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.
| 753 x 753 + 247 x 247 - 753 x 247 | = ? |
| 753 x 753 x 753 + 247 x 247 x 247 |
| Given Exp. = | (a2 + b2 - ab) | = | 1 | = | 1 | = | 1 |
| (a3 + b3) | (a + b) | (753 + 247) | 1000 |
1400 x x = 1050  x = |
1050 | = | 3 |
| 1400 | 4 |
x + 3699 + 1985 - 2047 = 31111
x + 3699 + 1985 = 31111 + 2047
x + 5684 = 33158
x = 33158 - 5684 = 27474.
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.
| 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) |
| 4.036 | = | 403.6 | = 100.9 |
| 0.04 | 4 |
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 two-digit number whose first digit is a and the second digit is b?. The number is greater than 9. | |
I. | The number is multiple of 51. |
II. | The sum of the digits a and b is 6. |
From statement I:
A two digit number, greater than 9 and multiple of 51 should be 51 itself.
Because, 2 x 51 = 102 (3 digit number). Therefore, I alone sufficient to answer.
From statement II:
A two digit number, greater than 9 and sum of the digit is 6.
It can be 15, 24, 33, 42, 51. So we cannot determine the required answer from the statement II alone.
Thus, I alone give the answer while II alone not sufficient to answer.
Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.
| Then, | (6x + 6) + 4 | = | 11 |
| (5x + 6) + 4 | 10 |
10(6x + 10) = 11(5x + 10)
5x = 10
x = 2.
Sagar's present age = (5x + 6) = 16 years.
After 10 days : 150 men had food for 35 days.
Suppose 125 men had food for x days.
Now, Less men, More days (Indirect Proportion)
125 : 150 :: 35 : x 
125 x x = 150 x 35
| x = |
150 x 35 |
| 125 |
x = 42.
of its actual speed covers 42 km in 1 hr 40 min 48 sec. Find the actual speed of the car.
| Time taken = 1 hr 40 min 48 sec = 1 hr 40 | 4 | min = 1 | 51 | hrs = | 126 | hrs. |
| 5 | 75 | 75 |
Let the actual speed be x km/hr.
| Then, | 5 | x x | 126 | = 42 |
| 7 | 75 |
| x = |
 |
42 x 7 x 75 |  |
= 35 km/hr. |
| 5 x 126 |
Let the speed of the stream x mph. Then,
Speed downstream = (10 + x) mph,
Speed upstream = (10 - x) mph.
| |
36 | - | 36 | = | 90 |
| (10 - x) | (10 + x) | 60 |
72x x 60 = 90 (100 - x2)
x2 + 48x - 100 = 0
(x+ 50)(x - 2) = 0
x = 2 mph.
S.P. of 1 kg of the mixture = Rs. 68.20, Gain = 10%.
| C.P. of 1 kg of the mixture = Rs. | |
100 | x 68.20 | |
= Rs. 62. |
| 110 |
By the rule of alligation, we have:
| Cost of 1 kg tea of 1st kind. Cost of 1 kg tea of 2nd kind. | ||
| Rs. 60 | Mean Price Rs. 62 | Rs. 65 |
| 3 | 2 | |
Required ratio = 3 : 2.
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 |
times as fast as B. If A gives B a start of 80 m, how far must the winning post be so that A and B might reach it at the same time?
| Ratio of the speeds of A and B = | 5 | : 1 = 5 : 3. |
| 3 |
Thus, in race of 5 m, A gains 2 m over B.
2 m are gained by A in a race of 5 m.
| 80 m will be gained by A in race of | |
5 | x 80 | m |
= 200 m. |
| 2 |
Winning post is 200 m away from the starting point.
100 years contain 5 odd days.
Last day of 1st century is Friday.
200 years contain (5 x 2) 
3 odd days.
Last day of 2nd century is Wednesday.
300 years contain (5 x 3) = 15 1 odd day.
Last day of 3rd century is Monday.
400 years contain 0 odd day.
Last day of 4th century is Sunday.
This cycle is repeated.
Last day of a century cannot be Tuesday or Thursday or Saturday.
| Angle traced by hour hand in | 13 | hrs = | |
360 | x | 13 | |
° | = 130°. |
| 3 | 12 | 3 |
| Angle traced by min. hand in 20 min. = | |
360 | x 20 | |
° | = 120°. |
| 60 |
Required angle = (130 - 120)° = 10°.
55 min. spaces are covered in 60 min.
| 60 min. spaces are covered in | |
60 | x 60 | min. |
= 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 | min. |
= | 32 | 8 | min. |
| 11 | 64 | 11 |
| The banker's discount on a certain sum due 2 years hence is | 11 | of the true discount. |
| 10 |
The rate percent is:
Let T.D. be Re. 1.
| Then, B.D. = Rs. | 11 | = Rs. 1.10. |
| 10 |
| Sum = Rs. |
|
1.10 x 1 | |
= Rs. | |
110 | |
= Rs. 11. |
| 1.10 - 1 | 10 |
S.I. on Rs. 11 for 2 years is Rs. 1.10
| Rate = |
|
100 x 1.10 | % |
= 5%. |
| 11 x 2 |
Terms at odd places are 5, 6, 7, 8 etc. and each term at even place is 16.
So, 9 is wrong.
Go on subtracting 24, 21, 18, 15, 12, 9 from the numbers to get the next number.
190 - 24 = 166 166 - 21 = 145 145 - 18 = 127 [Here, 128 is placed instead of 127] 127 - 15 = 112 112 - 12 = 100 ... and so on.
Therefore, 128 is wrong.