Online Aptitude Test - Aptitude Test 1
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- Total number of questions: 20.
- Time allotted: 30 minutes.
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Marks : 2/20
Test Review : View answers and explanation for this test.
Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)
= H.C.F. of 48, 92 and 140 = 4.
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 |
| So, | 7 | is the largest. |
| 8 |
| .009 | = .01 |
| ? |
| Let | .009 | = .01; Then x = | .009 | = | .9 | = .9 |
| x | .01 | 1 |
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
| If x = | 3 + 1 | and y = | 3 - 1 | , then the value of (x2 + y2) is: |
| 3 - 1 | 3 + 1 |
| 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
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.
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. |
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).
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 |
| x% of y = | |
x | x y | |
= | |
y | x x | |
= y% of x |
| 100 | 100 |
A = B.
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.
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.
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.
Video Explanation: https://youtu.be/lMFnNB3YQOo
| Principal = Rs. | |
100 x 5400 | |
= Rs. 15000. |
| 12 x 3 |
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 |
% stock at 64, one earns Rs. 1500. The investment made is:
| To earn Rs. | 50 | , investment = Rs. 64. |
| 3 |
| To earn Rs. 1500, investment = Rs. | |
64 x | 3 | x 1500 | |
= Rs. 5760. |
| 50 |
| S.P. | = P.W. of Rs. 2200 due 1 year hence | |||||
|
||||||
| = Rs. 2000. |
Gain = Rs. (2000 - 1950) = Rs. 50.
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.
Numbers are alternatively multiplied by 3 and divided by 2.
So, the next number = 54 ÷ 2 = 27.