Online Aptitude Test - Aptitude Test - Random
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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.
(xn + 1) will be divisible by (x + 1) only when n is odd.
(6767 + 1) will be divisible by (67 + 1)
(6767 + 1) + 66, when divided by 68 will give 66 as remainder.
6) 4456 (742
42
---
25
24 Therefore, Required number = (6 - 4) = 2.
---
16
12
---
4
The smallest 6-digit number 100000.
111) 100000 (900
999
-----
100
---
Required number = 100000 + (111 - 100)
= 100011.
N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
2|64009( 253
|4
|----------
45|240
|225
|----------
503| 1509
| 1509
|----------
| X
|----------
64009 = 253.
Each of the questions given below consists of a statement and / or a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statement(s) is / are sufficient to answer the given question. Read the both statements and
- Give answer (A) if the data in Statement I alone are sufficient to answer the question, while the data in Statement II alone are not sufficient to answer the question.
- Give answer (B) if the data in Statement II alone are sufficient to answer the question, while the data in Statement I alone are not sufficient to answer the question.
- Give answer (C) if the data either in Statement I or in Statement II alone are sufficient to answer the question.
- Give answer (D) if the data even in both Statements I and II together are not sufficient to answer the question.
- Give answer(E) if the data in both Statements I and II together are necessary to answer the question.
Divya is twice as old as Shruti. What is the difference in their ages? | |
I. | Five years hence, the ratio of their ages would be 9 : 5. |
II. | Ten years back, the ratio of their ages was 3 : 1. |
Let Divya's present age be D years and Shruti's present age b S years
Then, D = 2 x S 
D - 2S = 0 ....(i)
| I. | D + 5 | = | 9 | ....(ii) |
| S + 5 | 5 |
| II. | D - 10 | = | 3 | ....(iii) |
| S - 10 | 1 |
From (ii), we get : 5D + 25 = 9S + 45 5D - 9S = 20 ....(iv)
From (iii), we get : D - 10 = 3S - 30 D - 3S = -20 ....(v)
Thus, from (i) and (ii), we get the answer.
Also, from (i) and (iii), we get the answer.
I alone as well as II alone give the answer. Hence, the correct answer is (C).
of a job in 10 days. At this rate, how many more days will it takes him to finish the job?
| Work done = | 5 |
| 8 |
| Balance work = |  |
1 - | 5 |  |
= | 3 |
| 8 | 8 |
Let the required number of days be x.
| Then, | 5 | : | 3 | = :: 10 : x | |
5 | x x = | 3 | x 10 |
| 8 | 8 | 8 | 8 |
 x = |
|
3 | x 10 x | 8 | |
| 8 | 5 |
x = 6.
| 2(A + B + C)'s 1 day's work = | |
1 | + | 1 | + | 1 | |
= | 15 | = | 1 | . |
| 30 | 24 | 20 | 120 | 8 |
| Therefore, (A + B + C)'s 1 day's work = | 1 | = | 1 | . |
| 2 x 8 | 16 |
| Work done by A, B, C in 10 days = | 10 | = | 5 | . |
| 16 | 8 |
| Remaining work = | |
1 - | 5 | |
= | 3 | . |
| 8 | 8 |
| A's 1 day's work = | |
1 | - | 1 | |
= | 1 | . |
| 16 | 24 | 48 |
| Now, | 1 | work is done by A in 1 day. |
| 48 |
| So, | 3 | work will be done by A in | |
48 x | 3 | |
= 18 days. |
| 8 | 8 |
Each of the questions given below consists of a statement and / or a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statement(s) is / are sufficient to answer the given question. Read the both statements and
- Give answer (A) if the data in Statement I alone are sufficient to answer the question, while the data in Statement II alone are not sufficient to answer the question.
- Give answer (B) if the data in Statement II alone are sufficient to answer the question, while the data in Statement I alone are not sufficient to answer the question.
- Give answer (C) if the data either in Statement I or in Statement II alone are sufficient to answer the question.
- Give answer (D) if the data even in both Statements I and II together are not sufficient to answer the question.
- Give answer(E) if the data in both Statements I and II together are necessary to answer the question.
Two cars pass each other in opposite direction. How long would they take to be 500 km apart? | |
I. | The sum of their speeds is 135 km/hr. |
II. | The difference of their speed is 25 km/hr. |
I gives, relative speed = 135 km/hr.
| Time taken = |
500 | hrs. |
| 135 |
II does not give the relative speed.
I alone gives the answer and II is irrelevant.
Correct answer is (A).
| Speed = | |
72 x | 5 | m/sec |
= 20 m/sec. |
| 18 |
Time = 26 sec.
Let the length of the train be x metres.
| Then, | x + 250 | = 20 |
| 26 |
x + 250 = 520
x = 270.
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 the speed of the train? | |
I. | The train crosses a signal pole in 18 seconds. |
II. | The train crosses a platform of equal length in 36 seconds. |
III. | Length of the train is 330 metres. |
Let the speed of the train be x metres/sec.
| Time taken to cross a signal pole = | Length of the train |
| Speed of the train |
| Time taken to cross a platform = | (Length of the train + Length of the Platform) |
| Speed of the train |
Length of train = 330 m.
| I and III give, 18 = | 330 | x = |
330 | m/sec = | 55 | m/sec. |
| x | 18 | 3 |
| II and III give, 36 = | 2 x 330 | x = |
660 | m/sec = | 55 | m/sec. |
| x | 36 | 3 |
Correct answer is (D).
Man's rate in still water = (15 - 2.5) km/hr = 12.5 km/hr.
Man's rate against the current = (12.5 - 2.5) km/hr = 10 km/hr.
S.P. of 1 kg of the mixture = Rs. 68.20, Gain = 10%.
| C.P. of 1 kg of the mixture = Rs. | |
100 | x 68.20 | |
= Rs. 62. |
| 110 |
By the rule of alligation, we have:
| Cost of 1 kg tea of 1st kind. Cost of 1 kg tea of 2nd kind. | ||
| Rs. 60 | Mean Price Rs. 62 | Rs. 65 |
| 3 | 2 | |
Required ratio = 3 : 2.
S.I. for 3 years = Rs. (12005 - 9800) = Rs. 2205.
| S.I. for 5 years = Rs. | |
2205 | x 5 | |
= Rs. 3675 |
| 3 |
Principal = Rs. (9800 - 3675) = Rs. 6125.
| Hence, rate = | |
100 x 3675 | % |
= 12% |
| 6125 x 5 |
Video Explanation: https://youtu.be/UYwiBCRN39s
log10 5 + log10 (5x + 1) = log10 (x + 5) + 1
log10 5 + log10 (5x + 1) = log10 (x + 5) + log10 10
log10 [5 (5x + 1)] = log10 [10(x + 5)]
5(5x + 1) = 10(x + 5)
5x + 1 = 2x + 10
3x = 9
x = 3.
Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.
Area of each tile = (41 x 41) cm2.
| Required number of tiles = |
|
1517 x 902 | |
= 814. |
| 41 x 41 |
x weeks x days = (7x + x) days = 8x days.
Subtract 20, 25, 30, 35, 40, 45 from successive numbers.
So, 0 is wrong.