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At Ivan’s petting zoo, visitors can pay to ride yaks and zebras. A yak
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14 Feb 2017, 00:39
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67% (02:15) correct 33% (02:23) wrong based on 61 sessions
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Re: At Ivan’s petting zoo, visitors can pay to ride yaks and zebras. A yak
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14 Feb 2017, 01:21
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. AnswerA (0)



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Re: At Ivan’s petting zoo, visitors can pay to ride yaks and zebras. A yak
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14 Feb 2017, 03:50
since 703 is not divisible by 7, answer A Sent from my Mi 4i using GMAT Club Forum mobile app



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Re: At Ivan’s petting zoo, visitors can pay to ride yaks and zebras. A yak
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14 Feb 2017, 09:49
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
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15 Feb 2017, 15:17
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
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15 Feb 2017, 22:25
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 AThanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts
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Re: At Ivan’s petting zoo, visitors can pay to ride yaks and zebras. A yak &nbs
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