# BSE Odisha 7th Class Maths Solutions Chapter 5 ପରିମେୟ ସଂଖ୍ୟା Ex 5.2

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:
$$=\frac{4 \times 2+2 \times(-1)+1 \times(-3)}{4}=\frac{8+(-2)+(-3)}{4}=\frac{3}{4}$$