Pat bought 5 lbs. of apples. How many pounds of pears could Pat have bought for the same amount of money?
(1) One pound of pears costs $0.50 more than one pound of apples.
(2) One pound of pears costs 1 1/2 times as much as one pound of apples.
Thanks for your help!
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A = spending on apples
Qa = quantity of apples
Pa = price of apples
Z = spending on pears
Qz = quantity of pears
Pz = price of pears
so...
A = Qa * Pa <--- this is the amount of money he will have to spend on pears
Z=A <--- given in the stem
Qz = (5 * Pa ) / Pz <--- we are given Qa=5 in stem.
1) Pz = Pa + .50 <-- Insufficient because we can't solve for Pz or Pa
2) Pz = (3/2)Pa <-- Insufficient because we can't solve for Pz or Pa
Together we have the system of equations and can plug in Pa and Pz to find Qz
Ans C
Qa = quantity of apples
Pa = price of apples
Z = spending on pears
Qz = quantity of pears
Pz = price of pears
so...
A = Qa * Pa <--- this is the amount of money he will have to spend on pears
Z=A <--- given in the stem
Qz = (5 * Pa ) / Pz <--- we are given Qa=5 in stem.
1) Pz = Pa + .50 <-- Insufficient because we can't solve for Pz or Pa
2) Pz = (3/2)Pa <-- Insufficient because we can't solve for Pz or Pa
Together we have the system of equations and can plug in Pa and Pz to find Qz
Ans C
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My choice is D both are sufficient
A. we know that pears are fifty cents more a pound. If she bought 5 pounds of apples then lets says apples are $2 a pound, so she spent 10 dollars. That means pears are 2.50 and she can buy 4 pounds worth. SUFFICIENT
B. we know basically the same thing, but this time it's 1.5 times as much per pound, so we can still figure out how many pounds she can buy with the amount she spent on apples, whatever that amount is. SUFFICIENT
What is the OA?
A. we know that pears are fifty cents more a pound. If she bought 5 pounds of apples then lets says apples are $2 a pound, so she spent 10 dollars. That means pears are 2.50 and she can buy 4 pounds worth. SUFFICIENT
B. we know basically the same thing, but this time it's 1.5 times as much per pound, so we can still figure out how many pounds she can buy with the amount she spent on apples, whatever that amount is. SUFFICIENT
What is the OA?
Good catch...
I'm not so sure about C now
1) I still think this is insufficient. If you plug in some more values for price of apples you will get different quantities of pears.
Mathematically you can see how this is the case
Going back to my earlier post:
Qz = (5 * Pa) / Pz
Qz = (5 * Pa) / (Pa + .50)
2) I see how this can be sufficient:
You want to solve Qz = (5 * Pa) / Pz
and statement 2 tells you that Pz = (3/2) * Pa so...
Qz = (5 * Pa) / (3/2) * Pa which can be simplified to find Qz
Looks like B may be the answer after all.
I'm not so sure about C now
1) I still think this is insufficient. If you plug in some more values for price of apples you will get different quantities of pears.
Mathematically you can see how this is the case
Going back to my earlier post:
Qz = (5 * Pa) / Pz
Qz = (5 * Pa) / (Pa + .50)
2) I see how this can be sufficient:
You want to solve Qz = (5 * Pa) / Pz
and statement 2 tells you that Pz = (3/2) * Pa so...
Qz = (5 * Pa) / (3/2) * Pa which can be simplified to find Qz
Looks like B may be the answer after all.
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I would say A is insufficient becoz the number of pears that can be bought differs with the dollar amount . With the same example you had, if the apples were $4 a pound, then the number of pounds of pears that can be bought will be different. Without knowing exactly how much amount, it will be INSUFF. At least, that's my understanding.Mclaughlin wrote:My choice is D both are sufficient
A. we know that pears are fifty cents more a pound. If she bought 5 pounds of apples then lets says apples are $2 a pound, so she spent 10 dollars. That means pears are 2.50 and she can buy 4 pounds worth. SUFFICIENT
B. we know basically the same thing, but this time it's 1.5 times as much per pound, so we can still figure out how many pounds she can buy with the amount she spent on apples, whatever that amount is. SUFFICIENT
What is the OA?
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from statement 1, we get that pears cost 0.50 more than the what 1lb of apples cost. If X$ was the cost of 5lbs of apple, then x/5 is 1lb of apples.
Pears cost of 1 lb = x/5 + 0.5. The total numbers of lbs that can be bought for X$ is, X/(x/5 + 0.5) = 5x/(x+ 2.5); without knowing X, its insuffcient.
With statement b, 1lb of pears cost 3/2 * (x/5) = 3x/10.
Total num of lbs is, x / (3x/10) = 10/3 ~ 3 lbs.
That's the math I used to solve it. Not sure if I missed something.
Pears cost of 1 lb = x/5 + 0.5. The total numbers of lbs that can be bought for X$ is, X/(x/5 + 0.5) = 5x/(x+ 2.5); without knowing X, its insuffcient.
With statement b, 1lb of pears cost 3/2 * (x/5) = 3x/10.
Total num of lbs is, x / (3x/10) = 10/3 ~ 3 lbs.
That's the math I used to solve it. Not sure if I missed something.
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Hilarious.beeparoo wrote:Hahahahahaha! Schrute Beets! I couldn't even look at the question before bursting out laughing.
"QUESTION: Did my shoes come off in the plane crash?"
I think it's B. We're only looking for a ratio/relationship between the price of apples/lb and the price of pears/lb. We get that in B.
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Canman wrote:A = spending on apples
Qa = quantity of apples
Pa = price of apples
Z = spending on pears
Qz = quantity of pears
Pz = price of pears
so...
A = Qa * Pa <--- this is the amount of money he will have to spend on pears
Z=A <--- given in the stem
Qz = (5 * Pa ) / Pz <--- we are given Qa=5 in stem.
1) Pz = Pa + .50 <-- Insufficient because we can't solve for Pz or Pa
2) Pz = (3/2)Pa <-- Insufficient because we can't solve for Pz or Pa
Together we have the system of equations and can plug in Pa and Pz to find Qz
Ans C
Qz = (5 * Pa ) / Pz or Qz = 5 * (Pa / Pz)
From one we cant find Pa/Pz value
from 2 we have Pa/Pz value as 2/3 .... Therefore 2 alone is sufficient
Answer is B
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I would go with B.
A is insufficient cause the ammount of peers will change depending of te cost of apples.
B establishes a relationship:
5 pounds of apples -----> x $
y pounds of peers -------> 1.5x $
Solving we have: 5x/1/3x/2---->10/3---> 3 1/3 pounds of peers
A is insufficient cause the ammount of peers will change depending of te cost of apples.
B establishes a relationship:
5 pounds of apples -----> x $
y pounds of peers -------> 1.5x $
Solving we have: 5x/1/3x/2---->10/3---> 3 1/3 pounds of peers