Online Aptitude Test - Aptitude Test - Random



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Instruction:

  • Total number of questions : 20.
  • Time alloted : 30 minutes.
  • Each question carry 1 mark, no negative marks.
  • DO NOT refresh the page.
  • All the best :-).

1.

Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit's age. After further 8 years, how many times would he be of Ronit's age?

A.
2 times
B.
1 times
2
C.
3 times
4
D.
3 times

Your Answer: Option (Not Answered)

Correct Answer: Option A

Explanation:

Let Ronit's present age be x years. Then, father's present age =(x + 3x) years = 4x years.

(4x + 8) = 5 (x + 8)
2

8x + 16 = 5x + 40

3x = 24

x = 8.

Hence, required ratio = (4x + 16) = 48 = 2.
(x + 16) 24

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2.

Q is as much younger than R as he is older than T. If the sum of the ages of R and T is 50 years, what is definitely the difference between R and Q's age?

A.
1 year
B.
2 years
C.
25 years
D.
Data inadequate
E.
None of these

Your Answer: Option (Not Answered)

Correct Answer: Option D

Explanation:

Given that:

1. The difference of age b/w R and Q = The difference of age b/w Q and T.

2. Sum of age of R and T is 50 i.e. (R + T) = 50.

Question: R - Q = ?.

Explanation:

R - Q = Q - T

(R + T) = 2Q

Now given that, (R + T) = 50

So, 50 = 2Q and therefore Q = 25.

Question is (R - Q) = ?

Here we know the value(age) of Q (25), but we don't know the age of R.

Therefore, (R-Q) cannot be determined.

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3.

The last day of a century cannot be

A.
Monday
B.
Wednesday
C.
Tuesday
D.
Friday

Your Answer: Option (Not Answered)

Correct Answer: Option C

Explanation:

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.

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4.

The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team?

A.
23 years
B.
24 years
C.
25 years
D.
None of these

Your Answer: Option (Not Answered)

Correct Answer: Option A

Explanation:

Let the average age of the whole team by x years.

11x - (26 + 29) = 9(x -1)

11x - 9x = 46

2x = 46

x = 23.

So, average age of the team is 23 years.

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5.

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?

A.
4
B.
10
C.
15
D.
16

Your Answer: Option (Not Answered)

Correct Answer: Option D

Explanation:

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.

So, the bells will toll together after every 120 seconds(2 minutes).

In 30 minutes, they will toll together 30 + 1 = 16 times.
2

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6.

If 2994 ÷ 14.5 = 172, then 29.94 ÷ 1.45 = ?

A.
0.172
B.
1.72
C.
17.2
D.
172

Your Answer: Option (Not Answered)

Correct Answer: Option C

Explanation:

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

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Direction (for Q.No. 7):
Insert the missing number.
7.

2, 4, 12, 48, 240, (....)

A.
960
B.
1440
C.
1080
D.
1920

Your Answer: Option (Not Answered)

Correct Answer: Option B

Explanation:

Go on multiplying the given numbers by 2, 3, 4, 5, 6.

So, the correct next number is 1440.

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8.

The true discount on Rs. 2562 due 4 months hence is Rs. 122. The rate percent is:

A.
12%
B.
13 1 %
3
C.
15%
D.
14%

Your Answer: Option (Not Answered)

Correct Answer: Option C

Explanation:

P.W. = Rs. (2562 - 122) = Rs. 2440.

S.I. on Rs. 2440 for 4 months is Rs. 122.

Rate = 100 x 122 % = 15%.
2440 x 1
3

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Direction (for Q.No. 9):

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.

9.

What is the area of the hall?

I. 

Material cost of flooring per square metre is Rs. 2.50

II. 

Labour cost of flooring the hall is Rs. 3500

 III. 

Total cost of flooring the hall is Rs. 14,500.

A.
I and II only
B.
II and III only
C.
All I, II and III
D.
Any two of the three
E.
None of these

Your Answer: Option (Not Answered)

Correct Answer: Option C

Explanation:

  I. Material cost = Rs. 2.50 per m2

 II. Labour cost = Rs. 3500.

III. Total cost = Rs. 14,500.

Let the area be A sq. metres.

Material cost = Rs. (14500 - 3500) = Rs. 11,000.

5A = 11000     A = 11000 x 2 = 4400 m2.
2 5

Thus, all I, II and III are needed to get the answer.

Correct answer is (C).

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Direction (for Q.No. 10):
Find out the wrong number in the series.
10.

64, 71, 80, 91, 104, 119, 135, 155

A.
71
B.
80
C.
104
D.
119
E.
135

Your Answer: Option (Not Answered)

Correct Answer: Option E

Explanation:

Go on adding 7, 9, 11, 13, 15, 17, 19 respectively to obtain the next number.

So, 135 is wrong.

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11.

A, B and C can complete a piece of work in 24, 6 and 12 days respectively. Working together, they will complete the same work in:

A.
1 day
24
B.
7 day
24
C.
3 3 days
7
D.
4 days

Your Answer: Option (Not Answered)

Correct Answer: Option C

Explanation:

Formula: If A can do a piece of work in n days, then A's 1 day's work = 1 .
n

(A + B + C)'s 1 day's work = 1 + 1 + 1 = 7 .
24 6 12 24

Formula: If A's 1 day's work = 1 , then A can finish the work in n days.
n

So, all the three together will complete the job in 24 days = 3 3 days.
7 7

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12.

4 men and 6 women can complete a work in 8 days, while 3 men and 7 women can complete it in 10 days. In how many days will 10 women complete it?

A.
35
B.
40
C.
45
D.
50

Your Answer: Option (Not Answered)

Correct Answer: Option B

Explanation:

Let 1 man's 1 day's work = x and 1 woman's 1 day's work = y.

Then, 4x + 6y = 1 and 3x + 7y = 1 .
8 10

Solving the two equations, we get: x = 11 , y = 1
400 400

1 woman's 1 day's work = 1 .
400

10 women's 1 day's work = 1 x 10 = 1 .
400 40

Hence, 10 women will complete the work in 40 days.

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13.

A works twice as fast as B. If B can complete a work in 12 days independently, the number of days in which A and B can together finish the work in :

A.
4 days
B.
6 days
C.
8 days
D.
18 days

Your Answer: Option (Not Answered)

Correct Answer: Option A

Explanation:

Ratio of rates of working of A and B = 2 : 1.

So, ratio of times taken = 1 : 2.

B's 1 day's work = 1 .
12

Therefore A's 1 day's work = 1 ; (2 times of B's work)
6

(A + B)'s 1 day's work = ( 1 + 1 ) = 3 = 1 .
6 12 12 4

So, A and B together can finish the work in 4 days.

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Direction (for Q.No. 14):
Find out the wrong number in the given sequence of numbers.
14.

6, 13, 18, 25, 30, 37, 40

A.
25
B.
30
C.
37
D.
40

Your Answer: Option (Not Answered)

Correct Answer: Option D

Explanation:

The differences between two successive terms from the beginning are 7, 5, 7, 5, 7, 5.

So, 40 is wrong.

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15.

A, B and C jointly thought of engaging themselves in a business venture. It was agreed that A would invest Rs. 6500 for 6 months, B, Rs. 8400 for 5 months and C, Rs. 10,000 for 3 months. A wants to be the working member for which, he was to receive 5% of the profits. The profit earned was Rs. 7400. Calculate the share of B in the profit.

A.
Rs. 1900
B.
Rs. 2660
C.
Rs. 2800
D.
Rs. 2840

Your Answer: Option (Not Answered)

Correct Answer: Option B

Explanation:

For managing, A received = 5% of Rs. 7400 = Rs. 370.

Balance = Rs. (7400 - 370) = Rs. 7030.

Ratio of their investments = (6500 x 6) : (8400 x 5) : (10000 x 3)

   = 39000 : 42000 : 30000

   = 13 : 14 : 10

B's share = Rs. 7030 x 14 = Rs. 2660.
37

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16.

Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number.

A.
3
B.
10
C.
17
D.
20

Your Answer: Option (Not Answered)

Correct Answer: Option A

Explanation:

Let the number be x.

Then, x + 17 = 60
x

x2 + 17x - 60 = 0

(x + 20)(x - 3) = 0

x = 3.

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17.

A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?

A.
10
B.
20
C.
21
D.
25

Your Answer: Option (Not Answered)

Correct Answer: Option C

Explanation:

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x - 7 x 9 litres = 7x - 21  litres.
12 4

Quantity of B in mixture left = 5x - 5 x 9 litres = 5x - 15  litres.
12 4

7x - 21
4
= 7
5x - 15  + 9
4
9

28x - 21 = 7
20x + 21 9

252x - 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 litres of A.

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18.

Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, the price of the third variety per kg will be:

A.
Rs. 169.50
B.
Rs. 170
C.
Rs. 175.50
D.
Rs. 180

Your Answer: Option (Not Answered)

Correct Answer: Option C

Explanation:

Since first and second varieties are mixed in equal proportions.

So, their average price = Rs. 126 + 135 = Rs. 130.50
2

So, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find x.

By the rule of alligation, we have:

Cost of 1 kg of 1st kind Cost of 1 kg tea of 2nd kind
Rs. 130.50 Mean Price
Rs. 153
Rs. x
(x - 153) 22.50

x - 153 = 1
22.50

x - 153 = 22.50

x = 175.50

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19.

In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?

A.
810
B.
1440
C.
2880
D.
50400
E.
5760

Your Answer: Option (Not Answered)

Correct Answer: Option D

Explanation:

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.

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20.

In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?

A.
360
B.
480
C.
720
D.
5040
E.
None of these

Your Answer: Option (Not Answered)

Correct Answer: Option C

Explanation:

The word 'LEADING' has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

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