# Quant Test for SBI PO 2018 Prelim Exam Set – 34

Hello Aspirants

State Bank of India (SBI) is going to conduct examination for its recruitment for the post of Probationary Officers (SBI PO 2018) for a total of 2000 vacancies.

The examination will be held in three phases i.e. Preliminary Examination, Main Examination and Group Exercise & Interview. The Preliminary Exam is scheduled on 1st, 7th & 8th of July 2018. Details of the exam are as under: Practice the questions so as to familiarize yourself with the pattern of questions to be asked in the exam.

Directions(1-5): What will come in place of question mark (?) in the following question?

1. 75% of 1240 = 35% of 1560 + ?
202
285
300
384
400
Option D
75% of 1240 = 35% of 1560 + ?
=> 930 = 546 + ?
=> ? = 384

2. (15% of 2200)% of 240 = (? × 4 ) + (25% of 2450)
66.231
44.875
52.311
40.125
33.231
Option B
(15% of 2200)% of 240 = (? × 4 ) + (25% of 2450)
=> 330% of 240 = 4 × ? + 612.5
=> 792 – 612.5 = 4 × ?
=> ? = 179.5/4
=> ? = 44.875

3. 15% of 2520 + 42% of 2500 – ? = (42 × 22 ) ÷ 3.5 + 40% of 3420
– 204
– 85
– 100
– 75
– 66
Option A
15% of 2520 + 42% of 2500 – ? = (42 × 22 ) ÷ 3.5 + 40% of 3420
=> 378 + 1050 – ? = 924/3.5 + 1368
=>1428 – ? = 264 + 1368
=> 1428 – 1632 = ?
=> ? = – 204

4. (61)^2 + (51)^2 – 60% of 2/3 of 40% of 13200 = ?
4210
3010
4450
3500
2588
Option A
(61)^2 + (51)^2 – 60% of 2/3 of 40% of 13200 = ?
=> 3721 + 2601 – 2112 = ?
=> ? = 4210

5. 42% of 3000 + (? × 14 ) – (21 ÷ 3.5 ) = (13)^2 + (21 × 2.5)
– 60.06
– 65.65
– 33.12
– 52.02
-73.75
Option E
42% of 3000 + (? × 14 ) – (21 ÷ 3.5 ) = (13)^2 + (21 × 2.5)
=> 1260 + (14 × ? ) – 6 = 169 + 52.5
=> 14 × ? = 221.5 + 6 – 1260
=> ? = – 1032.5/14
=> ? = – 73.75

6. The amount of interest for a certain sum at 5•5% annual rate of interest for one year is Rs. 676•5. How much the interest will be less on the same sum at 5% annual rate of interest for one year ?
Rs.48.5
Rs.68.2
Rs.55.5
Rs.61.5
Rs.35.3
Option D
Let the certain sum be Rs. P
=> 676•5 = P *5•5*1/100
and second interest = (P *5 *1)/100 
Required sum = 676•5 – Second interest
= (P *5•5*1)/100 – (P*5*1)/100
= P (5•5 – 5)/100 = P*0•5/100
= (676•5 *100 *0•5)/(5•5*1*100)
= 61•5
Hence, Required sum = Rs. 61•5

7. Village A has a population of 6800, which is decreasing at the rate of 120 per year. Village B has a population of 4200, which is increasing at the rate of 80 per year. In how many years will the population of the two villages be equal ?
11 yr.
10 yr.
12 yr.
13 yr.
15 yr.
Option D
After 13 yr population of village A
= 6800 – 120 × 13
= 5240
After 13 yr population of village B
= 4200 + 80 × 13
= 5240

8. Taps A, B and C are connected to a water tank and the rate of flow of water is 42 litres/hr., 56 litres/ hr. and 48 litres/hr. respectively. Taps A and B fill the tank while tap C empties the tank. If all the three taps are opened simultaneously, the tank gets completely filled up in 16 hours. What is the capacity of the tank ?
500 l
800 l
600 l
900 l
700 l
Option B
Net amount of water filled in the tank in 1 hour
when all three taps are opened simultaneously = 42 + 56 – 48 litres = 50 litres.
The tank gets completely filled in 16 hours. Capacity of the tank = 16 × 50 = 800 litres.

9. Anu owes Biresh Rs 1120 payable 2 year hence, Biresh owes Anu Rs 1081.50 payable 6 months. If they decide to settle their accounts forthwith by payment of ready money and the rate of interest be 6% per annum, they who should pay and how much ?
Rs. 20, Anu
Rs. 80 Anu
Rs.50 , Brijesh
Rs. 60 , Brijesh
Rs. 40 Brijesh
Option C
1120 – P = (P*6*2)/100
=>P = Rs. 1000
Present worth of money for Biresh 1081.50 – P
= P*6*1/2*100
=> 108150 – 100 P = 3P
P = Rs 1050
Biresh should pay Rs 50

10. A bag contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the bag, if at least one black ball is to be included in the draw?
48
55
50
64
43
Option D
Number of ways to select 3 black balls = 3C3
Number of ways to select 2 black balls and 1 non-black ball = 3C2 × 6C1
Number of ways to select 1 black ball and 2 non-black balls = 3C1 × 6C2
Total number of ways = 3C3 + 3C2 × 6C1 + 3C1 × 6C2 = 3C3 + 3C1 × 6C1 + 3C1 × 6C2
=1+3×6+3×6×52×1
=1+18+45
=64

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