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Question

Difficulty: MediumLinear Equations in One and Two Variables

A boutique bakery sells custom gift baskets containing two types of pastries: almond tarts and chocolate croissants. Basket A contains 4 almond tarts and 3 chocolate croissants and costs 62.BasketBcontains3almondtartsand4chocolatecroissantsandcosts62. Basket B contains 3 almond tarts and 4 chocolate croissants and costs 57. What is the combined cost of 1 almond tart and 1 chocolate croissant?

  1. A
    $15
  2. $17Answer
  3. C
    $19
  4. D
    $30
  5. E
    $34

Answer

The combined cost of 1 almond tart and 1 chocolate croissant is $17.
By representing the prices of an almond tart and a chocolate croissant as tt and cc, we form the system 4t+3c=624t + 3c = 62 and 3t+4c=573t + 4c = 57. Adding both equations yields 7t+7c=1197t + 7c = 119. Dividing both sides by 7 gives t+c=17t + c = 17, which directly provides the combined price of 1 almond tart and 1 chocolate croissant.

Step-by-Step Solution

1
Set up a system of two linear equations using variables for the prices of the pastries.
Let tt be the price of one almond tart and cc be the price of one chocolate croissant.
Equation 1: 4t+3c=624t + 3c = 62
Equation 2: 3t+4c=573t + 4c = 57
Translating the problem statement into standard linear algebraic equations.
2
Add the two linear equations together.
(4t+3c)+(3t+4c)=62+57  ⟹  7t+7c=119(4t + 3c) + (3t + 4c) = 62 + 57 \implies 7t + 7c = 119
Symmetric coefficients allow finding the sum of t+ct + c without needing to solve for individual variables first.
3
Divide the combined equation by 7 to solve for (t+c)(t + c).
t+c=1197=17t + c = \frac{119}{7} = 17
Factoring out 7 gives 7(t+c)=1197(t + c) = 119, which simplifies directly to the requested sum.

Key Concept

Solving Systems of Linear Equations via Symmetric Coefficient Addition

Alternative Method

Alternatively, solve for one variable first: multiply Equation 1 by 3 (12t+9c=18612t + 9c = 186) and Equation 2 by 4 (12t+16c=22812t + 16c = 228). Subtracting the equations gives 7c=42  ⟹  c=67c = 42 \implies c = 6. Substituting c=6c = 6 into Equation 1 gives 4t+18=62  ⟹  4t=44  ⟹  t=114t + 18 = 62 \implies 4t = 44 \implies t = 11. Thus, t+c=11+6=17t + c = 11 + 6 = 17.
Estimated Time:1m 15s
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Topics

Linear Equations in One and Two Variables

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