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A total of X bikes and cars were sold by a dealer. If the number of ca

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A total of X bikes and cars were sold by a dealer. If the number of ca  [#permalink]

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25 Sep 2018, 04:12
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Difficulty:

25% (medium)

Question Stats:

78% (02:02) correct 22% (01:59) wrong based on 32 sessions

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A total of X bikes and cars were sold by a dealer. If the number of cars is (1/4)th the number of bikes, and 1/3 of the bikes are 200cc bikes, how many 200cc bikes, in terms of X, were sold by the dealer?

(A) 3X/20

(B) X/5

(C) 7X/11

(D) 8X/15

(E) 4X/15

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A total of X bikes and cars were sold by a dealer. If the number of ca  [#permalink]

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25 Sep 2018, 17:58
Bunuel wrote:
A total of X bikes and cars were sold by a dealer. If the number of cars is (1/4)th the number of bikes, and 1/3 of the bikes are 200cc bikes, how many 200cc bikes, in terms of X, were sold by the dealer?

(A) 3X/20

(B) X/5

(C) 7X/11

(D) 8X/15

(E) 4X/15

In steps, algebraically, is probably easiest.
Find the # of ALL bikes sold in terms of X. $$\frac{1}{3}$$ of that "general" number of bikes = the # of 200cc bikes sold in terms of X

$$B$$ = # of bikes sold
$$C$$ = # of cars sold
The number of cars is $$\frac{1}{4}$$ the number of bikes:
$$C =\frac{1}{4}B$$

(1) B overall in terms of X
$$(B + C)=X$$
$$B + \frac{1}{4}B=X$$
$$\frac{5}{4}B=X$$
$$B=\frac{4}{5}X$$

(2) 200cc bikes in terms of X?
200cc bikes = 1/3 of ALL bikes, so
200cc bikes = $$\frac{1}{3}B$$

Bikes generally in terms of $$X$$:
$$B=\frac{4}{5}X$$

200cc bikes in terms of X:

$$\frac{1}{3}*B=(\frac{1}{3}*\frac{4}{5})*X$$

$$\frac{1}{3}B=\frac{4X}{15}$$

*$$B=\frac{4}{5}X$$
is an equation. We need $$\frac{1}{3}$$ of B on LHS. What we do to one side of the equation (multiply by $$\frac{1}{3}$$), we do to the other.
A total of X bikes and cars were sold by a dealer. If the number of ca &nbs [#permalink] 25 Sep 2018, 17:58
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