Class Six Math Chapter 1: Exercise 1.4 Complete Solution

Class six math chapter 1: exercise 1.4 complete solution

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Question 1: Determine if the following pairs of fractions are equivalent:

\[ \left(a\right);\frac{5}{8}, \frac{15}{24} \]

Solution: We know that two fractions will be equivalent if the product of the numerator of the first fraction and the denominator of the second fraction is equal to the product of the denominator of the first fraction and the numerator of the second fraction.

First fraction’s numerator × Second fraction’s denominator = \[ 5 \times 24 = 120 \]
First fraction’s denominator × Second fraction’s numerator = \[ 8 \times 15 = 120 \]

Since both products are equal, the fractions are equivalent.

∴ The fraction pair \[ \frac{5}{8}, \frac{15}{24} \] is equivalent.

\[ \left(b\right);\frac{7}{11}, \frac{14}{33} \]

First fraction’s numerator × Second fraction’s denominator = \[ 7 \times 33 = 231 \]
First fraction’s denominator × Second fraction’s numerator = \[ 11 \times 14 = 154 \]

Since the two products are not equal, the fractions are not equivalent.

∴ The fraction pair \[ \frac{7}{11}, \frac{14}{33} \] is not equivalent.

\[ \left(c\right);\frac{38}{50}, \frac{114}{150} \]

First fraction’s numerator × Second fraction’s denominator = \[ 38 \times 150 = 5700 \]
First fraction’s denominator × Second fraction’s numerator = \[ 50 \times 114 = 5700 \]

Since both products are equal, the fractions are equivalent.

∴ The fraction pair \[ \frac{38}{50}, \frac{114}{150} \] is equivalent.

Question 2: Express the following fractions as equivalent fractions with a common denominator:

\[ \left(a\right);\frac{2}{5}, \frac{7}{10} , \frac{9}{40} \]

Solution: The denominators of the given fractions are 5, 10, and 40. The LCM of 5, 10, and 40 is 40.

\[ \left[∴ 40\div5=8\right];;\frac{2}{5}=\frac{2\times8}{5\times8}=\frac{16}{40}\]
\[ \left[∴ 40\div10=4\right];;\frac{7}{10}=\frac{7\times4}{10\times4}=\frac{28}{40}\]
\[ \left[∴ 40\div40=1\right];;\frac{9}{40}=\frac{9\times1}{40\times1}=\frac{9}{40}\]

Answer: The equivalent fractions of \[ \frac{2}{5}, \frac{7}{10} , \frac{9}{40} \] are \[ \frac{16}{40}, \frac{28}{40}, \frac{9}{40} \].

\[ \left(b\right);\frac{17}{25}, \frac{23}{40} , \frac{67}{120} \]

Solution: The denominators of the given fractions are 25, 40, and 120. The LCM of 25, 40, and 120 is 600.

\[ \left[∴ 600\div25=24\right];;\frac{17}{25}=\frac{17\times24}{25\times24}=\frac{408}{600}\]
\[ \left[∴ 600\div40=15\right];;\frac{23}{40}=\frac{23\times15}{40\times15}=\frac{345}{600}\]
\[ \left[∴ 600\div120=5\right];;\frac{67}{120}=\frac{67\times5}{120\times5}=\frac{335}{600}\]

Answer: The equivalent fractions of \[ \frac{17}{25}, \frac{23}{40}, \frac{67}{120} \] are \[ \frac{408}{600}, \frac{345}{600}, \frac{335}{600} \].

Question 3: Arrange the following fractions in ascending order of value:

\[ \left(a\right);\frac{6}{7}, \frac{7}{9}, \frac{16}{21}, \frac{50}{63} \]

Solution: Here, the denominators of the fractions are 7, 9, 21, and 63. The LCM of 7, 9, 21, and 63 is 63.

\[ \left[∴ 63\div7=9\right];;\frac{6}{7}=\frac{6\times9}{7\times9}=\frac{54}{63}\]
\[ \left[∴ 63\div9=7\right];;\frac{7}{9}=\frac{7\times7}{9\times7}=\frac{49}{63}\]
\[ \left[∴ 63\div21=3\right];;\frac{16}{21}=\frac{16\times3}{21\times3}=\frac{48}{63}\]
\[ \left[∴ 63\div63=1\right];;\frac{50}{63}=\frac{50\times1}{63\times1}=\frac{50}{63}\]

Since 48 < 49 < 50 < 54, we have:

\[ ∴ \frac{16}{21} < \frac{7}{9} < \frac{50}{63} < \frac{6}{7} \]

Answer: The fractions in ascending order are \[ \frac{16}{21} < \frac{7}{9} < \frac{50}{63} < \frac{6}{7} \].

\[ \left(b\right);\frac{65}{72}, \frac{31}{36}, \frac{53}{60}, \frac{17}{24} \]

Solution: The denominators of the fractions are 72, 36, 60, and 24. The LCM of 72, 36, 60, and 24 is 360.

\[ \left[∴ 360\div72=5\right];;\frac{65}{72}=\frac{65\times5}{72\times5}=\frac{325}{360}\]
\[ \left[∴ 360\div36=10\right];;\frac{31}{36}=\frac{31\times10}{36\times10}=\frac{310}{360}\]
\[ \left[∴ 360\div60=6\right];;\frac{53}{60}=\frac{53\times6}{60\times6}=\frac{318}{360}\]
\[ \left[∴ 360\div24=15\right];;\frac{17}{24}=\frac{17\times15}{24\times15}=\frac{255}{360}\]

Since 255 < 310 < 318 < 325, we have:

\[ ∴ \frac{17}{24} < \frac{31}{36} < \frac{53}{60} < \frac{65}{72} \]

Answer: The fractions in ascending order are \[ \frac{17}{24} < \frac{31}{36} < \frac{53}{60} < \frac{65}{72} \].

Question 4: Arrange the following fractions in descending order of value:

\[ \left(a\right);\frac{3}{4}, \frac{6}{7}, \frac{7}{8}, \frac{5}{12} \]

Solution: Here, the denominators of the fractions are \[4\], \[7\], \[8\], and \[12\]. The LCM of \[4\], \[7\], \[8\], and \[12\] is \[168\].

\[ \left[∴ 168\div4=42\right];;\frac{3}{4}=\frac{3\times42}{4\times42}=\frac{126}{168}\]

\[ \left[∴ 168\div7=24\right];;\frac{6}{7}=\frac{6\times24}{7\times24}=\frac{144}{168}\]

\[ \left[∴ 168\div8=21\right];;\frac{7}{8}=\frac{7\times21}{8\times21}=\frac{147}{168}\]

\[ \left[∴ 168\div12=14\right];;\frac{5}{12}=\frac{5\times14}{12\times14}=\frac{70}{168}\]

Since \[147 > 144 > 126 > 70\], we have:

\[ ∴ \frac{7}{8} > \frac{6}{7} > \frac{3}{4} > \frac{5}{12} \]

Answer: \[ \frac{7}{8} > \frac{6}{7} > \frac{3}{4} > \frac{5}{12} \]

\[ \left(b\right);\frac{17}{25}, \frac{23}{40}, \frac{51}{65}, \frac{67}{130} \]

Solution: Here, the denominators of the fractions are \[25\], \[40\], \[65\], and \[130\]. The LCM of \[25\], \[40\], \[65\], and \[130\] is \[2600\].

\[ \left[∴ 2600\div25=104\right];;\frac{17}{25}=\frac{17\times104}{25\times104}=\frac{1768}{2600}\]

\[ \left[∴ 2600\div40=65\right];;\frac{23}{40}=\frac{23\times65}{40\times65}=\frac{1495}{2600}\]

\[ \left[∴ 2600\div65=40\right];;\frac{51}{65}=\frac{51\times40}{65\times40}=\frac{2040}{2600}\]

\[ \left[∴ 2600\div130=20\right];;\frac{67}{130}=\frac{67\times20}{130\times20}=\frac{1340}{2600}\]

Since \[2040 > 1768 > 1495 > 1340\], we have:

\[ ∴ \frac{51}{65} > \frac{17}{25} > \frac{23}{40} > \frac{67}{130} \]

Answer: \[ \frac{51}{65} > \frac{17}{25} > \frac{23}{40} > \frac{67}{130} \]

Question 5: Add:

(a) \[ \frac{1}{5} + \frac{3}{16}\]

Solution:

\[ \frac{1}{5} + \frac{3}{16} = \frac{10+3}{16} = \frac{13}{16} \]

Answer: \[ \frac{13}{16} \]

(b) \[6 + 1\frac{6}{7}\]

Solution:

\[6 + 1\frac{6}{7} = 6 + 1 + \frac{6}{7} = 7 + \frac{6}{7} = \frac{49 + 6}{7} = \frac{55}{7} = 7\frac{6}{7} \]

Answer: \[ 7\frac{6}{7} \]

Solution:

\[ 8\frac{5}{13} + 12\frac{7}{26} = 8 + \frac{5}{13} + 12 + \frac{7}{26} = 20 + \frac{10 + 7}{26} = 20 \frac{17}{26} \]

Answer: \[ 20 \frac{17}{26} \]

(d) \[70\] meters \[ 9\frac{7}{10} \] centimeters + \[80\] meters \[ 17\frac{3}{50} \] centimeters + \[40\] meters \[ 27\frac{9}{25} \] centimeters.

Solution:

\[70\] meters \[ 9\frac{7}{10} \] centimeters + \[80\] meters \[ 17\frac{3}{50} \] centimeters + \[40\] meters \[ 27\frac{9}{25} \] centimeters
= \[70\] meters + \[80\] meters + \[40\] meters + \[ 9\frac{7}{10} \] centimeters + \[ 17\frac{3}{50} \] centimeters + \[ 27\frac{9}{25} \] centimeters
= \[190\] meters + \[ 54\frac{3}{25} \] centimeters

Answer: \[190\] meters + \[ 54\frac{3}{25} \] centimeters

Question 6: Subtract:

(a) \[ \frac{3}{8} – \frac{1}{7} \]
Solution: \[ \frac{3}{8} – \frac{1}{7} \]
= \[ \frac{21 – 8}{56} \]
= \[ \frac{13}{56} \]
Answer: \[ \frac{13}{56} \]

(b) \[ 8\frac{4}{15} – 7\frac{13}{45} \]
Solution: \[ 8\frac{4}{15} – 7\frac{13}{45} \]
= \[ \frac{124}{15} – \frac{328}{45} \]
= \[ \frac{372 – 328}{45} \]
= \[ \frac{44}{45} \]
Answer: \[ \frac{44}{45} \]

(c) \[ 20 – 9\frac{20}{21} \]
Solution: \[ 20 – 9\frac{20}{21} \]
= \[ 20 – \frac{209}{21} \]
= \[ \frac{420 – 209}{21} \]
= \[ \frac{211}{21} \]
= \[ 10\frac{9}{21} \]
Answer: \[ 10\frac{9}{21} \]

(d) \[ 25 \] kg \[ 10\frac{1}{5} \] g – \[ 17 \] kg \[ 7\frac{7}{25} \] g
Solution: \[ 25 \] kg \[ 10\frac{1}{5} \] g – \[ 17 \] kg \[ 7\frac{7}{25} \] g
= (\[ 25 – 17 \]) kg + (\[ 10\frac{1}{5} – 7\frac{7}{25} \]) g
= \[ 8 \] kg + (\[ \frac{51}{5} – \frac{182}{25} \]) g
= \[ 8 \] kg + (\[ \frac{255 – 182}{25} \]) g
= \[ 8 \] kg + \[ \frac{73}{25} \] g
= \[ 8 \] kg + \[ 2\frac{23}{25} \] g
= \[ 8 \] kg \[ 2\frac{23}{25} \] g
Answer: \[ 8 \] kg \[ 2\frac{23}{25} \] g

Question 7: Simplify:

(a) \[ 7 – \frac{3}{8} + 8 – \frac{4}{7} \]
Solution: \[ 7 – \frac{3}{8} + 8 – \frac{4}{7} \]
= \[ 7 + 8 – \frac{3}{8} – \frac{4}{7} \]
= \[ 15 – \frac{21 + 32}{56} \]
= \[ 15 – \frac{53}{56} \]
= \[ \frac{840 – 53}{56} \] (∴ LCM of \[ 8 \] and \[ 7 \] is \[ 56 \])
= \[ \frac{787}{56} \]
= \[ 14\frac{3}{56} \]
Answer: \[ 14\frac{3}{56} \]

(b) \[ 9 – 3\frac{15}{16} – 2\frac{7}{8} + \frac{9}{32} \]
Solution: \[ 9 – 3\frac{15}{16} – 2\frac{7}{8} + \frac{9}{32} \]
= \[ 9 – \frac{63}{16} – \frac{23}{8} + \frac{9}{32} \]
= \[ \frac{288 – 126 – 92 + 9}{32} \]
= \[ \frac{297 – 218}{32} \]
= \[ \frac{79}{32} \]
= \[ 2\frac{15}{32} \]
Answer: \[ 2\frac{15}{32} \]

(c) \[ 2\frac{1}{2} – 4\frac{3}{5} – 11 + 17\frac{7}{15} \]
Solution: \[ 2\frac{1}{2} – 4\frac{3}{5} – 11 + 17\frac{7}{15} \]
= \[ \frac{5}{2} – \frac{23}{5} – 11 + \frac{262}{15} \]
= \[ \frac{750 – 138 – 330 + 524}{30} \]
= \[ \frac{599 – 468}{30} \]
= \[ \frac{131}{30} \]
= \[ 4\frac{11}{30} \]
Answer: \[ 4\frac{11}{30} \]

Question 8: Mr. Azmain obtained \[ 20\frac{1}{10} \] quintals of Aman, \[ 30\frac{1}{20} \] quintals of IRRI, and \[ 10\frac{1}{50} \] quintals of Aush rice from his land in one year. How many quintals of rice did he obtain from his land in one year?

Solution: Total rice obtained by Mr. Azmain in one year = (\[ 20\frac{1}{10} + 30\frac{1}{20} + 10\frac{1}{50} \]) quintals
= (\[ 20 + 30 + 10 + \frac{1}{10} + \frac{1}{20} + \frac{1}{50} \]) quintals
= (\[ 60 + \frac{10 + 5 + 2}{100} \]) quintals
= (\[ 60 + \frac{17}{100} \]) quintals
= \[ 60\frac{17}{100} \] quintals
Answer: Total rice obtained in one year is \[ 60\frac{17}{100} \] quintals.

Question 9: A bamboo of length \[ 25 \] meters was painted \[ 5\frac{4}{25} \] meters black, \[ 7\frac{1}{4} \] meters red, and \[ 4\frac{3}{10} \] meters yellow. How much of the bamboo remains unpainted?

Solution: Total length of black, red, and yellow paint = (\[ 5\frac{4}{25} + 7\frac{1}{4} + 4\frac{3}{10} \]) meters
= (\[ \frac{129}{25} + \frac{29}{4} + \frac{43}{10} \]) meters
= \[ \frac{1671}{100} \] meters

∴ Unpainted length of the bamboo = Total length – painted length
= (\[ 25 – \frac{1671}{100} \]) meters
= \[ \frac{2500 – 1671}{100} \] meters
= \[ \frac{829}{100} \] meters
= \[ 8\frac{29}{100} \] meters
Answer: \[ 8\frac{29}{100} \] meters of the bamboo remains unpainted.

Question 10: Amina received \[ 105\frac{7}{10} \] grams of gold from her mother and \[ 98\frac{3}{5} \] grams from her brother. How much does she need from her father to have a total of 400 grams of gold?

Solution: Amina received gold from her mother and brother totaling (\[ 105\frac{7}{10} + 98\frac{3}{5} \]) grams
= (\[ 105 + 98 + \frac{7}{10} + \frac{3}{5} \]) grams
= (\[ 203 + \frac{7 + 6}{10} \]) grams
= (\[ 203 + \frac{13}{10} \]) grams
= (\[ 203 + 1\frac{3}{10} \]) grams
= \[ 204\frac{3}{10} \] grams

∴ She needs to receive (400 – \[ 204\frac{3}{10} \]) grams from her father
= (\[ 400 – \frac{2043}{10} \]) grams
= (\[ \frac{4000 – 2043}{10} \]) grams
= \[ \frac{1957}{10} \] grams
= \[ 195\frac{7}{10} \] grams

Answer: Amina needs to receive \[ 195\frac{7}{10} \] grams of gold from her father.

Question 11: Zabeed traveled \[ \frac{3}{10} \] of the total distance by rickshaw, \[ \frac{2}{5} \] by bicycle, \[ \frac{1}{5} \] on foot, and the remaining 2 kilometers by horse-drawn carriage. It takes him an average of 5 minutes to travel each kilometer by rickshaw or bicycle.

(a) Arrange \[ \frac{3}{10}, \frac{2}{5}, \frac{1}{5} \] in ascending order.

(b) Find the total distance traveled.

Solution:

(a) The denominators of the given fractions are 10, 5, and 5. The LCM of 10 and 5 is 10.

First fraction = \[ \frac{3}{10} \]
Second fraction = \[ \frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10} \]
Third fraction = \[ \frac{1}{5} = \frac{1 \times 2}{5 \times 2} = \frac{2}{10} \]

Arranging the numerators in ascending order, we get \[ 2 < 3 < 4 \]
∴ \[ \frac{2}{10} < \frac{3}{10} < \frac{4}{10} \]
Thus, \[ \frac{1}{5} < \frac{3}{10} < \frac{2}{5} \]

Answer: \[ \frac{1}{5} < \frac{3}{10} < \frac{2}{5} \].

(b) The total distance Zabeed traveled by rickshaw, bicycle, and on foot is (\[ \frac{1}{5} + \frac{3}{10} + \frac{2}{5} \]) of the total distance
= (\[ \frac{3 + 4 + 2}{10} \])
= \[ \frac{9}{10} \] of the total distance

∴ Distance traveled by horse-drawn carriage = (\[ 1 – \frac{9}{10} \])
= (\[ \frac{10 – 9}{10} \])
= \[ \frac{1}{10} \] of the total distance

Since \[ \frac{1}{10} \] of the total distance is 2 kilometers, the entire distance is
= \[ 2 \div \frac{1}{10} \] kilometers
= \[ 2 \times 10 \] kilometers
= 20 kilometers

Answer: Total distance traveled is 20 kilometers.

(c) From part (b), the total distance is 20 kilometers.
∴ Distance traveled by rickshaw = \[ \frac{3}{10} \] of 20 kilometers = 6 kilometers
Distance traveled by bicycle = \[ \frac{2}{5} \] of 20 kilometers = 8 kilometers

∴ Total distance traveled by rickshaw and bicycle = (6 + 8) kilometers = 14 kilometers

Since Zabeed takes 5 minutes per kilometer by rickshaw or bicycle, he spends (5 × 14) minutes
= 70 minutes
= 1 hour 10 minutes

Answer: Zabeed spent 1 hour 10 minutes traveling by rickshaw and bicycle.

অনুশীলনী ১.৪ সমাধান দেখুন: বাংলা ভার্সন

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Class six math chapter 1: exercise 1.4 complete solution