Important for :

1

C

Total score of 40 students = (40 × 76) = 3040

Total score of top 3 scorers = 3040 – (37 × 75) = 265

To minimize the score of the top scorer, we assume the other two top scorers score the maximum they can = 85 marks each.

So, the top scorer scored = 265 – 170 = 95 marks.

Hence, option C is correct.

2

D

Total weight of Amar, Visera and Daman = 70 × 3 = 210 kg

Again, Amar + Visera + Daman + Vishal = 66 × 4 = 264 kg ......... (i)

∴ Weight of Vishal = 264 – 210 = 54 kg

∴ Weight of Tahir = 54 + 6 = 60 kg

Now, as per the question

Visera + Daman + Vishal + Tahir = 75 × 4 = 300 kg. ................... (ii)

Subtracting (i) from (ii), we get

Tahir – Amar = 60 – Amar = 300 – 264 = 36

Therefore, weight of Amar = 60 – 36 = 24 kg

Hence, option D is correct.

3

D

The smallest cube divisible by 16 is 64.

Lets assume the average salary before the manager's salary is included is x

After addition of Manager's salary the average increases by 100

We can write down the above information in form of an equation as:

64x + 9500 = 65 × (x + 100)

Solving for x, we get x = 3000

The final average is 3000 + 100 = Rs. 3100

Hence, option D is correct.

4

D

Number of applicants that have been counted extra = 1857 − 1747 = 110

Hence, decrease in average = 110/222 = 0.495

∴ Correct average = 1560 − 0.495 = 1559.505 = 1559.51

Hence, option D is correct.

5

A

Rohit’s average score in the first 3 exams = 120Let the scores in the 5 exams be denoted by M1, M2, M3, M4, and M5

M1 + M2 + M3 = 120 × 3 = 360

Average of last 4 match = 140

⇒ | M2 + M3 + M4 + M5 | = 140 |

4 |

⇒ M2 + M3 + M4 + M5 = 560

Average of all the exams

⇒ | M1 + M2 + M3 + M4 + M5 | = 122 |

5 |

From solving above equation, we get M1 + M4 + M5 = 300

Required average runs = | 300 | = 100 |

3 |

Hence correct option (A) is correct.