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A retailer originally bought 50 equally priced phones for a total of z

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A retailer originally bought 50 equally priced phones for a total of z  [#permalink]

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New post 24 Jan 2019, 02:51
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  45% (medium)

Question Stats:

50% (01:28) correct 50% (01:27) wrong based on 15 sessions

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A retailer originally bought 50 equally priced phones for a total of z dollars. If he sold each phone for 25% more than he paid for it, then in terms of z, how much was each phone sold for?

A z/50

B z/40

C 5z/4

D 4z/5

E 62.5z

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Re: A retailer originally bought 50 equally priced phones for a total of z  [#permalink]

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New post 24 Jan 2019, 03:01
Bunuel wrote:
A retailer originally bought 50 equally priced phones for a total of z dollars. If he sold each phone for 25% more than he paid for it, then in terms of z, how much was each phone sold for?

A z/50

B z/40

C 5z/4

D 4z/5

E 62.5z


IMO B

50 mobiles were bought for 5000 dollars, just chose a value
1 mobile will be bought for 100 dollars

Now he sold for each phone for 25% greater value 125

Plug value back for z in the options to get value of each phone

5000/40 = 125
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Re: A retailer originally bought 50 equally priced phones for a total of z  [#permalink]

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New post 24 Jan 2019, 03:02
Bunuel wrote:
A retailer originally bought 50 equally priced phones for a total of z dollars. If he sold each phone for 25% more than he paid for it, then in terms of z, how much was each phone sold for?

A z/50

B z/40

C 5z/4

D 4z/5

E 62.5z



Cost price of each set = z/50

Selling price = z/50 * 125/100

= z/40.

B is the correct answer.
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Re: A retailer originally bought 50 equally priced phones for a total of z  [#permalink]

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New post 24 Jan 2019, 04:07

Solution


Given:
    • Original number of phones bought = 50, at a total price of z dollars
    • Price of each phone is same
    • Each phone is sold at a price of 25% more than the price at which he bought it

To find:
    • Selling price of each phone

Approach and Working:
    • Cost price of each phone = \(\frac{z}{50}\)
    • Implies, selling price of each phone = \(\frac{z}{50}\) + 25% of \(\frac{z}{50} = 1.25 * \frac{z}{50} = \frac{z}{40}\)

Hence, the correct answer is Option B

Answer: B

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Re: A retailer originally bought 50 equally priced phones for a total of z  [#permalink]

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New post 24 Jan 2019, 04:17
Bunuel wrote:
A retailer originally bought 50 equally priced phones for a total of z dollars. If he sold each phone for 25% more than he paid for it, then in terms of z, how much was each phone sold for?

A z/50

B z/40

C 5z/4

D 4z/5

E 62.5z


original price z/50
sp = 2*1.25/50 ; z/40
IMO B
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Re: A retailer originally bought 50 equally priced phones for a total of z   [#permalink] 24 Jan 2019, 04:17
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