MCQ's Ratio and Proportion, Indices, Logarithms - 1 - Quantitative Aptitude for CA

Q1: If p : q = r : s, implies q : p = s : r, then the process is called.
(a) Componendo
(b) Invertendo
(c) Alternendo
(d) Dividendo
Ans: (b)

Sol:
We know, 
p : q = r : s implies q : s : r, is the process of Invertendo.

Q2: 3x - 2/5x + 6 is the duplicate ratio of 2/3 , then find the value of x.
(a) 6
(b) 2
(c) 5
(d) 9
Ans: (a)

Sol:
Given, 3x - 2/5x + 6  is the duplicate ratio of 2/3
We know that, a² : b² is the duplicate ratio of a : b.
⇒ The duplicate ratio of  

According to the given problem, we have 
On cross multiplication, we get
9(3x – 2) = 4(5x + 6)
⇒ 27x – 18 = 20x + 24
⇒ 27x – 20x = 24 + 18
⇒ 7x = 42
⇒ x = 6
Therefore, the value of x is 6.
Hence, the correct option is (a).

Q3: What is the value of  P + q/p - q if p/q = 7?
(a) 2/3
(b) 4/3
(c) 2/6
(d) 7/8
Ans: (b)

Sol:
Given, p/q = 7
p/q = 7/1
Let p = 7x and q = x, then

Q4: If x : y = 7 : 8, then for 6x + 5y : 4x + 3y = ?
(a) 11 : 7 
(b) 30 : 12
(c) 35 : 24
(d) 41 : 26
Ans: (d)

Sol:
Given, x : y = 7 : 8 
Let x = 7a and y = 8a, then 6x + 5y/4x + 3y

= 82a/52a
= 41/26
= 41 : 26

Q5: If x : y = 4 : 6 and 2 : x = 1 : 2, then y = ? 
(a) 4 
(b) 6 
(c) 1/2
 (d) 3/2
Ans: (b)

Sol:
 Given,  2 : x = 1 : 2 
⇒ 2/x = 1/2 
⇒ x = 4 
Also, x : y = 4 : 6 
=x/y = 4/6
⇒ = 4/y = 4/6
⇒ y = 6

Q6: If x : y = 2 : 3, then (5x + 2y) : (3x - y)
(a) 19 : 3 
(b) 16 : 3 
(c) 7 : 2
(d) 7 : 3
Ans: (b)

Sol:
Given, x : y = 2 : 3
Let x = 2a and y = 3a, then
(5x + 2y) : (3x – y)
= 5(2a) + 2(3a)/3(2a) - 3a
= 10a + 6a /6a - 3a

= 16 : 3

Sol:
Given, A : B = 2 : 5 
Let A = 2x and B = 5x 
Thus, (10A + 3B) : (5A + 2B) 
= 10A + 3B/5A + 2B

= 7 : 4

Sol:
Given that, Salaries of A, B, C are in the ratio 2 : 3 : 5. 
Let A = 2K, B = 3K and C = 5K 
After the increment of 15%, 10% and 20% then
A's new salary = 2K + 15% of 2K = 115/110 of 2K =

B's new salary = 110/100 of 3K == 6K
C's new salary = 120/100 of 5K = = 6K
New Ratio = = 23 : 33 : 60

Sol:
Given, Ratio of 25 paise, 10 paise and 5 paise coins = 3 : 2 : 1 
Total value = ₹40 
Let the number of 25 paise coins, 10 paise coins and 5 coins be 3x, 2x and x respectively. 
Then, the value = 0.25 × 3x + 0.10 × 2x + 0.05 × x 
⇒ 0.75x + 0.20x + 0.05x = 40 
⇒ x = 40 
Therefore, the number of 5 paise coins = x = 4

Ans: (a)

Sol:
Let the number of 1 rupee coins = x, then 
Number of 50 paise coins = 4x 
Number of 25 paise coins = 2x 
Since, total amount = ₹56 
Thus, 1 × x + 0.50 × 4x + 0.25 × 2x = 56
 ⇒ x + 2x + 0.5x = 56 
⇒ 3.5x = 56 
⇒ x = 16 
Therefore, the number of 50 paise coins = 4x 4(16) = 64.

Sol:
We know, 
Sub-duplicate of 4 : a is √4 : √a = 2 : √a
Since, ratio compounded of 4 : 5 and sub-duplicate of 4 : a = 8 : 15 ie.

⇒ a = 3²
⇒ a = 9

Sol:
Let the income of A and B be 5x and 4x respectively. 
We know, Expenditure = Income – Savings
According to question, we have

⇒ 2(5x - 1600) = 3(4x - 1665)
 ⇒ 10x - 3200 = 12x - 4800
⇒ 12x - 10x = 4800 - 3200 
⇒ 2x = 1600 
⇒ x = 800

Therefore, the income of A will be 5(800) = ₹4000

Sol:
Let b be the mean proportional between 8 and 32. 
Then, b2 = 8 × 32 

⇒ b = 16

Sol: 
Let the third proportional be c, then 49 : 21 = 21 : x 
⇒ 49/21 = 21/x 
⇒ (21)² = 49x 
⇒ x = (21)²/49 ⇒ x = 9

Sol: 
Let the fourth proportional to x, 2x, (x + 1) be y, then

⇒ y = 2(x + 1)
⇒ y = 2x + 2

Sol: 
Let the mean proportional between 12x² and 27y² be 'b', then
b² = 12x² × 27y ²
⇒ b² = 324x² y²

⇒ b = 18xy

Sol: 
Given, D is 50% more than E 
Let E = x, then D = x + 50% of x 
⇒ D = x + 0.5x 
⇒ D = 1.5x 
Also, A : B = 3 : 4, B : C = 7 : 9, C : D = 2 : 3 
Therefore, the ratio between A and E is given by

⇒ A/E = 7/12

Sol: 
Given, 1/2,1/3, 1/5 and 1/x  are in proportion, then


⇒ 3/2 =  5/ x 
⇒ x = 15/2

Sol:
Given, Ratio of shares of wife, son and daughter = 3 : 2 : 1 
Total assets value = ₹1,48,200
Therefore, share of son =

2/6 x 148200 = 49400
Hence, the share of his son is  ₹49,400.

Sol: 
According to the question, we have X = 3Y and Y = 2/3 Z

⇒ X : Y : Z = 6 : 2 : 3

Sol: 
Let the number be x, y and z. According to question, we have
x : y = 2 : 3 and y : z = 5 : 8

⇒ x : y : z = 10 : 15 : 24
Sum of the ratios = 10 + 15 + 24 = 48 
Therefore, the second number × (15/49) x 98 = 30

Sol:
Given, Ratio of students in three classes = 2 : 3 : 5
Let the students in three classes be 2x, 3x and 5x respectively. 
According to question
(2x + 40) : (3x + 40) : (5x + 40) = 4 : 5 : 7

⇒ 5(2x + 40) = 4(3x + 40) 
⇒ 10x + 200 = 12x + 160 
⇒ 2x = 40 
⇒ x = 20
Therefore, the total students originally were, 
2x + 3x + 5x = 10x = 10(20) = 200

Sol: 
Since, 3 times Anil's share, 6 times Sunil's share and 8 times Nimal's share are all equal i.e.,
3A = 6B = 8C

Therefore, Anil's share is given by:

Sol: 
Let the quantity of mixture be x, then Quantity of acid = (100 – 64)% of x = 0.36x 
Now, quantit of acid in 4 litres of mixture = 4 × 36% = 1.44
Since, 4 litres of the solution were taken out of the vessel, thus 
Remaining acid = 0.36x – 1.44 
Also, same quantity of water was added, thus 
Total quantity = x – 4 + 4 = x litres 
Therefore, the percentage of acid is 

⇒ 0.36x - 1.44 = 0.3x 
⇒ 0.06x = 1.44 
⇒ x = 24 litres

Q25: If x = y^a , y = z^b , z = x ^c, then the value of abc is
(a) 1
(b) 2
(c) 3
(d) 4
Ans: (a)

Sol:
Given, x = y^a , y = z^b , z = x ^c
⇒ x = y^a & y = z^b
⇒ x = (z^b)^a
⇒ x = z^ab 
Also, z = x^c 
⇒ x = (x^c)^ab 
⇒ x = x^abc 
⇒ abc = 1

Q26:
(a) x^m
(b) x^–m
(c) xⁿ
(d) x^–n
Ans: (b)

Sol: 

x^{5m–6n–(6m–6n)} = x^–m

Q27: Find the value of  
(a)

(b)

(c)

(d)

Ans: (a)

Sol: 
To simplify:

We know that,
a^m/aⁿ = a^{m - n}


Hence, the correction answer is option (a) i.e.,