Aptitude

A car owner buys petrol at Rs.7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of petrol if he spends Rs. 4000 each year?

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Question: A car owner buys petrol at Rs.7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of petrol if he spends Rs. 4000 each year?
[A].

Rs. 7.98

[B].

Rs. 8

[C].

Rs. 8.50

[D].

Rs. 9

Answer: Option A

Explanation:

Total quantity of petrol
consumed in 3 years
= 4000 + 4000 + 4000  litres
7.50 8 8.50
= 4000 2 + 1 + 2  litres
15 8 17
= 76700  litres
51

Total amount spent = Rs. (3 x 4000) = Rs. 12000.

Average cost = Rs. 12000 x 51 = Rs. 6120   = Rs. 7.98
76700 767

The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class.

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Question:
The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class.

[A].

47.55 kg

[B].

48 kg

[C].

48.55 kg

[D].

49.25 kg

Answer: Option C

Explanation:

Required average
= 50.25 x 16 + 45.15 x 8
16 + 8
= 804 + 361.20
24
= 1165.20
24
= 48.55

A library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is:

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Question:
A library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is:

[A].

250

[B].

276

[C].

280

[D].

285

Answer: Option D

Explanation:

Since the month begins with a Sunday, to there will be five Sundays in the month.

Required average
= 510 x 5 + 240 x 25
30
= 8550
30
= 285

A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. What is the average age of the family?

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Question: A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. What is the average age of the family?
[A].

28  4 years
7

[B].

31  5 years
7

[C].

32  1 years
7

[D].

None of these

Answer: Option B

Explanation:

Required average
= 67 x 2 + 35 x 2 + 6 x 3
2 + 2 + 3
= 134 + 70 + 18
7
= 222
7
= 31  5 years.
7

Video Explanation: https://youtu.be/OXLnoItd0MA

In Arun’s opinion, his weight is greater than 65 kg but less than 72 kg. His brother doest not agree with Arun and he thinks that Arun’s weight is greater than 60 kg but less than 70 kg. His mother’s view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Arun?

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Question: In Arun’s opinion, his weight is greater than 65 kg but less than 72 kg. His brother doest not agree with Arun and he thinks that Arun’s weight is greater than 60 kg but less than 70 kg. His mother’s view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Arun?
[A].

67 kg.

[B].

68 kg.

[C].

69 kg.

[D].

Data inadequate

Answer: Option A

Explanation:

Let Arun’s weight by X kg.

According to Arun, 65 < X < 72

According to Arun’s brother, 60 < X < 70.

According to Arun’s mother, X <= 68

The values satisfying all the above conditions are 66, 67 and 68.

Required average = 66 + 67 + 68 = 201 = 67 kg.
3 3

A tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paise per sq. m, is:

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Question:
A tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paise per sq. m, is:

[A].

Rs. 456

[B].

Rs. 458

[C].

Rs. 558

[D].

Rs. 568

Answer: Option C

Explanation:

Area to be plastered = [2(l + b) x h] + (l x b)
= {[2(25 + 12) x 6] + (25 x 12)} m2
= (444 + 300) m2
= 744 m2.
Cost of plastering = Rs. 744 x 75 = Rs. 558.
100

An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is:

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Question:
An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is:

[A].

2%

[B].

2.02%

[C].

4%

[D].

4.04%

Answer: Option D

Explanation:

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

A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges?

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Question:
A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges?

[A].

20

[B].

24

[C].

30

[D].

33

Answer: Option C

Explanation:

Let the side of the square(ABCD) be x metres.

Then, AB + BC = 2x metres.

AC = 2x = (1.41x) m.

Saving on 2x metres = (0.59x) m.

Saving % = 0.59x x 100 % = 30% (approx.)
2x

What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?

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Question:
What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?

[A].

814

[B].

820

[C].

840

[D].

844

Answer: Option A

Explanation:

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.

Area of each tile = (41 x 41) cm2.

Required number of tiles = 1517 x 902 = 814.
41 x 41

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

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Question: A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?
[A].

2.91 m

[B].

3 m

[C].

5.82 m

[D].

None of these

Answer: Option B

Explanation:

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