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

Question

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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nick1816 · Accepted Answer

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

shridhar786 · Answer

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

Archit3110 · Answer

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

CareerGeek · Answer

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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anupkumar.patel · Answer

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

ScottTargetTestPrep · Answer

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

hero_with_1000_faces · Answer

3x : 4x turns into 1z : 4z

we use the LCM from both to make the total common, x = 5; y = 7
J = 15 and A = 20
after transfer we get J = 7 and A = 28

Hence, 8 has been tansferred

LeoN88 · Answer

Let J:A :: 3x:4x

Now new equation:-

(3x-c)/(4x+c) = 1/4

8x = 5c (c must be an integer, so automatically x is also an integer)

For min. c; x must also be min.
Then x = 5 and C=  8

Ravixxx · Answer

J:A
3:4 
next
1:4

We should combine ratio of J turning J in the same number, 
so: 1*3:4*3 --> 3:12

12-4 = 8 --> Minimum possible number of coins that Jack gave Anne.

Answer E.

Bunuel

twobagels · Answer

Below, how did you know to take LCM(7,5). Your method seems very good. Thank you.

gmatpro15 · Answer

before coin transfer
let Jack = 3x coins
Anne = 4x

Jack gave t coins to Anne

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

=

12x-4t = 4x+t
8x = 5t
t = 8x/5 
So minimum possible coins are 8

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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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