Online Aptitude Test - Aptitude Test - Random
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- Total number of questions: 20.
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Marks : 2/20
Test Review : View answers and explanation for this test.
Let x = 296q + 75
= (37 x 8q + 37 x 2) + 1
= 37 (8q + 2) + 1
Thus, when the number is divided by 37, the remainder is 1.
8796 x 223 + 8796 x 77 | = 8796 x (223 + 77) [Ref: By Distributive Law ] |
= (8796 x 300) | |
= 2638800 |
(6n2 + 6n) = 6n(n + 1), which is always divisible by 6 and 12 both, since n(n + 1) is always even.
Reduce | 128352 | to its lowest terms. |
238368 |
128352) 238368 ( 1 128352 --------------- 110016 ) 128352 ( 1 110016 ------------------ 18336 ) 110016 ( 6 110016 ------- x ------- So, H.C.F. of 128352 and 238368 = 18336. 128352 128352 ÷ 18336 7 Therefore, ------ = -------------- = -- 238368 238368 ÷ 18336 13
1 ÷ 0.2 = | 1 | = | 10 | = 5; |
0.2 | 2 |
0.2 = 0.222...;
(0.2)2 = 0.04.
0.04 < 0.2 < 0.22....<5.
Since 0.04 is the least, so (0.2)2 is the least.
4.2 x 4.2 - 1.9 x 1.9 | is equal to: |
2.3 x 6.1 |
Given Expression = | (a2 - b2) | = | (a2 - b2) | = 1. |
(a + b)(a - b) | (a2 - b2) |
Which of the following fractions is greater than | 3 | and less than | 5 | ? |
4 | 6 |
3 | = 0.75, | 5 | = 0.833, | 1 | = 0.5, | 2 | = 0.66, | 4 | = 0.8, | 9 | = 0.9. |
4 | 6 | 2 | 3 | 5 | 10 |
Clearly, 0.8 lies between 0.75 and 0.833.
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4 | lies between | 3 | and | 5 | . |
5 | 4 | 6 |
29.94 | = | 299.4 |
1.45 | 14.5 |
= | ![]() |
2994 | x | 1 | ![]() |
[ Here, Substitute 172 in the place of 2994/14.5 ] |
14.5 | 10 |
= | 172 |
10 |
= 17.2
3.14 x 106 = 3.14 x 1000000 = 3140000.
Let total number of children be x.
Then, x x | 1 | x = | x | x 16 ![]() |
8 | 2 |
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1 | x2 = | ![]() |
1 | x 64 x 64 | ![]() |
= 512. |
8 | 8 |
Let the numbers be x and y.
Then, xy = 9375 and | x | = 15. |
y |
xy | = | 9375 |
(x/y) | 15 |
y2 = 625.
y = 25.
x = 15y = (15 x 25) = 375.
Sum of the numbers = x + y = 375 + 25 = 400.
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 two-digit number? | |
I. | Sum of the digits is 7. |
II. | Difference between the number and the number obtained by interchanging the digits is 9. |
III. | Digit in the ten's place is bigger than the digit in the unit's place by 1. |
Let the tens and units digit be x and y respectively.
I. x + y = 7.
II. (10x + y) - (10y + x) = 9 x - y = 1.
III. x - y = 1.
Thus, I and II as well as I and III give the answer.
Correct answer is (E).
Number of pages typed by Ravi in 1 hour = | 32 | = | 16 | . |
6 | 3 |
Number of pages typed by Kumar in 1 hour = | 40 | = 8. |
5 |
Number of pages typed by both in 1 hour = | ![]() |
16 | + 8 | ![]() |
= | 40 | . |
3 | 3 |
![]() |
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110 x | 3 | ![]() |
hours |
40 |
= 8 | 1 | hours (or) 8 hours 15 minutes. |
4 |
Suppose A, B and C take x, | x | and | x | days respectively to finish the work. |
2 | 3 |
Then, | ![]() |
1 | + | 2 | + | 3 | ![]() |
= | 1 |
x | x | x | 2 |
![]() |
6 | = | 1 |
x | 2 |
x = 12.
So, B takes (12/2) = 6 days to finish the work.
Let the distance travelled on foot be x km.
Then, distance travelled on bicycle = (61 -x) km.
So, | x | + | (61 -x) | = 9 |
4 | 9 |
9x + 4(61 -x) = 9 x 36
5x = 80
x = 16 km.
Let the speed of the second train be x km/hr.
Relative speed | = (x + 50) km/hr | |||||||
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||||||||
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Distance covered = (108 + 112) = 220 m.
![]() |
220 | = 6 | ||
|
250 + 5x = 660
x = 82 km/hr.
55 min. spaces are covered in 60 min.
60 min. spaces are covered in | ![]() |
60 | x 60 | ![]() |
= 65 | 5 | min. |
55 | 11 |
Loss in 64 min. = | ![]() |
65 | 5 | - 64 | ![]() |
= | 16 | min. |
11 | 11 |
Loss in 24 hrs = | ![]() |
16 | x | 1 | x 24 x 60 | ![]() |
= | 32 | 8 | min. |
11 | 64 | 11 |
Angle traced by hour hand in | 17 | hrs = | ![]() |
360 | x | 17 | ![]() |
° | = 255°. |
2 | 12 | 2 |
Angle traced by min. hand in 30 min. = | ![]() |
360 | x 30 | ![]() |
° | = 180°. |
60 |
Required angle = (255 - 180)° = 75°.
Required number of ways = (7C5 x 3C2) = (7C2 x 3C1) = | ![]() |
7 x 6 | x 3 | ![]() |
= 63. |
2 x 1 |
Total number of balls = (2 + 3 + 2) = 7.
Let S be the sample space.
Then, n(S) | = Number of ways of drawing 2 balls out of 7 | |||
= 7C2 ` | ||||
|
||||
= 21. |
Let E = Event of drawing 2 balls, none of which is blue.
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= Number of ways of drawing 2 balls out of (2 + 3) balls. | |||
= 5C2 | ||||
|
||||
= 10. |
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n(E) | = | 10 | . |
n(S) | 21 |