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Contractor has 2 complete work within a fixed time. If he is late he has to pay fine. The first day it is Rs 200 and from then on it increases by 25 per day. He has to pay Rs 9450. How many days he got late.
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- answer is 21.....200,225,250,....,9250.
- 10 years agoHelpfull: Yes(4) No(1)
- 9450-200=9250
Now 2 nd day he will pay 225...then 250....up to 9250
Here common difference=25
First term =225
Last=9250
So 9250=225+(n-1)25
(N-1)25=9025
N-1=361
N=362
Delay=363 - 10 years agoHelpfull: Yes(2) No(6)
- the answer is 21
- 10 years agoHelpfull: Yes(1) No(1)
- 200+225+250+275+300+325+350+375+400+425+450+475+500+525+550+575+600+625+650+675+700
just add the answer will be 9450
- 9 years agoHelpfull: Yes(1) No(1)
- the solution is sum of ap
a=200 d=25 n=? sum=9450
if we use formula s=n/2[2a+(n-1)d]
then we get quadratic equation on solving it answer is 21 - 8 years agoHelpfull: Yes(1) No(0)
- can any one explain how 21 came?
- 9 years agoHelpfull: Yes(0) No(0)
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