Quantitative Aptitude - Arithmetic Ability Questions

Maximum earning will be only when he will won on the maximum yielding table.

     A ----> 10:1

     B ----> 20:1

     C ----> 30:1

 i.e, he won on B and C but lost on A

 20 x 200 + 30 x 200 -1 x 200 = 9800

 minimum earning will be when he won on  table A and B and lose on that table 3.

Therefore, 10 x 200 + 20 x 200 - 1 x 200  =  6000-200 = 5800

Therefore, Difference= 9800 - 5800 = 4000

As per given 

 40100×250 =50100×x

⇒x=200

P = Rs. 15225, n = 9 months = 3 quarters, R = 16% p.a. per quarter.

Amount = 15225x1+41003

= (15225 x 26/25 x 26/25 x 26/25) = Rs. 17126.05

=> C.I. = 17126 - 15625 = Rs. 1901.05.

Length of the thread is 36 cms.
length wise twice breath wise once then totally thrice.
so 36/3 = 12.
so length is 12 x 2 = 24 cms
breath is 12 cms
Area of the book = length x breath = 24 x 12 = 288 sq.cm

Let the length of the train be 'L' mts

let the speed of the train be 'S' m/s

Given it crosses a pole in 10 sec=> L/S = 10 ......(1)
Given it takes 20 sec (double of pole) to cross a platform of length 200 mts

=> (L + 200)/S = 20

=> L/S + 200/S = 20

But from (1) L/S = 10

=> 200/S = 20 - 10

=> S = 20 m/s

Then, from (1)

=> L = 10 x 20 = 200 mts.

Hence, the length of the train = 200 mts.

2 * 5 = 10. Sum of decimal places =  4.

.002  * .5 = 0.0010 = 0.001 

60(x + 1) = 64x

X = 15

60 x 16 = 960 km

According to the given data,

P + Q = 41 ......(1)

R - 1 = P + 2

R = P + 3

and

P + 4 = Q - 1

=> Q = P + 5 ....(2)

From (1)&(2)

P = 18

Q = 18 + 5 = 23

R = 18 + 3 = 21

=> P/S = 3/4

=> S = 4/3 x 18 = 24

Required Difference = S - R = 24 - 21 = 3.