A certain compound X has a ratio of 2 oxygen for every 5 carbon. Another
compound Y has a ratio of 1 oxygen for every 4 carbon. If a mixture of X and
Y has a ratio of 3 oxygen for every 10 carbon, what is the ratio of compound
X to compound Y in the mixture?
A)1 to 10
B)1 to 3
C)1 to 2
D)2 to 5
E)2 to 3
PLS HELP
A certain compound X has a ratio of 2 oxygen for every
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The following approach is called ALLIGATION.vikkimba17 wrote:A certain compound X has a ratio of 2 oxygen for every 5 carbon. Another
compound Y has a ratio of 1 oxygen for every 4 carbon. If a mixture of X and
Y has a ratio of 3 oxygen for every 10 carbon, what is the ratio of compound
X to compound Y in the mixture?
A)1 to 10
B)1 to 3
C)1 to 2
D)2 to 5
E)2 to 3
Step 1: Convert the ratios to FRACTIONS.
X:
Since oxygen:carbon = 2:5, and 2+5=7, oxygen/total = 2/7.
Y:
Since oxygen:carbon = 1:4, and 1+4=5, oxygen/total = 1/5.
M:
Since oxygen:carbon = 3:10, and 3+10=13, oxygen/total= 3/13.
Step 2: Put the fractions over a COMMON DENOMINATOR.
X = 2/7 = (2*5*13)/(5*7*13) = 130/(5*7*13).
Y = 1/5 = (1*7*13)/(5*7*13) = 91/(5*7*13).
Mixture = 3/13 = (3*5*7)/(5*7*13) = 105/(5*7*13).
Step 3: Plot the 3 numerators on a number line, with the numerators for X and Y on the ends and the numerator for the mixture in the middle.
X 130-----------105-----------91 Y
Step 4: Calculate the distances between the numerators.
X 130-----25-----105-----14-----91 Y
Step 5: Determine the ratio in the mixture.
The ratio of X to Y in the mixture is the RECIPROCAL of the distances in red.
X/Y = 14/25.
The required ratio is not among the answer choices.
Feel free to ignore this problem.
More practice will alligation:
https://www.beatthegmat.com/ratios-fract ... 15365.html
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Let's calculate the Oxygen from compound X, Y and the mixture.vikkimba17 wrote:A certain compound X has a ratio of 2 oxygen for every 5 carbon. Another
compound Y has a ratio of 1 oxygen for every 4 carbon. If a mixture of X and
Y has a ratio of 3 oxygen for every 10 carbon, what is the ratio of compound
X to compound Y in the mixture?
A)1 to 10
B)1 to 3
C)1 to 2
D)2 to 5
E)2 to 3
PLS HELP
The quantity of Oxygen from compound X + The quantity of Oxygen from compound Y = The quantity of Oxygen in the mixture
The fraction of Oxygen from compound X = 2 /(2+5) = 2/7
The fraction of Oxygen from compound Y = 1 /(1+4) = 1/5
The fraction of Oxygen in the mixture = 3 /(3+10) = 3/13
Say compound X and compound Y were mixed in the ratio of 1 : y. Thus, the mixture has 3(1 + y)/13 part of Oxygen
The quantity of Oxygen from compound X = 2/7
The quantity of Oxygen from compound y = (1*y)/5 = y/5
The quantity of Oxygen from compound X and the quantity of Oxygen from compound Y = The quantity of Oxygen in the mixture
=> 2/7 + y/5 = 3(1 + y)/13
=> 130 + 91y = 105 + 105y
=> y = 25/14
Thus, the required ratio = 1 : y = 1 : (25/14) = [spoiler]14 : 25.[/spoiler]
As Mitch rightly said that there is no option.
Hope this helps!
Relevant book: Manhattan Review GMAT Word Problems Guide
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This is actually a mixture problem that many algebra 1 students have exp with.
First mixture is (2/7) oxygen
Second is 1/5 oxygen
After mixing it becomes 3/13 Oxygen
(2/7)x+(1/5)y=(3/13)(x+y)
After doing some algebra (multiply both sides by (13X7x5) we get x/y=14/25
In terms of carbon the equation would be (3/5)x+(4/5)y=(10/13)(x+y). This gives the same result.
First mixture is (2/7) oxygen
Second is 1/5 oxygen
After mixing it becomes 3/13 Oxygen
(2/7)x+(1/5)y=(3/13)(x+y)
After doing some algebra (multiply both sides by (13X7x5) we get x/y=14/25
In terms of carbon the equation would be (3/5)x+(4/5)y=(10/13)(x+y). This gives the same result.
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Easier, I think:
(2/7)x + (1/5)y = (3/13)*(x + y)
multiply both sides by 7*5*13:
2*5*13x + 7*13*y = 3*7*5*(x + y)
130x + 91y = 105x + 105y
25x = 14y
and we're done!
(2/7)x + (1/5)y = (3/13)*(x + y)
multiply both sides by 7*5*13:
2*5*13x + 7*13*y = 3*7*5*(x + y)
130x + 91y = 105x + 105y
25x = 14y
and we're done!
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Also, I would guess that the OA is C, using this logic:
(2/5)x + (1/4)y = (3/10)*(x + y)
which reduces to a ratio of 1:2. Easy mistake to make!
(2/5)x + (1/4)y = (3/10)*(x + y)
which reduces to a ratio of 1:2. Easy mistake to make!
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Since compound X has a ratio of 2 oxygen for every 5 carbon, we can create the following ratio:vikkimba17 wrote: ↑Sun Mar 12, 2017 7:54 pmA certain compound X has a ratio of 2 oxygen for every 5 carbon. Another
compound Y has a ratio of 1 oxygen for every 4 carbon. If a mixture of X and
Y has a ratio of 3 oxygen for every 10 carbon, what is the ratio of compound
X to compound Y in the mixture?
A)1 to 10
B)1 to 3
C)1 to 2
D)2 to 5
E)2 to 3
PLS HELP
oxygen : carbon = 2x : 5x
Since compound Y has a ratio of 1 oxygen for every 4 carbon, we can create the following ratio:
oxygen : carbon = y : 4y
Since a mixture of X and Y has a ratio of 3 oxygen for every 10 carbon, we can create the following proportion:
(2x + y)/(5x + 4y) = 3/10
10(2x + y) = 3(5x + 4y)
20x + 10y = 15x + 12y
5x = 2y
x/y = 2/5
Notice that in the mixture, the amount of compound X is 2x + 5x = 7x and the amount of compound Y is y + 4y = 5y. We see that the ratio of compound X to compound Y in the mixture is 7x/5y = (7/5)*(x/y) = (7/5)*(2/5) = 14/25.
Answer: 14/25
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