Question

Chris and Evan are selling bags of popcorn to raise money for a club at school. Chris started with 10 bags and he sells 4 bags each day. Evan began with 8 bags and sells 2 bags each day. How many days will it take before they both have the same number of bags remaining?
a.2 days
b 4days
c..1 day

Answer 1

First, let us look at how Chris (C) and Evan (E) sell their respective bags of popcorn.

C had 10 bags and sells at the rate of 4 bags/day

E had 8 bags and sells at the rate of 2 bags/day;

So, on Day 1;

C sells 4 bags out of 10

∴ (10 - 4) = 6 bags are remaining after Day 1

On the other hand,

E sells 2 bags out of 8

∴ (8 - 2) = 6 bags remain at the end of Day 1

Thus, it is obvious that on the very first day of sale, Chris and Evan have the same number of bags remaining.

Ans) (c) 1 day

Answer 2

SOLUTION

GIVEN

• Chris and Evan are selling bags of popcorn to raise money for a club at school.

• Chris started with 10 bags and he sells 4 bags each day.

• Evan began with 8 bags and sells 2 bags each day.

TO DETERMINE

The number of days it will take before they both have the same number of bags remaining

EVALUATION

Let it will take d days before Chris and Evan both have the same number of bags remaining

Now Chris started with 10 bags and he sells 4 bags each day.

So after d days number of bags of Chris

= 10 - 4d

Evan began with 8 bags and sells 2 bags each day.

So after d days number of bags of Evan

= 8 - 2d

So by the given condition

$\sf{10 - 4d = 8 - 2d}$

$\sf{ \implies \: - 4d + 2d = 8 - 10}$

$\sf{ \implies \: - 2d = - 2}$

$\sf{ \implies \: d = 1}$

Hence it will take 1 day before Chris and Evan both have the same number of bags remaining

FINAL ANSWER

It will take 1 day before Chris and Evan both have the same number of bags remaining

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