Important for :

1

D

Sum of temperature on 1^{st}, 2^{nd}, 3^{rd}, and 4^{th} days = (52 × 4) = 208----------(I)

Sum of temperature on 2^{nd}, 3^{rd}, 4^{th}, and 5^{th} days = (55 × 4) = 220---------(II)

Subtracting equation (I) from (II)

Temperature on 5^{th} June 2017 – temperature on 1^{st} June 2017 = 12 degrees.

Let the temperature on 1^{st} and 5^{th} June 2017 be 5x and 7x degrees respectively.

Then, 7x – 5x = 12

2x = 12

X = 6

Since temperature on the 5^{th} June 2017 in Delhi = 7x = 7 × 6 = 42 degrees.

Hence, option (D) is correct.

2

B

Using Allegation method,Average salary of | Average salary of | |||

Labourers (Rs. 75) | Supervisors (Rs. 600) | |||

\ | / | |||

Mean (Rs. 100) | ||||

/ | \ | |||

500 | 25 |

Required ratio is 500 : 25 or 20 : 1

No. of Labourers | = | 20 |

No. of Supervisors | 1 |

⇒ | 840 | = | 20 |

No. of supervisors | 1 |

Hence, no. of supervisors = 42

Therefore, option (B) is correct.

3

D

Let the total number of matches Dravid has played, till the start of world cup be X

So, total runs scored by Dravid till the beginning of the world cup be 35 × X

After the world cup, the new average will be | 35 × X + 355 |

X + 6 |

According to question,

35 × X + 355 | = 37.5 |

X + 6 |

⇒ 37.5 × X + 225 = 35 × X + 355

⇒ 2.5 × x = 130

⇒ X = 52

Hence, the total number of matches played by Dravid till date = X + 6 = 52 + 6 = 58

Therefore, option (D) is correct.

Therefore, option (D) is correct.

4

D

Canberra 8:00 am (let Mon) 2:00 pm (Tue) |
Distance = 7800km |
Pretoria 3:00 am (Tue) 8:00 am (Tue) |

Let the plane speed with respect to the wind be u km/hr

The wind speed is 25 km/hr

Also, let Pretoria be t hours ahead of Canberra

Now,

7800 |
= 19 – t ………. (i) |

u – 25 |

7800 |
= 6 + t ………….(ii) |

u + 25 |

Adding (i) and (ii), we get

7800 | + | 7800 | = 25 | |

u – 25 | u + 25 |

⇒ 7800(2u) = 25(u^{2} – 625)

⇒ 312(2u) = (u^{2} – 625)

⇒ u^{2} – 624u – 625 = 0

⇒ (u – 625) (u + 1) = 0

As, u > 0 then

u – 625 = 0 ⇒ u = 625

Substituting in (i), we get

7800 | = 19 – t ⇒ 13 = 19 – t ⇒ t = 6 |

600 |

Thus, is the time difference between Canberra and Pretoria is 6 hours.

Hence, option (D) is correct.

5

D

Let the weight of four friends A, B, C, D and E are a, b, c, d and e kgs respectively

From the question,

3a = b + c + d ----------- (i)

3b = a + c + d ----------(ii)

Subtracting equation (i) and (ii)

3a – 3b = b – a

b = a ---- (iii)

Adding equation (i) and (ii)

3a + 3b = a + b + 2c + 2d

a + b = c + d ---- (iv)

The average weight of A and C is equal to the average weight of C and D

It means, a + c = c + d

a = d

Put a = d in equation (iv)

We get b = c

It means, b = a = c = d

The average weight of B and D is 60 kgs

b + d = 60 × 2 = 120 kgs

So, b = d = 60 kgs = a = c

E is 30 kg heavier than C

e = (60 + 30) kgs

By solving, e = 90 kg

The sum of the weight of A, B, C, D and E = 60 + 60 + 60 + 60 + 90 = 330 kgs

The average = | 330 | = 66 kgs |

5 |

Hence, option D is correct.