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What will come in place of question mark...

What will come in place of question mark (?) .
`1/4 + 3 1/2 + 1 1/4 + 1 1/2 = ?`

A

`2 1/2 `

B

`5//21`

C

3

D

4

Text Solution

AI Generated Solution

The correct Answer is:
To solve the equation \( \frac{1}{4} + 3 \frac{1}{2} + 1 \frac{1}{4} + 1 \frac{1}{2} = ? \), we will follow these steps: ### Step 1: Convert Mixed Numbers to Improper Fractions First, we need to convert the mixed numbers into improper fractions. 1. For \( 3 \frac{1}{2} \): \[ 3 \frac{1}{2} = \frac{3 \times 2 + 1}{2} = \frac{6 + 1}{2} = \frac{7}{2} \] 2. For \( 1 \frac{1}{4} \): \[ 1 \frac{1}{4} = \frac{1 \times 4 + 1}{4} = \frac{4 + 1}{4} = \frac{5}{4} \] 3. For \( 1 \frac{1}{2} \): \[ 1 \frac{1}{2} = \frac{1 \times 2 + 1}{2} = \frac{2 + 1}{2} = \frac{3}{2} \] ### Step 2: Rewrite the Expression Now we can rewrite the entire expression using improper fractions: \[ \frac{1}{4} + \frac{7}{2} + \frac{5}{4} + \frac{3}{2} \] ### Step 3: Find a Common Denominator The denominators are 4 and 2. The least common multiple (LCM) of 4 and 2 is 4. We will convert all fractions to have a denominator of 4. 1. \( \frac{1}{4} \) remains \( \frac{1}{4} \). 2. Convert \( \frac{7}{2} \) to a denominator of 4: \[ \frac{7}{2} = \frac{7 \times 2}{2 \times 2} = \frac{14}{4} \] 3. \( \frac{5}{4} \) remains \( \frac{5}{4} \). 4. Convert \( \frac{3}{2} \) to a denominator of 4: \[ \frac{3}{2} = \frac{3 \times 2}{2 \times 2} = \frac{6}{4} \] ### Step 4: Add the Fractions Now we can add all the fractions together: \[ \frac{1}{4} + \frac{14}{4} + \frac{5}{4} + \frac{6}{4} = \frac{1 + 14 + 5 + 6}{4} = \frac{26}{4} \] ### Step 5: Simplify the Fraction Now we simplify \( \frac{26}{4} \): \[ \frac{26 \div 2}{4 \div 2} = \frac{13}{2} \] ### Step 6: Convert to Mixed Number To convert \( \frac{13}{2} \) to a mixed number: \[ 13 \div 2 = 6 \quad \text{(remainder 1)} \] So, \( \frac{13}{2} = 6 \frac{1}{2} \). ### Final Answer Thus, the final answer is: \[ \boxed{6 \frac{1}{2}} \]
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UPKAR PUBLICATION -PRACTICE SET-2-PART-III Quantitative Aptitude
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