Question

Which is false?
(a) If the mean of 4, 6, x, 8, 10, 13 is 8, then x = 7.
(b) If the median of the following array 59, 62, 65, x, x + 2, 72, 85, 99 is 67, then x = 66.
(c) If the mode of 1, 3, 5, 7, 5, 2, 7, 5, 9, 3, p, 11 is 5, then the value of p is 7.
(d) If the mean of 10 observations is 15 and that of other 15 observations is 18, then the mean of all the 25 observations is 16.8.

Answer 1

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{Mean}=\frac{\mathrm{Sum}\mathrm{of}\mathrm{all}\mathrm{observations}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{observations}} \\ =\frac{4+6+x+8+10+13}{6} \\ \mathrm{We}\mathrm{have}: \\ \mathrm{Mean}=8 \\ \Rightarrow \frac{4+6+x+8+10+13}{6}=8 \\ \Rightarrow 41+x=48 \\ \Rightarrow x=7 \\ \mathrm{Hence},\left(\mathrm{a}\right)\mathrm{is}\mathrm{true}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{b}\right)\mathrm{Here},\mathrm{n}=8,\mathrm{which}\mathrm{is}\mathrm{an}\mathrm{even}\mathrm{number}. \\ \therefore \mathrm{Median}=\mathrm{Mean}\mathrm{of}\left(\frac{n}{2}\right)\mathrm{th}\mathrm{term}\&\left(\frac{n}{2}+1\right)\mathrm{th}\mathrm{term} \\ =\frac{1}{2}\left(4\mathrm{th}\mathrm{term}+5\mathrm{th}\mathrm{term}\right) \\ =\frac{1}{2}(x+x+2) \\ =(x+1) \\ \mathrm{Now}, \\ x+1=67 \\ \Rightarrow x=66 \\ \mathrm{Hence},\left(\mathrm{b}\right)\mathrm{is}\mathrm{true}. \end{gathered}$$

(c) When p = 7, mode cannot be 5.
Hence, (c) is false.

(d) Mean of 10 observations = 15
Sum of these 10 observations = 15$\times$10 = 150
Mean of other 15 observations = 18
Sum of these 15 observations = 18$\times$15 = 270
Now,

$$\begin{gathered} \mathrm{Mean}=\frac{\mathrm{Sum}\mathrm{of}\mathrm{all}\mathrm{observations}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{observations}} \\ =\frac{150+270}{25} \\ =16.8 \\ \mathrm{Hence},\left(\mathrm{d}\right)\mathrm{is}\mathrm{true}. \end{gathered}$$