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 7589 -x = 3434
Then, x = 7589 - 3434 = 4155
Required sum = (2 + 4 + 6 + ... + 30)
This is an A.P. in which a = 2, d = (4 - 2) = 2 and l = 30.
Let the number of terms be n. Then,
tn = 30 
a + (n - 1)d = 30
2 + (n - 1) x 2 = 30
n - 1 = 14
n = 15
 Sn = |
n | (a + l) | = | 15 | x (2 + 30) = 240. |
| 2 | 2 |
| Given Exp. = | 800 | x | 1296 | = 450 |
| 64 | 36 |
99 = 1 x 3 x 3 x 11
101 = 1 x 101
176 = 1 x 2 x 2 x 2 x 2 x 11
182 = 1 x 2 x 7 x 13
So, divisors of 99 are 1, 3, 9, 11, 33, .99
Divisors of 101 are 1 and 101
Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176
Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.
Hence, 176 has the most number of divisors.
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.
The average age of P, Q, R and S is 30 years. How old is R? | |
I. | The sum of ages of P and R is 60 years. |
II. | S is 10 years younger than R. |
P + Q + R + S = (30 x 4) P + Q + R + S = 120 .... (i)
I. P + R = 60 .... (ii)
II. S = (R - 10) .... (iii)
From (i), (ii) and (iii), we cannot find R.
Correct answer is (D)
Let the fourth proportional to 5, 8, 15 be x.
Then, 5 : 8 : 15 : x
5x = (8 x 15)
| x = | (8 x 15) | = 24. |
| 5 |
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.
How much did Rohit get as profit at the year-end in the business done by Nitin, Rohit and Kunal? | |
I. |
Kunal invested Rs. 8000 for nine months, his profit was |
II. | Nitin and Rohit invested for one year in the proportion 1 : 2 respectively. |
III. | The three together got Rs. 1000 as profit at the year end. |
times that of Rohit's and his investment was four times that of Nitin.
I and II give:
K = Rs. (8000 x 9) for 1 month = Rs. 72000 for 1 month.
| N = Rs. |  |
1 | x 8000 x 12 |  |
for 1 month = Rs. 24000 for 1 month. |
| 4 |
R = Rs. 48000 for 1 month.
K : N : R = 72000 : 24000 : 48000 = 3 : 1 : 2.
III gives, total profit = Rs. 1000.
| Rohit's share = Rs. |
|
1000 x | 2 | |
= Rs. 333 | 1 |
| 6 | 3 |
Correct answer is (D).
Let the required weight be x kg.
Less weight, Less cost (Direct Proportion)
250 : 200 :: 60 : x 
250 x x = (200 x 60)
| x= |
(200 x 60) |
| 250 |
x = 48.
| Work done by X in 8 days = | |
1 | x 8 | |
= | 1 | . |
| 40 | 5 |
| Remaining work = | |
1 - | 1 | |
= | 4 | . |
| 5 | 5 |
| Now, | 4 | work is done by Y in 16 days. |
| 5 |
| Whole work will be done by Y in | |
16 x | 5 | |
= 20 days. |
| 4 |
| X's 1 day's work = |
1 | , Y's 1 day's work = | 1 | . |
| 40 | 20 |
| (X + Y)'s 1 day's work = | |
1 | + | 1 | |
= | 3 | . |
| 40 | 20 | 40 |
| Hence, X and Y will together complete the work in | |
40 | |
= 13 | 1 | days. |
| 3 | 3 |
Let the man's rate upstream be x kmph and that downstream be y kmph.
Then, distance covered upstream in 8 hrs 48 min = Distance covered downstream in 4 hrs.
| |
|
x x 8 | 4 | |
= (y x 4) |
| 5 |
| |
44 | x =4y |
| 5 |
| y = |
11 | x. |
| 5 |
| Required ratio = |
|
y + x | |
: | |
y - x | |
| 2 | 2 |
| = | |
16x | x | 1 | |
: | |
6x | x | 1 | |
| 5 | 2 | 5 | 2 |
| = | 8 | : | 3 |
| 5 | 5 |
= 8 : 3.
% p.a for 2 years. Find his gain in the transaction per year.
| Gain in 2 years |
|
||||||||||||||||
| = Rs. (625 - 400) | |||||||||||||||||
| = Rs. 225. |
| Gain in 1 year = Rs. |
|
225 | |
= Rs. 112.50 |
| 2 |
Video Explanation: https://youtu.be/zBjcwqIcmL4
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.
What is the rate of simple interest? | |
I. | The total interest earned was Rs. 4000. |
II. | The sum was invested for 4 years. |
| We know that, R = | |
100 x S.I. | |
| P x T |
Now, I gives, S.I. = Rs. 4000.
II gives, T = 4 years.
But, P is unknown. So, we cannot find R.
So, given data is insufficient to get R.
Correct answer is (D).
Let original length = x and original breadth = y.
Original area = xy.
| New length = | x | . |
| 2 |
New breadth = 3y.
| New area = | |
x | x 3y | |
= | 3 | xy. |
| 2 | 2 |
| Increase % = |
|
1 | xy x | 1 | x 100 | % |
= 50%. |
| 2 | xy |
h = 14 cm, r = 7 cm.
So, l = (7)2 + (14)2 = 245 = 75 cm.
| Total surface area |
=  rl + r2 |
|||||||||
|
||||||||||
| = [154(5 + 1)] cm2 | ||||||||||
| = (154 x 3.236) cm2 | ||||||||||
| = 498.35 cm2. |
A : B = 60 : 45.
A : C = 60 : 40.
| |
B | = | |
B | x | A | |
= | |
45 | x | 60 | |
= | 45 | = | 90 | = 90 : 80. |
| C | A | C | 60 | 40 | 40 | 80 |
B can give C 10 points in a game of 90.
B runs 35 m in 7 sec.
| B covers 200 m in |
|
7 | x 200 | |
= 40 sec. |
| 35 |
B's time over the course = 40 sec.
A's time over the course (40 - 7) sec = 33 sec.
| For an income of Re. 1 in 9% stock at 96, investment = Rs. | |
96 | |
= Rs. | 32 |
| 9 | 3 |
| For an income Re. 1 in 12% stock at 120, investment = Rs. | |
120 | |
= Rs. 10. |
| 12 |
| Ratio of investments = |
32 | : 10 = 32 : 30 = 16 : 15. |
| 3 |
To earn Rs. 10, money invested = Rs. 100.
| To earn Rs. 12, money invested = Rs. | |
100 | x 12 | |
= Rs. 120. |
| 10 |
Market value of Rs. 100 stock = Rs. 120.
Here, S = {1, 2, 3, 4, ...., 19, 20}.
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
| P(E) = |
n(E) | = | 9 | . |
| n(S) | 20 |