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
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Let the numbers 13a and 13b.
Then, 13a x 13b = 2028
ab = 12.
Now, the coprimes 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  = 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3  6  9  21  15  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 11^{2} = 121.
Putting m = 11 and n = 2, we get:
(m  1)^{n + 1} = (11  1)^{(2 + 1)} = 10^{3} = 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  



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 log_{a} a = 1, so log_{10} 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 log_{a} 1 = 0, so log_{10} 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 cm^{2}
l x b = 216
l x  2  l  = 216 
3 
l^{2} = 324
l = 18 cm.
Area of the park = (60 x 40) m^{2} = 2400 m^{2}.
Area of the lawn = 2109 m^{2}.
Area of the crossroads = (2400  2109) m^{2} = 291 m^{2}.
Let the width of the road be x metres. Then,
60x + 40x  x^{2} = 291
x^{2}  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) m^{3} 
= 0.06 m^{3}. 
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.