Solve the following expression:

[tex]\[ \frac{2}{3} + \frac{1}{4} \times \frac{2}{5} - \frac{3}{4} + \frac{5}{2} = \][/tex]

A. [tex]\[ -\frac{67}{50} \][/tex]

B. [tex]\[ \frac{13}{15} \][/tex]

C. [tex]\[ -\frac{23}{50} \][/tex]

D. [tex]\[ \frac{7}{15} \][/tex]



Answer :

Let's solve this step-by-step:

Given expression:
[tex]\[ \frac{2}{3} + \frac{1}{4} \times \frac{2}{5} - \frac{3}{4} + \frac{5}{2} \][/tex]

Step 1: Evaluate the multiplication
[tex]\[ \frac{1}{4} \times \frac{2}{5} = \frac{1 \cdot 2}{4 \cdot 5} = \frac{2}{20} = \frac{1}{10} \][/tex]

Step 2: Simplify the expression by replacing the multiplication result
[tex]\[ \frac{2}{3} + \frac{1}{10} - \frac{3}{4} + \frac{5}{2} \][/tex]

Step 3: Find a common denominator for the addition and subtraction
The denominators we have are 3, 10, 4, and 2. The least common multiple (LCM) of these numbers is 60.

Step 4: Convert all fractions to have the common denominator of 60
[tex]\[ \frac{2}{3} = \frac{2 \times 20}{3 \times 20} = \frac{40}{60} \][/tex]
[tex]\[ \frac{1}{10} = \frac{1 \times 6}{10 \times 6} = \frac{6}{60} \][/tex]
[tex]\[ \frac{3}{4} = \frac{3 \times 15}{4 \times 15} = \frac{45}{60} \][/tex]
[tex]\[ \frac{5}{2} = \frac{5 \times 30}{2 \times 30} = \frac{150}{60} \][/tex]

Step 5: Perform the addition and subtraction
[tex]\[ \frac{40}{60} + \frac{6}{60} - \frac{45}{60} + \frac{150}{60} \][/tex]
Combine all the fractions:
[tex]\[ \frac{40 + 6 - 45 + 150}{60} = \frac{151}{60} \][/tex]

Step 6: Simplify if possible (it isn't in this case since 151 is a prime number and does not divide evenly by 60, and 60 already has its prime factors 2, 3, and 5)
So, the result is:
[tex]\[ \frac{151}{60} \][/tex]

Step 7: Compare the resulting fraction with the given options:
The decimal representation of this fraction is [tex]\(\approx 2.5166666666666666\)[/tex], which is clearly none of the given answer choices [tex]\(-\frac{67}{50}\)[/tex], [tex]\(\frac{13}{15}\)[/tex], [tex]\(-\frac{23}{50}\)[/tex], or [tex]\(\frac{7}{15}\)[/tex].

Therefore, the original problem seems to have no correct option based on our calculations.