DAV Class 7 Maths Chapter 12 Brain Teasers
Data Handling Brain Teasers
1. A. Tick the correct option.
(a) If the mean of \( 7,4,x \) and \( 10 \text{ is } 8 \), find the value of \( x \)- (i) 13 (ii) 11 (iii) 12 (iv) 9
Answer \( \color{red}11 \)
\[ \begin{align*} \text{Mean} &= \frac{7+4+x+10}{4} \\ \\ 8 &= \frac{21+x}{4} \\ \\ 21 + x &= 32 \\ x &= 32 - 21 \\ x & = \boxed{\color{red}11} \end{align*} \]
(b) The observation that occurs maximum number of times in a data is called- (i) range (ii) median (iii) mode (iv) mean
Answer \(\color{red} mode \)
(c) The heights (in cm) of six students in a class are \( \color{black} 145, 152, 148, 150, 153, 146 \). The median of the height is- (i) \( 149 \,cm \) (ii) \( 151\,cm \) (iii) \( 148 \,cm\) (iv) \( 150 \,cm \)
Answer \(\color{red} 149 \,cm\)
\[ \begin{align*} \text{Ascending order}& = 145, 146, 148, 150, 152, 153 \\ \text{No of data} & = 6\, (even) \\ \text{Median} & = \text{Mean of two middle most data} \\ \\ & = \frac{148 +150}{2} \\ \\ & = \frac{298}{2} \\ \\ \text{Median} & = \color{red} 149 \,cm \end{align*} \]
(d) The mean weight of five students is 51 kg and the mean weight of four students is 54 kg. The mean weight of nine students will be- (i) less than 52 kg (ii) more than 53 kg (iii) between 52 kg and 53 kg (iv) equal to 51 kg
Answer \( \color{red} \text{between 52 kg and 53 kg} \)
\[ \begin{align*} \text{Mean weight of 5 students} & = 51 \, kg \\ \text{Sum of weight of 5 students} & = 51 \times 5 \\ &= 255 \, kg \\ \\ \text{Mean weight of 4 students} & = 54 \, kg \\ \text{Sum of weight of 4 students} & = 54 \times 4 \\ &= 216 \, kg \\ \\ \text{Total weight of 9 students} & = 255 + 216 \\ &= 471 \, kg \\ \\\text{Mean} & = \frac{471}{9} \\ \\ \text{Mean weight of 9 students} & = \color{red} 52.3 \, kg \\ \\\end{align*} \]
(e) The mean of first ten odd numbers is - (i) \( 10 \) (ii) \( 13 \) (iii)\( 15 \) (iv) \( 9 \)
Answer \(\color{red} 10 \)
\[ \begin{align*} \text{Mean} & = \frac{1+3+5+7+9+11+13+15+17+19}{10} \\ \\ & = \frac{100}{10} \\ \\ \text{Mean}& = \color{red}10 \end{align*} \]
B. Answer the following questions.
(a) Calculate the mode of \( 1.6, 1.3, 1.8, 1.6, 1.3, 1.1, 1.2, 1.3 \)
\[ \begin{align*} \text{Ascending order} &= 1.1,1.2,{\color{green}1.3,1.3,1.3},1.6,1.6,1.8 \\ \text{Mode} &= \boxed{\color{red}1.3} \end{align*} \]
Answer \(\color{red} Mode = \boxed{1.3} \)
(b) Name the three measures of central tendency.
Answer \(\color{red} Mean, Median, Mode \)
(c) Find the mean of prime numbers between 70 and 10.
\[ \begin{align*} & \text{Prime numbers between 70 and 10}\\ &= 71,73,79,83,89,97 \\ \\ \text{Mean} &= \frac{71+73+79+83+89+97}{6} \\ \\ &= \frac{492}{6} \\ \\ \text{Mean} &= \color{red} 82 \\ \\ \end{align*} \]
Answer \( \color{red} Mean = \boxed{82} \)
(d) What is the median of the observations 36, 48, 29, 62, 71 and 84?
\[ \begin{align*} \text{Ascending order} &= 29,36,48,62,71,84 \\ \text{Number of observations} &= 6 \, (even) \\ \text{Median} &= \text{Mean of two middle most data} \\ &= \frac{48 +62}{2} \\ \\ &= \frac{110}{2} \\ \\ \text{Median} &= \color{red}55 \\ \end{align*} \]
Answer \( \color{red} Median = \boxed{55} \)
(e) The mean of 16 numbers is 8. If 2 is added to every number, find the new mean.
Answer \( \color{red} New \, Mean = \boxed{10} \)
2. Find the mean, median, and mode of the given data: 35, 32, 35, 42, 35, 32, 34.
Solution
\[ \begin{align*} & \boxed{\color{green}\text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}}} \\ \\ & = \frac{35 + 32 + 35 + 42 + 35 + 32 + 34}{7} \\ \\ & = \frac{245}{7} \\ \\ & \text{Mean}= \boxed{\color{green}35} \\ \\ \end{align*} \]
\[ \begin{align*} &\text{Ascending order} = 32, 32, 34, 35, 35, 35, 42 \\ \\ &\text{Number of observations} = 7 \ (\text{odd}) \\ \\ \color{green}\text{Median} &= \color{green} \left(\frac{n+1}{2}\right)^{th} \text{ observation} \\ \\ &= \left(\frac{7+1}{2}\right)^{th} \text{ observation} \\ \\ &= 4^{th} \text{ observation} \\ \\ \text{Median}&= \boxed{\color{green}35} \\ \\ \end{align*} \]
\[ \begin{align*} & \color{green} \text{Mode} = \text{Most frequently occurring observation} \\ \\ &\text{Observations: } 32, 32, 34, \boxed{\color{brown}35, 35, 35}, 42 \\ \\ &\text{Mode} = \boxed{\color{green}35} \\ \\ \end{align*} \]
Answer \(Mean = \boxed{\color{red}35} ,Median = \boxed{\color{red}35} , Mode = \boxed{\color{red}35} \)
3. The mean of 40 observations was 160. It was detected on rechecking that the value 165 was wrongly copied as 125 for computation of mean. Find the correct mean.
Solution
\[ \begin{align*} \text{Mean of 40 observations} &= 160 \\ \text{Sum of 40 observations} &= 160 \times 40 \\ &= 6400 \\ \\ \text{Correct sum of observations} &= 6400 + 165 - 125 \\ &= 6400+ 40 \\ &= 6440 \\ \\ \text{Correct mean} &= \frac{6440}{40} \\ \\ &= 161 \end{align*} \]
Answer Correct Mean \( = \boxed{\color{red}161} \)
4. The mean of 10 numbers is 20. If 5 is subtracted from every number, what will be the new mean?
Solution
\[ \begin{align*} \text{No. of observations} &= 10 \\ \text{Mean of 10 numbers} &= 20 \\ \text{Sum of 10 numbers} &= 20 \times 10 \\ &= 200 \\ \\ \text{Subtracting 5 from each number} &= 5 \times (\text{No. of observations}) \\ &= 5 \times 10 \\ &= 50 \\ \\ \text{New Sum}&= 200 - 50 \\ &= 150 \\ \\ \text{New mean} &= \frac{150}{10} \\ \\ &= 15 \\ \\ \end{align*} \]
Answer New Mean \( = \boxed{\color{red}15} \)
5. Find the median of the following observations: 46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33. If 92 is replaced by 99 and 41 by 43 in the above data, find the new median.
Solution
\[ \begin{align*} &\text{Ascending order:} \\ &33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92 \\ &\text{Number of observations:} \ 11 \ (\text{odd}) \\ \text{Median} &= \text{Middle most observation} \\ \text{Median} &= \color{green}58 \\ \\ &\text{Ascending order after replacing the data:} \\ & 33, 35, {\color{green}43}, 46, 55, 58, 64, 77, 87, 90, {\color{green}99} \\ \text{Median} &= \text{Middle most observation} \\ \text{New Median} &= \color{green}58 \\ \\ \end{align*} \]
Answer New Median \( = \boxed{\color{red}58} \)
6. Find the mode of \( \color{black} 2\frac{1}{4}, 3.5, 4.1, 3\frac{1}{2}, 2\frac{1}{2}, 0.5\)
Solution
\[ \begin{align*} &\text{Convert mixed fractions to decimals} \\ & = 2.25, 3.5, 4.1, 3.5, 2.5, 0.5 \\ \text{Ascending order} & = 0.5, 2.25, 2.5, \boxed{ \color{brown}3.5, 3.5}, 4.1 \\ \text{Mode} & = \text{Most frequently occurring observation} \\ \text{Mode} & = \boxed{\color{green}3.5} \end{align*} \]
Answer Mode \( = \boxed{\color{red}3.5} \)
7. The results of pass percentage of Class-X and XII in CBSE examination for five years are given below.
\[ \color{brown} \begin{array}{|c|c|c|c|c|c|} \hline \text{Year } & 2004-05 & 2005-06 & 2006-07 & 2007-08 & 2008-09 \\ \hline \text{X} & 90 & 95 & 90 & 80 & 98 \\ \hline \text{XII} & 95 & 80 & 85 & 90 & 95 \\ \hline \end{array} \]
Answer