# 100+ Alligation or Mixture Questions and Answers Pdf - 1

Question: 1

A vessel is filled with liquid, 3 parts of which are water and 4 parts syrup. How much of the mixture must be drawn off and replaces with water so that the mixture may be half water and half syrup?

(A) \$\${1}/{3}\$\$

(B) \$\${1}/{4}\$\$

(C) \$\${1}/{5}\$\$

(D) \$\${1}/{7}\$\$

Ans: C

Suppose the vessel initially contains 8 litres of liquid.

Let x litres of the liquid be replaced with water.

Quantity of water in new mixture = \$\$(3 - {3x} / {8} + x)\$\$ litres.

Quantity of syrup in new mixture = \$\$(5 - {5x} / {8})\$\$ litres.

∴ \$\$(3 - {3x}/{8} + x)\$\$ = \$\$(5 - {5x}/{8})\$\$

⇒ 5x + 24 = 40 – 5x

⇒ 10x = 16

⇒ x = \$\${8}/{5}\$\$.

So, part of the mixture replaced = \$\$({8}/{5} × {1}/{8})\$\$ = \$\${1}/{5}\$\$.

Question: 2

A milkman mixed some water with milk to gain 25% by selling the mixture at the cost price. The ratio of water and milk is respectively.

(A) 1 : 4

(B) 1 : 5

(C) 4 : 5

(D) 5 : 4

Ans: A

C.P. of 1 litres of milk = Rs. 100

∴ Mixture sold for Rs. 125 = \$\${125}/{100}\$\$ = \$\${5}/{4}\$\$ litres

∴ Quantity of mixture = \$\${5}/{4}\$\$ litres

∴ Quantity of water = 1 litres

∴ Quantity of water = \$\${5}/{4} - 1\$\$ = \$\${1}/{4}\$\$ litres

∴ Required ratio = \$\${1}/{4} : 1\$\$ = 1 : 4.

Question: 3

A dishonest trader professes to sell his article at cost price but he uses 900 g weight instead of 1 kg. What is the percentage gain of the trader in this trade?

(A) \$\$11{1}/{9}\$\$%

(B) \$\${110}/{9}\$\$%

(C) \$\${100}/{9}\$\$%

(D) \$\${80}/{9}\$\$%

Ans: A

%gain of the trader = \$\${100}/{900} × 100\$\$

= \$\${100}/{9}\$\$% = \$\$11{1}/{9}%\$\$.

Question: 4

8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of water is 16:65. How much wine did the cask hold

(A) 18 litres

(B) 24 litres

(C) 32 litres

(D) 42 litres

Ans: B

Let the quantity of the wine in the cask originally be x litres.

Then, the quantity of the wine left in case after 4 operations = \$\$[x(1 - {8}/{x})^4]\$\$ litres.

∴ \$\$x(1- {8/x})4 / {x}\$\$ = \$\${16}/{81}\$\$

⇒ \$\$(1 – {8}/{x})\$\$4 = \$\$({2}/{3})\$\$2

⇒ \$\$({x – 8} / {x})\$\$ = \$\${2}/{3}\$\$

⇒ 3x - 24 = 2x

⇒ x = 24.

Question: 5

A container contains 40litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?

(A) 26.34 litres

(B) 27.36 litres

(C) 28 litres

(D) 29.16 litres

Ans: D

Amount of milk left after 3 operations

=\$\$[40(1 – {4}/{40})]\$\$3litres

=\$\$ (40 × {9/10} × {9/10} × {9/10})\$\$ = 29.16 litres.

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