Author name: Prasanna

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 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.6

Question 1.
The circumference of the base of a cylinderical vessel is 132 cm and its height is 25 cm. How many liters of water can it hold? (1000 cm³ = 1L)
Solution:
We have given that

Circumference of base of cylinder = 132 cm
2πr = 132
⇒ r = \(\frac{132 \times 7}{22 \times 2}\) = 21 cm
and height of cylinderical vessel = 25 cm
Volume of cylindrical vessel = πr²h
= \(\frac {22}{7}\) × 21 × 21 × 25
= 34650 cm³
Now, we know that 1000 cm³ = 1L
34650 cm³ = 34.65 L
Therefore, the cylindrical vessel can hold 34.65 liters of water.

Question 2.
The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm³ is wood has a mass of 0.6 g.
Solution:

Volume of required wood = volume of outer cylinder – volume of inner cylinder.
= π(14)² × 35 – π(12)² × 35
= \(\frac {22}{7}\) × 35((14)² – (12)²)
= \(\frac {22}{7}\) × 35 × (196 – 144)
= \(\frac {22}{7}\) × 35 × 52
= 5720 cm³
Now, we have given that, mass of 1 cm³ of wood = 0.6
Therefore, mass of 5720 cm³ of wood = 0.6 × 5720 g.
= 3432g
= 3.432 kg.

Question 3.
A soft drink is available in two packs — (i) a tin can with a rectangular base of length 5 cm and wide 4 cm having a height of 15 cm and (ii) a plastic cylinder with a circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much.
Solution:
In the first case,
The tin can with a rectangular base of length 5 cm and wide 4 cm having a height of 15 cm.
∴ The volume of this can = l × b × h
= 15 × 4 × 5
= 300 cm³ …….(i)

In the second case,
The circular plastic cylinder has a base diameter is 7 cm and the height of the cylinder is 10 cm.

∴ Volume of plastic cylinder = πr²h
= \(\frac {22}{7}\) × 3.5 × 3.5 × 10
= 385 cm³ ……(ii)
From (i) and (ii), it is clear that
The volume of the circular cylinder has a greater capacity of 85 cm³.

Question 4.
If the lateral surface of a cylinder is 94.2 cm³ and its height is 5 cm, then find
(i) radius of its base
(ii) volume of the cylinder.
Solution:
(i) We have given that lateral surface of cylinder = 94.2 cm³ and height is 5 cm.

Now, we know that, lateral surface area of cylinder = 2πrh
94.2 = 2 × 3.14 × r × 5
r = \(\frac{94.2}{2 \times 3.14 \times 5}\) = 3 cm
(ii) We know that
Volume of cylinder = 2πrh
= 3.14 × 3 × 3 × 5
= 141.3 cm³.

Question 5.
It costs Rs. 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs. 20 per m², Find.
(i) Inner curved surface area of the vessel.
(ii) radius of the base.
(iii) capacity of the vessel.
Solution:
(i) We have given that

Rate of painting = Rs. 20 per m²
and total cost of curved surface area of cylindrical vessel = Rs. 2200
It means that inner curved surface area of cylindrical vessel = \(\frac{2200}{20}\) = 110 m²

(ii) we know that curved surface ara of cylinder = 2πrh
⇒ 110 = 2 × \(\frac {22}{7}\) × r × 10
⇒ r = \(\frac{110 \times 7}{2 \times 22 \times 10}\) = 1.75 m

(iii) We know that volume of cylinder = πr²h
= \(\frac {22}{7}\) × 1.75 × 1.75 × 10
= 96.25 m³
Therefore, Capacity of the vessel = 96.25 m³ = 96.25 kl
(1 m³ = 1000 l =1 kl)

Question 6.
The capacity of a closed cylindrical vessel of height 1 m is 15.4 liters. How many square meters of the metal sheet would be needed to make it?
Solution:
We have given that
Capacity of closed cylindrical vessel = 15.41 = \(\frac{15.4}{1000}\) m³

and height of the cylindrical vessel = 1 m.
We know that,
volume of cylinder = πr²h
⇒ \(\frac{15.4}{1000}\) = \(\frac {22}{7}\) × r² × 1
⇒ r² = 0.0049
⇒ r = 0.07m
Now, Total surface area of cylindrical vessel = 2πr(h + r)
= 2 × \(\frac {22}{7}\) × 0.07 (1 + 0.07)
= 0.4708 m²
Therefore, a metal sheet would be needed to make the required cylinder is 0.4708 m².

Question 7.
A lead pencil consists of a cylinder of wood will a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.
Solution:
We have given that
Diameter of pencil = 7 mm
Radius of pencil = 3.5 mm
and height of pencil = 14 cm = 140 mm

Volume of pencil = πr²h
= \(\frac {22}{7}\) × 3.5 × 3.5 × 140
= 5390 mm³
Agian, we have
Diamter of lead = 1 mm
Radius of lead = 0.5 mm
and height of lead = 14 cm = 140 mm
Volume of lead = πr²h
= \(\frac {22}{7}\) × 0.5 × 0.5 × 140
= 110 mm³
Volume of graphite = 110 mm³ = 0.11 cm³
and volume of wood = 5390 – 110
= 5280 mm³
= 5.28 cm³

Question 8.
A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?
Solution:

We have given that
Diameter of pencil = 7 mm
Radius of cylindrical bowl = 3.5 cm
and height of soup to serve = 4 cm.
Volume of soup to serve one patient = πr²h
= \(\frac {22}{7}\) × 3.5 × 3.5 × 4
= 154 cm³
Therefore, volume of soup to serve 250 patients are = 250 × 154
= 38500 cm³.
= 38.5 litre.
Hence, Hospital has to prepare daily 38.5 litres of soup to serve 250 patients.

These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.5

Question 1.
A matchbox measure 4 cm × 2.5 cm × 1.5 cm. What will be the volume of packets containing 12 such boxes?
Solution:

We have given the dimension of match box = 4 cm × 2.5 cm × 1.5 cm
Volume of matchbox = l × b × h
= 4 × 2.5 × 1.5
= 15 cm³
Volume of 12 such matchbox = 12 × 15 = 180 cm³

Question 2.
A cuboidal water tank 6 m long, 5 m, wide, and 4.5 m deep. How many liters of water can it hold? (1 m³ = 10001)
Solution:
We have given that
length of cuboidal tank = 6m

Breadth of cubodical tank = 5 m
and height of cubodical tank = 4.5 m
Volume of cubodical tank = l × b × h
= 6 × 5 × 4.5
= 135 m³
Again, we have given that
1 m² = 10001
135m³ = 135 × 10001 = 1350001
Therefore, the cuboidal tank holds 135000 liters of water.

Question 3.
A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic meters of a liquid?
Solution:
We have given that
length of cuboidal vessel = 10 m
Breadth of cuboidal vessel = 8 m
and volume of cuboidal vessel = 380 m³
We know that,
Volume of cuboidal Vessel = l × b × h
⇒ 380 = 10 × 8 × h
⇒ h = 4.75 m
Therefore, the height of required cuboidal vessels = 4.75m.

Question 4.
Find the cost of digging a cuboidal pit 8m long 6 m broad and 3 m deep at the rate of Rs. 30 per m³.
Solution:
We have given that length of cuboidal pit = 8m

Breadth of cubodical pit = 6 m
and height of cubodical pit = 4 m
Volume of cubodical pit = l × b × h
= 8 × 6 × 3
= 144 m³
Rate od digging = Rs. 30 per m³
Total cost of digging tire cubodical pit = 30 × 144 = Rs. 4,320

Question 5.
The capacity of a cuboidal tank is 50,000 liters of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.
Solution:
We know that
1000 liters = 1 m³
50,000 liters = 50 m³

Now, we have given that length of cuboidal tank = 2.5m
and height of cuboidal tank = 10m
and volume of cuboidal tank = 50 m³
We know that
The volume of cuboidal tank = l × b × h
⇒ 50 = 2.5 × b × 10
⇒ b = 2 m
The breadth of the required tank = 2m.

Question 6.
A village, having a population of 4000, requires 1500 liters of water per head per day. It has a tank measuring 20m × 15m × 6m. For how many days will the water of this tank last.
Solution:
Volume of water tank = l × b × h
= 20 × 15 × 6
= 1800 m³

We know that
1 m³ = 1000 litres
1800 m³ = 1800 × 1000 = 18,00,000 litres
Therefore, capacity of water tank = 18,00,000 litres.
Now, One person requires 150 litres per day
4000 person requires = 4000 × 150 = 6,00,000 litres per day
No. of days will the water of tank last = \(\frac{18,00,000}{6,00,000}\) = 3 days.

Question 7.
A godown measures 40 m × 25 m × 10 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.
Solution:
Volume of godown = 40 × 25 × 10 = 10,000 m³
Therefore, total number of crates can be stored in the godown = \(\frac{10,000}{0.9375}\) = 10666.66 or 10,666.

Question 8.
A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.
Solution:
Volume of larger cube = (side)³
= (12)³
= 1728 cm³
Now, volume of each smaller cube = \(\frac {1}{8}\) × 1728
(side)³ = 216
side = \(\sqrt[3]{216}\) = 6 cm
The side of’new cube = 6 cm
Again, Surface area of larger cube = 6 × (side)²
= 6 × (12)²
= 864 cm²
and surface area of smaller cube = 6 × (side)²
= 6 × (6)²
= 216 cm²
Ratio between them is = 984 : 216 = 4 : 1

Question 9.
A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a min?
Solution:

We have given that
Rate of flowing = 2 km/h 100 m
= \(\frac{2 \times 1000}{60}\) m/min
= \(\frac{100}{3}\) m/min.
∴ Volume of water will fall into the sea in one min is = 40 × 3 × \(\frac{100}{3}\) = 4,000 m³.

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

These NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 1 Integers Exercise 1.4

Question 1.
Evaluate each of the following:
(a) (-30) ÷ 10
(b) 50 ÷ (-5)
(c) (-36) ÷ (-9)
(d) (-49) ÷ (49)
(e) 13 ÷ [(-2) + 1]
(f) 0 ÷ (-12)
(g) [(-31) ÷ [(-30) + (-1)]
(h) [(-36) ÷ 12] ÷ 3
(i) = (-6 + 5) ÷ [(-2 + 1)]
Answer:
(a) (-30) ÷ 10 = – (30 ÷ 10) = – (3) = -3

(b) 50 ÷ (-5) = – (50 ÷ 5) = – (10) = -10

(c) (-36) ÷ (- 9) = + (36 ÷ 9) = + 4

(d) (-49) ÷ 49 = – (49 ÷ 49) = – (1) = -1

(e) 13 ÷ [(-2) + 1] = 13 ÷ (-2 + 1)
= 13 ÷ (-1) = – (13 ÷ 1) = -13

(f) 0 ÷ (-12) = – (0 ÷ 12) = 0

(g) (-31) ÷ [(-30) + (-1) = (-31) ÷ [-30 -1]
-31 ÷ (-31) = + (31 ÷ 31) = + 1

(h) [(-36) ÷ 12] ÷ 3 = [-36 ÷ 12] ÷ 3
= – (36 ÷ 12) ÷ 3 = – (3) ÷ 3
= -3 ÷ 3 = – 1

(i) [(-6) + 5] ÷ [(-2) + 1]
= (-6 + 5) ÷ (-2 + 1) = – 1 ÷ (-1)
= + (1 ÷ 1) = 1

Question 2.
Verify that a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c) for each of the following values of a, b and c.
(a) a = 12, b = -4, c = 2
(b) a = (-10), b = 1, c = 1
Answer:
(a) Given a = 12, b = – 4 and c = 2
L.H.S = a + (b + c)
= 12 ÷ (-4 + 2) = 12 ÷ (- 2)
= – [12 ÷ 2] = – 6
R.H.S = (a ÷ b) + (a ÷ c)
= (12 + -4) + [12 ÷ 2]
= -(12 ÷ 4) + (6) = -3 + 6 = 3
∴ L.H.S ≠ R.H.S

(b) Here a = – 10, b = 1, c = 1
L.H.S = a ÷ (b + c)
= (-10) ÷ (1 + 1) = -10 ÷ 2
= -(10 ÷ 2) = -5
R.H.S = (a ÷ b) + (a ÷ c)
= (-10 ÷ 1) + (-10 ÷ 1)
= -(10 ÷ 1) + (-10) ÷ 1
= -10 -10 = -20
L.H.S ≠ R.H.S

Question 3.
Fill in the blanks :
Answer:
(a) 369 ÷ 1 = 369 (a ÷ 1 = a)
(b) (-75) ÷ 75 = -1 (-a ÷ a = -1)
(c) (-206) ÷ 206 = 1 (- a) ÷ (-a) = 1
(d) -87 ÷ -1 = 87 (-a ÷ – 1 = a)
(e) -87 ÷ = -87 [(-a) ÷ 1 = -a]
(f) -48 ÷ 48 = -1 (-a ÷ a = -1)
(g) 20 ÷ -10 = -2
(h) -12 ÷ 4 = -3

Question 4.
Write five pairs of integers (a, b) such that a ÷ b = -3. One such pair is (6, -2) because 6 ÷ (-2) = (-3).
Answer:
Five pairs of integers (a, b) such that a ÷ b = -3
(1) (3, -1)
(2) (-3, 1)
(3) (9, -3)
(4) (-9, 3)
(5) (-12, 4)

Question 5.
The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2°C per hour until midnight, at what time would the temperature be 8°C below zero? What would be the temperature at mid-night?
Answer:
Temperature at 12 noon = + 10°C
Rate of change in temperature = -2°C per hour
Number of hours from 12 noon to midnight = 12
∴ Change in temperature in 12 hours = 12° × (-2) = -24°C
Temperatureat mid-night= 10°C + (-24°C)
(ie 12 hours after 12 noon) = -14°C
Temperature difference between + 10°C and -8°C = + 10°C – (-8°C)
= 10°C + 8°C = 18°C
Time taken = 18 ÷ 2 = 9 hours
Temperature change of 18°C will take place in 9 hours from 12 noon
Time after 9 hours from 12 noon = 9 P.M.

Question 6.
In a class test (+3) marks are given for every correct answer and (-2) marks are given for every incorrect answer and no marks for not attempting any question.
(i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly?
(ii) Mohini scores -5 marks in this
test, though she has got 7 correct answers. How many questions has she attempted incorrectly?
Answer:
(i) Let ‘x’ be the number of incorrect question attempted by Radhika
According to the given question, we get
(+3) × 12 + x × (-2) = 20
36 – 2x = 20
2x = 36 – 20
2x = 16
x = 8
∴ Radhika attempted 8 incorrect questions.

(ii) Let ‘x’ be the number of incorrect question attempted by Mohini
According to the given question, we get
(+3) × 7 + x × (-2) = -5
21 – 2x = -5
2x = 21 + 5
2x = 26
x = \(\frac{26}{2}\) = 13
∴ Mohini attempted 13 incorrect questions.

Question 7.
An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach – 350 m?
Answer:
Present position of the elevator is at 10 m above the ground level;
Distance moved by the elevator below the ground level = 350 m
∴ Total distance moved by the elevator = 350 m + 10 m = 360 m
∴ The rate of descend = 6 m/min
Time taken = \(\frac{360}{6}\) minutes = 60 minutes 6 = 1 hour
Required time = 1 hour

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

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes InText Questions

NCERT In-text Question Page No. 277

Question 1.
Match the shape with the name:


Answer:
(i) – (a) Cuboid
(ii) – (d) Cuboid
(iii) – (a) Cone
(iv) – (c) Sphere
(v) – (f) Cube
(vi) – (e) Pyramid

NCERT In-text Question Page No. 281

Question 1.
Here you find four nets. There are two correct nets among them to make a tetrahedron. See if you can work out which nets will make a tetrahedron.

Answer:
(i) and (iii) are correct nets to make a tetrahedron.

NCERT In-text Question Page No. 286

Question 1.
Try to guess the number of cubes in the following arrangements.

Answer:
(i) There are 24 cubes.
(ii) There are 8 cubes.
(ii) There are 9 cubes.

NCERT In-text Question Page No. 287

Question 1.
Two dice are placed side by side as shown: Can you say what the total would be on the face opposite to

(a) 5 + 6 (b) 4 + 3
(Remember that in a die sum of numbers on opposite faces is 7.)
Answer:
(a) The total opposite to face 5 + 6 is 2 + 1, i.e. 3.
(b) The total opposite to face 4 + 3 is 3 + 4, i.e. 7.

Question 2.
Three cubes each with 2 cm edge are placed side by side to form a cuboid. Try to make an oblique sketch and say what would be its length, breadth and height.

Answer:
Length of the cuboid
= 2 cm + 2 cm + 2 cm = 6 cm.
Breadth of the cuboid = 2 cm.
Height of the cuboid = 2 cm.

NCERT In-text Question Page No. 291

Question 1.
For each solid, the three views (1), (2), (3) are given. Identify for each solid the corresponding top, front and side views.
Answer:


(a) (1) → Front (2) → Side (3) → Top
(b) (1) → Top (2) → Side (3) → Front
(a) (1) → Side (2) → Front (3) → Top
(d) (1) → Side (2) → Top (3) → Front

Question 2.
Draw a view of each solid as seen from the direction indicated by the arrow.

Answer: