# Aptitude Test Paper

Topics: Orders of magnitude Pages: 22 (4685 words) Published: February 20, 2011
Aptitude Test 1

1. A can finish a work in 24 days, B in 9 days and C in 12 days. B and C start the work but are forced to leave after 3 days. The remaining work was done by A in: | A.| 5 days| |
| | B.| 6 days| |
|
| C.| 10 days| |
| | D.| 10| 1| days|
| 2| |
| |
|
Explanation:
(B + C)'s 1 day's work =| | 1| +| 1| | =| 7| .|
| | 9| | 12| | | 36| |

Work done by B and C in 3 days =| | 7| x 3| | =| 7| .| | | 36| | | | 12| |

Remaining work =| | 1 -| 7| | =| 5| .|
| | | 12| | | 12| |

Now,| 1| work is done by A in 1 day.|
| 24| |

So,| 5| work is done by A in| | 24 x| 5| | = 10 days.| | 12| | | | 12| | |

1. . The square root of 64009 is:
| A.| 253| |
| | B.| 347| |
|
| C.| 363| |
| | D.| 803| |
|
Explanation:
2|64009( 253
|4
|----------
45|240
|225
|----------
503| 1509
| 1509
|----------
| X
|----------
* 64009 = 253.
* 3.
How many marks did Tarun secure in English? |
I. | The average mark obtained by Tarun in four subjects including English is 60. |  II. | The total marks obtained by him in English and Mathematics together are 170. |  III. | The total marks obtained by him in Mathematics and Science together are 180. | *

| A.| I and II only| |
| | B.| II and III only| |
|
| C.| I and III only| |
| | D.| All I, II and III| |
|
| E.| None of these| |
| | | |
* Explanation:
* I gives, total marks in 4 subjects = (60 x 4) = 240.
* II gives, E + M = 170
* III gives, M + S = 180.
* Thus, none of (A), (B), (C), (D) is true.
The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:| | A.| 40%| |
| | B.| 42%| |
|
| C.| 44%| |
| | D.| 46%| |
|
Answer: Option CExplanation:Let original length = x metres and original breadth = y metres.Original area = (xy) m2. New length =| | 120| x| m| =| | 6| x| m.| | | 100| | | | | 5| | |

New breadth =| | 120| y| m| =| | 6| y| m.|
| | 100| | | | | 5| | |
New Area =| | 6| x x| 6| y| m2| =| | 36| xy| m2.| | | 5| | 5| | | | | 25| | |
The difference between the original area = xy and new-area 36/25 xy is= (36/25)xy - xy= xy(36/25 - 1) = xy(11/25) or (11/25)xy Increase % =| | 11| xy x| 1| x 100| %| = 44%.| | | 25| | xy| | | |

5.| Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.? | | | A.| 8 kmph| |

| | B.| 11 kmph| |
|
| C.| 12 kmph| |
| | D.| 14 kmph| |
|
Answer: Option CExplanation:Let the distance travelled by x km. Then,| x| -| x| = 2| | 10| | 15| |
3x - 2x = 60x = 60 km. Time taken to travel 60 km at 10 km/hr =| | 60| hrs| = 6 hrs.| | | 10| | |
So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M. Required speed =| | 60| kmph.| = 12 kmph.| | | 5| | |

Direction (for Q.No. 6):Find out the wrong number in the given sequence of numbers.| 6.| 8, 13, 21, 32, 47, 63, 83 |
| | A.| 47| |
| | B.| 63| |
|
| C.| 32| |
| | D.| 83| |
|
Answer: Option AExplanation:Go on adding 5, 8, 11, 14, 17, 20.So, the number 47 is wrong...