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# Company R's annual profit has increased by a constant amount each cale

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Re: Company R's annual profit has increased by a constant amount each cale [#permalink]
Company R's annual profit has increased by a constant amount each calendar year since 1985 - arithmetic progression.
If profit in 1985 = a(1), then every profit (on 7th year) in 1991 a(7)= a(1)+(7-1)*K =a(1)+6K=?

(1) In 1985 Company R's annual profit was $212,000; in 1989 Company R's annual profit was$242,000.

a(1)=$212,000 and a(5)=$242,000 - from this point you may stop calculations because when we know two numbers of arithmetic progression, we can find any other number. Suff

(2) Company R's annual profit has increased by $7,500 each year since 1985 K=$7,500 and we need to find a(7)=a(1)+6K - clearly Not Suff because we don't know starting ammount.

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Re: Company R's annual profit has increased by a constant amount each cale [#permalink]
niks18 wrote:
Bunuel wrote:
Company R's annual profit has increased by a constant amount each calendar year since 1985. What was Company R's annual profit in 1991?

(1) In 1985 Company R's annual profit was $212,000; in 1989 Company R's annual profit was$242,000.
(2) Company R's annual profit has increased by $7,500 each year since 1985. As the AMOUNT of profit increases every year hence the Total Profit is Compounded. Let $$p$$ be the profit in 1985. Time period $$=1991-1985=16$$ years. Let $$r$$ be the growth rate. Hence Profit in 1991 $$= p*(1+\frac{r}{100})^{16}$$ we need the value of $$p$$ & $$(1+\frac{r}{100})$$ Statement 1: implies $$p=212,000$$ Profit in 1989 i.e $$4$$ years since 1985 will be $$=p*(1+\frac{r}{100})^4=242000$$ $$=>(1+\frac{r}{100})^4=\frac{242000}{212000}=\frac{121}{106}$$. Now raise both sides to the power $$4$$ to get, Hence $$(1+\frac{r}{100})^{16}=(\frac{121}{106})^4$$ This statement gives all the required values. Sufficient Statement 2: we know the increase in profit but we do not know the profit amount. hence $$p$$ & $$r$$ cannot be calculated. Insufficient Option A May I know the thought behind using compound interest to calculate the amount when it clearly says that the profit has increased by a constant amount each year? Retired Moderator Joined: 25 Feb 2013 Posts: 892 Own Kudos [?]: 1577 [0] Given Kudos: 54 Location: India GPA: 3.82 Re: Company R's annual profit has increased by a constant amount each cale [#permalink] tanya0630 wrote: niks18 wrote: Bunuel wrote: Company R's annual profit has increased by a constant amount each calendar year since 1985. What was Company R's annual profit in 1991? (1) In 1985 Company R's annual profit was$212,000; in 1989 Company R's annual profit was $242,000. (2) Company R's annual profit has increased by$7,500 each year since 1985.

As the AMOUNT of profit increases every year hence the Total Profit is Compounded.

Let $$p$$ be the profit in 1985. Time period $$=1991-1985=16$$ years. Let $$r$$ be the growth rate.

Hence Profit in 1991 $$= p*(1+\frac{r}{100})^{16}$$

we need the value of $$p$$ & $$(1+\frac{r}{100})$$

Statement 1: implies $$p=212,000$$

Profit in 1989 i.e $$4$$ years since 1985 will be $$=p*(1+\frac{r}{100})^4=242000$$

$$=>(1+\frac{r}{100})^4=\frac{242000}{212000}=\frac{121}{106}$$. Now raise both sides to the power $$4$$ to get,

Hence $$(1+\frac{r}{100})^{16}=(\frac{121}{106})^4$$

This statement gives all the required values. Sufficient

Statement 2: we know the increase in profit but we do not know the profit amount. hence $$p$$ & $$r$$ cannot be calculated. Insufficient

Option A

May I know the thought behind using compound interest to calculate the amount when it clearly says that the profit has increased by a constant amount each year?

Hi tanya0630

Apologies for the late reply. My previous solution was incorrect and hence I have edited the solution. It seems I had mistaken the profit increase with growth in profit.
I hope your doubt is clear now.

Thanks
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Re: Company R's annual profit has increased by a constant amount each cale [#permalink]
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Re: Company R's annual profit has increased by a constant amount each cale [#permalink]
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