MATH SOLVE

2 months ago

Q:
# Please help.Beth is working on a sculpture and needs to transport iron plates to her workshop. Each plate weighs 12 3/4 pounds. She has a cloth bag that can carry less than 243 1/3 pounds without tearing. The bag already contains 24 1/2 pounds of supplies. Assuming she can transport fractional parts of plates, how many plates, x, can she place into the bag? 12 3/4x + 24 1/2 < 243 1/3A. x < 21 1/153B. x < 17 25/153C. x > 17 25/153D. x > 21 1/153

Accepted Solution

A:

The answer is B) x < 17 25/153.

The first thing we do to solve this equation is rewrite all of the fractions using a common denominator. The LCD for 2, 4 and 3 is 12:

(12 9/12)x + 24 6/12 < 243 4/12

Subtract 24 6/12 from both sides:

(12 9/12)x + 24 6/12 - 24 6/12 < 243 4/12 - 24 6/12

(12 9/12)x < 218 10/12 (we lack 2/12 from being able to subtract, so we borrow)

Divide both sides by 12 9/12:

(12 9/12)x ÷ (12 9/12) < (218 10/12) ÷ (12 9/12)

Convert both mixed numbers to improper fractions:

x < (2626/12) ÷ (153/12)

Multiply by the reciprocal:

x < (2626/12) × (12/153)

x < 2626/153

x < 17 25/153

The first thing we do to solve this equation is rewrite all of the fractions using a common denominator. The LCD for 2, 4 and 3 is 12:

(12 9/12)x + 24 6/12 < 243 4/12

Subtract 24 6/12 from both sides:

(12 9/12)x + 24 6/12 - 24 6/12 < 243 4/12 - 24 6/12

(12 9/12)x < 218 10/12 (we lack 2/12 from being able to subtract, so we borrow)

Divide both sides by 12 9/12:

(12 9/12)x ÷ (12 9/12) < (218 10/12) ÷ (12 9/12)

Convert both mixed numbers to improper fractions:

x < (2626/12) ÷ (153/12)

Multiply by the reciprocal:

x < (2626/12) × (12/153)

x < 2626/153

x < 17 25/153