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These NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Exercise 8.1

Question 1.
Find the ratio of
(a) ₹ 5 to 50 paise
(b) 15 kg to 210 g
(c) 9 m to 27 cm
(d) 30 days to 36 hours
Answer:
(a) ₹ 5 = 5 x 100 paise
0 = 500 paise
₹ 5 : 50 paise = 500 paise : 50 paise
= 500 : 50
= 10 : 1

(b) 15 kg = 15×1000 g
= 15000g
15 kg to 210 g = 15000: 210 g
= 1500 : 21
= 500 : 7

(c) 9 m = 9 x 100 cm
= 900 cm
9 m : 27 cm= 900 : 27
= 100:3 (÷9)

(d) 30 days = 30 x 24 hours
= 720 hours
30 days : 36 hours = 720 : 36
= 20:1 (÷36)

Question 2.
In a computer lab there are 3 computers for every6 students. How many computers will be needed for 24 students?
Ans: Let x be the required number of computers.
3 : x = 6:24
\(\frac{3}{x}=\frac{6}{24}\)
6x = 3 x 24
x = \(\frac{3 \times 24}{6}\) = 3 x 4 = 12
Hence, 12 computers will be needed for 24 students.

Question 3.
Population of Rajasthan = 570 lakhs and population of UP = 1660 lakhs. Area of Rajasthan = 3 lakh km2 and area of UP = 2 lakh km².
(i) How many people are there per km2 in both these states?
(ii) Which state is less populated?
Answer:
(i) Population of Rajasthan
= 570 lakhs
Area of Rajasthan
= 3 lakhs km²
Population per km²
= \(\frac{570}{3}\) lakhs km²
= 190 lakhs per km²
In Rajasthan, there are 190 lakh people per km²
For UP
population = 1660 lakhs
Area = 2 lakhs per km²
population per km²
= \(\frac{1660}{2}\) per km²
= 830 lakhs per km²
In UP, there are 830 lakhs people per km²

(ii) 190 lakhs < 830 lakhs
∴ Rajasthan is less populated state.

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

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers InText Questions

NCERT In-text Question Page No. 250

Question 1.
Find five more such examples, where a number is expressed in exponential form. Also identify the base and the exponent in each case.
Answer:

Note: 1. xxxxxxx = x4is read as ‘x raised to the power 4’ or ‘4^th power of x’.
2. x²y⁵ is read as ‘x squared into y raised to power 5’.
3. p⁶q³ is reas as ‘p raised to the power 6 into q cubed’.

NCERT In-text Question Page No. 251

Question 1.
Express:
(i) 729 as a power of 3
(ii) 128 as a power of 2
(iii) 343 as a power of 7
Answer:
(i) 729
We have: 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3⁶
Thus, 729 = 3⁶

(ii) 128
We have: 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2⁷
Thus, 128 = 2⁷

(iii) 343
We have: 343 = 7 × 7 × 7 × 7= 7³
Thus, 343 = 7³

NCERT In-text Question Page No. 254

Question 1.
Simplify and write in exponential form:
(i) 2⁵ × 2³
(ii) p³ × p²
(iii) 4³ × 4²
(iv) a³ × a² × a⁷
(v) 5³ × 5⁷ × 5¹²
(vi) (-4)¹⁰⁰ × (-4)²⁰
Answer:
(i) 2⁵ × 2³
We have: 2⁵ × 2³ = 2^{5 + 3} = 28

(ii) p³ × p²
We have: p³ × p² = p^{3 + 2} = p⁵

(iii) 4³ × 4²
We have: 4³ × 4^{2<.sup> = 43+2 = p5}

(iv) a³ × a² × a⁷
We have: a³ × a² × a⁷ = a^{3 + 2 + 7} = a¹²

(v) 5³ × 5⁷ × 5¹²
We have: 5³ × 5⁷ × 5¹² = 5^{3 + 7 + 12} = 5²²

(vi) (-4)¹⁰⁰ × (-4)²⁰
We have: (-4)¹⁰⁰ × (-4)²⁰ = (-4)¹⁰⁰⁺² = (4)¹²⁰

Note: The above rule is possible only for same bases. It is not true for different bases. Thus, 2³ x 3² will no obev this rule.

NCERT In-text Question Page No. 255

Question 1.
Simplify and write in exponential form: (eg., 11⁶ ÷ 11² = 11⁴)
(i) 2⁹ ÷ 2³
(ii) 10⁸ ÷ 10⁴
(iii) 9¹¹ ÷ 9⁷
(iv) 20¹⁵ ÷ 20¹³
(v) 7¹³ ÷ 7¹⁰
Answer:
Since a^m aⁿ = a^m-n, therefore;
(i) 2⁹ ÷ 2³
We have: 2⁹ ÷ 2³ = 2^9-3 = 2⁶

(ii) 10⁸ ÷ 10⁴
We have: 10⁸ ÷ 10⁴ = 10^8-4 = 10⁴

(iii) 9¹¹ ÷ 9⁷
We have: 9¹¹ ÷ 9⁷ = 9^11-7= 9⁴

(iv) 20¹⁵ ÷ 20¹³
We have: 20¹⁵ ÷ 20¹³ = 20^{15 -13} = 20²

(v) 7¹³ ÷ 7¹⁰
We have: 7¹³ ÷ 7¹⁰ = 7^13-10 = 7³

Question 15.
Simplify and write the answer in exponential form:
(i) (6²)⁴
(ii) (2²)¹⁰⁰
(iii) (7⁵⁰)²
(iv) ( 5³)⁷
Answer:
Since (a^m)ⁿ = a^mxn = a^mn, therefore;
(i) (6²)⁴
We have: (6²)⁴ = 6^{2 × 4} = 6⁸

(ii) (2²)¹⁰⁰
We have: ( 2²)¹⁰⁰ = 2^{2 × 100} = 2²⁰⁰

(iii) (7⁵⁰)²
We have: (7⁵⁰ )² = 7^{50 × 2} = 7¹⁰⁰

(iv) ( 5³)⁷
We have: (5³)⁷ = 5^{3 × 7} = 5²¹

NCERT In-text Question Page No. 256

Question 1.
Put into another form using a^m × b^m= (ab)^m
(i) 4³ × 2³
(ii) 2⁵ × b⁵
(iii) a² × t²
(iv) 5⁶ × (-2)⁶
(v) (-2)⁴ × (-3)⁴
Answer:
Since a^m ÷ aⁿ = a^m-n, therefore;
(i) 4³ × 2³
We have: 4³ × 2³ = (4 × 2)³ = 8³

(ii) 2⁵ x b⁵
We have: 2⁵ × b⁵ = (2 × b)⁵= (2b)⁵

(iii) a² × t²
We have: a² x t² = (a × t)² = (at)²

(iv) 5⁶ x (-2)⁶
We have: 5⁶ × (-2)⁶ = [5 × (-2)]⁶ = (-10)⁶

(v) (-2)⁴ x (-3)⁴
We have: (-2)⁴ × (-3)⁴ = [(-2) × (-3)]⁴ = (6)⁴

NCERT In-text Question Page No. 257

Question 2.
Put into another form using a^m ÷ b^m
(i) 4⁵ ÷ 3⁵
(ii) (-2)⁵ ÷ b⁵
(iii) (-2)³ ÷ b³
(iv) p⁴ ÷ q⁴
(v) 5⁶ ÷ (-2)⁶
Answer:
(i) 4⁵ ÷ 3⁵
We have: 4⁵ ÷ 3⁵ = \(\left(\frac{4}{3}\right)^{5}\)

(ii) (-2)⁵ ÷ b⁵
We have: (-2)⁵ ÷ b⁵ = \(\left(\frac{2}{b}\right)^{5}\)

(iii) (-2)³ ÷ b³
We have: (-2)³ ÷ b³ = \(\left(\frac{-2}{b}\right)^{3}\)

(iv) p⁴ ÷ q⁴
We have: p⁴ ÷ q⁴ = \(\left(\frac{p}{q}\right)^{4}\)

(v) 5⁶ ÷ (-2)⁶
We have: 5⁶ ÷ (-2)⁶ = \(\left(\frac{5}{-2}\right)^{6}\) = \(\left(-\frac{5}{2}\right)^{6}\)

NCERT In-text Question Page No. 261

Question 1.
Expand by expressing powers of 10 in the exponential form:
(i) 172
(ii) 5,643
(iii) 56,439
(iv) 1,76,428
Answer:
(i) 172:
We have:
172 =(1 × 100) + ( 7 × 10) + (2 × 1)
= 1 × 10² + 7 × 10¹ + 2 × 1)
(∵10⁰ = 1)

(ii) 5,643:
We have:
5,643 = 5 × 1000 + 6 × 100 + 4 × 10 + 3 × 1
= 5 × 10³ + 6 × 10² + 4 × 10¹ + 3 × 10⁰

(iii) 56,439:
We have:
56,439 = 5 × 10000 + 6 × 1000 + 4 × 100 + 3 × 10 + 9 × 1
= 5 × 10⁴ + 6 × 10³ + 4 × 10² + 3 × 10¹ + 9 × 10⁰

(iv) 1,76,428:
We have:
1,76,428 = 1,00,000 + 7 × 10,000 + 6 × 1000 + 4 × 100 + 2 × 10 + 8 × 1
= 1 × 10⁵ + 7 × 10⁴ + 6 × 10³ + 4 × 10² + 2 × 10¹ + 8 × 10⁰

These NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 14 Symmetry

Exercise 14.3

Question 1.
Name any two figures that have both line symmetry and rotational symmetry.
Answer:

Question 2.
Draw, wherever possible, a rough sketch of
(i) a triangle with both line and rotational symmetries of order more than 1.
(ii) a triangle with only line symmetry and no rotational symmetry of order more than 1.
(iii) a quadrilateral with a rotational symmetry of order more than 1 but not a line symmetry.
(iv) a quadrilateral with line symmetry but not a rotational symmetry of order more than 1.
Answer:
(i) An equilateral triangle has three lines of symmetry and rotational symmetry of order 3.

(ii) An isosceles triangle has only one line ofD symmetry but no rotational symmetry of order more than 1.
(iii) A parallelogram has no line of symmetry but I has a rotational symmetry of order 2. A
(iv) An isosceles trapezium has one line of symmetry but no rotational symmetry ofordermore than 1.

Question 3.
If a figure has two or more lines of symmetry, should it have rotational symmetry of order more than 1.
Answer:
Yes

Question 4.
Fill in the blanks:
Answer:

Question 5.
Name the quadrilaterals which have both line and rotational symmetry of order more than 1.
Answer:
The square, rectangle and a rhombus are the quadrilaterals having both line and rotational symmetry.

Question 6.
After rotating by 60° about a centre, a figure looks exactly the same as its original position.
At what other angles will this happen for the figure.
Answer:
The figure will look exactly the same as its original position at 120°, 180°, 240°, 300° and 360°.

Question 7.
Can we have a rotational symmetry of order more than 1 whose angle of rotation is (i) 45° (ii) 17°
Answer:
(i) 45° – yes
(ii) 17° – No

These NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Exercise 15.3

Question 1:
What cross-sections do you get when you give a
(i) vertical cut
(ii) horizontal cut to the following solids?
(a) A brick
(b) A round apple
(c) A die
(d) A circular pipe
(e) An ice cream cone
Answer:

These NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Exercise 15.2

Question 1.
Use isometric dot paper and make an isometric sketch for each one of the given shapes:

Answer:

Question 2.
The dimensions of a cuboid are 5 cm, 3 cm and 2 cm. Draw three different isometric sketches of this cuboid.
Answer:

Question 3.
Three cubes each with 2 cm edge are placed side by side to form a cuboid. Sketch an oblique
Answer:

Question 4.
Make an oblique sketch for each one of the given isometric shapes :

Answer:

Question 5.
Give (i) an oblique sketch and (ii) an isometric sketch for each of the following :
(a) A cuboid of dimensions 5 cm, 3 cm and 2 cm. (Is your sketch unique?)
(b) A cube with an edge 4 cm long.
An isometric sheet is attached at the end of the book. You could try to make on it some cubes or cuboids of dimensions specified by your friend.
Answer:
Sketch is not unique. There can be more shapes in different positions.