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24 Jul 2024, 01:30
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Difficulty:
45%
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Question Stats:
72% (02:17) correct
28%
(02:25)
wrong
based on 265
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All the 585 students of a school are divided into three different groups such that the half the number of students in first group, one-third the number of students in second group and one-fourth the number of students in third group are equal. Find the number of students in each group.
A. 130, 195, 260
B. 260, 195, 130
C. 270, 130, 135
D. 135, 180, 270
E. None of these
A. 130, 195, 260
B. 260, 195, 130
C. 270, 130, 135
D. 135, 180, 270
E. None of these
24 Jul 2024, 06:42
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\(x+y+z = 585\)
\(\frac{x}{2} = \frac{y}{3} = \frac{z}{4} = k\)
now, \(x=2k\), \(y=3k\) and \(z=4k.\)
after substituting,
\(9k = 585\)
\(k=65\)
Therefore values for x, y and z are 130, 195 and 260.
Answer is Option A.
\(\frac{x}{2} = \frac{y}{3} = \frac{z}{4} = k\)
now, \(x=2k\), \(y=3k\) and \(z=4k.\)
after substituting,
\(9k = 585\)
\(k=65\)
Therefore values for x, y and z are 130, 195 and 260.
Answer is Option A.
General Discussion
24 Jul 2024, 02:38
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Let group 1 =\(x\), group 2 = \(y\) and group 3 = \(z\)
We are given: \(x + y + z = 585\)
We are also given: \(\frac{1}{2}x = \frac{1}{3}y = \frac{1}{4}z\)
Looking at the answer choices one sees that the values for group 2 has a repeat value of 195. For this reason one should solve for either group 1 or group 3's value to find the answer.
Solving for y in terms of x: \(\frac{1}{2}x = \frac{1}{3}y\)
\(y = \frac{3}{2}\)
Solving for z in terms of x: \(\frac{1}{2}x = \frac{1}{4}z\)
\(z = 2x\)
Plugging the values back into the origianl equation and solving for x:
\(x + \frac{3}{2}x + 2x = 585\)
\(2x + 3x + 4x = 1170\)
\(9x = 1170\)
\(x = 130\)
Only answer choice A has 130 as a value for the first group.
ANSWER A
We are given: \(x + y + z = 585\)
We are also given: \(\frac{1}{2}x = \frac{1}{3}y = \frac{1}{4}z\)
Looking at the answer choices one sees that the values for group 2 has a repeat value of 195. For this reason one should solve for either group 1 or group 3's value to find the answer.
Solving for y in terms of x: \(\frac{1}{2}x = \frac{1}{3}y\)
\(y = \frac{3}{2}\)
Solving for z in terms of x: \(\frac{1}{2}x = \frac{1}{4}z\)
\(z = 2x\)
Plugging the values back into the origianl equation and solving for x:
\(x + \frac{3}{2}x + 2x = 585\)
\(2x + 3x + 4x = 1170\)
\(9x = 1170\)
\(x = 130\)
Only answer choice A has 130 as a value for the first group.
ANSWER A
