Quantitative Aptitude - Arithmetic Ability Questions

Principal = Rs.16,000;

Time=9 months = 3 quarters;

Amount

=Rs.[16000x(1+5/100)³] =[16000x21/20x21/20x21/20]

= Rs.18522.

C.I

= Rs.(18522 - 16000)

= Rs.2522.

The distance is constant in this case.

Let the time taken for travel with a speed of 10 kmph be 't'.

Now the speed of 15 kmph is 3/2 times the speed of 10 kmph.

Therefore, time taken with the speed of 15 kmph will be 2t/3 (speed is inversely proportional to time)

Extra time taken = t - 2t/3 = t/3

=> 1pm - 11am = 2hrs

=> t/3 = 2h

=> t = 6 hrs.

Now, Distance = speed x time = 10 x 6 = 60 kms

Time he takes to reach at noon = 6 - 1 = 5 hrs

Now, Speed = 60/5 = 12 kmph.

Let the speed of the boat in still water is 'w'

Speed of the current is 'c'

Let the distance between two places is 'd'

According to the question, motorboat takes half time to cover a certain distance downstream than upstream.

dw+c = 12dw-c

=> 2w - 2c = w + c

=> w = 3c

=> c : w = 1 : 3

Hence, the ratio between rate of current(c) and rate of boat in still water(w) = 1 : 3

Speed of train relative to man = 67 - 7 = 60 km/hr.
= 60 x 5/18 = 50/3 m/sec.

Time taken to pass the man = 600 x 3/50 = 36 sec.

PW = 100×Amount100+R×T

Let the actual distance be t kms

Now, According to the data,

$$\begin{gathered} 5t + 860 = 6t \\ => 560t + 860 = 6t \\ => 60t + 8 = 72t \\ => 8 = 12t \\ => t = 8/12 = 2/3 hrs \end{gathered}$$

Now, the total distance = 6t = 6 x 2/3 = 2 x 2 = 4 kms.

Speed of tractor = 360/12 = 30 kmph

Speed of jeep = 250 x 30/100 = 75 kmph

But, given ratio of speed of car, jeep and tractor is 3 : 5 : 2

Speed of car = 3 x 75/5 = 45 kmph

Required, average speed of car and jeep = 75 + 45/2 = 60 kmph.

Let M litres milk be added

=> 60 + M40 = 31

=> 60 + M = 120

=> M = 60 lit.