Three different liquids which have 10% water, 20% water and x% of water are…
2022
Three different liquids which have 10% water, 20% water and x% of water are mixed in the ratio of their quantity 2 : 3 : 4 respectively. If 12% of water is present in final mixture, calculate value of x.
- A.
9%
- B.
20%
- C.
7%
- D.
15%
Attempted by 2 students.
Show answer & explanation
Correct answer: C
Concept
When several solutions are blended, the amount of a component (here, water) is conserved. The water in the mixture equals the sum of the water contributed by each part. Expressing each contribution as concentration x quantity and dividing by the total quantity gives the final concentration. This is the weighted average (alligation) principle: final % = (sum of conc x parts) / (total parts).
Application
Take the quantities in the given ratio 2 : 3 : 4, so the parts are 2, 3 and 4 and the total is 2 + 3 + 4 = 9 parts. Setting the conserved water equal to 12% of the whole:
Water from the first liquid = 10% of 2 parts = 0.10 x 2 = 0.20 parts.
Water from the second liquid = 20% of 3 parts = 0.20 x 3 = 0.60 parts.
Water from the third liquid = x% of 4 parts = (x/100) x 4 = 0.04x parts.
Total water = 12% of 9 parts = 0.12 x 9 = 1.08 parts.
Form the equation: 0.20 + 0.60 + 0.04x = 1.08.
Combine the known terms: 0.80 + 0.04x = 1.08.
Isolate the unknown: 0.04x = 1.08 - 0.80 = 0.28.
Solve: x = 0.28 / 0.04 = 7.
So x = 7%.
Cross-check
Substitute x = 7 back: water = 10x2 + 20x3 + 7x4 = 20 + 60 + 28 = 108 percent-parts over 9 parts, i.e. 108 / 9 = 12%, which matches the given final concentration. The third liquid (7%) is leaner than the 12% target while the first (10%) is also below it; only the 20% liquid is richer, and because it carries fewer parts than the third, the blend settles below the simple midpoint, consistent with a low value for the unknown.