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Joe bought only twenty cent stamps and thirty cent stamps. How many

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Joe bought only twenty cent stamps and thirty cent stamps. How many  [#permalink]

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New post 13 Dec 2010, 06:53
20
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

60% (01:22) correct 40% (00:48) wrong based on 804 sessions

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Joe bought only twenty cent stamps and thirty cent stamps. How many twenty cent stamps did he buy?

(1) Joe bought more than 8 twenty cent stamps.
(2) Joe bought a total of $2.50 worth of stamps.

The OA says C but I am getting B ..guys can you please explain

I applied my reasoning that .2x+.3y=2.50
so .2*5=1.0
.3*5=1.5
Hence we add and we can get 2.50 ..so B should be sufficient..I tried but there are no other integer choices for this combination.

Please help .... :(
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Re: Joe bought only twenty cent stamps and thirty cent stamps. How many  [#permalink]

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New post 13 Dec 2010, 07:26
2
10
rite2deepti wrote:
Joe bought only twenty cent stamps and thirty cent stamps. How many twenty cent stamps did he buy?

(1) Joe bought more than 8 twenty cent stamps.
(2) Joe bought a total of $2.50 worth of stamps.

The OA says C but I am getting B ..guys can you please explain

I applied my reasoning that .2x+.3y=2.50
so .2*5=1.0
.3*5=1.5
Hence we add and we can get 2.50 ..so B should be sufficient..I tried but there are no other integer choices for this combination.

Please help .... :(


Actually there are more integer solutions possible to satisfy statement (2):

Joe bought only twenty cent stamps and thirty cent stamps. How many twenty cent stamps did he buy?

(1) Joe bought more than 8 twenty cent stamps --> clearly insufficient.

(2) Joe bought a total of $2.50 worth of stamps --> \(2x+3y=25\): as \(x\) and \(y\) must be an integers we must check whether this equation has unique solution (for more on this check below links) --> \(2x=25-3y\), so 25 minus multiple of 3 must be multiple of 2, following pairs of (x,y) are possible: (2, 7), (5, 5), (8, 3), (11, 1). Not sufficient.

(1)+(2) As from (1) \(x>8\) then from (2) only one pair is left: \(x=11\) and \(y=1\). Sufficient.

Answer: C.

For more on this type of questions check:
eunice-sold-several-cakes-if-each-cake-sold-for-either-109602.html
martha-bought-several-pencils-if-each-pencil-was-either-a-100204.html
a-rental-car-agency-purchases-fleet-vehicles-in-two-sizes-a-105682.html
joe-bought-only-twenty-cent-stamps-and-thirty-cent-stamps-106212.html
a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-101966.html
joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-101743.html
at-an-amusement-park-tom-bought-a-number-of-red-tokens-and-126814.html
collections-confused-need-a-help-81062.html

P.S. Also you should have spotted that x=5 and y=5 for (2) is not correct solution as (1) says that x>8. So if x=5 were correct solution then statements would clearly contradict each other, but on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other.

Hope it's clear.
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Re: Joe bought only twenty cent stamps and thirty cent stamps. How many  [#permalink]

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New post 13 Dec 2010, 08:20
Agreed--the x>8 is the key.

C for me.
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Re: Joe bought only twenty cent stamps and thirty cent stamps. How many  [#permalink]

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New post 13 Dec 2010, 10:36
Let no.of 0. 20 cent stamps "x", and 0.30 cent stamps "y"
stmnt (1) is not sufficient because:
x(0.20)+(x+8)(0.30)=?. we do not know x and total amount. Eliminate it.

stmnt (2) is not sufficient because:
x(0.20) +y(0.30)= 2.50. two unknowns in one equation. eliminate it.

(1)+(2)
x(0.20)+(x+8)(0.30)= 2.50.
we can easily solve "x" and"y" values. so answer is C
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Re: Joe bought only twenty cent stamps and thirty cent stamps. How many  [#permalink]

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New post 13 Dec 2010, 10:48
4
TomB wrote:
Let no.of 0. 20 cent stamps "x", and 0.30 cent stamps "y"
stmnt (1) is not sufficient because:
x(0.20)+(x+8)(0.30)=?. we do not know x and total amount. Eliminate it.

stmnt (2) is not sufficient because:
x(0.20) +y(0.30)= 2.50. two unknowns in one equation. eliminate it.

(1)+(2)
x(0.20)+(x+8)(0.30)= 2.50.
we can easily solve "x" and"y" values. so answer is C


About the red part: generally such kind of linear equations (ax+by=c) have infinitely many solutions for x and y, and we cannot get single numerical values for the variables. But since x and y represent # of stamps then they must be non-negative integers and in this case 2x+3y=25 is no longer a simple linear equation it's Diophantine equation (equations whose solutions must be integers only) and for such kind on equations there might be only one combination of x and y possible to satisfy it. When you encounter such kind of problems you must always check by trial and error whether it's the case.

So statement (2) is not sufficient not because there is one equation and two variables but because there exist more than one pair of integers x and y for which this equation holds true: (2, 7), (5, 5), (8, 3), (11, 1).

For more on such questions check:
equations-100204.html
gmat-prep2-92785.html
car-dealer-data-sufficiency-105682.html
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Joe bought only twenty cent stamps and thirty cent stamps. How many  [#permalink]

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New post 13 Dec 2010, 12:44
Bunnel, thanks for the explanation.
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Re: Joe bought only twenty cent stamps and thirty cent stamps. How many  [#permalink]

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New post 13 Dec 2010, 22:23
For detailed explanation on how to solve for integral solutions, check out the following post:

http://gmatquant.blogspot.com/2010/11/integral-solutions-of-ax-by-c.html
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Re: Joe bought only twenty cent stamps and thirty cent stamps. How many  [#permalink]

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New post 05 Sep 2014, 23:59
rite2deepti wrote:
Joe bought only twenty cent stamps and thirty cent stamps. How many twenty cent stamps did he buy?

(1) Joe bought more than 8 twenty cent stamps.
(2) Joe bought a total of $2.50 worth of stamps.

The OA says C but I am getting B ..guys can you please explain

I applied my reasoning that .2x+.3y=2.50
so .2*5=1.0
.3*5=1.5
Hence we add and we can get 2.50 ..so B should be sufficient..I tried but there are no other integer choices for this combination.

Please help .... :(



1--> no suff.
2--> .20x+ .30y= 2.5 muliple by 5 to remove clumsy.

1x +1.5y= 12.5 ---> now if x= 11 then y=1 or if x= 6 then y= 5 and similar way you could get different count for 20 cent stamps.

now 1+2 --> from one we got more than 8 hence it must be suff cuz, if x= 9 or 10 , you wont get proper value for y as y should be an integer.

Hence 11 - 20 cents stamps
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Re: Joe bought only twenty cent stamps and thirty cent stamps. How many  [#permalink]

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New post 22 Jan 2018, 22:46
rite2deepti wrote:
Joe bought only twenty cent stamps and thirty cent stamps. How many twenty cent stamps did he buy?

(1) Joe bought more than 8 twenty cent stamps.
(2) Joe bought a total of $2.50 worth of stamps.

The OA says C but I am getting B ..guys can you please explain

I applied my reasoning that .2x+.3y=2.50
so .2*5=1.0
.3*5=1.5
Hence we add and we can get 2.50 ..so B should be sufficient.I tried but there are no other integer choices for this combination.

Please help .... :(


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Assume a and b are number of stamps with twenty cents and thirty cents, respectively.

Since we have 2 variables (a and b) and 0 equations, C is most likely to be the answer. So, we should consider 1) & 2) first.

Conditions 1) & 2):

a > 8
0.2a + 0.3b = 2.5
2a + 3b = 25

Case 1: a = 9
We have 18 + 3b = 25 or 3b = 7. It doesn't have an integer solution.

Case 2: a = 10
We have 20 + 3b = 25 or 3b = 7. It doesn't have an integer solution.

Case 3: a = 11
We have 22 + 3b = 25 or 3b = 3. Thus the pair of a = 11, b = 1 is a solution.

Case 4: a = 12
We have 24 + 3b = 25 or 3b = 1. It doesn't have an integer solution.

Thus we have a unique solution.

Both conditions together are sufficient.

Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT(Common Mistake Type) 4(A).

Condition 1) a > 8.
We can't specify the value of a from the condition 1).
The condition 1) is not sufficient.

Condition 2) 2a + 3b = 25
We have solutions a = 2, b = 7 and a = 5, b = 5.
Since it doesn't have a unique solution, the condition 2) is not sufficient.

Therefore, the answer is C.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: Joe bought only twenty cent stamps and thirty cent stamps. How many &nbs [#permalink] 22 Jan 2018, 22:46
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