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faust18 [17]
3 years ago
10

Jacks mother gave him 50 chocolates to give to his friends at his birthday party. He gave 3 chocolates to each of his friends an

d still had 2 chocolates left. write an equation to determine the number of friends (z) at Jack's party. Find the number of friends at Jacks party, friends Answer Show me Stuck? Wat Two-step equa
Mathematics
2 answers:
nevsk [136]3 years ago
7 0

16

Explanation:

Okay so Jack started with 50 chocolates, and ended with 2. 
The simple way to calculate it would be by realising that Jack only distributed 48 chocolates. We can find how many times 3 fits into 48 by dividing <span>48÷3=16</span>.

Using algebra, we substitute the value we want to find with x. Here what we want to find is the number of friends that were at Jack's party.

We know that he started with 50 chocolates, then distributed <span>3×</span> the number of friends present (which is x). 
We write that down as <span>50−3x</span> 
(It's minus because when chocolates are distributed, Jack is taking away from what he has.)

We know that after this, there were only 2 chocolates left, so it's 
<span>50−3x=2</span>

Then we proceed by moving all the numbers to the right until only x is left:

<span>−3x=2−50</span>
<span>−3x=−48</span>

<span>x=<span><span>−48</span><span>−3</span></span></span>

<span>x=16</span>

Conclusion: The number of people that attended the party was 16

lawyer [7]3 years ago
6 0
Your answer is 16... ^-^
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Step-by-step explanation:

x = number of quarters.

y = number of dimes.

z = number of pennies.

0.25x + 0.1y + 0.01× = 1.05

but it is not possible to find a combination of a subset of the coins to create exactly 1 (= the change for $1).

like to have only 3 quarters and 3 dimes (and no pennies). that makes $1.05, and no combination can create exactly $1.

as a consequence

x < 4 (because 4 quarters is $1, and more than 4 quarters is already more than $1.05).

z < 5 (because with 5 or more pennies, since the total sum is $1.05, if I simply remove 5 pennies, I get $1 with the other coins, no matter what combination they are).

with 1 or 2 or 3 or 4 pennies I cannot build a total sum of $1.05, as the other coins will can only build a sum that ends in 0 or in 5.

so, the sum of these other coins and 1, 2, 3, 4 pennies can never end in 5.

that means, we have 0 pennies.

that rules out to have 0 or 2 quarters, because then together with the dimes the sum can only end in 0 (and actually also give change for $1).

so, we can have only 1 or 3 quarters.

that means the solutions are

1 quarters and 8 dimes

or

3 quarters and 3 dimes.

1 quarter and 8 dimes (and 0 pennies) is the desired solution, because it contains more coins.

for the maximum amount of money built by coins not providing change for $1 we need to remember :

4 quarters are $1 (and would provide change for $1). so, we always need less than 4 quarters. but also at least 1 quarter to keep the sum from ending with a 0 (which would allow again to build $1).

10 dimes are $1 (and provide change for $1). we always need less than 10 dimes.

5 or more pennies would always allow to take away or fill up to create a sum that ends in 0 and therefore build $1. so, we always need less than 5 pennies.

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max. 9 dimes

4 pennies.

if the the maximum possible sum would be higher than 1.19, then we could only increase in steps of 0.10 (dimes) : 1.29, 1.39, ...

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if it is 1.39, I could remove 1 quarter, 1 dime and 4 pennies to make $1.

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$1.19 can only be built by 1 quarter and 9 dimes and 4 pennies, so that we cannot build $1 out of a subset, or by 3 quarters, 4 dimes and 4 pennies.

and 1 quarter, 9 dimes and 4 pennies has more coins.

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