0
5968

# Ratio And Proportion Questions For SSC MTS

Download Top-20 SSC MTS Ratio and Proportion Questions PDF. Ration and Proportion questions based on asked questions in previous year exam papers very important for the SSC MTS exam.

Question 1: The ratio of two numbers is 5 : 4. If the sum of both the numbers is 180, then what is the smaller number among both the numbers?

a) 100

b) 80

c) 60

d) 75

Question 2: Rs 6300 is divided among A, B and C in the ratio of 1/2: 1: 3/5. What is the share (in Rs) of B?

a) 3300

b) 2700

c) 3000

d) 4200

Question 3: S, T and U together can complete a work in 30 days. If the ratio of efficiency of S, T and U is 20: 15: 12 respectively, then in how many days U alone can complete the same work?

a) 195/2

b) 235/2

c) 225/2

d) 215/2

Question 4: In what ratio must a mixture of 20% milk strength be mixed with that of 60% milk strength so as to get a new mixture of 25% milk strength ?

a) 7: 1

b) 4: 1

c) 5: 2

d) 9: 2

Question 5: Rs 18200 is divided among X, Y and Z in the ratio of 1/3: 1/4: 1/2. What is the share (in Rs) of X ?

a) 7000

b) 4400

c) 4200

d) 5600

Question 6: The ratio of milk and water in three samples is 1: 3, 3: 5 and 11: 5. A mixture containing equal quantities of all three samples is made. What will be the ratio of milk and water in the new mixture?

a) 7: 9

b) 5: 7

c) 15: 13

d) 9: 11

Question 7: Of three positive numbers, the ratio of first and second is 5: 2, ratio of second and third is 5: 4. The product of first and third is 1800. What is the sum of the three numbers?

a) 43

b) 133

c) 119

d) 129

Question 8: Two whole numbers are such that the square of first number exceeds the second by 112 and the ratio of the numbers is 4:3. What is the value of smaller number?

a) 3

b) 4

c) 12

d) 36

Question 9: The sum of three numbers is 162. If the ratio of the first number to the second number is 5: 7 and that of the second number to the third number is 5: 3, then what is the second number?

a) 35

b) 70

c) 80

d) 40

Question 10: The milk and water in two vessels are in the ratio of 3: 1 and 7: 11 respectively. In what ratio should the liquid in both the vessels be mixed to obtain a new mixture containing half milk and half water?

a) 5:7

b) 4:9

c) 1:1

d) 4:7

Question 11: The ratio of three numbers is 5: 6: 8. If the sum of the three numbers is 380, then
what is the smallest among the three numbers?

a) 75

b) 80

c) 120

d) 100

Question 12: If in the population of a town there is an increase of 44% in the first year and 75% in the second year, then what is the average rate of increase in the population?

a) 45%

b) 59.5%

c) 74%

d) 76%

Question 13: In a mixture of milk and water, there is 19% milk. If the quantity of milk in the mixture is 437 mililitres, then what is the total quantity (in millilitres) of mixture?

a) 2300

b) 2200

c) 2350

d) 2400

Question 14: The ratio of two numbers is 6 : 11 and their difference is 65. What will be the smaller number among both the numbers?

a) 58

b) 65

c) 52

d) 78

Question 15: If 35% of (x + y) = 40% of (x – y), then x is what percentage of y?

a) 1500

b) 150

c) 15

d) 105

Question 16: The ratio of three numbers is 2 : 3 : 5. If the sum of the three numbers is 275, then what is the largest among the three numbers?

a) 142

b) 82.5

c) 137.5

d) 152

Question 17: The sum of three numbers is 67. If the ratio of the first number to the second number is 3 : 5 and that of the second number to the third number is 4 : 7, then what is the second number?

a) 20

b) 24

c) 18

d) 16

Question 18: A 90-litre mixture of milk and water contains 80% milk. How much water (in litres) must be added to make the amount of water in the mixture 40%?

a) 30

b) 40

c) 15

d) 25

Question 19: A dishonest shopkeeper mixes vegetable oil with ghee in the ratio 1:3. It is also known that the ratio of the cost price of vegetable oil and ghee is 2:5. If the shopkeeper sells the mixture at 10% more then cost price of pure ghee, find out his profit percentage in this transaction.

a) $29\frac{7}{17}$%

b) $27\frac{3}{17}$%

c) $27\frac{7}{17}$%

d) $27\frac{9}{17}$%

Question 20: Price of a diamond is directly proportional to the cube of its weight. A man broke the diamond accidentally in three pieces in the ratio of 1 : 2 : 3 and thus, the net valuation of all 3 diamonds now is Rs. 21600. What was the original price (in Rs) of the diamond?

a) 10000

b) 15000

c) 169200

d) 129600

Let the numbers be $5x$ and $4x$

Sum = $5x+4x=180$

=> $9x=180$

=> $x=\frac{180}{9}=20$

$\therefore$ Smaller number = $4\times20=80$

=> Ans – (B)

Total amount = Rs. 6300

Ratio of amount with A : B : C = $\frac{1}{2}:1:\frac{3}{5}$

=> Share of B = $\frac{1}{(\frac{1}{2}+1+\frac{3}{5})}\times6300$

= $\frac{10}{5+10+6}\times6300$

= $10\times300=Rs.$ $3000$

=> Ans – (C)

Let efficiencies of S, T and U be $20x,15x$ and $12x$ units/day respectively.

S, T and U together can complete the work in 30 days

=> Total work to be done = $30\times(20x+15x+12x)=1410x$ units

$\therefore$ Time taken by U alone to complete the work = $\frac{1410x}{12x}=\frac{235}{2}$ days

=> Ans – (B)

Let the ratio in which 20% milk be mixed with that of 60% milk = $x:y$

According to ques,

=> $20x+60y=25(x+y)$

=> $20x+60y=25x+25y$

=> $25x-20x=60y-25y$

=> $5x=35y$

=> $\frac{x}{y}=\frac{35}{5}=7$

$\therefore$ Required ratio = 7 : 1

=> Ans – (A)

Total amount = Rs. 18,200

Ratio of amount with X : Y : Z = $\frac{1}{3}:\frac{1}{4}:\frac{1}{2}$

=> Share of X = $\frac{\frac{1}{3}}{(\frac{1}{3}+\frac{1}{4}+\frac{1}{2})}\times18,200$

= $(\frac{1}{3}\div\frac{4+3+6}{12})\times18,200$

= $\frac{1}{3}\times\frac{12}{13}\times18,200$

= $4\times1400=Rs.$ $5600$

=> Ans – (D)

Let quantity of each sample is L.C.M. (4,8,16) = 16 unit

=> Milk in first sample = $\frac{1}{(1+3)}\times16=4$ unit

and water in first sample = $16-4=12$ units

Similarly, milk in second sample = $\frac{3}{(3+5)}\times16=6$ units and water = $16-6=10$ units

Milk in third sample = $\frac{11}{(11+5)}\times16=11$ units and water = $16-11=5$ units

$\therefore$ Ratio of milk and water in the new mixture = $\frac{(4+6+11)}{(12+10+5)}$

= $\frac{21}{27}=\frac{7}{9}$

=> Ans – (A)

Let the numbers be $x,y,z$

Thus, $x:y=5:2$ ———–(i)

and $y:z=5:4$ ————-(ii)

Multiplying equation (i) by 5 and equation (ii) by 2

=> $x:y:z=25:10:8$

Let $x=25k$, $y=10k$ and $z=8k$

=> Product = $25k\times8k=1800$

=> $k^2=\frac{1800}{200}=9$

=> $k=\sqrt9=3$

$\therefore$ Sum of three numbers = $25k+10k+8k=43k$

= $43\times3=129$

=> Ans – (D)

Let the numbers be $4x$ and $3x$

According to ques,

=> $(4x)^2-(3x)^2=112$

=> $16x^2-9x^2=7x^2=112$

=> $x^2=\frac{112}{7}=16$

=> $x=\sqrt{16}=4$

$\therefore$ Smaller number = $3\times4=12$

=> Ans – (C)

Let the numbers be $x,y,z$

Thus, $x:y=5:7$ ———–(i)

and $y:z=5:3$ ————-(ii)

Multiplying equation (i) by 5 and equation (ii) by 7

=> $x:y:z=25:35:21$

Let $x=25k$, $y=35k$ and $z=21k$

=> Sum of three numbers = $25k+35k+21k=81k=162$

=> $k=\frac{162}{81}=2$

$\therefore$ Second number = $35\times2=70$

=> Ans – (B)

Milk in 1 litre mixture of vessel 1 = $\frac{3}{4}$

Milk in 1 litre mixture of vessel 2 = $\frac{7}{18}$

Milk in 1 litre of the new mixture = $\frac{1}{2}$

Let ratio of liquid from both vessels be $x:y$

=> $(\frac{3}{4}x)+(\frac{7}{18}y)=\frac{1}{2}(x+y)$

Multiply both sides by L.C.M. (4,18,2) = 36

=> $27x+14y=18x+18y$

=> $27x-18x=18y-14y$

=> $9x=4y$

=> $\frac{x}{y}=\frac{4}{9}=$

$\therefore$ Liquid of both the vessels should be mixed in the ratio 4 : 9 such that a new mixture containing half milk and half water is obtained.

=> Ans – (B)

Let three numbers be $5x,6x$ and $8x$ respectively.

Sum = $5x+6x+8x=380$

=> $x=\frac{380}{19}=20$

$\therefore$ Smallest number = $5\times20=100$

=> Ans – (D)

Let population of town initially = $100x$

Population after 44% increase in 1st year = $100x+\frac{44}{100}\times100x=144x$

Population after 75% increase in 2nd year = $144x+\frac{75}{100}\times144x=252x$

Thus, rate of increase in population = $\frac{(252x-100x)}{100x}\times100=152\%$

=> Average rate of increase in the population in 2 years = $\frac{152}{2}=76\%$

=> Ans – (D)

Let total quantity of mixture (milk+water) = $100x$ ml

=> Quantity of milk = $\frac{19}{100}\times100x=19x$ ml

Also, it is given that quantity of milk = 437 ml

=> $19x=437$

=> $x=\frac{437}{19}=23$

$\therefore$ Total quantity of mixture = $100\times23=2300$ ml

=> Ans – (A)

Let the numbers be $6x$ and $11x$

Difference = $11x-6x=65$

=> $x=\frac{65}{5}=13$

$\therefore$ Smaller number = $6\times13=78$

=> Ans – (D)

Expression : 35% of (x + y) = 40% of (x – y)

=> $\frac{35}{100}\times(x+y)=\frac{40}{100}\times(x-y)$

=> $35x+35y=40x-40y$

=> $40x-35x=35y+40y$

=> $5x=75y$

=> $\frac{x}{y}=\frac{75}{5}=15$

$\therefore$ Required % = $15\times100=1500\%$

=> Ans – (A)

Let three numbers be $2x,3x$ and $5x$ respectively.

Sum = $2x+3x+5x=275$

=> $x=\frac{275}{10}=27.5$

$\therefore$ Largest number = $5\times27.5=137.5$

=> Ans – (C)

Ratio of first number to second number = 3 : 5 and ratio of the second number to the third number = 4 : 7

Multiplying first equation by 4 and second equation by 5 to make the second (common) term equal.

Let first number be = $12x$, second number = $20x$ and third number = $35x$

=> Sum = $12x+20x+35x=67x=67$

=> $x=\frac{67}{67}=1$

$\therefore$ Second number = $20\times1=20$

=> Ans – (A)

Quantity of mixture = 90 litres
Quantity of milk = $\frac{80}{100}\times90=72$ litres
$\Rightarrow$ Quantity of Water = $90-72=18$ litres
Let us assume that $x$ litres of water is to be added.
According to problem statement,
$\Rightarrow$ $\frac{40}{100}(x+90)=18+x$
$\Rightarrow$ $2*(x+90) = 5*(18+x)$
$\Rightarrow$ $5x – 2x = 180 – 90$
$\Rightarrow$ $x = 30$.
Therefore, option A is the right answer.

Let us assume that the cost of pure ghee is Rs. 5X/kg.

So, the cost of vegetable oil = $\frac{2}{5} \times 5X = Rs. 2X/kg$

Let us assume that the shopkeeper used 3Y kg of pure ghee and Y kg of vegetable oil to prepare the mixture.

Hence, the total cost incurred by the shopkeeper = 5X * 3Y + 2X * Y = 17XY

Total weight of mixture = 3Y + Y = 4Y kg

Selling price of mixture = $\frac{100+10}{100} \times 5X = Rs 5.5X/kg$

Total revenue generated by selling the mixture = 5.5X * 4Y = 22XY.

Hence, we can see that the shopkeeper made profit of 22XY – 17XY = 5XY.

Profit percentage of the shopkeeper in the transaction = $\frac{5XY}{17XY} \times 100 = 29\frac{7}{17}$%.

Therefore, option A is the right answer.

Let weight of original diamond = $6x$ and price of diamond = Rs. $P$.
According to question, => $P \propto (6x)^3$.
$\Rightarrow$ $P=216kx^3$ ————–(i)
Let weight of each broken piece be $x,2x$ and $3x$ and price $P_{1}, P_{2} and P_{3}$ respectively.
So $P_{1}=k(x)^3=kx^3$.
$P_{2}=k(2x)^3=8kx^3$.
$P_{3}=k(3x)^3=27kx^3$.
So, the total price of all 3 diamonds combined = $kx^3+ 8kx^3+ 27kx^3= 36kx^3$.
It is given that $36kx^3 = 21600.$
$\Rightarrow kx^3 = \frac{21600}{36} = 600$
Substituting the value in equation (i)
$\Rightarrow P=216 * 600$ = Rs. 129600.
Therefore, option D is the right answer.