Practice | Hashtable

content Computer Science Arithmetic Aptitude Data Interpretation Logical Reasoning Verbal Reasoning Non-Verbal Reasoning

About About Privacy Policy

Hashtable

Home
Practice
PYQ
Mock Test
Register Free

Home Practice PYQ Mock Test Register Free

Home > Practice > Arithmetic Aptitude > Alligation or Mixture > Miscellaneous

11. Weights of two friends Ram and Shyam are in the ratio 4 : 5. If Ram's weight is increased by 10% and total weight of Ram and Shyam become 82.8 kg, with an increases of 15%. By what percent did the weight of Shyam has to be increased?

A. 19%

B. 10%

C. 21%

D. 16%

Discuss

Solution:

Now, given ratio of Ram and Shayam's weight = 4 : 5
Hence,

\frac{x - 15}{15 - 10} = \frac{4}{5}

Or, x = 19%

12. A lump of two metals weighing 18 g is worth Rs. 87 but if their weight is interchanged, it would be worth Rs. 78.60. If the price of one metal be Rs. 6.70 per gram, find the weight of the other metal in the mixture.

A. 8g

B. 12g

C. 15g

D. 18g

Discuss

Solution:

Cost of (18g of 1^st metal + 18g of 2^nd metal) = Rs. 165.60
Cost of (1g of 1^st metal + 1g metal of 2^nd metal) = Rs. 9.20
Hence cost of 1g of 2^nd metal,
= 9.20 - 6.70
= Rs. 2.5
Mean price = Rs.

\frac{87}{18}

Now,

\frac{quantity of 1^{st} metal}{quantity of 2^{nd} metal}

=

\frac{14}{6} : \frac{56}{30} = 5 : 4

Quantity of 2^nd metal =

\frac{18 \times 4}{9}

= 8g

13. A vessel contains milk and water in the ratio 3 : 2. The volume of the contents is increased by 50% by adding water to it. From this resultant solution 30 L is withdrawn and then replaced with water. The resultant ratio of milk water in the final solution is 3 : 7. Find the original volume of the solution.

A. 80 L

B. 65 L

C. 75 L

D. 82 L

Discuss

Solution:

Let the original volume be x.
then, quantity of milk and water,
=

\frac{3 x}{5}

and

\frac{2 x}{5}

respectively.
After adding water to it, the volume becomes 150%, the quantity of milk and water,
=

\frac{3 x}{5}

and

\frac{9 x}{10}

\frac{\frac{3 x}{5} - 12}{\frac{9 x}{10} + 12} = \frac{3}{7}

14(3x - 60) = 3(9x + 120)
Or, x = 80 L

14. 3 L water is taken out from vessel full of water and substituted by pure milk. This process is repeated two more times. Finally, the ratio of milk and water in the solution becomes 1728 : 27. Find the volume of the original solution.

A. 3 L

B. 5 L

C. 4 L

D. 9 L

Discuss

Solution:

We use formula for it where
n = number of time process of out going repeated.
y = Out going milk in process.
x = Volume of original solution.
F = Final milk left in mixture.

F = x {(1 - \frac{y}{x})}^{n}

Final quantity of water,

F = x {(1 - \frac{3}{x})}^{3}

\frac{F}{x} = {(\frac{x - 3}{x})}^{3}

\frac{27}{1728} = {(\frac{x - 3}{x})}^{3}

{(\frac{3}{12})}^{3} = {(\frac{x - 3}{x})}^{3}

\frac{3}{12} = \frac{x - 3}{x}

$x = 4 L$

15. One quantity of wheat at Rs 9.30 per Kg is mixed with another quality at a certain rate in the ratio 8 : 7. If the mixture so formed be worth Rs 10 per Kg, what is the rate per Kg of the second quality of wheat?

A. Rs. 12.47

B. Rs. 10.80

C. Rs. 15.17

D. Rs. 47.66

Discuss

Solution:

Let the rate of second quality be Rs. x per Kg.
C.P of 1 Kg wheat of 1st kind = 930 p
C.P of 1 Kg wheat of 2nd kind = 100x p
Mean price = 1000 p
By rule of alligation we have required ratio 8 : 7
930 x / (Mean Price) (1000) / (x-10) : 0.7
So we get required ratio, (x-10) : 0.7 :: 8 : 7
⇒ x = 10.80 per Kg

16. In a zoo, there are Rabbits and Pigeons. If heads are counted, there are 200 and if legs are counted, there are 580. How many pigeons are there?

A. 90

B. 100

C. 110

D. 120

Discuss

Solution:

Heads Count = 200
Legs count = 580
Average Legs count for per head =

\frac{580}{200}

\frac{29}{10}

Rabbits : Pigeons =

\frac{9}{10}

\frac{11}{10}

= 9 : 11
Number of Pigeons =

200 \times \frac{11}{20}

= 110

17. A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

A. $\frac{1}{3}$

B. $\frac{1}{4}$

C. $\frac{1}{5}$

D. $\frac{1}{7}$

Discuss

Solution:

Suppose the vessel initially contains 8 litres of liquid.
Let x litres of this liquid be replaced with water.
Quantity of water in new mixture =

(3 - \frac{3 x}{8} + x)

litres
Quantity of syrup in new mixture =

(5 - \frac{5 x}{8})

litres

\begin{aligned} ∴ 3 - \frac{3 x}{8} + x = 5 - \frac{5 x}{8} \\ \Rightarrow 5 x + 24 = 40 - 5 x \\ \Rightarrow 10 x = 16 \\ \Rightarrow x = \frac{8}{5} \end{aligned}

So, part of the mixture replaced

\begin{aligned} = \frac{8}{5} \times \frac{1}{8} \\ = \frac{1}{5} \end{aligned}

18. Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, the price of the third variety per kg will be:

A. Rs. 169.50

B. Rs. 170

C. Rs. 175.50

D. Rs. 180

Discuss

Solution:

Since first and second varieties are mixed in equal proportions.
So, their average price

\begin{aligned} = Rs . (\frac{126 + 135}{2}) \\ = Rs . 130.50 \end{aligned}

So, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find x.
By the rule of alligation, we have:
Alligation mcq solution image

\begin{aligned} ∴ \frac{x - 153}{22.50} = 1 \\ \Rightarrow x - 153 = 22.50 \\ \Rightarrow x = 175.50 \end{aligned}

19. A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?

A. 10

B. 20

C. 21

D. 25

Discuss

Solution:

\begin{aligned} Suppose the can initially contains \\ 7 x and 5 x of mixtures A and B respectively . \\ Quantity of A in mixture left \\ = (7 x - \frac{7}{12} \times 9) litres \\ = (7 x - \frac{21}{4}) litres \\ Quantity of B in mixture left \\ = (5 x - \frac{5}{12} \times 9) litres \\ = (5 x - \frac{15}{4}) litres \\ ∴ \frac{(7 x - \frac{21}{4})}{(5 x - \frac{15}{4}) + 9} = \frac{7}{9} \\ \Rightarrow \frac{28 x - 21}{20 x + 21} = \frac{7}{9} \\ \Rightarrow 252 x - 189 = 140 x + 147 \\ \Rightarrow 112 x = 336 \\ \Rightarrow x = 3 \end{aligned}

So, the can contained 21 litres of A

20. A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?

A. 4 litres, 8 litres

B. 6 litres, 6 litres

C. 5 litres, 7 litres

D. 7 litres, 5 litres

Discuss

Solution:

Let the cost of 1 litre milk be Rs. 1
Milk in 1 litre mixture in 1^st can =

\frac{3}{4}

litre, C.P. of 1 litre mixture in 1^st can Rs.

\frac{3}{4}

Milk in 1 litre mixture in 2^nd can =

\frac{1}{2}

litre, C.P. of 1 litre mixture in 2^nd can Rs.

\frac{1}{2}

Milk in 1 litre of final mixture =

\frac{5}{8}

litre, Mean price = Rs.

\frac{5}{8}

By the rule of alligation, we have:
Alligation mcq solution image

∴ Ratio of two mixtures =

\frac{1}{8}

\frac{1}{8}

= 1 : 1
So, quantity of mixture taken from each can =

(\frac{1}{2} \times 12)

= 6 litres

1 2 3 4 5 6 7 8 9 10

Sign up for our newsletter

Are you passionate about opportunities in the realm of Computer Science within the government sector? Don't miss out on crucial updates, job openings, and career prospects tailored specifically straight to your inbox.

Email address

By subscribing, you agree with our Terms of Service and Privacy Policy.

Resources

Practice PYQ Mock Test

Company

About us Contact us Privacy policy Terms of service

Hashtable