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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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Math Expert
Joined: 02 Sep 2009
Posts: 64134
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:

77% (02:21) correct 23% (02:36) wrong based on 83 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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Posts: 3848
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.

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

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20 Sep 2019, 01:36
let total cars+ bikes= 10
given
cars= 1/4 * bike
so 5bike= 40 ; bike = 8 and cars = 2
200 cc bikes ; 8/3
for given options put X=10
E ; 4X/15 ; 4*10/15 ; 8/3 sufficient

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
Intern
Joined: 14 Jan 2020
Posts: 2
Re: A total of X bikes and cars were sold by a dealer. If the number of ca  [#permalink]

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09 Feb 2020, 20:20
Let total bikes be 12
No of cars=1/4*12=3 cars
No of 200 cc bikes =1/3*12=4
X=12+3=15
In terms of X=4X/15
Manager
Joined: 14 Sep 2019
Posts: 222
Re: A total of X bikes and cars were sold by a dealer. If the number of ca  [#permalink]

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09 Feb 2020, 21:41
Let, the number of cars = C and the number of bikes = B
As per question, C + B = X, or, C = X - B
C = (1/4) *B
or, X – B = (1/4)*B
or, X = (1/4)*B +B = (5/4)* B
or, B = (4/5) *X

The number of 200cc bikes sold = (4/5)*x*(1/3) = 4x/15(E)
Intern
Joined: 26 Jun 2017
Posts: 24
Re: A total of X bikes and cars were sold by a dealer. If the number of ca  [#permalink]

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24 May 2020, 06:30
Number of Bikes is 4 time Number of Cars

4B = C

so when we assume that X is 100 then

No. of Bikes = 80
No. of Cars = 20

No of 200cc Bikes is 80/3

so put the value of X i.e 100 in the options and look for the option that becomes 80/3 when simplified.

only option e gives the value 80/3
Re: A total of X bikes and cars were sold by a dealer. If the number of ca   [#permalink] 24 May 2020, 06:30