\displaystyle \textbf{Question 1. }\text{If the mean of }x,y,z\text{ is }y,\text{ then mean of }x\text{ and }z\text{ is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \frac{x+y+z}{3}=y\Rightarrow x+z=2y
\displaystyle \therefore\ \frac{x+z}{2}=y
\displaystyle \therefore\ \text{The required mean is }y.
\displaystyle \\

\displaystyle \textbf{Question 2. }\text{The mode of the data }2,x,3,4,5,2,4,6\text{ where }x>2\text{, is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \therefore\ \text{Mode}=4.
\displaystyle \\

\displaystyle \textbf{Question 3. }\text{The mean of first }673\text{ natural numbers is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Mean}=\frac{1+673}{2}=\frac{674}{2}=337
\displaystyle \therefore\ \text{The required mean is }337.
\displaystyle \\

\displaystyle \textbf{Question 4. }\text{The mean of the observations }425,430,435,\ldots,495\text{ is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Mean}=\frac{\text{First term}+\text{Last term}}{2}
\displaystyle =\frac{425+495}{2}=\frac{920}{2}=460
\displaystyle \therefore\ \text{The required mean is }460.
\displaystyle \\

\displaystyle \textbf{Question 5. }\text{If the mean and median of a unimodal data are }34.5\text{ and }32.5
\displaystyle \text{respectively, then mode of the data is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Mode}=3\text{ Median}-2\text{ Mean}
\displaystyle =3(32.5)-2(34.5)
\displaystyle =97.5-69=28.5
\displaystyle \therefore\ \text{Mode}=28.5.
\displaystyle \\

\displaystyle \textbf{Question 6. }\text{The mean of first }n\text{ odd natural numbers is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Sum of first }n\text{ odd natural numbers}=n^2
\displaystyle \text{Mean}=\frac{n^2}{n}=n
\displaystyle \therefore\ \text{The required mean is }n.
\displaystyle \\

\displaystyle \textbf{Question 7. }\text{The mean of the observations }1,3,5,7,9,\ldots,99\text{ is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Mean}=\frac{\text{First term}+\text{Last term}}{2}
\displaystyle =\frac{1+99}{2}=\frac{100}{2}=50
\displaystyle \therefore\ \text{The required mean is }50.
\displaystyle \\

\displaystyle \textbf{Question 8. }\text{If the mean of first }n\text{ natural numbers is }20,\text{ then }n=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Mean of first }n\text{ natural numbers}=\frac{n+1}{2}
\displaystyle \frac{n+1}{2}=20\Rightarrow n+1=40\Rightarrow n=39
\displaystyle \therefore\ n=39.
\displaystyle \\

\displaystyle \textbf{Question 9. }\text{If a mode exceeds a mean by }12,\text{ then the mode exceeds median by }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Let mean}=M.\quad\therefore\ \text{Mode}=M+12
\displaystyle \text{Mode}=3\text{ Median}-2\text{ Mean}
\displaystyle M+12=3\text{ Median}-2M
\displaystyle \Rightarrow3\text{ Median}=3M+12\Rightarrow\text{Median}=M+4
\displaystyle \therefore\ \text{Mode}-\text{Median}=(M+12)-(M+4)=8
\displaystyle \therefore\ \text{The mode exceeds the median by }8.
\displaystyle \\

\displaystyle \textbf{Question 10. }\text{If the median of the observations }x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8\text{ is }m,
\displaystyle \text{then the median of }x_3,x_4,x_5,x_6\text{, where }x_1<x_2<x_3<x_4<x_5<x_6<x_7<x_8,
\displaystyle \text{is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle m=\frac{x_4+x_5}{2}
\displaystyle \text{Median of }x_3,x_4,x_5,x_6=\frac{x_4+x_5}{2}=m
\displaystyle \therefore\ \text{The required median is }m.
\displaystyle \\

\displaystyle \textbf{Question 11. }\text{If mode}-\text{median}=2,\text{ then median}-\text{mean}=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Mode}=3\text{ Median}-2\text{ Mean}
\displaystyle \Rightarrow\text{Mode}-\text{Median}=2(\text{Median}-\text{Mean})
\displaystyle 2=2(\text{Median}-\text{Mean})
\displaystyle \therefore\ \text{Median}-\text{Mean}=1.
\displaystyle \\

\displaystyle \textbf{Question 12. }\text{If the average of }a,b,c,d\text{ is the average of }b\text{ and }c,\text{ then}
\displaystyle \text{the value of }a-b-c+d\text{ is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \frac{a+b+c+d}{4}=\frac{b+c}{2}
\displaystyle \Rightarrow a+b+c+d=2b+2c
\displaystyle \Rightarrow a-b-c+d=0
\displaystyle \therefore\ \text{The required value is }0.
\displaystyle \\

\displaystyle \textbf{Question 13. }\text{Given that }a,b,c,d\text{ are non-zero integers such that }a<b<c<d.
\displaystyle \text{If the mean and median of }a,b,c,d\text{ are equal to zero, then }a=\underline{\hspace{1.5cm}}
\displaystyle \text{and }b=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Mean}=0\Rightarrow a+b+c+d=0
\displaystyle \text{Median}=0\Rightarrow\frac{b+c}{2}=0\Rightarrow b+c=0
\displaystyle \therefore\ b=-c
\displaystyle a+b+c+d=0\Rightarrow a+d=0\Rightarrow a=-d
\displaystyle \therefore\ a=-d\text{ and }b=-c.
\displaystyle \\

\displaystyle \textbf{Question 14. }\text{If }a<b<2a\text{ and mean and median of }a,b\text{ and }2a\text{ are }15
\displaystyle \text{and }12\text{ respectively, then }a=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Since }a<b<2a,\text{ the median is }b.
\displaystyle \therefore\ b=12
\displaystyle \frac{a+b+2a}{3}=15
\displaystyle \frac{3a+12}{3}=15\Rightarrow3a+12=45\Rightarrow a=11
\displaystyle \therefore\ a=11.
\displaystyle \\

\displaystyle \textbf{Question 15. }\text{If the mean of }26,19,15,24\text{ and }x\text{ is }x,\text{ then median of the}
\displaystyle \text{data is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \frac{26+19+15+24+x}{5}=x
\displaystyle \frac{84+x}{5}=x\Rightarrow84+x=5x\Rightarrow x=21
\displaystyle \text{Arranging the data: }15,19,21,24,26
\displaystyle \therefore\ \text{Median}=21.
\displaystyle \\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.