1
$\begingroup$

Anne and Hestor are both reading an 1100 page historical novel. Anne reads 50 pages a day. Hestor reads 10 pages the first day , 20 the second, 30 the third and so on. After how many days will they be on the same page. When I solve this problem I get 9 days. The books says the answer is 14 days. Which answer is correct?

$$50 \times 9 = 5(9)^2 + 5(9) = 450$$

$\endgroup$

3 Answers 3

3
$\begingroup$

$10+20+30+40+50+60+70+80+90=450=9\times 50$, so you are correct.

$\endgroup$
3
  • 3
    $\begingroup$ Downvoters should note that these sort of answers are also welcome on this site. Sometimes these sort of answers somehow look more beautiful. $\endgroup$ Mar 12, 2016 at 17:45
  • 1
    $\begingroup$ What about $$10+20+30+40+50+60+70+80+90\\=\\50+50+50+50+50+50+50+50+50$$ ? $\endgroup$
    – user65203
    Mar 12, 2016 at 17:47
  • 1
    $\begingroup$ @YvesDaoust Everything in moderation, including moderation. $\endgroup$ Mar 12, 2016 at 17:49
2
$\begingroup$

You are correct. The pages that Anne reads after k days is given by: $50k$.

The pages that Hector reads after k days is given by: $10\cdot \sum\limits_{n=1}^{k} n$ = $10 \cdot \frac{k(k+1)}{2}$ = $5k(k+1)$.

If you set these two equal, you will get $10 = k+1$, and therefore k = 9, meaning after 9 days they will have read the same amount.

$\endgroup$
2
$\begingroup$

Arithmetic progression for Hestor:

$$a_1=10\;,\;\;d=10\implies a_n=10+(n-1)10=10n$$

Thus the question is when

$$\overbrace{50n}^{\text{No. of pages Anne read}}=\overbrace{S_n}^{\text{No. of pages Hestor read}}=\frac n2(2a_1+(n-1)d)\implies 100=20+(n-1)10\implies$$

$$n-1=8\implies n=9$$

You are correct, the book is wrong...or you read the wrong question's answer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.