Plugging in answer choices is probably the easiest way to solve this, but boxing506's algebraic solution is good too. I'll explain both.
METHOD ONE:
We're looking for the number of books she read in 2010, but we've also been told she read twice as many French as English. That tells us two things.
1) She's fancy.
2) The number of books she read in 2010 must be divisible by 3!
Notice only D and E are possible. So we're down to a 50/50 shot. Now let's plug in D. If she read 39 books in 2010, then 2/3 of them, or 26 of them, must've been French, and 1/3 of them, or 13, English. Add the books from 2009, and we find that she read 23 English and 33 French. That doesn't look like a nice 40 to 60 ratio to me, so it must be E.
Let's try it though. If she read 48 books in 2010, then she read 32 French and 16 English. Adding the books from 2009 gets us 39 French and 26 English. This does give us a 40 to 60 English to French ratio, so the answer must be E.
METHOD TWO:
As boxing506 pointed out, we've been told that 60% of the total books she read were French, so we know there's a 40% to 60% English to French ratio over the two years. So we can make a ratio. I'll call the books she read in 2010 E and F for English and French. Here's the ratio:
10 + E 40 7 + F = 60
But since we know that the French books doubled the English books, we can replace the F with 2E and solve:
10 + E 40 7 + 2E = 60
Crossmultiply:
60(10 + E) = 40(7 + 2E)
Expand:
600 + 60E = 280 + 80E
Collect terms:
320 = 20E
You get it:
E = 16
Since the French books doubled that, she must've read a total of 16 + 32 = 48 books. So answer E.
(Note to kkaur423: you plugged in x and got 65 but that's the total number of books she read both years. We just want 2010's books, so you shouldn't include the 17 books from 2009.)
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