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Difficulty: Very hardDividing in a Ratio

A military supply depot distributes a shipment of Meals Ready-to-Eat (MREs) among three battalions: Alpha, Bravo, and Charlie. Battalion Alpha receives 13\frac{1}{3} of the total shipment. The remaining MREs are then divided between Battalion Bravo and Battalion Charlie in a ratio of 3:53 : 5 respectively. If Battalion Charlie receives exactly 1,2001,200 more MREs than Battalion Bravo, what is the total number of MREs in the entire shipment?

Answer: 7200 MREs

Answer

The total number of MREs in the entire shipment is 7,200.
The correct total shipment is found by reasoning backward from the difference. Bravo and Charlie split the remaining 2/3 of the shipment in a 3:5 ratio, meaning there are 8 total parts for that remainder. Charlie receives 2 more parts than Bravo (5−3=25 - 3 = 2). Since these 2 parts equal 1,200 MREs, 1 part equals 600 MREs. The entire remainder (all 8 parts) is therefore 8×600=4,8008 \times 600 = 4,800 MREs. Since this 4,800 represents the 2/3 of the total shipment left after Alpha's portion, the entire original shipment must be (4,800÷2)×3=7,200(4,800 \div 2) \times 3 = 7,200 MREs.

Step-by-Step Solution

1
Define a variable for the total shipment and calculate the remaining amount after Alpha's share is removed.
Let the total shipment be T. Alpha receives 13T\frac{1}{3}T, so the remaining portion for Bravo and Charlie is T−13T=23TT - \frac{1}{3}T = \frac{2}{3}T.
We must determine exactly what portion of the whole shipment is being divided between the remaining two battalions.
2
Determine the fractional share of the remainder that Bravo and Charlie each receive based on their 3:53:5 ratio.
The ratio 3:53:5 has 3+5=83 + 5 = 8 total parts. Bravo gets 38\frac{3}{8} of the remainder, and Charlie gets 58\frac{5}{8} of the remainder.
Ratios represent parts of a whole; in this step, the 'whole' is the remaining 23T\frac{2}{3}T of the shipment.
3
Express Bravo's and Charlie's individual shares in terms of the total original shipment T.
Bravo's share is 38×23T=14T\frac{3}{8} \times \frac{2}{3}T = \frac{1}{4}T. Charlie's share is 58×23T=512T\frac{5}{8} \times \frac{2}{3}T = \frac{5}{12}T.
Converting their shares back to fractions of the original total allows us to set up an algebraic equation for the entire shipment.
4
Set up an equation using the known difference between Charlie's and Bravo's shares.
Charlie receives 1,200 more than Bravo, so: 512T−14T=1200\frac{5}{12}T - \frac{1}{4}T = 1200.
This translates the word problem's condition into a solvable algebraic equation.
5
Solve the equation for T.
Convert 14T\frac{1}{4}T to 312T\frac{3}{12}T. The equation becomes 212T=1200\frac{2}{12}T = 1200, which simplifies to 16T=1200\frac{1}{6}T = 1200. Multiplying both sides by 6 yields T=7200T = 7200.
Isolating the variable gives the final total number of MREs.

Key Concept

Dividing a remaining quantity into ratio parts and utilizing the algebraic difference between parts to find an original whole.
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Topics

Dividing in a Ratio

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