Quantitative Aptitude - Arithmetic Ability Questions

(24 + 31 + 18) days = 73 days = 1/5 year .

P = Rs 3000 and R = 18 % p.a.

 S.I=3000*18*15100

We have: l = 20 ft and lb = 680 sq. ft.So, b = 34 ft.

Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

Speed =[ 78 x ( 5/18 ) ] m/sec = 65/3 m/sec.

Time = 1 minute = 60 sec.

Let the length of the tunnel be x metres.

Then, [ (800 + x )/ 60 ]= 65/3

3(800 + x) = 3900

 x = 500.

Let work done by 1 man in i day be m

and Let work done by 1 boy in 1 day be b

From the given data,

4(5m + 3b) = 23

20m + 12b = 23....(1)

2(3m + 2b) = 7

6m + 4b = 7 ....(2)

By solving (1) & (2), we get

m = 1, b = 1/4

Let the number of required boys = n

6(7 1 + n x 1/4) = 45

=> n = 2.

Speed of train 1 = 48 kmph
Let the length of train 1 = 2x meter

Speed of train 2 = 42 kmph
Length of train 2 = x meter (because it is half of train 1's length)

Distance = 2x + x = 3x
Relative speed= 48+42 = 90 kmph = (90*5/18)  m/s = 25 m/s 

Time = 12 s

Distance/time = speed 

=>3x/12 = 25 (25*12)/3 

Length of the first train = 2x = 200 meter
Time taken to cross the platform= 45 s

Speed of train 1 = 48 kmph = 480/36 = 40/3 m/s

Distance = 200 + y     [where y is the length of the platform] 

x =100m => 200+y = 45* 40/3
y = 400m

We have P = 8748, R = -10 and n = 3
Therefore, the purchase price of the machine
= P1+R100n= 87481-101003
= 8748 x (100/90) x (100/90) x (100/90)
= 12,000.

Therefore, the purchase price of the machine was Rs. 12,000.