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Math Expert
Joined: 02 Sep 2009
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03 Aug 2018, 05:44
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Difficulty:

35% (medium)

Question Stats:

77% (01:44) correct 23% (01:42) wrong based on 43 sessions

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Sarah receives 10z coins in addition to what she already had. She now has 5y + 1 times as many coins as she had originally. In terms of y and z, how many coins did Sarah have originally?

A. $$10z(5y + 1)$$

B. $$\frac{5y+1}{10z}$$

C. $$\frac{2z}{y}$$

D. $$\frac{10}{5y + 1}$$

E. $$\frac{10}{5y + 10}$$

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03 Aug 2018, 06:06
1
Ans: C

X + 10z = (5y+1) X
X+ 10z = 5Xy + X
=X = 2z/y

Bunuel wrote:
Sarah receives 10z coins in addition to what she already had. She now has 5y + 1 times as many coins as she had originally. In terms of y and z, how many coins did Sarah have originally?

A. $$10z(5y + 1)$$

B. $$\frac{5y+1}{10z}$$

C. $$\frac{2z}{y}$$

D. $$\frac{10}{5y + 1}$$

E. $$\frac{10}{5y + 10}$$

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03 Aug 2018, 06:56
O + 10z = (5y + 1) * O
O + 10z = 5yO + O
10z = 5yO
O = 10z / 5y
O = 2z / y

Hence, C.
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Joined: 23 Aug 2017
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03 Aug 2018, 11:37
Lets say that A is the number of coins Sarah already had.

A + 10Z = (5Y + 1)A
A + 10Z = 5YA + A
10Z = 5YA
A = $$\frac{10Z}{5Y}$$
A = $$\frac{2Z}{Y}$$

Option C
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