The DAV Class 7 Maths Book Solutions Pdf and DAV Class 7 Maths Chapter 2 Worksheet 2 Solutions of Operations on Rational Numbers offer comprehensive answers to textbook questions.
DAV Class 7 Maths Ch 2 WS 2 Solutions
Question 1.
Find the value of:
(i) \(\frac{6}{7}-\frac{-5}{7}\)
Answer:
\(\frac{6}{7}-\frac{-5}{7}\) = \(\frac{6-(-5)}{7}\)
= \(\frac{6+5}{7}\)
= \(\frac{11}{7}\)
(ii) \(\frac{5}{24}-\frac{7}{36}\)
Answer:
\(\frac{5}{24}-\frac{7}{36}\)
= \([\frac{5 \times 3-7 \times 2}{72}/latex]
= [latex]\frac{15-14}{72}\)
= \(\frac{1}{72}\)
(iii) \(\frac{9}{10}-\frac{7}{-15}\)
Answer:
\(\frac{9}{10}-\frac{7}{-15}\)
= \(\frac{9}{10}-\frac{-7}{15}\)
= \(\frac{9 \times 3-(-7) \times 2}{30}\)
= \(\frac{27+14}{30}\)
= \(\frac{31}{30}\)
(iv) \(\frac{-3}{8}-\frac{(-6)}{20}\)
Answer:
\(\frac{-3}{8}-\frac{(-6)}{20}\)
= \(\frac{-3}{8}+\frac{6}{20}\)
= \(\frac{-3 \times 5+6 \times 2}{40}\)
= \(\frac{-15+12}{40}\)
= \(\frac{-3}{40}\)
Question 2.
Subtract:
(i) \(\frac{5}{9}\) from \(\frac{-7}{9}\)
Answer:
\(\frac{-7}{9}-\frac{5}{9}\)
= \(\frac{-7-5}{9}\)
= \(\frac{-12}{9}=\frac{-4}{3}\)
(ii) \(\frac{-5}{7}\) from 0
Answer:
0 – \(\left(-\frac{5}{7}\right)\)
= 0 + \(\frac{5}{7}\)
= \(\frac{5}{7}\)
(iii) \(\frac{5}{11}\) from \(\frac{-8}{23}\)
Answer:
\(\frac{-8}{23}-\frac{5}{11}\)
= \(\frac{-8 \times 11-5 \times 23}{253}\)
= \(\frac{-88-115}{253}\)
= \(\frac{-203}{253}\)
(iv) \(\frac{-2}{9}\) from \(\frac{7}{6}\)
Answer:
\(\frac{7}{6}-\left(-\frac{2}{9}\right)\)
= \(\frac{7}{6}+\frac{2}{9}\)
= \(\frac{3 \times 7+2 \times 2}{18}\)
= \(\frac{21+4}{18}\)
= \(\frac{25}{18}\)
Question 3.
The sum of two rational numbers is – 5. If one of the number is \(\frac{2}{3}\), find the other.
Answer:
The required number = Sum of the numbers – the given number
= 5 – \(\frac{2}{3}\)
= \(\frac{-5}{1}-\frac{2}{3}\)
= \(\frac{-5 \times 3-2 \times 1}{3}\)
= \(\frac{-15-2}{3}\)
= \(\frac{-17}{3}\)
Hence the other number = \(\frac{-17}{3}\).
Question 4.
What number should be added to \(\frac{-3}{7}\) to get 1?
Answer:
Here sum of the two numbers is 1 and one of them is \(\frac{-3}{7}\)
∴ The other number = 1 – (\(\frac{-3}{7}\) )
= 1 + \(\frac{3}{7}\)
= \(\frac{1}{1}+\frac{3}{7}\)
= \(\frac{1 \times 7+3 \times 1}{7}\)
= \(\frac{7+3}{7}\)
= \(\frac{10}{7}\)
Hence the required number is \(\frac{10}{7}\).
Question 5.
What number should be subtracted from – 1 so as to get \(\frac{5}{3}\)?
Answer:
Here sum of the two numbers = – 1
and one of them is \(\frac{5}{3}\)
∴The required number = – 1 – \(\frac{5}{3}\)
= \(\frac{-1}{1}-\frac{5}{3}\)
= \(\frac{-1 \times 3-5 \times 1}{3}\)
= \(\frac{-3-5}{3}\)
= \(\frac{-8}{3}\)
Hence the required number = \(\frac{-8}{3}\)
Question 6.
Simplify:
(i) \(\frac{-4}{5}-\frac{3}{15}+\frac{7}{20}\)
Answer:
\(\frac{-4}{5}-\frac{3}{15}+\frac{7}{20}\)
= \(\frac{-4 \times 12-3 \times 4+7 \times 3}{60}\)
= \(\frac{-48-12+21}{60}\)
= \(\frac{-60+21}{60}\)
= \(-\frac{39}{60}=-\frac{13}{20}\)
(ii) \(\frac{-5}{13}-\frac{-3}{26}-\frac{9}{-52}\)
Answer:
\(\frac{-5}{13}-\frac{-3}{26}-\frac{9}{-52}\)
= \(\frac{-5}{13}+\frac{3}{26}+\frac{9}{52}\)
= \(\frac{-5 \times 4+3 \times 2+9 \times 1}{52}\)
= \(\frac{-20+6+9}{52}\)
= \(\frac{-20+15}{52}\)
= \(\frac{-5}{52}\)
(iii) \(\frac{7}{24}+\frac{5}{12}-\frac{11}{18}\)
Answer:
\(\frac{7}{24}+\frac{5}{12}-\frac{11}{18}\)
= \(\frac{7 \times 3+5 \times 6-11 \times 4}{72}\)
= \(\frac{21+30-44}{72}\)
= \(\frac{51-44}{72}\)
= \(\frac{7}{72}\)
(iv) \(\frac{-11}{30}-\frac{8}{15}+\frac{7}{6}+\frac{-2}{5}\)
Answer:
\(\frac{-11}{30}-\frac{8}{15}+\frac{7}{6}+\frac{-2}{5}\)
= \(\frac{-11 \times 1-8 \times 2+7 \times 5-2 \times 6}{30}\)
= \(\frac{-11-16+35-12}{30}\)
= \(\frac{-39+35}{30}\)
= \(\frac{-4}{30}= \frac{-2}{15}\)
Question 7.
Find the value of x – y and y – x for x = \(\frac{2}{3}\), y = \(\frac{5}{9}\). Are they equal?
Answer:
Here x = \(\frac{2}{3}\), y = \(\frac{5}{9}\)
x – y = \(\frac{2}{3}\) – \(\frac{5}{9}\)
= \(\frac{2 \times 3-5 \times 1}{9}\)
= \(\frac{6-5}{9}=\frac{1}{9}\)
and y – x = \(\frac{5}{9}-\frac{2}{3}\)
= \(\frac{5 \times 1-2 \times 3}{9}\)
= \(\frac{5-6}{9}=\frac{-1}{9}\)
∴ \(\frac{1}{9} \neq-\frac{1}{9}\)
So they are not equal.
Question 8.
For x = \(\frac{1}{10}\), y = \(\frac{-3}{5}\), z = \(\frac{7}{20}\), find the values of expressions (x – y) – z and x – (y – z). Are they equal?
Answer:
Here, x = \(\frac{1}{10}\),
y = \(\frac{-3}{5}\),
z = \(\frac{7}{20}\)
(x – y) – z = \(\left(\frac{1}{10}-\frac{-3}{5}\right)-\frac{7}{20}\)
= \(\left(\frac{1}{10}+\frac{3}{5}\right)-\frac{7}{20}\)
= \(\left(\frac{1 \times 1+3 \times 2}{10}\right)-\frac{7}{20}\)
= \(\left(\frac{1+6}{10}\right)-\frac{7}{20}=\frac{7}{10}-\frac{7}{10}\)
= 0
and x – (y – z) = \(\frac{1}{10}-\left(\frac{-3}{5}-\frac{7}{20}\right)\)
= \(\frac{1}{10}-\left(\frac{-3 \times 4-7 \times 1}{20}\right)\)
= \(\frac{1}{10}-\left(\frac{-12-7}{20}\right)\)
= \(\frac{1}{10}-\left(\frac{-19}{20}\right)\)
= \(\frac{1}{10}+\frac{19}{20}\)
= \(\frac{1 \times 2+19 \times 1}{20}\)
= \(\frac{2+19}{20}\)
= \(\frac{21}{20}\)
∴ 0 ≠ \(\frac{21}{20}\)
Hence they are not equal.