Fractions and Decimals - Solutions

 CBSE Class –VII Mathematics

NCERT Solutions
Chapter 2 Fractions and Decimals (Ex. 2.1)

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Question 1. Solve:

(i)2-35$\frac{3}{5}$          (ii) 4 + 78$\frac{7}{8}$       (iii)35+27$\frac{3}{5}+\frac{2}{7}$    (iv) 911−415$\frac{9}{11}-\frac{4}{15}$

(v)710+25+32$\frac{7}{10}+\frac{2}{5}+\frac{3}{2}$    (vi)i223+312$2\frac{2}{3}+3\frac{1}{2}$ 

(vii)812−358$8\frac{1}{2}-3\frac{5}{8}$

Answer: (i) 2−35$2-\frac{3}{5}$ = 10−35=75$\frac{10-3}{5}=\frac{7}{5}$ = 125$1\frac{2}{5}$

(ii)4+78$4+\frac{7}{8}$ = 32+78=398$\frac{32+7}{8}=\frac{39}{8}$ = 478$4\frac{7}{8}$

(iii)35+27$\frac{3}{5}+\frac{2}{7}$ = 21+1035$\frac{21+10}{35}$ = 3135$\frac{31}{35}$

(iv)911−415$\frac{9}{11}-\frac{4}{15}$ = 135−44165=91165$\frac{135-44}{165}=\frac{91}{165}$

(v) 710+25+32$\frac{7}{10}+\frac{2}{5}+\frac{3}{2}$ = 7+4+1510$\frac{7+4+15}{10}$ = 2610=135$\frac{26}{10}=\frac{13}{5}$ = 235$2\frac{3}{5}$

(vi) 223+312$2\frac{2}{3}+3\frac{1}{2}$ = 83+72$\frac{8}{3}+\frac{7}{2}$ = 16+216$\frac{16+21}{6}$ = 376$\frac{37}{6}$ = 616$6\frac{1}{6}$

(vii)812−358$8\frac{1}{2}-3\frac{5}{8}$ = 172−298$\frac{17}{2}-\frac{29}{8}$ = 68−298$\frac{68-29}{8}$ = 398$\frac{39}{8}$ = 478$4\frac{7}{8}$

Question 2. Arrange the following in descending order:

(i) 29,23,821$\frac{2}{9},\frac{2}{3},\frac{8}{21}$

(ii) 15,37,710$\frac{1}{5},\frac{3}{7},\frac{7}{10}$

Answer: (i) 29,23,821$\frac{2}{9},\frac{2}{3},\frac{8}{21}$ ⇒$\Rightarrow$ 1463,4263,2463$\frac{14}{63},\frac{42}{63},\frac{24}{63}$ [Converting into like fractions]

⇒$\Rightarrow$ 4263>2463>1463$\frac{42}{63}>\frac{24}{63}>\frac{14}{63}$ [Arranging in descending order]

Therefore, 23>821>29$\frac{2}{3}>\frac{8}{21}>\frac{2}{9}$

(ii) 15,37,710$\frac{1}{5},\frac{3}{7},\frac{7}{10}$ ⇒$\Rightarrow$1470,3070,4970$\frac{14}{70},\frac{30}{70},\frac{49}{70}$ [Converting into like fractions]

⇒$\Rightarrow$ 4970>3070>1470$\frac{49}{70}>\frac{30}{70}>\frac{14}{70}$ [Arranging in descending order]

Therefore, 710>37>15$\frac{7}{10}>\frac{3}{7}>\frac{1}{5}$

Question 3. In a “magic square”, the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square?

411$\frac{4}{11}$	911$\frac{9}{11}$	211$\frac{2}{11}$
311$\frac{3}{11}$	511$\frac{5}{11}$	711$\frac{7}{11}$
811$\frac{8}{11}$	111$\frac{1}{11}$	611$\frac{6}{11}$

(Along the first row 411+911+211=1511)$\left(\text{Along the first row }\frac{4}{11}+\frac{9}{11}+\frac{2}{11}=\frac{15}{11}\right)$

Answer: Sum of first row = 411+911+211=1511$\frac{4}{11}+\frac{9}{11}+\frac{2}{11}=\frac{15}{11}$ [Given]

Sum of second row = 311+511+711=3+5+711=1511$\frac{3}{11}+\frac{5}{11}+\frac{7}{11}=\frac{3+5+7}{11}=\frac{15}{11}$

Sum of third row = 811+111+611=8+1+611=1511$\frac{8}{11}+\frac{1}{11}+\frac{6}{11}=\frac{8+1+6}{11}=\frac{15}{11}$

Sum of first column = 411+311+811=4+3+811=1511$\frac{4}{11}+\frac{3}{11}+\frac{8}{11}=\frac{4+3+8}{11}=\frac{15}{11}$

Sum of second column = 911+511+111=9+5+111=1511$\frac{9}{11}+\frac{5}{11}+\frac{1}{11}=\frac{9+5+1}{11}=\frac{15}{11}$

Sum of third column = 211+711+611=2+7+611=1511$\frac{2}{11}+\frac{7}{11}+\frac{6}{11}=\frac{2+7+6}{11}=\frac{15}{11}$

Sum of first diagonal (left to right) = 411+511+611=4+5+611=1511$\frac{4}{11}+\frac{5}{11}+\frac{6}{11}=\frac{4+5+6}{11}=\frac{15}{11}$

Sum of second diagonal (left to right) = 211+511+811=2+5+811=1511$\frac{2}{11}+\frac{5}{11}+\frac{8}{11}=\frac{2+5+8}{11}=\frac{15}{11}$

Since, the sum of fractions in each row, in each column and along the diagonals are same, therefore, it s a magic square.

Question 4.  A rectangular sheet of paper is 1212$12\frac{1}{2}$ cm long and 1023$10\frac{2}{3}$ cm wide. Find its perimeter.

Answer: Given: The sheet of paper is in rectangular form.

Length of sheet = 1212$12\frac{1}{2}$ cm and Breadth of sheet = 1023$10\frac{2}{3}$ cm

Perimeter of rectangle = 2 (length + breadth)

= 2(1212+1023)$2\left(12\frac{1}{2}+10\frac{2}{3}\right)$ = 2(252+323)$2\left(\frac{25}{2}+\frac{32}{3}\right)$

= 2(25×3+32×26)$2\left(\frac{25\times 3+32\times 2}{6}\right)$ = 2(75+646)$2\left(\frac{75+64}{6}\right)$

= 2×1396$2\times \frac{139}{6}$ = 1393=4613$\frac{139}{3}=46\frac{1}{3}$ cm.

Thus, the perimeter of the rectangular sheet is 4613$46\frac{1}{3}$ cm.

Question 5. Find the perimeter of (i) Δ$\mathrm{\Delta }$ABE, (ii) the rectangle BCDE in this figure. Whose perimeter is greater?

Answer: (i) In ΔABE,$\mathrm{\Delta }\text{ABE,}$ AB = 52$\frac{5}{2}$ cm, BE = 234$2\frac{3}{4}$ cm, AE = 335$3\frac{3}{5}$ cm

The perimeter of ΔABE$\mathrm{\Delta }\text{ABE}$ = AB + BE + AE

= 52+234+335$\frac{5}{2}+2\frac{3}{4}+3\frac{3}{5}$ = 52+114+185$\frac{5}{2}+\frac{11}{4}+\frac{18}{5}$

= 50+55+7220$\frac{50+55+72}{20}$ = 17720$\frac{177}{20}$ = 81720$8\frac{17}{20}$ cm

Thus, the perimeter of ΔABE$\mathrm{\Delta }\text{ABE}$ is 81720$8\frac{17}{20}$ cm.

(ii) In rectangle BCDE, BE = 234$2\frac{3}{4}$ cm, ED = 76$\frac{7}{6}$ cm

Perimeter of rectangle = 2 (length + breadth)

= 2(234+76)$2\left(2\frac{3}{4}+\frac{7}{6}\right)$ = 2(114+76)$2\left(\frac{11}{4}+\frac{7}{6}\right)$

= 2(33+1412)$2\left(\frac{33+14}{12}\right)$ = 476$\frac{47}{6}$ = 756$7\frac{5}{6}$ cm

Thus, the perimeter of rectangle BCDE is 756$7\frac{5}{6}$ cm.

Comparing the perimeter of triangle and that of rectangle,

81720$8\frac{17}{20}$ cm >756$7\frac{5}{6}$ cm

Therefore, the perimeter of triangle ABE is greater than that of rectangle BCDE.

Question 6. Salil wants to put a picture in a frame. The picture is 735$7\frac{3}{5}$ cm wide. To fit in the frame the picture cannot be more than 7310$7\frac{3}{10}$ cm wide. How much should the picture be trimmed?

Answer: Given: The width of the picture = 735$7\frac{3}{5}$ cm and the width of picture frame = 7310$7\frac{3}{10}$ cm

Therefore, the picture should be trimmed = 735−7310$7\frac{3}{5}-7\frac{3}{10}$ = 385−7310$\frac{38}{5}-\frac{73}{10}$

= 76−7310$\frac{76-73}{10}$ = 310$\frac{3}{10}$ cm

Thus, the picture should be trimmed by 310$\frac{3}{10}$ cm.

Question 7. Ritu ate 35$\frac{3}{5}$ part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat? Who had the larger share? By how much?

Answer: The part of an apple eaten by Ritu = 35$\frac{3}{5}$

The part of an apple eaten by Somu = 1−35=5−35=25$1-\frac{3}{5}=\frac{5-3}{5}=\frac{2}{5}$

Comparing the parts of apple eaten by both Ritu and Somu 35>25$\frac{3}{5}>\frac{2}{5}$

Larger share will be more by 35−25=15$\frac{3}{5}-\frac{2}{5}=\frac{1}{5}$ part.

Thus, Ritu’s part is 15$\frac{1}{5}$ more than Somu’s part.

Question 8. Michael finished colouring a picture in 712$\frac{7}{12}$ hour. Vaibhav finished colouring the same picture in 34$\frac{3}{4}$ hour. Who worked longer? By what fraction was it longer?

Answer: Time taken by Michael to colour the picture = 712$\frac{7}{12}$ hour

Time taken by Vaibhav to colour the picture = 34$\frac{3}{4}$ hour

Converting both fractions in like fractions, 712$\frac{7}{12}$ and 3×34×3=912$\frac{3\times 3}{4\times 3}=\frac{9}{12}$

Here, 712$\frac{7}{12}$<912$\frac{9}{12}$ ⇒$\Rightarrow$712$\frac{7}{12}$<34$\frac{3}{4}$

Thus, Vaibhav worked longer time.

Vaibhav worked longer time by 34−712=9−712=212=16$\frac{3}{4}-\frac{7}{12}=\frac{9-7}{12}=\frac{2}{12}=\frac{1}{6}$ hour.

Thus, Vaibhav took 16$\frac{1}{6}$ hour more than Michael.