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 4300731 - x = 2535618
Then x, = 4300731 - 2535618 = 1765113
2 x 5 = 10.
Sum of decimal places = 4
0.002 x 0.5 = 0.001
| 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
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 number? | |
I. | The sum of the two digits is 8. The ratio of the two digits is 1 : 3. |
II. | The product of the two digit of a number is 12. The quotient of two digits is 3. |
Let the tens and units digit be x and y respectively. Then,
| I. x + y = 8 and | x | = | 1 |
| y | 3 |
I gives, 4y = 24 
y = 6.
So, x + 6 = 8 x = 2.
| II. xy = 12 and | x | = | 3 |
| y | 1 |
II gives, x2 = 36 x = 6.
So, 3y = 6 y = 2.
Therefore, Either I or II alone sufficient to answer.
In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is 
of the number of students of 8 years of age which is 48. What is the total number of students in the school?
Let the number of students be x. Then,
Number of students above 8 years of age = (100 - 20)% of x = 80% of x.
| 80% of x = 48 + |
2 | of 48 |
| 3 |
 |
80 | x = 80 |
| 100 |
x = 100.
Video Explanation: https://youtu.be/yPfocU6DA2M
| Work done by the waste pipe in 1 minute = | 1 | - | |
1 | + | 1 | |
| 15 | 20 | 24 |
| = | |
1 | - | 11 | |
| 15 | 120 |
| = - | 1 | . [-ve sign means emptying] |
| 40 |
| Volume of |
1 | part = 3 gallons. |
| 40 |
Volume of whole = (3 x 40) gallons = 120 gallons.
| Relative speed = | = (45 + 30) km/hr | |||||||
|
||||||||
|
We have to find the time taken by the slower train to pass the DRIVER of the faster train and not the complete train.
So, distance covered = Length of the slower train.
Therefore, Distance covered = 500 m.
| Required time = |
|
500 x | 6 | |
= 24 sec. |
| 125 |
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 speed of the boat in still water? | |
I. | The speed downstream is 12 kmph. |
II. | The speed upstream is 4 kmph. |
III. | In a to and fro journey between two points, the average speed of the boat was 6 kmph. |
| From I and II, speed of boat in still water = | 1 | (12 + 4) km/hr = 8 km/hr. |
| 2 |
From II and III, we get:
| Using average speed = | 2xy | , we get: | 2 x 4 x y | = 6 |
| x + y | 4 + y |
8y = 24 + 6y
y = 12.
| Required speed = |
1 | (12 + 4) km/hr = 8 km/hr. |
| 2 |
Similarly, I and III also give the answer.
Correct answer is (D).
| Amount of milk left after 3 operations = |  |
40 | |
1 - | 4 | |
3 |  litres |
| 40 |
| = | |
40 x | 9 | x | 9 | x | 9 | |
= 29.16 litres. |
| 10 | 10 | 10 |
Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x).
| Then, | |
x x 14 x 2 | |
+ | |
(13900 - x) x 11 x 2 | |
= 3508 |
| 100 | 100 |
28x - 22x = 350800 - (13900 x 22)
6x = 45000
x = 7500.
So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.
Video Explanation: https://youtu.be/Xi4kU9y6ppk
100 cm is read as 102 cm.
A1 = (100 x 100) cm2 and A2 (102 x 102) cm2.
(A2 - A1) = [(102)2 - (100)2]
= (102 + 100) x (102 - 100)
= 404 cm2.
| Percentage error = |
|
404 | x 100 | % |
= 4.04% |
| 100 x 100 |
Each of these questions is followed by three statements. You have to study the question and all the three statements given to decide whether any information provided in the statement(s) is redundant and can be dispensed with while answering the given question.
What is the area of the given rectangle? | |
I. | Perimeter of the rectangle is 60 cm. |
II. | Breadth of the rectangle is 12 cm. |
III. | Sum of two adjacent sides is 30 cm. |
From I and II, we can find the length and breadth of the rectangle and therefore the area can be obtained.
So, III is redundant.
Also, from II and III, we can find the length and breadth and therefore the area can be obtained.
So, I is redundant.
Correct answer is "II and either I or III".
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 volume of a cube? | |
I. | The area of each face of the cube is 64 square metres. |
II. | The length of one side of the cube is 8 metres. |
Let each edge be a metres. Then,
I. a2 = 64 a = 8 m Volume = (8 x 8 x 8) m3 = 512 m3.
Thus, I alone gives the answer.
II. a = 8 m Volume = (8 x 8 x 8) m3 = 512 m3.
Thus, II alone gives the answer.
Correct answer is (C).
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.
To be together between 9 and 10 o'clock, the minute hand has to gain 45 min. spaces.
55 min. spaces gained in 60 min.
| 45 min. spaces are gained in | |
60 | x 45 | min or 49 |
1 | min. |
| 55 | 11 |
| The hands are together at 49 |
1 | min. past 9. |
| 11 |
The hands of a clock coincide 11 times in every 12 hours (Since between 11 and 1, they coincide only once, i.e., at 12 o'clock).
AM
12:00
1:05
2:11
3:16
4:22
5:27
6:33
7:38
8:44
9:49
10:55
PM
12:00
1:05
2:11
3:16
4:22
5:27
6:33
7:38
8:44
9:49
10:55
The hands overlap about every 65 minutes, not every 60 minutes.
The hands coincide 22 times in a day.
| Angle traced by the hour hand in 6 hours = | |
360 | x 6 | |
° | = 180°. |
| 12 |
In 12 hours, they are at right angles 22 times.
In 24 hours, they are at right angles 44 times.
S.I. on Rs. 750 = T.D. on Rs. 960.
This means P.W. of Rs. 960 due 2 years hence is Rs. 750.
T.D. = Rs. (960 - 750) = Rs. 210.
Thus, S.I. on R.s 750 for 2 years is Rs. 210.
| Rate = |
|
100 x 210 | % |
= 14% |
| 750 x 2 |
Let AB be the lighthouse and C and D be the positions of the ships.
Then, AB = 100 m, 
ACB = 30° and ADB = 45°.
| AB | = tan 30° = | 1 | AC = AB x 3 = 1003 m. |
| AC | 3 |
| AB | = tan 45° = 1 AD = AB = 100 m. |
| AD |
| CD = (AC + AD) |
= (1003 + 100) m |
| = 100(3 + 1) | |
| = (100 x 2.73) m | |
| = 273 m. |