Author name: Prasanna

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

These NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers Ex 1.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers Exercise 1.1

Question 1.
Using appropriate properties, find:
(i) \(\frac{-2}{3} \times \frac{3}{5}+\frac{5}{2}-\frac{3}{5} \times \frac{1}{6}\)
(ii) \(\frac{2}{5} \times\left(\frac{-3}{7}\right)-\frac{1}{6} \times \frac{3}{2}+\frac{1}{14} \times \frac{2}{5}\)
Solution:

Question 2.
Write the additive inverse of each of the following:
(i) \(\frac{2}{8}\)
(ii) \(\frac{-5}{9}\)
(iii) \(\frac{-6}{-5}\)
(iv) \(\frac{2}{-9}\)
(v) \(\frac{19}{-6}\)
Solution:
(i) Additive inverse of \(\frac{2}{8}\) is \(\frac{-2}{8}\)
(ii) Additive inverse of \(\frac{-5}{9}\) is \(\frac{5}{9}\)
(iii) Additive inverse of \(\frac{-6}{-5}\) is \(\frac{-6}{5}\)
(iv) Additive inverse of \(\frac{2}{-9}=\left(\frac{-2}{9}\right)\) is \(\frac{2}{9}\)
(v) Additive inverse of \(\frac{19}{-6}=\left(\frac{-19}{6}\right)\) is \(\frac{19}{6}\)

Question 3.
Verify that -(-x) = x for:
(i) x = \(\frac{11}{15}\)
(ii) x = \(\frac{-13}{17}\)
Solution:

Question 4.
Find the multiplicative inverse of the following:
(i) -13
(ii) \(\frac{-13}{19}\)
(iii) \(\frac{1}{5}\)
(iv) \(\frac{-5}{8} \times \frac{-3}{7}\)
(v) \(-1 \times \frac{-2}{5}\)
(vi) -1
Solution:
(i) Multiplicative inverse of -13 is \(\frac{-1}{13}\)
(ii) Multiplicative inverse of \(\frac{-13}{19}\) is \(\frac{-19}{13}\)
(iii) Multiplicative inverse of \(\frac{1}{5}\) is 5.
(iv) \(\left(\frac{-5}{8} \times \frac{-3}{7}\right)=\frac{(-5) \times(-3)}{8 \times 7}=\frac{56}{15}\)
Multiplicative inverse of \(\frac{-5}{8} \times \frac{-3}{7}\) is \(\frac{15}{56}\)
(v) \(-1 \times \frac{-2}{5}=\frac{(-1) \times(-2)}{5}=\frac{2}{5}\)
Multiplicative inverse of \(-1 \times \frac{-2}{5}\) is \(\frac{5}{2}\)
(vi) Multiplicative inverse of -1 is -1.

Question 5.
Name the property under multiplication used in each of the following:
(i) \(\frac{-4}{5} \times 1=1 \times \frac{-4}{5}=\frac{-4}{5}\)
(ii) \(\frac{-13}{17} \times \frac{-2}{7}=\frac{-2}{7} \times \frac{-13}{17}\)
(iii) \(\frac{-19}{29} \times \frac{29}{-19}=1\)
Solution:
(i) \(\frac{-4}{5} \times 1=1 \times \frac{-4}{5}=\frac{-4}{5}\)
1 is the multiplicative identity.

(ii) \(\frac{-13}{17} \times \frac{-2}{7}=\frac{-2}{7} \times \frac{-13}{17}\)
Multiplication is commutative.

(iii) \(\frac{-19}{29} \times \frac{29}{-19}=1\)
Multiplicative inverse.

Question 6.
Multiply \(\frac{6}{13}\) by the reciprocal of \(\frac{-7}{16}\).
Solution:

Question 7.
Tell what property allows you to compute \(\frac{1}{3} \times\left(6 \times \frac{4}{3}\right) \text { as }\left(\frac{1}{3} \times 6\right) \times \frac{4}{3}\)
Solution:
\(\frac{1}{3} \times\left(6 \times \frac{4}{3}\right) \text { as }\left(\frac{1}{3} \times 6\right) \times \frac{4}{3}\)
In the given question, we use the associative property.

Question 8.
Is \(\frac{8}{9}\) the multiplicative inverse of -1\(\frac{1}{8}\)? Why or why not?
Solution:
\(\frac{8}{9} \times-\frac{9}{8}=-1\) but is not equal to 1.
So, \(\frac{8}{9}\) is not the multiplicative inverse of -1\(\frac{1}{8}\)

Question 9.
Is 0.3 the multiplicative inverse of 3\(\frac{1}{3}\)? Why or why not?
Solution:
0.3 = \(\frac{3}{10}\)
3\(\frac{1}{3}\) = \(\frac{10}{3}\)
\(\frac{3}{10} \times \frac{10}{3}=1\)
∴ Multiplicative inverse of 3\(\frac{1}{3}\) is 0.3.

Question 10.
Write:
(i) The rational number that does not have a reciprocal.
(ii) The rational numbers that are equal to their reciprocals.
(iii) The rational number that is equal to its negative.
Solution:
(i) The rational number ‘0’ does not have a reciprocal.
(ii) The rational numbers 1 and (-1) are equal to their reciprocal.
(iii) The rational number ‘0’ is equal to its negative [(0) + (0) = 0]
∴ The negative of 0 is 0.

Question 11.
Fill in the blanks:
(i) Zero has ________ reciprocal.
(ii) The numbers ________ and ________ are their own reciprocals.
(iii) The reciprocal of -5 is ________
(iv) Reciprocal of \(\frac{1}{\mathrm{x}}\) when x ≠ 0 is ________
(v) The product of two rational numbers is always a ________
(vi) The reciprocal of a positive rational number is ________
Solution:
(i) no
(ii) 1 and -1
(iii) \(\frac{-1}{5}\)
(iv) x
(v) rational number
(vi) positive

These NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.7

Question 1.
Find:
(i) 0.4 ÷ 2
(ii) 0.35 ÷ 5
(iii) 2.48 ÷ 4
(iv) 65.4 ÷ 6
(v) 651.2 ÷ 4
(vi) 14.49 ÷ 7
(vii) 3.96 ÷ 4
(viii) 0.80 ÷ 5

Question 2.
Find:
(1) 4.8 ÷ 10
(ii) 52.5 ÷ 10
(iii) 0.7 ÷ 10
(iv) 33.1 ÷ 10
(V) 272.23 ÷ 10
(vi) 0.56 ÷ 10
(vii) 3.97 ÷ 10
Answer:

Question 3.
Find:
(i) 2.7 ÷ 100
(ii) 0.3 ÷ 100
(iii) 0.78 ÷ 100
(iv) 432.6 ÷ 100
(v) 23.6 ÷ 100
(vi) 98.53 ÷ 100
Answer:
(i) 2.7 ÷ 100 \(\frac{2.7}{100}\)
There are two zeros in 100
∴ The decimal point in the quotient is shifted to the left by two places
∴ \(\frac{2.7}{100}\) = 0.027

(ii) 0.3 ÷ 100
There are two zeros in 100
∴ The decimal point in the quotient is shifted to the left by two places
∴ 0.3 ÷ 100 = \(\frac{0.3}{100}\) = 0.003

(iii) 0.78 ÷ 100
There are two zeros in 100
∴ The decimal point in the quotient is shifted to the left by two places
∴ 0.78 ÷ 100 = \(\frac{0.78}{100}\) = 0.0078

(iv) 432.6 ÷ 100
There are two zeros in 100
∴ The decimal point in the quotient is shifted to the left by two places
∴ 432.6 ÷ 100 = \(\frac{432.6}{100}\) = 4.326

(v) 23.6 ÷ 100
There are two zeros in 100
The decimal point in the quotient is shifted to the left by two places
∴ 23.6 ÷ 100 = \(\frac{23.6}{100}\) = 0.236

(vi) 98.53 ÷ 100
There are two zeros in 100
∴ The decimal point in the quotient is shifted to the left by two places
∵ 98.53 ÷ 100 = \(\frac{98.53}{100}\) = 0.9853

Question 4.
(i) 7.9 ÷ 1000
(ii) 26.3 ÷ 1000
(iii) 38.53 ÷ 1000
(iv) 128.9 ÷ 1000
(v) 0.5 ÷ 1000
Answer:
(i) 7.9 ÷ 1000 = \(\frac{7.9}{1000}\) = 0.0079

(ii) 26.3 ÷ 1000 = \(\frac{26.3}{1000}\) = 0.0263

(iii) 38.53 ÷ 1000 = \(\frac{38.53}{1000}\) = 0.03853

(iv) 128.9 ÷ 1000 = \(\frac{128.9}{1000}\) = 0.1289

(v) 0.5 ÷ 1000 = \(\frac{0.5}{1000}\) = 0.0005

Question 5.
(i) 7 ÷ 3.5
(ii) 36 ÷ 0.2
(iii) 3.25 ÷ 0.5
(iv) 30.94 ÷ 0.7
(v) 0.5 ÷ 0.25
(vi) 7.75 ÷ 0.25
(vii) 76.5 ÷ 0.15
(viii) 37.8 ÷ 1.4
(ix) 2.73 ÷ 1.3
Answer:

Question 6.
A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance
will it cover in one litre of petrol?
Answer:
Total distance covered 43.2 km
Quantity of petrol used = 2.4 litres
∴ Distance covered in one litre petrol
= \(\frac{\text { Total distance covered }}{\text { Total quantity of petrol }}\)
= \(\frac{43.2}{2.4}\) km
= \(\frac{432 \times 10}{24 \times 10}=\frac{432}{24}\) = l8km
Distance covered in one litre of petrol = 18km

These NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 15 Probability Exercise 15.1

Question 1.
In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.
Solution:
Total number of balls she plays = 30
Total number of balls that she hits a boundary = 6
The total number of balls that she does not hit a boundary = 30 – 6 = 24 balls.
Let us denote the event that she does not hit a ball is E: So,
P(E) = \(\frac{24}{30}=\frac{4}{5}\)

Question 2.
1500 families with two children were selected randomly, and the following data were recorded.

Compute the probability of a family, chosen at random having.
(i) 2 girls
(ii) 1 girl
(iii) No girls
Also, check whether the sum of these probabilities is 1.
Solution:
Total number of families = 1500
(i) Total number of families which have 2 girls are 475.
∴ P(2 girls in a family) = \(\frac{475}{1500}=\frac{19}{60}\)

(ii) Total number of families which have 1 girl is 814.
∴ P(1 girl in a family) = \(\frac{814}{1500}=\frac{407}{750}\)

(iii) Total number of families which ahve no girl is 211.
∴ P(no girl in a family) = \(\frac{211}{1500}\)

Check:
Sum of probabilities = \(\frac{9}{60}+\frac{407}{750}+\frac{211}{1500}\)
= \(\frac{475+814+211}{1500}\)
= \(\frac{1500}{1500}\)
= 1

Question 3.
Refer to example 5, section 14.4, chapter 14 final the probability that a student of the class was born in August.
Solution:
Total number of students who were born in August = 6
Total number of students = 40
Therefore,
P(Number of students bom in August) = \(\frac{6}{40}\) = \(\frac{3}{20}\)

Question 4.
Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:

If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.
Solution:
Since the coin is tossed 200 times, so the total number of trials is 200.
Let us call the events of getting 2 heads is E.
Then the number of times E happens, i.e. two heads coming up is 72.
So, P(E) = \(\frac{72}{200}=\frac{9}{25}\)

Question 5.
An organization selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a home. The information gathered is listed in the table below:

Suppose a family is chosen. Find the probability that the family chosen is:
(i) earning Rs. 10,000 – 13,000 per month and owning exactly 2 vehicles.
(ii) earning Rs. 16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than Rs. 7000 per month and does not own any vehicle.
(iv) earning Rs. 13,000 – 16,000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.
Solution:
The total number of families = 2400.
(i) The number of families earning Rs. 10,000 – 13,000 per month and owning exactly two vehicles is 29.
So, P(family earning Rs. 10,000 – 13,000 per month and owning exactly two vehicles) = \(\frac{29}{2400}\)

(ii) The number of families earning Rs. 16,000 or more per month and owning exactly one vehicle is 579.
So, P(earning Rs. 16,000 or more per month and owning exactly one vehicle) = \(\frac{579}{2400}\)

(iii) The number of families earning less than Rs. 7000 per month and does not own any vehicle is 10.
So, P(earning less than Rs. 7000 per month and does not own any vehicle) = \(\frac{10}{2400}=\frac{1}{240}\)

(iv) The number of families earning Rs. 13,000 – 16,000 per month and owning more than two vehicles is 25.
So, P(earning Rs. 13,000 – 16,000 per month and owning more than two vehicles) = \(\frac{25}{2400}=\frac{1}{96}\)

(v) The number of families owning not more than 1 vehicle is 2062 (family owning exactly one vehicle + family owning no vehicle)
So, P(owning not more than one vehicle) = \(\frac{2062}{2400}=\frac{1031}{1200}\)

Question 6.
Refer to table 14.7, chapter 14
(i) Find the probability that a student obtained less than 20% in the mathematics test.
(ii) Find the probability that a student obtained marks 60 or above.
Solution:
(i) Total number of students = 90
Number of students who gets less than 20% marks in mathematics test = 7.
∴ P(student obtained less than 20% marks) = \(\frac{7}{90}\)

(ii) Number of student getting 60 or above marks is 23.
∴ P(student obtained 60 or above marks) = \(\frac{23}{90}\)

Question 7.
To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table.

Find the probability that a student is chosen at random
(i) like statistics
(ii) does not like it.
Solution:
The total number of students = 200.
(i) Total number of students like statistics = 135
So, P(like statistics) = \(\frac{135}{200}=\frac{27}{40}\)

(ii) Total number of students does not like statistics = 65.
So, P(does not like statistics) = \(\frac{65}{200}=\frac{13}{40}\)

Question 8.
Refer to Q.2, Exercise-14.2. What is the empirical probability that an engineer lives:
(i) less than 7 km from her place of work?
(ii) more than or equal to 7 km from her place of work?
(iii) within \(\frac {1}{2}\) km. from her place of work?
Solution:
(i) Total number of female engineers whose residence to their place of work given is 40.
The number of engineers whose residence is less than 7 km from their place of work = 9.
∴ P(less than 7 km from her place of work) = \(\frac{9}{40}\)

(ii) Number of engineer who’s residence is more than or equal to 7 km from her place of work = 31.
∴ P(more than or equal to 7 km from place of work) = \(\frac{31}{40}\)

(iii) Number of engineer who’s residence is within \(\frac {1}{2}\) km. from her place of work = 0.
∴ P(within \(\frac {1}{2}\) km. from place of work) = \(\frac {0}{40}\) = 0

Question 9.
Activity: Note the frequency of two-wheelers, three-wheelers, and four-wheelers going past during the time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler.
Answer:
Let the person saw the vehicles for half an hour.
The number of vehicles seen by the person.
Two wheelers = 5
Three wheelers = 10
Four wheelers = 15
So,total number of space = 5 + 10 + 15 = 30
Number of events when two wheeler is seen = 5
Probability = \(\frac{\text { no. of event }}{\text { no. of space }}\)
= \(\frac{5}{30}\)
= \(\frac{1}{5}\)

Question 10.
Activity: Ask all the students in your class to write a three-digit number. Choose any student from the room at random, what is the probability that the number written by her/him is divisible by 3? Remember that a number is divisible by 3 if the sum of its digits is divisible by 3.
Solution:
The total number of students in my class = 40.
Out of 40 students, 10 students write a three-digit number which is divisible by 3.
∴ P(student write a number which is divisible by 3) = \(\frac{10}{40}=\frac{1}{4}\)

Question 11.
Eleven bags of wheat flour, each marked 5 kg. actually contained the following weights of flour (in kg): 4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00
Find the probability that any of these bags chosen at random contains more than 5 kg. of flour.
Solution:
The total number of wheat flour bags is 11.
The total number of wheat flour bags which contain more than 5 kg. of flour is 7.
∴ P(bags contain more than 5 kg of flour) = \(\frac{7}{11}\)

Question 12.
In Q.5 Exercise 14.2, you were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12 – 0.16 on any of these days.
Solution:
The total number of observations = 30.
Number of observation lie in the interval 0.11 – 0.16 is 2.
∴ P(The concentration of sulphur dioxide in the interval 0.12 – 0.16 = \(\frac{2}{30}=\frac{1}{15}\)

Question 13.
In Q. 1, Exercise 14.2, you were asked to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group AB.
Solution:
Total number of students whose blood group are recorded = 30.
Number of students which have blood group AB is 3
∴ P(student has blood group AB) = \(\frac{3}{30}=\frac{1}{10}\)

These NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.6

Question 1.
Find:
(i) 0.2 × 6
(ii) 8 × 4.6
(iii) 2.71 × 5
(iv) 20.1 × 4
(v) 0.05 × 7
(vi) 211.02 × 4
(vii) 2 × 0.86
Answer:
(i) 0.2 × 6
Since 2 × 6 = 12 and there is one digit to the right of decimal point in 0.2
∴ 0.2 × 6 = 1.2

(ii) 8 × 4.6
Since 8 × 46 = 368 and there is one digit to the right of decimal point in
∴ 8 × 4.6 = 36.8

(iii) 2.71 × 5
Since 271 × 5 = 1355 and there are two digits to the right of decimal point in 2.71
∴ 2.71 × 5 = 13.55

(iv) 20.1 × 4
Since 201 × 4 = 804 and there is one digit to the right of decimal point in 20.1
∴ 20.1 × 4 = 80.4

(v) 0.05 × 7
Since 5 × 7 = 35 and there are two digits to the right of decimal point
∴ 0.05 × 7 = 0.35

(vi) 211.02 × 4
Since 21102 × 4 = 84408 and there are two digits to the right of the decimal point
∴ 211.02 × 4 = 844.08

(vii) 2 × 0.86
Since 2 × 86 = 172 and there are two digits to the right of the decimal point
∴ 2 × 0.86 = 1.72

Question 2.
Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm.
Answer:
Length of a rectangle = 5.7 cm
Breadth of a rectangle = 3 cm
Area of the rectangle = Length × Breadth
= 5.7 cm × 3 cm = 17.1cm²

Question 3.
Find:
(i) 1.3 × 10
(ii) 36.8 × 10
(iii) 153.7 × 10
(iv) 168.07 × 10
(v) 31.1 × 100
(vi) 156.1 × 100
(vii) 3.62 × 100
(viii) 43.07 × 100
(ix) 0.5 × 10
(x) 0.08 × 10
(xi) 0.9 × 100
(xii) 0.03 × 1000
Answer:
(i) 1.3 × 10 = 13
(ii) 36.8 × 10 = 368
(iii) 153.7 × 10 = 1537
(iv) 168.07 × 10 = 1680.7
(v) 31.1 × 100 = 3110
Note:
(1) If there is one zero, the decimal point is shifted to the right by one place.
(2) If there are two zeroes, the decimal point is shifted to the right by two places.
(vi) 156.1 × 100 = 15610
(vii) 3.62 × 100 = 362
(viii) 43.07 × 100 = 4307
(ix) 0.5 × 10 = 5
(x) 0.08 × 10 = 0.8
(xi) 0.9 × 100 = 90
(xii) 0.03 × 1000 = 30

Question 4.
A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol?
Answer:
Distance covered in one litre of petrol = 55.3 km
Distance covered in 10 litres of petrol = 55.3 × 10 km = 553 km

Question 5.
Find:
(i) 2.5 × 0.3
(ii) 0.1 × 51.7
(iii) 0.2 × 316.8
(iv) 1.3 × 3.1
(v) 0.5 × 0.05
(vi) 11.2 × 0.15
(vii) 1.07 × 0.02
(viii) 10.05 × 1.05
(ix) 101.01 × 0.01
(x) 100.01 × 1.1
Answer:
(i) 25 × 3 = 75
∴ 2.5 × 0.3 = 0.75

(ii) 1 × 517 = 517
∴ 0.1 × 51.7 = 5.17

(iii) 2 × 3168 = 6336
∴ 0.2 × 316.8 = 63.36

(iv) 13 × 31 = 403
∴ 1.3 × 3.1= 4.03

(v) 5 × 5 = 25
∴ 0.5 × 0.05 = 0.025

(vi) 112 × 15 = 1680
∴ 11.2 × 0.15 = 1.680

(vii) 107 × 2 = 214
∴ 1.07 × 0.02 = 0.0214

(viii) 1005 × 105 = 105525
∴ 10.05 × 1.05 = 10.5525

(ix) 10101 × 1 = 10101
∴ 101.01 × 0.01 = 1.0101

(x) 10001 × 11 = 110011
∴ 100.01 × 1.1 = 110.011

These NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Exercise 13.1

Question 1.
Find the value of
(i) 2⁶
(ii) 9³
(iii) 11²
(iv) 5⁴
Answer:
(i) 2⁶ = 2 x 2 x 2 x 2 x 2 x 2 = 64
(ii) 9³ = 9 x 9 x 9 = 729
(iii) 11² = 11 x 11 = 121
(iv) 5⁴ = 5 x 5 x 5 x 5 = 625

Question 2.
Express the following in exponential form:
(i) 6 x 6 x 6 x 6
(ii) t x t
(iii) b x b x b x b
(iv) 5 x 5 x 7 x 7 x 7
(v) 2 x 2 x a x a
(vi) a x a x a x c x c x c x c x d
Answer:
(i) 6 x 6 x 6 x 6 = 6⁴
(ii) t x t = t²
(iii) b x b x b x b = b⁴
(iv) 5 X 5 X 7 X 7 X 7 = 5² X 7³
(v) 2 x 2 x a x a = 2² x a²
(vi) a x a x a x c x c x c x c x d = a³ x c⁴ x d

Question 3.
Express each of the following numbers using exponential notation.
(i) 512
(ii) 343
(iii) 729
(iv) 3125
Answer:
(i) 512

512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2⁹

(ii) 343

343 = 7 x 7 x 7 = 7³

(iii) 729

729 = 3 x 3 x 3x 3 x 3 x 3 = 3⁶

(iv) 3125

3125 = 5 x 5 x 5 x 5 x 5 = 5⁵

Question 4.
Identify the greater number wherever
possible in each of the following.
(i) 4³ or 3⁴ (ii) 5³ or 3⁵
(iii) 2⁸ or 8²
(iv) 100² or 2¹⁰⁰
(v) 2¹⁰ or 10²
Answer:
(i) 4³ or 3⁴
4³ = 4 x 4 x 4 = 64
3⁴ = 3x3x3x3 = 81
81 > 64
3⁴ > 4³
3⁴ is greater

(ii) 5³ or 3⁵
5³ = 5 x 5 x 5 = 125
3⁵ = 3x3x3x3x3 = 243
243 > 125
3⁵ > 5³
3⁵ is greater.

(iii) 2⁸ or 8²
2⁸ = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
= 256
8² = 8 x 8 = 64
256 > 64
2⁸ > 8²
2⁸ is greater.

(iv) 100² or 2¹⁰⁰
100² = 100 x 100 = 10000
2¹⁰⁰ = (2¹⁰)¹⁰
= (2x2x2x2x2x 2 x 2 x 2 x 2 x 2)¹⁰
= (1024)¹⁰ = [(1024)²]⁵
= (1024 x 1024)⁵ = (1048576)⁵
.’.(1048576) > 10000
or (1048576)⁵ > 100²
2¹⁰⁰ is greater

(v) 2¹⁰ or 10²
2¹⁰ = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 1024
10² = 10 x 10 = 100
1024 > 100
2¹⁰ > 10²
2¹⁰ is greater.

Question 5.
Express each of the following as product of powers of their prime factors.
(i) 648
(ii) 405
(iii) 540
(iv) 3600
Ans:
(i) 648

648 = 2 x 2 x 3 x 3 x 3 x 3
= 2³ x 3⁴

(ii) 405

405 = 3 x 3 x 3 x 3 x 5 = 3⁴ x 5

(iii) 540

540 = 2 x 2 x 3 x 3 x 3 x 5
= 2² x 3³ x 5

(iv) 3600

3600 = 2 x 2 x 2 x 2 x 3 x 3 x 5 x 5
= 2⁴ x 3² x 5²

Question 6.
Simplify
(i) 2 x 10³
(ii) 7² x 2²
(iii) 2³ x 5
(iv) 3 x 4⁴
(v) 0 x 10²
(vi) 5²x 3³
(vii) 2⁴ x 3²
Answer:
(i) 2 x 10³ = 2 x 10 x 10 x 10 = 2000
(ii) 7² x 2² = 7 x 7 x 2 x 2 = 49 x 4 = 196
(iii) 2³ x 5 = 2 x 2 x 2 x 5 = 8 x 5 = 40
(iv) 3 x 4⁴ = 3 x 4 x 4 x 4 x 4 = 3 x 256 = 768
(v) 0 x 10² = 0 x 10 x 10 = 0
(vi) 5² x 3³ = 5 x 5 x 3 x 3 x 3 = 25 x 27 = 675
(vii) 2⁴ x 3² = 2 x 2 x 2 x 2 x 3 x 3 = 16 x 9 = 144
(viii) 3² x 10⁴ = 3 x 3 x 10 x 10 x 10 x 10 = 9 x 10000 = 90000

Question 7.
Simplify:
(i) (-4)³
(ii) (-3) x (-2)³
(iii) (- 3)² x (- 5)²
(iv) (-2)³ x (- 10)³
Answer:
(i) (-4)³ = (-4) x (-4) x (-4) = – 64
(ii) (-3) x (-2)³ = (- 3) x (- 2) x (- 2) x (- 2) = (- 3) x (- 8) = 24
(iii) (- 3)² x (- 5)² = (-3) x (-3) x (- 5) x (- 5) = 9 x 25 = 225
(iv) (-2)³ x (- 10)³ = (- 2) x (- 2) x (- 2) x (-10) x (-10) x (- 10) =
(- 8) x (- 1000) = 8000

Question 8.
Compare the following numbers:
(i) 2.7 x 10¹²; 1.5 x 10⁸
(ii) 4 x 10¹⁴ ; 3 x 10¹⁷
Answer:
(i) 2.7 x 10¹² = \(\frac{27}{10}\) x 10¹²
= 27 x 10¹¹
(It contains 13 digits) (\(\) = a^m-n)
1.5 x 10⁸ = \(\frac{15}{10}\) x 10⁸
= 15 x 10⁷ (\(\) = a^m-n)
(It contains 9 digits)
27 x 10¹¹ > 15 x 10⁷
.’. 2.7 x 10¹² > 1.5 x 10⁸
2.7 x 10¹² is greater.

(ii) 4 x 10¹⁴ It contains 15 digits
3 x 10¹⁷ It contain 18 digits
.-. 3 x 10¹⁷ > 4 x 10¹⁴
3 x 10¹⁷ is greater.