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IBPS PO Mains 30 Nov 2024 Shift 2 Paper

Question Numbers: 153-155

Directions: The following questions are based on the information given below. Analyze the data carefully to answer the questions.

There are three mi×tures A, B, and C, each consisting of only milk and water. The ratio of the total quantity of mi×tures P, Q, and R is given by p:q:r

The table below shows the partial composition of these mi×tures.

Mi×ture	Milk (Litres)	Water (Litres)
A	24	m
B	n - 4	20
C	3m	j

Notes:

(i) m and n are positive integers such that m +

n = 58

(ii) pn - qm = 16p

(iii) p and q are single-digit positive integers, both are less than 5, and q > p.

Question : 155 of 155

Marks: +1, -0

If r is a single digit integer greater than 5, which of the following could be a possible value for j (the quantity of water in mi×ture C)?

m+10
m+20
m+30
m+40
m+50

Validate

Solution:

Concept:

The problem uses the ratio of total quantities of mixtures to find a possible value of

j

(water in mixture

C

) and then compares it with expressions in terms of

m

Explanation:

From the decoded values, we know

p=3

q=4

m=18

, and

n=40

Total quantity of mixture

A

24 + m = 24 + 18 = 42

litres.

Mixture

C

has

54

litres of milk, so its total quantity is

54 + j

The ratio of total quantities of

A:C

p:r = 3:r

Thus,

\frac{42}{54+j} = \frac{3}{r}

Cross-multiplying gives

42r = 3(54+j)

, so

14r = 54 + j

Therefore,

j = 14r - 54

Since

r

is a single digit integer greater than

5

, possible values are

6, 7, 8, 9

For

r=6

j = 14(6)-54 = 30

For

r=7

j = 14(7)-54 = 44

For

r=8

j = 14(8)-54 = 58

For

r=9

j = 14(9)-54 = 72

Now, using

m=18

, evaluate the options:

m+10 = 28

m+20 = 38

m+30 = 48

m+40 = 58

m+50 = 68

Only

m+40 = 58

matches a possible value of

j

Answer:

Hence, the possible value of

j

m+40

, which is option D.

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