CBSE Class 7

These NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Exercise 3.3

Question 1.
Use the bar graph to answer the following questions.
Data Handling
(a) Which is the most popular pet?
(b) How many students have dog as a pet?

Answer:
(a) Since the bar representing 10 is the longest, the most popular pet is cat.
(b) 8 students have dog as a pet.

Question 2.
Read the bar graph which shows the number of books sold by a book store
during five consecutive years and answer the following questions:

(i) About how many books were sold in 1989? 1990? 1992?
(ii) In which year were about 475 books sold? About 225 books sold?
(iii) In which years were fewer than 250 books sold?
(iv) Can you explain how you would estimate the number of books sold in 1989?
Answer:
(i) Number of books sold:
In 1989 were about 180
In 1990 were about 475
In 1992 were about 225
(ii) About 475 books were sold in 1990 and about 225 books were sold in 1992.
(iii) Fewer than 250 books were sold in 1989 and 1992.
(iv) The height of the bar for 1989 is slightly less than that for 200 books.
Therefore, about 180 books were sold in 1989.

Question 3.
Number of children in six different classes are given below. Represent the data on a bar graph.

(a) How would you choose a scale?
(b) Answer the following questions:
(i) Which class has the maximum number of children? And the minimum?

(ii) Find the ratio of students of class sixth to the students of class eighth.
Answer:
(a) Start the scale at 0. The greatest value in the data is 135, so end the scale at a
value greater than 135 such as 140.
Use equal divisions along the axes such as increment of 10.
(i) We know that all the bars would lie between 0 and 140.
(ii) We choose the scale such that the length between 0 and 140 is neither too long nor too small.
(iii) Here we take one unit for 10 children.

(b) (i) Fifth class has the maximum number of children. Tenth class has the minimum number of
children.
(ii) Ratio of students of class sixth to
Eighth = 120 : 100 = 12 : 10 = 6 : 5

Question 4.
The performance of a student in 1^st Term and 2^nd Term is given. Draw a double bar graph choosing appropriate scale and answer the following:

(i) In which subject, has the child improved his performance the most?
(ii) In which subject is the improvement the least?
(iii) Has the performance gone down in any subject?
Answer:
The required double bar graph is given below:

This graph shows the performance of students in ^st term and 2^nd term.
(i) The child improved his performance the most in Maths.
(ii) The improvement is the least in the subject of social science.
(iii) Yes, the performance has gone down in the subject of Hindi.

Question 5.
Consider this data collected from a survey of a colony.

(i) Draw a double bar graph choosing an appropriate scale.
What do you infer from the bar graph?
(ii) Which sport is most popular?
(iii) Which is more preferred, watching or participating in sports?
Answer:
(i) The required ‘double bar graph’ is given below :
From the bar graph it is inferred that the people of the colony like cricket the most and athletics the least.
(ii) The most popular sport is cricket.
(iii) Watching sports is more preferred than participating in sports.

Question 6.
Take the data giving the minimum and the maximum temperature of various cities given in the beginning of this Chapter. Plot a double bar graph using the data and answer the following:
(i) Which city has the largest difference in the minimum and maximum temperature on the given date?
(ii) Which is the hottest city and which is the coldest city?
(iii) Name two cities where maximum temperature of one was less than the minimum temperature of the other.
(iv) Name the city which has the least difference between its minimum and the maximum temperature.
Temperature of cities as on 20.06.2006

Answer:
Plotting a double bar graph, we get.
City Difference in the minimum and maximum temperature on 20-06-2006

(i) Jammu has the largest difference in the minimum and maximum temperatures on the given data.
(ii) Jammu is the hottest city and Bengaluru is the coldest city.
(iii) Bangaluru and Jaipur or Bengaluru and Ahmedabad.
(iv) Mumbai has the least difference between its maximum temperature and minimum temperature.

These NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Exercise 3.2

Question 1.
The scores in mathematics test (out of 25) of 15 students are as follows:
19, 25, 23, 20, 9, 20,15, 10, 5, 16, 25, 20, 24, 12, 20
Find the mode and median of this data. Are they same?
Answer:
Arrange the given data in ascending order, we get, 5, 9, 10,12, 15,16, 19, 20, 20, 20, 20, 23, 24, 25, 25
(a) Mode of the data = 20
20 is repeated 4 times

(b) Median of the data = 20
Since the middle value of the data is 20
Obviously, here the mode and median are the same.

Question 2.
The runs scored in a cricket match by 11 players are as follows:
6,15,120, 50, 100, 80, 10,15, 8,10, 15
Find the mean, mode and median of this data. Are the three same?
Answer:
Arrange the given data in ascending order, we get, 6, 8, 10, 10, 15, 15, 15, 50, 80, 100, 120
(i) Mean of the data

(ii) Since 15 is repeated 3 times, the mode of the data = 15
(iii) The value of the middle observation is 15. The Median of the data =15.
Here mean, median and mode are not the same.

Question 3.
The weights (in kg) of 15 students of a class are:
38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38,43, 38, 47
(i) Find the mode and median of this data.
(ii) Is there more than one mode?
Answer:
Arrange the given data in ascending order, we get,
32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50
(i) Since 38 and 43 are repeated 3 times, the mode of the data = 38 and 43
The value of the middle observation is 40.
∴ Median of the data = 40

(ii) Yes, there are more than one mode.

Question 4.
Find the mode and median of the data: 13,16, 12,14, 19,12,14, 13,14
Answer:
Arrange the given data in ascending order, we get, 12, 12, 13, 13, 14, 14, 14, 16, 19
Since 14 is repeated 3 times, the mode of the data = 14
Again the value of the middle observation is 14, median of the data = 14

Question 5.
Tell whether the following statements is true or false:
(i) The mode is always one of the numbers in a data.
(ii) The mean is one of the numbers in a data.
(iii) The median is always one of the numbers in a data.
(iv) The data 6, 4, 3, 8, 9, 12, 13, 9 has mean 9.
Answer:
(i) True
(ii) False
(iii) True
(iv) False

These NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Exercise 3.1

Question 1.
Find the range of heights of any ten students of your class.
Answer:
Do it yourself, one simple solution may be to consider the heights (in cm) of ten students as follows:
111, 118, 110, 120, 114, 113, 118, 117, 115, 119
Average height

= 115.5 cm

Question 2.
Organise the following marks in a class assessment in a tabular form.
4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6,7
(i) Which number is the highest?
(ii) Which number is the lowest?
(iii) What is the range of the data?
(iv) Find the arithmetic mean.
Answer:

(i) The highest number = 9
(ii) The lowest number = 1
(iii) The range = Highest number – Lowest number = 9 – 1 = 8
(iv) Arithmetic Mean
= \(\frac{\text { Sum of the marks }}{\text { Number of students }}\)

= \(\frac{100}{20}\) = 5
Thus, the mean marks = 5.

Question 3.
Find the mean of the first five whole numbers.
Answer:
First five whole numbers are 0, 1, 2, 3 and 4
Sum 0 + 1 + 2 + 3 + 4 = 10
Mean = \(\frac{\text { Sum of the number }}{\text { Number of whole numbers }}\)
= \(\frac{10}{5}\) = 2

Question 4.
A cricketer scores the following runs in eight innings.
58, 76, 40, 35, 46, 45, 0, 100
Find the mean score.
Answer:
Mean score = \(\frac{\text { Sum of the scores }}{\text { Number of innings }}\)
= \(\frac{58+76+40+35+46+45+0+100}{8}\)
= \(\frac{400}{8}\) = 50
Mean score = 50

Question 5.
Following table shows the points of each player scored in four games:

Now answer the following questions:
(i) Find the mean to determine A’s average number of points scored per game.
(ii) To find the mean number of points per game for C, would you divide the total points by 3 or by 4?Why?
(iii) B played in all the four games. How would you find the mean?
(iv) Who is the best performer?
Answer:
(i) Mean = \(\frac{\text { Sum of the observations }}{\text { Number of observations }}\)
= \(\frac{14+16+10+10}{4}\) = \(\frac{50}{4}\) = 12.5
A’s average score per game is 12.5

(ii) Since C played only 3 games (He did not play the 3rd game)
Total will be divided by 3

(iii) Mean score of B
= \(\frac{\text { Sum of the all observations }}{\text { Number of observations }}\)
= \(\frac{0+8+6+4}{4}\) = \(\frac{18}{4}\) = 4.5
Thus, th average’tnumber of points scored by is B is 4.5

(iv) Mean score of C
= \(\frac{\text { Sum of the all observations }}{\text { Number of observations }}\)
= \(\frac{8+11+13}{4}\) = \(\frac{32}{3}\) = 10.67
Thus, mean number of points scored by C is 10.67
A’s average score is = 12.5
B’s average score is = 4.5
C’s average score is = 10.67
The best performer is A.

Question 6.
The marks (out of 100) obtained by a group of students in a science test are 85, 76, 90, 85, 39,48, 56, 95, 81, and 75. Find the:
(i) Highest and the lowest marks obtained by the students.
(ii) Range of the marks obtained.
(iii) Mean marks obtained by the group.
Answer:
Write the given marks in ascending order, we get 39, 48, 56, 75, 76, 81, 85, 85, 90, 95
(i) Highest marks = 95, Lowest marks = 39

(ii) Range = Highest marks – Lowest marks = 95 – 39 = 56

(iii) Mean = \(\frac{\text { Sum of all the marks }}{\text { Number of students }}\)

= \(\frac{730}{10}\) = 73
Mean marks obtained by the group is 73.

Question 7.
The enrolment in a school during six consecutive years was as follows:
1555,1670,1750,2013,2540,2820 Find the mean enrolment of the school for this period.
Answer:

∴ The mean enrolment is 2058 per year.

Question 8.
The rainfall (in mm) in a city on 7days of a certain week was recorded as follows:

(i) Find the range of the rainfall in the above data.
(ii) Find the mean rainfall for the week.
(iii) On how many days was the rainfall less than the mean rainfall.
Answer:
(i) Range = Highest rainfall – Lowest rainfall = 20.5 – 0.0 = 20.5 mm.

= \(\frac{41.3}{7}\) = 5.9mm
Mean rainfall = 5.9 mm

(iii) The rainfall was less than the mean rainfall on 5 days.

Question 9.
The heights of 10 girls were measured in cm and the results are as follows:
135, 150, 139, 128, 151, 132, 146, 149, 143,141.
(i) What is the height of the tallest girl?
(ii) What is the height of the shortest girl?
(iii) What is the range of the data?
(iv) What is the mean height of the girls?
(v) How many girls have heights more than the mean height.
Answer:
Arrange the heights of girls in ascending order, we get
128, 132, 135, 139, 141, 143, 146, 149, 150, 151
(i) Height of the tallest girl = 151cm
(ii) Height of the shortest girl = 128cm
(iii) Range= Highest height – Lowest height = 151cm – 128cm = 23cm

The mean height of the girls is 141.4cm
(v) The heights of 5 girls are more than the mean height.

These NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals InText Questions

NCERT In-text Question Page No. 34
Question 1.
Find:
(a) \(\frac { 1 }{ 2 }\) × 3
(b) \(\frac { 9 }{ 7 }\) × 6
(c) 3 × \(\frac { 1 }{ 8 }\)
(d) \(\frac { 13 }{ 11 }\) × 6
if the product is an improper fraction express it as a mixed fraction
Answer:

Question 2.
Represent pictorially: 2 × \(\frac{2}{5}=\frac{4}{5}\)
Answer:

NCERT In-text Question Page No. 34
Question 1.
(i) 5 × 2 \(\frac{3}{7}\)
(ii) 1 \(\frac{4}{9}\) × 6
Answer:

NCERT In-text Question Page No. 35
Question 1.
Can you tell, what is (i) \(\frac{1}{2}\) of 10?
(i) \(\frac{1}{4}\) of 16? (iii) (i) \(\frac{2}{5}\) of 25?
Answer:

NCERT In-text Question Page No. 39
Question 1.
Fill in these boxes:

NCERT In-text Question Page No. 40
Question 1.
Find:
\(\frac{1}{3} \times \frac{4}{5} ; \frac{2}{3} \times \frac{1}{5}\)
Answer:

Question 2.
Find:
\(\frac{8}{3} \times \frac{4}{7} \times \frac{3}{4} \times \frac{2}{3}\)
Answer:

NCERT In-text Question Page No. 44
Question 1.
(i) Will the reciprocal of a proper fraction be again a proper fraction?
(ii) Will the reciprocal of an improper fraction be again an improper fraction?
Answer:
(i) No, the reciprocal of an improper fraction is a improper fraction.
(ii) No, the reciprocal of an improper fraction is a proper fraction.
Now, we can say that
(a) 1 ÷ \(\frac { 1 }{ 2 }\) = 1 × \(\frac { 2 }{ 1 }\) = 1 × reciprocal of \(\frac { 1 }{ 2 }\)

(b) 3 ÷ \(\frac { 1 }{ 4 }\) = 3 × \(\frac { 4 }{ 1 }\) = 3 × reciprocal of \(\frac { 1 }{ 4 }\)

(c) 3 ÷ \(\frac { 1 }{ 2 }\) = ………….. = …………….. 3 ÷ \(\frac { 1 }{ 2 }\) = 3 × \(\frac { 2 }{ 1 }\) = 3 × reciprocal of \(\frac { 1 }{ 2 }\)
And, 2 ÷ \(\frac { 3 }{ 4 }\) = 2 × reciprocal of \(\frac { 3 }{ 4 }\) = 2 × \(\frac { 4 }{ 3 }\)

(d) 5 ÷ \(\frac { 2 }{ 9 }\) = 5 × …………… = 5 × ……………..

∴ 5 ÷ \(\frac { 2 }{ 9 }\) = 5 × \(\frac { 9 }{ 2 }\) = 5 × reciprocal of \(\frac { 2 }{ 9 }\)

NCERT In-text Question Page No. 45
Question 1.
Find:
(i) 7 ÷ \(\frac { 2 }{ 5 }\)
(ii) 7 ÷ \(\frac { 4 }{ 7 }\)
(iii) 7 ÷ \(\frac { 8 }{ 9 }\)
Answer:

NCERT In-text Question Page No. 45
Question 1.
Find:
(i) 6 ÷ 5\(\frac { 1 }{ 3 }\)
(ii) 7 ÷ 2\(\frac { 4 }{ 7 }\)
Answer:

NCERT In-text Question Page No. 45
Question 1.
Find:
(i) \(\frac{3}{5} \div \frac{1}{2}\)
(ii) \(\frac{1}{2} \div \frac{3}{5}\)
(iii) 2\(\frac{1}{3} \div \frac{3}{5}\)
(iv) 5\frac{1}{6} \div \frac{9}{2}
Answer:

NCERT In-text Question Page No. 50
Question 1.
Find:
(i) 2.7 × 4
(ii) 1.8 × 1.2
(iii) 2.3 × 4.35
Answer:
(i) 27 × 4 = 108 and there is one digit to the right of the decimal point in 27 2.7 × 4 = 10.8
(ii) 18 × 12 = 216 and number of digits to the right of decimal point is (1 + 1) 2 = 2
1.8 × 1.2 = 2.16

(iii) 23 × 435 = 10005 and there are 1 + 2 = 3 digits to the right of decimal point
2.3 × 4.35 = 10.005

Question 2.
Arrange the products obtained in Question in descending order.
Answer:
The products are 10.8, 2.16, 10.005. Comparing 10.8 and 10.005, we have :
10 = 10, 8 > 0, i.e. 10.005 < 10.8
Here, the smallest number = 2.16
and, the largest number = 10.8
Thus the required descending order is : 10.8, 10.005,2.16.

NCERT In-text Question Page No. 51
Question 1.
Find:
(i) 0.3 × 10
(ii) 1.2 × 100
(iii) 56.3 × 1000
Answer:
(i) 0.3 × 10
There is one zero in 10
∵ The decimal point is shifted to the right by one place
Thus, 0.3 × 10 = 3

(ii) There are 2 zeroes in 100
∵ The decimal point is shifted to the right by 2 places
= 1.20 × 100= 120
Thus, 1.2 × 100 = 120

(iii) There are 3 zeros in 1000
∵ The decimal point is shifted to the right by 3 places
Thus, 56.3 × 1000 = 56.300 × 1000 = 56300

NCERT In-text Question Page No. 53
Question 1.
Find:
(i) 235.4 ÷ 10
(ii) 235.4 ÷ 100
(iii) 235.4 ÷ 1000
Answer:
(i) 235.4 ÷ 10
Since, there is one zero in 10.
∴ The decimal point in the quotient is shifted to the left by one place.
∴ 235.4 ÷ 10 = 23.54

(ii) 235.4 ÷ 100
Since, there are two zeros in 100
∴ The decimal point in the quotient is
shifted to the left by three places.
∴ 235.4 H- 1000 = 0.2354

NCERT In-text Question Page No. 53
Question 1.
Find:
(i) 35.7 ÷ 3 = ?
(ii) 25.5 ÷ 3 = ?
Answer:
(i) 35.7 ÷ 3
Since, \(\frac{357}{3}\) = 119 and there is one digit in the decimal part of the given decimal number.
∴ The decimal point is placed in the quotient after one digit from the right most digit.
∴ 35.7 ÷ 3 = 11.9

(ii) 25.5 ÷ 3
Since, 255 ÷ 3 = 85 and there is one digit in the decimal part of the given decimal number.
∴ The decimal is placed in the quotient after one digit from the right most digit.
∴ 25.5 ÷ 3 = 8.5

NCERT In-text Question Page No. 53

Question 1.
(i) 43.15 ÷ 5 = ?
(ii) 82.44 ÷ 6 = ?
Answer:

NCERT In-text Question Page No. 53
Question 1.
Find:
(i) 15.5 ÷ 5
(ii) 126.35 ÷ 7
Answer:
(i) 15.5 ÷ 5
Since 155 ÷ 5 = 31 and there is one digit in the decimal part of the given decimal number.
∴ Place the decimal point in 31 such that there is one digit to its right.
∴ 15.5 + 5 = 3.1

(ii) 126.37 + 7
Since 12635 + 7 = 1805 and there are two digits in the decimal part of the given decimal number.
∴ Place the decimal point in 1805 such that there are two digits to its right.
∴ 126.35 ÷ 7 = 18.05

NCERT In-text Question Page No. 54
Question 1.
Find:
(i) \(\frac{7.75}{0.25}\)
(ii) \(\frac{42.8}{0.02}\)
(iii) \(\frac{5.6}{1.4}\)
Answer:

These NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Exercise 13.2

Question 1.
Using laws of exponents, simplify and write the answer in exponential form:
(i) 3² × 3⁴ × 3⁸
(ii) 6¹⁵ ÷ 6¹⁰
(iii) a³ × a²
(iv) 7^x × 7²
(v) (5²)³ ÷ 5³
(vi) 2⁵ × 5⁵
(vii) a⁴ × b⁴
(viii) (3⁴)³
(ix) (2²⁰ ÷ 2¹⁵) × 2³
(x) 8^t ÷ 8²
Answer:
(i) 3² × 3⁴ × 3⁸ = 3^{2 + 4 + 8}
(a^m × aⁿ × ^r = a^m+n+r) = 3¹⁴
(ii) 6¹⁵ ÷ 6¹⁰ = \(\frac{6^{15}}{6^{10}}\) = 6^15-10
(\(\frac{a^{\mathrm{m}}}{\mathrm{a}^{\mathrm{n}}}\) = a^m-n) = 6⁵
(iii) a³ × a² = a³⁺²
(a^m × aⁿ = a^m+n ) = a⁵
(iv) 7^x × 7² = 7^x+2 (a^m × aⁿ = a^m+n )
(v) (5²)³ ÷ 5³

(vi) 2⁵ × 5⁵ = (2 × 5)⁵
[a^m × b^m = (ab)^m]
= (ab)⁴
(vii) a⁴ x b⁴
(viii) (3⁴)³ = 3¹²
[(a^m)ⁿ = a^mn]
(ix) (2²⁰ ÷ 2¹⁵) x 2³ = (\(\frac{2^{20}}{2^{15}}\)) x 2³
= (2^20-15) × 2³ [ \(\frac{a^{\mathrm{m}}}{\mathrm{a}^{\mathrm{n}}}\) = a^m-n]
= 2⁵ × 2³
= 2^{5 + 3} (a^m x aⁿ = a^m+n) = 2⁸
(x) 8^t ÷ 8² = \(\frac{8^{t}}{8^{2}}=8^{t-2}\left(\frac{\mathrm{a}^{\mathrm{m}}}{\mathrm{a}^{\mathrm{n}}}=\mathrm{a}^{\mathrm{m}-\mathrm{n}}\right)\)

Question 2.
Simplify and express each of the following in exponential form:
(i) \(\)
(ii) [(5²)³ × 5⁴)] ÷ 5⁷
(iii) 25⁴ ÷ 5³
(iv) \(\frac{3 \times 7^{2} \times 11^{8}}{21 \times 11^{3}}\)
(v) \(\frac{3^{7}}{3^{4} \times 3^{3}}\)
(vi) 2°+ 3°+ 4°
(vii) 2° × 3° × 4°
(viii) (3° + 2°) × 5°
(ix) \(\frac{2^{8} \times \mathrm{a}^{5}}{4^{3} \times \mathrm{a}^{3}}\)
(x) \(\left(\frac{a^{5}}{a^{3}}\right) \times a^{8}\)
(xi) \(\frac{4^{5} \times a^{8} b^{3}}{4^{5} \times a^{5} b^{2}}\)
(xii) (2³ × 2)²
Answer:

= 2° × 3³
= 1 × 3³ = 3³
(a° = 1)

(iv) \(\frac{3 \times 7^{2} \times 11^{8}}{21 \times 11^{3}}=\frac{3 \times 7^{2} \times 11^{8}}{3 \times 7 \times 11^{3}}\)
= 3^1-1 × 7^2-1 × 11^8-3
(\(\frac{\mathrm{a}^{\mathrm{m}}}{\mathrm{a}^{\mathrm{n}}}\) = a^m-n)
= 3° × 7¹ × 11⁵
= 1 × 7 × 11⁵
= 7 ×11⁵

(vi) 2° + 3°+ 4° = 1 + 1 + 1 = 3 (a° = 1)
(vii) 2° x 3° × 4° = 1 × 1 × 1 = 3 (a° = 1)
(viii) (3° + 2°) × 5° = (1 + 1) × 1 = 3 (a° = 1)
= 2 × 1 = 2

(ix) \(\frac{2^{8} \times a^{5}}{4^{3} \times a^{3}}=\frac{2^{8} \times a^{5}}{\left(2^{2}\right)^{3} \times a^{3}}\)
= \(\frac{2^{8} \times a^{5}}{2^{6} \times a^{3}}\) [(a^m)ⁿ = a^mn]
= 2^8-6 × a ^5-3 (\(\frac{\mathrm{a}^{\mathrm{m}}}{\mathrm{a}^{\mathrm{n}}}\) = a^m-n)
= 2² × a²
= 2a²

(x) \(\left(\frac{a^{5}}{a^{3}}\right)\) x a⁸ = (a^5-3) x a⁸
[(a^m)ⁿ = a^mn]
= a² × a⁸ (a^m x aⁿ = a^m+n)
= a²⁺⁸
= a¹⁰

(xi) \(\frac{4^{5} a^{8} b^{3}}{4^{5} a^{5} b^{2}}=\frac{4^{5-5} \times a^{8} \times b^{3}}{a^{5} \times b^{2}}\)
= 4° × a^8-5 × b^3-2 ( \(\frac{\mathrm{a}^{\mathrm{m}}}{\mathrm{a}^{\mathrm{n}}}\) = a^m-n)
= 1 × a³ × b
= a³b (a⁰ = 1)

(xii) (2³ x 2)² = (2³⁺¹)²
(a^m x aⁿ = a^m+n)
= (2⁴)²
= 2⁸ (a^m)ⁿ = a^mn

Question 3.
Say true or false and justify your answer:
(i) 10 × 10¹¹ = 100¹¹
(ii) 2³ > 5²
(iii) 2³ × 3² = 6⁵
(iv) 3° = (1000)°
Answer:
(i) 10 × 10¹¹ = 100¹¹
L.H.S = 10 x 10¹¹
= 10⁽¹⁺¹¹⁾= 10¹²
(a^m × aⁿ = a^m+n)
R.H.S = 100¹¹
= (10²)H= 10²²
(a^m)ⁿ = a^mn
L.H.S ≠ R.H.S.
∴ 10 × 10¹¹ ≠ 100¹¹
This statement is false.

(ii) 2³ > 5²
2³ = 2 × 2 × 2 = 8
5² = 5 × 5 = 25
8 < 25
i. e. 2³ > 5²
This statement is false.

(iii) 2³ × 3² = 6⁵
L.H.S = 2³ × 3²
= 2 × 2 × 2 × 3 × 3 = 72
R.H.S = 6⁵ = 6 × 6 × 6 × 6 × 6
= 36 × 36 × 6
= 7776
L.H.S ≠ R.H.S.
72 ≠ 7776
∴ 2³ × 3² ≠ 6⁵
This statement is false.

(iv) 3° = (1000)°
L.H.S = 3° = 1
R.H.S = (1000)°= 1
3° = (1000)°
L.H.S = R.H.S
3° = (1000)°
∴ This statement is true.

Question 4.
Express each of the following as a product of prime factors only in exponential form:
(i) 108 × 192
(ii) 270
(iii) 729 × 64
(iv) 768
Answer:
(i) 108 × 192
= 2 x 2 x 3 x 3 x 3 x 2 x 2

= 2⁸ × 3⁴

(ii) 270 = 2 × 3 × 3 × 3 × 5

= 2 × 3³ × 5

(iii) 729 × 64
= 3 × 3 × 3 × 3 × 3 × 3 × 2 × 2 × 2 × 2 × 2 × 2

= 3⁶ × 2⁶

(iv) 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2⁸ × 3

Question 5.
Simplify:
(i) \(\frac{\left(2^{5}\right)^{2} \times 7^{3}}{8^{3} \times 7}\)
(ii) \(\frac{25 \times 5^{2} \times t^{8}}{10^{3} \times t^{4}}\)
(iii) \(\frac{3^{5} \times 10^{5} \times 25}{5^{7} \times 6^{5}}\)
Answer:
(i) \(\frac{\left(2^{5}\right)^{2} \times 7^{3}}{8^{3} \times 7}=\frac{2^{10} \times 7^{3}}{\left(2^{3}\right)^{3} \times 7}\)
[(a^m)ⁿ = a^mn ]
= \(\frac{2^{10} \times 7^{3}}{2^{9} \times 7}\)
= 2^10-9 × 7^3-1 [\(\frac{a^{m}}{a^{n}}\) = a^m-n]
= 2¹ × 7²
= 2 × 49 = 98
Thus, \(\frac{\left(2^{5}\right)^{2} \times 7^{3}}{8^{3} \times 7}\) = 98

= 3^5-5 × 2^5-5 × 5^7-7
= 3° × 2° × 5°
= 1 × 1 × 1 = 1