What was the day of the week on 28^{th} May, 2006?

**A. ** Thursday

**B. ** Friday

**C. ** Saturday

**D. ** Sunday

**Answer : ****Option D**

**Explaination / Solution: **

28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)

Odd days in 1600 years = 0

Odd days in 400 years = 0

5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) 6 odd days

Jan. Feb. March April May

(31 + 28 + 31 + 30 + 28 ) = 148 days

148 days = (21 weeks + 1 day) 1 odd day.

Total number of odd days = (0 + 0 + 6 + 1) = 7 0 odd day.

Given day is Sunday.

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In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?

**A. ** 6.25

**B. ** 6.5

**C. ** 6.75

**D. ** 7

**Answer : ****Option A**

**Explaination / Solution: **

Required run rate = | 282 - (3.2 x 10) | = | 250 | = 6.25 | ||

40 | 40 |

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A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500?

**A. ** Rs. 4991

**B. ** Rs. 5991

**C. ** Rs. 6001

**D. ** Rs. 6991

**Answer : ****Option A**

**Explaination / Solution: **

Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.

Required sale = Rs. [ (6500 x 6) - 34009 ]

= Rs. (39000 - 34009)

= Rs. 4991.

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In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of ground is:

**A. ** 75 cu. m

**B. ** 750 cu. m

**C. ** 7500 cu. m

**D. ** 75000 cu. m

**Answer : ****Option B**

**Explaination / Solution: **

1 hectare = 10,000 m^{2}

So, Area = (1.5 x 10000) m^{2} = 15000 m^{2}.

Depth = | 5 | m | = | 1 | m. |

100 | 20 |

Volume = (Area x Depth) = | 15000 x | 1 | m^{3} | = 750 m^{3}. | |

20 |

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A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:

**A. ** 720

**B. ** 900

**C. ** 1200

**D. ** 1800

**Answer : ****Option C**

**Explaination / Solution: **

2(15 + 12) x *h* = 2(15 x 12)

h = | 180 | m = | 20 | m. |

27 | 3 |

Volume = | 15 x 12 x | 20 | m^{3} | = 1200 m^{3}. | |

3 |

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Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

**A. ** 4

**B. ** 7

**C. ** 9

**D. ** 13

**Answer : ****Option A**

**Explaination / Solution: **

Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)

= H.C.F. of 48, 92 and 140 = 4.

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The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

**A. ** 276

**B. ** 299

**C. ** 322

**D. ** 345

**Answer : ****Option C**

**Explaination / Solution: **

Clearly, the numbers are (23 x 13) and (23 x 14).

Larger number = (23 x 14) = 322.

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(17)^{3.5} x (17)^{?} = 17^{8}

**A. ** 2.29

**B. ** 2.75

**C. ** 4.25

**D. ** 4.5

**Answer : ****Option D**

**Explaination / Solution: **

Let (17)^{3.5} x (17)^{x} = 17^{8}.

Then, (17)^{3.5 + x} = 17^{8}.

3.5 + *x* = 8

*x* = (8 - 3.5)

*x* = 4.5

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Given that 10^{0.48} = *x*, 10^{0.70} = *y* and *x*^{z} = *y*^{2}, then the value of *z* is close to:

**A. ** 1.45

**B. ** 1.88

**C. ** 2.9

**D. ** 3.7

**Answer : ****Option C**

**Explaination / Solution: **

*x*^{z} = *y*^{2} 10^{(0.48z)} = 10^{(2 x 0.70)} = 10^{1.40}

0.48*z* = 1.40

z = | 140 | = | 35 | = 2.9 (approx.) |

48 | 12 |

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3 pumps, working 8 hours a day, can empty a tank in 2 days. How many hours a day must 4 pumps work to empty the tank in 1 day?

**A. ** 9

**B. ** 10

**C. ** 11

**D. ** 12

**Answer : ****Option D**

**Explaination / Solution: **

Let the required number of working hours per day be *x*.

*More pumps, Less working hours per day (Indirect Proportion)*

*Less days, More working hours per day (Indirect Proportion)*

Pumps | 4 | : | 3 | :: 8 : x | |

Days | 1 | : | 2 |

4 x 1 x *x* = 3 x 2 x 8

x = | (3 x 2 x 8) |

(4) |

*x* = 12.

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