2 Mark Question

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	12	13	15	18	25
Frequency:	5	1	3	1	1	1	1

Since, maximum frequency 5 corresponds to value 7 then mode = 7

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Question 22 Marks

Find the mean of first five multiples of 3

Answer

First five multiples of 3 are 3, 6, 9, 12, 15$\therefore\text{Mean}=\frac{\text{Sum of numbers}}{\text{Total numbers}}$
$=\frac{3+6+9+12+15}{5}$
$=\frac{45}{5}$
$=9$
$\text{Mean} = 9$

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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

Marks:	4	5	6	7	8	9	10
No. of students:	2	2	2	4	2	2	1

Since, the maximum frequency corresponds to the value 7 then mode = 7 marks.

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Question 42 Marks

Find the mean of x, x + 2, x + 4, x + 6, x + 8

Answer

Numbers are x, x + 2, x + 4, x + 6, x + 8$\therefore\text{Mean}=\frac{\text{Sum of numbers}}{\text{Total numbers}}$
$=\frac{\text{x}+\text{x}+2+\text{x}+4+\text{x}+6+\text{x}+8}{5}$
$=\frac{\text{5x}+20}{5}$
$=5\Big(\frac{\text{x}+4}{5}\Big)$
$=\text{x}+4$

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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

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Question 62 Marks

Find the mean of first five natural numbers.

Answer

The first five odd numbers are 1, 2, 3, 4, 5.$\therefore\text{Mean}=\frac{\text{Sum of numbers}}{\text{Total numbers}}$
$=\frac{1+2+3+4+5}{5}$
$=\frac{15}{5}=3$
$\text{Mean} = 3$

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Question 72 Marks

The mean weight of 8 numbers is 15. If each number is multiplied by 2, what will be the new mean?

Answer

We have, The mean weight of 8 numbers is 15 Then, the sum of 8 numbers = 8 × 15 = 120 If each number is multiplied by 2 Then, new mean = 120 × 2 = 240$\therefore$ new mean $=\frac{240}{8}=30$

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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.

Answer

The mean of 200 items = 50 Then the sum of 200 items = 200 × 50 = 10,000 Correct values = 192 and 88. Incorrect values = 92 and 8.$\therefore$ correct sum = 10000 - 92 - 8 + 192 + 88 = 10180
$\therefore\text{correct mean}=\frac{10180}{200}$ $=\frac{101.8}{2}=50.9$

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Question 92 Marks

Find the mean of first ten even natural numbers.

Answer

The first five even natural numbers are 2, 4, 6, 8, 10, 12, 14, 16,18, 20.$\therefore\text{Mean}=\frac{\text{Sum of numbers}}{\text{Total numbers}}$
$=\frac{2+4+6+8+10+12+14+16+18+20}{10}$
$=11$
$\text{Mean} = 11$

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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.

Answer

Mean marks of 100 students = 40 Sum of marks of 100 students = 100 × 40 = 4000 Correct value = 53 Incorrect value = 83 Correct sum = 4000 - 83 + 53 = 3970$\therefore$ correct mean $=\frac{3970}{100}=39.7$

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Question 112 Marks

The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.

Answer

The mean of 5 numbers is 18 Then, the sum of 5 numbers = 5 × 18 = 90 If one number is excluded Then, the mean of 4 numbers = 16$\therefore$ sum of 4 numbers = 4 × 16 = 64
Excluded number = 90 - 64 = 26.

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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

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Question 132 Marks

If the heights of 5 persons are 140cm, 150cm, 152cm, 158cm and 161cm respectively. Find the mean height.

Answer

Given: The heights of 5 persons are 140cm, 150cm, 152cm, 158cm and 161cm.$\therefore\text{Mean Weight}=\frac{\text{Sum of height}}{\text{Total no. of persons}}$
$=\frac{140+150+152+158+161}{5}$
$=\frac{761}{5}=152.2$
$\text{Mean} = 152.2$

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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

Answer

Arranging 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.

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Question 152 Marks

Find the mean of 994, 996, 998, 1000, 1002.

Answer

Numbers are 994, 996, 998, 1000, 1002.$\therefore\text{Mean}=\frac{\text{Sum of numbers}}{\text{Total numbers}}$
$=\frac{994+996+998+1000+1002}{5}$
$=\frac{4990}{5}=998$
$\text{Mean} = 998$

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Question 162 Marks

The demand of different shirt sizes, as obtained by a survey, is given below:

Size:	38	39	40	41	42	43	44	Total
Number of persons (wearing it):	26	39	20	15	13	7	5	125

Find the modal shirt sizes, as observed from the survey.

Answer

Size:	38	39	40	41	42	43	44	Total
Number of persons:	26	39	20	15	13	7	5	125

Since, maximum frequency 39 corresponds to value 39 then model size = 39

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Question 172 Marks

Find the mean of all factors of 10.

Answer

All factors of 6 are 1, 2, 5, 10.$\therefore\text{Mean Weight}=\frac{\text{Sum of factors}}{\text{Total factors}}$
$=\frac{1+2+5+10}{4}$
$=4.5$
$\text{Mean} = 4.5$

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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.

Answer

The numbers of children in 10 families are : 2, 4, 3, 4, 2, 3, 5, 1, 1, 5$\therefore\text{Mean}=\frac{\text{Total no. children}}{\text{Total families}}$
$=\frac{2+ 4+ 3+ 4+ 2+ 3+ 5+ 1+1+ 5}{10}$
$=3$

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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 $\overline{\text{X}}.$

Answer

The 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$\therefore\text{Mean Weight}=\frac{\text{Sum of weights}}{\text{Total no. of babies}}$
$=\frac{3.4+ 3.6+ 4.2+ 4.5+ 3.9+ 4.1+ 3.8+ 4.5+4.4+ 3.6}{10}$
$=4\text{kg}$

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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.

Answer

The 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.
$\therefore\text{Correct Mean}=\frac{52+58+54+65+44+47+60+62+57+53}{10}$
$=\frac{552}{10}=55.2\ \text{km/hr}$

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Question 212 Marks

The mean of five numbers is 27. If one number is excluded, their mean is 25. Find the excluded number.

Answer

The mean of five numbers is 27 The sum of five numbers = 5 × 27 = 135 If one number is excluded, the new mean is 25$\therefore$ Sum of 4 numbers = 4 × 25 = 100
$\therefore$ Excluded number = 135 - 100 = 35

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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.

Answer

The 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

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