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Jack and Anne had coins in the ratio 3 : 4. Jack gave a few coins to A

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Jack and Anne had coins in the ratio 3 : 4. Jack gave a few coins to A  [#permalink]

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New post 05 Nov 2019, 01:56
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A
B
C
D
E

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Question Stats:

61% (02:01) correct 39% (02:32) wrong based on 56 sessions

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Jack and Anne had coins in the ratio 3 : 4. Jack gave a few coins to A  [#permalink]

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New post Updated on: 05 Nov 2019, 07:07
As total coins has an integral value

Minimum possible coins= LCM(7,5)=35

Coins Jack has before transfer= \(\frac{3}{7}*35\) = 15

Coins Jack has after transfer= \(\frac{1}{5}*35\)= 7

minimum possible number of coins that Jack gave Anne= 15-7=8

Bunuel wrote:
Jack and Anne had coins in the ratio 3 : 4. Jack gave a few coins to Anne and the ratio became 1 : 4. What is the minimum possible number of coins that Jack gave Anne?

A. 1
B. 3
C. 6
D. 7
E. 8


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Originally posted by nick1816 on 05 Nov 2019, 05:48.
Last edited by nick1816 on 05 Nov 2019, 07:07, edited 1 time in total.
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Re: Jack and Anne had coins in the ratio 3 : 4. Jack gave a few coins to A  [#permalink]

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New post 05 Nov 2019, 06:57
before coin transfer
let Jack = 3x coins
Anne = 4x

Jack gave t coins to Anne

now, Jack = 3x-t
Anne = 4x+t

\(\frac{3x-t}{4x+t}\) = \(\frac{1}{4}\)

12x-4t = 4x+t
8x = 5t
t = \(\frac{8x}{5}\)

the minimum value of t will be attained at x = 5 (since number of coin transferred should be an integer)
the minimum possible number of coins that Jack gave Anne = 8

E is the answer
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Re: Jack and Anne had coins in the ratio 3 : 4. Jack gave a few coins to A  [#permalink]

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New post 05 Nov 2019, 07:19
ratio Jack : Anne = 3/4
they have 3x and 4x coins each
later
3x+y/4x-y = 1/4
12x-4y=4x+y
y=8x/5
if x= 5 ; min value of y=8
IMO E

Bunuel wrote:
Jack and Anne had coins in the ratio 3 : 4. Jack gave a few coins to Anne and the ratio became 1 : 4. What is the minimum possible number of coins that Jack gave Anne?

A. 1
B. 3
C. 6
D. 7
E. 8


Are You Up For the Challenge: 700 Level Questions
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Re: Jack and Anne had coins in the ratio 3 : 4. Jack gave a few coins to A  [#permalink]

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New post 05 Nov 2019, 11:00
Bunuel wrote:
Jack and Anne had coins in the ratio 3 : 4. Jack gave a few coins to Anne and the ratio became 1 : 4. What is the minimum possible number of coins that Jack gave Anne?

A. 1
B. 3
C. 6
D. 7
E. 8


Are You Up For the Challenge: 700 Level Questions


Given, J : A = 3 : 4
—> Let J = 3k and A = 4k for any positive integer k

Let jack gave n number of coins to Anne
—> New ratio of J : A = 1 : 4
—> (3k - n)/(4k + n) = 1/4
—> 12k - 4n = 4k + n
—> 5n = 8k
—> n = 8k/5

—> n has to be a multiple of 8 always
—> Minimum value of n = 8

IMO Option E

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Re: Jack and Anne had coins in the ratio 3 : 4. Jack gave a few coins to A  [#permalink]

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New post 05 Nov 2019, 12:43
Given that J / A = 3 / 4
if Jack gives some coins to Anne, he has fewer coins now....let that be x
(J - x) / (A - x) = 1 / 4

On solving we get A = (5/2)x
And J = 15x / 8

that mean at the minimum he can give 8 coins
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Re: Jack and Anne had coins in the ratio 3 : 4. Jack gave a few coins to A  [#permalink]

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New post 11 Nov 2019, 14:34
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Bunuel wrote:
Jack and Anne had coins in the ratio 3 : 4. Jack gave a few coins to Anne and the ratio became 1 : 4. What is the minimum possible number of coins that Jack gave Anne?

A. 1
B. 3
C. 6
D. 7
E. 8


Are You Up For the Challenge: 700 Level Questions


We can express the ratio of the number of Jack’s coins to the number of Anne’s coins as 3x : 4x. Let’s let n = the number of coins that Jack gave to Anne. Thus, Jack now has (3x - n) coins and Anne has (4x + n) coins, and this new ratio is 1 : 4, as shown in this equation:
(3x - n)/(4x + n) = 1/4
4(3x - n) = 4x + n
12x - 4n = 4x + n
8x = 5n
x = 5n/8
Since x is an integer and 8 does not divide 5, then 8 must divide n, and thus the minimum value for n is 8.
Answer: E
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Re: Jack and Anne had coins in the ratio 3 : 4. Jack gave a few coins to A   [#permalink] 11 Nov 2019, 14:34
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