## Algebra

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### Algebra

by RSK » Sat Feb 25, 2012 2:03 am
Hi guys,

Does anyone know how to solve such a problem quickly?

Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamps did she buy? (1) She bought$4.40 worth of stamps.
(2) She bought an equal number of $0.15 stamps and$0.29 stamps.

Thanks!

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by sanju09 » Sat Feb 25, 2012 4:17 am
RSK wrote:Hi guys,

Does anyone know how to solve such a problem quickly?
Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamps did she buy? (1) She bought$4.40 worth of stamps.
(2) She bought an equal number of $0.15 stamps and$0.29 stamps.

Thanks!
It depends on how promptly we raison d'Ãªtre the obvious resolutions.

Let's say Joanna bought x number of $0.15 stamps and y number of$0.29 stamps. The wordings only suggest that x and y are some positive integers. We have to find x.

I. If 0.15 x + 0.29 y = 4.40; or 15 x + 29 y = 440 and hence x = (440 - 29 y)/15. For x to be a positive integer, 440 - 29 y must be a positive multiple of 15, which may have a 0 or 5 in unit's place. It means that 440 - 29 y, or 29 y must have a 0 or 5 in unit's place, or y is a multiple of 5. Let's take y = 5 k, where k is a positive integer, then 440 - 145 k is a multiple of 15, or 88 - 29 k is a multiple of 3. This is close, as 29 times 3 is 87 and 1 is not a multiple of 3, so leave it and come down to take k = 2, which makes 88 - 29 k a multiple of 3. Hence, k must be 2 and hence y must be 10, and hence x must be 10. Sufficient

II. If x = y, then 15 x + 29 x = 440, or 44 x = 440, or x = 10. Sufficient

[spoiler]Take D[/spoiler]
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by RSK » Sat Feb 25, 2012 5:03 am
Hi Sanju,

A is sufficient on its own but B is not sufficient on its own so the answer is A.

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by sanju09 » Sat Feb 25, 2012 5:19 am
RSK wrote:Hi Sanju,

A is sufficient on its own but B is not sufficient on its own so the answer is A.
Agreed, thanks
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by kul512 » Sat Feb 25, 2012 11:44 pm
Main stem information provided-
x stamp of 0.15
y stamp of 0.29

Statement 1)

0.15x+0.29y=4.4
15x+29y=440
Now we have two options to rewrite this equation-
x=(440-29y)/15 OR y=(440-15x)/29=5(88-3X)/29
Which one preferable. CHOOSE THE FORM IN WHICH NUMERATOR AND DENOMINATOR ARE COMPARATIVELY CLOSE TO EACH OTHER, BECAUSE WE NEED TO FIND THE VALUE OF X FOR WHICH NUMERATOR IS DIVISIBLE BY DENOMINATOR. SO, CLOSER THE NUMERATOR AND DENOMINATOR ARE, LESSER NUMBER OF ATTEMPTS WE REQUIRE.
according to this thumb-rule, we will choose the second option
y=5(88-3x)/29...................(1)
now check for 88-3x=0,29,58,84
we get solution for 88-3x=58; x=10.
BUT REMEMBER THAT AN EQUATION OF THE FORM.....(1) DOES NOT NECESSARILY MEANS THAT STATEMENT IS SUFFICIENT. BECAUSE SOMETIME SUCH EQUATION CAN RETURN YOU MORE THAN ONE SOLUTION.

Statement 2)

x=y.
but that is of no use, till we know either total number of tickets OR total cost of tickets.

Sometimes there is very fine line between right and wrong: perspective.

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by [email protected] » Tue Nov 12, 2019 5:24 pm
RSK wrote:Hi guys,

Does anyone know how to solve such a problem quickly?

Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamps did she buy? (1) She bought$4.40 worth of stamps.
(2) She bought an equal number of $0.15 stamps and$0.29 stamps.

Thanks!
Target question: What is the value of C?

Statement 1: She bought $4.40 worth of stamps We can write the equation 0.15C + 0.29E = 4.40 IMPORTANT: In high school we learned that, if we're given 1 equation with 2 variables, we cannot find the value of either variable. However, if we restrict the variables to positive integers within a certain range of values, then there are times when we can find the value of a variable if we're given 1 equation with 2 variables. To determine whether this is the case here, let's examine all possible values of E. If E = 0, then the entire$4.40 was spent on $0.15 stamps. Since 0.15 does NOT divide evenly into 4.40, it cannot be the case that E = 0 If E = 1, then$0.29 was spent on $0.29 stamps, leaving the remaining$4.11 to be spent on $0.15 stamps. Since 0.15 does NOT divide evenly into 4.11, it cannot be the case that E = 1 If E = 2, then$0.58 was spent on $0.29 stamps, leaving the remaining$3.82 to be spent on $0.15 stamps. Since 0.15 does NOT divide evenly into 3.82, it cannot be the case that E = 2 If E = 3, then$0.87 was spent on $0.29 stamps, leaving the remaining$3.53 to be spent on $0.15 stamps. Since 0.15 does NOT divide evenly into 3.53, it cannot be the case that E = 3 IMPORTANT: At this point, we might speed up our solution by recognizing that, in order for 0.15 to divide evenly into a number, that number must end with 5 or 0. Also recognize that, in order for the resulting value to end with a 5 or 0, E must be divisible by 5 So, from this point on, we'll just check values of E that are divisible by 5. If E = 5, then$1.45 was spent on $0.29 stamps, leaving the remaining$2.95 to be spent on $0.15 stamps. NICE!$2.95 ends with a 5. So this MIGHT work. Unfortunately, 0.15 does NOT divide evenly into 2.95. So, it cannot be the case that E = 5

Keep going!

If E = 10, then $2.90 was spent on$0.29 stamps, leaving the remaining $1.50 to be spent on$0.15 stamps. 1.50/0.15 = 10 = C. So, one possible solution is E = 10 and C = 10
If E = 15, then $4.35 was spent on$0.29 stamps, leaving the remaining $0.05 to be spent on$0.15 stamps. Doesn't work.
If E = 20, then $5.80 was spent on$0.29 stamps. Hmmm. Looks like we can stop here!

So, there is only one possible scenario that meets the given conditions.
So, it MUST be the case that E = 10 and C = 10
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: She bought an equal number of $0.15 stamps and$0.29 stamps.
We have no idea how much money Joanna spent on stamps.
As such, there are infinitely many scenarios that satisfy statement 2. Here are two:
Case a: She bought 3 $0.29 stamps and 3$0.15 stamps. In this case, the answer to the target question is C = 3
Case b: She bought 8 $0.29 stamps and 8$0.15 stamps. In this case, the answer to the target question is C = 8
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Cheers,
Brent
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by SampathKp » Wed Dec 18, 2019 3:13 am
RSK wrote:Hi guys,

Does anyone know how to solve such a problem quickly?

Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamps did she buy? (1) She bought$4.40 worth of stamps.
(2) She bought an equal number of $0.15 stamps and$0.29 stamps.

Thanks!
Hi RSK

As you know, we dont need to get the specific answer for the question asked in the DS questions. In this case, we dont really need to answer the "how many $0.15 stamps did Joanna buy?" to solve this question. We can address this question without putting pen to paper. From data given let number of$0.15 stamps purchased by Joanna be X and number of $0.29 stamps purchased be Y So the given equation is 0.15X + 0.29Y = T , let T be total$ amount spent by Joanna on purchasing both the $0.15 and$0.29 stamps.

Now in (1) we are given value of T so the equation becomes 0.15X+ 0.29Y = 4.40. We still have 1 equation and 2 variable 1 and 2. So we cannot solve for either of the variables X and Y. This is clearly NOT Sufficient.

Now in (2), we are given that Joanna purchased equal number of $0.15 and$0.29 stamps. The equation becomes 0.15X +0.29X = ?? Although there is only one variable here the equation is not complete so we cannot solve for X. So even (2) in itself is NOT Sufficient to answer the question.

Quite clearly combining 1 and 2 we get an equation 0.15X+0.29X, = 4.40. Now we can solve for X and hence we have sufficient information to solve this question.

So Answer is C both statements together sufficient to answer the question and neither alone is sufficient to answer the question.

PS - Solving for X 0.44X = 4.40 X = 10 so Joanna purchased 10 stamps of \$0.15

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### Algebra

by regor60 » Wed Dec 18, 2019 8:07 am
SampathKp wrote:
Now in (1) we are given value of T so the equation becomes 0.15X+ 0.29Y = 4.40. We still have 1 equation and 2 variable 1 and 2. So we cannot solve for either of the variables X and Y. This is clearly NOT Sufficient.
Just because the equation cannot be solved by the process of elimination of variables doesn't mean it can't be solved. See the prior posts for various ways of arriving at the correct answer.

Said more clearly, when the solution can take on only integer values, it may be solvable, as is the case here

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### Re: Algebra

by gmatknight » Thu Feb 27, 2020 10:23 pm
I don't know if there is a single WAY to solve questions such as this. You have to find a method that works for you AND IN less than 2 minutes. Some people may hate working with all the decimals and multiply everything by 100. Some may love working with small values and keeps things anchored nearer to the actual reality. My go-to approach to solve this would not be to use algebra, though I daresay most on here would.

Just a Q50 guy.
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