Question

One day my assistant Rajesh went to a stationary shop to purchase marker pens. He purchased the pens for Rs. 52 and he gave a Rs. 100 note to the shopkeeper. The shopkeeper offered Rajesh to take back Rs. 48 in the denominations of only Re. 1 , Rs. 2 and Rs. 5 . So, Rajesh received total 26 coins from the shopkeeper. Minimum number of Re. 1 coins which Rajesh has with him if he has at least one coin of each denomination offered by the shopkeeper : (a) 4 (b) 5 (c) 7 (d) 10

Short answer

Answer: 19

Step-by-step solution

Write the equations in terms of x, y, and z.

Let's represent the given information as equations: 1. x + y + z = 26 (total number of coins) 2. x + 2y + 5z = 48 (total value of coins) 3. x, y, z > 0 (at least one coin of each denomination)

Solve for one variable.

From the first equation (1), we can write the equation for z in terms of x and y: z = 26 - x - y

Substitute the value of z in the second equation.

Now, let's substitute the expression for z from step 2 into the second equation (2): x + 2y + 5(26 - x - y) = 48

Simplify and solve for x.

Now, we will simplify the equation from step 3 and solve for x: x + 2y + 130 - 5x - 5y = 48 4x + 3y = 130 - 48 4x + 3y = 82 Now let's isolate x: x = (82 - 3y)/4

Find the minimum value of x by substituting values of y.

We will now substitute the possible values of y into the equation from step 4 and find the smallest integer value of x, where x, y, and z are all greater than 0. y=1: x = (82 - 3(1))/4 = 79/4 (not an integer) y=2: x = (82 - 3(2))/4 = 76/4 = 19 (integer value of x) With y = 2, we find the minimum value of x is 19. Now we also need to find z. We can substitute the values of x and y in the equation for z from step 2: z = 26 - x - y = 26 - 19 - 2 = 5 Now we know that Rajesh has 19 Re. 1 coins, 2 Rs. 2 coins, and 5 Rs. 5 coins. Summing these, he does receive a total of 26 coins with a value of Rs. 48. All three values are greater than 0, so the constraints from the problem are satisfied. Thus, the answer is: Minimum number of Re.1 coins Rajesh has with him is 19, which is not listed among the given options (a), (b), (c), and (d).

Key concepts

Linear Equations

Linear equations are fundamental in algebra and represent relationships with one or more variables. Each term in a linear equation is either a constant or the product of a constant and a single variable. This particular exercise uses linear equations to find the number of coins of each type Rajesh received.
Linear equations in this context include:

• x + y + z = 26: This represents the total number of coins Rajesh received.
• x + 2y + 5z = 48: This shows the total monetary value of these coins.

By solving these equations, we can determine the quantities of each coin.
Linear equations are a staple in solving many types of algebraic problems. They often serve as the starting point for more complex mathematical tasks.

Coins Problem

Coins problems in algebra involve determining the number of coins of specific denominations that add up to a certain total number or monetary value. In our problem, we need to determine how many Re. 1, Rs. 2, and Rs. 5 coins Rajesh has.
These type of problems usually require setting up equations that represent the conditions given in the problem. Setting up equations such as those in the given exercise helps to break down the problem into manageable parts.

• The first equation deals with the total number of coins.
• The second equation considers the total value of these coins.

These equations aid in modeling the scenario mathematically, allowing us to solve for the unknowns.
Successfully tackling coins problems often involves logical thinking and strategic substitution to simplify equations.

Optimization

Optimization in mathematics involves finding the most efficient or optimal solution under given constraints. In this exercise, we are dealing with optimizing the number of Re. 1 coins Rajesh received while satisfying the total number and value constraints.
The approach typically involves:

• Substituting one variable in terms of others to reduce complexity.
• Finding integer values that satisfy all conditions.

By substituting and rearranging equations, we can easily identify when variables reach minimum or maximum permissible values.
Optimization problems are common in real-world scenarios where resources must be managed efficiently under restrictions.

Word Problems

Word problems are narrative-based exercises in math that involve translating words into mathematical expressions or equations. The given coins problem is an example, requiring us to decode a practical scenario into solvable equations.
When tackling word problems:

• Identify the key pieces of data and how they relate to each other.
• Translate these relationships into mathematical expressions or equations.

In this exercise, we used Rajesh's shopping scenario to derive equations for the number and value of the coins.
Word problems enhance critical thinking by requiring us to make connections between the real world and mathematical concepts. They develop a deeper understanding of math beyond abstract symbols and numbers.