Online Aptitude Test - Logical Reasoning Test 4
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- Total number of questions: 20.
- Time allotted: 30 minutes.
- Each question carries 1 mark; there are no negative marks.
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Marks : 2/20
Test Review : View answers and explanation for this test.
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.
Largest 4-digit number = 9999
88) 9999 (113
88
----
119
88
----
319
264
---
55
---
Required number = (9999 - 55)
= 9944.
Required number = (L.C.M. of 12, 15, 20, 54) + 8
= 540 + 8
= 548.
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 many marks did Tarun secure in English? | |
I. | The average mark obtained by Tarun in four subjects including English is 60. |
II. | The total marks obtained by him in English and Mathematics together are 170. |
III. | The total marks obtained by him in Mathematics and Science together are 180. |
I gives, total marks in 4 subjects = (60 x 4) = 240.
II gives, E + M = 170
III gives, M + S = 180.
Thus, none of (A), (B), (C), (D) is true.
Correct answer is (E).
Let the number be x.
| Then, | 1 | of | 1 | of x = 15  x = 15 x 3 x 4 = 180. |
| 3 | 4 |
| So, required number = |  |
3 | x 180 |  |
= 54. |
| 10 |
Video Explanation: https://youtu.be/z49OUnzTnwY
Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit.
Let ten's and unit's digits be 2x and x respectively.
Then, (10 x 2x + x) - (10x + 2x) = 36
9x = 36
x = 4.
Required difference = (2x + x) - (2x - x) = 2x = 8.
| If x = 3 + 22, then the value of | |
x | - | 1 | |
is: |
| x |
|
x | - | 1 | |
2 | = x + | 1 | - 2 |
| x | x |
| = (3 + 22) + | 1 | - 2 |
| (3 + 22) |
| = (3 + 22) + | 1 | x | (3 - 22) | - 2 |
| (3 + 22) | (3 - 22) |
= (3 + 22) + (3 - 22) - 2
= 4.
| |
|
x | - | 1 | |
= 2. |
| x |
| Principal |
|
|||||
|
||||||
| = Rs. 8925. |
Let breadth = x metres.
Then, length = (x + 20) metres.
| Perimeter = | |
5300 | |
m = 200 m. |
| 26.50 |
2[(x + 20) + x] = 200
2x + 20 = 100
2x = 80
x = 40.
Hence, length = x + 20 = 60 m.
Volume of the large cube = (33 + 43 + 53) = 216 cm3.
Let the edge of the large cube be a.
So, a3 = 216 a = 6 cm.
| Required ratio = |
|
6 x (32 + 42 + 52) | |
= | 50 | = 25 : 18. |
| 6 x 62 | 36 |
1st April, 2001 = (2000 years + Period from 1.1.2001 to 1.4.2001)
Odd days in 1600 years = 0
Odd days in 400 years = 0
Jan. Feb. March April
(31 + 28 + 31 + 1) = 91 days 
0 odd days.
Total number of odd days = (0 + 0 + 0) = 0
On 1st April, 2001 it was Sunday.
In April, 2001 Wednesday falls on 4th, 11th, 18th and 25th.
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
| Required number of ways |
= (7C3 x 6C2) + (7C4 x 6C1) + (7C5) | |||||||||||
|
||||||||||||
|
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| = (525 + 210 + 21) | ||||||||||||
| = 756. |