## RD Sharma Solutions Class 7 Chapter 23 Exercise 23.1

**Exercise 23.1**

*Â **Q1) Ashish studies for 4 hours, 5 hours and 3 hours on three consecutive days. How many hours does he study daily on an average? *

**Solution:**

Average number of study hours = (4 + 5 + 3) \(\div\)

= 12 \(\div\)

= 4 hours

Thus, Ashish studies for 4 hours on an average.

*Q2) A cricketer scores the following runs in 8 innings: 58, 76, 40, 35, 48, 45, 0, 100. *

*Find the mean score. *

**Solution:**

We have:

The mean score = (58 + 76 + 40 + 35 + 48 + 45 + 0 + 100) \(\div\)

= 402 \(\div\)

= 50.25 runs.

*Q3) The marks (out of 100) obtained by a group of students in science test are 85, 76, 90, 84, 39, 48, 56, 95, 81 and 75. Find the *

*(i) highest and the lowest marks obtained by the students. *

*(ii) range of marks obtained. *

*(iii) mean marks obtained by the group. *

**Solution:**

In order to find the highest and lowest marks, let us arrange the marks in ascending order as follows:

39, 48, 56, 75, 76, 81, 84, 85, 90, 95

(i) Clearly, the highest mark is 95 and the lowest is 39.

(ii) The range of the marks obtained is: (95 â€“ 39) = 56.

(iii) We have:

Mean marks = Sum of the marks \(\div\)

=> Mean marks = (39 + 48 + 56 + 75 + 76 + 81 + 84 + 85 + 90 + 95) \(\div\)

= 729 \(\div\)

= 72.9.

Hence, the mean mark of the students is 72.9.

*Q4) The enrolment of a school during six consecutive years was as follows: *

*1555, 1670, 1750, 2019, 2540, 2820 *

*Find the mean enrollment of the school for this period. *

**Solution:**

The mean enrolment = Sum of the enrolments in each year \(\div\)

The mean enrolment = (1555 + 1670 + 1750 + 2019 + 2540 + 2820) \(\div\)

= 12354 \(\div\)

= 2059.

Thus, the mean enrolment of the school for the given period is 2059.

*Â **Q5) The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows: *

*Day Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â MonÂ Â Â Tue Â Â Â Wed Â Â Â Thu Â Â Â Fri Â Â Â Â Â Sat Â Â Â Â Â Sun *

*Rainfall (in mm) Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.0 Â Â Â Â Â 12.2 Â Â Â 2.1 Â Â Â Â Â 0.0 Â Â Â Â Â 20.5 Â Â Â 5.3 Â Â Â Â Â 1.0 *

*(i) Find the range of the rainfall from the above data. *

*(ii) Find the mean rainfall for the week. *

*(iii) On how many days was the rainfall less than the mean rainfall. *

**Solution:**

(i) The range of the rainfall = Maximum rainfall â€“ Minimum rainfall

= 20.5 â€“ 0.0

= 20.5 mm.

(ii) The mean rainfall Â = (0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.3 + 1.0) \(\div\)

= 41.1 \(\div\)

= 5.87 mm.

(iii) Clearly, there are 5 days (Mon, Wed, Thu, Sat and Sun), when the rainfall was less than the mean, i.e., 5.87 mm.

*Q6) If the heights of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm respectively, find the mean height.*

**Solution:**

The mean height = Sum of the heights \(\div\)

= (140 + 150 + 152 + 158 + 161) \(\div\)

= 761 \(\div\)

= 152.2 cm.

*Q7) Find the mean of 994, 996, 998, 1002 and 1000. *

**Solution:**

Mean = Sum of the observations \(\div\)

Mean = (994 + 996 + 998 + 1002 + 1000) \(\div\)

= 4990 \(\div\)

= 998.

*Q8) Find the mean of first five natural numbers. *

**Solution:**

The first five natural numbers are 1, 2, 3, 4 and 5.

Let \(\overline{X}\)

\(\overline{X}\)

= 15 \(\div\)

= 3.

*Q9) Find the mean of all factors of 10.*

**Solution:**

The factors of 10 are 1, 2, 5 and 10.

Let \(\overline{X}\)

\(\overline{X}\)

= 18 \(\div\)

= 4.5.

*Q10) Find the mean of first 10 even natural numbers. *

**Solution:**

The first 10 even natural numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20.

Let \(\overline{X}\)

\(\overline{X}\)

= 110 \(\div\)

= 11.

*Q11) Find the mean of x, x + 2, x + 4, x + 6, x + 8 *

**Solution:**

Mean = Sum of observations \(\div\)

=> Mean = (x + x + 2 + x + 4 + x + 6 + x + 8) \(\div\)

=> Mean = (5x + 20) \(\div\)

=> Mean = \(\frac{5(x+4)}{5}\)

=> Mean = x + 4

*Q12) Find the mean of first five multiples of 3. *

**Solution:**

The first five multiples of 3 are 3, 6, 9, 12 and 15.

Let \(\overline{X}\)

\(\overline{X}\)

= 45 \(\div\)

= 9.

*Â *

*Q13) Following are the weights (in kg) 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 Find the mean \(\overline{X}\). *

**Solution:**

We have:

\(\overline{X}=\frac{Sum\;of\;observations}{Number\;of\;observations}\)

=> \(\overline{X}=\frac{3.4+3.6+4.2+4.5+3.9+4.1+3.8+4.5+4.4+3.6}{10}\)

=> \(\overline{X}=\frac{40}{10}\)

=> \(\overline{X}\)

*Q14) The percentage of marks obtained by students of a class in mathematics are: *

*64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1 Find their mean. *

**Solution:**

We have:

Mean = \(\frac{Sum\;of\;the\;marks\;obtained}{Total\;number\;of\;students}\)

=> Mean = \(\frac{64+36+47+23+0+19+81+93+72+35+3+1}{12}\)

=> Mean = \(\frac{474}{12}\)

*Q15) The numbers of children in 10 families of a locality are: *

*2, 4, 3, 4, 2, 3, 5, 1, 1, 5 Find the mean number of children per family. *

**Solution:**

The mean number of children per family = \(\frac{Sum\;of\;the\;total\;number\;of\;children}{Total\;number\;of\;families}\)

Mean = \(\frac{2+4+3+4+2+3+5+1+1+5}{10}\)

= \(\frac{30}{10}\)

= 3.

Thus, on an average there are 3 children per family in the locality.

*Q16) 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. *

**Solution:**

We have:

n = The number of observations = 100, Mean = 40

Mean = \(\frac{Sum\;of\;the\;observations}{Total\;number\;of\;observations}\)

=> 40 = \(\frac{Sum\;of\;the\;observations}{100}\)

=> Sum of the observations = 40 x 100

Thus, the incorrect sum of the observations = 40 x 100 = 4000.

Now,

The correct sum of the observations = Incorrect sum of the observations – Incorrect observation +

Correct observation

=> The correct sum of the observations = 4000 – 83 + 53

=> The correct sum of the observations = 4000 – 30 = 3970

\(âˆ´ Correct\;mean=\frac{Correct\;sum\;of\;the\;observations}{Number\;of\;observations}=\frac{3970}{100}=39.7\)

*Q17) The mean of five numbers is 27. If one number is excluded, their mean is 25. Find the excluded number. *

**Solution:**

We have:

Mean = \(\frac{sum\;of\;the\;five\;numbers}{5}=27\)

So, sum of the five numbers = 5 x 27 = 135.

Now,

The mean of four numbers = \(\frac{sum\;of\;the\;four\;numbers}{4}=25\)

So, sum of the four numbers = 4 x 25 = 100.

Therefore, the excluded number = Sum of the five number â€“ Sum of the four numbers

=> The excluded number = 135 â€“ 100 = 35.

*Q18) The mean weight per student in a group of 7 students is 55 kg. The individual weights of 6 of them (in kg) are 52, 54, 55, 53, 56 and 54. Find the weight of the seventh student. *

**Solution:**

We have:

Mean = \(\frac{sum\;of\;the\;weights\;of\;the\;students}{Number\;of\;students}\)

Let the weight of the seventh student be x kg.

Mean = \(\frac{52+54+55+53+56+54+x}{7}\)

55 = \(\frac{52+54+55+53+56+54+x}{7}\)

=> 385 = 324 + x

=> x = 385 â€“ 324

=> x = 61 kg.

Thus, the weight of the seventh student is 61 kg.

*Q19) The mean weight of 8 numbers is 15 kg. If each number is multiplied by 2, what will be the new mean? *

**Solution:**

Let \(x_{1},x_{2},x_{3}…x_{8}\)

15 = \(\frac{x_{1}+x_{2}+x_{3}+…+x_{8}}{8}\)

\(x_{1}+x_{2}+x_{3}+…+x_{8}=15\times 8\)

=>\(x_{1}+x_{2}+x_{3}+…+x_{8}=120.\)

Let the new numbers be \(2x_{1},2x_{2},2x_{3}…2x_{8}\)

Then,

M = \(\frac{2x_{1}+2x_{2}+2x_{3}+…+2x_{8}}{8}\)

=> M = \(\frac{2(x_{1}+x_{2}+x_{3}+…+x_{8})}{8}\)

=> M = \(\frac{2\times 120}{8}\)

=> M = 30

*Q20) The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number. *

**Solution:**

Let \(x_{1},x_{2},x_{3},x_{4}\;Â and\; x_{5}\)

18 = Sum of five numbers \(\div\)

âˆ´ Sum of five numbers = 18 x 5 = 90

Now, if one number is excluded, then their mean is 16.

So,

16 = Sum of four numbers \(\div\)

âˆ´ Sum of four numbers = 16 x 4 = 64.

The excluded number = Sum of five observations â€“ Sum of four observations

âˆ´ The excluded number = 90 â€“ 64

âˆ´ The excluded number = 26.

*Q21) The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88. Find the correct mean.*

**Solution:**

n = Number of observations = 200

Mean = \(\frac{Sum\;of\;the\;observations}{Number\;of\;observations}\)

=> 50 = \(\frac{Sum\;of\;the\;observations}{200}\)

=> Sum of the observations = 50 x 200 = 10,000.

Thus, the incorrect sum of the observations = 50 x 200

Now,

The correct sum of the observations = Incorrect sum of the observations â€“ Incorrect observations + Correct observations

=> Correct sum of the observations = 10,000 â€“ (92 + 8) + (192 + 88)

=> Correct sum of the observations = 10,000 â€“ 100 + 280

=> Correct sum of the observations = 9900 + 280

=> Correct sum of the observations = 10,180.

âˆ´ Correct Mean = \(\frac{Correct\;sum\;of\;the\;observations}{Number\;of\;observations}=\frac{10180}{200}\)

*Q22) The mean of 5 numbers is 27. If one more number is included, then the mean is 25. Find the included number.*

**Solution:**

We have:

Mean = Sum of five numbers \(\div\)

=> Sum of the five numbers = 27 x 5 = 135.

Now, New mean = 25

25 = Sum of six numbers \(\div\)

=> Sum of the six numbers = 25 x 6 = 150.

The included number = Sum of the six numbers – Sum of the five numbers

=> The included number = 150 â€“ 135

=> The included number = 15.

*Q23) The mean of 75 numbers is 35. If each number is multiplied by 4, find the new mean.*

**Solution:**

Let \(x_{1},x_{2},x_{3}…x_{75}\)

=> 35 = \(\frac{x_{1}+x_{2}+x_{3}+…+x_{75}}{75}\)

\(x_{1}+x_{2}+x_{3}+…+x_{75}=35\times 75\)

=>\(x_{1}+x_{2}+x_{3}+…+x_{75}=2625\)

The new numbers are \(4x_{1},4x_{2},4x_{3}…4x_{75}\)

M = \(\frac{4x_{1}+4x_{2}+4x_{3}+…+4x_{75}}{75}\)

=> M = \(\frac{4(x_{1}+x_{2}+x_{3}+…+x_{75})}{75}\)

=> M = \(\frac{4\times 2625}{75}\)

=> M = 140.