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A dealer owns a group of station wagons and motorcycles. If the number

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A dealer owns a group of station wagons and motorcycles. If the number  [#permalink]

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New post 26 Oct 2016, 22:31
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A dealer owns a group of station wagons and motorcycles. If the number of tires (excluding spare tires) on the vehicles is 30 more than twice the number of vehicles, then the number of station wagons the dealer owns is

A) 10
B) 15
C) 20
D) 30
E) 45

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Re: A dealer owns a group of station wagons and motorcycles. If the number  [#permalink]

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New post 26 Oct 2016, 23:53
Bunuel wrote:
A dealer owns a group of station wagons and motorcycles. If the number of tires (excluding spare tires) on the vehicles is 30 more than twice the number of vehicles, then the number of station wagons the dealer owns is

A) 10
B) 15
C) 20
D) 30
E) 45


Answer: B

Assume X station wagons, Y motorcycles

Framing the equation.

2(X+Y)+30 = 4X + 2Y
=> X = 15
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Re: A dealer owns a group of station wagons and motorcycles. If the number  [#permalink]

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New post 27 Oct 2016, 00:00
4b+2m = 30+ 2(b+m)

2b=30

b=15

IMO B


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Re: A dealer owns a group of station wagons and motorcycles. If the number  [#permalink]

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New post 27 Oct 2016, 00:07
Bunuel wrote:
A dealer owns a group of station wagons and motorcycles. If the number of tires (excluding spare tires) on the vehicles is 30 more than twice the number of vehicles, then the number of station wagons the dealer owns is

A) 10
B) 15
C) 20
D) 30
E) 45


IMO B
Let the no. of station wagons be 'x' and no.of motorcycles be 'y'.
no. of tires on station wagon excluding spare tire is 4 and no. of tires on motorcycle excluding spare tire is 2.
Total number of tires at the dealer's depot is 4x+2y.
Given No. of Tires on the vehicles is 30 more than twice the number of vehicles,i.e., 30+2(x+y)=4x+2y
solving we get x=15
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Re: A dealer owns a group of station wagons and motorcycles. If the number  [#permalink]

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New post 27 Oct 2016, 10:06
Bunuel wrote:
A dealer owns a group of station wagons and motorcycles. If the number of tires (excluding spare tires) on the vehicles is 30 more than twice the number of vehicles, then the number of station wagons the dealer owns is

A) 10
B) 15
C) 20
D) 30
E) 45

Attachment:
Capture.PNG
Capture.PNG [ 5.27 KiB | Viewed 822 times ]

4w + 2c = 2 (w + c ) + 30

Or, 4w + 2c = 2w + 2c + 30

Or, 2w = 30

So, w = 15

Hence, the number of station wagons the dealer owns is (B) 15

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Re: A dealer owns a group of station wagons and motorcycles. If the number  [#permalink]

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New post 05 Dec 2017, 18:15
1
Bunuel wrote:
A dealer owns a group of station wagons and motorcycles. If the number of tires (excluding spare tires) on the vehicles is 30 more than twice the number of vehicles, then the number of station wagons the dealer owns is

A) 10
B) 15
C) 20
D) 30
E) 45


To start we can define some variables:

w = the number of station wagons

m = the number of motorcycles

Since there are 4 tires on a station wagon and 2 tires on a motorcycle, the number of tires per station wagon is 4w and the number of tires per motorcycle is 2m.

We are given that the number of tires (excluding spare tires) on the vehicles is 30 more than twice the number of vehicles so we can create the following equation:

4w + 2m = 30 + 2(w + m)

4w + 2m = 30 + 2w + 2m

2w = 30

w = 15

Thus, there are 15 station wagons at the dealership.

Answer: B
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Re: A dealer owns a group of station wagons and motorcycles. If the number  [#permalink]

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New post 24 Jan 2018, 20:13
Hi All,

Although the prompt does not state it, you are expected to know that each station wagon has 4 tires and each motorcycle has 2 tires. You might find it easiest to treat this algebraically:

W = number of station wagons
M = number of motorcycles

Total vehicles = W + M
Total tires = 4W + 2M

The number of tires is 30 more than twice the number of vehicles...

4W + 2M = 30 + 2(W+M)

Now we can combine like terms and simplify:

4W + 2M = 30 + 2W + 2M
2W = 30
W = 15

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Re: A dealer owns a group of station wagons and motorcycles. If the number &nbs [#permalink] 24 Jan 2018, 20:13
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