Quantitative Aptitude - Arithmetic Ability Questions

Area of a square SxS = 1024 => S = 32 cm
Length of the rectangle = 32x2 cm = 64 cm
Breadth of the rectangle = 32 - 12 = 20 cm
Required ratio = 64 : 20 = 16 : 5

Compound ratio of A:B:C
A:B = 3:4
B:C = 5:6
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A:B:C = 15:20:24

We can divide Rs.118000 in this ratio.

Share of A = 15/59 x 118000 = 30000

Share of B = 20/59 x 118000 = 40000

Share of C = 24/59 x 118000 = 48000

Let average for 10 innings be x. Then,

(10x + 108)/11 = x + 6 => 11x + 66 = 10x + 108  =>  x = 42.

New average = (x + 6) = 48 runs.

As given in the question, Marked price is 25% more than the Cost price.

=> C.P of the article = 34 x 400 = 300

Now, 

Let the original S.P of the article be Rs. P

Now the new S.P = P + 16.666 +13.333300 x P

=> S.P = 7P6

According to the question,

7P6 - 300 = 2P - 300

=> 5P = 1800

=> P = Rs. 360

Hence, the increased S.P = 360 x 7/6 = Rs. 420. 

Sum of the decimal places = 7 .

Since  the last digit to the extreme right  wil be zero ( 5 * 4 = 20),  so there will be  6  significant digits to the right of the decimal point....

1 Oct, 1994 = (1993 years + Period from 1.1.1993 to 1.10.1993)

Odd days in 1600 years = 0

Odd days in 300 years = 1

93 years = (70 ordinary years + 23 leap year) = (70 x 1 + 23 x 2)= 4 odd days

Jan  Feb  March  April  May  June  July  Aug  sept  Oct

(31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 1 ) = 274 days

274 days = (39 weeks + 1 day) =1 odd day.

Total number of odd days = (0 + 1 + 4 + 1) = 6 odd days.

Given day is Saturday.

To cover a distance of 800 kms using a 800 cc engine, the amount of diesel required = 80 litres.

=> the amount of diesel required to cover a distance of 1000 kms = 1000 x 80/800 = 100 litres

However, the vehicle uses a 1200 cc engine and it is given that the amount of diesel required varies directly as the engine capacity.

i.e., for instance, if the capacity of engine doubles, the diesel requirement will double too.

Therefore, with a 1200 cc engine, quantity of diesel required = 1200/800 x 100 = 150 litres.

We know that

Time = Distance/speed
Required time = (10 1/4)/2 + (10 1/4)/6
= 41/8 + 41/6
= 287/24 = 11.9 hours.