Question

In a laboratory, two bottles contain mixture of acid and water in the ratio 2 : 5 in the first bottle and 7 : 3 in the second. The ratio in which the contents of these two bottles be mixed such that the new mixture has acid and water in the ratio 2 : 3 is

(A) 4 : 15
(B)9 : 8
(C) 21 : 8
(D)1 : 2

Answer 1

Cant seem to get it

Answer 2

in 1st bottle,
every unit contains
2/7 acid and 5/7 water

in 2nd bottle,
every unit contains
7/10 acid and 3/10 water

lets take p units from 1st bottle and q units from 2nd bottle and mix together,
thus we have on mixing
2p/7 + 7q/10 acid
and
5p/7 + 3q/10 water

we need to find p/q where we are given that
(2p/7 + 7q/10) / (5p/7 + 3q/10) = 2/3

let p/q = x
so we are left with on simplifying,

(20 x + 49)/(50x + 21) = 2/3

60x + 147 = 100x + 42
or 40x = 105
or x = 21/8 <--------ans..

Answer 3

Thank you :) @shubhamsrg

Answer 4

glad to have helped :)

Answer 5

How did you simplify down to find p/q?

Answer 6

took q common from both numerator and denominator ,
or you can also see that as substituting p = xq

Answer 7

ofcourse

Answer 8

you asked something else? @Skaematik