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# In 1999, Diana read 10 English books and 7 French books. In 2000, she

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In 1999, Diana read 10 English books and 7 French books. In 2000, she  [#permalink]

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03 May 2018, 02:28
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45% (medium)

Question Stats:

70% (02:29) correct 30% (02:56) wrong based on 43 sessions

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In 1999, Diana read 10 English books and 7 French books. In 2000, she reads twice as many French books as English books. If 60% of the books that she read during the two years were French, how many books did she read in 2000?

A. 16
B. 26
C. 32
D. 39
E. 48

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Joined: 28 Feb 2018
Posts: 8
Re: In 1999, Diana read 10 English books and 7 French books. In 2000, she  [#permalink]

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03 May 2018, 02:35
1
60% is french of the total 100% all books.
(7+2x)/(17+3x) = .6
7+2X = 0.6(17+3X)
7+2X = 10.2 +1.8X
2x-1.8X= 10.2-7
.2x = 3.2
x = 16
oops the total books = 3x= 48

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In 1999, Diana read 10 English books and 7 French books. In 2000, she  [#permalink]

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04 May 2018, 14:12
Bunuel wrote:
In 1999, Diana read 10 English books and 7 French books. In 2000, she reads twice as many French books as English books. If 60% of the books that she read during the two years were French, how many books did she read in 2000?

A. 16
B. 26
C. 32
D. 39
E. 48

Priyanka8793 , belated welcome and nice solution! +1

French:English books during the year 2,000 = 2x:1x
Total of ratio parts: 3x. The answer must be a multiple of 3

Eliminate A, B, C

Try D) 39 books read during year 2000
3x = 39, so x, multiplier, = 13

# of French books: 2x = 26
# of English books: x = 13

Two-year total of all books?
English: 13 + 10 = 23
French: 26 + 7 = 33
Overall: 23 + 33 = 46

French/Total should = $$\frac{3}{5}$$

$$\frac{33}{46}$$. Stop. That fraction cannot equal $$\frac{3}{5}$$

If unsure, check. No need to divide.
-- estimate: $$\frac{3}{5}$$ of 45 = 27 French
-- or multiply (46 * 0.6) = 27.6 French
In answer D, French = 33. Too great.

Check? E) 48 total books in year 2000
3x = 48, so x = 16
French in 2000: 2x = 32
English in 2000: 1x = 16

Total E: 16 + 10 = 26
Total F: 32 + 7 = 39
Overall total: (26 + 39) = 65

$$\frac{39}{65} = \frac{3}{5} = 60$$ percent
That's a match.

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Re: In 1999, Diana read 10 English books and 7 French books. In 2000, she  [#permalink]

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08 Feb 2019, 07:54
Top Contributor
Bunuel wrote:
In 1999, Diana read 10 English books and 7 French books. In 2000, she reads twice as many French books as English books. If 60% of the books that she read during the two years were French, how many books did she read in 2000?

A. 16
B. 26
C. 32
D. 39
E. 48

In 2000, she read twice as many French books as English books.
Let x = number of English books read in 2000
So, 2x = number of French books read in 2000

If 60% of the books that she read during the two years were French, how many books did she read in 2000? ​
Diana read 17 books (10 English and 7 French) in 1999, and she read 3x books (x English and 2x French) in 2000
So, 17 + 3x = the TOTAL number of books she read in two years

FRENCH BOOKS
Diana read 7 French books in 1999 and 2x French books in 2000
So, she read a TOTAL of 2x + 7 French books

If French books comprise 60% of the books she read in 2 years, we can write: (2x + 7)/(17 + 3x) = 60/100
Cross multiply to get: 100(2x + 7) = 60(17 + 3x)
Expand both sides to get: 200x + 700 = 1020 + 180x
Subtract 180x from both sides to get: 20x + 700 = 1020
Subtract 700 from both sides to get: 20x = 320
Solve: x = 320/20 = 16

BE CAREFUL! x = number of English books read in 2000, and we want the TOTAL number of books read in 2000

3x = TOTAL number of books read in 2000
3(16) = 48, so Diana read a TOTAL of 48 books in 2000

Cheers,
Brent
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Re: In 1999, Diana read 10 English books and 7 French books. In 2000, she   [#permalink] 08 Feb 2019, 07:54
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