Quantitative Aptitude - Arithmetic Ability Questions

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

Let the sum(principal) received by A and B are x and y.

(1+r) =  =

Then,  =

Hence, the ratio in which the sum is divided =121:100

Let the original number is 82 i.e 8 + 2 = 10 and when the digits are reversed i.e 28
The difference is 82 - 28 = 54 i.e, the number is decreased by 54.

Number of applicants = 5
On a day, only 1 leave is approved.
Now favourable events =  1 of 5 applicants is approved
Probability that Laxmi priya's leave is granted = 1/5.

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

Let number of days Charan can do the same work alone is 'd' days.

According to the given data,

$$\begin{gathered} 512 + 716 + 14d = 1 \\ 14d = 1 - 20 + 2148 \\ 14d = 48 - 4148 \\ d = 14 x 487 \\ d = 96 days \end{gathered}$$

Therefore, Charan alone can complete the work in 96 days.

Let the distance between home and school is 'x'.
Let actual time to reach be 't'.
Thus, x/4 = t + 4 ---- (1)
and x/6 = t - 30 -----(2)
Solving equation (1) and (2)
=> x = 98 mts.