At a fruit stand, apples can be purchased for $0.15 each and pears for $0.20 each. At these rates a bag of apples and pears was purchased for $3.80. If the bag contained exactly 21 pieces of fruit how many were pears?
I tried to use alligation method for solving this problem.
First, I divide the prices of apples and fruits to find the ratio of prices. The ratio of prices of peach to apples is 4:3.So,
apple------------peach
3 ----3.8------- 4
or 1:4
Therefore, p:p+a is 4:5=p:21.Therefore p is 84/5. My solution leads to the wrong answer. Would you please explain the reason?
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alligation
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talaangoshtari
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Brent@GMATPrepNow
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Here's an algebraic solution:talaangoshtari wrote:At a fruit stand, apples can be purchased for $0.15 each and pears for $0.20 each. At these rates a bag of apples and pears was purchased for $3.80. If the bag contained exactly 21 pieces of fruit how many were pears?
Let A = the number of apples purchased
Let P = the number of pears purchased
A bag of apples and pears was purchased for $3.80
So, we get: 0.15A + 0.2B = 3.80
The bag contained exactly 21 pieces of fruit
So, A + P = 21
We now have a system of two equations:
0.15A + 0.2B = 3.80
A + P = 21
For the top equation, multiply both sides by 100.
For the bottom equation, multiply both sides by 15.
We get:
15A + 20P = 380
15A + 15P = 315
Subtract the bottom equation from the top equation to get:
5P = 65
Solve: P = 13
Answer: 13 pears
Cheers,
Brent
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GMATGuruNY
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The GMAT would include answer choices:
When the correct answer choice is plugged in, the total price will be 380 cents.
Answer choice D: 12 pears, implying 9 apples
Total price for 12 pears and 9 apples = 12(20) + 9(15) = 375 cents.
Too small.
Eliminate D.
For the total price to increase, more PEARS must be purchased, since pears are more expensive than apples.
The correct answer is E.
We can PLUG IN THE ANSWERS, which represent the number of pears.At a fruit stand, apples can be purchased for $0.15 each and pears for $0.20 each. At these rates a bag of apples and pears was purchased for $3.80. If the bag contained exactly 21 pieces of fruit how many were pears?
8
10
11
12
13
When the correct answer choice is plugged in, the total price will be 380 cents.
Answer choice D: 12 pears, implying 9 apples
Total price for 12 pears and 9 apples = 12(20) + 9(15) = 375 cents.
Too small.
Eliminate D.
For the total price to increase, more PEARS must be purchased, since pears are more expensive than apples.
The correct answer is E.
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DavidG@VeritasPrep
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talaangoshtari,
If you wanted to use the approach you outlined, you'd have to find the overall average price for the fruit purchased. In this case that number ends up being a decidedly unfriendly: 3.80/21 = .18 (rounded to the nearest hundredths)
Now you'd have
apples--------- pears
.15-----.18----.20
Giving you, approximately, a 2:3 ratio in favor of pears.
Apples: 2x
Pears: 3x
Total: 21
2x + 3x = 21
5x = 21. x = a little more than 4.
Pears: 3 * (a little more than 4) = a little more than 12. Round up to 13.
If you wanted to use the approach you outlined, you'd have to find the overall average price for the fruit purchased. In this case that number ends up being a decidedly unfriendly: 3.80/21 = .18 (rounded to the nearest hundredths)
Now you'd have
apples--------- pears
.15-----.18----.20
Giving you, approximately, a 2:3 ratio in favor of pears.
Apples: 2x
Pears: 3x
Total: 21
2x + 3x = 21
5x = 21. x = a little more than 4.
Pears: 3 * (a little more than 4) = a little more than 12. Round up to 13.
Last edited by DavidG@VeritasPrep on Mon May 18, 2015 6:42 am, edited 1 time in total.
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GMATGuruNY
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Alternate approach:
Price for each apple = 15.
Price for each pear = 20.
Average price for the MIXTURE of apples and pears = 380/21.
The following approach is called ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS.
Step 1: Put the prices over a COMMON DENOMINATOR
Price for each apple = (15*21)/21 = 315/21.
Price for each pear = (20*21)/21 = 420/21.
Average price for the MIXTURE of apples and pears = 380/21.
Step 2: Plot the 3 numerators on a number line, with the numerators for A and P on the ends and the numerator for the mixture in the middle.
A 315-----------380-----------420 W
Step 3: Calculate the distances between the numerators.
A 315----65-----380----40-----420 P
Step 4: Determine the ratio in the mixture.
The required ratio of A to P is equal to the RECIPROCAL of the distances in red.
A : P = 40:65 = 8:13.
Thus, A=8 and P=13, for a total of 21 pieces of fruit.
For two similar problems, check here:
https://www.beatthegmat.com/ratios-fract ... 15365.html
Price for each apple = 15.
Price for each pear = 20.
Average price for the MIXTURE of apples and pears = 380/21.
The following approach is called ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS.
Step 1: Put the prices over a COMMON DENOMINATOR
Price for each apple = (15*21)/21 = 315/21.
Price for each pear = (20*21)/21 = 420/21.
Average price for the MIXTURE of apples and pears = 380/21.
Step 2: Plot the 3 numerators on a number line, with the numerators for A and P on the ends and the numerator for the mixture in the middle.
A 315-----------380-----------420 W
Step 3: Calculate the distances between the numerators.
A 315----65-----380----40-----420 P
Step 4: Determine the ratio in the mixture.
The required ratio of A to P is equal to the RECIPROCAL of the distances in red.
A : P = 40:65 = 8:13.
Thus, A=8 and P=13, for a total of 21 pieces of fruit.
For two similar problems, check here:
https://www.beatthegmat.com/ratios-fract ... 15365.html
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Mo2men
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GMATGuruNY wrote:The GMAT would include answer choices:
Alternative way to think about questionAt a fruit stand, apples can be purchased for $0.15 each and pears for $0.20 each. At these rates a bag of apples and pears was purchased for $3.80. If the bag contained exactly 21 pieces of fruit how many were pears?
8
10
11
12
13
Price/apple(A)=15 cent
Price/Pear (P)= 20 cent
Total = 380 cent
15A+ 20P =380
(15A) must be multiple of 10 as (20P) is multiple of 10 and total is multiple of 10.
so (15A) could be multiple of 10 when multiplied by even number. It implied that pear is odd number (21-even=odd) So Eliminate choices A,B and D.
Check C....15*10 + 20*11=375 does nt equal to 380. Eliminate C
Only answer: E
Is my reasoning correct?
Last edited by Mo2men on Sat May 13, 2017 4:18 pm, edited 1 time in total.
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Nice work!Mo2men wrote:Alternative way to think about questionAt a fruit stand, apples can be purchased for $0.15 each and pears for $0.20 each. At these rates a bag of apples and pears was purchased for $3.80. If the bag contained exactly 21 pieces of fruit how many were pears?
8
10
11
12
13
Price/apple(A)=15 cent
Price/Pear (P)= 20 cent
Total = 380 cent
15A+ 20P =380
(15A) must be multiple of 10 as (20P) is multiple of 10 and total is multiple of 10.
so (15A) could be multiple of 10 when multiplied by even number. It implied that pear is odd number (21-even=odd) So Eliminate choices A,B and D.
Check D....15*10 + 20*11=375 does nt equal to 380. Eliminate C
Only answer: E
Is my reasoning correct?
One typo: the letter in red should be not D but C.
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