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I usually solve SAT questions easily and fast, but this one got me thinking for several minutes and I cannot seem to find an answer.

Here it is:

In 1995, Diana read $10$ English and $7$ French books. In 1996, she read twice as many French books as English books. If 60% of the books that she read during the 2 years were French, how many English and French books did she read in 1996?

(A) $16$, (B) $26$, (C) $32$, (D) $39$, (E) $48$

Could you please help me by either giving hints or explaining how to solve the problem?

I cannot find, in any way, the number of books of any language she read in 1996. I have tried a lot of operations with percentages, but no results.

Sorry for this question, as I am not very good at Maths.

Thank you.

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    $\begingroup$ -.- I thought this question was about SAT. $\endgroup$
    – Bakuriu
    May 23, 2015 at 13:31
  • $\begingroup$ Oh, hahaha! :D I would have never thought. $\endgroup$
    – Veo
    May 23, 2015 at 20:55

5 Answers 5

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The total number of books in the second year (call it $x$) is a multiple of $3$, so only D and E remain. Now since "60% of the total number of books" should be a whole number, we find that $x+17$ should be a multiple of $5$. Now only E remains.

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  • $\begingroup$ Thanks. This is true SAT logic! :D $\endgroup$
    – Veo
    May 23, 2015 at 11:15
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I don't know what SAT is - can we use algebra?

Let $x$ = no. of French books in 1996.

French books = 0.6(total books)

$x + 7 = 0.6(x+0.5x+10+7)$

Then, solve for $x$, and $1.5x$ is total books in 1996

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  • $\begingroup$ SAT is the name of a test... $\endgroup$
    – user90667
    May 23, 2015 at 10:59
  • $\begingroup$ Yes, but what level test? $\endgroup$ May 23, 2015 at 10:59
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    $\begingroup$ Oh, sorry--it's a test taken by 11th grade students in the US, so yes, algebra should be OK. $\endgroup$
    – user90667
    May 23, 2015 at 11:04
  • $\begingroup$ A test that students usually sit for in grades 10-12 $\endgroup$
    – Veo
    May 23, 2015 at 11:04
  • $\begingroup$ @SMF Okay, we have a similarly named test in the UK called SATS, taken by much younger students - hence my confusion! $\endgroup$ May 23, 2015 at 11:05
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Total English books: $x=10+E$
Total French books: $y=7+F$

Where $E$ and $F$ are English and French books in 1996.

We know $F=2E$

Thus:

Total books: $T=x+y=17+E+F$

We know that $0.6t=y$ that is

$0.6(17+E+F)=7+F\implies 0.6(17+3E)=7+2E$ - solve this for $E$

Which is sufficient to work it out :)

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Total books in 1995 = 17

Let no. Of books studied in 1996 be x

Then no.of French books be 2x

Total books in 2 years = 2x+x+17= 3x+17

60%of total books are French= (3x+17)*6/10

French books in 1995 + French books in 1996 = 2x+7

By above two equations 2x+7=(3x+17)*6/10

Solving this we get x value 16

Total books 3x value is 48.

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@user99542: 48 isn't the total of books, because 16 are the English books read in the second year. (Read bnosnehpets' answer) Thus the total of English books is 10+16=26. And the total of French books is 7+32=39. Therefore the overall total of books is 65. A simple check: Total French books=0.6*65=39 Total English books=0.4*65=26

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