Online Aptitude Test - Aptitude Test - Random
- This is a FREE online test. Beware of scammers who ask for money to attend this test.
- Total number of questions: 20.
- Time allotted: 30 minutes.
- Each question carries 1 mark; there are no negative marks.
- DO NOT refresh the page.
- All the best!
Marks : 2/20
Test Review : View answers and explanation for this test.
The square of a natural number nerver ends in 2.
143642 is not the square of natural number.
The smallest 3-digit number is 100, which is divisible by 2.
100 is not a prime number.
101 < 11 and 101 is not divisible by any of the prime numbers 2, 3, 5, 7, 11.
101 is a prime number.
Hence 101 is the smallest 3-digit prime number.
This is an A.P. in which a = 51, l = 100 and n = 50.
Sum = | n | (a + l) | = | 50 | x (51 + 100) = (25 x 151) = 3775. |
2 | 2 |
106 x 106 - 94 x 94 | = (106)^{2} - (94)^{2} |
= (106 + 94)(106 - 94) [Ref: (a^{2} - b^{2}) = (a + b)(a - b)] | |
= (200 x 12) | |
= 2400. |
Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.
So, the numbers 12 and 16.
L.C.M. of 12 and 16 = 48.
3 - | 1 | 2 | simplifies to: | ||
3 |
3 - | 1 | 2 | = (3)^{2} + | 1 | 2 | - 2 x 3 x | 1 | ||||
3 | 3 | 3 |
= 3 + | 1 | - 2 |
3 |
= 1 + | 1 |
3 |
= | 4 |
3 |
If a = 0.1039, then the value of 4a^{2} - 4a + 1 + 3a is:
4a^{2} - 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
The square root of (7 + 35) (7 - 35) is
(7 + 35)(7 - 35) | = | (7)^{2} - (35)^{2} | = 49 - 45 = 4 = 2. |
If 5 = 2.236, then the value of | 5 | - | 10 | + 125 is equal to: |
2 | 5 |
5 | - | 10 | + 125 | = | (5)^{2} - 20 + 25 x 55 |
2 | 5 | 25 |
= | 5 - 20 + 50 |
25 |
= | 35 | x | 5 |
25 | 5 |
= | 355 |
10 |
= | 7 x 2.236 |
2 |
= 7 x 1.118 |
= 7.826 |
Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
5x = 20
x = 4.
Age of the youngest child = x = 4 years.
Let the amount taxable purchases be Rs. x.
Then, 6% of x = | 30 |
100 |
x = | 30 | x | 100 | = 5. | |
100 | 6 |
Cost of tax free items = Rs. [25 - (5 + 0.30)] = Rs. 19.70
Let B's capital be Rs. x.
Then, | 3500 x 12 | = | 2 | ||
7x | 3 |
14x = 126000
x = 9000.
Let distance = x km and usual rate = y kmph.
Then, | x | - | x | = | 40 | 2y(y + 3) = 9x ....(i) |
y | y + 3 | 60 |
And, | x | - | x | = | 40 | y(y - 2) = 3x ....(ii) |
y -2 | y | 60 |
On dividing (i) by (ii), we get: x = 40.
Distance = (240 x 5) = 1200 km.
Speed = Distance/Time
Speed = 1200/(5/3) km/hr. [We can write 1 hours as 5/3 hours]
Required speed = | 1200 x | 3 | km/hr | = 720 km/hr. | |
5 |
1 hectare = 10,000 m^{2}
So, Area = (1.5 x 10000) m^{2} = 15000 m^{2}.
Depth = | 5 | m | = | 1 | m. |
100 | 20 |
Volume = (Area x Depth) = | 15000 x | 1 | m^{3} | = 750 m^{3}. | |
20 |
Video Explanation: https://youtu.be/dDbVuBzYop8
A : B = 100 : 75
B : C = 100 : 96.
A : C = | A | x | B | = | 100 | x | 100 | = | 100 | = 100 : 72. | ||||
B | C | 75 | 96 | 72 |
A beats C by (100 - 72) m = 28 m.
28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)
Odd days in 1600 years = 0
Odd days in 400 years = 0
5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) 6 odd days
Jan. Feb. March April May (31 + 28 + 31 + 30 + 28 ) = 148 days
148 days = (21 weeks + 1 day) 1 odd day.
Total number of odd days = (0 + 0 + 6 + 1) = 7 0 odd day.
Given day is Sunday.
Here, n(S) = 52.
Let E = event of getting a queen of club or a king of heart.
Then, n(E) = 2.
P(E) = | n(E) | = | 2 | = | 1 | . |
n(S) | 52 | 26 |
Let the given numbers be A, B, C, D, E, F, G.
Then, A, A x 1 + 1, B x 2 + 2, C x 3 + 3, D x 4 + 4, E x 5 + 5, F x 6 + 6 are the required numbers.
Clearly, 228 is wrong.