# BSE Odisha 7th Class Maths Solutions Chapter 5 ପରିମେୟ ସଂଖ୍ୟା Ex 5.4

Odisha State Board BSE Odisha 7th Class Maths Solutions Chapter 5 ପରିମେୟ ସଂଖ୍ୟା Ex 5.4 Textbook Exercise Questions and Answers.

## BSE Odisha Class 7 Maths Solutions Chapter 5 ପରିମେୟ ସଂଖ୍ୟା Ex 5.4

Question 1.
ନିମ୍ନଲିଖ ପରିମେୟ ସଂଖ୍ୟାମାନଙ୍କର ଗୁଣଫଳ ନିର୍ଣ୍ଣୟ କର ।

(କ) $$\frac{7}{24}$$ × -16
ସମାଧାନ:
$$=\frac{7}{24} \times \frac{(-16)}{1}=\frac{-16 \times 7}{24 \times 1}=\frac{-8 \times 2 \times 7}{8 \times 3}=\frac{-2 \times 7} {3}=\frac{-14}{3}$$

(ଖ) $$\frac{-3}{5}$$ × 2
ସମାଧାନ:
= $$-\frac{3}{5} \times \frac{2}{1}=\frac{-3 \times 2}{5 \times 1}=\frac{-6}{5}=-1 \frac{1}{5}$$

(ଗ) $$\frac{-7}{6}$$ × (-24)
ସମାଧାନ:
= $$\frac{(-7)}{6} \times \frac{(-24)}{1}=\frac{(-7) \times(-24)}{6 \times 1}=\frac{7 \times 24}{6}=\frac{7 \times 4 \times 6}{6}$$ = 7 × 4 = 28

(ଘ) $$\frac{5}{7} \times\left(\frac{-2}{3}\right)$$
ସମାଧାନ:
= $$\frac{5 \times(-2)}{7 \times 3}=-\frac{-10}{21}$$

(ଙ) $$\frac{9}{8} \times \frac{32}{7}$$
ସମାଧାନ:
= $$\frac{9 \times 32}{8 \times 7}=\frac{9 \times 8 \times 4}{8 \times 7}=\frac{9 \times 4}{7}=\frac{36}{7}=5 \frac{1}{7}$$

(ଚ) $$\frac{50}{7} \times \frac{14}{7}$$
ସମାଧାନ:
= $$\frac{50}{7} \times \frac{2}{1}=\frac{50 \times 2}{7 \times 1}=\frac{100}{7}=14 \frac{2}{7}$$

(ଛ) $$\frac{4}{7} \times \frac{2}{7}$$
ସମାଧାନ:
= $$\frac{4 \times 2}{7 \times 7}=\frac{8}{49}$$

(ଜ) $$\frac{13}{15} \times \frac{25}{26}$$
ସମାଧାନ:
$$=\frac{13 \times 25}{15 \times 26}=\frac{13 \times 5 \times 5}{3 \times 5 \times 2 \times 13}=\frac{5}{3 \times 2}=\frac{5}{6}$$

Question 2.
ସରଳ କର ।

(କ) $$\left(\frac{-16}{15} \times \frac{20}{8}\right)-\left(\frac{15}{5} \times \frac{35}{5}\right)$$
ସମାଧାନ:

(ଖ) $$\left(\frac{13}{8} \times \frac{12}{13}\right)+\left(\frac{-4}{9} \times \frac{3}{2}\right)$$
ସମାଧାନ:

Question 3.
ପ୍ରମାଣ କର x × y = y × x ଯେତେବେଳେ

(କ) x = $$\frac{1}{2}$$, y = $$\frac{3}{5}$$
ସମାଧାନ:

(ଖ) x = $$\frac{2}{7}$$, y = $$\frac{-11}{8}$$
ସମାଧାନ:

(ଗ) x = $$\frac{3}{5}$$, y = $$\frac{2}{9}$$
ସମାଧାନ: