Connie has a number of gold bars, all of different weights. She gives the 24 lightest bars, which weigh 45% of the total weight, to Brennan. She gives the 13 heaviest bars, which weigh 26% of the total weight, to Maya. She gives the rest of the bars to Blair. How many bars did Blair receive?

Guest Jun 24, 2020

#1**0 **

Because Brennan has the lightest bars, and Maya has the heaviest bars, the average weight of Blair's bars must be between their amounts. We can then write an inequality and solve it. Set the total weight as w and the number of bars Blair received as x.

Average weight = Total weight/Number of bars

So Brennan's average weight = 0.45W/24

And Maya's average weight = 0.26W/13

Since the rest of the bars are given to Blair, she must have (1-0.45-0.26)W as her total weight.

Blair's average weight = 0.29W/x

Now we write our inequality:

\(\frac{0.45W}{24} < \frac{0.29W}{x} < \frac{0.26W}{13}\)

We can divide everyone by W.

\(\frac{0.45}{24} < \frac{0.29}{x} < \frac{0.26}{13}\)

Then split it into two equations.

\(\frac{0.45}{24} < \frac{0.29}{x}\) and \(\frac{0.29}{x} < \frac{0.26}{13}\)

Simplify by cross-multiplying to get:

\(0.45x<0.29*24\) and \(0.26x>0.29*13\)

Isolate x to get:

x < 15.47 and x > 14.5

14.5 < x < 15.47

The only integer value that satisfies this is 15, which is our final answer.

thelizzybeth Jun 24, 2020