#### IBPS SO CWE Exam Quantitative Aptitude Study Material

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b) 38 2/5

c) 32 2/5

d) 42 6/7

= Rs. (420 + 210) = Rs. 630

S.P of 9 kg of mixture = 9×100 = Rs. 900

Therefore Profit percent = [(900-630)/630]×100

= 2700/63

= 300/7

= 42 6/7 %

b) 48

c) 52

d) 56

Milk = (72×70)/100 = 50.4 litres

Water = 72 - 50.4 = 21.6 litres On adding 18 litress of water,

Required percentage of milk = (50.4/90)×100

= 504/9

= 56%

b) 62 litres

c) 41 litres

d) 84 litres

b) 35 litres

c) 30 litres

d) 32 litres

b) 13 : 22

c) 14 : 25

d) 12 : 23

4kg of second alloy could be mixed together.

Therefore, In 3kg of mixture,

Tin = 1kg

Iron = 2kg.

In 4kg of mixture,

Tin=(2/5)×4 = 8/5 kg=1.6kg

Iron=(3/5)×4=12/5=2.4kg

Therefore,Required ratio

=(1+1.6) : (2+2.4) =2.6 : 4.4

13 : 22

b) 58.23 L

c) 85.23 L

d) 58.32 L

b) 40

c) 42

d) 45

(x+5)/3x=1/2

=>2x+10=3x

=>x=10

Therefore, quantity of new mixture

=4x+5=45litres.

b) 60

c) 62.5

d) 64

water = 60 cc

Glycerine = 180 cc

Let x cc of water be mixed.

Therefore (60+x)/180 = 2/3.

==> 180+3x =360

⇒3x= 360 -180 = 180

Hence x=180/3 = 60 cc

b) 71 ml

c) 52 ml

d) 81 ml

Quantity of water =(2/9)*729 =162ml

Let `x` be the quantity that should be added to make the ratio 7:3.

According to the question, 567/(162+x) =7/3

=>1701 =1134+7x

=> 7x =1701 -1134

=>x=81ml

b) Rs. 525

c) Rs. 485

d) Rs. 450