Question

HI THERE

Acid and water are mixed in the ratio 4:3 to make a
liquid. On adding 10 more litres of acid, the ratio changed
to 3:1. How many litres of acid and water does the liquid
contain now?
pls send me the ans plssssssssssssssss

Answer 1

Answer:

Step-by-step explanation:

let a = acid and w = water

a/w=4/3 =

3a=4w -1

10+A/w=3/1=

3w=10+a -2

multiply 1 by 3 and 2 by 4

we get

9a=12w

40a+4a=12w

equating them

40a+4a=9a

a=8

putting a in 1 we get

w=6

acid=8L and water=6L

Answer 2

1st approach:-

let initial quantity of Acid=4A,

initial quantity of water=3A,

after adding 10 ltr acid,

(4A+10)/3A = 3/1,

4A+10=9A,

then

10=9A-4A,

10=5A,

then
A=2,

then
initial acid=4×2=8 ltr,

initial water=3×2=6 ltr,

final acid=8 ltr+10 ltr=18 ltr,

and final water will be 6 ltr coz there is no change in the quantity of water.



short approach:- (Orally approach to solve ratios based questions),

Initial ratio of Acid and water = 4 : 3,

final ratio of Acid and water = 3 : 1,

but it is clearly mentioned in the question that the quantity of water has no changed then just equate the ratios of water in initial and final conditions so just multiply the final ratios by 3 so that the ratio of water will be same mean 3.

hence

new final ratio of Acid and water = (3 : 1) × 3,

=9 : 3,

so it's clear that the gap between the initial ratio and new final ratio of Acid is 5 unit but according to the question actuall difference is 10 ltr,

then

5 units = 10 ltr,

so

1 unit = 2 ltr,

therefore
final quantity of Acid = 9 unit,

= 18 ltr,

final quantity of water = 3 unit,

= 6 ltr