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Three friends divided a bag.
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Updated on: 17 Dec 2021, 04:15
Updated on: 17 Dec 2021, 04:15
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Question Stats:
80% (02:21) correct
20% (01:39) wrong based on 10 sessions
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Three friends divided a bag of chocolates so that David received a fifth the number of chocolates that Fouad did, and Stina received 80 percent of the total number of chocolates. What is the ratio of the number of chocolates Stina received to the number that David received?
(A) 4 : 3
(B) 8 : 5
(C) 8 : 1
(D) 24 : 1
(E) 80 : 1
(A) 4 : 3
(B) 8 : 5
(C) 8 : 1
(D) 24 : 1
(E) 80 : 1
Originally posted by Sharifkhan757575 on 17 Dec 2021, 03:30.
Last edited by Carcass on 17 Dec 2021, 04:15, edited 1 time in total.
Last edited by Carcass on 17 Dec 2021, 04:15, edited 1 time in total.
Formatted
Re: Three friends divided a bag.
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17 Dec 2021, 04:14
17 Dec 2021, 04:14
Expert Reply
Let the number that Stina received be 80, David and Fouda received be D and F, and the total be 100.
Thus \(D+F=20.........\frac{F}{5}+F=20...........6F=100..........F=\frac{100}{6}..........D=\frac{F}{5}=\frac{100}{6*5}=\frac{10}{3}\)
Ratio = \(S:D=80:\frac{10}{3}=240:10=24:1\)
D
Thus \(D+F=20.........\frac{F}{5}+F=20...........6F=100..........F=\frac{100}{6}..........D=\frac{F}{5}=\frac{100}{6*5}=\frac{10}{3}\)
Ratio = \(S:D=80:\frac{10}{3}=240:10=24:1\)
D
Re: Three friends divided a bag.
[#permalink]
17 Dec 2021, 04:15
17 Dec 2021, 04:15
Expert Reply
Total = S + D + F(1)
\(D = \frac{1}{5} F\)
\(5D = F\)(2)
\(S = \frac{80}{100} T = \frac{4}{5} T\)(3)
So, \(F + D = \frac{1}{5} T\)(4)
From (2):
\(6D = \frac{T}{5}, D = \frac{T}{30}\)
\(\frac{S}{D} = 4/5T / T/30\)
\(\frac{4}{5 T} * \frac{30}{T}\)
24:1,
D
\(D = \frac{1}{5} F\)
\(5D = F\)(2)
\(S = \frac{80}{100} T = \frac{4}{5} T\)(3)
So, \(F + D = \frac{1}{5} T\)(4)
From (2):
\(6D = \frac{T}{5}, D = \frac{T}{30}\)
\(\frac{S}{D} = 4/5T / T/30\)
\(\frac{4}{5 T} * \frac{30}{T}\)
24:1,
D
