# Quant Test for NIACL AO Prelims Exam Set – 7

Directions(1-5): What will be the approximate value come in place of question mark “?” in the following question.

1. 29.98% of 1780.21 + ? + 19.72% of 699.23 = 64.83% of 2359.69
785
800
755
820
860
Option E
29.98% of 1780.21 + ? + 19.72% of 699.23 = 64.83% of 2359.69
=> 30% of 1780 + ? + 20% of 700 = 65% of 2360
=> 534 + ? + 140 = 1534
=>? = 860

2. (20.99)^2 â€“ (13.01)^2 + (291.12 + 76.92)/? = 55.92% of 499.88
40
35
46
39
30
Option C
(20.99)^2 â€“ (13.01)^2 + (291.12 + 76.92)/? = 55.92% of 499.88
=> 21^2 â€“ 13^2 + (291+77)/? = 56% of 500
=> 441 â€“ 169 + 368/? = 280
=> ? = 46

3. (2210)^1/2 + 979.90 / 13.927 = ?^2 â€“ 5.099% of 1579.99
14
12
10
15
16
Option A
(2210)^1/2 + 979.90 / 13.927 = ?^2 â€“ 5.099% of 1579.99
=> (2210)^1/2 + 979/14 = ?^2 â€“ 5% of 1580
=> 47 + 70 = ?^2 â€“ 79
=> ? = 14

4. 11.92% of 1199.98 – ?* 13.9*20.99 / (196)^1/2 = 43.21 â€“ 297.78
22
19
20
14
16
Option B
11.92% of 1199.98 – ?* 13.9*20.99 / (196)^1/2 = 43.21 â€“ 297.78
=> 12% of 1200 – ?*14*221/(196)^1/2 = 43 â€“ 298
=> 144 – ?*14*21/(196)^1/2 = – 255
=> 21 * ? = 399
=> ? = 19

5. (2.197)^4 / (1.69)^2 *(1.3)^2 = (1.3)^(? – 2)
11
8
7
10
12
Option E
(2.197)^4 / (1.69)^2 *(1.3)^2 = (1.3)^(? – 2)
=> (1.3)^12/(1.3)^4 *(1.3)^2 = (1.3)^(?-2)
=> (1.3)^10 = (1.3)^(?-2)
=> ? = 12

6. Directions(6-10): The table shows the marks scored by five students in five different subjects.

7. Marks obtained by D in all the six subjects together are approximately how much percent more than the marks scored by A in all the six subjects together?
7%
3%
4%
6%
5%
Option D
Marks obtained by A in all subjects = 485
Marks obtained by D in all subjects = 513
Difference = 513 â€“ 485 = 28
Required% = 28/485*100 = 5.77% == 6%

8. Marks obtained by B,C and D in Science together are approximately how much percent less than marks obtained by A,B and E in G.K. together.
22.11%
31.52%
18.41%
32.23%
30.04%
Option B
Marks obtained by B,C and D in Science together = 120+105+90 = 315
Marks obtained by A,B and D in G.K. together = 180+120+160 = 460
Required% = (460-315)/460*100 = 31.52%

9. Ram scored 20% more marks than A in all the subjects together and Shyam scored 25% less than C in all the subjects together. Find the difference between the total marks of Ram and Shyam.
185
200
220
231
193
Option D
Marks obtained by A = 485
Marks obtained by Ram = 1.2*485 = 582
Marks obtained by C = 468
Marks obtained by Shyam = 0.75*468 = 351
Difference = 582 â€“ 351 = 231

10. How many number of students scored more than 65% of maximum marks.
6
3
4
5
2
Option C
Maximum marks = 100+80+150+120+75+200 = 725
65% of maximum marks = 0.65*725 = 471.75
Marks obtained by A = 485
Marks obtained by B = 480
Marks obtained by C = 468
Marks obtained by D = 513
Marks obtained by E = 545
Only 4 students = A,B,D and E.

11. Average percentage of marks per student scored by the five students in which subject are the highest?
Science
English
Hindi
G.K.
Maths
Option E
Average percentage per student for Maths
= (70+72+73+89+91)/5*100*100 = 79%