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7 times the number of coins that A has is equal to 5 times the number

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7 times the number of coins that A has is equal to 5 times the number  [#permalink]

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New post 15 Apr 2020, 07:48
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E

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62% (02:15) correct 38% (02:31) wrong based on 77 sessions

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7 times the number of coins that A has is equal to 5 times the number of coins B has while 6 times the number of coins B has is equal to 11 times the number of coins C has. What is the minimum number of coins with A, B and C put together?

110
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174


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Re: 7 times the number of coins that A has is equal to 5 times the number  [#permalink]

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New post 15 Apr 2020, 07:59
A:B = 5:7

B:C= 11:6

A:B:C = 5*11: 7*11: 6*7 = 55 :77 : 42

Minm of a+b+c = 55+77+42= 174




Bunuel wrote:
7 times the number of coins that A has is equal to 5 times the number of coins B has while 6 times the number of coins B has is equal to 11 times the number of coins C has. What is the minimum number of coins with A, B and C put together?

110
120
154
165
174


Are You Up For the Challenge: 700 Level Questions
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Re: 7 times the number of coins that A has is equal to 5 times the number  [#permalink]

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New post 15 Apr 2020, 08:00
1
Bunuel wrote:
7 times the number of coins that A has is equal to 5 times the number of coins B has while 6 times the number of coins B has is equal to 11 times the number of coins C has. What is the minimum number of coins with A, B and C put together?

110
120
154
165
174


Are You Up For the Challenge: 700 Level Questions


Let us get all the values in terms of ONE variable...

We are looking for A+B+C...
7 times the number of coins that A has is equal to 5 times the number of coins B has => \(7A=5B...A=\frac{5B}{7}\)
while 6 times the number of coins B has is equal to 11 times the number of coins C has.=> \(11C=6B...C=\frac{6B}{11}\)

\(A+B+C=\frac{5B}{7}+B+\frac{6B}{11}=\frac{174B}{77}\)
As \(\frac{174B}{77}\) has to be an integer, the LEAST value of B is 77, and SUM = \(\frac{174*77}{77}=174\)

E
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Re: 7 times the number of coins that A has is equal to 5 times the number  [#permalink]

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New post 15 Apr 2020, 08:03
1
7A=5B
A/B=5/7

(A:B=5:7=55:77)

6B=11C
B/C=11/6

(B:C=11:6=77:42)

A:B:C=55:77:42

Min value of A+B+C=55+77+42=174

Answer is (E)

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Re: 7 times the number of coins that A has is equal to 5 times the number  [#permalink]

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New post 15 Apr 2020, 08:10
\(7A=5B -----(1)\)
\(6B=11C -----(2)\)
From \((1)\) \(A=\frac{5B}{7}\) and from \((2)\) \(C=\frac{6B}{11}\)
\(A+B+C=\) Total number of coins
\(= \frac{5B}{7}+B+\frac{6B}{11}\)
\(=\frac{55B+77B+42B}{77}\)
\(=\frac{174B}{77}\)
For the minimum number of coins possible, \(B\) must be \(77\)
Hence minimum \(A+B+C=174\)

The answer is E.
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Re: 7 times the number of coins that A has is equal to 5 times the number  [#permalink]

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New post 24 Apr 2020, 04:20
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Bunuel

Please check OA

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Re: 7 times the number of coins that A has is equal to 5 times the number  [#permalink]

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New post 24 Apr 2020, 05:07
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Re: 7 times the number of coins that A has is equal to 5 times the number  [#permalink]

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New post 25 Apr 2020, 14:52
Bunuel wrote:
7 times the number of coins that A has is equal to 5 times the number of coins B has while 6 times the number of coins B has is equal to 11 times the number of coins C has. What is the minimum number of coins with A, B and C put together?

110
120
154
165
174


Are You Up For the Challenge: 700 Level Questions


We can create the equations:

7A = 5B

A = 5B/7

and

6B = 11C

6B/11 = C

Thus, the total number of coins is:

5B/7 + B + 6B/11

55B/77 + 77B/77 + 42B/77

174B/77

Since 174 and 77 are relatively prime, B must be divisible by 77. If B = 77 (notice that is the least positive value of B), then the total number of coins is 174(77)/77 = 174 (which is also the least total number of coins there can be).

Answer: E
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Re: 7 times the number of coins that A has is equal to 5 times the number  [#permalink]

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New post 30 Apr 2020, 04:08
Can someone point out why I was wrong please, my approach is below:
From stimulus, ->
7A = 5B ==> 42A = 30B
6B = 11C ==> 30B = 55C
==> 42A = 30B = 55C ==> 42+30+55 = 127 total minium number of coins
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Re: 7 times the number of coins that A has is equal to 5 times the number  [#permalink]

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New post 30 May 2020, 22:08

Solution



Given
    • 7 times the number of coins that A has is equal to 5 times the number of coins B has
    • 6 times the number of coins B has is equal to 11 times the number of coins C has.

To find
    • The minimum number of coins with A, B and C put together.

Approach and Working out
Though it is a 700-level question, but it can be easily solved by applying the ‘process skill of inference’.

Let number of coins with A = x, with B = y, with C = z. (applying the process skill of ‘translation’)
    • 7x = 5y
    • 6y = 11z

Simplifying the ratios in terms of z alone
    • \(7x = 5*(\frac{11z}{6})\)
      o \(x = \frac{55z}{42} = \frac{11 * 5}{7 * 3 * 2}\)

    • Using second equation y = \(\frac{11z}{6}\)

Using the above information, we have following inferences.
    • Inference 1: since x =\(\frac{55z}{42}\) and there is no common factor between 55 and 42, thus z should be a multiple of 42. (since x has to be an integer)
    • Inference 2: Since y = \(\frac{11z}{6}\) and there is no common factor between 11 and 6, thus z should be a multiple of 6. (since y has to be an integer)
    • Inference 3: Since sum = x + y + z = \(\frac{55z}{42} + \frac{11z}{6} + z = \frac{29z}{7}\), thus z should be a multiple of 7

Using the inferences found above, we can have one more inference as follows.
    • z should be a multiple of 7, 42 and 6.
    • Thus z = k * LCM of (6,7,42)
    • z (min) = 42

Now to evaluate the sum,
    • Sum = \(\frac{29*z}{7} = \frac{29*42}{7} = 29*6 = 174\)

Correct Answer: Option E

Thus, the above question could be solved easily using the inferences found above. Essentially, we did the following.
    • Applied the process skill of inference
    • Used the conceptual knowledge of LCM

If we didn’t apply the process skills, the solution would have been longer and would have involved more complexity. Application of process skills help us solve the GMAT questions by reducing the complexity of the problems thereby improving both the efficiency as well as the accuracy.

To know more about process skills... visit this link
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Re: 7 times the number of coins that A has is equal to 5 times the number   [#permalink] 30 May 2020, 22:08

7 times the number of coins that A has is equal to 5 times the number

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