Simple Ratios MCQ Quiz - Objective Question with Answer for Simple Ratios - Download Free PDF

Given:

Ratio of A to B = 8 : 11

A = 168

Calculation:

A / B = 8 / 11

168 / B = 8 / 11

168 × 11 = 8 × B

1848 = 8B

B = 1848 / 8

B = 231

∴ The value of B is: 231

Given:

Ratio of boys to girls = 5 : 6

Total number of students = 66

Calculation:

The total parts in the ratio

Total parts = 5 + 6 = 11 parts

One part = Total number of students / Total parts

One part = 66 / 11 = 6

Number of girls = 6 × 6 = 36

∴ The number of girls in the class is: 36

Given:

x : 7 = 56 : 40

Method:

Use the property of proportions:

If a : b = c : d, then a × d = b × c

Calculation:

⇒ x × 40 = 7 × 56

⇒ x × 40 = 392

⇒ x = 392 ÷ 40

⇒ x = 9.8

∴ The correct answer is: 9.8

Given:

Ratio of monthly incomes of A and B = 4 : 3

Ratio of expenditures of A and B = 3 : 2

Calculation:

Let the incomes of A and B be '4x' and '3x' .

So, \(\dfrac{4x-600}{3x -600} \) = \(\dfrac{3}{2}\)

⇒ 2 × (4x - 600) = 3 × (3x - 600)

⇒ 8x - 1200 = 9x -1800

⇒ x = 600

The monthly income of B = 3 × 600 = Rs. 1,800

∴The answer is Rs.1,800 .

Given:

Sum of two parts = 141

One-eighth part of the first part and one-ninth part of the second part are in the ratio 5:6.

Calculation:

Let the first part be x and the second part be (141 - x).

According to the given condition:

\(\frac{\frac{x}{8}}{\frac{141 - x}{9}} = \frac{5}{6}\)

Cross multiplying gives:

⇒ 9x / 8 × 6 = 5(141 - x)

⇒ 54x / 8 = 705 - 5x

⇒ 54x = 5640 - 40x

⇒ 94x = 5640

⇒ x = 60

The first part is 60.

Given:

A = 75% of B

Calculation:

A = 3/4 of B

⇒ A/B = 3/4

Let the value of A be 3x and B be 4x

So (2B – A)/A = (2 × 4x – 3x)/3x

⇒ (2B – A)/A = 5x/3x

∴ (2B – A)/A = 5/3

Short Trick:

Ratio of A : B = 3 : 4

∴ (2B – A)/A = (8 – 3) /3 = 5/3

Given:

x : y = 5 : 4

Explanation:

(x/y) = (5/4)

(y/x) = (4/5)

Now, \(\left( {\frac{x}{y}} \right):\left( {\frac{y}{x}} \right)\) = (5/4)/(4/5) = 25/16

∴ \(\left( {\frac{x}{y}} \right):\left( {\frac{y}{x}} \right)\) = 25 : 16

Given :

Ratio of two numbers is 4 : 7 

Calculations :

Let the number added to denominator and numerator be 'x' 

Now according to the question 

(4 + x)/(7 + x) = 2 : 3 

⇒ 12 + 3x = 14 + 2x 

⇒ x = 2 

∴ 2 will be added to make the term in the ratio of 2 : 3.

Given:

Ratio of two numbers is 14 : 25

Difference between them is 264

Calculation:

Let the numbers be 14x and 25x

⇒ 25x – 14x = 264

⇒ 11x = 264

∴ x = 24

⇒ Smaller number = 14x = 14 × 24 = 336

∴ The smaller of the two numbers is 336.

Given:

The ratio of the salaries of Ravi and Sarita is 3 ∶ 5.

If the salary of each is increased by ₹ 5,000, the new ratio becomes 29 ∶ 45.

Formula Used:

Initial salaries: R = 3x and S = 5x.

New salaries: R + 5000 and S + 5000.

New ratio: (R + 5000) / (S + 5000) = 29/45.

Calculation:

Substituting the values of R and S in the new ratio equation:

(3x + 5000) / (5x + 5000) = 29 / 45

Cross multiplying to solve for x:

⇒ 45 × (3x + 5000) = 29 × (5x + 5000)

⇒ 135x + 225000 = 145x + 145000

⇒ 145x - 135x = 225000 - 145000

⇒ 10x = 80000

⇒ x = 8000

Now, finding the current salary of Sarita:

S = 5x = 5 × 8000

S = 40000

The present salary of Sarita is ₹ 40,000.

Shortcut Trick 

Given:

x : y = 6 : 5

And z : y = 9 : 25

Calculation :

x/y = 6/5     ---- (i)

And z/y = 9/25

⇒ y/z = 25/9     ---- (ii)

Multiply equation (i) and (ii) we get,

(x/y) × (y/z) = (6/5) × (25/9)

⇒ x/z = 10/3

∴ x : z = 10 : 3

Alternate Method 

x : y = 6 : 5     ----- (i)

And z : y = 9 : 25     ---- (ii)

As y is in both the ratios, Multiply (i) × 5 to make equal value of y in both the ratios

x : y = (6 : 5) × 5 = 30 : 25    ---- (iii)

from (ii) and (iii), Since y is same in both the ratios

x : z = 30 : 9 = 10 : 3
 

Given: 

5p : 10p : 25p = 3 : 2 : 1 = 3x : 2x : x

Concept:

1 Rupee = 100 paise

Calculation:

60 Rupees = 60 × 100 = 6000 paise

⇒ 5 × 3x + 10 × 2x + 25 × 1x = 6000

⇒ 15x + 20x + 25x = 6000

⇒ 60x = 6000

⇒ x = 100

∴ Number of 5 paise coins = 3x = 3 × 100 = 300

Given:

Ratio of Speed of Deepak and Vinod = 19 : 12

Let the speeds of Deepak and Vinod be 19x km/hr and 12x km/hr

Speed of Vinod = 84 km/hr

Calculations:

Speed of Vinod = 84 km/hr

⇒ 12x = 84

⇒ x = 7

Speed of Deepak = 19x = 19 × 7 = 133 km/hr

∴ The speed of Deepak is 133 km/hr.

Shortcut Trick

P : Q : R = 5 : 3 : 6

Le P be 5x, Q be 3x and R be 6x

Then, (P/Q) ∶ (Q/R) ∶ (R/P) = (5x/3x) ∶ (3x/6x) ∶ (6x/5x)

Let us take the LCM (3, 6, 5) = 30

So, (P/Q) ∶ (Q/R) ∶ (R/P) = (5x/3x) × 30 ∶ (3x/6x) × 30 ∶ (6x/5x) × 30

∴ Required ratio is 50 ∶ 15 ∶ 36

Alternate Method

Given:

P : Q : R = 5 : 3 : 6

Le P be 5x, Q be 3x and R be 6x.

Concept:

If N is divided into a : b, then

First part = N × a/(a + b)

Second part = N × b/(a + b)

Calculations:

The required ratio = P/Q : Q/R : R/P

Multiplying the above ratio with PQR

⇒ Required ratio = P²R : Q²P : R²Q

Putting values of P,Q and R in above ratio, we get

⇒ Required ratio = (5x)²(6x) : (3x)²(5x) : (6x)²(3x)

⇒ Required ratio = (25x²)(6x) : (9x²)(5x): (36x²)(3x)

⇒ Required ratio = (25)(2) : (3)5: (36)

⇒ Required ratio = 50 : 15 : 36

∴ Required ratio is 50 ∶ 15 ∶ 36.

Given:

Total Rs. = Rs.550

Calculation:

Let the number of denominations of 50p = 2x

the number of denominations of 25p = 3x

the number of denominations of 20p = 5x

Total paisa = 550 × 100 = 55000 paise

According to the question:

⇒ (50 × 2x) + (25 × 3x) + (20 × 5x) = 55000

⇒ 100x + 75x + 100x = 55000

⇒ 275x = 55000

⇒ x = 55000/275 = 200

Amount in (Rs.) of 50p = 100x = (100 × 200) = 20000 paise = Rs.200

Amount in (Rs.) of 20p = 100x = (100 × 200) = 20000 paise = Rs.200

Required difference = 200 - 200 = 0

∴ The correct answer is 0.