DXC Quants Ratio and Proportion Quiz 4

Answer: c.
Explanation:
Let the quantity of each fuels be ‘a’
Then quantity of Ethanol after mixing is,
(\frac{2}{3}+\frac{3}{10}+\frac{4}{15}+\frac{11}{20})a
By solving this equation:
We get,
(\frac{91}{60})a
and quantity of Petrol after mixing is
=(\frac{3}{5}+\frac{7}{10}+\frac{11}{15}+\frac{9}{20})a
By solving the equation:
=(\frac{149}{60})a

Required ratio of Ethanol to Petrol after mixing,
=(\frac{\frac{91}{60}a}{\frac{149}{60}a})= 91:149

Answer: b.
Explanation:
Ratio of 1st and 2nd numbers = 3 : 4
Ratio of 2nd and 3rd numbers = 5 : 6
Let the 2nd number = 5x, third number = 6x
Product of 2nd and 3rd numbers = 4320
5x × 6x = 4320
x2 = 144
x = 12
2nd number = 60, 3rd number = 72,
=(\frac{60}{4}\times3)= 45
Sum of three numbers = 60 + 72 + 45 = 177

Answer: c.
Explanation:
Let the cost price of one curtain = x Rs., the cost price of one table cloth = y Rs.
According to the question,
2x + 3y = 2200 ....1
2y + 4x = 2400
2x + y = 1200 ....2
After solving these 2 equation,
x = 350 Rs. , y = 500 Rs.
Ratio = 500 : 350 = 10 : 7

Answer: c.
Explanation:
Let the income of Tony in 2018 and 2017 be 3x and 5x respectively.
Let the income of Stephan in 2018 and 2017 be 3y and 2y respectively
Since, the ratio of their income in the year 2017 was 5 : 4
5x : 2y = 5 : 4
2x = y
The sum of their incomes in 2018 is Rs. 10242
3x + 3y = 10, 242
9x = 10, 242
x = 1,138 and y = 2276
Stephan’s income for the year 2017 = 2y = Rs. 4552

Answer: c.
Explanation:
Amount of Red wine poured into C from vessel A and B,
=2\times \frac{3}{5}+3\times \frac{2}{3}=\frac{16}{5}litres

Also, amount of Grape Wine poured into C from vessels A and B,
=5-\frac{16}{5}=\frac{9}{5}litres

Given, 5 litres rePresent 40% of the capacity of vessel C, vessel C has a capacity of,
=5\times \frac{5}{2} = 12.5 litres

To make the quantities of Grape Wine and Red wine same in the vessel C, quantity of Grape Wine to be added,
=\frac{16}{5}-\frac{9}{5}= \frac{7}{5} litres

Therefore, the required ratio is,
=\frac{\frac{7}{5}}{12.5}= \frac{14}{125}litres

Answer: d.
Explanation:
Let the score of the topper be A.
Total score of the 52 students = 52 × 85 = 4420
Total score of the remaining 47 students after scores of the best five performers are removed,
= 47 × 83 = 3901
Total score of the top five students = 4420 – 3901 = 519

A + (total score of the next 4 top scores) = 519
A is the maximum when the total score of the next 4 top scorers is minimum.

Total score of the next 4 top scorers has a minimum value of 80 + 81 + 82 + 83 = 326
(since all the top 5 scores are distinct) and the least is 80.

A has a maximum value of 519 – 326 = 193

Answer: b.
Explanation:
Ratio of the number of coins of denominations Rs. 1 : Rs. 2 : Rs. 5 = 5 : 7 : 6
Let, the number of coins of Rs. 1, Rs. 2 and Rs. 5 in the bag be 5x, 7x and 6x.
Since, the total value of Rs. 1 coin in the bag is Rs. 15
So, the number of coins of Rs. 1 in a bag = 15
5x = 15 , ⇒ x = 3
Therefore, number of Rs. 5 coins in the bag
= 6x = 18
So, required value of Rs. 5 coins,
= 6 × 3 × 5 = Rs.90
Therefore, Donna has the coins worth Rs 90.

Answer: b.
Explanation:
Let the children be P, Q, R and S and Mr John be F
candies with P : Q : R = 3 : 7 : 11
Let the number of candies be 3X, 7X and 11X
Total candies with three eldest children = 21X
candies with F and S = 3 × 21X = 63X
Total candies = (21X + 63X) = 84X
candy with F : (P + Q + R + S) = 3 : 4
Total 7 units of candy = 84 X
1 unit = 12X
Candy with F = 3 × 12X = 36X
Candy with S = (63X – 36X) = 27X
27 X = 81 → X=3
Total number of candies = 84X = 84 × 3 = 252

Answer: d.
Explanation:
Let, marks obtained by Vin Dieselin theory and total marks of theory be ‘7x’ and ‘10x’ respectively

So, total marks (theory + practical) = 10x + 2x = 12x
Marks obtained by Vin (theory + practical) = 7x + 2x = 9x
Required Ratio = \frac{9x}{12x}=\frac{3}{4}

Answer: c.
Explanation:
Let the strength of A = 16X and that of = 8X

Section A ratio of Boys: girls
= 3 : 1 → boys = 12X and girls = 4X

Section B ratio of boys: girls .
= 3 : 5 → boys = 3X and girls = 5X
A + B – boys = 15X and girls = 9X

Maximum number of students in a row = HCF (15X, 9X) = 3X → 3X = 96 → X=32
Difference between boys of section A and B = 12X – 3X = 9X → 9 × 32 = 288