Online Aptitude Test - Aptitude Test - Random
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Marks : 2/20
Test Review : View answers and explanation for this test.
Required numbers are 24, 30, 36, 42, ..., 96
This is an A.P. in which a = 24, d = 6 and l = 96
Let the number of terms in it be n.
Then tn = 96
a + (n - 1)d = 96
24 + (n - 1) x 6 = 96
(n - 1) x 6 = 72
(n - 1) = 12
n = 13
Required number of numbers = 13.
19657 Let x - 53651 = 9999 33994 Then, x = 9999 + 53651 = 63650 ----- 53651 -----
For every natural number n, (xn - an) is completely divisible by (x - a).
36 = 22 x 32
84 = 22 x 3 x 7
H.C.F. = 22 x 3 = 12.
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.
Let the numbers be a, b and c.
Then, a2 + b2 + c2 = 138 and (ab + bc + ca) = 131.
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 138 + 2 x 131 = 400.
(a + b + c) = 400 = 20.
Video Explanation: https://youtu.be/qmJ-0X8j_xQ
| (0.04)-1.5 = | ![]() |
4 | ![]() |
-1.5 |
| 100 |
| = | ![]() |
1 | ![]() |
-(3/2) |
| 25 |
= (25)(3/2)
= (52)(3/2)
= (5)2 x (3/2)
= 53
= 125.
| C.P. of 1 orange = Rs. | ![]() |
350 | ![]() |
= Rs. 3.50 |
| 100 |
| S.P. of 1 orange = Rs. | ![]() |
48 | ![]() |
= Rs. 4 |
| 12 |
Gain% = |
![]() |
0.50 | x 100 | % |
= | 100 | % = 14 | 2 | % |
| 3.50 | 7 | 7 |
For managing, A received = 5% of Rs. 7400 = Rs. 370.
Balance = Rs. (7400 - 370) = Rs. 7030.
Ratio of their investments = (6500 x 6) : (8400 x 5) : (10000 x 3)
= 39000 : 42000 : 30000
= 13 : 14 : 10
B's share = Rs. |
![]() |
7030 x | 14 | ![]() |
= Rs. 2660. |
| 37 |
| 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 |
| Total time taken = | ![]() |
160 | + | 160 | hrs. |
= | 9 | hrs. |
| 64 | 80 | 2 |
Average speed = |
![]() |
320 x | 2 | km/hr |
= 71.11 km/hr. |
| 9 |
Suppose the vessel initially contains 8 litres of liquid.
Let x litres of this liquid be replaced with water.
| Quantity of water in new mixture = | ![]() |
3 - | 3x | + x | ![]() |
litres |
| 8 |
| Quantity of syrup in new mixture = | ![]() |
5 - | 5x | ![]() |
litres |
| 8 |
|
![]() |
3 - | 3x | + x | ![]() |
= | ![]() |
5 - | 5x | ![]() |
| 8 | 8 |
5x + 24 = 40 - 5x
10x = 16
x = |
8 | . |
| 5 |
| So, part of the mixture replaced = | ![]() |
8 | x | 1 | ![]() |
= | 1 | . |
| 5 | 8 | 5 |
| Sum = Rs. | ![]() |
50 x 100 | ![]() |
= Rs. 500. |
| 2 x 5 |
| Amount |
|
||||||||||
|
|||||||||||
| = Rs. 551.25 |
C.I. = Rs. (551.25 - 500) = Rs. 51.25
ax = by
log ax = log by
x log a = y log b
|
log a | = | y | . |
| log b | x |
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 capacity of the cylindrical tank? | |
I. | The area of the base is 61,600 sq. cm. |
II. | The height of the tank is 1.5 times the radius. |
III. | The circumference of base is 880 cm. |
Capacity =
r2h.
I gives,
r2 = 61600. This gives r.
II gives, h = 1.5 r.
Thus, I and II give the answer.
Again, III gives 2
r = 880. This gives r.
So, II and III also give the answer.
Correct answer is (E).
To earn Rs. 135, investment = Rs. 1620.
| To earn Rs. 8, investment = Rs. | ![]() |
1620 | x 8 | ![]() |
= Rs. 96. |
| 135 |
Market value of Rs. 100 stock = Rs. 96.
In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
Number of ways of arranging these letters = |
8! | = 10080. |
| (2!)(2!) |
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
| Number of ways of arranging these letters = | 4! | = 12. |
| 2! |
Required number of words = (10080 x 12) = 120960.
Total number of balls = (8 + 7 + 6) = 21.
| Let E | = event that the ball drawn is neither red nor green |
| = event that the ball drawn is blue. |
n(E) = 7.
P(E) = |
n(E) | = | 7 | = | 1 | . |
| n(S) | 21 | 3 |
One of AB, AD and CD must have given.

So, the data is inadequate.
2nd term = (1st term) x 1 + 1 = 15 x 1 + 1 = 16.
3rd term = (2nd term) x 2 + 2 = 16 x 2 + 2 = 34.
4th term = (3th term) x 3 + 3 = 34 x 3 + 3 = 105.
5th term = (4th term) x 4 + 4 = 105 x 4 + 4 = 424
6th term = (5th term) x 5 + 5 = 424 x 5 + 5 = 2125
6th term should 2125 instead of 2124.



