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Problem Solving — algebra and arithmetic
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by Noman » Sun Jan 31, 2016 8:47 pm
Max can have his mobile phone repaired for $4000, or he can trade it in & receive $2000 credit toward the purchase of new mobile that sells for $8400. if Max trades in his mobile phone, the cost of the purchasing the new mobile is what percent greater than the cost of having his current mobile phone repaired?
A.44 B.60 C.70 D.75 E. 80
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by sanju09 » Thu Feb 04, 2016 11:26 pm
Noman wrote:Max can have his mobile phone repaired for $4000, or he can trade it in & receive $2000 credit toward the purchase of new mobile that sells for $8400. if Max trades in his mobile phone, the cost of the purchasing the new mobile is what percent greater than the cost of having his current mobile phone repaired?
A.44 B.60 C.70 D.75 E. 80
The story tells us that Max has two options for the spending, spend $4000 or spend $6400; and the question simply asks, “6400 is what percent greater than 4000?”, let’s solve

[(6400 - 4000)/4000] × 100 = [(2400)/4000] × 100 = [spoiler]60% (B)[/spoiler]
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by sana.noor » Thu Feb 11, 2016 6:01 am
Its an easy one!
the cost of repair is $4000
if he purchase new one by giving the old one he will have to pay $2000 less than the original amount of new mobile= the price of new mobile is $8400
$8400-$2000= he has to pay $6400 which is $2400 more than the repair cost.
the percentage change will be = 2400/4000 * 100 = 60%
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by Brent@GMATPrepNow » Thu Feb 11, 2016 7:42 am
For additional practice, try this similar question - https://www.beatthegmat.com/ann-s-new-bi ... 72691.html

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