# Quant Test for IBPS Clerk 2018 Main Exam Set – 28

Directions(1-5): The table given shows the number of newspaper sold daily by different newspaper firms (A,B,C,D and E) in Delhi and Lucknow. It shows the percentage of English newspapers sold daily in Delhi and the percentage of English newspapers sold daily in Delhi and the percentage of Hindi newspapers sold daily in lucknow by the respective newspaper firms.
Each newspaper firm sells both English and Hindi newspaper.
Total number of newspapers sold in a city = Number of (English+Hindi) newspapers sold in a city.

1. Find the ratio of the number of English newspapers sold by firm A in Delhi to the number of English newspaper sold by firm E in Delhi.
303:403
301:399
291:400
301:408
307:403
Option D
Required ratio = 10836:14688 = 301:408

2. Find the difference between the number of English newspapers sold by firm C in Delhi and the number of English newspapers sold by firm C in Lucknow.
3222
4100
3333
3225
3200
Option D
Required difference = 21285 – 18060 = 3225

3. Find the percentage of the total number of Hindi newspapers sold by firm D in both cities together with respect to the total number of newspapers sold in both cities.
62%
40%
58%
49%
50%
Option C
Required% = [28210+24820]/[54250+36500]*100 = 58.4% == 58%

4. Find the difference between the numbers of Hindi newspapers sold by firm A in Lucknow and the numbers of Hindi newspapers sold by firm B daily in Lucknow.
5550
5000
5780
5460
5985
Option D
Required difference = 17460 – 12000 = 5460

5. Find the average number of English newspaper sold each firm in Lucknow.
11250
10952
11250
10000
21500
Option B
Required average = (6750+6790+18060+11680+11480)/5 = 10952

6. A shopkeeper marks up an item of cost price Rs.1800 by (x+2)% and sells it at a discount of (x-3)% and in the whole transaction, he makes neither profit nor loss. Find the value of x.
10
20
15
12
23
Option E
MP = 1800*(100+x+2)% = 18(102+x)
SP = MP – Discount = 18(102+x) – 18(102+x)*(x-3)% Now, 18*(102+x){1-(x-3)% } – 1800 = 0
=> (102+x){1-(x-3)%} = 0
=>x^2 – x -506 = 0
=> x^2 – 23x + 22x – 506 = 0
=> x = 23

7. The ratio of the number of males and females in a company is 9:5 resp. If the number of males is increased by (x+10)% and the number of females is increased by (x+5)%, then the ratio of males and females in the company becomes 15:8 resp. Find the value of x.
11
10
15
12
13
Option C
[9*(100+x+10)%]/[5*(100+x+5)%] = 15/8
=> [3*(110+x)]/[5(105+x)] = 5/8
=> x = 15

8. A 240 m long train can cross a platform of 319 m in 43 seconds. If ‘x’ new number of bogies of length 69 m each are attached to the train, then the train will be able to cross a man standing on the platform in 45 seconds. Find the value of x.
2
6
5
4
3
Option C
[240+319]/speed = 43
=>speed = 13 m/s Also, [240+x*69]/13 = 45
=> x = 5

9. The average weight of a class of (x+6) students is (x+1) kg. If 7 new students of average weight (x+7) kg join the class, the average weight of the class is increased by 0.7 kg. Find the value of x.
30
47
36
44
46
Option B
Total weight of the class = (x+6)*(x+1) kg
Weight of new students = 7*(x+7) kg
Now, [(x+6)(x+1)+7(x+7)]/[(x+6)+7] = (x+1)+0.7
=> x = 47

10. In a 1000 m race , A beats B by 100 m and A beats C by 200 m or 50 seconds. Find the speed of B.
4.7m/s
6.2 m/s
3.2 m/s
4.5 m/s
5.5 m/s
Option D
1000/A = (1000-100)/B => A:B = 10:9 And 1000/A = (1000-200)/C
=> A:C = 5:4
So, A:B:C = 10:9:8 1000/A = 1000/C -50
=> 1000/(10/8)C = 1000/C – 50
=> C = 200/50 = 4 m/s
Speed of B = 9/8*4 = 4.5 m/s

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