BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

A shop sells two variants of chocolates - one that costs

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Moderator
Posts: 2505
Joined: Sun Oct 15, 2017 1:50 pm
Followed by:6 members

A shop sells two variants of chocolates - one that costs

by BTGmoderatorLU » Thu May 17, 2018 4:32 pm

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

A shop sells two variants of chocolates - one that costs $3 and the other that costs $5. If the shop sold $108 chocolates on a given day, how many different combinations of (number of $3 sold, number of $5 sold) exist?

A. 6
B. 7
C. 8
D. 12
E. 15

The OA is C.

I solve this PS questions as follows,

Let the variant with $3 as cost be x and $5 be y. With this assumption, we can make the equation 3x + 5y = 108.

The number of combinations will be different values of (x, y) which satisfy the above equation.

Now we can rewrite the equation as 5y = 108 - 3x. For y to be an integer for a value of x, 108 - 3x needs to be divisible by 5.

A number is divisible by 5 if its last digit is either 0 or 5. Therefore 108 - 3x needs to end with 0 or 5. This would be possible only when 3x ends with 3 or 8.

First value of x which satisfies this condition is x = 1 and next is x = 6. This forms a series with x= 1, 6, 11 ... till 36. [ 108/3 =36]. Hence 8 values option D.\

Can anyone explain another way to solve this question? Thanks!
Top
Source: — Problem Solving |
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Thu May 17, 2018 6:19 pm
BTGmoderatorLU wrote:A shop sells two variants of chocolates - one that costs $3 and the other that costs $5. If the shop sold $108 chocolates on a given day, how many different combinations of (number of $3 sold, number of $5 sold) exist?

A. 6
B. 7
C. 8
D. 12
E. 15
Rule:
MULTIPLE OF X + MULTIPLE OF X = MULTIPLE OF X.
MULTIPLE OF X + NON-MULTIPLE OF X = NON-MULTIPLE OF X,

Since the price of each chocolate is either $3 or $5, and the total amount sold = $108, we get:
3x + 5y = 108.
Since 108 is an integer whose digits sum to a multiple of 3, 108 is a multiple of 3.
In accordance with the rule above:
Since 3x is a multiple of 3, and 108 is a multiple of 3, 5y must also be a multiple of 3.
Replacing 5y with 5(3k), we get:
3x + 5(3k) = 108
3(x + 5k) = 108
x + 5k = 36
x = 36 - 5k.

Since x must be NONNEGATIVE, 5k must be less than or equal to 36:
5k ≤ 36
k ≤ 7.2.
Nonnegative integer options for k:
0, 1, 2, 3, 4, 5, 6, 7.
Total options = 8.

The correct answer is C.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top

GMAT/MBA Expert

User avatar
Elite Legendary Member
Posts: 10392
Joined: Sun Jun 23, 2013 6:38 pm
Location: Palo Alto, CA
Thanked: 2867 times
Followed by:511 members
GMAT Score:800

by [email protected] » Fri May 18, 2018 10:09 am
Hi All,

We're told that a shop sells two variants of chocolates - one that costs $3 and the other that costs $5 and the shop sold $108 of chocolates on a given day. We're asked for the number of different combinations of (number of $3 sold, number of $5 sold) that could have been sold. This question comes down to basic Arithmetic, but there's a built-in pattern that you can use to save some time.

To start, $108 is NOT a multiple of $5, so you can't hit that total with just $5 chocolates. $105 IS a multiple of $5 though, so we could have....
One $3 and twenty-one $5 chocolates

At this point, it's worth noting that five $3 chocolates and three $5 chocolates both cost $15, so we can use this pattern to 'exchange' groups of chocolates and quickly determine all of the possibilities....

Four $3s and eighteen $5s
Nine $3s and fifteen $5s
Fourteen $3s and twelve $5s
Nineteen $3s and nine $5s
Twenty four $3s and six $5s
Twenty nine $3s and three $5s
Thirty four $3s and zero $5s

Total options = 8

Final Answer: C

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
Image
Top
User avatar
GMAT Instructor
Posts: 1462
Joined: Thu Apr 09, 2015 9:34 am
Location: New York, NY
Thanked: 39 times
Followed by:22 members

by Jeff@TargetTestPrep » Mon May 21, 2018 3:53 pm
BTGmoderatorLU wrote:A shop sells two variants of chocolates - one that costs $3 and the other that costs $5. If the shop sold $108 chocolates on a given day, how many different combinations of (number of $3 sold, number of $5 sold) exist?

A. 6
B. 7
C. 8
D. 12
E. 15
Letting x = the number of $5 chocolates sold and y = the number of $3 chocolates sold, we can create the equation:

5x + 3y = 108

5x = 108 - 3y

5x = 3(36 - y)

x = 3(36 - y)/5

We know that x must be an integer; since 3 is not evenly divisible by 5, we see that (36 - y) must therefore be evenly divisible by 5. In order for 3(36 - y)/5 = integer, y can be 1, 6, 11, 16, 21, 26, 31, and 36; so there are 8 possible combinations.

Answer: C

Jeffrey Miller
Head of GMAT Instruction
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews
Top
Post new topic Post Reply
• Page 1 of 1