Quantitative Aptitude - Arithmetic Ability Questions

17th June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)

Odd days in 1600 years = 0

Odd days in 300 years = (5 x 3) 1

97 years has 24 leap years + 73 ordinary years.

Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.

Jan. Feb. March April May June

(31 + 28 + 31 + 30 + 31 + 17) = 168 days

168 days = 24 weeks = 0 odd day.

Total number of odd days = (0 + 1 + 2 + 0) = 3.

Given day is Wednesday

This can be solved as

$$\begin{gathered} 20 1 - 4202 = 20 1 - 152 \\ = 20452 \\ = 20 1625 \\ = 645 \\ = 12.8 \end{gathered}$$

Water required = 4000 x 140 = 560000 lit = 560 cu.m (1 cu.m = 1000 lit)

Volume of tank = 18 x 12 x 8 = 1728 cu.m

Water of this tank will suffice for = 1728/560 = 3 days.

74.99% of 1299 + 8.98% of 1899 = ? 

$$\begin{gathered} ? = 75 x 1400100 + 9 x 1800100 \\ ? = 1050 + 162 \\ ? = 1212. \end{gathered}$$

As the work is same
M1D1 = M2D2

Here Let the number of men employed was M
=> 11M = 7(M+4)
=> 11M - 7M = 28
=> M = 7

PW = 100×Amount100+R×T

Let the rquired number be 'P'

According to the question,

$$\begin{gathered} 100 x P100 = 20 \\ => P = 20 x 100100 = 20 \end{gathered}$$

Hence, 100 percent of the number 20 is 20.

Now, you can also try for

20 percent of what number is 100

$$\begin{gathered} 20 x P100 = 100 \\ => P = 100 x 10020 = 500 \\ => P = 500 \end{gathered}$$

Hence, 20 percent of the number 500 is 100.

17x12x19.5x12x 2660 = 2660266 = 10