Question

conversions
word\tfraction\tdecimal
one whole\t1/1 = 1\t1
one half\t½\t0.5
two halves\t2/2\t1
one fourth\t¼\t0.25
two fourths\t2/4 = ½\t0.5
three fourths\t¾\t0.75
four fourths\t4/4 = 1\t1

7a. (2 pts.) one fourth of a chocolate cake plus 1 vanilla cake contains 745 calories. one half of a chocolate cake and three fourths of a vanilla cake contains 765 calories.
what do you know?\twhat do you need to know?
how many calories are in one whole chocolate cake? ______ calories
how many calories are in one whole vanilla cake? ______ calories

7b. (2 pts.) one half of a cup of tea and three fourths cup of cocoa contains 635 calories. one fourth of a cup of tea and 1 cup of cocoa contain 625 calories.
what do you know?\twhat do you need to know?
how many calories are in one cup of tea? ______ calories
how many calories are in one cup of cocoa? ______ calories

Explanation:

7a Solution:

Let \( c \) be calories in 1 chocolate cake, \( v \) be calories in 1 vanilla cake.

Step1: Set up equations

From "One fourth of a chocolate cake plus 1 vanilla cake contains 745 calories":
\( \frac{1}{4}c + v = 745 \) --- (1)

From "One half of a chocolate cake and three fourths of a vanilla cake contains 765 calories":
\( \frac{1}{2}c + \frac{3}{4}v = 765 \) --- (2)

Step2: Eliminate \( c \) (multiply (1) by 2)

\( \frac{1}{2}c + 2v = 1490 \) --- (3)

Subtract (2) from (3):
\( (\frac{1}{2}c + 2v) - (\frac{1}{2}c + \frac{3}{4}v) = 1490 - 765 \)
\( \frac{5}{4}v = 725 \)

Step3: Solve for \( v \)

\( v = 725 \times \frac{4}{5} = 580 \)

Step4: Substitute \( v = 580 \) into (1)

\( \frac{1}{4}c + 580 = 745 \)
\( \frac{1}{4}c = 745 - 580 = 165 \)
\( c = 165 \times 4 = 660 \)

7b Solution:

Let \( t \) be calories in 1 cup tea, \( co \) be calories in 1 cup cocoa.

Step1: Set up equations

From "One half of a cup of tea and three fourths cup of cocoa contains 635 calories":
\( \frac{1}{2}t + \frac{3}{4}co = 635 \) --- (1)

From "One fourth of a cup of tea and 1 cup of cocoa contain 625 calories":
\( \frac{1}{4}t + co = 625 \) --- (2)

Step2: Eliminate \( t \) (multiply (2) by 2)

\( \frac{1}{2}t + 2co = 1250 \) --- (3)

Subtract (1) from (3):
\( (\frac{1}{2}t + 2co) - (\frac{1}{2}t + \frac{3}{4}co) = 1250 - 635 \)
\( \frac{5}{4}co = 615 \)

Step3: Solve for \( co \)

\( co = 615 \times \frac{4}{5} = 492 \)

Step4: Substitute \( co = 492 \) into (2)

\( \frac{1}{4}t + 492 = 625 \)
\( \frac{1}{4}t = 625 - 492 = 133 \)
\( t = 133 \times 4 = 532 \)

Answer:

7a:

• Chocolate cake: \( \boldsymbol{660} \) calories
• Vanilla cake: \( \boldsymbol{580} \) calories

7b:

• Tea: \( \boldsymbol{532} \) calories
• Cocoa: \( \boldsymbol{492} \) calories