+24 Connie has a number of gold bars, all of different masses. She gives the 17 lightest bars, which represent 30% of the total mass, to Brennan. She gives the 25 heaviest bars, which represent 48% of the total mass, to Maya. She gives the rest of the bars to Blair. How many bars did Blair receive?
The total mass of the gold bars is 100%. So, 30% of the total mass is 0.3 * 100% = 30%. This means that the 17 lightest bars have a total mass of 30%.
Similarly, the 25 heaviest bars have a total mass of 48%. So, the remaining 22% of the total mass is the mass of the bars that Blair received.
Since the mass of each bar is the same, the number of bars that Blair received is 22% / 1% = 22 bars.
Therefore, Blair received 22 bars.
+2487 You are absolutely full of Intentional moronic bullshit, WAT (Wrong Answer Troll).
Since the mass of each bar is the same <--- This statement is proof you intentionally answered this wrong.
The question clearly says, Connie has a number of gold bars, all of different masses
You are an unethical and destructive asshole!
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Here’s a descriptive solution for this problem:
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\(\text {Let B = the total number of bars.}\\ \text {Then }\\ 0.30(B) = 17 \text {(lightest bars)} | \small B = \dfrac {17}{0.30} \approx 56 \text { total bars }\\ 0.48(B) = 25 \text { (heaviest bars)} | \small B= \dfrac {25} {0.48} \approx 52 \text { total bars }\\ \text {Weighted average estimate of total number of bars: } \small \dfrac {(17 * 56 + 25 * 52)} {(17 + 25)} \approx 54 \text { bars } \\ \text {Fraction of total mass remaining:} (0.22) \\ 0.22(54) \approx 12 \text { bars }\\ \text {Total bars: } 54 \\ \text {Blair receives 12 of the 54 bars.}\\ \)
GA
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