Question

4) Out of 100 sums, Rita could do 60 sums correctly. Out of 80 of those sums,
Binoy could do 50 sums correctly. Let's express them in ratio to find, who got
more sums correctly.



chapter Ratio​

Answer 1

Given :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• Out of 100 sums, Rita could do 60 sums correctly. Out of 80 of those sums, Binoy could do 50 sums correctly.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Need to calculate :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• Who got more sums correctly ?

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Method of solving : ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• Here, we are given that Out of 100 sums, Rita could do 60 sums correctly. Out of 80 of those sums, Binoy could do 50 sums correctly.

• So, firstly take Ritu's ratio of solving sums and then take Binoy's ratio of solving sums.

$\dag$ Here the ratios of Ritu and Binoy will be :

• Ritu's ratio = $\sf \dfrac{60}{100}$

• Binoy's ratio = $\sf \dfrac{50}{80}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Calculation :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➨ $\sf \dfrac{60}{100}$ = $\sf \dfrac{50}{80}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➨ 60 × 80 = 50 × 100

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➨ 4800 < 5000

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Therefore, Binoy solved more sums correctly.