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.

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.

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

(Men 14 : 16) : (Hrs/Day 12 : 7) :: 48 : x
14 * 12 * x = 16 * 7 * 48 or x = 32

Let Q complete that work in 'L' days

=> 1L + 1L-5 = 9100

=> 9L2-245L+500=0

L = 25 days.

Clearly total persons are increased in  28/35 :: 52/65 = 4:5.

As time is inversely proportional to men, so total time will decrease in the ratio 5:4.

Hence, 22.5 x 4/5 = 18 days.

Let C.P. of each Magazine be Rs. 1

C.P. of P articles = Rs. P. 

S.P. of P articles = Rs. 40.

Profit = Rs. (40 - P)

Now, Gain% = ProfitC.P×100 = 40-PP×100 

Here 40-PP×100= 25

-->P = 32.

The given 3b - 7 < 32 can be solved as

3b - 7 < 32

3b < 39

b < 13.