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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.
Re: If Randy has twice as many coins as Alice, and if Maria has 7 times as
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23 May 2020, 06:42
If Randy has twice as many coins as Alice, R = 2(A) ----(1) and if Maria has 7 times as many coins as Alice, M=7(A) -----(2)
what is the combined number of coins that all three of them have?
(1) Alice has 4 fewer coins than Randy. A=R-4 ----(3) Putting value in equation 1 R=2(R-4) R=2R-8 8=R Randy have 8 coins and Alice have 4. What about Maria? (Not sufficient)
(2) Maria has 20 more coins than Randy M=R+20 What about Alice (Not sufficient)
Statement 1 and 2: A=4, R=8, M=R+20 => M=8+20 => 28 Total coins = A+R+M = 8+4+28= 40 coins (sufficient)
Re: If Randy has twice as many coins as Alice, and if Maria has 7 times as
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23 May 2020, 21:39
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
Let the no. of coins that Alice has = x. => Randy has 2x coins and Maria has 7x coins.
Total no. of coins = x + 2x + 7x = 10x
Statement 1: Alice has 4 fewer coins than Randy. =>2x- x = 4 => x = 3. => 10x = 30 coins in total. Thus we are able to obtain the value of the total no. of coins that all 3 of them have.
So Sufficient.
Statement 2: Maria has 20 more coins than Randy. => 7x = 2x + 20 => 5x = 20 => x = 4, 10x = 40. Again we have the total no. of coins all 3 of them have.
Re: If Randy has twice as many coins as Alice, and if Maria has 7 times as
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24 May 2020, 10:05
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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
Given: Randy has twice as many coins as Alice, and if Maria has 7 times as many coins as Alice
Target question:What is the combined number of coins that all three of them have? This is a good candidate for rephrasing the target question.
Let A = the number of coins Alice has So 2A = the number of coins Randy has And 7A = the number of coins Maria has So, the COMBINED number of coins they have = A + 2A + 7A = 10A
REPHRASED target question:What is the value of 10A?
Aside: the video below has tips on rephrasing the target question
Statement 1: Alice has 4 fewer coins than Randy. In other words: (the number of coins Alice has) = (the number of coins Randy has) - 4 Substitute to get: A = 2A - 4 Solve to get: A = 4, which means 10A = 40 Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT
Statement 2: Maria has 20 more coins than Randy. In other words: (the number of coins Maria has) = (the number of coins Randy has) + 20 Substitute to get: 7A = 2A + 20 Solve to get: A = 4, which means 10A = 40 Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT
Answer: D
Cheers, Brent VIDEO ON REPHRASING THE TARGET QUESTION:
Re: If Randy has twice as many coins as Alice, and if Maria has 7 times as
[#permalink]
24 May 2020, 10:15
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
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