Exercise: 3-D
Q1: \(6 + \left\{\frac{4}{3} + \left(\frac{3}{4} – \frac{1}{3}\right)\right\}\)
Step 1: Start by solving the innermost part of the expression:
\[
\frac{3}{4} – \frac{1}{3}
\]
To subtract fractions, we need to find a common denominator. The least common denominator (LCD) of 4 and 3 is 12.
Rewrite both fractions with a denominator of 12:
\[
\frac{3}{4} = \frac{9}{12}, \quad \frac{1}{3} = \frac{4}{12}
\]
Now subtract the fractions:
\[
\frac{9}{12} – \frac{4}{12} = \frac{5}{12}
\]Step 2: Now add this result to \(\frac{4}{3}\):
\[
\frac{4}{3} + \frac{5}{12}
\]
To add these fractions, we need a common denominator. The least common denominator of 3 and 12 is 12.
Rewrite \(\frac{4}{3}\) as:
\[
\frac{4}{3} = \frac{16}{12}
\]
Now add the fractions:
\[
\frac{16}{12} + \frac{5}{12} = \frac{21}{12}
\]Step 3: Finally, add 6 to the result:
\[
6 + \frac{21}{12}
\]
To add these, express 6 as a fraction with denominator 12:
\[
6 = \frac{72}{12}
\]
Now add:
\[
\frac{72}{12} + \frac{21}{12} = \frac{93}{12}
\]Step 4: Simplify the fraction \(\frac{93}{12}\).
Divide both the numerator and the denominator by their greatest common divisor (GCD), which is 3:
\[
\frac{93}{12} = \frac{93 \div 3}{12 \div 3} = \frac{31}{4}
\]Answer: \(\frac{31}{4}\) or \(7\frac{3}{4}\)
Q2: \(8 – \left\{\frac{3}{2} + \left(\frac{3}{5} – \frac{1}{2}\right)\right\}\)
Step 1: Start by solving the innermost part of the expression:
\[
\frac{3}{5} – \frac{1}{2}
\]
To subtract fractions, we need to find a common denominator. The least common denominator (LCD) of 5 and 2 is 10.
Rewrite both fractions with a denominator of 10:
\[
\frac{3}{5} = \frac{6}{10}, \quad \frac{1}{2} = \frac{5}{10}
\]
Now subtract the fractions:
\[
\frac{6}{10} – \frac{5}{10} = \frac{1}{10}
\]Step 2: Now add this result to \(\frac{3}{2}\):
\[
\frac{3}{2} + \frac{1}{10}
\]
To add these fractions, we need a common denominator. The least common denominator of 2 and 10 is 10.
Rewrite \(\frac{3}{2}\) as:
\[
\frac{3}{2} = \frac{15}{10}
\]
Now add the fractions:
\[
\frac{15}{10} + \frac{1}{10} = \frac{16}{10}
\]Step 3: Now subtract this result from 8:
\[
8 – \frac{16}{10}
\]
To perform the subtraction, express 8 as a fraction with denominator 10:
\[
8 = \frac{80}{10}
\]
Now subtract:
\[
\frac{80}{10} – \frac{16}{10} = \frac{64}{10}
\]Step 4: Simplify the fraction \(\frac{64}{10}\).
Divide both the numerator and the denominator by their greatest common divisor (GCD), which is 2:
\[
\frac{64}{10} = \frac{64 \div 2}{10 \div 2} = \frac{32}{5}
\]Answer: \(\frac{32}{5}\) or \(6\frac{2}{5}\)
Q3: \(\frac{1}{4}\left(\frac{1}{4} + \frac{1}{3}\right) – \frac{2}{5}\)
Step 1: Start by solving the expression inside the parentheses: \[ \frac{1}{4} + \frac{1}{3} \] To add these fractions, we need to find a common denominator. The least common denominator (LCD) of 4 and 3 is 12. Rewrite both fractions with a denominator of 12: \[ \frac{1}{4} = \frac{3}{12}, \quad \frac{1}{3} = \frac{4}{12} \] Now add the fractions: \[ \frac{3}{12} + \frac{4}{12} = \frac{7}{12} \]Step 2: Now multiply the result by \(\frac{1}{4}\): \[ \frac{1}{4} \times \frac{7}{12} = \frac{7}{48} \]Step 3: Now subtract \(\frac{2}{5}\) from \(\frac{7}{48}\): \[ \frac{7}{48} – \frac{2}{5} \] To subtract these fractions, we need to find a common denominator. The least common denominator (LCD) of 48 and 5 is 240. Rewrite both fractions with a denominator of 240: \[ \frac{7}{48} = \frac{35}{240}, \quad \frac{2}{5} = \frac{96}{240} \] Now subtract the fractions: \[ \frac{35}{240} – \frac{96}{240} = \frac{-61}{240} \]Answer: \(\frac{-61}{240}\)
Q4: \(2\frac{3}{4} – \left[3\frac{1}{8} \div \left\{5 – \left(4\frac{2}{3} – \frac{11}{12}\right)\right\}\right]\)
Step 1: Start by solving the innermost part of the expression:
\[
4\frac{2}{3} – \frac{11}{12}
\]
Convert \(4\frac{2}{3}\) into an improper fraction:
\[
4\frac{2}{3} = \frac{14}{3}
\]
Now subtract \(\frac{11}{12}\) from \(\frac{14}{3}\). To subtract these fractions, we need a common denominator. The least common denominator (LCD) of 3 and 12 is 12.
Rewrite \(\frac{14}{3}\) with denominator 12:
\[
\frac{14}{3} = \frac{56}{12}
\]
Now subtract:
\[
\frac{56}{12} – \frac{11}{12} = \frac{45}{12}
\]
Simplify \(\frac{45}{12}\):
\[
\frac{45}{12} = \frac{15}{4}
\]Step 2: Now subtract \(\frac{15}{4}\) from 5:
\[
5 – \frac{15}{4}
\]
Express 5 as a fraction with denominator 4:
\[
5 = \frac{20}{4}
\]
Now subtract:
\[
\frac{20}{4} – \frac{15}{4} = \frac{5}{4}
\]Step 3: Now divide \(3\frac{1}{8}\) by \(\frac{5}{4}\):
\[
3\frac{1}{8} \div \frac{5}{4}
\]
Convert \(3\frac{1}{8}\) into an improper fraction:
\[
3\frac{1}{8} = \frac{25}{8}
\]
To divide by a fraction, multiply by the reciprocal:
\[
\frac{25}{8} \times \frac{4}{5} = \frac{100}{40} = \frac{5}{2}
\]Step 4: Finally, subtract \(\frac{5}{2}\) from \(2\frac{3}{4}\):
\[
2\frac{3}{4} – \frac{5}{2}
\]
Convert \(2\frac{3}{4}\) into an improper fraction:
\[
2\frac{3}{4} = \frac{11}{4}
\]
Now find the common denominator. The least common denominator of 4 and 2 is 4.
Rewrite \(\frac{5}{2}\) with denominator 4:
\[
\frac{5}{2} = \frac{10}{4}
\]
Now subtract:
\[
\frac{11}{4} – \frac{10}{4} = \frac{1}{4}
\]Answer: \(\frac{1}{4}\)
Q5: \(12\frac{1}{2} – \left[8\frac{1}{2} + \left\{9 – \left(5 – \overline{3-2}\right)\right\}\right]\)
Step 1: Start by solving the innermost part of the expression, \(\overline{3 – 2}\). The overline indicates that the expression inside should be solved first: \[ 3 – 2 = 1 \]Step 2: Now, subtract 1 from 5: \[ 5 – 1 = 4 \]Step 3: Now, subtract 4 from 9: \[ 9 – 4 = 5 \]Step 4: Now, add \(8\frac{1}{2}\) to 5: \[ 8\frac{1}{2} + 5 = \frac{17}{2} + 5 = \frac{17}{2} + \frac{10}{2} = \frac{27}{2} \]Step 5: Now subtract \(\frac{27}{2}\) from \(12\frac{1}{2}\). First, convert \(12\frac{1}{2}\) to an improper fraction: \[ 12\frac{1}{2} = \frac{25}{2} \] Now subtract: \[ \frac{25}{2} – \frac{27}{2} = \frac{25 – 27}{2} = \frac{-2}{2} = -1 \]Answer: \(-1\)
Q6: \(1\frac{1}{5} \div \left\{2\frac{1}{3} – \left(5 + \overline{2-3}\right)\right\} – 3\frac{1}{2}\)
Step 1: Start by solving the innermost part of the expression, \(\overline{2 – 3}\). The overline indicates that the expression inside should be solved first:
\[
2 – 3 = -1
\]Step 2: Now, add \(-1\) to 5:
\[
5 + (-1) = 4
\]Step 3: Now, subtract 4 from \(2\frac{1}{3}\). First, convert \(2\frac{1}{3}\) to an improper fraction:
\[
2\frac{1}{3} = \frac{7}{3}
\]
Now subtract:
\[
\frac{7}{3} – 4 = \frac{7}{3} – \frac{12}{3} = \frac{-5}{3}
\]Step 4: Now, divide \(1\frac{1}{5}\) by \(\frac{-5}{3}\). First, convert \(1\frac{1}{5}\) to an improper fraction:
\[
1\frac{1}{5} = \frac{6}{5}
\]
Now divide by \(\frac{-5}{3}\), which is the same as multiplying by the reciprocal:
\[
\frac{6}{5} \times \frac{3}{-5} = \frac{18}{-25} = -\frac{18}{25}
\]Step 5: Finally, subtract \(3\frac{1}{2}\) from \(-\frac{18}{25}\). First, convert \(3\frac{1}{2}\) to an improper fraction:
\[
3\frac{1}{2} = \frac{7}{2}
\]
Now subtract:
\[
-\frac{18}{25} – \frac{7}{2}
\]
To subtract these fractions, we need to find the common denominator. The least common denominator (LCD) of 25 and 2 is 50.
Rewrite \(-\frac{18}{25}\) and \(\frac{7}{2}\) with denominator 50:
\[
-\frac{18}{25} = -\frac{36}{50}, \quad \frac{7}{2} = \frac{175}{50}
\]
Now subtract:
\[
-\frac{36}{50} – \frac{175}{50} = \frac{-36 – 175}{50} = \frac{-211}{50}
\]Answer: \(\frac{-211}{50} = 4\frac{11}{50}\)
Q7: \(\left(\frac{1}{2} + \frac{2}{3}\right) \div \left(\frac{3}{4} – \frac{2}{9}\right)\)
Step 1: First, solve the addition inside the parentheses: \(\frac{1}{2} + \frac{2}{3}\). To add these fractions, we need to find a common denominator. The least common denominator (LCD) of 2 and 3 is 6.
Rewrite \(\frac{1}{2}\) and \(\frac{2}{3}\) with denominator 6:
\[
\frac{1}{2} = \frac{3}{6}, \quad \frac{2}{3} = \frac{4}{6}
\]
Now, add the fractions:
\[
\frac{3}{6} + \frac{4}{6} = \frac{7}{6}
\]Step 2: Now, solve the subtraction inside the other parentheses: \(\frac{3}{4} – \frac{2}{9}\). To subtract these fractions, we need to find a common denominator. The least common denominator (LCD) of 4 and 9 is 36.
Rewrite \(\frac{3}{4}\) and \(\frac{2}{9}\) with denominator 36:
\[
\frac{3}{4} = \frac{27}{36}, \quad \frac{2}{9} = \frac{8}{36}
\]
Now, subtract the fractions:
\[
\frac{27}{36} – \frac{8}{36} = \frac{19}{36}
\]Step 3: Now, divide \(\frac{7}{6}\) by \(\frac{19}{36}\). Dividing by a fraction is the same as multiplying by its reciprocal:
\[
\frac{7}{6} \div \frac{19}{36} = \frac{7}{6} \times \frac{36}{19}
\]
Multiply the fractions:
\[
\frac{7 \times 36}{6 \times 19} = \frac{252}{114}
\]
Now simplify \(\frac{252}{114}\). The greatest common divisor (GCD) of 252 and 114 is 6. Divide both the numerator and the denominator by 6:
\[
\frac{252}{114} = \frac{252 \div 6}{114 \div 6} = \frac{42}{19}
\]Answer: \(\frac{42}{19} = 2\frac{4}{19}\)
Q8: \(\frac{6}{5} \text{ of } \left(3\frac{1}{3} – 2\frac{1}{2}\right) \div \left(2\frac{5}{21} – 2\right)\)
Step 1: First, simplify the expression inside the first parentheses: \(\left(3\frac{1}{3} – 2\frac{1}{2}\right)\). Convert both mixed fractions to improper fractions:
\[
3\frac{1}{3} = \frac{10}{3}, \quad 2\frac{1}{2} = \frac{5}{2}
\]
Now, subtract the fractions:
\[
\frac{10}{3} – \frac{5}{2}
\]
To subtract, we need to find a common denominator. The least common denominator (LCD) of 3 and 2 is 6. Rewrite the fractions with denominator 6:
\[
\frac{10}{3} = \frac{20}{6}, \quad \frac{5}{2} = \frac{15}{6}
\]
Now, subtract the fractions:
\[
\frac{20}{6} – \frac{15}{6} = \frac{5}{6}
\]Step 2: Next, simplify the expression inside the second parentheses: \(\left(2\frac{5}{21} – 2\right)\). Convert \(2\frac{5}{21}\) to an improper fraction:
\[
2\frac{5}{21} = \frac{47}{21}
\]
Now subtract \(2\) from \(\frac{47}{21}\). Convert \(2\) to \(\frac{42}{21}\) and subtract:
\[
\frac{47}{21} – \frac{42}{21} = \frac{5}{21}
\]Step 3: Now, divide the results of the two expressions:
\[
\frac{5}{6} \div \frac{5}{21}
\]
Dividing by a fraction is the same as multiplying by its reciprocal:
\[
\frac{5}{6} \times \frac{21}{5}
\]
Multiply the fractions:
\[
\frac{5 \times 21}{6 \times 5} = \frac{105}{30}
\]
Now simplify \(\frac{105}{30}\). The greatest common divisor (GCD) of 105 and 30 is 15. Divide both the numerator and the denominator by 15:
\[
\frac{105}{30} = \frac{105 \div 15}{30 \div 15} = \frac{7}{2}
\]Step 4: Now, multiply the result by \(\frac{6}{5}\):
\[
\frac{6}{5} \times \frac{7}{2}
\]
Multiply the fractions:
\[
\frac{6 \times 7}{5 \times 2} = \frac{42}{10}
\]
Simplify \(\frac{42}{10}\) by dividing both the numerator and the denominator by 2:
\[
\frac{42}{10} = \frac{42 \div 2}{10 \div 2} = \frac{21}{5}
\]Answer: \(\frac{21}{5} = 4\frac{1}{5}\)
Q9: \((10\frac{1}{8}\ of\ \frac{4}{5})\ \div(\frac{35}{36}\ of\ \frac{20}{49})\)
Step 1: First, simplify \(10\frac{1}{8}\) as an improper fraction:
\[
10\frac{1}{8} = \frac{81}{8}
\]
Now, find \(10\frac{1}{8} \text{ of } \frac{4}{5}\):
\[
\frac{81}{8} \times \frac{4}{5} = \frac{81 \times 4}{8 \times 5} = \frac{324}{40}
\]
Simplify \(\frac{324}{40}\) by dividing both the numerator and the denominator by 4 (their GCD):
\[
\frac{324}{40} = \frac{324 \div 4}{40 \div 4} = \frac{81}{10}
\]Step 2: Next, simplify \(\frac{35}{36} \text{ of } \frac{20}{49}\):
\[
\frac{35}{36} \times \frac{20}{49} = \frac{35 \times 20}{36 \times 49} = \frac{700}{1764}
\]
Simplify \(\frac{700}{1764}\) by dividing both the numerator and the denominator by 28 (their GCD):
\[
\frac{700}{1764} = \frac{700 \div 28}{1764 \div 28} = \frac{25}{63}
\]Step 3: Now, divide the two results obtained in Step 1 and Step 2:
\[
\frac{81}{10} \div \frac{25}{63}
\]
Dividing by a fraction is the same as multiplying by its reciprocal:
\[
\frac{81}{10} \times \frac{63}{25}
\]
Multiply the fractions:
\[
\frac{81 \times 63}{10 \times 25} = \frac{5103}{250}
\]
Simplify \(\frac{5103}{250}\). Since 5103 and 250 have no common factors, the fraction is already in its simplest form.
Answer: \(\frac{5103}{250} = 20\frac{103}{250}\)
Q10: \(5\frac{3}{4}-\frac{3}{7}\times15\frac{3}{4}+2\frac{2}{35}\div1\frac{11}{25}\)
Step 1: First, convert the mixed fractions into improper fractions:
\[
5\frac{3}{4} = \frac{23}{4}, \quad 15\frac{3}{4} = \frac{63}{4}, \quad 2\frac{2}{35} = \frac{72}{35}, \quad 1\frac{11}{25} = \frac{36}{25}
\]Step 2: Now, evaluate each part of the expression step by step.Start with the multiplication \(\frac{3}{7} \times 15\frac{3}{4}\):
\[
\frac{3}{7} \times \frac{63}{4} = \frac{3 \times 63}{7 \times 4} = \frac{189}{28}
\]
Simplify \(\frac{189}{28}\) by dividing both the numerator and the denominator by 7:
\[
\frac{189}{28} = \frac{189 \div 7}{28 \div 7} = \frac{27}{4}
\]Step 3: Now, evaluate the division part \(\frac{72}{35} \div \frac{36}{25}\):
\[
\frac{72}{35} \div \frac{36}{25} = \frac{72}{35} \times \frac{25}{36}
\]
Multiply the fractions:
\[
\frac{72 \times 25}{35 \times 36} = \frac{1800}{1260}
\]
Simplify \(\frac{1800}{1260}\) by dividing both the numerator and the denominator by 180:
\[
\frac{1800}{1260} = \frac{1800 \div 180}{1260 \div 180} = \frac{10}{7}
\]Step 4: Now, substitute the results back into the original expression:
\[
5\frac{3}{4} – \frac{3}{7} \times 15\frac{3}{4} + 2\frac{2}{35} \div 1\frac{11}{25} = \frac{23}{4} – \frac{27}{4} + \frac{10}{7}
\]
Now, combine the terms:
\[
\frac{23}{4} – \frac{27}{4} = \frac{-4}{4} = -1
\]
So, the expression becomes:
\[
-1 + \frac{10}{7}
\]
To add these fractions, convert \(-1\) into \(\frac{-7}{7}\):
\[
\frac{-7}{7} + \frac{10}{7} = \frac{-7 + 10}{7} = \frac{3}{7}
\]Answer: \(\frac{3}{7}\)
Q11: \(\frac{3}{4}\ of\ 7\frac{3}{7}-5\frac{3}{5}\div3\frac{4}{15}\)
Step 1: First, convert the mixed fractions into improper fractions:
\[
7\frac{3}{7} = \frac{52}{7}, \quad 5\frac{3}{5} = \frac{28}{5}, \quad 3\frac{4}{15} = \frac{49}{15}
\]Step 2: Start with the multiplication part \(\frac{3}{4} \ of\ 7\frac{3}{7}\):
\[
\frac{3}{4} \times \frac{52}{7} = \frac{3 \times 52}{4 \times 7} = \frac{156}{28}
\]
Simplify \(\frac{156}{28}\) by dividing both the numerator and the denominator by 4:
\[
\frac{156}{28} = \frac{156 \div 4}{28 \div 4} = \frac{39}{7}
\]Step 3: Now, evaluate the division part \(\frac{28}{5} \div 3\frac{4}{15}\):
\[
\frac{28}{5} \div \frac{49}{15} = \frac{28}{5} \times \frac{15}{49}
\]
Multiply the fractions:
\[
\frac{28 \times 15}{5 \times 49} = \frac{420}{245}
\]
Simplify \(\frac{420}{245}\) by dividing both the numerator and the denominator by 35:
\[
\frac{420}{245} = \frac{420 \div 35}{245 \div 35} = \frac{12}{7}
\]Step 4: Now, substitute the results back into the original expression:
\[
\frac{39}{7} – \frac{12}{7}
\]
Since the denominators are the same, subtract the numerators:
\[
\frac{39 – 12}{7} = \frac{27}{7}
\]Answer: \(\frac{27}{7} = 3\frac{6}{7}\)
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