Quantitative Aptitude - Arithmetic Ability Questions

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.

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.

0.127 expressed as a fraction = 127/1000

1/8 can also be expressed as (1 x 125) / (8 x 125) = 125/1000

The difference is 2/1000, which is 1/500

(20 x 18) men can complete the work in in one day.

one man's one day work = 1/360

(18 x 15) women can complete the work in 1 day

1 woman's one day work = 1/270

So, required ratio = 1270:1360= 4:3

Angle between hands of a clock

When the minute hand is behind the hour hand, the angle between the two hands at M minutes past H 'o clock

=> 30H-M5+M2 degrees

Here H = 4, M = 15 and the minute hand is behind the hour hand.

Hence the angle

 = 30[4-(15/5)]+15/2 = 30(1)+7.5 = 37.5 degrees

The amount of petrol left after 4 operations  

= 200 × 1-402004  

= 200 × 454  

= 200 × 256625  

= 81.92 litres

Hence the amount of kerosene = 200 - 81.92 = 118. 08 litres

It is explicitly given that all the 4 black balls are different, all the 3 red balls are different and all the 5 blue balls are different. Hence this is a case where all are distinct objects.

Initially let's find out the number of ways in which we can select the black balls. Note that at least 1 black ball must be included in each selection.

Hence, we can select 1 black ball from 4 black balls
or 2 black balls from 4 black balls.
or 3 black balls from 4 black balls.
or 4 black balls from 4 black balls.

Hence, number of ways in which we can select the black balls

= 4C1 + 4C2 + 4C3 + 4C4
= 24-1 ........(A)

Now let's find out the number of ways in which we can select the red balls. Note that at least 1 red ball must be included in each selection.

Hence, we can select 1 red ball from 3 red balls
or 2 red balls from 3 red balls
or 3 red balls from 3 red balls

Hence, number of ways in which we can select the red balls
= 3C1 + 3C2 + 3C3
=23-1........(B)

Hence, we can select 0 blue ball from 5 blue balls (i.e, do not select any blue ball. In this case, only black and red balls will be there)
or 1 blue ball from 5 blue balls
or 2 blue balls from 5 blue balls
or 3 blue balls from 5 blue balls
or 4 blue balls from 5 blue balls
or 5 blue balls from 5 blue balls.

Hence, number of ways in which we can select the blue balls
= 5C0 + 5C1 + 5C2 + … + 5C5
= 25..............(C)

From (A), (B) and (C), required number of ways
=  2524-123-1