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Ankita J. Sharma

· started a discussion

· 1 Months ago

value of petrol (Question:968482.0)

part of petrol in mixture X=415, how it happened ?

Question:

DIRECTIONS: Read the data given in line graphs carefully and answer the questions given below. 

There are 5 mixtures A, B, C, D and E of Petrol and Water and another 5 mixtures P, Q, R, S and T of Diesel and Water.

First line graph shows the amount of Petrol as per cent of amount of water in mixtures A, B. C, D and E. While second line graph shows the amount of Diesel as per cent of amount of water in mixtures P, Q, R, S and T.

  

Anil mixes mixtures A and B in the ratio 4: 3 and form mixture X. next he mixes mixtures P and Q in the ratio 10: 11 and form mixture Y. Finally, he mixes mixture X and Y in the ratio 5: 6 along with 15 litres of Diesel Now, the ratio of Petrol, Diesel and Water in the final mixture becomes 7: 8: 5, then what is the difference between quantity of mixture A and mixture P?

Options:
A)

0

B)

15 litres

C)

30 litres

D)

45 litres

Solution:

 

Ans: (a)

Mixtures

Ratio (Petrol: Water)

Mixtures

Ratio (Diesel: Water)

A

400: 100 = 4: 1

P

150: 100 = 3: 2

B

400: 100 = 4: 1

Q

340: 100 = 17: 5

C

300: 100 = 3: 1

R

250: 100 = 5: 2

D

200: 100 = 2: 1

S

300: 100 = 3: 1

E

360:100 =18:5

T

150:100 = 3:2

Since the ratio of Petrol to Water in mixture A and B are same and equal to 4: 1. Hence, after mixing A and B in any ratio, ratio of petrol to water will remains same

So, part of petrol in mixture X = 415

Mixture P and Q are mixed in the ratio 10: 11-

Part of Diesel in P 3/5

Part of Diesel in Q = 17/22

Let part of Diesel in Y = Y

By the rule of allegation:

  

[(17/22) – Y] : [Y - (3/5)] = 10 : 11

8.5- 11 Y= 10 Y - 6

21Y = 14.5

Y = 29/42

Let quantity of mixtures X and Y are 5 m and 6 m respectively.

After mixing mixtures X and Y along with 16 litres of diesel amount of petrol = 5 m * (4/5) = 4 m

Amount of Diesel = 6m * (29/42) + 15 = (29m/7) + 15

Amount of Water = [5m * (115)] + [6m * (13/42)] = 20m/7

Ratio of Petrol, Diesel and Water 4m : [(29 m/7) + 15]: [20 mm = 7: 8: 5

4m/[(29 m/7) + 15] = 7/8

32 m = 29 m + 105

m = 35

Quantity of X = 5 m = 175 litres

Quantity of Y = 6 m 210 litres

Quantity of A = 175 * (4/7) = 100 litres

Quantity of A = 210 * (10/21) = 100 litres

Required difference 100 - 100 = 0

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