Odisha State Board BSE Odisha 7th Class Maths Solutions Chapter 5 ପରିମେୟ ସଂଖ୍ୟା Ex 5.2 Textbook Exercise Questions and Answers.
BSE Odisha Class 7 Maths Solutions Chapter 5 ପରିମେୟ ସଂଖ୍ୟା Ex 5.2
Question 1.
ନିମ୍ନଲିଖୂତ ପରିମେୟ ସଂଖ୍ୟାମାନଙ୍କୁ ଯୋଗ କର ।
(କ) \frac{2}{9}, \frac{5}{9}
ସମାଧାନ:
= \frac{2+5}{9}=\frac{7}{9}
(ଖ) \frac{-3}{7}, \frac{5}{7}
ସମାଧାନ:
= \frac{-3+5}{7}=\frac{2}{7}
(ଗ) \frac{5}{4}, \frac{-7}{4}
ସମାଧାନ:
= \frac{5+(-7)}{4}=\frac{-2}{4}=-\frac{1}{2}
(ଘ) \frac{-17}{6}, \frac{-13}{6}
ସମାଧାନ:
=\frac{(-17)+(-13)}{6}=\frac{-30}{6}
Question 2.
ମାନ ନିର୍ଣ୍ଣୟ କର
(କ) \frac{11}{2}+\frac{5}{4}
ସମାଧାନ:
= \frac{11 \times 2+5 \times 1}{4}=\frac{22+5}{4}=\frac{27}{4}=6 \frac{3}{4}
(ଖ) \frac{-3}{7}+\frac{7}{17}
ସମାଧାନ:
= \frac{-3 \times 17+7 \times 7}{7 \times 17}=\frac{-51+49}{119}=\frac{-2}{119}
(ଗ) \frac{5}{4}+\frac{-4}{3}
ସମାଧାନ:
= \frac{5 \times 3+(-4) \times 4}{4 \times 3}=\frac{15+(-16)}{12}=-\frac{1}{12}
(ଘ) \frac{-1}{2}+\frac{-2}{7}
ସମାଧାନ:
= \frac{(-1) \times 7+2 \times(-2)}{14}=\frac{-7+(-4)}{14}=-\frac{11}{14}
Question 3.
x ଓ y ର ନିମ୍ନଲିଖତ ମାନ ପାଇଁ ପ୍ରମାଣ କର x + y = y + x
(କ) x = \frac{5}{7}, y = \frac{-3}{2}
ସମାଧାନ:
(ଖ) x = -8, y = \frac{9}{2}
ସମାଧାନ:
Question 4.
ମାନ ନିର୍ଣ୍ଣୟ କର ।
(କ) \frac{-3}{10}+\frac{12}{-10}+\frac{14}{10}
ସମାଧାନ:
= \frac{-3}{10}+\frac{-12}{10}+\frac{14}{10}=\frac{(-3)+(-12)+14}{10}=\frac{-1}{10}
(ଖ) \frac{-9}{11}+\frac{2}{3}+\frac{-3}{4}
ସମାଧାନ:
=\frac{12 \times(-9)+44 \times 2+33 \times(-3)}{132}=\frac{(-108)+88+(-99)}{132}=\frac{-119}{132}
(ଗ) 2 + \frac{-1}{2}+\frac{-3}{4}
ସମାଧାନ:
=\frac{4 \times 2+2 \times(-1)+1 \times(-3)}{4}=\frac{8+(-2)+(-3)}{4}=\frac{3}{4}