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Quantitative Aptitude

ALLIGATION MCQs

Total Questions : 48 | Page 5 of 5 pages
Question 41.

In what ratio must water be mixed with milk to gain 16 \(\frac{2}{3}\) % on selling the mixture at cost price?

  1.    1 : 6
  2.    6 : 1
  3.    2 : 3
  4.    4 : 3
 Discuss Question
Answer: Option A. -> 1 : 6

Let C.P. of 1 litre milk be Re. 1.


S.P. of 1 litre of mixture = Re.1, Gain  =  \(\frac{50}{3}\)  %


 So, C.P. of 1 litre of mixture =  \(\left(100\times\frac{3}{350}\times1\right)=\frac{6}{7}\)


By the rule of alligation, we have:


C.P. of 1 litre of water  = 0


C.P. of 1 litre of milk =  Rs.1


Main Price = \(\frac{6}{7}\)


So, Ratio of water and milk =  \(\frac{1}{7}:\frac{6}{7}=1:6.\)

Question 42.

Find the ratio in which rice at Rs. 7.20 a kg be mixed with rice at Rs. 5.70 a kg to produce a mixture worth Rs. 6.30 a kg.

  1.    1 : 3
  2.    2 : 3
  3.    3 : 4
  4.    4 : 5
 Discuss Question
Answer: Option B. -> 2 : 3

By the rule of alligation:


Cost of 1 kg of 1st kind = 720 p


Cost of 1 kg of 2nd kind = 570 p


Main Price = 630 p


So, Required ratio = 60 : 90 = 2 : 3.

Question 43.

In what ratio must a grocer mix two varieties of tea worth Rs. 60 a kg and Rs. 65 a kg so that by selling the mixture at Rs. 68.20 a kg he may gain 10%?

  1.    3 : 2
  2.    3 : 4
  3.    3 : 5
  4.    4 : 5
 Discuss Question
Answer: Option A. -> 3 : 2

S.P. of 1 kg of the mixture = Rs. 68.20, Gain = 10%.


C.P. of 1 kg of the mixture = Rs. \(\left(\frac{100}{110}\times68.20\right)=Rs. 62.\)


By the rule of alligation, we have:


Cost of 1 kg tea of 1st kind = Rs. 60


Cost of 1 kg tea of 2nd kind. = Rs. 65


Main Price = Rs, 62


So, Required ratio = 3 : 2.

Question 44.

The cost of Type 1 rice is Rs. 15 per kg and Type 2 rice is Rs. 20 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2 : 3, then the price per kg of the mixed variety of rice is:

  1.    Rs. 18
  2.    Rs. 18.50
  3.    Rs. 19
  4.    Rs. 19.50
 Discuss Question
Answer: Option A. -> Rs. 18

Let the price of the mixed variety be Rs. x per kg.


By rule of alligation, we have


Cost of 1 kg of Type 1 rice = Rs.15


Cost of 1 kg of Type 2 rice = Rs. 20


Main Price = Rs. x


So, \(\left(\frac{20-x}{x-15}\right)=\frac{2}{3}\) 


60 - 3x = 2x - 30


 5x = 90


 x = 18.

Question 45.

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 originally?

  1.    18 litres
  2.    24 litres
  3.    32 litres
  4.    42 litres
 Discuss Question
Answer: Option B. -> 24 litres

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


Then, quantity of wine left in cask after 4 operations = \(\left[x\left(1-\frac{8}{x}\right)^{4}\right]liters\)


So, \(\left(\frac{x\left(1-\frac{8}{x}\right)^{4}}{x}\right)=\frac{16}{81}\)


  \({\left(1-\frac{8}{x}\right)^{4}}=\left(\frac{2}{3}\right)^{4}\)


 \({\left(\frac{x-8}{x}\right)}=\frac{2}{3}\)  


3x - 24 = 2x


x = 24

Question 46.

A merchant has 1000 kg of sugar, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The quantity sold at 18% profit is:

  1.    400 kg
  2.    560 kg
  3.    600 kg
  4.    640 kg
 Discuss Question
Answer: Option C. -> 600 kg

By the rule of alligation, we have:


Profit on 1st part   = 8%


Profit on 2nd part = 18%


Main Profit = 14%


Ration of 1st and 2nd parts = 4 : 6 = 2 : 3


So, Quantity of 2nd kind = \({\left(\frac{3}{5}\times1000\right)}kg=600 kg.\)

Question 47. The acid and water in two vessels A and B are in the ratio 4 : 3 and 2 : 3. In what ratio should the liquid in both the vessels be mixed to obtain a new mixture in vessel C containing half acid and half water ?
  1.    7 : 5
  2.    5 : 7
  3.    7 : 3
  4.    5 : 3
 Discuss Question
Answer: Option A. -> 7 : 5
Question 48. The acid and water in two vessels A and B are in the ratio 4 : 3 and 2 : 3. In what ratio should the liquid in both the vessels be mixed to obtain a new mixture in vessel C containing half acid and half water ?
  1.    7 : 5
  2.    5 : 7
  3.    7 : 3
  4.    5 : 3
 Discuss Question
Answer: Option A. -> 7 : 5
According to the question,
$$\eqalign{
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Acid}}\,\,\,\,\,\,{\text{Water}} \cr
& {\text{Vessel A}}\,\,{\text{4}}\,\,\,:\,\,\,\,{\text{3}} \cr
& {\text{Vessel B}}\,\,\,2\,\,\,\,{\text{:}}\,\,\,\,\,{\text{3}} \cr} $$
Now using alligation,

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