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

Suppose he buys each share for Rs. x.

Then, Rs.25*9100=x*1010 or x = Rs. 22.50.

Cost of each share = Rs. 22.50.

Cost price of the article is given by

= 400x100/(20+20)

= Rs.1000

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.

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

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

K's work in a day(1st day) = 1/30
L's work in a day(2nd day)= -1/60(demolishing)
hence in 2 days, combined work= 1/30 - 1/60 =1/60
since both works alternatively, K will work in odd days and L will work in even days.
1/60 unit work is done in 2 days
58/60 unit work will be done in 2 x 58 days = 116 days
Remaining work = 1-58/60 = 1/30
Next day, it will be K's turn and K will finish the remaining 1/30 work.
hence total days = 116 + 1 = 117.

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.