A man is employed to count Rs. 10710. He counts at the rate of Rs. 180 per 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.

Academic Mathematics NCERT Class 10

Given:

A man is employed to count Rs. 10710. He counts at the rate of Rs. 180 per minute for half an hour. After this he counts at the rate of Rs. 3 less every minute than the preceding minute.
To do:
We have to find the time taken by him to count the entire amount.

Solution:

Total amount to be counted \( =Rs.\ 10710 \).

Amount counted for first half an hour at the rate of \( Rs.\ 180 \) per minute \( = Rs.\ 180 \times 30= Rs.\ 5400 \).
Remaining amount to be counted $=Rs.\ (10710-5400) = Rs.\ 5310$
After half an hour, he counts at the rate of Rs. 3 less every minute than the preceding minute.

Rate of counting per minute each successive minute is,

$= (180−3),(180−3−3), (180−3−3−3).....$

$=177, 174, 171.....$

This form an A.P. whose first term =177 and difference =−3

Let \( a \) be the first term and \( d \) the common difference.

$a=180-3=177, d=-3$ and $S_{n}=5310$

We know that,

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow 5310=\frac{n}{2}[2 \times 177+(n-1) \times-3]$

$10620=n[354-3 n+3]$
$10620=n(357-3 n)$

$10620=357 n-3 n^{2}$

$3 n^{2}-357 n+10620=0$

$3(n^2-119n+3540)=0$

$n^{2}-119 n+3540=0$

$n^{2}-59 n-60 n+3540=0$
$n(n-59)-60(n-59)=0$
$(n-60)(n-59)=0$

$n=60$ or $n=59$

$59<60$ (59 minutes comes earlier than 60 minutes)

$\therefore n=59$, the total time required $=59+30=89$ minutes.

The time taken by him to count the entire amount is 89 minutes.

Simply Easy Learning

Updated on: 10-Oct-2022

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