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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # 7 times the number of coins that A has is equal to 5 times the number

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Math Expert V
Joined: 02 Sep 2009
Posts: 65290
7 times the number of coins that A has is equal to 5 times the number  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 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
120
154
165
174

Are You Up For the Challenge: 700 Level Questions

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DS Forum Moderator V
Joined: 19 Oct 2018
Posts: 1998
Location: India
Re: 7 times the number of coins that A has is equal to 5 times the number  [#permalink]

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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
Math Expert V
Joined: 02 Aug 2009
Posts: 8757
Re: 7 times the number of coins that A has is equal to 5 times the number  [#permalink]

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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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Director  D
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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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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

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CR Forum Moderator D
Joined: 18 May 2019
Posts: 811
Re: 7 times the number of coins that A has is equal to 5 times the number  [#permalink]

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$$7A=5B -----(1)$$
$$6B=11C -----(2)$$
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 $$A+B+C=174$$

Director  D
Joined: 16 Jan 2019
Posts: 660
Location: India
Concentration: General Management
WE: Sales (Other)
Re: 7 times the number of coins that A has is equal to 5 times the number  [#permalink]

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1
Bunuel

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Math Expert V
Joined: 02 Sep 2009
Posts: 65290
Re: 7 times the number of coins that A has is equal to 5 times the number  [#permalink]

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firas92 wrote:
Bunuel

Posted from my mobile device

_______________________________
Edited the OA. Thank you for noticing!
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Target Test Prep Representative V
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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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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).

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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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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
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3410
Re: 7 times the number of coins that A has is equal to 5 times the number  [#permalink]

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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 = $$\frac{29*z}{7} = \frac{29*42}{7} = 29*6 = 174$$

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.

_________________ 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  