Mixture & Alligation Solved Questions For IBPS, SBI and Other Bank Exams - Page 1

You will find 8 problems in 2 pages..

Mixture & Alligation Solved Questions (Page 1 of 2)

Dear Reader,

Below you will find important formulas of mixture & alligation problems. Following that you will find solved questions. These questions will help you in preparation for all bank exams including IBPS, SBI, RRB and other banks.

Basic Formulas:

Alligation:

It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of desired price.

Mean Price:

The cost of a unit quantity of the mixture is called the mean price.

Rule of Alligation:

If two ingredients are mixed, then

(Quantity of Cheaper/Quantity of Dearer) = (CP of Dearer – Mean Price)/(Mean Price – CP of Cheaper).

Suppose a container contains x units of liquid, from which y units are replaced by water. If this process takes place n times, then the quantity of the pure liquid = {x (1-(y/x))n}.

Question 1

A fruit seller mix two qualities of fruits costing Rs.75 and Rs.50 per kg respectively. In what ratio he should mix the mixture to get worth Rs.65 per kg ?
a) 1:3 b) 1:2 c) 2:3 d) 3:2

Answer: c)2:3

Solution:

Cost of 1 kg fruit of 1st quality = Rs.75
cost of 1 kg fruit of 2nd quality = Rs.50
i.e.,
C.P. of dearer (d) = Rs.75
C.P. of cheaper(c) = Rs.50
And Mean price (m) = Rs.65

d - m = 75 - 65 = 10
And m - d = 65 - 50 = 15

Required rate = Quantity of cheaper / Quantity of dearer = d - m / m - c = 10/15 = 2/3

Hence the answer is 2:3


Question 2

Find the ratio in which milk at Rs.12.50 per litre be mixed with milk at Rs.10.70 per litre to make a mixture worth Rs.11.50 per litre.
a)5:4 b)2:3 c)1:5 d)2:5

Answer: a)5:4

Solution:

Cost of 1 litre milk of 1st type = Rs.12.50 = 1250 paise and
Cost of 1 litre milk of 2nd type = Rs.10.70 = 1070 paise

i.e,
C.P. of dearer (d) = 1250 p
C.P. of cheaper(c) = 1070 p
And Mean price (m) = 1150 p

Then, d - m = 1250 p - 1150 p = 100 p
and m - c = 1150 p - 1070 p = 80 p
Required rate = Quantity of cheaper / Quantity of dearer = d - m / m - c = 100/80 = 5:4
Hence the answer 5:4


Question 3

In what ratio should a seller mix oil of Rs.62 per litre with oil at Rs.72 per litre, so that the mixture would worth Rs.64.50 per litre?
a) 2:3 b) 1:3 c) 4:3 d) 3:1

Answer: d)3:1

Solution:

Cost of 1 litre oil of 1st type = Rs.62
Cost of 1 litre oil of 2nd type = Rs.72
Mean price (m) = Rs.64.50
i.e.,
C.P. of dearer (d) = Rs.72
C.P. of cheaper(c) = Rs.62
Then d - m = 7.50
and, m - c = 2.50
Now, the required rate = Quantity of cheaper / Quantity of dearer = d - m / m - c = 7.5 : 2.5 = 3:1
Hence the answer is 3:1


Question 4
The milk seller adds water to milk worth Rs.11 per litre. In what ratio should he mix to get a mixture of cost of Rs.9.90 per litre ?

a)1:3 b) 1:9 c) 2:5 d) 1:11

Answer: b)1:9

Solution:
Cost of 1 litre water = Rs.0
Cost of 1 litre milk = Rs.11
Mean price(m) = Rs.9.90

C.p of dearer = Rs.11
C.P of cheaper = Rs.0
Then d - m = 11 - 9.90 = Rs.1.10
And, m - c = Rs.9.90

Now, the required rate = Quantity of cheaper / Quantity of dearer = d - m / m - c = 1.1 : 9.9
= 11:99 = 1:9
Hence the answer is 1:9


Mixture & Alligation Solved Questions (Page 1 of 2)

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