# Quant Test for IBPS PO Prelims Exam set – 4

Directions(1-5): What approximate value will come in place of question mark ‘?’ in the following questions.

1. 42*5.12/2.99 + 49.89 = ?*1.99
54
84
60
50
78
Option C
42*5.12/2.99 + 49.89 = ?*1.99
=>42*5/3 + 50 = ?*2
=>70+50 = ?*2
=>? = 60

2. 18.18*3.99 *(135.003 / 26.87) – (15.001*3.89) = ?
356
378
313
300
389
Option D
18.18*3.99 *(135.003 / 26.87) – (15.001*3.89) = ?
=>18*4 *(135/27) – (15*4) = ?
=>72*5 – 60 = ?
=>? = 360 -60
=>? = 300

3. 30.01*(1024)^1/2 + 196 = ?^2 – 287
30
38
44
50
61
Option B
30.01*(1024)^1/2 + 196 = ?^2 – 287
=>30*32 + 196 = ?^2 – 287
=>960 + 196 = ?^2 – 287
=>?^2 = 1444
=>? = 38

4. 35.2% of 1599 = ? – (449.92/6.12)*2
759
700
647
652
770
Option B
35.2% of 1599 = ? – (449.92/6.12)*2
=>35% of 1600 = ? – (450/6)*2
=>560 = ? – 75*2
=>560 = ? – 150
=>? = 700

5. 16.12 * 15.94 + 654.92 – 344.84 = ?*5.93
88
82
90
78
94
Option C
16.12 * 15.94 + 654.92 – 344.84 = ?*5.93
=>16*16 + 655 – 345 = ?*6
=>256 + 655 – 345 = ?*6
=>? = 90

6. A boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes. How long will it take to go 5 km in stationary water?
3.05 hrs.
2.17 hrs.
1.25 hrs.
2.14 hrs.
3.05 hrs.
Option C
Downstream =(1x 60)/10 = 6 kmph
Rate upstream = 2 km/hr.
Speed in still water =(6 + 2) /2= 4 kmph
Time = 5/4 = 1.25 hrs.

7. In a right angled triangle smallest side of is 13 cm less than the side of a square of perimeter 72 cm. Second largest side of the right angled triangle is 2 cm less than the length of the rectangle of area 112 cm² and breadth 8 cm. What is the largest side of the right angled triangle?
17 cm
10 cm
13 cm
12 cm
19 cm
Option C
Side of square = 72/4 = 18 cm
Smallest side of the right angled triangle = 18 – 13 = 5 cm
Length of rectangle = 112/8 = 14 cm
Second side of the right angled triangle = 14 – 2 = 12 cm
Hypotenuse of the right angled triangle = √(25 + 144) = 13 cm

8. 3 years ago, the average age of A, B, and C was 27 years. Also 5 years ago, the average age of B and C was 20 years. What is the present age of A?
20
40
50
80
90
Option B
Sum of ages of A+B+C 3 yrs ago = 27*3 = 81
After 3 yrs, i.e. at present their total = 81 + 3*3 = 90
Similarly sum of present age of B&C = 20*2 +5*2 = 50
Present age of A = 90-50 = 40

9. If the difference between the simple interest and compound interests on some principal amount at 20% for 3 years is Rs. 48, then the principal amount is:
Rs. 327
Rs. 313
Rs. 320
Rs. 300
Rs. 375
Option E
Sum = Difference x (100)^3
r^2(300 + r)
= 48 x (100)^3
20^2 (300 + 20)
= Rs. 375

10. In a team of 6 boys and 8 girls, 5 students have to be selected. In how many ways it can be done so that at least 2 boys are included.
1748
1947
1379
1526
1500
Option D
No. of ways = 6c2*5c3 + 6c3*5c2 + 6c4*5c1 + 6c5 = 1526