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Maths 4 Marks Questions

Measures Of Central Tendency · Jharkhand Board (JAC) STD 9 (Jharkhand - English Medium). 22 questions.

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Question 14 Marks
Duration of sunshine(in hours) in Amritsar for first $10$ days of August $1997$ as reported by the Meterological Department are given as follows: $9.6, 5.2, 3.5, 1.5, 1.6, 2.4, 2.6, 8.4, 10.3, 10.9$
$i.$ Find the mean $\overline{\text{X}}$
$ii.$ Verify that $\sum\limits_{\text{i}=1}^{10}\Big(\text{X}-\overline{\text{X}}\Big)=0$
Answer
Duration of sunshine $($in hours$)$ for $10$ days are $= 9.6, 5.2, 3.5, 1.5, 1.6, 2.4, 2.6, 8.4, 10.3, 10.9$
$i.$ Mean $X=\frac{\text{Sum of numbers}}{\text{Total numbers}}$
$=\frac{9.6+5.2+3.5+1.5+1.6+2.4+2.6+8.4+10.3+10.9}{10}$
$=\frac{56}{10}=5.6$
$ii. \text{L.H.S.}=\sum\limits_{\text{i}=1}^{10}\Big(\text{x}^\text{i}-\overline{\text{X}}\Big)$
$=\Big(\text{X}_1-\overline{\text{X}}\Big)+\Big(\text{X}_2-\overline{\text{X}}\Big)+\Big(\text{X}_3-\overline{\text{X}}\Big)+\dots+\Big(\text{X}_{10}-\overline{\text{X}}\Big)$
$= (9.6-5.6) + (5.2 - 5.6) + (3.5 - 5.6) + (1.5 - 5.6) + (1.6 - 5.6) \\+ (2.4 - 5.6) + (2.6 - 5.6) + (8.4 - 5.6) + (10.3 - 5.6) + (10.9 - 5.6)$
$= 4 - 0.4 - 2.1 - 4.1 - 4 - 3.2 - 3 + 2.8 + 4.7 + 5.3$
$= 16.8-16.8 = 0$
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Question 24 Marks
If the mean of the following data is $15$, find $p$.
f:
$6$
$p$
$6$
$10$
$5$
x:
$5$
$10$
$15$
$20$
$25$
Answer
x
f
fx
$5$
$6$
$30$
$10$
$p$
$10p$
$15$
$6$
$90$
$20$
$10$
$200$
$25$
$5$
$125$
 
$N = p + 27$
$\sum\text{fx}=10\text{p}+445$
It is given that,
Mean $= 15$
$\Rightarrow\frac{\sum\text{fx}}{\text{N}}=15$
$\Rightarrow\text{10p}+\text{445p}+27=15$
$\Rightarrow\text{10p}+445=15\times(\text{p}+27)$
$\Rightarrow\text{10p}+445=\text{15p}+405$
$\Rightarrow\text{15p}-\text{10p}=445-405$
$\Rightarrow\text{5p}=40$
$\Rightarrow\text{p}=405=8$
$\Rightarrow\text{p}=8$
$\therefore\text{p}=8$
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Question 34 Marks
Find the value of $p$, if the mean of the following distribution is $20.$
x:
$15$
$17$
$19$
$20 + p$
$23$
f:
$2$
$3$
$4$
$5p$
$6$
Answer
x
f
fx
$15$
$2$
$30$
$17$
$3$
$51$
$19$
$4$
$76$
$20+p$
$5p$
$100p + 5p^2$
$23$
$6$
$138$
 
$N = 5p + 15$
$fx = 5p^2 + 100p + 295$
It is given that,
Mean $= 20$
$\Rightarrow\frac{\sum\text{fx}}{\text{N}}=20$
$\Rightarrow\frac{\text{5p}^2+\text{100p}+295}{\text{5p}+15}=20$
$\Rightarrow 5p^2 + 100p + 295 = 20(5p + 15)$
$\Rightarrow 5p^2 + 100p + 295 = 100p + 300$
$\Rightarrow 5p^2 = 300 − 295$
$\Rightarrow 5p^2 = 5$
$\Rightarrow p^2 = 1$
$\Rightarrow\text{p}=\pm1$
Frequency can’t be negative.
Hence, value of $p$ is $1$
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Question 44 Marks
Find the mean of the following distribution:
x:
$10$
$12$
$20$
$25$
$35$
f:
$3$
$10$
$15$
$7$
$5$
Answer
x
f
fx
$10$
$3$
$30$
$12$
$10$
$120$
$20$
$15$
$300$
$25$
$7$
$175$
$35$
$5$
$175$
 
N $= 40$
$\sum\text{fx}=800$
$\therefore\text{Mean }\bar{\text{x}}=\frac{\sum\text{fx}}{\text{N}}$
$=\frac{800}{40}=20$
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Question 54 Marks
Find the missing frequency $(p)$ for the following distribution whose mean is $7.68$
x:
$3$
$5$
$7$
$9$
$11$
$13$
f:
$6$
$8$
$15$
$p$
$8$
$4$
Answer
x
f
fx
$3$
$6$
$18$
$5$
$8$
$40$
$7$
$15$
$105$
$9$
$P$
$9p$
$11$
$8$
$88$
$13$
$4$
$52$
 
$N = p + 41$
$\sum\text{fx}=\text{9p}+303$
It is given that,
Mean $= 7.68$
$\Rightarrow\frac{\sum\text{fx}}{\text{N}}=7.68$
$\Rightarrow 9p + 303p + 41 = 7.68$
$\Rightarrow 9p + 303 = 7.68p + 314.88$
$\Rightarrow 9p − 7.68p = 314.88 - 303$
$\Rightarrow 1.32p = 11.88$
$\Rightarrow p = 11.881.32 = 9$
$\Rightarrow p = 9$
$\therefore$ $P = 9$
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Question 64 Marks
The mean of the following data is $20.6$ Find the value of $p$.
x:
$10$
$15$
$p$
$25$
$35$
f:
$3$
$10$
$25$
$7$
$5$
Answer
x
f
fx
$10$
$3$
$30$
$15$
$10$
$150$
$P$
$25$
$25p$
$25$
$7$
$175$
$35$
$5$
$175$
 
N $= 50$
$\sum\text{fx}=25\text{p}+530$
It is given that,
Mean $= 20.6$
$\Rightarrow\frac{\text{25p}+530}{50}=20.6$
$\Rightarrow\text{25p}+530=20.6\times50$
$\Rightarrow\text{25p}=1030-530$
$\Rightarrow\text{25p}=500$
$\Rightarrow\text{p}=\frac{500}{25}=20$
$\Rightarrow\text{p}=20$
$\therefore\text{p}=20$
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Question 74 Marks
Candidates of four schools appear in a mathematics test. The data were as follows:
Schools
No. of Candidates
Average Score
$I$
$60$
$75$
$II$
$48$
$80$
$III$
Not Available
$55$
$IV$
$40$
$50$
If the average score of the candidates of all four schools is $66$, Find the number of candidates that appeared from school III.
Answer
Schools
No. of Candidates
Average Score
$I$
$60$
$75$
$II$
$48$
$80$
$III$
$x$
$55$
$IV$
$40$
$50$
Given the average score of all schools $= 66$
$\Rightarrow\frac{\text{N}_1\overline{\text{X}}_1+\text{N}_2\overline{\text{X}}_2+\text{N}_3\overline{\text{X}}_3+\text{N}_4\overline{\text{X}}_4}{\text{N}_1+\text{N}_2+\text{N}_3+\text{N}_4}=66$
$\Rightarrow\frac{60\times75+48\times80+\text{x}\times55+40\times50}{60+48+\text{x}+40}=66$
$\Rightarrow\frac{4500+3840+\text{55x}+2000}{148+\text{x}}=66$
$\Rightarrow\frac{10340+\text{55x}}{148+\text{x}}=66$
$\Rightarrow10340+55\text{x}=66\text{x}+9768$
$\Rightarrow10340-9768=\text{66x}-\text{55x}$
$\Rightarrow\text{11x}=572$
$\Rightarrow\text{x}=\frac{572}{11}=52$
$\therefore$ No. of candidates appeared from school $III = 52$
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Question 84 Marks
Find the values of $n$ and $\overline{\text{X}}$ in the following case:$\sum\limits^\text{n}_{\text{i}=1}(\text{x}_\text{i}-10)=30$ and $\sum\limits^\text{n}_{\text{i}=1}(\text{x}_\text{i}-6)=150$
Answer
Given $\sum\limits^\text{n}_{\text{i}=1}(\text{x}_\text{i}-10)=30$$\Rightarrow(\text{x}_1-10)+(\text{x}_2-10)+\dots+(\text{x}_\text{n}-10)=30$
$\Rightarrow(\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4+\text{x}_5+\dots+\text{x}_\text{n})\\-(10+10+10+10+\dots+10)=30$
$\Rightarrow\sum\text{x}-\text{10n}=30\dots(1)$
And $\sum\limits^\text{n}_{\text{i}=1}(\text{x}_\text{i}-6)=150$
$\Rightarrow(\text{x}_1-6)+(\text{x}_2-6)+\dots+(\text{x}_\text{n}-6)=150$
$\Rightarrow(\text{x}_1+\text{x}_2+\dots+\text{x}_\text{n})-(6+6+6+\dots+6)=150$
$\Rightarrow\sum\text{x}-\text{6n}=150\dots(2)$
By subtracting equation $(1)$ from equation $(2)$, we get
$\sum\text{x}-\text{6n}-\sum\text{x}+\text{10n}=150-30$
$\Rightarrow\text{4n}=120$
$\Rightarrow\text{n}=\frac{120}{4}=30$
Put value of n in equation $(1)$
$\sum\text{x}-10\times30=30$
$\Rightarrow\sum\text{x}-300=30$
$\Rightarrow\sum\text{x}=30+300=330$
$\therefore\overline{\text{x}}=\frac{\sum\text{x}}{\text{n}}=\frac{330}{30}=11$
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Question 94 Marks
Find the median of the following data: $41, 43, 127, 99, 61, 92, 71, 58, 57$. If $58$ is replaced by $85$, what will be the new median?
Answer
Given the numbers are $41, 43, 127, 99, 61, 92, 71, 58, 57$ Arranging the numbers in ascending order $41, 43, 57, 58, 61, 71, 92, 99, 127$ $n = 9$ (odd)
$\therefore\text{New Median}=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{value}$
$=\Big(\frac{\text{9}+1}{2}\Big)^{\text{th}}\text{value}$
If $58$ is replaced by $85$ Then the new values be in order are: $41, 43, 57, 61, 71, 85, 92, 99, 127$
$\therefore\text{New Median}=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{value}$
$=\Big(\frac{\text{9}+1}{2}\Big)^{\text{th}}\text{value}$
$=5^{\text{th}}\text{value}=71$
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Question 104 Marks
If the ratio of mode and median of a certain data is $6 : 5$, then find the ratio of its mean and median.
Answer
Given that the ratio of mode and median of a certain data is $6 : 5$.
That is, MODE : MEDIAN $= 6$ $: 5$
$\Rightarrow\frac{\text{MODE}}{\text{MEDIAN}}=\frac{6}{5}$
$\Rightarrow5\times\text{MODE}=6\times\text{MEDIAN}$
$\Rightarrow\text{MODE}=\frac{6}{5}\text{MEDIAN}$
We know that,$\text{MODE}=3\times\text{MEDIAN}-2\times\text{MEAN}$
$\Rightarrow2\times\text{MEAN}=\Big(3-\frac{6}{5}\Big)\text{MEDIAN}$
$\Rightarrow2\times\text{MEAN}=\frac{9}{5}\text{MEDIAN}$
$\Rightarrow\text{MEAN}=\frac{9}{10}\text{MEDIAN}$
$\Rightarrow\frac{\text{MEAN}}{\text{MEDIAN}}=\frac{9}{10}$
$\Rightarrow\text{MEAN}:\text{MEDIAN}=9:10$
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Question 114 Marks
Find the value of $p$ for the following distribution whose mean is $16.6$.
x:
$8$
$12$
$15$
$p$
$20$
$25$
$30$
f:
$12$
$16$
$20$
$24$
$16$
$8$
$4$
Answer
x
f
fx
$8$
$12$
$96$
$12$
$16$
$192$
$15$
$20$
$300$
$P$
$24$
$24p$
$20$
$16$
$320$
$25$
$8$
$200$
$30$
$4$
$120$
 
N $= 100$
$\sum\text{fx} = 24\text{p} + 1228$
It is given that,
Mean $= 16.6$
$\Rightarrow\frac{\sum\text{fx}}{\text{N}}=16.6$
$\Rightarrow 24\text{p} + 1228 = 1660$
$\Rightarrow 24\text{p} = 1660 − 1228$
$\Rightarrow 24\text{p} = 432$
$\Rightarrow\text{p}=\frac{432}{24}=18$
$\Rightarrow \text{p} = 18$
$\therefore\text{p} = 18$
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Question 124 Marks
Five coins were simultaneously tossed $1000$ times and at each, toss the number of heads was observed. The number of tosses during which $0, 1, 2, 3, 4$ and $5$ heads were obtained are shown in the table below. Find the mean number of heads per toss.
No. of heads per toss
No. of tosses
$0$
$38$
$1$
$144$
$2$
$342$
$3$
$287$
$4$
$164$
$5$
$25$
Total
$1000$
Answer
No. of heads per toss (x)
No. of tosses (f)
fx
$0$
$38$
$0$
$1$
$144$
$144$
$2$
$342$
$684$
$3$
$287$
$861$
$4$
$164$
$656$
$5$
$25$
$125$
 
N $= 1000$
$\sum\text{fx}=2470$
$\therefore$ Mean number of heads per toss $=\frac{\sum\text{fx}}{\text{N}}$
$=\frac{2470}{1000}$
$=2.47$
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Question 134 Marks
Calculate the mean for the following distribution:
x:
$5$
$6$
$7$
$8$
$9$
f:
$4$
$8$
$14$
$11$
$3$
Answer
X
f
fx
$5$
$4$
$20$
$6$
$8$
$48$
$7$
$14$
$98$
$8$
$11$
$88$
$9$
$3$
$27$
 
N $= 40$
$\sum\text{fx}=281$
$\therefore\text{Mean}\ \bar{\text{x}}=\frac{\sum\text{fx}}{\text{N}}$
$=\frac{281}{40}=7.025$
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Question 144 Marks
Find the mean of the following data:
x:
$19$
$21$
$23$
$25$
$27$
$29$
$31$
f:
$13$
$15$
$16$
$18$
$16$
$15$
$13$
Answer
x
f
fx
$19$
$13$
$247$
$21$
$15$
$315$
$23$
$16$
$368$
$25$
$18$
$450$
$27$
$16$
$432$
$29$
$15$
$435$
$31$
$13$
$403$
 
N $= 106$
$\sum\text{fx}=2650$
$\therefore\text{Mean }\bar{\text{x}}=\frac{\sum\text{fx}}{\text{N}}$
$=\frac{2650}{106}=25$
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Question 154 Marks
Find the missing value of $p$ for the following distribution whose mean is $12.58$.
x:
$5$
$8$
$10$
$12$
$p$
$20$
$25$
f:
$2$
$5$
$8$
$22$
$7$
$4$
$2$
Answer
x
f
fx
$5$
$2$
$10$
$8$
$5$
$40$
$10$
$8$
$80$
$12$
$22$
$264$
$P$
$7$
$7p$
$20$
$4$
$80$
$25$
$2$
$50$
 
N $= 50$
$\sum\text{fx}=\text{7p}+524$
It is given that,
Mean $= 12.58$
$\Rightarrow\frac{\sum\text{fx}}{\text{N}}=12.38$
$\Rightarrow\frac{\text{7p}+524}{50}12.58$
$\Rightarrow 7p + 524 = 629$
$\Rightarrow 7p = 629 - 524$
$\Rightarrow 7p = 105$
$\Rightarrow p = 1057 = 15$
$\Rightarrow p = 15$
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Question 164 Marks
Find the values of $n$ and $\overline{\text{X}}$ in the following case:$\sum\limits^\text{n}_{\text{i}=1}(\text{x}_\text{i}-12)=-10$ and $\sum\limits^\text{n}_{\text{i}=1}(\text{x}_\text{i}-3)=62$
Answer
Given $\sum\limits^\text{n}_{\text{i}=1}(\text{x}_\text{i}-12)=-10$
$\Rightarrow(\text{x}_1-12)+(\text{x}_2-12)+\dots+(\text{x}_\text{n}-12)=-10$
$\Rightarrow(\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4+\text{x}_5+\dots+\text{x}_\text{n})\\-(12+12+12+12+\dots+12)=-10$
$\Rightarrow\sum\text{x}-\text{12n}=-10\dots(1)$
And $\sum\limits^\text{n}_{\text{i}=1}(\text{x}_\text{i}-3)=62$
$\Rightarrow(\text{x}_1-3)+(\text{x}_2-3)+\dots+(\text{x}_\text{n}-3)=62$
$\Rightarrow(\text{x}_1+\text{x}_2+\dots+\text{x}_\text{n})-(3+3+3+\dots+3)=62$
$\Rightarrow\sum\text{x}-\text{3n}=62\dots(2)$
By subtracting equation $(1)$ from equation $(2)$, we get
$\sum\text{x}-\text{3n}-\sum\text{x}+\text{12n}=62+10$
$\Rightarrow\text{9n}=72$
$\Rightarrow\text{n}=\frac{72}{9}=8$
Put value of n in equation $(1)$
$\sum\text{x}-12\times8=-10$
$\Rightarrow\sum\text{x}-96=-10$
$\Rightarrow\sum\text{x}=96-10=86$
$\therefore\overline{\text{x}}=\frac{\sum\text{x}}{\text{n}}=\frac{86}{8}=10.75$
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Question 174 Marks
Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is $50$
x:
$10$
$30$
$50$
$70$
$90$
f:
$17$
$f_1$
$32$
$f_2$
$19$
Answer
$x$
$f$
$fx$
$10$
$17$
$170$
$30$
$f_1$
$30f_1$
$50$
$32$
$1600$
$70$
$f_2$
$70f_2$
$90$
$19$
$1710$
 
N $= 120$
$\sum\text{fx}=3480+30\text{f}_1+70\text{f}_2$
It is given that
Mean $= 50$
$\Rightarrow\frac{\sum\text{fx}}{\text{N}}=50$
$\Rightarrow\frac{3480+30\text{f}_1+70\text{f}_2}{\text{N}}=50$
$\Rightarrow 3480 + 30f_1 + 70f_2 = 50 \times 120$
$\Rightarrow 30f_1 + 70f_2 = 6000 - 3480$
$\Rightarrow 10(3f_1 + 7f_2) = 10(252)$
$\Rightarrow 3f_1 + 7f_2 = 252 ...(1)$ $[\because$ Divide by 10$]$
And $N = 20$
$\Rightarrow 17 + f_1 + 32 + f_2 + 19 = 120$
$\Rightarrow 68 + f_1 + f_2= 120$
$\Rightarrow f_1+ f_2 = 120 - 68$
$\Rightarrow f_1 + f_2 = 52$
Multiply with 3 on both sides
$\Rightarrow 3f_1 + 3f_2 = 156 ...(2)$
Subtracting equation $(2)$ from equation $(1)$
$\Rightarrow 3f_1 + 7f_2 - 3f_1 - 3f_2 = 252 - 156$
$\Rightarrow 4f_2 = 96$
$\Rightarrow\text{f}_2=\frac{96}{4}=24$
Put the value of $f_2$ in equation $(1)$
$\Rightarrow 3f_1 + 7 \times 24 = 252$
$\Rightarrow 3f_1= 252 - 168$
$\Rightarrow\text{f}_1=\frac{84}{3}=28$
$\Rightarrow f_1 = 28$
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Question 184 Marks
The weights (in kg) of $15$ students are: $31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30$. Find the median. If the weight $44\ kg$ is replaced by $46\ kg$ and $27\ kg$ by $25\ kg$, find the new median.
Answer
Given the numbers are $31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30$
 Arranging the numbers in ascending order $27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 44, 45. n = 15$ (odd)
$\therefore\text{New Median}=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{value}$
$=\Big(\frac{\text{15}+1}{2}\Big)^{\text{th}}\text{value}$
$=8^{\text{th}}\text{value}=35\text{kg}$
If the weight $44\ kg$ is replaced by $46\ kg$ and $27\ kg$ is replaced by $25\ kg$
Then the new values be in order are: $25, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 45, 46$
$\therefore\text{New Median}=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{value}$
$=\Big(\frac{\text{15}+1}{2}\Big)^{\text{th}}\text{value}$
$=8^{\text{th}}\text{value}=35\text{kg}$
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Question 194 Marks
If the mean of $x + 2, 2x + 3, 3x + 4, 4x + 5 is x + 2$, find $x$.
Answer
The given data is $x + 2, 2x + 3, 3x + 4, 4x + 5$.
They are four in numbers.
 The mean is:$\frac{(\text{x}+2)+(\text{2x}+3)+(\text{3x}+4)+(\text{4x}+5)}{4}$
$=\frac{\text{x}+2+\text{2x}+3+\text{3x}+4+\text{4x}+5}{4}$
$=\frac{\text{10x}+14}{4}$
$=\frac{2(\text{5x}+7)}{4}$
$=\frac{\text{5x}+7}{2}$
But, it is given that the mean is $x + 2$.
Hence, we have$\frac{\text{5x}+7}{2}=\text{x}+2$
$\Rightarrow\text{5x}+7=2(\text{x}+2)$
$\Rightarrow\text{5x}+7=\text{2x}+4$
$\Rightarrow\text{5x}-\text{2x}=4-7$
$\Rightarrow\text{3x}=-3$
$\Rightarrow\text{x}=-\frac{3}{3}$
$\Rightarrow\text{x}=-1$
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Question 204 Marks
If the ratio of mean and median of a certain data is $2 : 3$, then find the ratio of its mode and mean.
Answer
Given that the ratio of mean and median of a certain data is $2 : 3$.
That is, MEAN : MEDIAN $= 2 : 3$
$\Rightarrow\frac{\text{MEAN}}{\text{MEDIAN}}=\frac{2}{3}$
$\Rightarrow3\times\text{MEAN}=2\times\text{MEDIAN}$
$\Rightarrow\text{MEDIAN}=\frac{3}{2}\text{MEAN}$
We know that,$\text{MODE}=3\times\text{MEDIAN}-2\times\text{MEAN}$
$\Rightarrow\text{MODE}=3\times\frac{3}{2}\text{MEAN}-2\times\text{MEAN}$
$\Rightarrow\text{MODE}=\Big(\frac{9}{2}-2\Big)\text{MEAN}$
$\Rightarrow\frac{\text{MODE}}{\text{MEAN}}=\frac{9}{5}-2$
$\Rightarrow\frac{\text{MODE}}{\text{MEAN}}=\frac{5}{2}$
$\Rightarrow\text{MODE}:\text{MEAN}=5:2$
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Question 214 Marks
If the median of $33, 28, 20, 25, 34, x$ is $29$, find the maximum possible value of $x.$
Answer
The given data is $33, 28, 20, 25, 34, x.$
The total number of values is $n = 6$, is an even number.
Hence the median depends on the $\Big(\frac{6}{3}\Big)=3^{\text{rd}}$
observation and $\Big(\frac{6}{2}+1\Big)=4^\text{th}$ observation.
Since we have to find the maximum possible value of $x$.
So we must put it in the $4^{th}$ position when ordering in ascending order.
​​​​​​​Arranging the data in ascending order,
we have $20, 25, 28, x, 33, 34$
 Hence, the median is:$\frac{\Big(\frac{\text{n}}{2}\Big)^{\text{th}}\text{observation}+\Big(\frac{\text{n}}{2}+1\Big)^{\text{th}}\text{observation}}{2}$
$=\frac{\Big(\frac{\text{6}}{2}\Big)^{\text{th}}\text{observation}+\Big(\frac{\text{6}}{2}+1\Big)^{\text{th}}\text{observation}}{2}$
$=\frac{3^\text{rd}\text{observation}+4^\text{th}\text{observation}}{2}$
$=\frac{28+\text{x}}{2}$
Here it is given that the median is $29$. So, we have$\frac{28+\text{x}}{2}=29$
$\Rightarrow28+\text{x}=58$
$\Rightarrow\text{x}=58-28$
$\Rightarrow\text{x}=30$
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Question 224 Marks
If the difference of mode and median of a data is $24$, then find the difference of median and mean.
Answer
Given that the difference of mode and median of a data is $24$.
That is, $\text{MODE} - \text{MEDIAN}$
$$$= 24 \text{MODE}$
$= \text{MEDIAN} + 24$
We have to find the difference between median and mean We know that $\text{MEDIAN} = 3 \times\ \text{MEDIAN} - 2\ \times \text{MEAN}$
$\Rightarrow\text{MEDIAN} + 24 = 3 \times\ \text{MEDIAN} - 2 \times \text{MEAN}$
$\Rightarrow 24 = 3 \times\ \text{MEDIAN}-\text{MEAN} - 2\ \times \text{MEAN}$
$\Rightarrow 24= 2\ \times \text{MEDIAN} - 2\ \times \text{MEAN}$
$\Rightarrow 2 \times\ \text{MEDIAN} - 2 \times\ \text{MEAN} = 24$
$\Rightarrow 2(\text{MEDIAN}-\text{MEAN}) = 24$
$\Rightarrow\text{MEDIAN}-\text{MEAN}=\frac{24}{2}$
$\Rightarrow\text{MEDIAN}-\text{MEAN} = 12$
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