ncert solutions for class 7 maths chapter 2 exercise 2.3
ncert solutions for class 7 maths chapter 2 exercise 2.3 in simple solutions are given here. ncert solutions for class 7 maths chapter 2 exercise 2.3 of NCERT Solutions for Maths Class 7 Chapter 2 contains all the topics containing the basic information of Fractions. ncert solutions for class 7 maths chapter 2 exercise 2.3 provide students the study of fractions including proper, improper and mixed fractions as well as their addition and subtraction. NCERT Solutions for ncert solutions for class 7 maths chapter 2 exercise 2.3 provides necessary material for solving different range of questions that test the students capability of understanding the concepts. It also helps the students of CBSE Class 7 Maths studentsQuestion 1.
Find:
(i) \(\dfrac{1}{4}\) of (a) \(\dfrac{1}{4}\) (b) \(\dfrac{3}{5}\) (c)\(\dfrac{4}{3}\)
Solution:-
(a) \(\dfrac{1}{4}\)
We have,
= \(\dfrac{1}{4}\) × \(\dfrac{1}{4}\)
= (1 × 1)/ (4 × 4) \(\dfrac{(1 × 1)}{(4 × 4)}\)
= \(\dfrac{1}{16}\)
(b) \(\dfrac{3}{5}\)
We have,
= \(\dfrac{1}{4}\) × \(\dfrac{3}{5}\)
= \(\dfrac{(1 × 3)}{(4 × 5)}\)
= \(\dfrac{3}{20}\)
(c) \(\dfrac{4}{3}\)
We have,
= \(\dfrac{1}{4}\) × (\(\dfrac{4}{3}\))
= \(\dfrac{1}{4}\) × (\(\dfrac{4}{3}\))
= \(\dfrac{(1 × 4)}{(4 × 3)}\)
=\(\dfrac{4}{12}\)
= \(\dfrac{1}{3}\)
(ii) \(\dfrac{1}{7}\) of (a) \(\dfrac{2}{9}\) (b) \(\dfrac{6}{5}\) (c) \(\dfrac{3}{10}\)
Solution:-
(a) \(\dfrac{2}{9}\)
We have,
= (\(\dfrac{1}{7}\) ) × \(\dfrac{2}{9}\)
= \(\dfrac{1 ×2}{7 ×9}\)
= \(\dfrac{2}{63}\)
(b) \(\dfrac{6}{5}\)
We have,
= \(\dfrac{1}{7}\) × \(\dfrac{6}{5}\)
= \(\dfrac{1 × 6}{7 × 5}\)
= \(\dfrac{6}{35}\)
(c) \(\dfrac{3}{10}\)
We have,
= \(\dfrac{1}{7}\) × \(\dfrac{3}{10}\)
= \(\dfrac{1 ×3}{7 ×10}\)
= \(\dfrac{3}{70}\)
Question 2.
Multiply and reduce to lowest form (if possible):
(i) \(\dfrac{2}{3}\) × \(2\dfrac{2}{3}\)
Solution:-
First convert the given mixed fraction into improper fraction.
\(2\dfrac{2}{3}\)= \(\dfrac{8}{3}\)
Now,
= \(2\dfrac{2}{3}\) × \(\dfrac{8}{3}\)
Then,
= \(\dfrac{2 × 8}{3 × 3}\)
= \(\dfrac{16}{9}\)
=\(1\dfrac{7}{9}\)
(ii) \(\dfrac{2}{7}\) × \(\dfrac{7}{9}\)
Solution:-
Then,
= \(\dfrac{(2 × 7)}{ (7 × 9)}\)
= \(\dfrac{2}{9}\)
(iii) \(\dfrac{3}{8}\) × \(\dfrac{6}{4}\)
Solution:-
= \(\dfrac{(3 × 6)}{(8 × 4)}\)
= \(\dfrac{9}{16}\)
(iv) \(\dfrac{9}{5}\) × \(\dfrac{3}{5}\)
Solution:-
\(\dfrac{9}{5}\) × \(\dfrac{3}{5}\)
Then,
=\(\dfrac{(9 × 3)}{(5 × 5)}\)
= \(\dfrac{27}{25}\)
=\(1\dfrac{2}{25}\)
(v) \(\dfrac{1}{3}\) × \(\dfrac{15}{8}\)
Solution:-
= \(\dfrac{(1 × 15)}{(3 × 8)}\)
= \(\dfrac{5}{8}\)
(vi) \(\dfrac{11}{2}\) × \(\dfrac{3}{10}\)
Solution:-
= \(\dfrac{(11 × 3)}{(2 × 10)}\)
= \(\dfrac{33}{20}\)
=\(1\dfrac{13}{20}\)
(vii) \(\dfrac{4}{5}\) × \(\dfrac{12}{7}\)
Solution:-
=\(\dfrac{4}{5}\) × \(\dfrac{12}{7}\)
=\(\dfrac{(4 × 12)}{(5 × 7)}\)
= \(\dfrac{48}{35}\)
=\(1\dfrac{13}{35}\)
Question 3.
Multiply the following fractions:
(i) \(\dfrac{2}{5} × 5 \dfrac{1}{4}\)
Solution:-
First convert the given mixed fraction into improper fraction.
= \(5\dfrac{1}{4} = \dfrac{21}{4}\)
Now,
= \(\dfrac{2}{5} × \dfrac{21}{4}\)
= \(\dfrac{(2 × 21)}{(5 × 4)}\)
= \(\dfrac{21}{10}\)
=\(2\dfrac{1}{10}\)
(ii) \(6\dfrac{2}{5} × \dfrac{7}{9}\)
Solution:-
=\(\dfrac{32}{5} × \dfrac{7}{9}\)
= \(\dfrac{(32 × 7)}{(5 × 9)}\)
=\(\dfrac{224}{45}\)
=\(4\dfrac{44}{45}\)
(iii) \(\dfrac{3}{2}\) × \(5\dfrac{1}{3}\)
Solution:-
= \(\dfrac{3}{2}\) × \(5\dfrac{16}{3}\)
=\(\dfrac{ (3 × 16)}{ (2 × 3)}\)
= 8
(iv) \(\dfrac{5}{6}\) × \(5\dfrac{16}{3}\)
Solution:-
=\(\dfrac{5}{6}\) × \(5\dfrac{17}{3}\)
= \(\dfrac{(5 × 17)}{(6 × 7)}\)
= \(\dfrac{85}{42}\)
=\(2\dfrac{1}{42}\)
(v) \(3\dfrac{2}{5} × \dfrac{4}{7}\)
Solution:-
=\(\dfrac{17}{5} × \dfrac{4}{7}\)
= \(\dfrac{17 × 4}{5 × 7}\)
= \(\dfrac{68}{35}\)
=\(1\dfrac{33}{35}\)
(vi)\(2\dfrac{3}{5} × 3\)
Solution:-
=\(\dfrac{13}{5} × \dfrac{3}{1}\)
= \(\dfrac{13 × 3}{5 × 1}\)
= \(\dfrac{39}{5}\)
=\(7\dfrac{4}{5}\)
(vii) \(3\dfrac{4}{7} × \dfrac{3}{5}\)
Solution:-
= \(\dfrac{25}{7} × \dfrac{3}{5}\)
= \(\dfrac{25 × 3 }{7 × 5}\)
= \(\dfrac{15 }{7 }\)
=\(2\dfrac{1}{7 }\)
Question 4.
Which is greater?
(i) \(\dfrac{2 }{7 }\) of \(\dfrac{3 }{4 }\) or \(\dfrac{3 }{5}\) of \(\dfrac{5}{8}\)
Solution:-
We have,
= \(\dfrac{2 }{7 }\)× \(\dfrac{3 }{4 }\) and \(\dfrac{3 }{5}\) × \(\dfrac{5}{8}\)
= \(\dfrac{2 }{7 }\) × \(\dfrac{3 }{4 }\)
= \(\dfrac{2 × 3 }{7 × 4 }\)
= (1 × 3)/ (7 × 2)
= \(\dfrac{3 }{14 }\) … [i]
And,
= \(\dfrac{3 }{5}\) × \(\dfrac{5}{8}\)
= \(\dfrac{3 × 5 }{5 × 8}\)
= \(\dfrac{3 }{8}\) … [ii]
Now, convert [i] and [ii] into like fractions,
LCM of 14 and 8 is 56
Now, let us change each of the given fraction into an equivalent fraction having 56 as the denominator.
[\(\dfrac{3 }{14 } × \dfrac{4 }{4 }] = \dfrac{12 }{56 }\)
\([(\dfrac{3 }{8} ×\dfrac{7 }{7 }] = \dfrac{21 }{56 }\)
Clearly,
\(\dfrac{12 }{56 } < \dfrac{21 }{56 }\)
Hence,
\(\dfrac{3 }{14 } < \dfrac{3 }{8}\)
(ii) \(\dfrac{1}{2}\) of \(\dfrac{6}{7}\) or \(\dfrac{2}{3}\) of \(\dfrac{3}{7}\)
Solution:-
We have,
= \(\dfrac{1}{2}\) × \(\dfrac{6}{7}\) and \(\dfrac{2}{3}\) × \(\dfrac{3}{7}\)
= \(\dfrac{1}{2}\) × \(\dfrac{6}{7}\)
= \(\dfrac{1 × 6}{2 × 7}\)
= \(\dfrac{3}{7}\) … [i]
And,
= \(\dfrac{2}{3} × \(\dfrac{3}{7}\)
= \(\dfrac{2 × 3}{3 × 7}\)
= \(\dfrac{2}{7}\) … [ii]
By comparing [i] and [ii],
Clearly,
\(\dfrac{3}{7} > \dfrac{2}{7}\)
Question 5.
Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is \(\dfrac{3}{4}\) m. Find the distance between the first and the last sapling.
Solution:-
The distance between two adjacent saplings = \(\dfrac{3}{4}\) m
Number of saplings planted by Saili in a row = 4
Then, number of gap in saplings = \(\dfrac{3}{4}\) × 4 = 3
∴The distance between the first and the last saplings = 3 × \(\dfrac{3}{4}\)
= \(\dfrac{9}{4}\) m
= 2 \(\dfrac{1}{4}\) m
Hence, the distance between the first and the last saplings is 2 \(\dfrac{1}{4}\) m.
Question 6
Lipika reads a book for 1 ¾ hours every day. She reads the entire book in 6 days. How many hours in all were required by her to read the book?
Solution:-
From the question, it is given that,
Lipika reads the book for = 1 \(\dfrac{3}{4}\) hours every day = \(\dfrac{7}{4}\) hours
Number of days she took to read the entire book = 6 days
∴Total number of hours required by her to complete the book = \(\dfrac{7}{4}\) × 6
= \(\dfrac{7}{2}\) × 3
= \(\dfrac{21}{2}\)
= 10 \(\dfrac{1}{2}\) hours
Hence, the total number of hours required by her to complete the book is 10 \(\dfrac{1}{2}\) hours.
Question 7.
A car runs 16 km using 1 litre of petrol. How much distance will it cover using 2 \(\dfrac{3}{4}\) litres of petrol.
Solution:-
From the question, it is given that,
The total number of distance travelled by a car in 1 liter of petrol = 16 km
Then,
Total quantity of petrol = 2 \(\dfrac{3}{4}\) liter = \(\dfrac{11}{4}\) liters
Total number of distance travelled by car in \(\dfrac{11}{4}\) liters of petrol = \(\dfrac{11}{4}\) × 16
= 11 × 4
= 44 km
∴Total number of distance travelled by car in \(\dfrac{11}{4}\) liters of petrol is 44 km.
Question 8.
(a) (i) provide the number in the box [ ], such that \(\dfrac{2}{3}\) × [ ] = \(\dfrac{10}{30}\).
Solution:-
Let the required number be x,
Then,
= \(\dfrac{2}{3} × (x) \) = \(\dfrac{10}{30}\)
By cross multiplication,
= x = \(\dfrac{10}{30}\) × \(\dfrac{3}{2}\)
= x = \(\dfrac{10 × 3}{ 30 × 2}\)
= x = \(\dfrac{5}{10}\)
∴The required number in the box is \(\dfrac{5}{20}\)
(ii) The simplest form of the number obtained in [ ] is
Solution:-
The number in the box is \(\dfrac{5}{10}\)
Then,
The simplest form of \(\dfrac{5}{10}\) is \(\dfrac{1}{2}\)
(b) (i) provide the number in the box [ ], such that \(\dfrac{3}{5}\) × [ ] = \(\dfrac{24}{75}\)
Solution:-
Let the required number be x,
Then,
= \(\dfrac{3}{5}\) × (x) = \(\dfrac{24}{75}\)
By cross multiplication,
= x = \(\dfrac{24}{75}\) × \(\dfrac{5}{3}\)
= x = \(\dfrac{24 × 5}{75 × 3}\)
= x = \(\dfrac{8}{15}\)
∴The required number in the box is \(\dfrac{8}{15}\).
(ii) The simplest form of the number obtained in [ ] is
Solution:-
The number in the box is \(\dfrac{8}{15}\).
Then,
The simplest form of \(\dfrac{8}{15}\) is \(\dfrac{8}{15}\).
