A man is employed to count Rs 10710 He counts at the rate of Rs 180/ minute for half an hour After this he counts at the rate of Rs 3 less every minute than the preceding minute Find the time taken by him to count the entire amount
2007-02-09 02:31:29 · 15 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
use diffrential equation to solve this
\
ans must b 30 hours
as
da/dt= 3
integrate
5310=3*t
t=29.5 hours
29.5+.5=30 hours
2007-02-09 02:42:33 · answer #1 · answered by n nitant 3 · 0⤊ 3⤋
The money for the first half hour is 5400(180*30)
let at any instant t the counting rate is 180-3t
so money counted at that instant is 180-3t
therefore if i sum this from time=0 to time=T
i should get the remaining amount equal to 5310
therefore Sigma(180-3t)=5310 (t from 0 to T)
we get an eqn in T as
T^2-119t+3540=0
solving for t we get T=59 minutes
so total time =30+59=89 minutes
Hope this is right
2007-02-12 22:39:48 · answer #2 · answered by Nikhil007 2 · 1⤊ 0⤋
At the end of 89 minutes he has counted all the 10710 when he counts 3 Rs per minute and this is the maximum amount of Rs he can count as the next minute he counts 0 Rs per minute
2007-02-09 02:40:45 · answer #3 · answered by Good Advice 2 · 0⤊ 1⤋
ans. is 89 mins.
since total amount to be counted =rs.10710
in first half an hour(30mins), he counted at the rate=rs. 180/min.
therefore, amount counted in half an hour=rs.180*30
=rs. 5400
acc to quest.
in first 31 mins he wud have counted rs. 5400+177
similarly, in 32 mins rs. 5400+177+174
and so on..............
we cn see that, it forms an progression
i.e., 177+174+171...........
the sum of the terms of a.p. will be the amount left 2 b counted after the 1st half an hour.
i.e.,rs. 10710-5400=rs.5310
and d sum of the terms of an a.p.=n/2[2(a)+(n-1)d]
where, a=first term=177
d=common difference= -3
n=no. of terms in d a.p.(to b calculated)
therefore, 5310=n/2[2(177)+(n-1)(-3)]
--->5310=n/2[354-3n+3]
--->5310=n/2[357-3n]
--->10620=357n-3n^2
--->3n^2-357n+10620=0
the rest of d calculation cn b done by d factrisation method.
the value of n will be=59 or 60
bt we will take d minimum time taken
hence n=59
therefore total time will be n+first half an hour
--->59+30 mins.
--->89 mins.
2007-02-09 23:16:44 · answer #4 · answered by $#Romeo Boy#$ 2 · 0⤊ 0⤋
he counts Rs. (180*30) i.e. Rs. 5400 in the first hour
after which he counts @Rs.3 less every minute..
=> {n[300-(n-1)3]}/2=5310 as the common difference in the arithmetic progression is -3
=>300n-3n^2+3n=10620
solve the quadratic eqn....
can't do it myself...no paper here nor a pen...
2007-02-09 02:45:12 · answer #5 · answered by s_d_sondhi 2 · 0⤊ 1⤋
the answer is '1'hr.
total amount=Rs 10710
1 minute=Rs180
30 min =Rs 180*30=Rs5400
money left to be counted after this is Rs(10710-5400)=Rs5310
this time
1 minute=Rs177 =(180-3)
Rs 5310=? minutes
=5310/177 =30 minutes
therefore the whole time taken is 30minutes+30 minutes= 1 hour
i think you got it.right!
2007-02-09 03:42:37 · answer #6 · answered by achyut s 1 · 0⤊ 2⤋
180*30=540
10710-540=10170
10170/177=57.46
57.46+30=87.46
87.46 Minutes = 87 Minutes and 27.6 Seconds
Regards
Praveen Kumar
2007-02-09 03:02:02 · answer #7 · answered by P Praveen Kumar 5 · 0⤊ 4⤋
The question is from the book of RD Sharma n the answer is 89 min
2014-02-05 04:09:14 · answer #8 · answered by Farooque Ali 1 · 0⤊ 0⤋
180*30=540
10710-540=10170
10170/177=57.46
57.46+30=87.46
Answer = 87 minutes 46 seconds
2007-02-09 02:37:42 · answer #9 · answered by JM 2 · 0⤊ 2⤋
180*30=5400
10710-5400=5310
if he lost 3 a min he would get out 5267 beofre he got to zero
so 5267+5310= 10577
he wouldnt make it he be 133 short
easy to do on excel.
2007-02-09 02:46:26 · answer #10 · answered by simsad31 2 · 0⤊ 1⤋