Aptitude Questions Y7

61. A takes 3 min 45 seconds to complete a kilometre. B takes 4 minutes to complete the same 1 km track. If A and B were to participate in a race of 2 kms, how much start can A give B in terms of distance?



Solution: A can give B a start of 15 seconds in a km race.
B takes 4 minutes to run a km. i.e = 250 m/min = m/sec
Therefore, B will cover a distance of = 62.5 meters in 15 seconds.
The start that A can give B in a km race therefore, is 62.5 meters, the distance that B

run in 15 seconds. Hence in a 2 km race, A can give B a start of 62.5 * 2 = 125 m or

30 seconds.



62. P can give Q a start of 20 seconds in a kilometer race. P can give R a start of 200 meters in the same kilometer race. And Q can give R a start of 20 seconds in the same kilometer race. How long does P take to run the kilometer?

Solution:
P can give Q a start of 20 seconds in a kilometer race. So, if Q takes 'x' seconds to run a kilometer, then P will take x – 20 seconds to run the kilometer.

Q can give R a start of 20 seconds in a kilometer race. So, if R takes 'y' seconds to run a kilometer, then Q will take y – 20 seconds to run the kilometer.

We know Q takes x seconds to run a kilometer
Therefore, x = y – 20

Therefore, P will take x – 20 = y – 20 – 20 = y – 40 seconds to run a kilometer.

i.e. P can give R a start of 40 seconds in a kilometer race, as R takes y seconds to run a kilometer and P takes only y – 40 seconds to run the kilometer.

We also know that P can give R a start 200 meters in a km race.
This essentially means that R runs 200 meters in 40 seconds.
Therefore, R will take 200 seconds to run a km.

If R takes 200 seconds to run a km, then P will take 200 – 40 = 160 seconds to run a km.



63. How many squares can be formed using the checkered 1 * 1 squares in a normal chessboard?

Solution:

The number of squares that can be formed using the 1 * 1 checkered squares of a chess board are given by the relation 12 + 22 + 32 + 42 + ... + 82 = 204

64. A and B enter in to a partnership and A invests Rs. 10,000 in the partnership. At the end of 4 months he withdraws Rs.2000. At the end of another 5 months, he withdraws another Rs.3000. If B receives Rs.9600 as his share of the total profit of Rs.19,100 for the year, how much did B invest in the company?

Solution:

The total profit for the year is 19100. Of this B gets Rs.9600. Therefore, A would

get (19100 – 9600) = Rs.9500.
The partners split their profits in the ratio of their investments.

Therefore, the ratio of the investments of A : B = 9500 : 9600 = 95 : 96.

A invested Rs.10000 initially for a period of 4 months. Then, he withdrew Rs.2000.
Hence, his investment has reduced to Rs.8000 (for the next 5 months).
Then he withdraws another Rs.3000. Hence, his investment will stand reduced to Rs.5000

during the last three months.


So, the amount of money that he had invested in the company on a money-month basis

will be = 4 * 10000 + 5 * 8000 + 3 * 5000 = 40000 + 40000 + 15000 = 95000
If A had 95000 money months invested in the company, B would have had 96,000

money months invested in the company (as the ratio of their investments is 95 : 96).


If B had 96,000 money-months invested in the company, he has essentially invested

96000/12 = Rs.8000



65. Four horses are tethered at 4 corners of a square field of side 70 metres so that they just cannot reach one another. The area left ungrazed by the horses is:

Sol: The length of the rope in which the horses tied should be equal to half of the side of the

square plot so that they just cannot reach one another.

Therefore, the length of the rope is 35m (70/2).

The area covered by each horse should be equal to the area of sector with radius of 70/2 =

35m(length of the rope).

Total area covered by the four horses = 4* area of sector of radius 35 metres = Area of

circle of radius 35m.

Area left ungrazed by the horses = Area of square field - Area covered by four horses.


= 702 - (22/7)*35*35 = 4900 - 3850 = 1050 sq.m.





66. The area of a square field is 24200 sq m. How long will a lady take to cross the field diagonally at the rate of 6.6 km/hr?



Sol: Let ‘a’ meters be the length of a side of the square field.

Therefore, its area = a2 square meters. --- (1)
We know that the length of the diagonal ‘d’ of a square whose side is ‘a’ meters =

a –-- (2)

From (1) and (2), we can deduce that the square of the diagonal = d2 = 2a2

Or d = meters.
The time taken to cross a length of 220 meters while traveling at 6.6 kmph is

given by (converting

1 km = 1000 meters and 1 hour = 60 minutes).

= 2 minutes



67. For what values of 'm' is y = 0, if y = x2 + (2m + 1)x + m2 - 1? x is a real number.

(1) m -2
(2) m < m =" 0" c =" 0)" d =" b2" mixture =" 0="" 2 =" 1" 1 =" 2" ratio =" 1" share =" 12" share =" 6"> A : B : C :: 3 : 2 : 4

The sum of the total wages = 3x + 2x + 4x = 432 => 9x = 432 or x = 48.
Hence A gets 3 * 48 = Rs. 144.