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05 Nov 2019, 01:56
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
71% (02:03) correct
29%
(02:18)
wrong
based on 750
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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
A. 1
B. 3
C. 6
D. 7
E. 8
Are You Up For the Challenge: 700 Level Questions
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
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
05 Nov 2019, 06:57
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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
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
