Courses

Test: Ratio And Proportion, Indices, Logarithms - 2


40 Questions MCQ Test Quantitative Aptitude for CA CPT | Test: Ratio And Proportion, Indices, Logarithms - 2


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 81/3 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

2½ .4¾ is equal to
Solution:
QUESTION: 22

 has simplified value equal to 
Solution:
QUESTION: 23

xa–b × xb–c × xc–a is equal to
Solution:
QUESTION: 24

The value of  is equal to 
Solution:
QUESTION: 25

{(33)2 × (42)3 × (53)2} / {(32)3 × (43)2 × (52)3} is
Solution:
QUESTION: 26

Which is True ?
Solution:
QUESTION: 27

If x1/p = y1/q = z1/r and xyz = 1, then the value of p+q+r is
Solution:
QUESTION: 28

The value of ya–b × yb–c × yc–a × y–a–b is
Solution:
QUESTION: 29

The True option is
Solution:
QUESTION: 30

The simplified value of 16x–3y2 × 8–1x3y–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√53 × 51/2 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–x2)–1}–1]–1/2 is equal to
Solution:

ANSWER :- a

Solution :- [1–{1–(1–x2)–1}–1]–1/2

=[1 - { 1- x2 -1/(1-x2) }-1]-1/2 here we used a-n = 1/an

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

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

= [ 1- (x2-1)/ x2]-1/2

= [ (x2 -x2+1)/ x2]-1/2

= [1/x2]-1/2

= [x2]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 a3–b3 = (a–b) (a2 + ab + b2), then the simplified form of 
Solution:
QUESTION: 38

 Using (a–b)3 = a3–b3–3ab(a–b) tick the correct of these when x = p1/3 – p–1/3
Solution:
QUESTION: 39

On simplification, 1/(1+am–n+am–p) + 1/(1+an–m+an–p) + 1/(1+ap–m+ap–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:

Recommended Content

Recommended Tests

Follow Us
© 2015-20 EduRev