CBSE Class 7

These NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.2

Question 1.
Which of the drawings (a) to (d) show:
(i) 2 × \(\frac { 1 }{ 5 }\)
(ii) 2 × \(\frac { 1 }{ 2 }\)
(iii) 3 × \(\frac { 2 }{ 3 }\)
(iv) 3 × \(\frac { 1 }{ 4 }\)

Answer:
(i) 2 × \(\frac{1}{5}=\frac{1}{5}+\frac{1}{5}\) which is represented by the drawing (d)
Thus (i) → (d)

(ii) 2 × \(\frac{1}{2}=\frac{1}{2}+\frac{1}{2}\) which is represented by the drawing (b)
Thus (ii) → (b)

(iii) which is represented by the drawing (a)
Thus (iii) → (a)

(iv) which is represented by the drawing (c)
Thus (iv) → (c)

Question 2.
Some pictures (a) to (c) are given below. Tell which of them show:

Answer:
(i) which is shown by the drawing (c)
∴ (i) → (c)

(ii) which is shown by the drawing (a)
∴ (ii) → (a)

(iii) which is shown by the drawing (b)
∴(iii) → (b)

Question 3.
Multiply and reduce to lowest form and convert into a mixed fraction :


Answer:

Question 4.
Shade:
(i) \(\frac{1}{2}\) of the circles in box (a)
(ii) \(\frac{2}{3}\) of the triangles in box (b)
(iii) \(\frac{3}{5}\) of the squares in box (c).

Answer:
(i)
\(\frac{1}{2}\) of the circles
\(\frac{1}{2}\) of 12 = \(\frac{1}{2}\) × 12
= 6
∴ So, we shade 6 circles.

(ii)
\(\frac{2}{3}\) of the triangles
\(\frac{2}{3}\) of 9 = \(\frac{2 \times 9}{3}\)
= 2 × 3 = 6
∴ So, we shade 6 triangles.

(iii)
\(\frac{3}{5}\) of the squares
\(\frac{3}{5}\) of 15
\(\frac{3}{5}\) × 15
\(\frac{3 \times 15}{5}\) = 3 × 3 = 9
∴ So, we shade 9 squares.

Question 5.
Find:
(a) \(\frac{1}{2}\) of (i) 24, (ii) 46
(b) \(\frac{2}{3}\) of (i) 18, (ii)27
(c) \(\frac{3}{4}\) of (i) 16, (ii) 36
(d) \(\frac{4}{5}\) of (i) 20, (ii) 35
Answer:

Question 6.
Multiply and express as a mixed fraction:

Answer:

Question 7.
(a) \(\frac{1}{2}\) of (i) 2\(\frac{3}{4}\),(ii) 4\(\frac{2}{9}\)
(b) \(\frac{5}{8}\) of (i) 3\(\frac{5}{6}\),(ii) 9 \(\frac{2}{3}\)
Answer:

Question 8.
Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of water. Vidya \(\frac{2}{5}\) consumed of the water.
Pratap consumed the remaining water.
(i) How much water did Vidya drink?
(ii) What fraction of the total quantity of water did Pratap drink?
Answer:
Total quantity of water = 5 litres
(i) Amount of water consumed by Vidya = \(\frac{2}{5}\) of litres
= \(\frac{2}{5}\) x 5 litres = 2 litres

(ii) Remaining water = water
consumed by Pratap
= 5 litres – 2 litres = 3 litres
∵ Fraction of water consumed by \(\frac{3}{5}\) Pratap =

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

NCERT Solutions for Class 7 Maths Chapter 14 Symmetry InText Questions

Question 1.
(a) Can you now tell the order of the rotational symmetry for an equilateral triangle? Look at the following figures.

(b) How many positions are there at which the triangle looks exactly the same, when rotated about its centre by 120°?
Answer:
(a) There are exactly three positions where the triangle looks the same?
∴ It has a rotational symmetry of order 3.
(b) There is only one position where the triangle looks exactly the same, when rotated about its centre through 120°?

Question 2.
Which of the following shapes have rotational symmetry about the marked point?

Answer:
All the above shapes (i) (ii) (iii) and (iv) have rotational symmetry about the marked point.

NCERT In-text Question Page No. 273

Question 1.
Give the order of the rotational symmetry of the given figures about the point marked x.

Answer:
The order of the rotational symmetry is 4.

The order of rotational symmetry is 3.

The order of the rotational symmetry is 3.
(iii)

The order of the rotational symmetry is 2.

NCERT In-text Question Page No. 275

Question 1.
Some of the English alphabets have fascinating symmetrical structures. Which capital letters have just one line of symmetry (like E)? Which capital letters have a rotational symmetry of order 2 (like I)? By attempting to think on such lines, you will be able to fill in the following table:
Answer:

These NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.1

Question 1.

Answer:

Question 2.
(i) \(\frac{2}{9}, \frac{2}{3}, \frac{8}{21}\)
(ii) \(\frac{1}{5}, \frac{3}{7}, \frac{7}{10}\)
Answer:
(i) \(\frac{2}{9}, \frac{2}{3}, \frac{8}{21}\)
L.C.M of 9, 3 and 21 = 63

∴ The descending order is \(\frac{2}{3}, \frac{8}{21}, \frac{2}{9}\)

(ii) \(\frac{1}{5}, \frac{3}{7}, \frac{7}{10}\)
L.C.M of 5, 7 and 10 = 70

∴ The descending order is \(\frac{7}{10}, \frac{3}{7}, \frac{1}{5}\)

Question 3.
In a ‘magic square the sum of the numbers in each row, in each column
and along the diagonals is the same. Is this a magic square?

(Along the first row \(\frac{4}{11}+\frac{9}{11}+\frac{2}{11}=\frac{15}{11}\))
Answer:
Sum of numbers along the
1^strow = \(\frac{4}{11}+\frac{9}{11}+\frac{2}{11}\) = \(\frac{4+9+2}{11}\)
\(\frac{15}{11}\)

Since, the sum of numbers in each case is the same therefore, it is a magic square.

Question 4.
A rectangular sheet of paper is 12\(\frac{1}{2}\) cm long and 10\(\frac{2}{3}\) cm wide. Find its perimeter.
Answer:
Length of a rectangle = 12\(\frac{1}{2}\) cm = \(\frac{25}{2}\) cm
Width of a rectangle = 10\(\frac{2}{3}\)cm = \(\frac{32}{3}\) cm
Perimeter of a rectangle = 2 (length + breadth)

Question 5.
Find the perimeters of (i) ΔABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?

Answer:
(i) Perimeter of ΔABE = AB + BE + AE

(ii) Perimeter of rectangle BCDE
= 2 [length + breadth]
= 2 [CD + DE]


Thus, the perimeter of ΔABE is greater.

Question 6.
Salil wants to put a picture in a frame. The picture is 7\(\frac { 3 }{ 5 }\) cm wide. To fit in the frame the picture cannot be more than 7\(\frac { 3 }{ 10 }\) cm wide. How much should the picture be trimmed?
Answer:
Width of the picture = 7\(\frac { 3 }{ 5 }\)cm
\(\frac { 38 }{ 5 }\)cm
Required width of the picture
= 7\(\frac { 3 }{ 10 }\) cm = \(\frac { 73 }{ 10 }\) cm
The picture should be trimmed by

The picture should be trimmed by \(\frac { 3 }{ 10 }\) cm.

Question 7.
Ritu ate \(\frac { 3 }{ 5 }\) part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat? Who had the larger share? By how much?
Answer:
Ritu ate \(\frac { 3 }{ 5 }\) part of an apple, therefore part of the apple ate by Somu = 1 – \(\frac { 3 }{ 5 }\)
= \(\frac{5-3}{5}=\frac{2}{5}\)
\(\frac{3}{5}>\frac{2}{5}\)
∴ Ritu had the larger share by = \(\frac{3}{5}-\frac{2}{5}\)
= \(\frac{3-2}{5}=\frac{1}{5}\)
Thus, Ritu had the larger share by \(\frac{1}{5}\) part of an apple.

Question 8.
Michael finished colouring a picture in \(\frac{7}{12}\) hour. Vaibhav finished colouring the
same picture in \(\frac{3}{4}\) hour. Who worked longer? By what fraction was it longer?
Answer:
Time taken by Michael = \(\frac{7}{12}\) hour
Time taken by Vaibhav = \(\frac{3}{4}\) hour.

Thus, Vaibhav worked longer by \(\frac{1}{6}\) hour.

These NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 14 Symmetry

Exercise 14.2

Question 1.
Which of the following figures have rotational symmetry of order more than 1:

Answer:
The figures (a), (b), (d), (e) and (f) have rotational symmetry of order more than 1.

Question 2.
Give the order of rotational symmetry for each figure:

Answer:
(a) → 2
(b) → 2
(c) → 3
(d) → 4
(e) → 4
(f) → 5
(g) → 6
(h) → 3
Explanation

Let us mark a point P as show in the figure (i) it requires two rotations each though 180° about the point (x) to come back to its original position.
∴ It has a rotational symmetry of order 2.

Mark a point P as shown in figure (i). It requires two rotations, each through an angle of 180° about the marked point (x) to come back to its original position.
Thus, it has a rotational symmetry of order 2.

(c) Mark a P as shown in figure (i). It requires three rotations each through an angle of 120° about the marked point (x) to come back to its original position.
Thus, it has a rotational symmetry of order 3.

(d) Mark a point P as shown in the figure. It requires four rotations, each through an angle of
900 about the marked point (x) to come back to its original position.
Thus it has a rotational symmetry of order 4.

(e) The figure requires four rotations each of 90°, about the marked point (x) to come back to its original position.
∴ It has a rotational symmetry of order 4.

(f) The figure is a regular pentagon. It requires five rotations, each though an angle of 72° about the marked point to come back to its original position.
∴ It has rotational symmetry of order 5.

(g) The given figure requires six rotations, each though angle of 60°; about the marked point (x) to come back to its original position.
∴ Thus, it has a rotational symmetry of order 6.

(h) The given figure requires three rotations each through an angle of 120°, about the marked point (x) to come back to its original position.
∴ It has a rotational symmetry of order 3.

These NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 14 Symmetry

Exercise 14.1

Question 1.
Copy the figures with punched holes and find the axes of symmetry for the following:
Answer:

Question 2.
Given the line (s) of symmetry. Find the other hole (s):
Answer:

Question 3.
In the following figures, the mirror line (i.e., the line of symmetry) is given as a dotted line. Complete each figure performing reflection in the dotted (mirror) line. (You might perhaps place a mirror along the dotted line and look into the mirror for the image). Are you able to recall the name of the figure you complete?
Answer:

Question 4.
The following figures have more than one line of symmetry. Such figures are said to have multiple lines of symmetry.

Identify multiple lines of symmetry, if any, in each of the following figures :
Answer:

Question 5.
Copy the figure given here.

Take any one diagonal as a line of symmetry and shade a few more squares to make the figure symmetric about a diagonal. Is there more than one way to do that? Will the figure be symmetric about both the diagonals?
Answer:
Yes, there is more than one way to make the figure symmetric.
(i) Let us take the diagonal BD and shade the squares as shown in the figure to make the figure symmetric about BD.(ii) Similarly, thefigureissymmetricaboutthediagonal AC. Thus, the figure is symmetric about both the diagonals.
(iii) The figure is symmetric about EF and GH also.

Question 6.
Copy the diagram and complete each shape to be symmetric about the mirror line (s)
Answer:

Question 7.
State the number of lines of symmetry for the following figures :
(a) An equilateral triangle
(b) An isosceles triangle
(c) A scalene triangle
(d) A square
(e) A rectangle
(f) A rhombus
(g) A parallelogram
(h) A quadrilateral
(i) A regular hexagon
(j) A circle
Answer:

Question 8.
What letters of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about.
(a) a vertical mirror
(b) a horizontal mirror
(c) both horizontal and vertical mirrors
Answer:

Question 9.
Give three examples of shapes with no line of symmetry.
Answer:

(b) A scalene triangle
(c) A parallelogram

Question 10.
What other name can you give to the line of symmetry of
(a) an isosceles triangle? (b) a circle?
Answer:
(a) Median of an isosceles triangle.
(b) Diameter of a circle.

These NCERT Solutions for Class 7 Maths Chapter 1 Integers InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 1 Integers InText Questions

NCERT In-text Question Page No. 2
Question 1.
A number line representing integers is given below:

-3 and -2 are marked by E and F respectively. Which integers are marked by B, D, H, J, M and O?
Answer:
Let us complete the given number line such that integers marked by various alphabets are shown

The integer marked by B = -6
The integer marked by D = -4
The integer marked by H = 0
The integer marked by J = 2
The integer marked by M = 5
The integer marked by O = 7

Question 2.
Arrange 7, -5, 4, 0 and -4 in ascending order and then mark them on a number line to check your answer.
Answer:
(i) Since every positive integer is greater than 0.
(ii) every negative integer is less than 0
-5 < (-4) < 0 < 4 < 7
ascending order is : -5, -4, 0, 4, 7

NCERT In-text Question Page No. 3
Question 1.
We have have done various patterns with numbers in our previous class. Can you find a pattern for each of the following? If yes, complete them.
(a) 7, 3, -1, -5, -9, -13, -17
(b) -2, -4, -6, -8, -10, -12, -14
(c) 15, 10, 5, 0, -5, -10, -15
(d) -11,-8, -5, -2, 1, 4, 7

NCERT In-text Question Page No. 8
Question 1.
Write a pair of integers whose sum gives
(a) a negative integer
(b) zero
(c) an integer smaller than both the integer.
(d) an integer smaller than only one of the integers.
(e) an integer greater than both the integers.
Answer:
(a) -7 and 2
sum = -7 + 2 = -5 (negative integer)

(b) -13 and 13
sum = -13 + 13 = 0

(c) – 13 and (- 7)
sum = – 13 + (-7) = – 13 – 7 = -20
(- 20 is smaller than – 13 and (-7))

(d) 7 and -5
sum = 7 + (-5) = 7 – 5 = 2
(2 is smaller than 7 only)

(e) 19 and 21
sum = 19 + 21 = 40
(40 is greater than both 19 and 21)

Question 2.
Write a pair of integers whose difference gives
(a) a negative integer.
(b) zero
(c) an integer smaller than both the integers.
(d) an integer greater than only one of the integers.
(e) an integer greater than both the integers.
Answer:
(a) 7 and -12
Difference = – 12 – (7) = -12 -7
= -19 (is a negative integer)

(b) (-3) and (-3)
Difference = -3 – (-3) = – 3 + 3 = 0

(c) 5 and 9 Difference = 9 – 5
= 4 (4 is smaller than 9 as well as 5)

(d) 16 and 5 Difference =16 – 5
= 11 (11 is greater than 5)

(e) 15 and-6
Difference = 15 – (-6) = 15 + 6
= 21 (21 is greater than 15 as well as -6)

NCERT Index Question Page No. 10
Question 1.
Using number line, find:
(i) 4 × (-8)
(ii) 8 (-2)
(iii) 3 × (-7)
(iv) 10 × (-1)
Answer:
(i) 4 × (-8)

From the number line, we have:
(- 8) + (-8) + (-8) + (-8) = -32
∴ 4 × (-8) = -32

(ii) 8 × (-2)

From the number line, we have:
(-2) + (-2) + (-2) + (-2) + (-2) + (-2) + (-2) + (-2) = -16
∴ 8 × (-2) = -16

(iii) 3 × (-7)

From the number line, we have:
(-7) + (-7) + (-7) = -21
3 × (-7) = -21

(iv) 10 × (-1)

From the number line, we have:
(-1) + (-1) + (-1) + (-1) + ( -1) + ( -1) + (-1) + ( -1) + (-1) + (-1) = -10
∴ 10 × (-1) =-10

NCERT In-text Question Page No. 10
Question 1.
Find:
(i) 6 × (-19)
(ii) 12 × (-32)
(iii) 7 × (-22)
Answer:
(i) 6 × (-19)= – (6 × 19)
= -(114) = -114

(ii) 12 × (-32) = – (12 × 32)
= -(384) = -384

(iii) 7 × (-22) = – (7 × 22)
= -(154) = – 154

NCERT In-text Question Page No. 11
Question 1.
Find:
(a) 15 × (-16)
(b) 21 × (- 32)
(c) (-42) × 12
(d) (-55) × 15
Answer:
(a) 15 × (-16) = -(15 × 16) = -[240] = -240
(b) 21 × (-32) = -(21 × 32) = -[672] = -672
(c) (-42) × 12 = -(42 × 12) = -[504] = -504
(d) (-55) × 15 = -(55 × 15) = -[825] = -825

Question 2.
Check if
(a) 25 × (-21) = (-25) × 21.
(b) (-23) × 20 = 23 × (-20).
Write five more such examples.
Answer:
(a) L.H.S.= 25 × (-21) – (25 × 21)
= – [525] = -525
R.H.S.= (-25) × 21 = – [25 × 21]
= – [525] = -525
∵ L.H.S.= R.H.S.
∴ 25 × (-21)= (-25) × 21

(b) L.H.S.= (-23) × 20 = – [23 × 20]
= – [460] = -460
R.H.S.= 23 × (-20) = – [23 × 20]
= – [460] = – 460
∵ L.H.S.= R.H.S.
∴ (-23) × 20= 23 × (-20)

Other examples:
(i) (-9) × 18 = 9 × (-18)
(ii) 51 × (-40)= (-51) × 40
(iii) (-12) = (-51) × 40
(iv) (-20) × 25= 20 × (-25)
(v) 16 × (-15) = (-16) × 15

NCERT In-text Question Page No. 12
Question 1.
(i) Starting from (-5) × 4, find (-5) × (-6)
(ii) Starting from (-6) × 3, find (-6) × (-7)
Answer:
(i) Look at the following pattern:
(-5) × 4 = – [5 × 4] = -20
(-5) × 3 = – [5 × 3] = -15 = -20 + 5
(-5) × 2 = – [5 × 2] = -10 = -15 + 5
(-5) × 1 = – [5 × 1] = -5 = -10 + 5
(-5) × 0 = – [5 × 0] = 0 = -5 + 5

From this pattern, we have:
(-5) × (-l) = 0 + 5 = 5
(-5) × (-2) = 5 + 5 = 10
(-5) × (-3) = 10 + 5 = 15
(-5) × (-4) = 15 + 5 = 20
(-5) × (-5) = 20 + 5 = 25
(-5) × (-6) = 25 + 5 = 30
Thus, (-5) × (-6) = 30

(ii) Look at the following pattern:
-6 × 3 = – [6 × 3] = – 18
-6 × 2 = – 12 = – 18 + 6
-6 × 1 = -6 = – 12 + 6
-6 × 0 = 0 = -6 + 6
From this pattern, we have:
-6 × (-1) = 0 + 6 = 5
-6 × (-2) = 6 + 6= 12
-6 × (-3) = 12 + 6= 18
-6 × (-4) = 18 + 6 = 24
-6 × (-5) = 24 + 6 = 30
-6 × (-6) = 30 + 6 = 36
-6 × (-7) = 36 + 6 = 42
Thus, (-6) × (-7) = 42

NCERT In-text Question Page No. 12
Question 1.
(i) Find: (-31) × (-100), (-25) × (-72), (-83) × (-28)
Answer:
We multiply the two negative integers as whole numbers and put the positive sign (+) before the product.
(-31) × (-100) = + [31 × 100] = + [3100] = 3100
(-25) × (-72) = + [25 × 72] = + [1800] = 1800
(-83) × (-28) = + [83 × 28] = + [2324] = 2324

NCERT In-text Question Page No. 14
Question 1.
(i) The product (-9) × (-5) × (-6) × (-3) is positive whereas the product (-9) × (-5) × 6 × (-3) is negative. Why?
(ii) What will be the sign of the product if we multiply together:
(a) 8 negative integers and 3 positive integers?
(b) 5 negative integers and 4 positive integers?
(c) (-1), twelve timers?
(d) (-1), 2m times, m is a natural number?
Answer:
We know that, if the number of negative integers in a product is even, then the product is a positive integer; if the number of negative integers in a product is odd, then the product is a negative integer.

∴ (i) The product (-9) × (-5) × (-6) × (-3) is positive because an even number of negative integers are multiplied.
The product (-9) × (-5) × 6 × (-3) is negative because an odd number of negative integers are multiplied.

(ii) (a) Positive [ ∵ Product of 8 negative integers is positive]
(b) Negative [ ∵ Product of 5 negative integers is positive]
(c) Positive [ ∵ 12 even and product of even number of negative]
(d) Positive [ ∵ 2m is an even number]

NCERT In-text Question Page No. 18
Question 1.
(i) Is 10 × [(6 + (-2)] = 10 × 6 + 10 × (-2)?
(ii) Is (-15) × [(-7 + (-1)] = (-15) × (-7) + (-15) × (-1)?
Answer:
(i) Yes, [ ∵ a × (b + c) = a × b + a × c]
10 × [6 + (-2) = 10 × [6 – 2]
= 10 × 4 = 40
And, 10 × 6 + 10 × (-2) = 60 – 20 = 40.
Thus, 10 × [6 + (-2) = 10 × 6 + 10 × (-2)

(ii) Yes, [ v a × (b + c) = a × b + a × c]
(-15) × [(-7 + (-1)] = (-15) × [-7 -1]
= (-15) × (-8)
= (+) (15 × 8)
= 120.
And, (-15) × (-7) + (-15) × (-1)
= (+) (15 × 7) + (+) (15 × 1)
= 105 + 15 = 120.
Thus, (-15) × [(-7) + (-1)]

Question 2.
(i) Is 10 × [6 – (-2)] = 10 × 6 – 10 × (-2)?
(ii) Is (- 15) × [(-7) – (- 1)]
= (-15) × (-7) – (-15) × (-1)?
Answer:
(i) L.H.S = 10 × [6 – (-2)]
= 10 × [6 + 2] = 10 × 8 = 80
R.H.S = 10 × 6 – 10 × (- 2)
= 60 – (-20) = 60 + 20 = 80
∴ L.H.S = R.H.S

(ii) L.H.S = (-15) × [(-7) – (- 1)]
= (-15) × [-7 + 1]
= (-15) × (- 6) = 90
R.H.S = (-15) × (-7) – (-15) × (- 1)
= 105 – (15) = 105 – 15 = 90
∴ L.H.S = R.H.S

NCERT In-text Question Page No. 19
Question 1.
By using distributive property, find: (-49) × 18; (-25) × (-31); 70 × (-19) + (-1) × 70.
Answer:
(i) (-49) × 18
∵ 18 = 10 + 8
∴ (-49) × 18
= (-49) × [10 + 8]
= (-49) × 10 + [-49) × 8
[using distributivity]
= -490 + (-49) [10 – 2] [∵ 8 = 10 – 2]
= -490 + (-49) × 10 – (-49) × 2
= -490 +(-490)+ 98
= -980 + 98 = -882

(ii) (-25) × (-31)
v -31 = (-30) + (-1)
.-. (-25) × (-31)
= (-25) × (-30) + (-1)]
= (- 5) × (-30) + (-25) × (-1)
[using distributivity] = + (25 × 30) + ((25 × 1)]
= 750 + 25 = 775

(iii) 70 × (-19) + (-1) × 70:
Y a × b + a × c = a × [b + c]
70 × (-19) + (-1) × 70
70 × [(-19) + (-1)] = 70 × [-20]
-[70 × 20] = -1400

NCERT In-text Question Page No. 22
Question 1.
Find:
(a) (-100)+ 5
(b) (-81) + 9
(c) (-75) + 5
(d) (-32) + 2
Answer:
(a) (-100) + 5 = -(100 + 5) = -(20) = -20
(b) (-81) + 9 = -(81 + 9) = -(9) = -9
(c) (-75) + 5 = -(75 + 5) = -(15) = -15
(d) (-32) + 2 = -(32 + 2) = -(16) = -16

NCERT In-text Question Page No. 23
Question 1.
Find:
(a) 125 ÷ (-25)
(b) 80 ÷ (-5)
(c) 64 ÷ + (-16)
Answer:
We know that to divide a positive integer by a negative integer, we first divide them as whole numbers and them put a minus sign (-) before the quotient.
(a) ∵ 125 ÷ 25 = 5

(b) ∵ 80 ÷ 5 = 16
∴ 125 ÷ (-25) = -5
∴ 80 ÷ (-5) = -16

(c) ∵ 64 ÷ 16 = 4
∴ 64 ÷ (-16) = -4

NCERT In-text Question Page No. 23
Question 1.
Find:
(a) (-36) ÷ (-4)
(b) (-201) ÷ (-3)
(c) (-325) ÷ (-13)
Answer:
To divide a negative integer by a negative integer, we first divide them as whole numbers and them put a positive sign (+) before the quotient.
(a) ∵ 36 ÷ 4 = 9
∴ (-36) ÷ (-4) = 9

(b) ∵ 201 ÷ 3 = 67
∴ (-201) ÷ (-3) = 67

(c) ∵ 325 ÷ 13 = 25
∴ (-325) ÷ (-13) = 25

NCERT In-text Question Page No. 24
Question 1.
Is (i) 1 ÷ a = 1 and (ii) a ÷ (-1) = -a for any integer a? Take different values of a and check.
Answer:
(i) Let us take a = -1, 1, 2, 3,…
For a = -1,
L.H.S. = 1 ÷ (-1) =-l [ ∵ 1 ÷ 1 = 1]
R.H.S. = 1
i. e. L.H.S. ≠ R.H.S.
For a = 2,
L.H.S. = 1 ÷ a = 1 ÷ 2 = \(\frac { 1 }{ 2 }\) ≠ R.H.S.
For a = 3,
L.H.S. = 1 ÷ a = 1 ÷ 3 = \(\frac { 1 }{ 3 }\) ≠ R.H.S.
Thus, 1 ÷ a = 1 is true only for a = 1

(ii) Let us take a = 1,2,3,…
For a = 1,
L.H.S. = a ÷ (-1) = 1 ÷ (-1) = -2
R.H.S. = -a = -1
i.e. L.H.S. = R.H.S.
For a = 2,
L.H.S. = a ÷ (-1) = 2 ÷ (-1) = -2
R.H.S. = a = -2
i.e. L.H.S. = R.H.S.
For a = 7,
L.H.S. = a ÷ (-1) = 3 ÷ (-1) = -3
R.H.S. = -a = -3
i.e. L.H.S. = R.H.S.
For a = 7,
L.H.S. = a ÷ (-1) = 7 + (-1) = -7
R.H.S. = -a = -7
i.e. L.H.S. = R.H.S.
i..e. For every integer, we have a ÷ (-1) = -a