Mass of the compound = 0.24g
Mass of boron = 0.096 g
Mass of oxygen = 0.144 g
Percentage of boron = \(\frac{\text{Mass of boron}}{\text{ Mass of compound}}\) x 100 = \(\frac{ 0.096g}{0.240 g}\) x 100 = 40%
Percentage of oxygen =\(\frac{\text{Mass of oxygen}}{\text{ Mass of compound }}\)x 100 = \(\frac{0.144g}{0.240g }\)x 100 = 60%
Alternative method
Percentage of oxygen =100 - percentage of boron
=100 - 40
= 60%

Section A | Section B | ||
|---|---|---|---|
Marks | Frequency | Marks | Frequency |
0 − 10 | 3 | 0 − 10 | 5 |
10 − 20 | 9 | 10 − 20 | 19 |
20 − 30 | 17 | 20 − 30 | 15 |
30 − 40 | 12 | 30 − 40 | 10 |
40 − 50 | 9 | 40 − 50 | 1 |
Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.