Gujarat Board GSEB Textbook Solutions Class 7 Maths Chapter 9 Rational Numbers Ex 9.2 Textbook Questions and Answers.
Gujarat Board Textbook Solutions Class 7 Maths Chapter 9 Rational Numbers Ex 9.2
Question 1.
Find the sum:
(i) \frac { 5 }{ 4 } + ( \frac { -11 }{ 4 } )
(ii) \frac { 5 }{ 3 } + \frac { 3 }{ 5 }
(iii) \frac { -9 }{ 10 } + \frac { 22 }{ 15 }
(iv) \frac { -3 }{ -11 } and \frac { 5 }{ 9 }
(v) \frac { -8 }{ 19 } + \frac { (-2) }{ 57 }
(vi) \frac { -2 }{ 3 } + 0
(vii) – 2\frac { 1 }{ 3 } + 4\frac { 3 }{ 5 }
Solution:
(i) \frac { 5 }{ 4 } + ( \frac { -11 }{ 4 } )
We have \frac { 5 }{ 4 } + ( \frac { -11 }{ 4 } ) = \frac { 5+(-11) }{ 4 } = \frac { -6 }{ 4 }
= \frac { -3 }{ 2 } or -1\frac { 1 }{ 2 }
(ii) \frac { 5 }{ 3 } + \frac { 3 }{ 5 }
∵ LCM of 3 and 5 is 15.
(iv) \frac { -3 }{ -11 } and \frac { 5 }{ 9 }
∵ LCM of 11 and 9 is 99.
∴ \frac { -3 }{ -11 } = \frac { (-3)×9 }{ (-11)×9 } = \frac { -27 }{ -99 } = \frac { 27 }{ 99 }
and \frac { 5 }{ 9 } = \frac { 5×11 }{ 9×11 } = \frac { 55 }{ 99 }
(vi) \frac { -2 }{ 3 } + 0
We have \frac { -2 }{ 3 } + 0 = \frac { -2 }{ 3 } + \frac { 0 }{ 3 } = \frac { -2+0 }{ 3 } = \frac { -2 }{ 3 }
Thus, \frac { -2 }{ 3 } + 0 = \frac { -2 }{ 3 }
(vii) – 2\frac { 1 }{ 3 } + 4\frac { 3 }{ 5 }
Question 2.
Find:
(i) \frac { 7 }{ 24 } – \frac { 17 }{ 36 }
(ii) \frac { 5 }{ 63 } – ( \frac { -6 }{ 21 } )
(iii) \frac { -6 }{ 13 } – \frac { -7 }{ 15 }
(iv) \frac { -3 }{ 8 } – \frac { 7 }{ 11 }
(v) -2\frac { 1 }{ 9 } – 6
Solution:
(i) \frac { 7 }{ 24 } – \frac { 17 }{ 36 }
∵ LCM of 24 and 36 is 72.
(ii) \frac { 5 }{ 63 } – ( \frac { -6 }{ 21 } )
∵ LCM of 63 and 21 is 63.
Thus, \frac { 5 }{ 63 } – ( \frac { -6 }{ 21 } = \frac { -13 }{ 72 }
(iii) \frac { -6 }{ 13 } – \frac { -7 }{ 15 }
∵ LCM of 13 and 15 is 195.
(v) -2\frac { 1 }{ 9 } – 6
Question 3.
Find the product:
(i) \frac { 9 }{ 2 } x ( \frac { -7 }{ 4 } )
(ii) \frac { 3 }{ 10 } x (-9)
(iii) \frac { -6 }{ 5 } x \frac { 9 }{ 11 }
(iv) \frac { 3 }{ 7 } x \frac { -2 }{ 5 }
(v) \frac { 3 }{ 11 } x \frac { 2 }{ 5 }
(vi) \frac { 3 }{ -5 } x \frac { -5 }{ 3 }
Solution:
Question 4.
Draw the number line and represent the following rational numbers on it:
(i) (- 4) ÷ \frac { 2 }{ 3 }
(ii) \frac { -3 }{ 5 } ÷ 2
(iii) \frac { -4 }{ 5 } ÷ (-3)
(iv) \frac { -1 }{ 8 } ÷ \frac { 3 }{ 4 }
(v) \frac { -2 }{ 13 } ÷ \frac { 1 }{ 7 }
(vi) \frac { -7 }{ 12 } ÷ ( \frac { -2 }{ 13 } )
(vii) \frac { 3 }{ 13 } ÷ ( \frac { -4 }{ 65 } )
Solution:
(i) (- 4) ÷ \frac { 2 }{ 3 }
Thus, (- 4) ÷ \frac { 2 }{ 3 } = – 6
(ii) \frac { -3 }{ 5 } ÷ 2
∵ The reciprocal of 2 is \frac { 1 }{ 2 }
∴ \frac { -3 }{ 5 } ÷ 2 = \frac { -3 }{ 5 } x \frac { 1 }{ 2 }
= \frac { (-3)×1 }{ 5×2 } = \frac { -3 }{ 10 }
Thus, \frac { -3 }{ 5 } ÷ 2 = \frac { -3 }{ 10 }
(iii) \frac { -4 }{ 5 } ÷ (-3)
(iv) \frac { -1 }{ 8 } ÷ \frac { 3 }{ 4 }
(vi) \frac { -7 }{ 12 } ÷ ( \frac { -2 }{ 13 } )
∵ The reciprocal of \frac { -2 }{ 13 } is ( \frac { 13 }{ 2 }
∴ \frac { -7 }{ 12 } ÷ ( \frac { -2 }{ 13 } ) = \frac { -7 }{ 12 } x ( \frac { 13 }{ 2 }
= \frac { (-7)×(-13) }{ 12×2 }
= \frac { 91 }{ 24 } or 3\frac { 19 }{ 24 }
Thus,
\frac { -7 }{ 12 } ÷ ( \frac { -2 }{ 13 } )
= 3\frac { 19 }{ 24 }
= \frac { 91 }{ 24 } or 3\frac { 19 }{ 24 }
Thus,
\frac { -7 }{ 12 } ÷ ( \frac { -2 }{ 13 } )
= 3\frac { 19 }{ 24 }
(vii) \frac { 3 }{ 13 } ÷ ( \frac { -4 }{ 65 } )