RRB NTPC Prelims Model Paper 4 with solutions

Question : 86 of 100

Marks: +1, -0

In a mixture of wine and soda, 75% soda is present. If 10 ml of wine is added, percentage of wine in the mixture becomes 50. What was the original quantity of mixture?

2 ml
20 ml
10 ml
25 ml

Validate

Solution:

(2) Let the quantity of mixture be xml.
Amount of wine = 0.25x ml
Amount of water = 0.75x ml
New per cent of wine is,

{0.25x +10}/{x+10 }×100

={25x +1000}/{x +10}=50

⇒25x =500 ⇒x =20ml

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