Quantitative Aptitude - Arithmetic Ability Questions

Let average expenditure was Rs. R

13 x 79 + 6x(R + 4) = 19R

=> R = Rs. 80.84

Total money = 19 x 80.84 = Rs. 1536.07.

let 't' be the time after which they met since L starts.

Given K is 50% faster than L

50 t + 1.5*50(t-1) = 300

50 t +75 t = 300 + 75

t = 375 / 125 = 3 hrs past the time that L starts

So they meet at (9 + 3)hrs = 12:00 noon.

 S.I=Rs.2600*203*1100*t

 S.I=Rs.5203*t

 t=3

Ayush's present age = 10 years.
His mother's present age = (10 + 20) years = 30 years.
Ayush's father's present age = (30 + 5) years = 35 years.
Ayush's father's age at the time of Ayush's birth = (35 - 10) years = 25 years.

Therefore Ayush's father's age at the time of marriage = (25 - 2) years = 23 years.

Amount = Rs(30000 + 4347) = Rs.34347.

Let the time be n years. Then,

 300001+7100n=34347

107100n=3434730000=1144910000=1071002

n= 2years.

Digits are 0,1,2,3,4,5,6,7,8,9. So no. of digits are 10

First all possible case => 9(0 excluded) x 10 x 10 = 900

Second no repetition allowed =>9 x 9 x 8 = 648

Third all digits are same => 9 (111,222,333,444,555,666,777,888,999)

Three digit numbers where two of the three digits are same = 900 - 648 - 9 = 243 ;

Speeds of two trains = 30 kmph and 58 kmph

=> Relative speed = 58 - 30 = 28 kmph = 28 x 5/18 m/s = 70/9 m/s

Given a man takes time to cross length of faster train = 18 sec

Now, required Length of faster train = speed x time = 70/9  x  18 = 140 mts.

Sum =Rs.(50 x 100/2x5)

= Rs. 500.

Amount

=[Rs.500x(1+5/100)²]

=Rs(500x21/20x21/20).

=Rs. 551.25

C.I

= Rs. (551.25 - 500)

= Rs. 51.25