Question 12 Marks
Find the mode of the following data in each case:$ 7, 9, 12, 13, 7, 12, 15, 7, 12, 7, 25, 18, 7$
Answer
|
Values:
|
$7$
|
$9$
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$12$
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$13$
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$15$
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$18$
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$25$
|
|
Frequency:
|
$5$
|
$1$
|
$3$
|
$1$
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$1$
|
$1$
|
$1$
|
Since, maximum frequency $5$ corresponds to value $7$ then mode $= 7$ View full question & answer→Question 22 Marks
Find the mean of first five multiples of $3$
AnswerFirst five multiples of $3$ are $3, 6, 9, 12, 15$ $\therefore\text{Mean}=\frac{\text{Sum of numbers}}{\text{Total numbers}}$
$=\frac{3+6+9+12+15}{5}$
$=\frac{45}{5}$
$=9$
$\text{Mean} = 9$
View full question & answer→Question 32 Marks
Find out the mode of the following marks obtained by $15$ students in a class: Marks: $4, 6, 5, 7, 9, 8, 10, 4, 7, 6, 5, 9, 8, 7, 7.$
Answer
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Marks:
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$4$
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$5$
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$6$
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$7$
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$8$
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$9$
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$10$
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No. of students:
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$2$
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$2$
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$2$
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$4$
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$2$
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$2$
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$1$
|
Since, the maximum frequency corresponds to the value $7$ then mode $= 7$ marks. View full question & answer→Question 42 Marks
Find the mean of $x, x + 2, x + 4, x + 6, x + 8$
AnswerNumbers are $x, x + 2, x + 4, x + 6, x + 8$
$\therefore\text{Mean}=\frac{\text{Sum of numbers}}{\text{Total numbers}}$
$=\frac{\text{x}+\text{x}+2+\text{x}+4+\text{x}+6+\text{x}+8}{5}$
$=\frac{\text{5x}+20}{5}$
$=5\Big(\frac{\text{x}+4}{5}\Big)$
$=\text{x}+4$
View full question & answer→Question 52 Marks
Find the mode from the following data: $125, 175, 225, 125, 225, 175, 325, 125, 375, 225, 125$
Answer
| Values: |
$125$ |
$175$ |
$225$ |
$325$ |
$375$ |
| Frequency: |
$4$ |
$2$ |
$3$ |
$1$ |
$1$ |
Since, the maximum frequency $4$ corresponds to the value $125$ then mode $= 125$ View full question & answer→Question 62 Marks
Find the mean of first five natural numbers.
AnswerThe first five odd numbers are $1, 2, 3, 4, 5$.
$\therefore\text{Mean}=\frac{\text{Sum of numbers}}{\text{Total numbers}}$
$=\frac{1+2+3+4+5}{5}$
$=\frac{15}{5}=3$
$\text{Mean} = 3$
View full question & answer→Question 72 Marks
The mean weight of $8$ numbers is $15$. If each number is multiplied by $2$, what will be the new mean?
AnswerWe have, The mean weight of $8$ numbers is $15$
Then, the sum of $8$ numbers $= 8 \times 15 = 120$
If each number is multiplied by $2$
Then, new mean $= 120 \times 2 = 240$
$\therefore$ new mean $=\frac{240}{8}=30$
View full question & answer→Question 82 Marks
The mean of $200$ items was $50$. Later on, it was on discovered that the two items were misread as $92$ and $8$ instead of $192$ and $88$. Find the correct mean.
AnswerThe mean of $200$ items $=50$ Then the sum of $200$ items $=200 \times 50=10,000$
Correct values $=192$ and $88$ .
Incorrect values $=92$ and $8 . $
$\therefore$ correct sum $=10000-92-8+192+88=10180 $
$\therefore$ correct mean $=\frac{10180}{200}=\frac{101.8}{2}=50.9$
View full question & answer→Question 92 Marks
Find the mean of first ten even natural numbers.
AnswerThe first five even natural numbers are $2, 4, 6, 8, 10, 12, 14, 16,18, 20$.$\therefore\text{Mean}=\frac{\text{Sum of numbers}}{\text{Total numbers}}$
$=\frac{2+4+6+8+10+12+14+16+18+20}{10}$
$=11$
$\text{Mean} = 11$
View full question & answer→Question 102 Marks
The mean of marks scored by $100$ students was found to be $40$. Later on, it was discovered that a score of $53$ was misread as $83$. Find the correct mean.
AnswerMean marks of $100$ students $= 40$
Sum of marks of $100$ students $= 100 \times 40 = 4000$
Correct value $= 53$
Incorrect value $= 83$
Correct sum $= 4000 - 83 + 53 = 3970 $
$\therefore$ correct mean $=\frac{3970}{100}=39.7$
View full question & answer→Question 112 Marks
The mean of 5 numbers is $18$. If one number is excluded, their mean is $16.$ Find the excluded number.
AnswerThe mean of $5$ numbers is $18$
Then, the sum of $5$ numbers $=5 \times 18=90$ If one number is excluded
Then, the mean of $4$ numbers $=16 $
$\therefore$ sum of $4$ numbers $=4 \times 16=64$
Excluded number $=90-64=26$.
View full question & answer→Question 122 Marks
Find the mode for the following series:$ 7.5, 7.3, 7.2, 7.2, 7.4, 7.7, 7.7, 7.5, 7.3, 7.2, 7.6, 7.2$
Answer
| Values: |
$7.2$ |
$7.3$ |
$7.4$ |
$7.5$ |
$7.6$ |
$7.7$ |
| Frequency: |
$4$ |
$2$ |
$1$ |
$2$ |
$1$ |
$2$ |
Since, the maximum frequency $4$ corresponds to the value $7.2$ then mode $= 7.2$ View full question & answer→Question 132 Marks
If the heights of $5$ persons are $140\ cm, 150\ cm, 152\ cm, 158\ cm$ and $161\ cm$ respectively. Find the mean height.
AnswerGiven: The heights of $5$ persons are $140\ cm, 150\ cm, 152\ cm, 158\ cm$ and $161\ cm.$
$\therefore\text{Mean Weight}=\frac{\text{Sum of height}}{\text{Total no. of persons}}$
$=\frac{140+150+152+158+161}{5}$
$=\frac{761}{5}=152.2$
$\text{Mean} = 152.2$
View full question & answer→Question 142 Marks
Find the mode of the following data in each case: $14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18$
AnswerArranging the data in an ascending order $14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28$
Here observation $14$ is having the highest frequency i.e. $4$ in given data.
So, mode of given data is $14.$
View full question & answer→Question 152 Marks
Find the mean of $994, 996, 998, 1000, 1002.$
AnswerNumbers are $994, 996, 998, 1000, 1002$
$\therefore\text{Mean}=\frac{\text{Sum of numbers}}{\text{Total numbers}}$
$=\frac{994+996+998+1000+1002}{5}$
$=\frac{4990}{5}=998$
$\text{Mean} = 998$
View full question & answer→Question 162 Marks
The demand of different shirt sizes, as obtained by a survey, is given below:
|
Size:
|
$38$
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$39$
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$40$
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$41$
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$42$
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$43$
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$44$
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Total
|
|
Number of persons (wearing it):
|
$26$
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$39$
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$20$
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$15$
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$13$
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$7$
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$5$
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$125$
|
Find the modal shirt sizes, as observed from the survey. Answer
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Size:
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$38$
|
$39$
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$40$
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$41$
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$42$
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$43$
|
$44$
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Total
|
|
Number of persons:
|
$26$
|
$39$
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$20$
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$15$
|
$13$
|
$7$
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$5$
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$125$
|
Since, maximum frequency $39$ corresponds to value $39$ then model size $= 39$ View full question & answer→Question 172 Marks
Find the mean of all factors of $10.$
AnswerAll factors of $6$ are $1, 2, 5, 10$
.$\therefore\text{Mean Weight}=\frac{\text{Sum of factors}}{\text{Total factors}}$
$=\frac{1+2+5+10}{4}$
$=4.5$
$\text{Mean} = 4.5$
View full question & answer→Question 182 Marks
The numbers of children in $10$ families of a locality are $2, 4, 3, 4, 2, 3, 5, 1, 1, 5$. Find the number of children per family.
AnswerThe numbers of children in $10$ families are : $2, 4, 3, 4, 2, 3, 5, 1, 1, 5$
$\therefore\text{Mean}=\frac{\text{Total no. children}}{\text{Total families}}$
$=\frac{2+ 4+ 3+ 4+ 2+ 3+ 5+ 1+1+ 5}{10}$
$=3$
View full question & answer→Question 192 Marks
Following are the weights of $10$ new born babies in a hospital on a particular day: $3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6$ (in kg). Find the mean $\overline{\text{X}}.$
AnswerThe weights (in kg) of $10$ new born babies are $: 3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6$
$\therefore\text{Mean Weight}=\frac{\text{Sum of weights}}{\text{Total no. of babies}}$
$=\frac{3.4+ 3.6+ 4.2+ 4.5+ 3.9+ 4.1+ 3.8+ 4.5+4.4+ 3.6}{10}$
$=4\text{kg}$
View full question & answer→Question 202 Marks
The traffic police recorded the speed (in km/hr) of $10$ motorists as $47, 53, 49, 60, 39, 42, 55, 57, 52, 48$. Later on, an error in recording instrument was found. Find the correct average speed of the motorists if the instrument is recorded $5 \ km/hr$ less in each case.
AnswerThe speed of $10$ motorists are $47, 53, 49, 60, 39, 42, 55, 57, 52, 48$ .
Later on it was discovered that the instrument recorded $5 \ km/hr$ less than in each case
$\therefore$ correct values are $= 52, 58, 54, 65, 44, 47, 60, 62, 57, 53.$
$\therefore\text{Correct Mean}=\frac{52+58+54+65+44+47+60+62+57+53}{10}$
$=\frac{552}{10}=55.2\ \text{km/hr}$
View full question & answer→Question 212 Marks
The mean of five numbers is $27$. If one number is excluded, their mean is $25$. Find the excluded number.
AnswerThe mean of five numbers is $27$ The sum of five numbers $=5 \times 27=135$ If one number is excluded, the new mean is $25 \therefore$ Sum of 4 numbers $=4 \times 25=100 \therefore$ Excluded number $=135-100=35$
View full question & answer→Question 222 Marks
The percentage marks obtained by students of a class in mathematics are as follows: $64, 36 , 47, 23, 0, 19, 81, 93, 72, 35, 3, 1.$ Find their mean.
AnswerThe percentage marks obtained by students are $64, 36 , 47, 23, 0, 19, 81, 93, 72, 35, 3, 1$
$\therefore\text{Mean Marks}=\frac{\text{Sum of marks}}{\text{Total numbers of marks}}$
$=\frac{64+ 36+ 47+ 23+ 0+ 19+ 81+ 93+72+ 35+3+1}{12}$
$=39.5 $
Mean Marks $= 39.5$
View full question & answer→