Mathnasium 13303 Shelbyville Road, #103, Middletown KY 40223 (502) 409-6284   middletownky@mathnasium.com

Contact Us for Complimentary Assessment

* indicates a required field
  • phone number format invalid
  • email format invalid
Problems detected, please review the form.
protected by reCAPTCHA
Privacy - Terms

News from Mathnasium of Middletown KY

Problem of the Week 05-31-16

May 31, 2016

Lower Elementary:
Question: Perry has $20.00. He spends $2.75 on breakfast, $10.00 on lunch, and $6.45 on dinner. How much money does he have left?
Answer:  80¢
Solution:  Perry starts with $20.00, then spends $2.75, so he has $20.00 – $2.75 = $17.25 left after breakfast. After lunch, he has $17.25 – $10.00 = $7.25 left. After dinner, he has $7.25 – $6.45 = $0.80, or 80¢.

Upper Elementary:
Question: Joseph owns and runs a bakery. He makes 10 batches of cookies. Each batch yields 18 cookies. If he sells the cookies in boxes of a dozen, how many boxes can he fill?
Answer:  15 boxes
Solution:   To find the total cookies, multiply the number of batches times the number of cookies per batch. 10 × 18 = 180 cookies. Each box gets a dozen (12) cookies, so we divide 180 cookies by 12 to get the number of boxes. 180 ÷ 12 = 15 boxes of cookies.

Middle School:
Question: It takes Cameron 2 minutes to make a ham sandwich. It takes Morgan 3 minutes to make a ham sandwich. If they work together, how many ham sandwiches can they make in an hour?
Answer:  50 ham sandwiches
Solution:  If it takes Cameron 2 minutes to make a sandwich, then he can make 60 ÷ 2 = 30 sandwiches in an hour. Morgan takes 3 minutes to make a sandwich, or 60 ÷ 3 = 20 sandwiches. If they work together, they make 30 + 20 = 50 sandwiches in an hour.

Algebra and Up:
Question: Table A orders two steaks and one salad. Table B orders three steaks and four salads.  If Table A’s bill is $53 and Table B’s bill is $112, what is the cost of one steak and one salad?
Answer:  $33
Solution:  Let’s call the cost of a steak x and the cost of a salad y. We can write the problem as two linear equations. 2x + y = 53 and 3x + 4y = 112. One way to solve this problem is to use substitution. We can rewrite the first equation as y = 53 – 2x and substitute that value of y into the second equation. Solving for x, we have x = 20 So, one steak will cost $20. To find the price for the salad, we substitute the value for x into the first equation. So, one salad will cost $13.