Online Aptitude Test - Aptitude Test 8
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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 the numbers 13a and 13b.
Then, 13a x 13b = 2028
ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).
Clearly, there are 2 such pairs.
L.C.M. of 12, 18, 21 30 2 | 12 - 18 - 21 - 30
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= 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 - 9 - 21 - 15
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Required number = (1260 รท 2) | 2 - 3 - 7 - 5
= 630.
Let the ten's digit be x and unit's digit be y.
Then, x + y = 15 and x - y = 3 or y - x = 3.
Solving x + y = 15 and x - y = 3, we get: x = 9, y = 6.
Solving x + y = 15 and y - x = 3, we get: x = 6, y = 9.
So, the number is either 96 or 69.
Hence, the number cannot be determined.
Let the ten's digit be x and unit's digit be y.
Then, number = 10x + y.
Number obtained by interchanging the digits = 10y + x.
(10x + y) + (10y + x) = 11(x + y), which is divisible by 11.
Video Explanation: https://youtu.be/lytJE8GqRvM
Let the ten's digit be x and unit's digit be y.
Then, (10x + y) - (10y + x) = 36
9(x - y) = 36
x - y = 4.
Video Explanation: https://youtu.be/7QOJjAmGVx0
We know that 112 = 121.
Putting m = 11 and n = 2, we get:
(m - 1)n + 1 = (11 - 1)(2 + 1) = 103 = 1000.
A shopkeeper sells some toys at Rs. 250 each. What percent profit does he make? To find the answer, which of the following information given in Statements I and II is/are necessary? | |
I. | Number of toys sold. |
II. | Cost price of each toy. |
S.P. = Rs. 250 each.
To find gain percent, we must know the C.P. of each.
Correct answer is (B).
Relative speed = (120 + 80) km/hr
| = |  |
200 x | 5 |  m/sec |
| 18 |
| = | |
500 | m/sec. |
| 9 |
Let the length of the other train be x metres.
| Then, | x + 270 | = | 500 |
| 9 | 9 |
x + 270 = 500
x = 230.
Let the speed of the second train be x km/hr.
| Relative speed | = (x + 50) km/hr | |||||||
|
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|
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Distance covered = (108 + 112) = 220 m.
| |
220 | = 6 | ||
|
250 + 5x = 660
x = 82 km/hr.
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 length of a running train crossing another 180 metre long train running in the opposite direction? | |
I. | The relative speed of the two trains was 150 kmph. |
II. | The trains took 9 seconds to cross each other. |
Let the two trains of length a metres and b metres be moving in opposite directions at u m/s and v m/s.
| Time taken to cross each other = | (a + b) | sec. |
| (u + v) |
| Now, b = 180, u + v = | |
150 x | 5 | m/sec |
= | 125 | m/sec. |
| 18 | 3 |
| 9 = |
a + 180 |
| (125/3) |
a = (375 - 180) = 195 m.
(a) Since loga a = 1, so log10 10 = 1.
(b) log (2 + 3) = log 5 and log (2 x 3) = log 6 = log 2 + log 3
log (2 + 3) 
log (2 x 3)
(c) Since loga 1 = 0, so log10 1 = 0.
(d) log (1 + 2 + 3) = log 6 = log (1 x 2 x 3) = log 1 + log 2 + log 3.
So, (b) is incorrect.
| 2(l + b) | = | 5 |
| b | 1 |
2l + 2b = 5b
3b = 2l
| b = | 2 | l |
| 3 |
Then, Area = 216 cm2
l x b = 216
| l x |
2 | l | = 216 |
| 3 |
l2 = 324
l = 18 cm.
Area of the park = (60 x 40) m2 = 2400 m2.
Area of the lawn = 2109 m2.
Area of the crossroads = (2400 - 2109) m2 = 291 m2.
Let the width of the road be x metres. Then,
60x + 40x - x2 = 291
x2 - 100x + 291 = 0
(x - 97)(x - 3) = 0
x = 3.
Video Explanation: https://youtu.be/R3CtrAKGxkc
| Volume of water displaced | = (3 x 2 x 0.01) m3 |
| = 0.06 m3. |
| Mass of man |
= Volume of water displaced x Density of water |
| = (0.06 x 1000) kg | |
| = 60 kg. |
| When B runs 25 m, A runs | 45 | m. |
| 2 |
| When B runs 1000 m, A runs | |
45 | x | 1 | x 1000 | m |
= 900 m. |
| 2 | 25 |
B beats A by 100 m.
In the word 'CORPORATION', we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
| Number of ways arranging these letters = | 7! | = 2520. |
| 2! |
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged
| in | 5! | = 20 ways. |
| 3! |
Required number of ways = (2520 x 20) = 50400.
Video Explanation: https://youtu.be/o3fwMoB0duw
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 |
The differences between two successive terms from the beginning are 7, 5, 7, 5, 7, 5.
So, 40 is wrong.
Go on multiplying 2 and adding 1 to get the next number.
So, 39 is wrong.