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505-555 Level|   Fractions and Ratios|               
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Gacarva
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Bunuel
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let manager=m , and worker =w
m/w=5/72
=5w=72m.....1
m/w+8=5/74
5w+40=74m...........2
by solving the equations.....we find m=20
so ans is D
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Gacarva
In a acertain company, the ratio of the numer of manager to the number of production-line workers is 5 to 72. If 8 additional Production-line workers were to be hired, the ratio of the number of managers to the number of production-line workers would be 5 to 74. How many managers does the company have?

A. 5
B. 10
C. 15
D. 20
E. 25

We are given that the ratio of managers to production-line workers is 5 to 72, or 5x to 72x. We can create the following equation:

(5x)/(72x + 8) = 5/74

74(5x) = (72x + 8)(5)

370x = 360x + 40

10x = 40

x = 4

The company has 4 x 5 = 20 managers.

Answer: D
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Bunuel
Gacarva
In a acertain company, the ratio of the numer of manager to the number of production-line workers is 5 to 72. If 8 additional Production-line workers were to be hired, the ratio of the number of managers to the number of production-line workers would be 5 to 74. How many managers does the company have?

A. 5
B. 10
C. 15
D. 20
E. 25

You can solve this problem the way discussed above or try the following:

Given: \(\frac{m}{w}=\frac{5x}{72x}\), now 8 workers were hired and new ratio became \(\frac{m}{w+8}=\frac{5x}{74x}=\frac{5x}{72x+2x}\), notice that nominator didn't change, so increase by 2 parts in denominator corresponds to increase by 8 workers (2 parts correspond to 8 workers = 4 times or in another words 2x=8 --> x=4), so 5 parts in nominator correspond to 5*4=20 managers.

Answer: D.

Hi Bunuel :-)

can please explain whats wrong with my solution and why :?

let manager be \(y\)

let line worker production be \(x\)

\(\frac{5y}{72x+4}\) =\(\frac{5y}{74x}\) cross multiply

\(360xy+20y=370xy\)

\(20y=370xy-360xy\)

\(20y=10xy\)
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Bunuel
Gacarva
In a acertain company, the ratio of the numer of manager to the number of production-line workers is 5 to 72. If 8 additional Production-line workers were to be hired, the ratio of the number of managers to the number of production-line workers would be 5 to 74. How many managers does the company have?

A. 5
B. 10
C. 15
D. 20
E. 25

You can solve this problem the way discussed above or try the following:

Given: \(\frac{m}{w}=\frac{5x}{72x}\), now 8 workers were hired and new ratio became \(\frac{m}{w+8}=\frac{5x}{74x}=\frac{5x}{72x+2x}\), notice that nominator didn't change, so increase by 2 parts in denominator corresponds to increase by 8 workers (2 parts correspond to 8 workers = 4 times or in another words 2x=8 --> x=4), so 5 parts in nominator correspond to 5*4=20 managers.

Answer: D.

Hi Bunuel :-)

can please explain whats wrong with my solution and why :?

let manager be \(y\)

let line worker production be \(x\)

\(\frac{5y}{72x+4}\) =\(\frac{5y}{74x}\) cross multiply

\(360xy+20y=370xy\)

\(20y=370xy-360xy\)

\(20y=10xy\)


It's because you are saying that the ratio is not 5 to 74. You are saying that managers are y and workers are x, you can say that but when you go to solve the problem you are saying that the ratios are 5y:74x thus the resolution is not correct.

Hope it's clear!
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Managers/ PLW= 5/72
new ratio= 5/74 after adding 8 PLW
5x/72x+8=5/74
x= 4
therefore Managers= 5*x= 5*4= 20

therefore D
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m/p = 5/72 ---1)
m/(p+8) = 5/74 --- 2)
putting values for p from 1) in 2)
m/(72m/5 + 8) = 5/74
cross multiplying
74m = 72m + 40
2m = 40
--> m = 20
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I think the simplest way...an Increase of 8 points in causing the number to increase from 72 to 74 ..means 2 points....
So .....we have 5 points in the numerator....and since 8 = 2 points.... 5 points will equal to 20..... (2 X 2.5 = 8 x 2.5)

Gacarva
In a acertain company, the ratio of the numer of manager to the number of production-line workers is 5 to 72. If 8 additional Production-line workers were to be hired, the ratio of the number of managers to the number of production-line workers would be 5 to 74. How many managers does the company have?

A. 5
B. 10
C. 15
D. 20
E. 25
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Bunuel
In a acertain company, the ratio of the numer of manager to the number of production-line workers is 5 to 72. If 8 additional Production-line workers were to be hired, the ratio of the number of managers to the number of production-line workers would be 5 to 74. How many managers does the company have?

A. 5
B. 10
C. 15
D. 20
E. 25

You can solve this problem the way discussed above or try the following:

Given: \(\frac{m}{w}=\frac{5x}{72x}\), now 8 workers were hired and new ratio became \(\frac{m}{w+8}=\frac{5x}{74x}=\frac{5x}{72x+2x}\), notice that nominator didn't change, so increase by 2 parts in denominator corresponds to increase by 8 workers (2 parts correspond to 8 workers = 4 times or in another words 2x=8 --> x=4), so 5 parts in nominator correspond to 5*4=20 managers.

Answer: D.
Bunuel,

Awesome approach!
One query - In this case the numerator hasn't changed... can we use the same approach even in cases where the numerator changes ? can't figure it out...
Yes/No and why so ?

Thank you in advance!
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rushimehta

Bunuel,

Awesome approach!
One query - In this case the numerator hasn't changed... can we use the same approach even in cases where the numerator changes ? can't figure it out...
Yes/No and why so ?

Thank you in advance!

No, you can’t use that shortcut when the numerator (managers) also changes.

That method works only because the numerator stays constant, so the change in the ratio directly reflects the change in the denominator (workers). If both parts change, you must form an equation with actual numbers rather than comparing “parts.” The parts approach loses validity once both numerator and denominator vary.
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