Quantitative Aptitude - Arithmetic Ability Questions

Let the total book be 1

Then, in the 1st week 613th of book is read

i.e remaining pages =  1-613=713

in 2nd week he read 59th of it

i.e 59×713=35117
Remaining pages after 2nd week = 713-35117=28117

But given remaining pages after 2nd week = 100

28x117=100

--> x = 418 

 The total pages in the book are 418

Let the required number of days be x.

Less cows, More days (Indirect Proportion)

Less bags, Less days (Direct Proportion)

$$\begin{gathered} cows 1:40 :40:x \\ bags 40:1 :40:x \\ \end{gathered}$$

1 x 40 x x = 40 x 1 x 40

=>  x = 40.

Let x = the youngest child. Each of the other four children will then be x+3, x+6, x+9, x+12. 

We know that the sum of their ages is 50 .

so, x+(x+3)+(x+6)+(x+9)+(x+12) = 50

=>  x= 4

 The youngest child is 4 years old.

loss % = common gain or loss102 %

           = (20/10)x(20/10)% = 4%

Here, k = 7 and l = 28
Therefore, first number = klx100 % of second number = 7/28 x 100% of second number = 25% of second number.

Sol : Time = Circumference Relative Speed = 60010 = 60 sec.

Actually Arjun has to make a lead of 600 m, because when Arjun will be 600m ahead of Bhishma, they will be together again as a person when completes the total length it starts retracing he same path and thus Arjun and Bhishma can be together again.

Since , they make  a difference of 10 m in 1 second. so, he will create 600m difference in 60 second.

The third proportional of two numbers p and q is defined to be that number r such that

p : q = q : r.

Here, required third proportional of 10 & 20, and let it be 'a'

=> 10 : 20 = 20 : a

10a = 20 x 20

=> a = 40

Hence, third proportional of 10 & 20 is 40.

Let the speeds of the two trains be x m/sec and y m/sec respectively.

Then, length of the first train = 27 x meters, and

length of the second train = 17 y meters.

(27 x + 17 y) / (x + y) = 23

=> 27 x + 17 y = 23 x + 23 y

=> 4 x = 6 y

=> x/y = 3/2.