If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 Swiss Francs enough to buy 12 notebooks and 12 pencils?
(1) 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils.
(2) 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils.
OA coming soon.
If 20 Swiss Francs is enough to buy 9 notebooks
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Given question is 12n+12p=40----------- I
9n+3p=20======= a
Statement 1
7n+5p=20 b *3
9n+3p=20 a *5
21n+15p=60
45n+15p=100
-24n = -40
n= -40/-24= 5/3
so substituting n=5/3 in equation b, we get
7*(5/3)+ 5p=20
So 20- (35/3)=5p
25/3=5p
p=25/3*5=5/3 so n and p is equal to 5/3
Subtituing these value in equation I
12(5/3) + 12(5/3)=40
20+20=40
Sufficient
Statement 2
4n+8p=20 c
9n+3p=20 Equation a
36n+72p=180
36n+12p=80
60p=100
p=5/3, substitute this in equation c
4n+8(5/3)=20
4n+40/3=20
4n=20-(40/3)
4n=20/3
n=(20/3)4
n=5/3
so again substitute these values in equation I
12*(5/3)+12(5/3)=40
20+20=40
Sufficient
OA is D
9n+3p=20======= a
Statement 1
7n+5p=20 b *3
9n+3p=20 a *5
21n+15p=60
45n+15p=100
-24n = -40
n= -40/-24= 5/3
so substituting n=5/3 in equation b, we get
7*(5/3)+ 5p=20
So 20- (35/3)=5p
25/3=5p
p=25/3*5=5/3 so n and p is equal to 5/3
Subtituing these value in equation I
12(5/3) + 12(5/3)=40
20+20=40
Sufficient
Statement 2
4n+8p=20 c
9n+3p=20 Equation a
36n+72p=180
36n+12p=80
60p=100
p=5/3, substitute this in equation c
4n+8(5/3)=20
4n+40/3=20
4n=20-(40/3)
4n=20/3
n=(20/3)4
n=5/3
so again substitute these values in equation I
12*(5/3)+12(5/3)=40
20+20=40
Sufficient
OA is D
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Question Stem : 20 = 9x + 3y
To find x & y
Statement 1:
20 = 7x + 5y
we can calculate x & y both ..
SUFFICIENT
Statement 2:
20 = 4x + 8 y
we can calculate x & y both ..
SUFFICIENT
Answer [spoiler]{D}[/spoiler]
To find x & y
Statement 1:
20 = 7x + 5y
we can calculate x & y both ..
SUFFICIENT
Statement 2:
20 = 4x + 8 y
we can calculate x & y both ..
SUFFICIENT
Answer [spoiler]{D}[/spoiler]
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Nice explanation. But answer is wrong. Correct answer is BtheCodeToGMAT wrote:Question Stem : 20 = 9x + 3y
To find x & y
Statement 1:
20 = 7x + 5y
we can calculate x & y both ..
SUFFICIENT
Statement 2:
20 = 4x + 8 y
we can calculate x & y both ..
SUFFICIENT
Answer [spoiler]{D}[/spoiler]
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OA is Bvinay1983 wrote:Given question is 12n+12p=40----------- I
9n+3p=20======= a
Statement 1
7n+5p=20 b *3
9n+3p=20 a *5
21n+15p=60
45n+15p=100
-24n = -40
n= -40/-24= 5/3
so substituting n=5/3 in equation b, we get
7*(5/3)+ 5p=20
So 20- (35/3)=5p
25/3=5p
p=25/3*5=5/3 so n and p is equal to 5/3
Subtituing these value in equation I
12(5/3) + 12(5/3)=40
20+20=40
Sufficient
Statement 2
4n+8p=20 c
9n+3p=20 Equation a
36n+72p=180
36n+12p=80
60p=100
p=5/3, substitute this in equation c
4n+8(5/3)=20
4n+40/3=20
4n=20-(40/3)
4n=20/3
n=(20/3)4
n=5/3
so again substitute these values in equation I
12*(5/3)+12(5/3)=40
20+20=40
Sufficient
OA is D
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Brakeshd347 wrote:If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 Swiss Francs enough to buy 12 notebooks and 12 pencils?
(1) 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils.
(2) 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils.
OA is .
Let us say Notebook as B and Pencils as P.
Given is 9B+3P<=20
Question is 12B+12P<=40 or 6B+6P<=20 ( here the question is Can we swap 3 Books for 3 Pencils).
1) 7B+5P<=20 Now here the word enough means equal to or less than. Suppose if someone tells you that $10 is enough to buy a burger it means less than or equal to. It doesn't mean equal so we can't just take it as two linear equations and solve it.
Now from this statement we can conclude that we can swap 2 books for a pencils9B+3P<=20
7B+5P<=20
When you compare two equation all it tells us that we can safely swap 2 books for 2 pencils. But we are not sure about 3. We can also conclude that we can swap 1 book for 1 pencil from it.
2) 4B+8P<=20 This statement tells us that we can swap 5 books for 5 pencils. So we can safely say that we can also swap 3 books for 3 pencils. So this statement is sufficient and the answer is B
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if we can swap 2 Pencils with 2 books then why we can't swap 3 pencils for 3 books. Is it because of inequality?rakeshd347 wrote:Brakeshd347 wrote:If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 Swiss Francs enough to buy 12 notebooks and 12 pencils?
(1) 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils.
(2) 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils.
OA is .
Let us say Notebook as B and Pencils as P.
Given is 9B+3P<=20
Question is 12B+12P<=40 or 6B+6P<=20 ( here the question is Can we swap 3 Books for 3 Pencils).
1) 7B+5P<=20 Now here the word enough means equal to or less than. Suppose if someone tells you that $10 is enough to buy a burger it means less than or equal to. It doesn't mean equal so we can't just take it as two linear equations and solve it.
Now from this statement we can conclude that we can swap 2 books for a pencils9B+3P<=20
7B+5P<=20
When you compare two equation all it tells us that we can safely swap 2 books for 2 pencils. But we are not sure about 3. We can also conclude that we can swap 1 book for 1 pencil from it.
2) 4B+8P<=20 This statement tells us that we can swap 5 books for 5 pencils. So we can safely say that we can also swap 3 books for 3 pencils. So this statement is sufficient and the answer is B
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This is a MAX/MIN problem.rakeshd347 wrote:If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 Swiss Francs enough to buy 12 notebooks and 12 pencils?
(1) 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils.
(2) 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils.
OA coming soon.
Try to MAXIMIZE and MINIMIZE the price of each item.
Nothing in the problem restrict the prices to integers.
The price of each notebook can be INFINITELY small, as can the price of each pencil.
Thus, we can simplify the problem by trying one case in which the price of each notebook = 0, and one case in which the price of each pencil = 0.
Statement 1:
Since 5 pencils can be purchased for 20 francs, the maximum possible price of each pencil = 20/5 = 4 francs.
Since 9 notebooks can be purchased for 20 francs, the maximum possible price of each notebook = 20/9 francs.
Case 1: Price of each notebook = 0 francs, price of each pencil = 4 francs
Total cost for 12 notebooks and 12 pencils = 12*(0) + 12(4) = 48 francs, which is MORE than 40 francs.
Case 2: Price of each notebook = 20/9 francs, price of each pencil = 0 francs
Total cost for 12 notebooks and 12 pencils = 12*(20/9) + 12(0) ≈ 24 francs, which is LESS than 40 francs.
Since 40 francs is NOT enough in Case 1 but IS enough in Case 2, INSUFFICIENT.
Statement 2:
Since 8 pencils can be purchased for 20 francs, the maximum possible price of each pencil = 20/8 = 5/2 francs.
Since 9 notebooks can be purchased for 20 francs, the maximum possible price of each notebook = 20/9 francs.
Case 1: Price of each notebook = 0 francs, price of each pencil = 5/2 francs
Total cost for 12 notebooks and 12 pencils = 12*(0) + 12(5/2) = 30 francs, which is LESS than 40 francs.
Case 2: Price of each notebook = 20/9 francs, price of each pencil = 0 francs
Total cost for 12 notebooks and 12 pencils = 12*(20/9) + 12(0) ≈ 24 francs, which is LESS than 40 francs.
Since 40 francs is enough in each case, SUFFICIENT.
The correct answer is B.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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As a tutor, I don't simply teach you how I would approach problems.
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GMATGuruNY wrote:This is a MAX/MIN problem.rakeshd347 wrote:If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 Swiss Francs enough to buy 12 notebooks and 12 pencils?
(1) 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils.
(2) 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils.
OA coming soon.
Try to MAXIMIZE and MINIMIZE the price of each item.
Nothing in the problem restrict the prices to integers.
The price of each notebook can be INFINITELY small, as can the price of each pencil.
Thus, we can simplify the problem by trying one case in which the price of each notebook = 0, and one case in which the price of each pencil = 0.
Statement 1:
Since 5 pencils can be purchased for 20 francs, the maximum possible price of each pencil = 20/5 = 4 francs.
Since 9 notebooks can be purchased for 20 francs, the maximum possible price of each notebook = 20/9 francs.
Case 1: Price of each notebook = 0 francs, price of each pencil = 4 francs
Total cost for 12 notebooks and 12 pencils = 12*(0) + 12(4) = 48 francs, which is MORE than 40 francs.
Case 2: Price of each notebook = 20/9 francs, price of each pencil = 0 francs
Total cost for 12 notebooks and 12 pencils = 12*(20/9) + 12(0) ≈ 24 francs, which is LESS than 40 francs.
Since 40 francs is NOT enough in Case 1 but IS enough in Case 2, INSUFFICIENT.
Statement 2:
Since 8 pencils can be purchased for 20 francs, the maximum possible price of each pencil = 20/8 = 5/2 francs.
Since 9 notebooks can be purchased for 20 francs, the maximum possible price of each notebook = 20/9 francs.
Case 1: Price of each notebook = 0 francs, price of each pencil = 5/2 francs
Total cost for 12 notebooks and 12 pencils = 12*(0) + 12(5/2) = 30 francs, which is LESS than 40 francs.
Case 2: Price of each notebook = 20/9 francs, price of each pencil = 0 francs
Total cost for 12 notebooks and 12 pencils = 12*(20/9) + 12(0) ≈ 24 francs, which is LESS than 40 francs.
Since 40 francs is enough in each case, SUFFICIENT.
The correct answer is B.
Mitch ,
Your solution is soo simple, clear and easy to understand,
I like to know for what kind of problems I can use this approach.
Could you explain with example if available.
Thanks in advance.
Regards,
Uva.
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Target question: Is 40 Swiss Francs enough to buy 12 notebooks and 12 pencils?rakeshd347 wrote:If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 Swiss Francs enough to buy 12 notebooks and 12 pencils?
(1) 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils.
(2) 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils.
OA coming soon.
This is a great candidate for rephrasing the target question.
Let N = the cost of 1 notebook (in Swiss francs)
Let P = the cost of 1 pencil (in Swiss francs)
So, 12 notebooks cost 12N and 12 pencils cost 12P
So, we want to know whether 12N + 12P ≤ 40 (francs)
We can divide both sides by 12 to get: N + P ≤ 40/12
Simplify to get: N + P ≤ 10/3
We can now REPHRASE the target question....
REPHRASED target question: Is N + P ≤ 10/3?
Given: 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils
We can write: 9N + 3P ≤ 20
Statement 1: 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils
We can write: 7N + 5P ≤ 20
We also know that 9N + 3P ≤ 20
Can we use these two inequalities to answer the REPHRASED target question?
It's hard to tell.
Let's add the inequalities to get: 16N + 8P ≤ 40
Divide both sides by 8 to get: 2N + P ≤ 5
In other words (N + P) + N ≤ 5
ASIDE: This inequality looks similar to our REPHRASED target question.
If we subtract N from both sides we get: (N + P) ≤ 5 - N
The REPHRASED target question asks Is N + P ≤ 10/3?
The answer to that question depends on the value of N.
So, let's test some (extreme) values that satisfy the given information:
Case a: N = 2 and P = 0 In this case, N + P = 2. So, the answer to the REPHRASED target question is YES, N + P ≤ 10/3
Case b: N = 0 and P = 4 In this case, N + P = 4. So, the answer to the REPHRASED target question is NO, N + P > 10/3
Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils
We can write: 4N + 8P ≤ 20
We also know that 9N + 3P ≤ 20
Let's add the inequalities to get: 13N + 11P ≤ 40
So close!!!
Too bad we don't have the same number of N's and P's!!
Wait. Perhaps we CAN have the same number of N's and P's if we create some EQUIVALENT inequalities.
First notice that 4N + 8P ≤ 20 is the same as 4(N + P) + 4P ≤ 20
And 9N + 3P ≤ 20 is the same as 3(N + P) + 6N ≤ 20
Now take 4(N + P) + 4P ≤ 20 and multiply both sides by 3 to get: 12(N + P) + 12P ≤ 60
And take 3(N + P) + 6N ≤ 20 and multiply both sides by 2 to get: 6(N + P) + 12N ≤ 40
ADD the two inequalities to get: 18(N + P) + 12P + 12N ≤ 100 [aha!! We now have the SAME number of P's and Q's]
Rewrite as: 18(N + P) + 12(P + N) ≤ 100
Simplify: 30(P + N) ≤ 100
Divide both sides by 30 to get: (P + N) ≤ 100/30
Simplify: (P + N) ≤ 10/3
Perfect! The answer to the REPHRASED target question is YES, N + P ≤ 10/3
Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT
Answer: B
Cheers,
Brent