Ms Shikongo, our mathematics teacher bought 40 calculators at N$ 8.20 each and a number of other calculators costing N$ 2.95 each. On all these items, she spent N$ 387.00. How many of the cheaper calculators did she buy?
The Correct Answer and Explanation is :
Let the number of the cheaper calculators be ( x ).
Step 1: Write an equation for the total cost
We know that:
- The price of one of the first type of calculators is N$ 8.20, and she bought 40 of them. Thus, the total cost for these calculators is ( 40 \times 8.20 = 328 ) N$.
- The price of one of the cheaper calculators is N$ 2.95, and she bought ( x ) of them. Thus, the total cost for these calculators is ( 2.95 \times x ).
The total amount spent on all calculators is N$ 387.00. So, the total cost of all calculators is the sum of the costs of the two types, which is given by:
[
328 + 2.95x = 387
]
Step 2: Solve for ( x )
To solve for ( x ), subtract 328 from both sides of the equation:
[
2.95x = 387 – 328
]
[
2.95x = 59
]
Now, divide both sides by 2.95 to find ( x ):
[
x = \frac{59}{2.95} = 20
]
Thus, Ms. Shikongo bought 20 of the cheaper calculators.
Explanation:
To verify, we can check the total cost with the calculated number of calculators:
- The cost of the 40 calculators at N$ 8.20 each is ( 40 \times 8.20 = 328 ) N$.
- The cost of the 20 cheaper calculators at N$ 2.95 each is ( 20 \times 2.95 = 59 ) N$.
Adding these together:
[
328 + 59 = 387
]
This matches the total cost of N$ 387.00, confirming the solution is correct. Therefore, Ms. Shikongo bought 20 of the cheaper calculators.