1. From the pictograph, the number of TV symbols against each day is :
Monday = $4\frac{1}{2}$, Tuesday = 4, Wednesday = $2\frac{1}{2}$, Thursday = 1, Friday = $5\frac{1}{2}$ and Saturday = 3.
The maximum number of symbols is against Friday, i.e. $5\frac{1}{2}$ symbols.
∴ Number of TVs sold on Friday = $5\frac{1}{2} \times 6$
= 33, which is the maximum.
Hence, option (c) is the correct option.
2. Maximum number of TVs sold in a day (i.e. on Friday) = $5\frac{1}{2} \times 6$
= 33
Minimum number of TVs sold in a day (i.e. on Thursday) = 1 × 6
= 6
∴ Required percentage = $\frac{33 - 6}{6} \times 100$%
= $\frac{27}{6} \times 100$%
= 450%
Hence, option (a) is the correct option.
3. Number of TVs sold on Monday = $4\frac{1}{2} \times 6$
= 27
Let the number of TVs sold on the required day be $y$.
Given,
$$\begin{aligned}\frac{27}{y} = \frac{3}{2} \\[1em] \Rightarrow 3y = 27 \times 2 \\[1em] \Rightarrow 3y = 54 \\[1em] \Rightarrow y = \frac{54}{3} \\[1em] \Rightarrow y = 18\end{aligned}$$
Number of TVs sold on Saturday = 3 × 6
= 18
∴ The required day is Saturday.
Hence, option (d) is the correct option.
4. Number of TVs sold on Tuesday = 4 × 6
= 24
Number of TVs sold on Wednesday = $2\frac{1}{2} \times 6$
= 15
Decline in sale = 24 − 15
= 9
∴ Percentage decline = $\frac{9}{24} \times 100$%
= 37.5%
Hence, option (b) is the correct option.