Quantitative Aptitude - Arithmetic Ability Questions

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.

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.

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.

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

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.

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

As we know that Speed = Distance / Time
for average speed = Total Distance / Total Time Taken
Thus, Total Distance = 325 + 470 = 795 km
Thus, Total Speed = 7.5 hrs
Average Speed = 795/7.5 => 106 kmph.