Math Problem Statement

A sum of ₹3000 is to be given in the form of 63 prizes. If a prize is of either ₹100 or ₹25, find the number of prizes of each type.

Solution

The question from the image reads:

"A sum of ₹3000 is to be given in the form of 63 prizes. If a prize is of either ₹100 or ₹25, find the number of prizes of each type."

Let's solve this step by step:

Let:

  • xx be the number of ₹100 prizes.
  • yy be the number of ₹25 prizes.

Step 1: Form two equations based on the given information.

  1. Total number of prizes:
    x+y=63x + y = 63

  2. Total amount of money:
    100x+25y=3000100x + 25y = 3000

Step 2: Solve the system of equations.

From the first equation, we have: x+y=63    y=63xx + y = 63 \implies y = 63 - x

Substitute y=63xy = 63 - x into the second equation: 100x+25(63x)=3000100x + 25(63 - x) = 3000

Simplify and solve for xx: 100x+25×6325x=3000100x + 25 \times 63 - 25x = 3000 100x25x+1575=3000100x - 25x + 1575 = 3000 75x+1575=300075x + 1575 = 3000 75x=3000157575x = 3000 - 1575 75x=142575x = 1425 x=142575=19x = \frac{1425}{75} = 19

Step 3: Find yy:

y=63x=6319=44y = 63 - x = 63 - 19 = 44

Final Answer:

  • The number of ₹100 prizes is 19.
  • The number of ₹25 prizes is 44.

Let me know if you would like further details or explanations!

Here are five related questions:

  1. What is a system of linear equations, and how do you solve it?
  2. Can you graphically represent the solution to this problem?
  3. What happens if the prize amounts change?
  4. How would the equation change if there were only 60 prizes?
  5. How can the solution be verified using substitution?

Tip: Always double-check your solution by substituting the values back into the original equations!

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Math Problem Analysis

Mathematical Concepts

Algebra
Linear Equations
Systems of Equations

Formulas

x + y = 63
100x + 25y = 3000

Theorems

Solving systems of linear equations by substitution

Suitable Grade Level

Grades 8-10