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Maths 3 Marks Question

Collection and Organisation of Data · CBSE STD 7 (CBSE - English Medium). 12 questions.

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3 Marks Question

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12 questions · timed · auto-graded

Question 13 Marks
Find the median of first $50$ whole numbers.
Answer
First 50 whole numbers $0, 1, 2, 3, 4, ....., 49$
 Here, number of terms $= 50, $
which is even $\therefore\ \text{Median}=\frac{1}{2}\Big\{\frac{\text{n}}{2}\text{th term}+\Big(\frac{\text{n}}{2}+1\Big)\text{th term}\Big\}$
$=\frac{1}{2}\Big\{\frac{50}{2}\text{th term}+\Big(\frac{50}{2}+1\Big)\text{th term}\Big\}$
$=\frac{1}{2}\{25\text{th}+26\text{th term}\}$
$=\frac{1}{2}(24+25)=\frac{1}{2}\times49=24.5$
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Question 23 Marks
The ages (in years) of $10$ teachers in a school are, $34, 37, 53, 46, 52, 43, 31, 36, 40, 50$. Find the median age.
Answer
Arranging is ascending order, $31, 34, 36, 37, 40, 43, 46, 50, 52, 53$
 Here, number of terms $= 10$,
which is even $\therefore\ \text{Median}=\frac{1}{2}\Big\{\frac{\text{n}}{2}\text{th term}+\Big(\frac{\text{n}}{2}+1\Big)\text{th term}\Big\}$
$=\frac{1}{2}\Big\{\frac{10}{2}\text{th term}+\Big(\frac{10}{2}+1\Big)\text{th term}\Big\}$
$=\frac{1}{2}\{5\text{th}+6\text{th term}\}$
$=\frac{1}{2}(40+43)=\frac{1}{2}\times83=41.5\ \text{years}$
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Question 33 Marks
The daily wages (in rupees) of $60$ workers in a factory are given below:
Daily wages (in Rs.) $280$ $300$ $320$ $360$ $380$
Number of workers $14$ $16$ $15$ $7$ $8$
Find the mean daily wages.
Answer
Daily wages (in Rs.) $x_i$ Number of workers $f_i$ $x_i \times f_i$
$280$ $14$ $3920$
$300$ $16$ $4800$
$320$ $15$ $4800$
$360$ $7$ $2520$
$380$ $8$ $3040$
Total $60$ $19080$
$\text{Mean}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{19080}{60}=318$
$\text{Mean}=\text{Rs}.318$
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Question 43 Marks
Find the median of: $10, 32, 17, 19, 21, 22, 9, 35$
Answer
Arranging in ascending order, $9, 10, 17, 19, 21, 22, 32, 35$
 Here, number of terms $= 8$
which is even $\therefore\ \text{Median}=\frac{1}{2}\Big\{\frac{\text{n}}{2}\text{th term}+\Big(\frac{\text{n}}{2}+1\Big)\text{th term}\Big\}$
$=\frac{1}{2}\Big\{\frac{8}{2}\text{th term}+\Big(\frac{8}{2}+1\Big)\text{th term}\Big\}$
$=\frac{1}{2}(4\text{th term}+5\text{th term})=\frac{1}{2}(19+21)$
$=\frac{1}{2}\times40=20$
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Question 53 Marks
The heights (in cm) of $40$ boys were measured and recorded as under:
Height (in cm) $165$ $170$ $175$ $180$
Number of boys $9$ $8$ $11$ $12$
Find the mean height.
Answer
Height (in cm) $x_i$ Number of boys $f_i$ $f_i \times x_i$
$165$ $9$ $1485$
$170$ $8$ $1360$
$175$ $11$ $1925$
$180$ $12$ $2160$
Total $40$ $6930$
$\text{Mean}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{6930}{40}=173.25\text{cm}$
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Question 63 Marks
Calculate the median for the following data:
Marks $17$ $20$ $22$ $15$ $30$ $25$
Number of students $5$ $9$ $4$ $3$ $10$ $6$
Answer
Arranging in order and preparing the cumulative frequency table.
Marks x Number of students f c.f
$15$ $3$ $3$
$17$ $5$ $8$
$20$ $9$ $17$
$22$ $4$ $21$
$25$ $6$ $27$
$30$ $10$ $37$
Here, number of terms $(N) = 37$ which is odd
$\text{Median}=\frac{\text{n}+1}{2}\text{th term}$
$=\frac{37+1}{2}=\frac{38}{2}\text{th term}$
$= 19th$ term $= 22$ (Value of $18$ to $21 = 22)$
Hence median $= 22$
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Question 73 Marks
Find the median of first $10$ even numbers.
Answer
First $10$ even numbers are $2, 4, 6, 8, 10, 12, 14, 16, 18, 20$
Here, number of terms $= 10$
which is even $\therefore\ \text{Median}=\frac{1}{2}\Big\{\frac{\text{n}}{2}\text{th term}+\Big(\frac{\text{n}}{2}+1\Big)\text{th term}\Big\}$
$=\frac{1}{2}\Big\{\frac{10}{2}\text{th term}+\Big(\frac{\text{10}}{2}+1\Big)\text{th term}\Big\}$
$=\frac{1}{2}\{5\text{th term}+6\text{th term}\}$
$=\frac{1}{2}(10+12)=\frac{1}{2}\times22=11$
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Question 83 Marks
The heights (in cm) of $90$ plants in a garden are given below:
Height (in cm) $58$ $60$ $62$ $64$ $66$ $74$
Number or plants $20$ $25$ $15$ $8$ $12$ $10$
Find the mean height.
Answer
Height (in cm) ($x_i$) Number of plants ($f_i$) $f_i \times x_i$
$58$ $20$ $1160$
$60$ $25$ $1500$
$62$ $15$ $930$
$64$ $8$ $512$
$66$ $12$ $792$
$74$ $10$ $740$
  $\sum\text{f}_\text{i}=90 $ $\sum(\text{f}_\text{i}\times\text{x}_\text{i})=5634$
$\text{Mean height}=\frac{\sum(\text{f}_\text{i}\times\text{x}_\text{i})}{\sum\text{f}_1}=\frac{5634}{90}=62.6\text{cm}$
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Question 93 Marks
Find the median of: $55, 60, 35, 51, 29, 63, 72, 91, 85, 82$
Answer
Arranging in ascending order,$29, 35, 51, 55, 60, 63, 72, 82, 85, 97$
Here, number of terms $= 10$
which is even $\therefore\ \text{Median}=\frac{1}{2}\Big\{\frac{\text{n}}{2}\text{th term}+\Big(\frac{\text{n}}{2}+1\Big)\text{th term}\Big\}$
$=\frac{1}{2}\Big\{\frac{10}{2}\text{th term}+\Big(\frac{10}{2}+1\Big)\text{th term}\Big\}$
$=\frac{1}{2}(5\text{th term}+6\text{th term})$
$\frac{1}{2}(60+63)=\frac{1}{2}\times123$
$=61.5$
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Question 103 Marks
The ages (in years) of $50$ players of a school are given below:
Age (in years) $14$ $15$ $16$ $17$ $18$
Number of players $15$ $14$ $10$ $8$ $3$
Find the mean age.
Answer
Age (24 years) $x_i$ Number of players $f_i$ $x_i \times f_i$
$14$ $15$ $210$
$15$ $14$ $210$
$16$ $10$ $160$
$17$ $8$ $136$
$18$ $3$ $54$
Total $50$ $770$
$\text{Mean}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{770}{50}=15.4\ \text{years}$
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Question 113 Marks
Find the median weight for the following data:
Weight (in kg) $45$ $46$ $48$ $50$ $52$ $54$ $55$
Number of boys $8$ $5$ $6$ $9$ $7$ $4$ $2$
Answer
Preparing the cumulative frequency table
Weight (in kg) x Number of boys (f) c.f
$45$ $8$ $8$
$46$ $5$ $13$
$48$ $6$ $19$
$50$ $9$ $28$
$52$ $7$ $35$
$54$ $4$ $39$
$55$ $2$ $41$
Here, number of terms $(N) = 41,$
which is odd $\text{Median}=\frac{\text{n}+1}{2}\text{th term}$
$=\frac{41+1}{2}=\frac{42}{2}\text{th}=21\text{th term}$
$50kg$ (value of $20$ to $28 = 50)$
Hence median $= 50kg$
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Question 123 Marks
The following table shows the weights (in kg) of $15$ workers in a factory.
Weight (in kg) $60$ $63$ $66$ $72$ $75$
Number of workers $4$ $5$ $3$ $1$ $2$
Calculate the mean weight.
Answer
Weight (intag) $x_i$ Number of workers $f_i$ $x_i \times f_i$
$60$ $4$ $240$
$63$ $5$ $315$
$66$ $3$ $198$
$72$ $1$ $72$
$75$ $2$ $150$
Total $15$ $975$
$\text{Mean}=\frac{\sum\text{f}_1\text{x}_1}{\sum\text{f}_1}=\frac{975}{15}=65\text{kg}$
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