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.
Let the two consecutive even integers be 2n and (2n + 2). Then,
(2n + 2)2 = (2n + 2 + 2n)(2n + 2 - 2n)
= 2(4n + 2)
= 4(2n + 1), which is divisible by 4.
| 106 x 106 - 94 x 94 | = (106)2 - (94)2 |
| = (106 + 94)(106 - 94) [Ref: (a2 - b2) = (a + b)(a - b)] | |
| = (200 x 12) | |
| = 2400. |
Let the number of hens be x and the number of cows be y.
Then, x + y = 48 .... (i)
and 2x + 4y = 140 
x + 2y = 70 .... (ii)
Solving (i) and (ii) we get: x = 26, y = 22.
The required answer = 26.
If a = 0.1039, then the value of 4a2 - 4a + 1 + 3a is:
4a2 - 4a + 1 + 3a = (1)2 + (2a)2 - 2 x 1 x 2a + 3a
= (1 - 2a)2 + 3a
= (1 - 2a) + 3a
= (1 + a)
= (1 + 0.1039)
= 1.1039
1.5625 = ?
1|1.5625( 1.25
|1
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22| 56
| 44
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245| 1225
| 1225
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| X
|-------
1.5625 = 1.25.
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Video Explanation: https://youtu.be/OXLnoItd0MA
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xb |  |
(b + c - a) | . | |
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(c + a - b) | . | |
xa | |
(a + b - c) | = ? |
| xc | xa | xb |
| Given Exp. |
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Let their marks be (x + 9) and x.
| Then, x + 9 = | 56 | (x + 9 + x) |
| 100 |
25(x + 9) = 14(2x + 9)
3x = 99
x = 33
So, their marks are 42 and 33.
| C.P. of 1st transistor = Rs. |  |
100 | x 840 |  |
= Rs. 700. |
| 120 |
| C.P. of 2nd transistor = Rs. | |
100 | x 960 | |
= Rs. 1000 |
| 96 |
So, total C.P. = Rs. (700 + 1000) = Rs. 1700.
Total S.P. = Rs. (840 + 960) = Rs. 1800.
| Gain % = |
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100 | x 100 | % |
= 5 | 15 | % |
| 1700 | 17 |
Let the required number of bottles be x.
More weavers, More mats (Direct Proportion)
More days, More mats (Direct Proportion)
| Wavers | 4 | : | 8 |  |
:: 4 : x |
| Days | 4 | : | 8 |
4 x 4 x x = 8 x 8 x 4
| x = |
(8 x 8 x 4) |
| (4 x 4) |
x = 16.
| Relative speed = | = (45 + 30) km/hr | |||||||
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We have to find the time taken by the slower train to pass the DRIVER of the faster train and not the complete train.
So, distance covered = Length of the slower train.
Therefore, Distance covered = 500 m.
| Required time = |
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500 x | 6 | |
= 24 sec. |
| 125 |
| Speed = | |
54 x | 5 | m/sec = 15 m/sec. |
| 18 |
Length of the train = (15 x 20)m = 300 m.
Let the length of the platform be x metres.
| Then, | x + 300 | = 15 |
| 36 |
x + 300 = 540
x = 240 m.
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.
Let the price of the mixed variety be Rs. x per kg.
By rule of alligation, we have:
| Cost of 1 kg of Type 1 rice Cost of 1 kg of Type 2 rice | ||
| Rs. 15 | Mean Price Rs. x | Rs. 20 |
| (20 - x) | (x - 15) | |
| |
(20 - x) | = | 2 |
| (x - 15) | 3 |
60 - 3x = 2x - 30
5x = 90
x = 18.
In each of the following questions, a question is asked and is followed by three statements. While answering the question, you may or may not require the data provided in all the statements. You have to read the question and the three statements and then decide whether the question can be answered with any one or two of the statements or all the three statements are required to answer the question. The answer number bearing the statements, which can be dispensed with, if any, while answering the question is your answer.
Mr. Gupta borrowed a sum of money on compound interest. What will be the amount to be repaid if he is repaying the entire amount at the end of 2 years? | |
I. | The rate of interest is 5 p.c.p.a. |
II. | Simple interest fetched on the same amount in one year is Rs. 600. |
III. | The amount borrowed is 10 times the simple interest in 2 years. |
I. gives, Rate = 5% p.a.
II. gives, S.I. for 1 year = Rs. 600.
III. gives, sum = 10 x (S.I. for 2 years).
Now I, and II give the sum.
For this sum, C.I. and hence amount can be obtained.
Thus, III is redundant.
Again, II gives S.I. for 2 years = Rs. (600 x 2) = Rs. 1200.
Now, from III, Sum = Rs. (10 x 1200) = Rs . 12000.
| Thus, Rate = | 100 x 1200 | = 5% p.a. |
| 2 x 12000 |
Thus, C.I. for 2 years and therefore, amount can be obtained.
Thus, I is redundant.
Hence, I or III redundant.
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 |
| When B runs 25 m, A runs | 45 | m. |
| 2 |
| When B runs 1000 m, A runs | |
45 | x | 1 | x 1000 | m |
= 900 m. |
| 2 | 25 |
B beats A by 100 m.
A : B = 100 : 90.
A : C = 100 : 72.
| B : C = | B | x | A | = | 90 | x | 100 | = | 90 | . |
| A | C | 100 | 72 | 72 |
When B runs 90 m, C runs 72 m.
| When B runs 100 m, C runs | |
72 | x 100 | m |
= 80 m. |
| 90 |
B can give C 20 m.
On 31st December, 2005 it was Saturday.
Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.
On 31st December 2009, it was Thursday.
Thus, on 1st Jan, 2010 it is Friday.
Go on subtracting 3 and dividing the result by 2 to obtain the next number.
Clearly, 46 is wrong.