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# If Randy has twice as many coins as Alice, and if Maria has 7 times as

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Re: If Randy has twice as many coins as Alice, and if Maria has 7 times as [#permalink]
Bunuel wrote:
If Randy has twice as many coins as Alice, and if Maria has 7 times as many coins as Alice, what is the combined number of coins that all three of them have?

(1) Alice has 4 fewer coins than Randy.
(2) Maria has 20 more coins than Randy.

DS21245

Asked: If Randy has twice as many coins as Alice, and if Maria has 7 times as many coins as Alice, what is the combined number of coins that all three of them have?
R = 2A ; M = 7A; R: M : A = 2: 7: 1;
Let Randy (R) have 2x , Maria( M ) have 7x & Alice( A) have x coins
Q. R + M + A = ?

(1) Alice has 4 fewer coins than Randy.
A = R - 4
x = 2x - 4
x = 4
R = 8 ; M = 28 ; A = 4; R + M + A = 40
SUFFICIENT

(2) Maria has 20 more coins than Randy.
M = R + 20
7x = 2x + 20
x = 4
R = 8 ; M = 28 ; A = 4; R + M + A = 40
SUFFICIENT

IMO D
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Re: If Randy has twice as many coins as Alice, and if Maria has 7 times as [#permalink]
Given: Let coins with Alice =x; then Randy =2x; Maria = 7x & total = 10x

We need to find x.

Statement I. Alice has 4 fewer coins than Randy ==> Alice' coins count = Randy's coins count - 4
i.e. x = 2x -4 or x=4 SUFFICIENT

A or D

Statement II. Maria has 20 more coins than Randy ==> Maria's coin count = Randy's coins count + 20
i.e. 7x = 2x + 20 ==> 5x =20 ==> x =4 SUFFICIENT

Since both statements gave a unique answer, Hence D
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Re: If Randy has twice as many coins as Alice, and if Maria has 7 times as [#permalink]
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Re: If Randy has twice as many coins as Alice, and if Maria has 7 times as [#permalink]
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