Examveda

In a 729 litres mixture of milk and water, the ratio of milk to water is 7 : 2. to get a new mixture containing milk and water in the ratio 7 : 3, the amount of water to be added is:

A. 81 litres

B. 71 litres

C. 56 litres

D. 50 litres

Solution(By Examveda Team)

Quantity of milk in 729 litre of mixture
= 7 × $$\frac{{729}}{{9}}$$ = 567 litre
Quantity of water
= 729 - 567
= 162 litres
Let x litre of water be added to become ratio 7 : 3

Or, $$\frac{{7}}{{3}}$$ = $$\frac{{567}}{{162 + {\text{x}}}}$$
Or, 162 × 7 + 7x = 567 × 3
Or, 7x = 1701 - 1134 = 567
Or, x = $$\frac{{567}}{{7}}$$ = 81 litres water is to be added.

={729/(7+2)}×(3-2)
=(729/9)×1
=81

2. 7x+2x=9x
9x=729

2X=162
162+x/576=3/7
X=81

3. Milk : Water = 7 : 2
Milk : Water = 7 : 3
Now, (7 + 2) i.e 9 units ≡ 729 L
So, (3 - 2) i.e 1 units ≡ 81L (Ans.)

4. 1st term 7:2 = 9
2nd term = 7:3 where only water difference 3-2 = 1
now
9x = 729
or x = 81

5. Qty of Milk = 729*(7/9) = 567 L
Qty of Water = 162 L
Had to mix water (3 - 2 ) = 1 unit
ATQ, 7 ≡ 567
Then, 1 ≡ (567/7) = 81 L

6. lets 1 part to be mixed so as per allegation method take only same qty ratio
so i am taking water ration
2/9 : 1(assume 1 part water to be mixed)
3/10
(1-3/10=7/10) (3/10-2/9= 7/90)
so ratio is 7/10 : 7/90 = 9:1
now water to be added 1/9 x729 = 81 litres

7. How about just taking 1 part of the ratio 7:2 which amounts to 81 litres. (1/9 of 729 = 81). Since 7:2 changes to 7:3. only 1 part of water is increasing.

8. Let 729=X
a:b = 7:2
c:d =7:3

ans 81

9. if the raito was 7:2,then we r calculating as 7*729/2.Then as the same way for 7:3 we can calculate
Quantity of milk in 729 litre of mixture
7*729/10 =510.3
Quantity of water needed
729-510.3=218.7
As by 7:2 ratio the water was 162 litres.
Now by subtracting 162 from 218.7 ans comes
218.7-162=56.7
As the most appropriate ans is 56

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