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In a city council election, Candidate X received more votes than Cand 04 Jan 2018, 23:51
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35% (medium)

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80% (02:38) correct 20% (02:59) wrong based on 228 sessions

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In a city council election, Candidate X received 1/3 more votes than Candidate Y, and Candidate Y received 1/4 fewer votes than Candidate Z. If the average (arithmetic mean) of votes per candidate was 22,000, how many votes did Candidate X receive?

A. 18,000
B. 20,000
C. 21,000
D. 22,000
E. 24,000
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E
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Schools: IIM (A) ISB '24
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In a city council election, Candidate X received more votes than Cand 05 Jan 2018, 05:39
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Bunuel
In a city council election, Candidate X received 1/3 more votes than Candidate Y, and Candidate Y received 1/4 fewer votes than Candidate Z. If the average (arithmetic mean) of votes per candidate was 22,000, how many votes did Candidate X receive?

A. 18,000
B. 20,000
C. 21,000
D. 22,000
E. 24,000
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Candidate X received 1/3 more votes than Candidate Y
ie. X = (4/3)Y

Candidate Y received 1/4 fewer votes than Candidate Z
i.e. Y = (3/4)Z

If the average (arithmetic mean) of votes per candidate was 22,000
i.e. (X+Y+Z)/3 = 22000

i.e. X+(3/4)X+(4/3)(3/4)X = 66000

i.e. (11/4)X = 66000
i.e. X = 24000

Answer: Option E
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In a city council election, Candidate X received more votes than Cand 05 Jan 2018, 07:00
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Bunuel
In a city council election, Candidate X received 1/3 more votes than Candidate Y, and Candidate Y received 1/4 fewer votes than Candidate Z. If the average (arithmetic mean) of votes per candidate was 22,000, how many votes did Candidate X receive?

A. 18,000
B. 20,000
C. 21,000
D. 22,000
E. 24,000
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z = z
y =3z/4
x = z

x+y+z = 66000
z+3z/4+z = 66000
11z/4 = 66000
z = 24000
X received 24000 votes
E
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In a city council election, Candidate X received more votes than Cand 05 Jan 2018, 15:41
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Bunuel
In a city council election, Candidate X received 1/3 more votes than Candidate Y, and Candidate Y received 1/4 fewer votes than Candidate Z. If the average (arithmetic mean) of votes per candidate was 22,000, how many votes did Candidate X receive?

A. 18,000
B. 20,000
C. 21,000
D. 22,000
E. 24,000
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Y = \(\frac{3}{4}\)Z => Z = \(\frac{4}{3}\)Y
X = \(\frac{4}{3}\)Y
X = Z = \(\frac{4}{3}\)Y

\(\frac{X + Y + Z}{3}\) = 22,000
X + Y + Z = 66,000

\(\frac{4}{3}\)Y + Y + \(\frac{4}{3}\)Y = 66,000

\(\frac{11}{3}\)Y = 66,000
Y = 66,000 * \(\frac{3}{11}\) = 18,000

X = \(\frac{4}{3}\) * 18,000 = 24,000

Answer E
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In a city council election, Candidate X received more votes than Cand 16 Oct 2020, 07:28
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Average votes received by all 3 candidates => x+y+z/3=22000;
Total votes received by all 3 candidates= 22000*3=66000;
X has 1/4 more votes than Y this makes X 4/3 times of Y;
Y has 1/4 times less votes than Z that makes Z 4/3 times of Y;
Therefore, (4/3)y+y+(4/3)y=66000.
Taking lcm; 11y/3=66000;
(66000*3)/11= 18000. Y received 18000 votes.
Therefore, X received (18000/3)*4 = 24000 votes.
Option E.
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In a city council election, Candidate X received more votes than Cand 16 Jan 2025, 19:00
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Bunuel MathRevolution
Can anyone help me understand how did we translate
1. Candidate X received 1/3 more votes than Candidate Y to X = (4/3)Y

2. Candidate Y received 1/4 fewer votes than Candidate Z to Y = (3/4)Z ?
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In a city council election, Candidate X received more votes than Cand 16 Jan 2025, 22:55
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Candidate X received 1/3 more votes than Candidate Y => \(X = Y + \frac{1}{3}Y = \frac{4}{3}Y\)

Candidate Y received 1/4 fewer votes than Candidate Z => \(Y = Z - \frac{1}{3}Z = \frac{3}{4}Z\)

Jujumamavudoo
Bunuel MathRevolution
Can anyone help me understand how did we translate
1. Candidate X received 1/3 more votes than Candidate Y to X = (4/3)Y

2. Candidate Y received 1/4 fewer votes than Candidate Z to Y = (3/4)Z ?
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Bunuel
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