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!
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A shop sells two variants of chocolates - one that costs
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BTGmoderatorLU
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Rule: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
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.
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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
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
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Jeff@TargetTestPrep
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Letting x = the number of $5 chocolates sold and y = the number of $3 chocolates sold, we can create the equation: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
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
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