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At Ivan’s petting zoo, visitors can pay to ride yaks and zebras. A yak

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Bunuel
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At Ivan’s petting zoo, visitors can pay to ride yaks and zebras. A yak [#permalink] New post   14 Feb 2017, 01:39
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At Ivan’s petting zoo, visitors can pay to ride yaks and zebras. A yak ride costs $91, whereas a zebra ride costs just $28. In how many ways can a visitor spend at total of exactly $703 on yak and/or zebra rides at the petting zoo?

A. 0
B. 1
C. 2
D. 3
E. 4
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Re: At Ivan’s petting zoo, visitors can pay to ride yaks and zebras. A yak [#permalink] New post   14 Feb 2017, 02:21
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Bunuel wrote:
At Ivan’s petting zoo, visitors can pay to ride yaks and zebras. A yak ride costs $91, whereas a zebra ride costs just $28. In how many ways can a visitor spend at total of exactly $703 on yak and/or zebra rides at the petting zoo?

A. 0
B. 1
C. 2
D. 3
E. 4


Hi

91x + 28y = 703

7(13x + 4y) = 703

13x + 4y = 703/7 =/= an integer.

We are adding two multiples if 7, so the resultant also must be a multiple of 7, but 703 is not divisible by 7.

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Re: At Ivan’s petting zoo, visitors can pay to ride yaks and zebras. A yak [#permalink] New post   14 Feb 2017, 04:50
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since 703 is not divisible by 7,
answer -A

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Re: At Ivan’s petting zoo, visitors can pay to ride yaks and zebras. A yak [#permalink] New post   14 Feb 2017, 10:49
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Bunuel wrote:
At Ivan’s petting zoo, visitors can pay to ride yaks and zebras. A yak ride costs $91, whereas a zebra ride costs just $28. In how many ways can a visitor spend at total of exactly $703 on yak and/or zebra rides at the petting zoo?

A. 0
B. 1
C. 2
D. 3
E. 4


91y + 28z = 703

If y = 1 , 28z = 612 ( Not divisible by 28 )
If y = 2 , 28z = 521 ( Not divisible by 28 )
If y = 3 , 28z = 430 ( Not divisible by 28 )
If y = 4 , 28z = 339 ( Not divisible by 28 )
If y = 5 , 28z = 248 ( Not divisible by 28 )
If y = 6 , 28z = 157 ( Not divisible by 28 )
If y = 7 , 28z = 66 ( Not divisible by 28 )

Thus, answer will be (A) 0
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Re: At Ivan’s petting zoo, visitors can pay to ride yaks and zebras. A yak [#permalink] New post   15 Feb 2017, 16:17
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Bunuel wrote:
At Ivan’s petting zoo, visitors can pay to ride yaks and zebras. A yak ride costs $91, whereas a zebra ride costs just $28. In how many ways can a visitor spend at total of exactly $703 on yak and/or zebra rides at the petting zoo?

A. 0
B. 1
C. 2
D. 3
E. 4


If we let y = the number of yak rides and z = the number of zebra rides, we can create the following equation:

91y + 28z = 703

Factoring out a 7, we have:

7(13y + 4z) = 703

13y + 4z = 703/7

Since 13y + 4z MUST be an integer, and since 703/7 IS NOT an integer, there is no way that a visitor can ride a zebra and/or yak with exactly 703 dollars.

Answer: A
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Re: At Ivan’s petting zoo, visitors can pay to ride yaks and zebras. A yak [#permalink] New post   15 Feb 2017, 23:25
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Bunuel wrote:
At Ivan’s petting zoo, visitors can pay to ride yaks and zebras. A yak ride costs $91, whereas a zebra ride costs just $28. In how many ways can a visitor spend at total of exactly $703 on yak and/or zebra rides at the petting zoo?

A. 0
B. 1
C. 2
D. 3
E. 4


    • The cost to ride a yak = $\(91\)
    • The cost to ride a zebra = $\(28\)
    • Total amount we can spend = $\(703\)
    • Therefore we can write the equation as -
    • \(91y + 28z = 703\)
      • Where y is the number of times one rides a yak
      • and z is the number of times one rides a zebra.

    • \(28z\) is always even, and \(91y\) + Even = Odd, therefore, we can conclude that y will be odd.

    • Therefore, \(y\) can take the value \(1,3,5\) and \(7\)

    • If we plug in these values in the equation \(91y + 28z = 703\) we will find that none of the above 4 values gives us an integer value of z.

    • Hence there are \(0\) ways.


Correct Answer is Option A

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Re: At Ivan’s petting zoo, visitors can pay to ride yaks and zebras. A yak [#permalink] New post   25 Feb 2022, 08:38
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Re: At Ivan’s petting zoo, visitors can pay to ride yaks and zebras. A yak [#permalink]
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