Quantitative Aptitude - Arithmetic Ability Questions

Let Siya's age = x

Then Riya's age = 4x

But given Riya's age = 24

=> 4x = 24 => x = 6

Hence Siya's age = 6 yrs

=> The age difference = 24 - 6 = 18 yrs.

Let the speed of the man in still water = p kmph

Speed of the current = s kmph

Now, according to the questions

(p + s) x 10 = (p - s) x 15

2p + 2s = 3p - 3s

=> p : s = 5 : 1

Hence, ratio of his speed to that of current = 5:1.

Ratio of efficiencies of P, Q and R = 2 : 3 : 4

From the given data,

Number of working days of P, Q, R = 5 : 10 : 5

Hence, ratio of amount of p, Q, R = 2x5 : 3x10 : 4x5 = 10 : 30 : 20

Amounts of P, Q, R = 200, 600 and 400.

The distance to be covered = Sum of their lengths = 200 + 300 = 500 m.

Relative speed = 72 -36 = 36 kmph = 36 x 5/18 = 10 mps.

Time required = d/s = 500/10 = 50 sec.

For solving this problem first we would break the whole range in 5 sections

1) From 1 to 9
Total number of zero in this range = 0

2) From 10 to 99
Total possibilities = 9*1 = 9 ( here 9 is used for the possibilities of a non zero integer)

3) From 100 to 999 - three type of numbers are there in this range
a) x00 b) x0x c) xx0 (here x represents a non zero number)
Total possibilities
for x00 = 9*1*1 = 9, hence total zeros = 9*2 = 18
for x0x = 9*1*9 = 81, hence total zeros = 81
similarly for xx0 = 81
total zeros in three digit numbers = 18 + 81 +81 = 180

4) From 1000 to 9999 - seven type of numbers are there in this range
a)x000 b)xx00 c)x0x0 d)x00x e)xxx0 f)xx0x g)x0xx
Total possibilities
for x000 = 9*1*1*1 = 9, hence total zeros = 9*3 = 27
for xx00 = 9*9*1*1 = 81, hence total zeros = 81*2 = 162
for x0x0 = 9*1*9*1 = 81, hence total zeros = 81*2 = 162
for x00x = 9*1*1*9 = 81, hence total zeros = 81*2 = 162
for xxx0 = 9*9*9*1 = 729, hence total zeros = 729*1 = 729
for xx0x = 9*9*1*9 = 729, hence total zeros = 729*1 = 729
for x0xx = 9*1*9*9 = 729, hence total zeros = 729*1 = 729
total zeros in four digit numbers = 27 + 3*162 + 3*729 = 2700
thus total zeros will be 0+9+180+2700+4 (last 4 is for 4 zeros of 10000)
= 2893

Let Price be Rs.100

1st CASE:

Then 100 x 70/100 = 70
=> price after discount = 100-70 = Rs.30

2nd CASE:

Then 100 x 40/100 = 40
=> price after discount = 100-40 = Rs.60
then
60 x 30/100 = 18
=> price after 2 consequent discount = 60-18 = Rs.42

=>from 1 and 2 cases, 70% discount is better.

P ( neither A nor B) = PA and B  

=  PA∩B=  PA∪B= 1-PA∪B  

= 1-61100=39100

250 numbers between 101 and 350 i.e. n(S)=250

n(E)=100th digits of 2 = 299−199 = 100

P(E)= n(E)/n(S) = 100/250 = 0.40