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Mathematics problem on value-based questions involving matrices to calculate total promotional expenses in two cities.

Matrix problem involving cost of promotional activities in two cities, showing matrices A and B with telephone, house call, and letter costs and contact numbers.

Matrix problem involving cost of promotional activities in two cities, showing matrices A and B with telephone, house call, and letter costs and contact numbers.

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Show Answer Key & Explanations Step-by-step solution for: kvasnapsmaths.blogspot.in: XII-MATHS: WORD PROBLEMS IN MATRICES ...
Final Answer:
Rs. 2,850,000

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Explanation:
We need to find the total amount spent by the party in both cities.

Matrix A tells us the cost per contact for each method:
- Telephone: Rs. 140
- House Call: Rs. 200
- Letters: Rs. 150

Matrix B tells us how many contacts were made in each city:
- City X: 1000 telephone, 500 house calls, 5000 letters
- City Y: 3000 telephone, 1000 house calls, 10000 letters

To find the total cost, we multiply the number of contacts by their cost and add them up — this is exactly what matrix multiplication does.

Let’s compute cost for each city:

City X:
Telephone: 1000 × 140 = Rs. 140,000
House Call: 500 × 200 = Rs. 100,000
Letters: 5000150 = Rs. 750,000
Total for City X = 140,000 + 100,000 + 750,000 = Rs. 990,000

City Y:
Telephone: 3000 × 140 = Rs. 420,000
House Call: 1000 × 200 = Rs. 200,000
Letters: 10000 × 150 = Rs. 1,500,000
Total for City Y = 420,000 + 200,000 + 1,500,000 = Rs. 2,120,000

Now add both cities:
990,000 + 2,120,000 = Rs. 2,850,000

(Alternatively, using matrix multiplication: A is a 1×3 row matrix, B is a 3×2 matrix, so A × B gives a 1×2 row matrix with totals for City X and City Y. Then sum those two numbers.)
Parent Tip: Review the logic above to help your child master the concept of matrix multiplication word problems worksheet.
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