# LIC HFL Associate/ Assistant Numerical Ability Questions – (Day-04)

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Directions (Q. 1 â€“ 5): What approximate value should come in place of question mark (?) in the following questions?

1) 26 % of 349.75 â€“ 32 % of 599.27 =? â€“ 157.94 â€“ (27.21)2

a) 820

b) 568

c) 786

d) 634

e) 912

2) 5 2/3 of 299.78 + 3 1/8 of 429.67 =? % of 749.85

a) 384

b) 360

c) 448

d) 412

e) 490

3) (553.672 + 959.325 â€“ 432.72) Ã· 14.85 =?

a) 72

b) 86

c) 65

d) 58

e) 95

4) [((149 Ã· 3) Ã· 9.97)2 + (64.13 Ã— 9.05)1/2] = (?)2

a) 12

b) 18

c) 7

d) 23

e) 30

5) ? Ã· 3.9 + 72.03 Ã— 15 = 4824.12 + 1598.23

a) 25682

b) 21368

c) 18924

d) 16780

e) 27950

Directions (Q. 6 â€“ 10) Study the following information carefully and answer the given questions.

The following table shows the total number of tyres (In lakhs) manufactured by 5 different companies in a certain year.

6) Total number of tyres rejected by the company P is approximately what percentage more/less than the total number of tyres rejected by the company T?

a) 7 % less

b) 15 % more

c) 15 % less

d) 24 % more

e) 7 % more

7) Find the average number of tyres manufactured by all the given companies together?

a) 32.4 lakhs

b) 28.5 lakhs

c) 30.8 lakhs

d) 34.6 lakhs

e) None of these

8) Find the difference between the total number of tyres sold by the company Q to that of company S?

a) 792600

b) 756800

c) 833200

d) 809270

e) None of these

9) Find the ratio between the total number of tyres rejected by the company Q to that of R?

a) 72: 55

b) 53: 115

c) 84: 137

d) 65: 96

e) None of these

10) Total number of tyres sold by the company P is approximately what percentage of total number of tyres sold by the company R?

a) 60 %

b) 75 %

c) 50 %

d) 85 %

e) 105 %

Directions (1-5):

26 % of 350 â€“ 32 % of 600 = x â€“ 158 â€“ 272

(26/100)*350 â€“ (32/100)*600 = x â€“ 158 â€“ 729

91 â€“ 192 + 158 + 729 = x

X = 786

6 of 300 + 3 of 430 = x% of 750

(300*6) + (3*430) = (x/100)*750

1800 + 1290 = 15x/2

(3090*2)/15 = x

X = 412

Â (554 + 959 â€“ 433)/15 = x

X = 1080/15 = 72

Â [((150 Ã· 3) Ã· 10)2 + (64 Ã— 9)1/2] = x2

(150/30)2 + (8*3) = x2

25 + 24 = x2

X2 = 49

X = 7

x Ã· 4 + 72 Ã— 15 = 4824 + 1598

(x/4) + 1080 = 6422

(x/4) = 5342

X = 5342*4 = 21368

Directions (6-10):

Total number of tyres rejected by the company P

= > 3200000*(2/100) = 64000

Total number of tyres rejected by the company T

= > 4000000*(1.5/100) = 60000

Required % = [(64000 â€“ 60000)/60000]*100 = 7 % more

The average number of tyres manufactured by all the given companies together

= > (32 + 26 + 48 + 16 + 40)/5

= > 162/5 = 32.4 lakhs

The total number of tyres sold by the company Q

= >2600000*(95/100)*(84/100) = 2074800

The total number of tyres sold by the company S

= > 1600000*(97/100)*(80/100) = 1241600

Required difference = 2074800 â€“ 1241600 = 833200

The total number of tyres rejected by the company Q

= > 26*(5/100) = 1.3 lakhs

The total number of tyres rejected by the company R

= > 48*(4/100) = 1.92 lakhs

Required ratio = 1.3: 1.92 = 130: 192 = 65: 96

Total number of tyres sold by the company P

= > 3200000*(98/100)*(78/100) = 2446080

Total number of tyres sold by the company R

= > 4800000*(96/100)*(72/100) = 3317760

Required % = (2446080/3317760)*100 = 73.726 = 75 %