NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.4

NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.4 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.4. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-1-ex-1-4/

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 1
Chapter NameReal Numbers
ExerciseEx 1.4
Number of Questions Solved3
CategoryNCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.4

Ex 1.4 Class 10 Question 1.
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.4 1
Solutions:
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.4 2
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.4 3

Ex 1.4 Class 10 NCERT Solutions Question 2.
Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.
Solutions:
In Question 1, (i), (ii), (iv), (vi), (viii) and (ix) are having terminating decimal expansion.
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.4 4
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.4 5

Class 10 Ex 1.4 Question 3.
The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form \(\frac { p }{ q }\), what can you say about the prime factors of q?
(i) 43.123456789
(ii) 0.120120012000120000…
(iii) 43. \(\overline { 123456789 }\)
Solutions:
(i) 43.123456789
Since the decimal expansion terminates,so the given real number is rational and therefore of the form \(\frac { p }{ q }\)
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.4 6
Here, q = 29 x 59, So prime factorisation of q is of the form 2n x 5m.

(ii) 0.120120012000120000…
Since the decimal expansion is neither terminating nor non-terminating repeating, therefore the given real number is not rational.

(iii) 43. \(\overline { 123456789 }\)
Since the decimal expansion is non-terminating repeating, therefore the given real number is rational and therefore of the form \(\frac { p }{ q }\)
Let x = 43. \(\overline { 123456789 }\) = 43.123456789123456789… ….(i)
Multiply both sides of (i) by 1000000000, we get
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.4 7

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

NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers Ex 1.1 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers Ex 1.1.

BoardCBSE
TextbookNCERT
ClassClass 8
SubjectMaths
ChapterChapter 1
Chapter NameRational Numbers
ExerciseEx 1.1
Number of Questions Solved11
CategoryNCERT Solutions

NCERT Solutions for Class 8 Maths will help you to score more marks in your CBSE board Examination.

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

Question 1.
Using appropriate properties find:
(i) \(-\frac { 2 }{ 5 } \times \frac { 3 }{ 5 } +\frac { 5 }{ 2 } -\frac { 3 }{ 5 } \times \frac { 1 }{ 6 } \)
(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.
NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers Ex 1.1 1
NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers Ex 1.1 2
NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers Ex 1.1 3

Question 2.
Write the additive inverse of each of the following:
NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers Ex 1.1 4
Solution.
(i) \(\frac { 2 }{ 8 } \)
Additive inverse of \(\frac { 2 }{ 8 } \) is \(\frac { 2 }{ 8 } \)

(ii) \(-\frac { 5 }{ 9 } \)
\(\frac { -6 }{ -5 } =\frac { 6 }{ 5 } \)
Additive inverse of \(\frac { -6 }{ -5 } \) is \(\frac { -6 }{ 5 } \)

(iii) \(\frac { -6 }{ -5 } \)
\(\frac { -6 }{ -5 } \)=\(\frac { 6 }{ 5 } \)
Additive inverse of \(\frac { -6 }{ -5 } \) is \(\frac { -6 }{ 5 } \)

(iv) \(\frac { 2 }{ -9 } \)
Additive inverse of \(\frac { 2 }{ -9 } \) is \(\frac { 2 }{ 9 }\)

(v) \(\frac { 19 }{ -6 } \)
Additive inverse of  \(\frac { 19 }{ -6 } \) is \(\frac { 19 }{ 6 }\)

Question 3.
Verify that – (-x) = x for :
(i) \(x=\frac { 11 }{ 15 } \)
(ii) \(x=-\frac { 13 }{ 17 } \)
Solution.
NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers Ex 1.1 5

Question 4.
Find the multiplicative inverse of the following:
NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers Ex 1.1 6
Solution.
NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers Ex 1.1 7

Question 5.
Name the property under multiplication used in each of the following:
(i) \(\frac { -4 }{ 5 } \times \left( 1 \right) =1\times \frac { -4 }{ 5 } =-\frac { 4 }{ 5 } \)
(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) 1 is the multiplicative identity
(ii) Commutativity of multiplication
(iii) Multiplicative inverse.

Question 6.
Multiply \(\frac { 6 }{ 13 } \) by the reciprocal of \(\frac { -7 }{ 16 } \)
Solution.
Reciprocal of \(\frac { -7 }{ 16 } \) is \(\frac { -16 }{ 7 } \)
Now,
\(\frac { 6 }{ 13 } \times \frac { -16 }{ 7 } =\frac { 6\times \left( -16 \right) }{ 13\times 7 } =\frac { -96 }{ 91 } \)

Question 7.
Tell what property allows you to compute : \(\frac { 1 }{ 3 } \times \left( 6\times \frac { 4 }{ 3 } \right) \) as \(\left( \frac { 1 }{ 3 } \times 6 \right) \times \frac { 4 }{ 3 } \)
Solution.
Associativity.

Question 8.
Is the \(\frac { 8 }{ 9 } \) multiplicative inverse of \(-1\frac { 1 }{ 8 } \) ? Why or why not?
Solution.
\(-1\frac { 1 }{ 8 } =-\frac { 9 }{ 8 } \)
Now, \(\frac { 8 }{ 9 } \times \frac { -9 }{ 8 } =-1\neq 1\)
So, No ; \(\frac { 8 }{ 9 } \) is not the multiplicative inverse of \(-1\frac { 1 }{ 8 } \left( =-\frac { 9 }{ 8 } \right) \) because the product of \(\frac { 8 }{ 9 } \) and -13(-) and \(-1\frac { 1 }{ 8 } \left( =-\frac { 9 }{ 8 } \right) \) is not 1.

Question 9.
Is 0.3 the multiplicative inverse of \(3\frac { 1 }{ 3 }\) ? Why or why not?
Solution.
Yes ; 0.3 is the multiplicative inverse of \(\frac { 10 }{ 3 } \) because
\(\frac { 3 }{ 10 } \times \frac { 10 }{ 3 } =\frac { 3\times 10 }{ 10\times 3 } =\frac { 30 }{ 30 } =1\)

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 own reciprocals.
(iii) The rational number 0 is equal to its negative.

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 \(\frac { 1 }{ x } \), where \(x\neq 0\)
(v) The product of two rational numbers is always a.………
(vi) The reciprocal of a positive rational number is……….
Solution.
(i) Zero has no reciprocal.
(ii) The numbers 1 and -1 are their own reciprocals.
(iii) The reciprocal of – 5 is \(-\frac { 1 }{ 5 } \)
(iv) Reciprocal of \(\frac { 1 }{ x } \), where \(x\neq 0\) is x.
(v) The product of two rational numbers is always a rational number.
(vi) The reciprocal of a positive rational number is positive.

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NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1

NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1.

BoardCBSE
TextbookNCERT
ClassClass 7
SubjectMaths
ChapterChapter 1
Chapter NameIntegers
ExerciseEx 1.1
Number of Questions Solved10
CategoryNCERT Solutions

NCERT Solutions for Class 7 Maths will help you to score more marks in your CBSE board Examination.

NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1

Question 1.
Following number line shows the temperature in degree Celsius (°C) at different places on a particular day:
NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1 1
(a) Observe this number line and write the temperature of the places marked on it.
(b) What is the temperature difference between the hottest and the coldest places among the above?
(c) What is the temperature difference between Lahul-Spiti and Srinagar?
(d) Can we say temperature of Srinagar and Shimla taken together is less than the temperature at Shimla? Is it also less than the temperature at Srinagar?
Solution:
(a)
NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1 2
(b) The hottest place is Bangalore (22 °C) and the coldest place is Lahulspiti (-8°C). The temperature difference between the hottest and the coldest places
= 22 °C – (-8 °C)
= 22 °C + 8 °C
= 30 °C

(c) Temperature difference between Lahul-Spiti and Srinagar
= Temperature of Srinagar
– Temperature of Lahul-Spiti
= – 2°C – (- 8°C)
= – 2°C + 8°C
= 6°C

(d) Yes, we can say that the temperature of Srinagar and Shimla took together is less than the temperature at Shimla as -2 + 5 = 3 and 3 < 5.
This temperature is not less than the temperature at Srinagar.

We have designed Class 7 Maths NCERT Solutions on LearnInsta.com, based on CBSE Syllabus. These Solutions are very helpful to score more marks in your exams.

Question 2.
In a quiz, positive marks are given for correct answers and negative marks are given for incorrect answers. If Jack’s scores in five successive rounds were 25, -5, -10, 15 and 10, what was his total at the end?
Solution:
Total at the end
= 25 + (- 5) + (- 10) + 15 + 10
= (25 + 15 + 10) + {(- 5) + (- 10))
= 50 + (- 15) = 35

Question 3.
At Srinagar, the temperature was – 5°C on Monday and then it dropped by 2°C on Tues¬day. What was the temperature of Srinagar on Tuesday? On Wednesday, it rose by 4°C. What was the temperature on this day?
Solution:
At Srinagar, the temperature was on Monday = -5 °C
Since the temperature was dropped by 2 °C on Tuesday, therefore, the temperature was on Tuesday = (-5-2) °C = -7°C
Also, on Wednesday the temperature rose by 4 °C.
∴ Temperature on Wednesday= (-7 + 4) °C = -3 °C

Question 4.
A plane is flying at the height of 5000 m above sea level. At a particular point, it is e×actly above a submarine floating 1200 m below sea level. What is the vertical distance between them?
NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1 3
Solution:
The vertical distance between the plane and submarine.
= 5000 m – (- 1200 m)
= 5000 m + 1200 m
= 6200 m.

Question 5.
Mohan deposits ₹ 2,000 in his bank account and withdraws ₹1,642 from it, the next day. If the withdrawal of the amount from the account is represented by a negative integer, then how will you represent the amount deposited?
Find the balance in Mohan’s account after the withdrawal.
Solution:
The amount deposited will be represented by a positive integer.
Balance in Mohan’s account after withdrawal
= (+ ₹ 2000) + (- ₹ 1642)
= ₹ (2000-1642)
= ₹ 358

Question 6.
Rita goes 20 km towards the east from point A to point B. From B, she moves 30 km towards the west along the same road. If the distance towards the east is represented by a positive integer then, how will you represent the distance travelled towards the west? By which integer will you represent her final position from A?
NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1 4
Solution:
The distance towards the west = -30 km
Her final position from A
= + 20 km + (- 30) km
= – (30 – 20) km
= – 10 km.

Question 7.
In a magic square each row, column, and diagonal have the same sum. Check which of the following is a magic square.
NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1 5
Solution:
NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1 6
I. Row, Sum = 5 + (- 1) + (- 4)
= 5 + (- 5) = 0
II. Row, Sum = (- 5) + (- 2) + 7
= (- 7) + 7 = 0
III. Row, Sum = 0 + 3 + (- 3)
= 0 + 0 = 0
I. Column, Sum = 5 + (- 5) + 0
= 0 + 0 = 0
II. Column, Sum = (- 1) + (- 2) + 3
= (- 3) + 3 = 0
III. Column, Sum = (- 4) + 7 + (- 3)
= 7 + (- 4) + (- 3)
= 7 + (- 7) = 0
One Diagonal, Sum = 5 + (- 2) + (- 3) = 5 + (- 5) = 0
Other Diagonal, Sum = 0 + (- 2) + (- 4) = 0 + (- 6) = -6 ≠ 0
Therefore, the given square is not a magic square.
NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1 7
I. Row, Sum = 1 + (- 10) + 0 = – 9
II. Row, Sum = (- 4) + (- 3) + (- 2) = – 9
III. Row, Sum = (- 6) + 4 + (- 7)
= 4 + (- 6) + (- 7)
= 4 + (- 13)
= – (13-4) = -9
I. Column, Sum = 1 + (- 4) + (- 6)
= 1 + (- 10)
= -(10 – 1) = – 9
II. Column, Sum = (- 10) + (- 3) + 4
= (-13)+ 4
= – (13-4) = -9
III. Column, Sum = 0 + (- 2) + (- 7)
= 0 + (- 9) = – 9
One Diagonal, Sum = 1 + (- 3) + (- 7)
= 1 + (- 10)
= -(10-1) = -9
Other Diagonal, Sum = (- 6) + (- 3) + 0
= (- 9) + 0 = – 9
Since each row, column, and diagonal ‘ have the same sum, therefore, the given square is a magic square.

Question 8.
Verify a – (- b) = a + b for the following values of a and b.
(i) a = 21, b = 18
(ii) a = 118, b = 125
(iii) a = 75, b = 84
(iv) a = 28, b = 11.
Solution:
(i) a = 21, b = 18
L.H.S. = a – (- b) = 21 – (- 18) = 21 + 18 = 39 …..(1)
R.H.S. = a + b = 21 + 18 = 39 …..(2)
From (1) and (2), we get a -(-b) = a + b

(ii) a = 118, b = 125
L.H.S. = a – (- b) = 118 – (- 125) = 118 + 125 = 243 …(1)
R.H.S. = a + b = 118 + 125 = 243 …(2)
From (1) and (2), we get a -(- b) = a + b

(iii) a = 75, b = 84
L.H.S. = a – (- b) = 75 – (- 84) = 75 + 84 = 159 …(1)
R.H.S. = a + b = 75 + 84 = 159 …(2)
From (1) and (2), we get a – (- b) = a + b

(iv) a = 28, b = 11
L.H.S. = a – (- b) = 28 -(-11) = 28 + 11 = 39 …(1)
R.H.S. = a + b = 28 + 11 = 39 …(2)
From (1) and (2), we get a – (- b) = a + b.

Question 9.
Use the sign of >, < or = in the bo× to make the statements true.
(a) (- 8) + (-4) …… (- 8) – (- 4)
(b) (-3) +7 – (19) …… 15-8 +(-9)
(c) 23 – 41 + 11 …… 23-41- 11
(d) 39 + (-24) – (15) …… 36+ (-52) – (- 36)
(e) – 231 + 79 + 51 …… -399 + 159 + 81.
Solution:
(a) L.H.S. = (- 8) + (- 4)
= – (8 + 4) = – 12
R.H.S. = (- 8) – (- 4)
= – 8 + 4 = – (8 – 4) = – 4
∴ (-8) + (-4) < (-8)-(-4)

(b) L.H.S. = (-3) + 7 – (19)
= + 4 – (19)
= + 4 – 19
= -15
R.H.S. = 15 – 8 + (- 9)
= 7 + (-9) = 7- 9 = -2
∴ (- 3) + 7 – (19) < 15 – 8 + (- 9)

(c) L.H.S. = 23 – 41 + 11
= 23 + 11- 41
= 34 – 41 = – (41 – 34)
= – 7
R.H.S. = 23 – 41 – 11
= 23 – (41 + 11)
= 23 – 52
= – (52 – 23)
= – 29
∴ 23 – 41 + > 23 – 41 – 11

(d) L.H.S. = 39 + (- 24) – (15)
= 39 – 24 – (15)
= 15 – (15) = 0
R.H.S. = 36 + (- 52) – (- 36)
= – (52 – 36) – (- 36)
= – 16 – (- 36)
= – 16 + 36 = 20
∴ 39 + (- 24) – (15) < 36 + (- 52) – (- 36)

(e) L.H.S. = – 231 + 79 + 51
= – 231 + 130
= – (231 – 130) = – 101
R.H.S. = – 399 + 159 + 81
= – 399 + 240
= – (399 – 240)
= – 159
∴ – 231 + 79 + 51 > – 399 + 159 + 81.

Question 10.
A water tank has stepped inside it. A monkey is sitting on the topmost step (i.e., the first step The water level is at the ninth step.
NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1 8
(i) He jumps 3 steps down and 2 steps up. In how many jumps will he reach the water level?
(ii) After drinking water he wants to go back. For this, he jumps 4 steps up and then jumps back 2 steps down in every move. In how many jumps will he reach back the top step?
(iii) If the number of steps moved down is represented by negative integers and the number of steps moved up by positive integers, represent his moves in part (i) and (ii) by completing the following;
(a) -3 + 2 – …………. = -8
(b) 4 – 2 + = 8.
In (a) the sum (-8) represents going down by eight steps. So, what will the sum 8 in (b) represent?
Solution:
(i) While going down the monkey jumps 3 steps down and then jumps back 2 steps up. To reach the water level he is to jump as under:
-3+ 2 -3 + 2 – 3 +2 – 3 + 2 – 3 + 2 – 3 = -8 Hence, he takes 11 jumps to reach the water level.
(ii) After drinking water, he jumps back as under to reach the top step as under : 4 – 2+ 4 – 2+ 4 = 8
Hence, he takes 5 jumps to reach back the top.
(iii) (a) – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 = -8
(b) 4 – 2 + 4 – 2 + 4 = 8
In (b) the sum 8 represents going up 8 steps.

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NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9.

BoardCBSE
TextbookNCERT
ClassClass 9
SubjectMaths
ChapterChapter 13
Chapter NameSurface Areas and Volumes
ExerciseEx 13.9
Number of Questions Solved3
CategoryNCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9

Question 1.
A wooden bookshelf has external dimensions as follows : Height = 110cm, Depth = 25cm, Breadth = 85cm (see figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing-is 20 paise per cm2 and the rate of pointing is 10 paise per cm2, find the total expenses required for palishing and painting the surface of the bookshelf.
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 img 1
Solution:
Here, l = 85 cm b = 25 cm and h = 110 cm
Area of the bookshelf of outer surface = 2 lb + 2bh + hl
= [2(85 x 25)+ 2(110 x 25)+ 85×110] cm2 = (4250 + 5500 + 9350) cm2 = 19100 cm2
Cost of polishing of the outer surface of bookshelf
= 19100 x \(\frac { 20 }{ 100 }\) = ₹ 3820
Thickness of the plank = 5 cm
Internal height of bookshelf = (110 – 2 x 5) = 100 cm
Internal depth of bookshelf = (25 – 5) = 20 cm
Internal breadth of bookshelf = 85 – 2 x 5 = 75 cm
Hence, area of the internal surface of bookshelf
= 2(75 x 20)+ 2(100x 20)+ 75×100
= 3000 + 4000 + 7500 = 14500 cm2
So, cost of painting of internal surface of bookshelf
= 14500 x \(\frac { 10 }{ 100 }\) = ₹1450
Hence, total costing of polishing and painting = 3820 + 1450= ₹ 5270

Question 2.
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in figure. Eight such spheres are-used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5 paise per cm2.
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 img 2
Solution:
It is obvious, we have to subtract the cost of the sphere that is resting on the supports while calculating the cost of silver paint.
Surface area to be silver paint
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 img 3

Question 3.
The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?
Solution:
Let d be the diameter of the sphere
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 img 4
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 img 5

We hope the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8.

BoardCBSE
TextbookNCERT
ClassClass 9
SubjectMaths
ChapterChapter 13
Chapter NameSurface Areas and Volumes
ExerciseEx 13.8
Number of Questions Solved10
CategoryNCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8

Question 1.
Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 cm
Solution:
(i) We have, r = 7 cm
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 img 1

Question 2.
Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm
(ii) 0.21 m
Solution:
(i) We have, d = 28 cm
⇒ r = 14 cm
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 img 2

Question 3.
The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3?
Solution:
We have, d = 4.2 cm ⇒ r = 2.1 cm
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 img 3
The density of the metal per cm3 = 8.9 g
The density of the metal 38.808 cm3 = 38.808 x 8.9 g = 345.39 g
Hence, the mass of the ball = 345.39 g

Question 4.
The diameter of the Moon is approximately one-fourth of the diameter of the Earth. What fraction of the volume of the Earth is the volume of the Moon?
Solution:
Let diameter of the Earth is d3.
We have, diameter of Moon (d1) = \(\frac { 1 }{ 4 }\) diameter of the Earth
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 img 4

Question 5.
How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Solution:
Diameter = 10.5 cm
⇒ Radius, r = 5.25 cm
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 img 5

Question 6.
A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
Solution:
Inner radius (r1) = 1 m -100 cm
Outer radius (r2) = (100 + 1)cm ( ∵ Inner radius + Thickness)
= 101 cm
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 img 6

Question 7.
Find the volume of a sphere whose surface area is 154 cm2.
Solution:
We have, surface area of a sphere = 154 cm2
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 img 7

Question 8.
A dome of a building is in the form of a hemisphere. From inside, it was white washed at the cost of ₹498.96. If the cost of white washing is ₹2.00 per square metre, find the
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.
Solution:
Cost of white washed = ₹ 498.96
Cost of white washing per square metre = ₹ 2.00
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 img 8

Question 9.
Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S’. Find the
(i) radius r’ of the new sphere,
(ii) ratio of S and S’.
Solution:
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 img 9

Question 10.
A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to fill this capsule?
Solution:
We have, diameter = 3.5 mm
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 img 10

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