Data Sufficiency

theCodeToGMAT 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]

vinay1983 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

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 is .

rakeshd347 wrote:

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 is .

B

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

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.

GMATGuruNY wrote:

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 MAX/MIN problem.
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.