| 1. | Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case. | Answer: Option A Explanation: Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43) = H.C.F. of 48, 92 and 140 = 4. Learn more problems on : Problems on H.C.F and L.C.M Discuss about this problem : Discuss in Forum |
| 2. | Which of the following fraction is the largest ? | Answer: Option A Explanation: L.C.M. of 8, 16, 40 and 80 = 80. | 7 | = | 70 | ; | 13 | = | 65 | ; | 31 | = | 62 | | 8 | 80 | 16 | 80 | 40 | 80 |
| Since, | 70 | > | 65 | > | 63 | > | 62 | , so | 7 | > | 13 | > | 63 | > | 31 | | 80 | 80 | 80 | 80 | 8 | 16 | 80 | 40 |
Learn more problems on : Problems on H.C.F and L.C.M Discuss about this problem : Discuss in Forum |
| 4. | The least perfect square, which is divisible by each of 21, 36 and 66 is: | Answer: Option A Explanation: L.C.M. of 21, 36, 66 = 2772. Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11 To make it a perfect square, it must be multiplied by 7 x 11. So, required number = 22 x 32 x 72 x 112 = 213444 Learn more problems on : Square Root and Cube Root Discuss about this problem : Discuss in Forum |
| 5. | | If x = | 3 + 1 | and y = | 3 - 1 | , then the value of (x2 + y2) is: | | 3 - 1 | 3 + 1 |
| Answer: Option C Explanation: | x = | (3 + 1) | x | (3 + 1) | = | (3 + 1)2 | = | 3 + 1 - 23 | = 2 + 3. | | (3 - 1) | (3 + 1) | (3 - 1) | 2 |
| y = | (3 - 1) | x | (3 - 1) | = | (3 - 1)2 | = | 3 + 1 - 23 | = 2 - 3. | | (3 + 1) | (3 - 1) | (3 - 1) | 2 |
 x2 + y2 = (2 + 3)2 + (2 - 3)2
= 2(4 + 3) = 14 Learn more problems on : Square Root and Cube Root Discuss about this problem : Discuss in Forum |
Direction (for Q.No. 6):
Each of the questions given below consists of a question followed by three statements. You have to study the question and the statements and decide which of the statement(s) is/are necessary to answer the question. | | 6. | What is Arun's present age? | I. | Five years ago, Arun's age was double that of his son's age at that time. | II. | Present ages of Arun and his son are in the ratio of 11 : 6 respectively. | III. | Five years hence, the respective ratio of Arun's age and his son's age will become 12 : 7. |
| Answer: Option D Explanation: II. Let the present ages of Arun and his son be 11x and 6x years respectively. I. 5 years ago, Arun's age = 2 x His son's age. | III. 5 years hence, | Arun's Age | = | 12 | | Son's age | 7 |
Clearly, any two of the above will give Arun's present age. Correct answer is (D). Learn more problems on : Problems on Ages Discuss about this problem : Discuss in Forum |
| 7. | If A = x% of y and B = y% of x, then which of the following is true? | Answer: Option E Explanation: | x% of y = |  | x | x y |  | = | | y | x x | | = y% of x | | 100 | 100 |
A = B. Learn more problems on : Percentage Discuss about this problem : Discuss in Forum |
| 8. | If 20% of a = b, then b% of 20 is the same as: | Answer: Option A Explanation: 20% of a = b  | 20 | a = b. | | 100 |
| b% of 20 = | | b | x 20 | | = | | 20 | a x | 1 | x 20 | | = | 4 | a = 4% of a. | | 100 | 100 | 100 | 100 |
Learn more problems on : Percentage Discuss about this problem : Discuss in Forum |
| 9. | 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? | Answer: Option D Explanation: 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. Learn more problems on : Chain Rule Discuss about this problem : Discuss in Forum |
| 10. | One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in: | Answer: Option C Explanation: Let the slower pipe alone fill the tank in x minutes. | Then, faster pipe will fill it in | x | minutes. | | 3 |
| 1 | + | 3 | = | 1 | | x | x | 36 |
| 4 | = | 1 | | x | 36 |
x = 144 min. Learn more problems on : Pipes and Cistern Discuss about this problem : Discuss in Forum |
| 11. | A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr) is: | Answer: Option B Explanation: Let the speed of the stream be x km/hr. Then, Speed downstream = (15 + x) km/hr, Speed upstream = (15 - x) km/hr. | 30 | + | 30 | = 4 | 1 | | (15 + x) | (15 - x) | 2 |
| 900 | = | 9 | | 225 - x2 | 2 |
9x2 = 225 x2 = 25 x = 5 km/hr. Learn more problems on : Boats and Streams Discuss about this problem : Discuss in Forum |
| 12. | A man took loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was: | Answer: Option C Explanation: | Principal = Rs. | | 100 x 5400 | | = Rs. 15000. | | 12 x 3 |
Learn more problems on : Simple Interest Discuss about this problem : Discuss in Forum |
| 13. | A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges? | Answer: Option C Explanation: Let the side of the square(ABCD) be x metres. Then, AB + BC = 2x metres. AC = 2x = (1.41x) m. Saving on 2x metres = (0.59x) m. | Saving % = | | 0.59x | x 100 | % | = 30% (approx.) | | 2x |
Learn more problems on : Area Discuss about this problem : Discuss in Forum |
| 14. | By investing in 16 % stock at 64, one earns Rs. 1500. The investment made is: | Answer: Option B Explanation: | To earn Rs. | 50 | , investment = Rs. 64. | | 3 |
| To earn Rs. 1500, investment = Rs. | | 64 x | 3 | x 1500 | | = Rs. 5760. | | 50 |
Learn more problems on : Stocks and Shares Discuss about this problem : Discuss in Forum |
| 15. | A man buys a watch for Rs. 1950 in cash and sells it for Rs. 2200 at a credit of 1 year. If the rate of interest is 10% per annum, the man: | Answer: Option B Explanation: | S.P. | = P.W. of Rs. 2200 due 1 year hence | | | = Rs. |  | 2200 x 100 |  | | 100 + (10 x 1) |
| | = Rs. 2000. |
Gain = Rs. (2000 - 1950) = Rs. 50. Learn more problems on : True Discount Discuss about this problem : Discuss in Forum |
Direction (for Q.No. 16):
Find the odd man out. | | 16. | 10, 25, 45, 54, 60, 75, 80 | |
Direction (for Q.Nos. 17 - 19):
Find out the wrong number in the given sequence of numbers. | | 17. | 582, 605, 588, 611, 634, 617, 600 | Answer: Option A Explanation: Alternatively 23 is added and 17 is subtracted from the terms. So, 634 is wrong. Learn more problems on : Odd Man Out and Series Discuss about this problem : Discuss in Forum |
| 18. | 36, 54, 18, 27, 9, 18.5, 4.5 | Answer: Option B Explanation: The terms are alternatively multiplied by 1.5 and divided by 3. However, 18.5 does not satisfy it. Learn more problems on : Odd Man Out and Series Discuss about this problem : Discuss in Forum |
| 19. | 56, 72, 90, 110, 132, 150 | Answer: Option D Explanation: The numbers are 7 x 8, 8 x 9, 9 x 10, 10 x 11, 11 x 12, 12 x 13. So, 150 is wrong. Learn more problems on : Odd Man Out and Series Discuss about this problem : Discuss in Forum |
Direction (for Q.No. 20):
Insert the missing number. | | 20. | 8, 24, 12, 36, 18, 54, (....) | Answer: Option A Explanation: Numbers are alternatively multiplied by 3 and divided by 2. So, the next number = 54 ÷ 2 = 27. Learn more problems on : Odd Man Out and Series Discuss about this problem : Discuss in Forum |
| | Marks : 0/20 | | Total number of questions | : | 20 | | Number of answered questions | : | 0 | | Number of unanswered questions | : | 20 |
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