Question

Can someone please answer the question?​

Answer 1

$\frac{31}{6}$Answer:

Step-by-step explanation:

(i) equation formed:

= 7$\frac{3}{4}$ + 2$\frac{1}{2}$ + 3$\frac{3}{5}$

convert mixed fraction into proper fraction.

= $\frac{31}{4}$ + $\frac{5}{2}$ + $\frac{18}{5}$

find the LCM

LCM = 20

= ($\frac{31}{4}$ × $\frac{5}{5}$ ) + ( $\frac{5}{2}$ × $\frac{10}{10}$ ) + ( $\frac{18}{5}$ × $\frac{4}{4}$ )

= $\frac{155}{20}$ + $\frac{50}{20}$ + $\frac{72}{20}$ = $\frac{277}{20}$ = 13.85kg

(ii) Equation formed:

total weight of the sugar - 13.85 = sugar left with the shopkeeper

= 20 - 13.85

= 6.15kg

(iii)

3$\frac{3}{5}$ kg sugar purchased by Raj.

2$\frac{1}{2}$ kg sugar purchased by Sunil.

Equation formed:

= 3$\frac{3}{5}$ - 2$\frac{1}{2}$

= $\frac{18}{5}$ -  $\frac{5}{2}$

LCM = 10

=  ($\frac{18}{5}$ × $\frac{2}{2}$) -  ($\frac{5}{2}$ × $\frac{5}{5}$)

= $\frac{36}{10}$ - $\frac{25}{10}$ = $\frac{11}{10}$ = 1$\frac{1}{10}$kg

(iv) Equation formed:

7$\frac{3}{4}$ × x = $\frac{31}{6}$

⇒ $\frac{31}{4}$ × x = $\frac{31}{6}$

∴ x =  $\frac{31}{6}$ ÷  $\frac{31}{4}$

x =  $\frac{31}{6}$ × $\frac{4}{31}$ = $\frac{3}{2}$

(v)

five rational between 2$\frac{1}{2}$ and 3$\frac{1}{2}$

$\frac{5}{2}$ and $\frac{7}{2}$

$\frac{5}{2}$ × $\frac{3}{3}$ and $\frac{7}{2}$ × $\frac{3}{3}$

$\frac{15}{6}$ and $\frac{21}{6}$

$\frac{16}{6}$, $\frac{17}{6} , \frac{18}{6} , \frac{19}{6} , \frac{20}{6}$ are five rational numbers between 2$\frac{1}{2}$ and 3$\frac{1}{2}$.

convert them all in decimals.

2.66; 2.83; 3; 3.16; 3.33

Answer 2

Step-by-step explanation:

please reply if seen.... I think you might forget me