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CAT 2020 Slot 1 — QA Question 13

MixturesEasy

A solution, of volume 40 litres, has dye and water in the proportion 2 : 3. Water is added to the solution to change this proportion to 2 : 5. If one-fourths of this diluted solution is taken out, how many litres of dye must be added to the remaining solution to bring the proportion back to 2 : 3?

Answer & solution

Answer: 8

Solution

Easy

Track the actual litres of dye and water through each step. Dye is unaffected by adding water, and removing one-fourth of the mixture scales both components by 34\tfrac34. Use the final ratio to solve for the dye added.

1

Initial amounts. 4040 L in ratio 2:32:3.

dye=25×40=16 Lwater=35×40=24 L\begin{aligned} &\text{dye}=\tfrac{2}{5}\times 40 = 16\ \text{L}\\ &\text{water}=\tfrac{3}{5}\times 40 = 24\ \text{L} \end{aligned}
2

Add xx litres of water to reach 2:52:5. Dye stays at 1616.

1624+x=25(dye : water)⇒ 80=2(24+x)⇒ x=16(litres of water added)\begin{aligned} &\frac{16}{24+x}=\frac{2}{5} \quad\text{(dye : water)}\\ &\Rightarrow\ 80 = 2(24+x)\\ &\Rightarrow\ x = 16 \quad\text{(litres of water added)} \end{aligned}

So now dye =16=16 L and water =40=40 L.

3

Remove one-fourth. Three-quarters of each component remains.

dye=34×16=12 Lwater=34×40=30 L\begin{aligned} &\text{dye}=\tfrac34\times 16 = 12\ \text{L}\\ &\text{water}=\tfrac34\times 40 = 30\ \text{L} \end{aligned}
4

Add yy litres of dye to restore 2:32:3. Water stays at 3030.

12+y30=23(dye : water)⇒ 3(12+y)=60⇒ y=8\begin{aligned} &\frac{12+y}{30}=\frac{2}{3} \quad\text{(dye : water)}\\ &\Rightarrow\ 3(12+y) = 60\\ &\Rightarrow\ y = 8 \end{aligned}
y=8 litresy = 8\ \text{litres}
CAT 2020 Slot 1 QA Q13: A solution, of volume 40 litres, has dye and water in the proportion 2 : 3. Water is added to the solution to — Solution | TheCATExam