RS Aggarwal Class 7 Mathematics Solutions Chapter 2 Fractions

Exercise 2A

Question 1. Which is greater in each of the following?
(i) \( \frac{5}{8} \) or \( \frac{7}{12} \)
(ii) \( \frac{5}{9} \) or \( \frac{11}{15} \)
(iii) \( \frac{11}{12} \) or \( \frac{15}{16} \)
Answer: Using cross multiplication to compare each pair:
(i) \( 5 \times 12 = 60 \) and \( 7 \times 8 = 56 \). Since 60 is bigger than 56, \( \frac{5}{8} > \frac{7}{12} \).
(ii) \( 5 \times 15 = 75 \) and \( 9 \times 11 = 99 \). Since 75 is smaller than 99, \( \frac{5}{9} < \frac{11}{15} \).
(iii) \( 11 \times 16 = 176 \) and \( 12 \times 15 = 180 \). Since 176 is smaller than 180, \( \frac{11}{12} < \frac{15}{16} \).
In simple words: To compare two fractions, multiply across diagonally and see which product is larger; the fraction on that side of the larger product is the bigger fraction.

Exam Tip: Always cross multiply carefully, numerator of one fraction with denominator of the other, and double check which side each product belongs to before concluding.

Question 2. Arrange the following fractions in ascending order.
(i) \( \frac{3}{4}, \frac{5}{6}, \frac{7}{9}, \frac{11}{12} \)
(ii) \( \frac{4}{5}, \frac{7}{10}, \frac{11}{15}, \frac{17}{20} \)
Answer:
(i) The LCM of 4, 6, 9 and 12 is 36. Changing each fraction to have 36 as its denominator: \( \frac{3}{4} = \frac{27}{36} \), \( \frac{5}{6} = \frac{30}{36} \), \( \frac{7}{9} = \frac{28}{36} \), \( \frac{11}{12} = \frac{33}{36} \). Since \( \frac{27}{36} < \frac{28}{36} < \frac{30}{36} < \frac{33}{36} \), the ascending order is \( \frac{3}{4} < \frac{7}{9} < \frac{5}{6} < \frac{11}{12} \).
(ii) The LCM of 5, 10, 15 and 20 is 60. Changing each fraction to have 60 as its denominator: \( \frac{4}{5} = \frac{48}{60} \), \( \frac{7}{10} = \frac{42}{60} \), \( \frac{11}{15} = \frac{44}{60} \), \( \frac{17}{20} = \frac{51}{60} \). Since \( \frac{42}{60} < \frac{44}{60} < \frac{48}{60} < \frac{51}{60} \), the ascending order is \( \frac{7}{10} < \frac{11}{15} < \frac{4}{5} < \frac{17}{20} \).
In simple words: Turn all fractions into ones with the same bottom number using the LCM, then it is easy to line them up from smallest to biggest just by looking at the top numbers.

Exam Tip: Finding the LCM of all denominators first makes comparing several fractions at once far quicker than comparing them two at a time.

Question 3. Arrange the following fractions in descending order.
(i) \( \frac{3}{4}, \frac{7}{8}, \frac{7}{12}, \frac{17}{24} \)
(ii) \( \frac{2}{3}, \frac{3}{5}, \frac{7}{10}, \frac{8}{15} \)
Answer:
(i) The LCM of 4, 8, 12 and 24 is 24. Changing each fraction: \( \frac{3}{4} = \frac{18}{24} \), \( \frac{7}{8} = \frac{21}{24} \), \( \frac{7}{12} = \frac{14}{24} \), \( \frac{17}{24} = \frac{17}{24} \). Since \( \frac{21}{24} > \frac{18}{24} > \frac{17}{24} > \frac{14}{24} \), the descending order is \( \frac{7}{8} > \frac{3}{4} > \frac{17}{24} > \frac{7}{12} \).
(ii) The LCM of 3, 5, 10 and 15 is 30. Changing each fraction: \( \frac{2}{3} = \frac{20}{30} \), \( \frac{3}{5} = \frac{18}{30} \), \( \frac{7}{10} = \frac{21}{30} \), \( \frac{8}{15} = \frac{16}{30} \). Since \( \frac{21}{30} > \frac{20}{30} > \frac{18}{30} > \frac{16}{30} \), the descending order is \( \frac{7}{10} > \frac{2}{3} > \frac{3}{5} > \frac{8}{15} \).
In simple words: Just like ascending order, convert every fraction to the same bottom number first, then arrange from biggest to smallest by comparing the top numbers.

Exam Tip: A quick sanity check: after converting, the fraction that was already closest to 1 (like 7/8 here) will usually come out on top in descending order.

Question 4. Reenu and Sonal had two apples of the same size. Reenu ate \( \frac{2}{7} \) of her apple and Sonal ate \( \frac{4}{5} \) of her apple. Who ate the larger part, and how much larger was it?
Answer: Comparing \( \frac{2}{7} \) and \( \frac{4}{5} \) by cross multiplication: \( 2 \times 5 = 10 \) and \( 4 \times 7 = 28 \). Since 10 is less than 28, \( \frac{2}{7} < \frac{4}{5} \), so Sonal ate the bigger portion. The difference is \( \frac{4}{5} - \frac{2}{7} = \frac{28 - 10}{35} = \frac{18}{35} \). Sonal's share was \( \frac{18}{35} \) more than Reenu's.
In simple words: Compare the two fractions first to see who ate more, then subtract the smaller from the bigger to find exactly how much extra was eaten.

Exam Tip: For "who got more" word problems, always state the winner clearly before showing the subtraction that finds the exact difference.

Question 5. Add the following:
(i) \( \frac{5}{9} + \frac{3}{9} \)
(ii) \( \frac{8}{9} + \frac{7}{12} \)
(iii) \( \frac{5}{6} + \frac{7}{8} \)
(iv) \( \frac{7}{12} + \frac{11}{16} + \frac{9}{24} \)
(v) \( 3\frac{4}{5} + 2\frac{3}{10} + 1\frac{1}{15} \)
(vi) \( 8\frac{3}{4} + 10\frac{2}{5} \)
Answer:
(i) \( \frac{5}{9} + \frac{3}{9} = \frac{8}{9} \)
(ii) LCM of 9 and 12 is 36. \( \frac{8}{9} + \frac{7}{12} = \frac{32}{36} + \frac{21}{36} = \frac{53}{36} = 1\frac{17}{36} \)
(iii) LCM of 6 and 8 is 24. \( \frac{5}{6} + \frac{7}{8} = \frac{20}{24} + \frac{21}{24} = \frac{41}{24} = 1\frac{17}{24} \)
(iv) LCM of 12, 16 and 24 is 48. \( \frac{28}{48} + \frac{33}{48} + \frac{18}{48} = \frac{79}{48} = 1\frac{31}{48} \)
(v) LCM of 5, 10 and 15 is 30. Writing as improper fractions and converting: \( \frac{19}{5} + \frac{23}{10} + \frac{16}{15} = \frac{114}{30} + \frac{69}{30} + \frac{32}{30} = \frac{215}{30} = 7\frac{5}{30} = 7\frac{1}{6} \)
(vi) LCM of 4 and 5 is 20. \( \frac{35}{4} + \frac{52}{5} = \frac{175}{20} + \frac{208}{20} = \frac{383}{20} = 19\frac{3}{20} \)
In simple words: Find the LCM of the bottom numbers first, rewrite every fraction with that same bottom number, then simply add the top numbers together.

Exam Tip: For mixed numbers, convert to improper fractions first before finding the LCM, otherwise it is easy to lose track of the whole-number part.

Question 6. Subtract:
(i) \( \frac{2}{7} \) from \( \frac{5}{7} \)
(ii) \( \frac{3}{4} \) from \( \frac{5}{6} \)
(iii) \( \frac{7}{10} \) from \( 3\frac{1}{5} \)
(iv) \( 4\frac{2}{3} \) from 7
(v) \( 1\frac{7}{15} \) from \( 3\frac{3}{10} \)
(vi) \( 1\frac{7}{15} \) from \( 2\frac{5}{9} \)
Answer:
(i) \( \frac{5}{7} - \frac{2}{7} = \frac{3}{7} \)
(ii) LCM of 6 and 4 is 12. \( \frac{5}{6} - \frac{3}{4} = \frac{10}{12} - \frac{9}{12} = \frac{1}{12} \)
(iii) \( 3\frac{1}{5} = \frac{16}{5} \). LCM of 5 and 10 is 10. \( \frac{32}{10} - \frac{7}{10} = \frac{25}{10} = \frac{5}{2} = 2\frac{1}{2} \)
(iv) \( 7 - 4\frac{2}{3} = \frac{7}{1} - \frac{14}{3} = \frac{21}{3} - \frac{14}{3} = \frac{7}{3} = 2\frac{1}{3} \)
(v) LCM of 10 and 15 is 30. \( \frac{33}{10} - \frac{22}{15} = \frac{99}{30} - \frac{44}{30} = \frac{55}{30} = 1\frac{5}{6} \)
(vi) LCM of 9 and 15 is 45. \( \frac{23}{9} - \frac{22}{15} = \frac{115}{45} - \frac{66}{45} = \frac{49}{45} = 1\frac{4}{45} \)
In simple words: Convert both mixed numbers into improper fractions, bring them to a shared bottom number using the LCM, then subtract the top numbers.

Exam Tip: Read subtraction word problems carefully to see which fraction is being taken away from which, the order matters.

Question 7. Simplify:
(i) \( \frac{2}{3} + \frac{5}{6} - \frac{1}{9} \)
(ii) \( 8 - 4\frac{1}{2} - 2\frac{1}{4} \)
(iii) \( 8\frac{5}{6} - 3\frac{3}{8} + 1\frac{7}{12} \)
Answer:
(i) LCM of 3, 6 and 9 is 18. \( \frac{12}{18} + \frac{15}{18} - \frac{2}{18} = \frac{25}{18} = 1\frac{7}{18} \)
(ii) LCM of 1, 2 and 4 is 4. \( \frac{32}{4} - \frac{18}{4} - \frac{9}{4} = \frac{5}{4} = 1\frac{1}{4} \)
(iii) LCM of 6, 8 and 12 is 24. \( \frac{212}{24} - \frac{81}{24} + \frac{38}{24} = \frac{169}{24} = 7\frac{1}{24} \)
In simple words: When a sum has several fractions with plus and minus signs, put them all over the LCM first and then work through the additions and subtractions together.

Exam Tip: Keep every term's sign attached to its numerator once you bring everything over the common denominator, so the final combined fraction stays correct.

Question 8. Aneeta bought \( 3\frac{3}{4} \) kg of apples and \( 4\frac{1}{2} \) kg of grapes. Find the total weight of fruits bought by her.
Answer: Total weight \( = 3\frac{3}{4} + 4\frac{1}{2} = \frac{15}{4} + \frac{9}{2} \). LCM of 2 and 4 is 4. \( \frac{15}{4} + \frac{18}{4} = \frac{33}{4} = 8\frac{1}{4} \) kg. Aneeta bought \( 8\frac{1}{4} \) kg of fruits in total.
In simple words: Simply add up the two weights, converting to a common denominator first, to get the combined weight of the fruits.

Exam Tip: For "total" word problems, always add the given quantities directly rather than overcomplicating the steps.

Question 9. The length and breadth of a rectangular park are \( 15\frac{3}{4} \) cm and \( 12\frac{1}{2} \) cm respectively. Find its perimeter.
Answer: Perimeter \( = 2 \times (\text{length} + \text{breadth}) = \left(15\frac{3}{4} + 12\frac{1}{2} + 15\frac{3}{4} + 12\frac{1}{2}\right) \) cm \( = \left(\frac{63}{4} + \frac{25}{2} + \frac{63}{4} + \frac{25}{2}\right) \) cm \( = \left(\frac{63+50+63+50}{4}\right) \) cm \( = \frac{226}{4} \) cm \( = 56\frac{1}{2} \) cm. The perimeter of the rectangle is \( 56\frac{1}{2} \) cm.
In simple words: Add up all four sides of the rectangle, two lengths and two breadths, to get the total distance around it.

Exam Tip: Remember the perimeter formula uses each side twice, do not just add length and breadth once and stop there.

Question 10. The actual width of a picture is \( 7\frac{3}{5} \) cm, but it is required to be \( 7\frac{3}{10} \) cm wide. By how much should the width be trimmed?
Answer: Actual width \( = 7\frac{3}{5} \) cm \( = \frac{38}{5} \) cm. Required width \( = 7\frac{3}{10} \) cm \( = \frac{73}{10} \) cm. Extra width \( = \frac{38}{5} - \frac{73}{10} = \frac{76}{10} - \frac{73}{10} = \frac{3}{10} \) cm. The width of the picture should be trimmed by \( \frac{3}{10} \) cm.
In simple words: Subtract the smaller required width from the bigger actual width to find out exactly how much needs to be cut off.

Exam Tip: Convert mixed numbers to a common denominator before subtracting so the "extra" amount comes out correctly.

Question 11. What number should be added to \( 7\frac{3}{5} \) to get 18?
Answer: Required number \( = 18 - 7\frac{3}{5} = \frac{18}{1} - \frac{38}{5} \). LCM of 1 and 5 is 5. \( \frac{90}{5} - \frac{38}{5} = \frac{52}{5} = 10\frac{2}{5} \). The required number is \( 10\frac{2}{5} \).
In simple words: Subtract the fraction from the target number to find what extra amount is still needed to reach it.

Exam Tip: "What number added to X gives Y" always means Y minus X, watch the order of subtraction.

Question 12. What number should be added to \( 7\frac{4}{15} \) to get \( 8\frac{2}{5} \)?
Answer: Required number \( = 8\frac{2}{5} - 7\frac{4}{15} = \frac{42}{5} - \frac{109}{15} \). LCM of 5 and 15 is 15. \( \frac{126}{15} - \frac{109}{15} = \frac{17}{15} = 1\frac{2}{15} \). The required number should be \( 1\frac{2}{15} \).
In simple words: Just as before, subtract the smaller mixed number from the bigger one to find the missing amount.

Exam Tip: Always convert both mixed numbers to improper fractions with the same bottom number before subtracting.

Question 13. A wire of length \( 3\frac{3}{4} \) m is cut into two pieces. One piece is \( 1\frac{1}{2} \) m long. Find the length of the other piece.
Answer: Required length \( = \left(3\frac{3}{4} - 1\frac{1}{2}\right) \) m \( = \left(\frac{15}{4} - \frac{3}{2}\right) \) m. LCM of 4 and 2 is 4. \( \left(\frac{15-6}{4}\right) \) m \( = \frac{9}{4} \) m \( = 2\frac{1}{4} \) m. The length of the other piece of wire is \( 2\frac{1}{4} \) m.
In simple words: Take away the length of the piece you already know from the total length to find the length of the remaining piece.

Exam Tip: For "cut into two pieces" problems, total minus one known piece always gives the other piece.

Question 14. A movie is scheduled to last \( 3\frac{2}{3} \) hours, but \( 1\frac{1}{2} \) hours of that is taken up by advertisements. Find the actual duration of the film.
Answer: Actual duration \( = \left(3\frac{2}{3} - 1\frac{1}{2}\right) \) hours \( = \left(\frac{11}{3} - \frac{3}{2}\right) \) hours. LCM of 3 and 2 is 6. \( \left(\frac{22-9}{6}\right) \) hours \( = \frac{13}{6} \) hours \( = 2\frac{1}{6} \) hours. The actual duration of the film was \( 2\frac{1}{6} \) hours.
In simple words: Subtract the advertisement time from the total scheduled time to find how long the actual film runs for.

Exam Tip: Watch out for problems that mention a "scheduled" or "total" time separately from the actual useful time, subtraction is the key step.

Question 15. Which is bigger, \( \frac{2}{3} \) or \( \frac{5}{9} \)? Also find by how much.
Answer: Comparing by cross multiplication: \( 2 \times 9 = 18 \) and \( 5 \times 3 = 15 \). Since 18 is bigger than 15, \( \frac{2}{3} > \frac{5}{9} \). Now, \( \frac{2}{3} - \frac{5}{9} \). LCM of 3 and 9 is 9. \( \frac{6}{9} - \frac{5}{9} = \frac{1}{9} \). So, \( \frac{2}{3} \) is \( \frac{1}{9} \) part more than \( \frac{5}{9} \).
In simple words: First figure out which fraction is bigger using cross multiplication, then subtract to find the exact gap between them.

Exam Tip: Always state which fraction is larger before calculating the difference, examiners look for that comparison step.

Question 16. The cost of a pen is Rs \( 16\frac{3}{5} \) and the cost of a pencil is Rs \( 4\frac{3}{4} \). How much more does the pen cost than the pencil?
Answer: Cost of pen \( = \) Rs \( 16\frac{3}{5} = \) Rs \( \frac{83}{5} \). Cost of pencil \( = \) Rs \( 4\frac{3}{4} = \) Rs \( \frac{19}{4} \). Comparing by cross multiplication: \( 83 \times 4 = 332 \) and \( 19 \times 5 = 95 \). Since 332 is bigger, the pen costs more. The difference \( = \) Rs \( \left(\frac{83}{5} - \frac{19}{4}\right) \). LCM of 4 and 5 is 20. Rs \( \left(\frac{332-95}{20}\right) = \) Rs \( \frac{237}{20} = \) Rs \( 11\frac{17}{20} \). The pen costs Rs \( 11\frac{17}{20} \) more than the pencil.
In simple words: Change both mixed prices into improper fractions, work out which is bigger, then subtract the smaller price from the larger one.

Exam Tip: Cost comparison word problems always need a "which is bigger" check before subtracting to get a sensible positive difference.

Exercise 2B

Question 1. Multiply:
(i) \( \frac{3}{5} \times \frac{7}{11} \)
(ii) \( \frac{5}{8} \times \frac{4}{7} \)
(iii) \( \frac{4}{9} \times \frac{15}{16} \)
(iv) \( \frac{2}{5} \times 15 \)
(v) \( \frac{8}{15} \times 20 \)
(vi) \( \frac{5}{8} \times 1000 \)
(vii) \( 3\frac{1}{8} \times 16 \)
(viii) \( 2\frac{4}{15} \times 12 \)
(ix) \( 3\frac{6}{7} \times 4\frac{2}{3} \)
(x) \( 9\frac{1}{2} \times 1\frac{9}{19} \)
(xi) \( 4\frac{1}{8} \times 2\frac{10}{11} \)
(xii) \( 5\frac{5}{6} \times 1\frac{5}{7} \)
Answer:
(i) \( \frac{3 \times 7}{5 \times 11} = \frac{21}{55} \)
(ii) \( \frac{5 \times 4}{8 \times 7} = \frac{5}{14} \) (after cancelling)
(iii) \( \frac{4 \times 15}{9 \times 16} = \frac{5}{12} \) (after cancelling)
(iv) \( \frac{2 \times 15}{5 \times 1} = 6 \)
(v) \( \frac{8 \times 20}{15 \times 1} = \frac{32}{3} = 10\frac{2}{3} \)
(vi) \( \frac{5 \times 1000}{8 \times 1} = 625 \)
(vii) \( \frac{25}{8} \times \frac{16}{1} = 50 \)
(viii) \( \frac{34}{15} \times \frac{12}{1} = \frac{136}{5} = 27\frac{1}{5} \)
(ix) \( \frac{27}{7} \times \frac{14}{3} = 18 \)
(x) \( \frac{19}{2} \times \frac{28}{19} = 14 \)
(xi) \( \frac{33}{8} \times \frac{32}{11} = 12 \)
(xii) \( \frac{35}{6} \times \frac{12}{7} = 10 \)
In simple words: Change any mixed or whole number into a fraction, multiply the top numbers together and the bottom numbers together, then cancel common factors to simplify.

Exam Tip: Cancel common factors between numerators and denominators before multiplying, it keeps the numbers much smaller and easier to work with.

Question 2. Multiply:
(i) \( \frac{2}{3} \times \frac{5}{44} \times \frac{33}{35} \)
(ii) \( \frac{12}{25} \times \frac{15}{28} \times \frac{35}{36} \)
(iii) \( \frac{10}{27} \times \frac{28}{65} \times \frac{39}{56} \)
(iv) \( 1\frac{4}{7} \times 1\frac{13}{22} \times 1\frac{1}{15} \)
(v) \( 2\frac{3}{17} \times 7\frac{2}{9} \times 1\frac{33}{52} \)
(vi) \( 3\frac{1}{16} \times 7\frac{3}{7} \times 1\frac{25}{39} \)
Answer:
(i) \( \frac{2 \times 5 \times 33}{3 \times 44 \times 35} = \frac{1}{14} \) (after cancelling)
(ii) \( \frac{12 \times 15 \times 35}{25 \times 28 \times 36} = \frac{1}{4} \) (after cancelling)
(iii) \( \frac{10 \times 28 \times 39}{27 \times 65 \times 56} = \frac{1}{9} \) (after cancelling)
(iv) Converting to improper fractions and cancelling: \( \frac{11}{7} \times \frac{35}{22} \times \frac{16}{15} = \frac{8}{3} = 2\frac{2}{3} \)
(v) \( \frac{37}{17} \times \frac{65}{9} \times \frac{85}{52} = 25 \)
(vi) \( \frac{49}{16} \times \frac{52}{7} \times \frac{64}{39} = \frac{112}{3} = 37\frac{1}{3} \)
In simple words: With three fractions being multiplied at once, look across all of them for common factors to cancel before working out the final product.

Exam Tip: Cancellation across three fractions can happen diagonally too, not only between a fraction and its own pair, always scan all numerators against all denominators.

Question 3. Find:
(i) \( \frac{1}{3} \) of 24
(ii) \( \frac{3}{4} \) of 32
(iii) \( \frac{5}{9} \) of 45
(iv) \( \frac{7}{50} \) of 1000
(v) \( \frac{3}{20} \) of 1020
(vi) \( \frac{5}{11} \) of Rs. 220
(vii) \( \frac{4}{9} \) of 54 m
(viii) \( \frac{6}{7} \) of 35 L
(ix) \( \frac{1}{6} \) of 1 h
(x) \( \frac{5}{6} \) of a year
(xi) \( \frac{7}{20} \) of a kg
(xii) \( \frac{9}{20} \) of 1 m
(xiii) \( \frac{7}{8} \) of a day
(xiv) \( \frac{3}{7} \) of a week
(xv) \( \frac{7}{50} \) of 1 L
Answer:
(i) \( 24 \times \frac{1}{3} = 8 \)
(ii) \( 32 \times \frac{3}{4} = 24 \)
(iii) \( 45 \times \frac{5}{9} = 25 \)
(iv) \( 1000 \times \frac{7}{50} = 140 \)
(v) \( 1020 \times \frac{3}{20} = 153 \)
(vi) Rs. \( 220 \times \frac{5}{11} = \) Rs. 100
(vii) \( 54 \times \frac{4}{9} = 24 \) m
(viii) \( 35 \times \frac{6}{7} = 30 \) L
(ix) 1 h \( = 60 \) min, so \( \frac{1}{6} \) of 60 \( = 10 \) min
(x) 1 year \( = 12 \) months, so \( \frac{5}{6} \) of 12 \( = 10 \) months
(xi) 1 kg \( = 1000 \) g, so \( \frac{7}{20} \) of 1000 \( = 350 \) g
(xii) 1 m \( = 100 \) cm, so \( \frac{9}{20} \) of 100 \( = 45 \) cm
(xiii) 1 day \( = 24 \) h, so \( \frac{7}{8} \) of 24 \( = 21 \) h
(xiv) 1 week \( = 7 \) days, so \( \frac{3}{7} \) of 7 \( = 3 \) days
(xv) 1 L \( = 1000 \) ml, so \( \frac{7}{50} \) of 1000 \( = 140 \) ml
In simple words: "Of" means multiply, so for units like hours, years, or kilograms, first switch to the smaller unit and then multiply by the given fraction.

Exam Tip: When a fraction of a unit like a year, day, or kg is asked, always convert to the standard smaller unit first before multiplying.

Question 4. The cost of 1 kg of apples is Rs \( 18\frac{2}{5} \). Find the cost of \( 3\frac{3}{4} \) kg of apples.
Answer: Cost of \( 3\frac{3}{4} \) kg \( = \) Rs \( \left(\frac{92}{5} \times \frac{15}{4}\right) = \) Rs \( \left(\frac{23 \times 3}{1 \times 1}\right) = \) Rs 69. The cost of \( 3\frac{3}{4} \) kg of apples is Rs 69.
In simple words: Multiply the price of one kilogram by the number of kilograms bought to get the total cost.

Exam Tip: For rate-based problems, always multiply rate per unit by the total quantity to find the total cost.

Question 5. The cost of 1 m of cloth is Rs \( 42\frac{1}{2} \). Find the cost of \( 5\frac{3}{5} \) m of cloth.
Answer: Cost of \( 5\frac{3}{5} \) m \( = \) Rs \( \left(\frac{85}{2} \times \frac{28}{5}\right) = \) Rs \( \left(17 \times 14\right) = \) Rs 238. The cost of \( 5\frac{3}{5} \) m of cloth is Rs 238.
In simple words: Just like before, multiply the price for one metre by the number of metres to get the total cost of cloth.

Exam Tip: Always convert the mixed number rate into an improper fraction before multiplying by the quantity.

Question 6. A car covers \( 66\frac{2}{3} \) km in 1 hour. How much distance will it cover in 9 hours?
Answer: Distance in 9 h \( = \left(\frac{200}{3} \times 9\right) \) km \( = \left(\frac{200 \times 9}{3}\right) \) km \( = (200 \times 3) \) km \( = 600 \) km. The car will cover 600 km in 9 hours.
In simple words: To find distance covered over several hours, multiply the speed per hour by the number of hours.

Exam Tip: Speed and time word problems almost always use the relation distance equals speed multiplied by time.

Question 7. The capacity of 1 tin is \( 12\frac{3}{4} \) L. Find the capacity of 26 such tins.
Answer: Capacity of 26 tins \( = \left(26 \times \frac{51}{4}\right) \) L \( = \left(\frac{26 \times 51}{4}\right) \) L \( = \left(\frac{13 \times 51}{2}\right) \) L \( = \left(\frac{663}{2}\right) \) L \( = 331\frac{1}{2} \) L. The 26 tins can hold \( 331\frac{1}{2} \) L of oil.
In simple words: Multiply the capacity of a single tin by the total number of tins to find the combined capacity.

Exam Tip: Cancel common factors between the whole number and the fraction's denominator early to keep the multiplication simple.

Question 8. The cost of 1 ticket is Rs \( 35\frac{1}{2} \). Find the cost of 308 tickets.
Answer: Cost of 308 tickets \( = \) Rs \( \left(\frac{71}{2} \times 308\right) = \) Rs \( \left(71 \times 154\right) = \) Rs 10934. 308 tickets were sold for Rs 10,934.
In simple words: Multiply the price of a single ticket by the number of tickets sold to get the total money collected.

Exam Tip: When multiplying large whole numbers by a fraction, simplify the fraction's denominator against the whole number first if possible.

Question 9. The thickness of 1 board is \( 3\frac{2}{3} \) cm. Find the height of a stack of 9 such boards.
Answer: Thickness of 9 boards \( = \left(9 \times \frac{11}{3}\right) \) cm \( = (3 \times 11) \) cm \( = 33 \) cm. The height of the stack is 33 cm.
In simple words: Multiply the thickness of one board by how many boards are stacked to get the total height.

Exam Tip: Stack height problems are direct multiplication, no need for addition of individual boards one by one.

Question 10. Rohit takes \( 4\frac{4}{5} \) minutes to complete one round of a circular park. How long will he take to complete 15 rounds?
Answer: Time for 15 rounds \( = \left(15 \times \frac{24}{5}\right) \) min \( = (3 \times 24) \) min \( = 72 \) min \( = 1 \) h 12 min (since 1 hour equals 60 minutes). Rohit will take 1 h 12 min to make 15 complete rounds of the circular park.
In simple words: Multiply the time for one round by the number of rounds, then convert any leftover minutes over 60 into hours.

Exam Tip: Always convert your final minutes answer into hours and minutes if it goes over 60, examiners expect the cleaner form.

Question 11. Amit weighs 35 kg. Kavita's weight is \( \frac{3}{5} \) of Amit's weight. Find Kavita's weight.
Answer: Kavita's weight \( = 35 \times \frac{3}{5} \) kg \( = (7 \times 3) \) kg \( = 21 \) kg. Kavita's weight is 21 kg.
In simple words: Multiply Amit's weight by the given fraction to find Kavita's weight, since her weight is a fraction of his.

Exam Tip: Whenever a problem says one quantity is "a fraction of" another, that always means multiplication.

Question 12. There are 42 students in a class. The number of boys is \( \frac{5}{7} \) of the total number of students. Find the number of girls in the class.
Answer: Number of boys \( = \frac{5}{7} \times 42 = 5 \times 6 = 30 \). Number of girls \( = 42 - 30 = 12 \). There are 12 girls in the class.
In simple words: First work out how many boys there are using the fraction, then subtract that from the total to find how many girls remain.

Exam Tip: Whenever a fraction gives one part of a group, subtracting from the total always gives the remaining part.

Question 13. Sapna's total monthly income is Rs 12,000. Her monthly expenditure is \( \frac{7}{8} \) of her income. How much does she save every month?
Answer: Monthly expenditure \( = \) Rs \( \frac{7}{8} \times 12000 = \) Rs \( (7 \times 1500) = \) Rs 10,500. Monthly savings \( = \) Rs \( 12000 - 10500 = \) Rs 1500. Sapna deposits Rs 1,500 in the bank every month.
In simple words: Work out the spending using the fraction of income, then subtract that from the total income to find the savings.

Exam Tip: Income minus expenditure equals savings, a relation worth remembering for this whole category of problems.

Question 14. The side of a square field is \( 4\frac{2}{3} \) m. Find its area.
Answer: Area of square \( = (\text{side})^2 = \left(\frac{14}{3}\right)^2 = \frac{14}{3} \times \frac{14}{3} = \left(\frac{14 \times 14}{3 \times 3}\right) \) m² \( = \frac{196}{9} \) m² \( = 21\frac{7}{9} \) m². The area of the square field is \( 21\frac{7}{9} \) m².
In simple words: To find the area of a square, multiply the side length by itself, after converting it to an improper fraction first.

Exam Tip: Remember area of a square is side squared, not side multiplied by 2, a common slip under exam pressure.

Question 15. The length of a rectangular park is \( 41\frac{2}{3} \) m and its breadth is \( 18\frac{3}{5} \) m. Find its area.
Answer: Length \( = \frac{125}{3} \) m, breadth \( = \frac{93}{5} \) m. Area \( = \text{length} \times \text{breadth} = \left(\frac{125}{3} \times \frac{93}{5}\right) \) m² \( = (25 \times 31) \) m² \( = 775 \) m². The area of the rectangular park is 775 m².
In simple words: Multiply the length by the breadth, after turning both mixed numbers into improper fractions, to find the rectangle's area.

Exam Tip: Area of a rectangle is always length times breadth, keep the two measurements in the same unit before multiplying.

Exercise 2C

Question 1. Find the reciprocal of each of the following:
(i) \( \frac{5}{8} \)
(ii) 7
(iii) \( \frac{1}{12} \)
(iv) \( 12\frac{3}{5} \)
Answer:
(i) Reciprocal of \( \frac{5}{8} \) is \( \frac{8}{5} \), since \( \frac{5}{8} \times \frac{8}{5} = 1 \)
(ii) Reciprocal of 7 is \( \frac{1}{7} \), since \( 7 \times \frac{1}{7} = 1 \)
(iii) Reciprocal of \( \frac{1}{12} \) is 12, since \( \frac{1}{12} \times 12 = 1 \)
(iv) \( 12\frac{3}{5} = \frac{63}{5} \), so its reciprocal is \( \frac{5}{63} \), since \( \frac{63}{5} \times \frac{5}{63} = 1 \)
In simple words: The reciprocal of a fraction is simply that fraction turned upside down, with the top and bottom numbers swapped.

Exam Tip: Convert mixed numbers to improper fractions first, then flip the fraction to get the reciprocal, whole numbers are treated as having 1 as the denominator.

Question 2. Divide:
(i) \( \frac{4}{7} \div \frac{9}{14} \)
(ii) \( \frac{7}{10} \div \frac{3}{5} \)
(iii) \( \frac{8}{9} \div 16 \)
(iv) \( 9 \div \frac{1}{3} \)
(v) \( 24 \div \frac{6}{7} \)
(vi) \( 3\frac{3}{5} \div \frac{4}{5} \)
(vii) \( 3\frac{3}{7} \div \frac{8}{21} \)
(viii) \( 5\frac{4}{7} \div 1\frac{3}{10} \)
(ix) \( 15\frac{3}{7} \div 1\frac{23}{49} \)
Answer: To divide by a fraction, multiply by its reciprocal.
(i) \( \frac{4}{7} \times \frac{14}{9} = \frac{8}{9} \)
(ii) \( \frac{7}{10} \times \frac{5}{3} = \frac{7}{6} = 1\frac{1}{6} \)
(iii) \( \frac{8}{9} \times \frac{1}{16} = \frac{1}{18} \)
(iv) \( 9 \times 3 = 27 \)
(v) \( 24 \times \frac{7}{6} = 4 \times 7 = 28 \)
(vi) \( \frac{18}{5} \times \frac{5}{4} = \frac{9}{2} = 4\frac{1}{2} \)
(vii) \( \frac{24}{7} \times \frac{21}{8} = 3 \times 3 = 9 \)
(viii) \( \frac{39}{7} \times \frac{10}{13} = \frac{30}{7} = 4\frac{2}{7} \)
(ix) \( \frac{108}{7} \times \frac{49}{72} = \frac{21}{2} = 10\frac{1}{2} \)
In simple words: Dividing by a fraction is the same as multiplying by that fraction flipped upside down, this trick works for whole numbers and mixed numbers too.

Exam Tip: Never divide fractions directly, always flip the second fraction to its reciprocal and multiply instead.

Question 3. Divide:
(i) \( \frac{11}{24} \div \frac{7}{8} \)
(ii) \( 6\frac{7}{8} \div \frac{11}{16} \)
(iii) \( 5\frac{5}{9} \div 3\frac{1}{3} \)
(iv) \( 32 \div 1\frac{3}{5} \)
(v) \( 45 \div 1\frac{4}{5} \)
(vi) \( 63 \div 2\frac{1}{4} \)
Answer:
(i) \( \frac{11}{24} \times \frac{8}{7} = \frac{11}{21} \)
(ii) \( \frac{55}{8} \times \frac{16}{11} = 5 \times 2 = 10 \)
(iii) \( \frac{50}{9} \times \frac{3}{10} = \frac{5}{3} = 1\frac{2}{3} \)
(iv) \( 32 \times \frac{5}{8} = 4 \times 5 = 20 \)
(v) \( 45 \times \frac{5}{9} = 5 \times 5 = 25 \)
(vi) \( 63 \times \frac{4}{9} = 7 \times 4 = 28 \)
In simple words: Again, convert any mixed numbers to improper fractions and multiply by the reciprocal of the number being divided by.

Exam Tip: Simplify by cancelling common factors before multiplying, it saves time and keeps the numbers manageable.

Question 4. A rope of length \( 13\frac{1}{2} \) m is cut into 9 equal pieces. Find the length of each piece.
Answer: Length of rope \( = \frac{27}{2} \) m. Length of each piece \( = \left(\frac{27}{2} \div 9\right) \) m \( = \left(\frac{27}{2} \times \frac{1}{9}\right) \) m \( = \frac{3}{2} \) m \( = 1\frac{1}{2} \) m. The length of each piece of rope is \( 1\frac{1}{2} \) m.
In simple words: Divide the total length of rope by the number of equal pieces needed to find the length of one piece.

Exam Tip: "Cut into equal pieces" always means dividing the total by the number of pieces.

Question 5. 18 boxes of nails weigh \( 49\frac{1}{2} \) kg in total. Find the weight of each box.
Answer: Total weight \( = \frac{99}{2} \) kg. Weight of 1 box \( = \left(\frac{99}{2} \div 18\right) \) kg \( = \left(\frac{99}{2} \times \frac{1}{18}\right) \) kg \( = \left(\frac{11}{4}\right) \) kg \( = 2\frac{3}{4} \) kg. The weight of each box is \( 2\frac{3}{4} \) kg.
In simple words: Divide the total weight by the number of boxes to find how much a single box weighs.

Exam Tip: "Weight of each" or "cost of each" almost always signals a division problem where the total is split evenly.

Question 6. A man sold oranges worth Rs 210 at the rate of Rs \( 3\frac{3}{4} \) per orange. How many oranges did he sell?
Answer: Cost of 1 orange \( = \) Rs \( \frac{15}{4} \). Required number of oranges \( = \left(210 \div \frac{15}{4}\right) = \left(210 \times \frac{4}{15}\right) = (14 \times 4) = 56 \). The man sold 56 oranges.
In simple words: Divide the total money earned by the price of one orange to find how many oranges were sold.

Exam Tip: Total amount divided by rate per unit gives the number of units, a pattern that repeats across many word problems.

Question 7. Mangoes worth Rs \( 157\frac{1}{4} \) were bought at the rate of Rs \( 18\frac{1}{2} \) per kg. Find the weight of the mangoes bought.
Answer: Cost of 1 kg \( = \) Rs \( \frac{37}{2} \). Total cost \( = \) Rs \( \frac{629}{4} \). Required weight \( = \left(\frac{629}{4} \div \frac{37}{2}\right) = \left(\frac{629}{4} \times \frac{2}{37}\right) = \left(\frac{17}{2}\right) = 8\frac{1}{2} \) kg. The weight of the mangoes available for Rs \( 157\frac{1}{4} \) is \( 8\frac{1}{2} \) kg.
In simple words: Divide the total cost paid by the rate per kilogram to find out how many kilograms were bought.

Exam Tip: Whenever a total cost and a rate per unit are given together, dividing one by the other finds the quantity.

Question 8. Vikas covers a distance of \( 20\frac{2}{3} \) km in \( 7\frac{3}{4} \) hours. Find the distance he covers in 1 hour.
Answer: Distance covered in 1 h \( = \left(\frac{62}{3} \div \frac{31}{4}\right) \) km \( = \left(\frac{62}{3} \times \frac{4}{31}\right) \) km \( = \left(\frac{2 \times 4}{3}\right) \) km \( = \frac{8}{3} \) km \( = 2\frac{2}{3} \) km. The distance covered by Vikas in 1 hour is \( 2\frac{2}{3} \) km.
In simple words: Divide the total distance travelled by the total time taken to find how far he travels in just one hour.

Exam Tip: Speed per hour is always total distance divided by total time, a fundamental relation worth memorising.

Question 9. \( 8\frac{1}{2} \) kg of sugar costs Rs \( 148\frac{3}{4} \). Find the cost of 1 kg of sugar.
Answer: Cost of 1 kg \( = \) Rs \( \left(\frac{595}{4} \div \frac{17}{2}\right) = \) Rs \( \left(\frac{595}{4} \times \frac{2}{17}\right) = \) Rs \( \left(\frac{35}{2}\right) = \) Rs \( 17\frac{1}{2} \). The cost of 1 kg of sugar is Rs \( 17\frac{1}{2} \).
In simple words: Divide the total cost by the total weight in kilograms to find the cost of just one kilogram.

Exam Tip: Cost per unit is always total cost divided by total quantity, keep this relation handy for rate problems.

Question 10. The cost of 1 notebook is Rs \( 7\frac{3}{4} \). How many notebooks can be purchased for Rs \( 69\frac{3}{4} \)?
Answer: Number of notebooks \( = \left(69\frac{3}{4} \div 7\frac{3}{4}\right) = \left(\frac{279}{4} \div \frac{31}{4}\right) = \left(\frac{279}{4} \times \frac{4}{31}\right) = \left(\frac{279}{31}\right) = 9 \). 9 notebooks can be purchased for Rs \( 69\frac{3}{4} \).
In simple words: Divide the total money available by the price of one notebook to find out how many can be bought.

Exam Tip: "How many can be bought" problems always divide the total budget by the unit price.

Question 11. The cost of 1 ticket for a charity show is Rs \( 10\frac{1}{2} \). A boy collected Rs \( 283\frac{1}{2} \) by selling tickets. Find the number of tickets he sold.
Answer: Number of tickets sold \( = \left(\frac{567}{2} \div \frac{21}{2}\right) = \left(\frac{567}{2} \times \frac{2}{21}\right) = \frac{567}{21} = 27 \). The boy sold 27 tickets of the charity show.
In simple words: Divide the total money collected by the price of one ticket to find how many tickets were sold in all.

Exam Tip: Whenever total collection and price per item are both given, dividing gives the count of items sold.

Question 12. Each student in a group contributed Rs \( 61\frac{1}{2} \) towards a fund, and the total amount collected was Rs \( 676\frac{1}{2} \). Find the number of students in the group.
Answer: Number of students \( = \left(\frac{1353}{2} \div \frac{123}{2}\right) = \left(\frac{1353}{2} \times \frac{2}{123}\right) = \frac{1353}{123} = 11 \). There are 11 students in the group.
In simple words: Divide the total amount collected by the amount each person paid to find how many people contributed.

Exam Tip: Total collected divided by contribution per person always gives the number of people involved.

Question 13. Each student in a hostel is given \( \frac{2}{5} \) L of milk daily, and 24 L of milk is distributed among all the students. Find the number of students in the hostel.
Answer: Number of students \( = \left(24 \div \frac{2}{5}\right) = \left(24 \times \frac{5}{2}\right) = (12 \times 5) = 60 \). There are 60 students in the hostel.
In simple words: Divide the total milk available by the amount given to each student to find the total number of students.

Exam Tip: Total quantity divided by the share given to each individual finds the number of individuals sharing it.

Question 14. The capacity of a small jug is \( \frac{3}{4} \) L, and the capacity of a bucket is \( 20\frac{1}{4} \) L. How many times must the small jug be filled to empty the bucket?
Answer: Required number of small jugs \( = \left(\frac{81}{4} \div \frac{3}{4}\right) = \left(\frac{81}{4} \times \frac{4}{3}\right) = \left(\frac{81}{3}\right) = 27 \). The small jug has to be filled 27 times to empty the water from the bucket.
In simple words: Divide the total capacity of the bucket by the capacity of the small jug to find how many times it needs to be filled.

Exam Tip: "How many times to fill/empty" problems always divide the bigger capacity by the smaller one.

Question 15. The product of two numbers is \( 15\frac{5}{6} \). One of the numbers is \( 6\frac{1}{3} \). Find the other number.
Answer: The other number \( = \left(\frac{95}{6} \div \frac{19}{3}\right) = \left(\frac{95}{6} \times \frac{3}{19}\right) = \left(\frac{5}{2}\right) = 2\frac{1}{2} \). The other number is \( 2\frac{1}{2} \).
In simple words: Divide the product by the known number to work out what the other number must be.

Exam Tip: If a product and one factor are given, dividing the product by the known factor always finds the missing factor.

Question 16. The product of two numbers is 42. One of the numbers is \( 9\frac{4}{5} \). Find the other number.
Answer: The other number \( = \left(42 \div \frac{49}{5}\right) = \left(42 \times \frac{5}{49}\right) = \left(\frac{6 \times 5}{7}\right) = \frac{30}{7} = 4\frac{2}{7} \). The required number is \( 4\frac{2}{7} \).
In simple words: Divide the known product by the given number to find the missing number that multiplies with it.

Exam Tip: The same product-and-factor rule applies whether the product is a whole number or a fraction.

Question 17. By what number should \( 6\frac{2}{9} \) be divided to get \( 4\frac{2}{3} \)?
Answer: Required number \( = \left(6\frac{2}{9} \div 4\frac{2}{3}\right) = \left(\frac{56}{9} \div \frac{14}{3}\right) = \left(\frac{56}{9} \times \frac{3}{14}\right) = \left(\frac{4}{3}\right) = 1\frac{1}{3} \). We have to divide \( 6\frac{2}{9} \) by \( 1\frac{1}{3} \) to get \( 4\frac{2}{3} \).
In simple words: Divide the starting number by the target result to find the exact number that produces that result.

Exam Tip: For "divided by what number gives" problems, divide the original number by the result, not the other way around.

Exercise 2D

Question 1. Which of the following is a vulgar fraction?
(a) \( \frac{3}{10} \)
(b) \( \frac{7}{100} \)
(c) \( \frac{10}{3} \)
(d) \( \frac{21}{1000} \)
Answer: (c) \( \frac{10}{3} \)
In simple words: A vulgar fraction is any fraction whose denominator is not 10, 100, 1000 or a similar power of ten, and \( \frac{10}{3} \) fits that description.

Exam Tip: Remember that decimal fractions have denominators of 10, 100, 1000 and so on, anything else is a vulgar fraction.

Question 2. Which of the following is an improper fraction?
(a) \( \frac{3}{7} \)
(b) \( \frac{5}{9} \)
(c) \( \frac{9}{7} \)
(d) \( \frac{4}{11} \)
Answer: (c) \( \frac{9}{7} \)
In simple words: An improper fraction is one where the top number is bigger than the bottom number, which is true only for \( \frac{9}{7} \) here.

Exam Tip: Just compare numerator and denominator directly, if the numerator wins, it is an improper fraction.

Question 3. Which of the following is a reducible fraction?
(a) \( \frac{105}{112} \)
(b) \( \frac{17}{35} \)
(c) \( \frac{29}{41} \)
(d) \( \frac{53}{79} \)
Answer: (a) \( \frac{105}{112} \) A fraction that is reducible can be simplified by dividing both the numerator and denominator by a common factor. \( \frac{105 \div 7}{112 \div 7} = \frac{15}{16} \). So, \( \frac{105}{112} \) is a reducible fraction.
In simple words: A reducible fraction has a common factor shared between its top and bottom numbers, letting both be divided down to a simpler form.

Exam Tip: Check quickly for common factors like 2, 3, 5, or 7 between numerator and denominator to spot reducible fractions fast.

Question 4. Which of the following describes fractions such as \( \frac{2}{3}, \frac{4}{6}, \frac{6}{9}, \frac{8}{12} \)?
(a) like fractions
(b) unlike fractions
(c) equivalent fractions
(d) improper fractions
Answer: (c) equivalent fractions Equivalent fractions are those which look different but represent the same value. Thus, \( \frac{2}{3}, \frac{4}{6} = \frac{2}{3}, \frac{6}{9} = \frac{2}{3}, \frac{8}{12} = \frac{2}{3} \) are all equivalent fractions.
In simple words: Equivalent fractions may look different on the surface, but they all simplify down to the exact same value.

Exam Tip: To check if fractions are equivalent, simplify each one fully and see if they all reduce to the same fraction.

Question 5. Which of the following is correct?
(a) \( \frac{9}{16} > \frac{13}{24} \)
(b) \( \frac{9}{16} = \frac{13}{24} \)
(c) \( \frac{9}{16} > \frac{13}{24} \)
(d) \( \frac{9}{16} < \frac{13}{24} \)
Answer: (c) \( \frac{9}{16} > \frac{13}{24} \) The two fractions are \( \frac{9}{16} \) and \( \frac{13}{24} \). By cross multiplication: \( 9 \times 24 = 216 \) and \( 13 \times 16 = 208 \). Since 216 is bigger than 208, \( \frac{9}{16} > \frac{13}{24} \).
In simple words: Cross multiply the two fractions and compare the results to decide which fraction is actually larger.

Exam Tip: Cross multiplication is the fastest way to compare two fractions without finding a common denominator first.

Question 6. What is the reciprocal of \( 1\frac{3}{4} \)?
(a) \( \frac{3}{4} \)
(b) \( \frac{4}{3} \)
(c) \( \frac{7}{4} \)
(d) \( \frac{4}{7} \)
Answer: (d) \( \frac{4}{7} \) Reciprocal of \( 1\frac{3}{4} \) is the reciprocal of \( \frac{7}{4} \), which is \( \frac{4}{7} \).
In simple words: Change the mixed number into an improper fraction first, and then flip it to get the reciprocal.

Exam Tip: Never try to find the reciprocal of a mixed number directly, always convert to an improper fraction first.

Question 7. What is \( \frac{3}{10} + \frac{8}{15} \)?
(a) \( \frac{11}{25} \)
(b) \( \frac{4}{5} \)
(c) \( \frac{5}{6} \)
(d) \( \frac{2}{3} \)
Answer: (c) \( \frac{5}{6} \) LCM of 10 and 15 is 30. \( \left(\frac{3}{10} + \frac{8}{15}\right) = \left(\frac{9+16}{30}\right) = \frac{25}{30} = \frac{5}{6} \).
In simple words: Bring both fractions to the same bottom number using the LCM, then add the top numbers and simplify.

Exam Tip: Always simplify the final fraction to its lowest terms before selecting an option in an MCQ.

Question 8. What is \( 3\frac{1}{4} - 2\frac{1}{3} \)?
(a) \( \frac{5}{12} \)
(b) \( \frac{7}{12} \)
(c) \( \frac{9}{12} \)
(d) \( \frac{11}{12} \)
Answer: (d) \( \frac{11}{12} \) \( \left(3\frac{1}{4} - 2\frac{1}{3}\right) = \left(\frac{13}{4} - \frac{7}{3}\right) \). LCM of 4 and 3 is 12. \( \left(\frac{39-28}{12}\right) = \frac{11}{12} \).
In simple words: Convert both mixed numbers to improper fractions, bring them to a common denominator, then subtract.

Exam Tip: Double check the LCM calculation, a wrong LCM is the most common source of error in these subtraction MCQs.

Question 9. What is \( 36 \div \frac{1}{4} \)?
(a) 9
(b) 36
(c) 40
(d) 144
Answer: (d) 144 \( 36 \div \frac{1}{4} = 36 \times 4 = 144 \) (since the reciprocal of \( \frac{1}{4} \) is 4).
In simple words: Dividing by a small fraction like a quarter is the same as multiplying by 4, which makes the number much bigger.

Exam Tip: Dividing by a fraction smaller than 1 always gives a bigger answer than the original number, a useful check on your final result.

Question 10. What is \( 1\frac{6}{7} \div 2\frac{3}{5} \)?
(a) \( \frac{3}{7} \)
(b) \( \frac{5}{7} \)
(c) \( \frac{6}{7} \)
(d) 1
Answer: (b) \( \frac{5}{7} \) Required number \( = \left(1\frac{6}{7} \div 2\frac{3}{5}\right) = \left(\frac{13}{7} \div \frac{13}{5}\right) = \left(\frac{13}{7} \times \frac{5}{13}\right) = \frac{5}{7} \).
In simple words: Convert both mixed numbers to improper fractions, then multiply the first by the reciprocal of the second.

Exam Tip: Watch for matching numerators or denominators after conversion, like the 13s here, since they often cancel neatly.

Question 11. What is \( 1\frac{1}{2} \div \frac{2}{3} \)?
(a) \( \frac{3}{4} \)
(b) \( \frac{4}{3} \)
(c) \( 2\frac{1}{4} \)
(d) \( 2\frac{1}{4} \)
Answer: (d) \( 2\frac{1}{4} \) Required number \( = \left(1\frac{1}{2} \div \frac{2}{3}\right) = \left(\frac{3}{2} \div \frac{2}{3}\right) = \left(\frac{3}{2} \times \frac{3}{2}\right) = \frac{9}{4} = 2\frac{1}{4} \).
In simple words: Flip the second fraction and multiply, remembering to convert the mixed number to an improper fraction first.

Exam Tip: Dividing by a fraction less than 1 always increases the value, a quick way to sanity check your MCQ answer.

Question 12. What is \( 1\frac{3}{5} \div \frac{2}{3} \)?
(a) \( \frac{3}{5} \)
(b) \( 1\frac{1}{5} \)
(c) \( 2\frac{2}{5} \)
(d) \( 3\frac{1}{5} \)
Answer: (c) \( 2\frac{2}{5} \) \( 1\frac{3}{5} \div \frac{2}{3} = \frac{8}{5} \div \frac{2}{3} = \frac{8}{5} \times \frac{3}{2} = \left(\frac{4 \times 3}{5}\right) = \frac{12}{5} = 2\frac{2}{5} \).
In simple words: Convert the mixed number, flip the divisor, multiply, and simplify to reach the final mixed number answer.

Exam Tip: Cancel any common factors between the numerators and denominators before the final multiplication step.

Question 13. What is \( 2\frac{1}{5} \div 1\frac{1}{5} \)?
(a) 1
(b) \( 1\frac{1}{5} \)
(c) \( 1\frac{2}{5} \)
(d) \( 1\frac{5}{6} \)
Answer: (d) \( 1\frac{5}{6} \) \( 2\frac{1}{5} \div 1\frac{1}{5} = \frac{11}{5} \div \frac{6}{5} = \frac{11}{5} \times \frac{5}{6} = \frac{11}{6} = 1\frac{5}{6} \).
In simple words: Turn both mixed numbers into improper fractions and multiply the first by the reciprocal of the second.

Exam Tip: When both fractions share the same denominator after conversion, it often cancels neatly during multiplication.

Question 14. What is the reciprocal of \( 1\frac{2}{3} \)?
(a) \( \frac{2}{3} \)
(b) \( \frac{3}{2} \)
(c) \( \frac{5}{3} \)
(d) \( \frac{3}{5} \)
Answer: (d) \( \frac{3}{5} \) Reciprocal of \( 1\frac{2}{3} \) is the reciprocal of \( \frac{5}{3} \), which is \( \frac{3}{5} \).
In simple words: Change the mixed number to an improper fraction first, then simply swap the numerator and denominator.

Exam Tip: A common mistake is flipping the mixed number without converting it first, always convert before flipping.

Question 15. Which is the correct ascending order of \( \frac{3}{5}, \frac{2}{3} \) and \( \frac{14}{15} \)?
(a) \( \frac{2}{3} < \frac{3}{5} < \frac{14}{15} \)
(b) \( \frac{3}{5} < \frac{2}{3} < \frac{14}{15} \)
(c) \( \frac{14}{15} < \frac{2}{3} < \frac{3}{5} \)
(d) \( \frac{3}{5} < \frac{14}{15} < \frac{2}{3} \)
Answer: (b) \( \frac{3}{5} < \frac{2}{3} < \frac{14}{15} \) The LCM of 5, 3 and 15 is 15. Converting: \( \frac{2}{3} = \frac{10}{15} \), \( \frac{3}{5} = \frac{9}{15} \), \( \frac{14}{15} = \frac{14}{15} \). Since \( \frac{9}{15} < \frac{10}{15} < \frac{14}{15} \), the order is \( \frac{3}{5} < \frac{2}{3} < \frac{14}{15} \).
In simple words: Convert each fraction to the same bottom number using the LCM, then it becomes easy to line them up smallest to biggest.

Exam Tip: In ascending-order MCQs, converting to a common denominator first is far more reliable than guessing from the original fractions.

Question 16. A car covers a distance of 44 km using \( 2\frac{3}{4} \) L of petrol. How far can the car travel using 1 L of petrol?
(a) 11 km
(b) 22 km
(c) 16 km
(d) 20 km
Answer: (c) 16 km Distance covered by the car on \( 2\frac{3}{4} \) L of petrol \( = \left(16 \times 2\frac{3}{4}\right) \) km \( = \left(16 \times \frac{11}{4}\right) \) km \( = (4 \times 11) \) km \( = 44 \) km. So the car covers 16 km on 1 L of petrol.
In simple words: Divide the total distance travelled by the amount of fuel used to find how far the car goes on just one litre.

Exam Tip: Mileage word problems always use total distance divided by total fuel to find the distance per litre.

Question 17. Lalit takes 6 hours to read \( 1\frac{3}{4} \) times the length of an average book. How long will he take to read the entire book?
(a) \( 9\frac{1}{2} \) hours
(b) \( 10\frac{1}{2} \) hours
(c) \( 11\frac{1}{2} \) hours
(d) \( 12\frac{1}{2} \) hours
Answer: (b) \( 10\frac{1}{2} \) hours Time taken by Lalit to read the entire book \( = \left(6 \times 1\frac{3}{4}\right) \) h \( = \left(6 \times \frac{7}{4}\right) \) h \( = \left(\frac{21}{2}\right) \) h \( = 10\frac{1}{2} \) h.
In simple words: Multiply the time taken for the given portion by the multiplying factor to find the time for the complete book.

Exam Tip: Read carefully whether a rate given is "per part" or "for the whole", it changes whether you multiply or divide.