Quantitative Aptitude - Arithmetic Ability Questions

Assume first child (the youngest) get = Rs. x

According to the question ;
each son having Rs. 30 more than the younger one

Second child will get = Rs. x + 30
Third child will get = Rs. x + 30 + 30 = x + 60
Forth child will get = Rs. x + 30 + 30 + 30 = x + 90
Fifth child will get = Rs. x + 30 + 30 + 30 + 30 = x + 120

Total amount they got = Rs. 2000

x + (x+30) + (x+60) + (x+90) + (x+120) = 2000
5x + 300 = 2000
5x = 1700
x = Rs. 340
So the youngest child will get Rs. 340.

From the given data,

When the price of rice is decreased by 10%, the value of Rs. 270 would become

270 x 100/90 = 300

Now, the value 270 becomes 300

And given, the girl could buy 1 kg extra for that value

Change in value = 300 - 270 = Rs. 30

Therefore, 1 kg extra rice she can get for Rs. 30

=> The original price of rice = Rs. 30/kg.

As she gets only 9 kgs for Rs. 270 before 10% discount, now after discount she gets 10 kgs instead of 9 kgs for Rs. 270.  

1 boy's 1 day work = 1/70

1 girl's 1 day work = 1/140

(5 boys + 10 girls)'s 1 day work = (5/70 + 10/140) = (1/14 + 1/14) = 1/7

5 boys and 10 girls will complete the work in 7 days.

Given length of the rectangle = 3 cm

Breadth of the rectangle = 4 cm

Then, the diagonal of the rectangle D = 32 + 42 = 25 = 5

Then, it implies side of square = 5 cm

We know that Area of square = S x S = 5 x 5 = 25 sq.cm.

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

Let the price of 1 litre of diesel be Rs.x and let Karthik initially buy 'y' litres of diesel.
Therefore, he would have spent Rs. xy on diesel.

When the price of diesel increases by 25%, the new price per litre of diesel is 1.25x.

Karthik intends to increase the amount he spends on diesel by 15%.
i.e., he is willing to spend xy + 15% of xy = 1.15xy

Let the new quantity of diesel that he can get be 'q'.
Then, 1.25x x q = 1.15xy
Or q = (1.15/1.25)y = 0.92y.

As the new quantity that he can buy is 0.92y, he gets 0.08y lesser than what he used to get earlier.

Or a reduction of 8%.

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.