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01 Dec 2015, 04:24
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For bringing each copper coin from the bottom of a river, a coin-diver gets 20 cents, and for each brass coin he gets 25 cents. If after one dive, he got $2.80, what is the minimum number of copper coins that he brought?
(A) 4
(B) 3
(C) 2
(D) 1
(E) 0
(A) 4
(B) 3
(C) 2
(D) 1
(E) 0
01 Dec 2015, 06:26
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Let number of copper coins = c
number of bronze coins = b
.20c + .25 b = 2.80
=> 20c + 25b = 280
=> 20c = 280 - 25b
=> c = 14 - (25/20)b = 14 - (5/4)b
To minimize c , we need to maximze b . b needs to be multiple of 4
b = 4 , c = 14 - 5 = 9
b = 8 , c = 14 - 10 = 4
b = 12 , c = 14 - 15 = -1 , Number of coins can be negative .
Therefore , minimum number of copper coins = 4
Answer A
Alternatively ,
we use backsolving here . It will be much quicker .
if b= 0 , 2.8 is not a multiple of .25 . Not possible
b = 1 , 2.6 is not a multiple of .25 . Not possible
b = 2 , 2.4 is not a multiple of .25 . Not possible
b= 3 , 2.2 is not a multiple of .25 . Not possible
b=4 , 2.0 is a multiple of .25 .
Answer A
number of bronze coins = b
.20c + .25 b = 2.80
=> 20c + 25b = 280
=> 20c = 280 - 25b
=> c = 14 - (25/20)b = 14 - (5/4)b
To minimize c , we need to maximze b . b needs to be multiple of 4
b = 4 , c = 14 - 5 = 9
b = 8 , c = 14 - 10 = 4
b = 12 , c = 14 - 15 = -1 , Number of coins can be negative .
Therefore , minimum number of copper coins = 4
Answer A
Alternatively ,
we use backsolving here . It will be much quicker .
if b= 0 , 2.8 is not a multiple of .25 . Not possible
b = 1 , 2.6 is not a multiple of .25 . Not possible
b = 2 , 2.4 is not a multiple of .25 . Not possible
b= 3 , 2.2 is not a multiple of .25 . Not possible
b=4 , 2.0 is a multiple of .25 .
Answer A
01 Dec 2015, 09:52
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0.2C + 0.25B = 2.8
For C to be minimum, we have to maximize B such that 2.8 - 0.25B is divisible by 0.2.
Max value of B can be 8
C = (2.8 - 2)/0.2 = 4
Answer: A
For C to be minimum, we have to maximize B such that 2.8 - 0.25B is divisible by 0.2.
Max value of B can be 8
C = (2.8 - 2)/0.2 = 4
Answer: A
