{3}–3ab(a–b) tick the correct of these when x = p^{1/3} – p^{–1/3}

Solution:

QUESTION: 39

On simplification, 1/(1+a^{m–n}+a^{m–p}) + 1/(1+a^{n–m}+a^{n–p}) + 1/(1+a^{p–m}+a^{p–n}) is equal to

Solution:

QUESTION: 40

If A:B = 2:3, B:C = 4:5 and C:D = 6:7, then find the value of A:B:C:D

Solution:

### Ratio and Proportion, Indices, Logarithms - 2

Doc | 11 Pages

### Ratio and Proportion, Indices, Logarithms - 1

Doc | 13 Pages

### Ratio and Proportion, Indices, Logarithms (Part - 2)

Doc | 11 Pages

### Ratio and Proportion, Indices, Logarithms (Part - 1)

Doc | 12 Pages

- Test: Ratio And Proportion, Indices, Logarithms - 2
Test | 40 questions | 40 min

- Test: Ratio And Proportion, Indices, Logarithms - 3
Test | 40 questions | 40 min

- Test: Ratio And Proportion, Indices, Logarithms - 4
Test | 40 questions | 40 min

- Test: Ratio And Proportion, Indices, Logarithms - 1
Test | 40 questions | 40 min

- Ratio And Proportion, Indices And Logarithm
Test | 70 questions | 140 min