Bhagya

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 1 Real Numbers Ex 1.4 Textbook Exercise Questions and Answers.

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

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 :
(i) \(\frac{13}{3125}\)

(ii) \(\frac{17}{8}\)

(iii) \(\frac{64}{455}\)

(iv) \(\frac{15}{1600}\)

(v) \(\frac{29}{343}\)

(vi) \(\frac{23}{2^{3} 5^{2}}\)

(vii) \(\frac{129}{2^{5} 5^{7} 7^{5}}\)

(viii) \(\frac{6}{15}\)

(ix) \(\frac{35}{50}\)

(x) \(\frac{77}{210}\)
Solution:
(i)Let x = \(\frac{13}{3125}\) ………..(1)
Compare (1) with x = \(\frac{p}{q}\)
Here p = 13 and q = 3125
Prime factors of q = 3125 = 5 × 5 × 5 × 5 × 5 = 5⁵ × 2⁰
which are of the form 2ⁿ × 5^m here n = 0, m = 5
which are non negative integers.
∴ x = \(\frac{13}{3125}\) have a terminating decimal expansion.

(ii) Let x = \(\frac{17}{8}\) …………………(1)
Compare (1) with x = \(\frac{p}{q}\)
Here p = 17 and q = 8
Prime factors of q = 8 = 2 × 2 × 2 = 2³ = 2³ × 5⁰

which are of the form 2ⁿ × 5^m here n = 3, m = 0
and these are non negative integers.
∴ x = \(\frac{17}{8}\) have a terminating decimal expansion.

(iii) Let x = \(\frac{64}{455}\) …………(1)
Compare (1) with x = \(\frac{p}{q}\)
Here p = 64, q = 455
Prime factors of q = 455 = 5 × 7 × 13 which are not of the form 2ⁿ × 5^m
∴ x = \(\frac{64}{455}\) has a non – terminating decimal expansion.

(iv) Let x = \(\frac{15}{1600}\) ……….(1)
Compare (1) with x = \(\frac{p}{q}\)
Here p = 15 and q = 1600
Prime factors of q = 1600
= 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 = 2⁶ × 5²
which are of the form 2ⁿ × 5^m, here n = 6, m = 2 and these are non negative integers.
∴ x = having terminating decimal expansion.

(v) Let x = \(\frac{29}{343}\) ………… (1)
Compare (1) with x = \(\frac{p}{q}\)
Here p = 29 and q = 343
prime factors of q = 343
= 7 × 7 × 7 = 7³
which are not of the form 2ⁿ × 5^m, here n = 0, m = 0
∴ x = \(\frac{29}{343}\) will have a non – terminating decimal expansion.

(vi) Let x = \(\frac{23}{2^{3} 5^{2}}\) ……….(1)
Compare (1) with x = \(\frac{p}{q}\)
Here p = 23, q = 2³5²
Prime factors of q = 2³5²
which are of the form 2ⁿ × 5^m, here n = 3, m = 2 and these are non negative integers.
∴ x = \(\frac{23}{2^{3} 5^{2}}\) will have a terminating decimal expansion.

(vii) Let x = \(\frac{129}{2^{5} 5^{7} 7^{5}}\) ……………….(1)
Compare (1) with x = \(\frac{p}{q}\)
Here p = 129 and q = 2⁵ 5⁷ 7⁵
Prime factors of q = 2⁵ 5⁷ 7⁵
which are not of the form 2ⁿ × 5^m,
∴ x = \(\frac{129}{2^{5} 5^{7} 7^{5}}\) have a non – terminating decimal expansion.

(ix) Let x = \(\frac{35}{50}=\frac{7}{10}\) …………. (1)
Compare (1) with x = \(\frac{p}{q}\)
Here p = 7, q = 10
Prime factors of q = 10 = 2 × 5 = 21 × 51
Which is of the form 2ⁿ × 5^m here n = 1, m = 1
both n and m are non negative integer.
∴ x = \(\frac{35}{50}\) have a terminating decimal expansion.

(x) Let x = \(\frac{77}{210}=\frac{11}{30}\) ……………..(1)
Compare (1) with x = \(\frac{p}{q}\)
Here p = 11, q = 30
Prime factors of q = 30 = 2 × 5 × 3
which are not of the form 2ⁿ × 5^m,
∴ x = \(\frac{77}{210}\) have a non – terminating decimal expansion.

Question 2.
Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.
Solution:
(i) Let x = \(\frac{13}{3125}\)
Compare (1) with x = \(\frac{p}{q}\)
Here p = 13,q = 3125
Prime factors of q = 3125 = 5 × 5 × 5 × 5 × 5 = 5⁵ × 2⁰
Which are of the form 2ⁿ × 5^m, where n = 0, m = 5 and these are non negative integers
∴ x = have a terminating decimal expansion.
To Express in Decimal form
x = \(\frac{13}{3125}=\frac{13}{5^{5} \times 2^{0}}\)

x = \(\frac{13 \times 2^{5}}{5^{5} \times 2^{5}}\)
[∵ we are to make 10 in the denominator so multiply and divide by 2⁵]

x = \(\frac{13 \times 32}{(2 \times 5)^{5}}\)

x = \(\frac{416}{(10)^{5}}=\frac{416}{100000}\)

x = 0.00416

(ii) Let x = \(\frac{17}{8}\) ………………..(1)
Compare (1) with x = \(\frac{p}{q}\)
Here p = 17, q = 8
Prime factors of q = 8 = 2 × 2 × 2 = 2³ × 5⁰
Which are of the form 2ⁿ × 5^m, where n = 3, m = 0 and these are non negative integers
∴ x = \(\frac{17}{8}\) can be expressed as a terminating decimal expansion.
To Express in Decimal form

x = \(\frac{17}{8}=\frac{17}{2^{3} \times 5^{0}}\)

x = \(\frac{17 \times 5^{3}}{2^{3} \times 5^{3}}\)
[Multiply and divide with 5³ to make the denominator 10]

x = \(\frac{17 \times 125}{(2 \times 5)^{3}}\)

x = \(\frac{2125}{(10)^{3}}=\frac{2125}{1000}\)

x = 2.125

⇒ \(\frac{17}{8}\) = 2.125

(iii) Let x = \(\frac{15}{1600}\) …………….(1)
Compare (1) with x = \(\frac{p}{q}\)
Here p = 15, q = 1600
Prime factors of q = 1600
= 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5
= 2⁶ × 5²
Which are of the form 2ⁿ × 5^m, where n = 6, m = 2 and these are non negative integers
∴ x = \(\frac{15}{1600}\) can be expressed as a terminating decimal expansion.
To Express in Decimal form

x = \(\frac{15}{1600}\)

x = \(\frac{15 \times 5^{4}}{2^{6} \times 5^{2} \times 5^{4}}\)
[To make denominator a power of 10 multiply and divide by 5⁴]

x = \(\frac{15 \times 625}{2^{6} \times 5^{6}}\)

x = \(\frac{9375}{(2 \times 5)^{6}}\)

x = \(\frac{9375}{(10)^{6}}=\frac{9375}{1000000}\) = 0.009375

In Decimal form, x = \(\frac{15}{1600}\) = 0.009375

(iv) Let x = \(\frac{23}{2^{3} 5^{2}}\) …………….(1)
Compare (1) with x = \(\frac{p}{q}\)
Here p = 15, q = 2³ 5²
Prime factors of q = 2³ 5²
Which are of the form 2ⁿ × 5^m, where n = 3, m = 2 and these are non negative integers
∴ x = \(\frac{23}{2^{3} 5^{2}}\) have a terminating decimal expansion.
To Express in Decimal form

x = \(\frac{23}{2^{3} 5^{2}}=\frac{23 \times 5}{2^{3} \times 5^{2} \times 5}=\frac{115}{2^{3} \times 5^{3}}\)

x = \(\frac{115}{(2 \times 5)^{3}}=\frac{115}{1000}\) = 0.115

In Decimal form,
x = \(\frac{23}{2^{3} 5^{2}}\) = 0.115

(v) Let x = \(\frac{6}{15}=\frac{2}{5}\)
Compare (1) with x = \(\frac{p}{q}\)
Here p = 2, q = 5
Prime factors of q = 5 = 2⁰ × 5¹
Which are of the form 2ⁿ × 5^m, where n = 0, m = 1 and these are non negative integers
∴ x = \(\frac{6}{15}\) have a terminating decimal expansion.
To Express in Decimal form

x = \(\frac{6}{15}=\frac{2}{5}\)

x = \(\frac{2 \times 2^{1}}{2^{1} \times 5^{1}}=\frac{4}{10}\) = 0.4

In Decimal form,
x = \(\frac{6}{15}\) = 0.4

(vi) Let x = \(\frac{35}{50}=\frac{7}{10}\)
Compare (1) with x = \(\frac{p}{q}\)
Here p = 7, q = 10
Prime factors of q = 5 = 2¹ × 5¹
Which are of the form 2ⁿ × 5^m, where n = 1, m = 1 and these are non negative integers
∴ x = \(\frac{7}{10}\) have a terminating decimal expansion.
To Express in Decimal form

x = \(\frac{35}{50}\)

x = \(\frac{7}{10}\)

x = \(\frac{7}{2^{1} \times 5^{1}}\)

x = \(\frac{7}{(2 \times 5)^{1}}=\frac{7}{(10)^{1}}\) = 0.7
Hence in decimal form, x = 0.7

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 f, what can you say about the prime factors of q?
(i) 43.123456789
(ii) O.120120012000120000……
(iii) 4.3.123456789
Solution:
(i) Let x= 43.123456789 ……….. (1)
It is clear from the number that x is rational number.
Now remove the decimal from the number

∴ x = \(\frac{43123456789}{1000000000}\)

= \(\frac{43123456789}{10^{9}}\) …………….(2)
From (2) x is a rational number and of the \(\frac{p}{q}\).

Where p = 43123456789 and q = 10⁹
Now, Prime factors of q = 10⁰ = (2 × 5)⁹
⇒ Prime factors of q are 2⁹ × 5⁹

(ii) Let x = 0.120120012000120000
It is clear from the number that it is an irrational number.

(iii) Let x = 43.123456789 …. (1)
It is clear that the given number is a rational number because it is non-terminating and repeating decimal.
To show that (i) is of the form \(\frac{p}{q}\)
Multiply (1) with 10⁹ on both sides,
10⁹ x = 43123456789.123456789 …………….(2)
Subtract (1) from (2), we get:

which is rational number of the form \(\frac{p}{q}\)
x = \(\frac{4791495194}{111111111}\)
Here p = 4791495194, q = 111111111
x = \(\frac{4791495194}{3^{2}(12345679)}\)
Hence, prime factors of q are 3² (123456789)

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Question 1.
The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Am and Biju Is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Solution:
Let Ani’s age = x years
and Biju’s age = y years
Dharam’s age = 2x years
Cathy’s age = years
According to 1st condition,
(Ani’s age) (Biju’s age) = 3
x – y = 3 ……………(1)
According to 2nd condition,
(Dharam’ age) – (Cathy’s age) = 30
2x – \(\frac{y}{2}\) = 30
or \(\frac{4 x-y}{2}\) = 30
or 4x – y = 60 ………….(2)
Now (2) – (1) gives,

Substitute this value of x in (1), we get:
19 – y = 3
or -y = 3 – 19
or -y = -16
or y = 16
Hence, Ani’s age = 19 years
Biju’s age =16 years.

Question 2.
One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “if you give me ten, I shall be six times as rich as you.” Tell me what is the amount of their (respective) capital ? [From the Bijaganita of Bhaskara II] [Hint: x + 100 = 2(y – 100), y + 10 = 6 (x – 10)].
Solution:
Let Capital of one friend = ₹ x
and capital of 2nd friend = ₹ y
According to 1st condition
x + 100 = 2(y – 100)
or x + 100 = 2y – 200
or x – 2y = -200 – 100
or x – 2y = -300 …………..(1)
According to 2nd condition
y + 10 = 6(x – 10)
or v-f 10 = 6x – 60
or 6x – y = 10 + 60
or 6x – y = 70 ……………(2)
Multiplying (1) by 6, we get
6x – 12y = – 1800 …………….(3)
Now, (3) – (2) gives

Substitute this value of y in (2), we get:
6x – 170 = 70
or 6x = 70 + 170
or 6x = 240
or x = \(\frac{240}{6}\) = 40
Hence, amount of their capital are 40 and 170 respectively.

Question 3.
A train covered a certain distance at a uniform speed. If the train would have been 10 km/b faster, It would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Solution:
Let speed of train =x km/hour
and time taken by train =y hour
∴ Distance covered by train = (Speed) (Time) = (xy) km
According to 1st condition,
(x + 10)(y – 2) = xy
or xy – 2x + 10y – 20 = y
or -2x + 10y – 20 = 0
or x – 5y + 10 = 0
According to 2nd condition,
(x – 10) (y + 3) = xy
or xy + 3x – 10y – 30 = xy
or 3x – 10y – 30 = 0
Multiplying (1) by 3, we get:
3x – 15y + 30 = 0
Now, (3) – (2) gives

Substitute this value of y in (1), we get:
x – 5 × 12 + 10 = 0
or x – 60 + 10 = 0
or x – 50 = 0
or x = 50
Speed of train = 50 km/hour
Time taken by train = 12 hour
Hence, distance covered by train = (50 × 12) km = 600 km.

Question 4.
The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Solution:
Let number of students in each row = x
and number of rows = y
Total number of students in the class = xy
According to 1st condition,
(x + 3) (y – 1) = xy
or xy – x + 3y – 3 = xy
or -x + 3y – 3 = 0
or x – 3y + 3 = 0
According to 2nd condition,
(x -3) (y + 2) = xy
or xy + 2x – 3y – 6= xy
or 2x – 3y – 6 = 0
Now, (2) – (1) gives

x = 9
Substitute this value of x in (1), we get:
9 – 3y + 3 = 0
or -3y + 12 = 0
or -3y = -12
or y = \(\frac{12}{3}\) = 4
∴ Number of students in each row = 9 and number of rows = 4
Hence, total number of students in the class = 9 × 4 = 36.

Question 5.
In a ∆ABC, ∠C = 3 ∠B = 2(∠A +∠B) find the three angles.
Solution:
In ∆ABC,
Given that, ∠C = 3∠B = 2(∠A + ∠B)
I II III
From II and III, we get:
3∠B =2(∠A+∠B)
or 3∠B = 2∠A + 2∠B
or 3∠B – 2∠B = 2∠A
or ∠B = 2∠A ……………..(1)
From I and II, we get:
∠C = 3∠B
or ∠C = 3(2∠A) [using (1)]
or ∠C = 6∠A …………….(2)
Sum of three angles of a triangle is 180°
∠A + ∠B + ∠C = 180°
or ∠A + 2∠A + 6∠A = 180°
or 9∠A = 180°
∠A = \(\frac{180^{\circ}}{9}\) = 20°
Hence, ∠A = 20°: ∠B =2 x 20° = 40°; ∠C = 6 x 20° = 120°.

Question 6.
Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle
formed by these lines and the y axis.
Solution:
Given pair of linear equation are 5x – y = 5 and 3x – y = 3
Consider,
5x – y = 5
or 5x = 5 + y
Putting y = 0 in (1), we get:
x = \(\frac{5+0}{5}=\frac{5}{5}\)
Putting y = – 5 in (1), we get:
x = \(\frac{5-5}{5}=\frac{0}{5}\) = 0
Putting y = 5 in (1), we get:
x = \(\frac{5+5}{5}=\frac{10}{5}\) = 2

Plotting A (1, 0); B (0, — 5); C (2, 5) on the graph, we get the equation of line 5x — y = 5
and 3x – y = 3
or 3x = 3 + y
or x = \(\frac{3+y}{3}\) ……………(2)
Putting y = 0 in (2), we get:
x = \(\frac{3+0}{3}=\frac{3}{3}\) = 1
Putting y = 3 in (2), we get:
x = \(\frac{3-3}{3}=\frac{0}{3}\) = 0
Putting y = 3 in (2), we get:
x = \(\frac{3+3}{3}=\frac{6}{3}\) = 2
Table:

Plotting A (1, 0); D (0, -3); E (2, 3) on the graph. we get the equation of line 3x – y = 3. From the graph, it is clear that given lines intersect at A (1, 0). Triangle formed by these lines and y axis are shaded in the graph i.e. ∆ABD. Coordinates of the vertices of ∆ABD are A(1, 0); B(0, -5) and D(0, -3).

Question 7.
Solve the following pair of linear equations:
(i) px + qy = p – q
qx – py = p + q

(ii) ax + by = c
bx + ay = 1 + c

(iii) \(\frac{x}{a}-\frac{y}{b}\) = 0
ax + by = a² + b²

(iv) (a – b)x + (a + b)y = a² – 2ab – b²
(a + b) (x + y) = a² + b²

(v) 152x – 378y = 74
-378x + 152y = -604
Solution:
(i) Given pair of linear equation are
px + qy = p – q …………(1)
and qx – py = p + q ………….(2)
Multiplying (1) by q and (2) by p, we get:

or y = -1
Substitute this value of y in (1), we get:
px + q(-1) = p – q
or px – q = p – q
or px = p – q + q
or px = p
or x = 1
Hence, x = 1 and y = -1.

(ii) Given pair of linear equation are
ax + by = c
bx + ay = 1 + c
ax + by – c = 0
bx + ay – (1 + c) = 0

From I and III, we get:
\(\frac{x}{-b-b c+a c}=\frac{1}{a^{2}-b^{2}}\)
or x = \(\frac{a c-b c-b}{a^{2}-b^{2}}\)

From II and III, we get:
\(\frac{y}{-b c+a+a c}=\frac{1}{a^{2}-b^{2}}\)
or y = \(\frac{a+a c-b c}{a^{2}-b^{2}}\)
Hence, x = \(\frac{a c-b c-b}{a^{2}-b^{2}}\) and y = \(\frac{a+a c-b c}{a^{2}-b^{2}}\).

(iii) Given pair of linear equation are
\(\frac{x}{a}-\frac{y}{b}\) = 0
or \(\frac{b x-a y}{a b}\) = 0
or bx – ay = 0
or bx—ay = O …(1)
and ax + by = a² + b²
or ax + by – (a² + b²) = 0 …………..(2)

\(\frac{x}{a\left(a^{2}+b^{2}\right)-0}=\frac{y}{0+b\left(a^{2}+b^{2}\right)}=\frac{1}{b^{2}+a^{2}}\)

or \(\frac{x}{a\left(a^{2}+b^{2}\right)}=\frac{y}{b\left(a^{2}+b^{2}\right)}=\frac{1}{a^{2}+b^{2}}\)

or \(\frac{x}{a}=\frac{y}{b}=\frac{1}{1}\)
I II III
From I and III, we get:
\(\frac{x}{a}=\frac{1}{1}\)
⇒ x = a
From II and III, we get:
\(\frac{y}{b}=\frac{1}{1}\)
⇒ y = b
Hence, x = a, y = b.

(iv) Given pair of linear equation are
(a – b)x + (a + b)y = a² – 2ab – b²
or ax – bx + ay + by = a² – 2ab – b² …………….(1)
and (a + b) (x + y) = a² + b²
or ax + bx + ay + by = a² + b²
Now, (1) – (2) gives

Substitute this value of x in (1), we get:
(a – b) (a + b) + (a + b) y = a² – 2ab – b²
or a² – b² + (a + b) y = a² – 2ab – b²
or (a + b) y = a² – 2ab – b² – a² + b²
or (a + b)y = -2ab
or y = \(\frac{-2 a b}{a+b}\)
Hence, x = a + b and y = a + b

(v) Given pair of linear equation are
152x – 378y = – 74
and -378x + 152y = -604
or 76x – 189y + 37 = 0
and -189x + 76y + 302 = 0

From I and III, we get:
\(\frac{x}{59890}=\frac{1}{29945}\)
⇒ x = \(\frac{59890}{29945}\)
⇒ x = 2

From II and III, we get:
\(\frac{y}{29945}=\frac{1}{29945}\)
⇒ y = \(\frac{29945}{29945}\)
⇒ y = 1
Hence x = 2 and y = 1.

Question 8.
ABCD is a cyclic quadrilateral (see Fig.). Find the angles of the cyclic quadrilateral.

Solution:
In cyclic quadrilateral ABCD,
∠A = (4y + 20); ∠B = 3y – 5; ∠C = 4x and ∠D = 7x + 5
Sum of opposite angles of a cyclic quadrilateral are of measure 180°.
∴ ∠A + ∠C = 180°
or 4y + 20 + (4x) = 180°
or 4x + 4y = 180° – 20
or 4x + 4y = 160
or x + y = 40
or y = 40 – x ……………..(1)
and ∠B + ∠D = 180°
or 3y – 5 + (7x + 5)= 180°
or 3y – 5 + 7x + 5 = 180°
or 7x + 3y = 180° …………….(2)
Substitute the value of y from (1) in (2). we get:
7x + 3(40 – x)= 180°
or 7x + 120 – 3x = 180°
or 4x = 180 – 120
or 4x = 60
x = \(\frac{60}{4}\) = 15
Substitute this value of x in (1 ), we get:
y = 40 – 15 = 25
∴ ∠A = 4y + 20 = 4 × 25 + 20 = 120°
∠B = 3y – 5 = 3 × 25 – 5 = 70°
∠C = 4x =4 × 15 = 60°
∠D = 7x + 5 = 7 × 15 + 5 = 110°
Hence, ∠A = 120°, ∠B = 70°; ∠C = 60° and ∠D = 110°.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 2 Fractions and Decimals Ex 2.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.3

1. Solve the following :

Question (i).
\(\frac {1}{3}\) of
(a) \(\frac {1}{5}\)
(b) \(\frac {2}{7}\)
(c) \(\frac {3}{2}\)
Answer:

Question (ii).
\(\frac {3}{4}\) of
(a) \(\frac {2}{9}\)
(b) \(\frac {4}{7}\)
(c) \(\frac {8}{3}\)
Answer:

2. Multiply the fractions and reduce it to the lowest form (If possible) :

Question (i).
\(\frac{2}{7} \times \frac{7}{9}\)
Answer:
\(\frac{2}{7} \times \frac{7}{9}\)
= \(\frac {2}{9}\)

Question (ii).
\(\frac{1}{3} \times \frac{15}{8}\)
Answer:
\(\frac{1}{3} \times \frac{15}{8}\)
= \(\frac {5}{8}\)

Question (iii).
\(\frac{12}{27} \times \frac{3}{9}\)
Answer:
\(\frac{12}{27} \times \frac{3}{9}\)
= \(\frac {4}{27}\)

Question (iv).
\(\frac{2}{5} \times \frac{6}{4}\)
Answer:
\(\frac{2}{5} \times \frac{6}{4}\)
= \(\frac {3}{5}\)

Question (v).
\(\frac{81}{100} \times \frac{6}{7}\)
Answer:
\(\frac{81}{100} \times \frac{6}{7}\)
= \(\frac {243}{350}\)

Question (vi).
\(\frac{3}{5} \times \frac{5}{27}\)
Answer:
\(\frac{3}{5} \times \frac{5}{27}\)
= \(\frac {1}{9}\)

3. Multiply the following fractions :

Question (i).
\(\frac{3}{2} \times 5 \frac{1}{3}\)
Answer:
\(\frac{3}{2} \times 5 \frac{1}{3}\)
= \(\frac{3}{2} \times \frac{16}{3}\)
= 8

Question (ii).
\(\frac{1}{7} \times 5 \frac{2}{3}\)
Answer:
\(\frac{1}{7} \times 5 \frac{2}{3}\)
= \(\frac{1}{7} \times \frac{17}{3}\)
= \(\frac {17}{21}\)

Question (iii).
\(2 \frac{5}{6} \times 4\)
Answer:
\(2 \frac{5}{6} \times 4\)
= \(\frac {17}{6}\) × 4
= \(\frac {34}{3}\)
= 11\(\frac {1}{3}\)

Question (iv).
\(4 \frac{1}{3} \times 9 \frac{1}{4}\)
Answer:
\(4 \frac{1}{3} \times 9 \frac{1}{4}\)
= \(\frac{13}{3} \times \frac{37}{4}\)
= \(\frac {481}{12}\)

Question (v).
\(2 \frac{2}{3} \times 3 \frac{5}{8}\)
Answer:
\(2 \frac{2}{3} \times 3 \frac{5}{8}\)
= \(\frac{8}{3} \times \frac{29}{8}\)
= 9\(\frac {2}{3}\)

Question (vi).
\(3 \frac{1}{5} \times 2 \frac{1}{4}\)
Answer:
\(3 \frac{1}{5} \times 2 \frac{1}{4}\)
= \(\frac{16}{5} \times \frac{9}{4}\)
= \(\frac {36}{5}\)
= 7\(\frac {1}{5}\)

4. Which fraction is greater in the following fractions ?

Question (i).
\(\frac {3}{2}\) of \(\frac {2}{7}\) or \(\frac {5}{2}\) of \(\frac {3}{8}\)
Answer:
\(\frac {3}{2}\) of \(\frac {2}{7}\)
= \(\frac{3}{2} \times \frac{2}{7}\)
= \(\frac {3}{7}\)
or
\(\frac {5}{2}\) of \(\frac {3}{8}\)
= \(\frac{5}{2} \times \frac{3}{8}\)
= \(\frac {15}{16}\)
Since, \(\frac {15}{16}\) > \(\frac {3}{7}\)
So, \(\frac {5}{2}\) of \(\frac {3}{8}\) is greater.

Question (ii).
\(\frac {1}{2}\) of \(\frac {6}{5}\) or \(\frac {1}{3}\) of \(\frac {4}{5}\)
Answer:
\(\frac {1}{2}\) of \(\frac {6}{5}\)
= \(\frac {1}{2}\) × \(\frac {6}{5}\)
= \(\frac {3}{5}\)
or
\(\frac {1}{3}\) of \(\frac {4}{5}\)
= \(\frac {1}{3}\) × \(\frac {4}{5}\)
= \(\frac {4}{15}\)
Since, \(\frac {3}{5}\) > \(\frac {4}{15}\)
So, \(\frac {1}{2}\) of \(\frac {6}{5}\) is greater.

5. If the speed of a car is 105 \(\frac {1}{3}\) km/hr, find the distance covered by it m 3\(\frac {2}{3}\) hours.
Answer:
Distance travelled in 1 hour= 105\(\frac {1}{3}\) km
= \(\frac {316}{3}\) km
Distance travelled in 3\(\frac {2}{3}\) hours i.e. in \(\frac {11}{3}\) hours
= \(\frac {11}{3}\) × \(\frac {316}{3}\) km
= \(\frac {3476}{9}\) km
= 386\(\frac {2}{9}\)km

6. The length of a rctangular plot of land is 29\(\frac {3}{7}\) m. If its breadth is 12\(\frac {8}{11}\)m, find its area.
Answer:
Length of rectangular plot (l)
= 29\(\frac {3}{7}\)m = \(\frac {206}{7}\)m
Breadth of rectangular plot (b)
= 12\(\frac {8}{11}\)m = \(\frac {140}{11}\)m
Area of rectangular plot
= l × b = \(\frac{206}{7} \mathrm{~m} \times \frac{140}{11} \mathrm{~m}\)
= \(\frac{4120}{11}\) sq.m
= 374\(\frac{6}{11}\) sq.m

7. If a cloth costs ₹ 120 \(\frac {1}{4}\) per metre, find the cost of 4\(\frac {1}{3}\) metre of this cloth.
Answer:
Per m cost of cloth = ₹ 120\(\frac {1}{4}\)
= ₹ \(\frac {481}{4}\)
Cost of 4\(\frac {1}{3}\) metre of this cloth
= \(\frac {1}{3}\) × ₹ \(\frac {481}{4}\)
= ₹ \(\frac {6253}{12}\)
= ₹ 521\(\frac {1}{12}\)

8. Multiple choice questions :

Question (i).
\(\frac {1}{4}\) of \(\frac {8}{3}\) is :
(a) \(\frac {9}{7}\)
(b) \(\frac {8}{4}\)
(c) \(\frac {2}{3}\)
(d) 1
Answer:
(c) \(\frac {2}{3}\)

Question (ii).
\(\frac {3}{2}\) × \(\frac {2}{3}\) = ?
(a) 1
(b) \(\frac {5}{6}\)
(c) 3
(d) \(\frac {6}{5}\)
Answer:
(a) 1

Question (iii).
The value of the product of two proper fractions is :
(a) greater than both of the proper fractions.
(b) lesser than both of the proper fractions.
(c) lie between the given two fractions.
(d) none of these.
Answer:
(b) lesser than both of the proper fractions.

9. Check if the following equations are ‘True’ or ‘False’ :

Question (i).
\(1 \frac{2}{3} \times 4 \frac{5}{7}=4 \frac{10}{21} ?\) (True/False)
Answer:
False

Question (ii).
\(\frac{3}{4} \times \frac{2}{3}=\frac{2}{3} \times \frac{3}{4} ?\) (True/False)
Answer:
True

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 1 Real Numbers Ex 1.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2

Question 1.
Express each number as a product of its prime factors :
(i) 140 (Pb, 2019)
(ii) 156
(iii) 3825
(iv) 5005 (Pb. 2019, Set-1, II, III)
(v) 7429.
Solution:
(i) Prime factorisation of 140 = (2)² (35) = (2)² (5) (7)

(ii) Prime factorisation of 156 = (2)² (39) = (2)² (3) (13)

(iii) Prime factorisation of 3825 = (3)² (425)
= (3)² (5) (85)
= (3)² (5)² (17)

(iv) Prime factorisation of 5005
= (5) (1001)
= (5) (7) (143)
= (5) (7) (11) (13)

(v) Prime factorisation of 7429
= (17) (437)
= (17) (19) (23)

Question 2.
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = Product of the two numbers.
(i) 26 and 91 [Pb. 2017 Set-C]
(ii) 510 and 92
(iii) 336 and 54.
Solution:
(i) Given numbers are 26 and 91 Prime factorisation of 26 and 91 are
26 = (2) (13) and
91 = (7) (13)
HCF (26, 91)
Product of least powers of common factors
∴ HCF (26, 91) = 13
and LCM (26, 91) = Product of highest powers of all the factors
= (2) (7) (13) = 182
Verification :
LCM (26, 91) × HCF (26, 91)
= (13) × (182) = (13) × (2) × (91)
= (26) × (91)
= Product of given numbers.

(ii) Given numbers are 510 and 92 Prime factorisation of 510 and 92 are
510 = (2) (255)
= (2) (3) (85)
= (2) (3) (5) (17)
and 92 = (2) (46) = (2)² (23)
HCF (510, 92) = Product of least powers of common factors = 2
LCM (510, 92) = Product of highest Powers of all the factors
= (2)² (3) (5) (17) (23) = 23460

Verification:
LCM (510, 92) × HCF (510, 92)
= (2) (23460)
= (2) × (2)² (3) (5) (17) (23)
= (2) (3) (5) (17) × (2)² (23)
= 510 × 92 = Product of given numbers.

(iii) Given numbers are 336 and 54
Prime factorisation of 336 and 54 are
336 = (2) (168)
= (2) (2) (84)
= (2) (2) (2) (42)
= (2) (2) (2) (2) (21)
= (2)⁴ (3) (7)

and 54 = (2) (27)
= (2) (3) (9)
= (2) (3) (3) (3)
= (2) (3)³
HCF (336, 54) = Product of least powers of common factors = (2) (3) = (6)
LCM (336, 54) = Product of highest powers of all the factors
= (2)⁴ (3)³ (7) = 3024

Verification :
LCM (336, 54) × H.C.F. (336, 54)
= 6 × 3024
= (2) (3) × (2)⁴ (3)³ (7)
= (2)⁴ (3) (7) × (2) (3)³
= 336 × 54
= Product of given numbers.

Question 3.
Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21 [MQP 2015, Pb. 2015 Set A, Set B, Set C]
(ii) 17, 23 and 29
(iii) 8, 9 and 25
Solution:
(i) Given numbers are 12, 15 and 21
Prime factorisation of 12, 15 and 21 are
12 = (2) (6)
= (2) (2) (3)
= (2)² (3)
15 = (3) (5)
21 = (3) (7)
HCF (12, 15 and 21) = 3
LCM (12, 15 and 21) = (2)² (3) (5) (7) = 420

(ii) Given numbers are 17, 23 and 29
Prime factorisation of 17, 23 and 29 are
17 = (17) (1)
23 = (23) (1)
29 = (29) (1)
HCF (17, 23 and 29) = 1
LCM (17, 23 and 29)
= 17 × 23 × 29 = 11339

(iii) Given numbers are 8, 9 and 25
Prime factorisation of 8, 9 and 25 are
8 = (2) (4)
= (2) (2) (2)
= (2)³ (1)
9 = (3) (3) = (3)² (1)
25 = (5) (5) = (5)² (1)
HCF (8, 9 and 25) = 1
LCM (8, 9 and 25) = (2)³ (3)² (5)² = 1800

Question 4.
Prove that HCF (306,657) = 9, find LCM (306,657). [Pb. 2016 Set-B, 2017 Set-B]
Solution:
Given numbers are 306 and 657 Prime factorisation of 306 and 657 are
306 = (2) (153)
= (2) (3) (51)
= (2) (3) (3) (17)
= (2) (3)2 (17)

657 = (3) (219)
= (3) (3) (73)
= (3)² (73)

HCF (306, 657) = (3)² = 9
∵HCF × LCM = Product of given number
∵9 × LCM (306, 657) = 306 × 657

= 34 × 657 = 22338

Question 5.
Check whether 6ⁿ can end with the digit 0 for any natural number n:
Solution:
Let us suppose that 6ⁿ ends with the digit 0 for some n ∈ N.
6ⁿ is divisible by 5.
But, prime factor of 6 are 2 and 3 Prime factor of (6)ⁿ are (2 × 3)ⁿ
⇒ It is clear that in prime factorisation of 6ⁿ there is no place for 5.
∵ By Fundamental Theorem of Arithmetic, Every composite number can be expressed as a product of primes and this factorisation is unique, apart from the order in which the prime factors occur.
∴ Our supposition is wrong.
Hence, there exists no natural number n for which 6ⁿ ends with the digit zero.

Question 6.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
Consider, 7 × 11 × 13 + 13 = 13 [7 × 11 + 1]
which is not a prime number because it has a factor 13. So, it is a composite number.
Also, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5[7 × 6 × 5 × 4 × 3 × 2 × 1 + 1], which is not a prime number because it has a factor 5. So it is a composite number.

Question 7.
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for die same. Suppose they both start at the same point and at file same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Solution:
Time taken by Sonia to drive one round of the field =18 minutes
Time taken by Ravi to drive one round of same field = 12 minutes
They meet again at the starting point = LCM (18, 12)
Now, Prime factorisation of 18 and 12 are 18 = (2) (9)
= (2) (3) (3)
= (2) (3)²

12 = (2) (6)
= (2) (2) (3)
= (2)² (3)
LCM (18, 12) = (2)² (3)² = 4 × 9 = 36
Hence, After 36 minutes Sonia and Ravi will meet again at the starting point.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 2 Fractions and Decimals Ex 2.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.2

1. Match the following :

Answer:
(a) (ii),
(b) (iii),
(c) (iv),
(d) (i).

2. Multiply whole number to a fraction and reduce it to the lowest form. Also convert into a mixed fraction :

Question (i).
4 × \(\frac {1}{3}\)
Answer:
4 × \(\frac {1}{3}\) = \(\frac{4 \times 1}{3}\)
= \(\frac {4}{3}\)
= 1\(\frac {1}{3}\)

Question (ii).
11 × \(\frac {4}{7}\)
Answer:
11 × \(\frac {4}{7}\) = \(\frac{11 \times 4}{7}\)
= \(\frac {44}{7}\)
= 6\(\frac {2}{7}\)

Question (iii).
\(\frac {3}{4}\) × 6
Answer:
\(\frac {3}{4}\) × 6 = \(\frac{3 \times 6}{4}\)
= \(\frac {9}{2}\)
= 4\(\frac {1}{2}\)

Question (iv).
\(\frac {9}{7}\) × 5
Answer:
\(\frac {9}{7}\) × 5 = \(\frac {45}{7}\)
= 6\(\frac {3}{7}\)

Question (v).
2\(\frac {5}{6}\) × 4
Answer:
2\(\frac {5}{6}\) × 4 = \(\frac{17}{6} \times 4\)
= \(\frac{17 \times 4}{6}\)
= \(\frac{17 \times 2}{3}\)
= \(\frac {34}{3}\)
= 11\(\frac {1}{3}\)

Question (vi).
10\(\frac {5}{6}\) × 5
Answer:
10\(\frac {5}{6}\) × 5
= \(\frac {65}{6}\) × 5
= \(\frac{65 \times 5}{6}\)
= \(\frac {325}{6}\)
= 54\(\frac {1}{6}\)

Question (vii).
5 × 6\(\frac {3}{4}\)
Answer:
5 × 6\(\frac {3}{4}\) = 5 × \(\frac {27}{4}\)
= \(\frac{5 \times 27}{4}\)
= \(\frac {135}{4}\)
= 33\(\frac {3}{4}\)

Question (viii).
3 \(\frac {2}{5}\) × 8
Answer:
3 \(\frac {2}{5}\) × 8 = \(\frac {17}{5}\) × 8
= \(\frac{17 \times 8}{5}\)
= \(\frac {136}{5}\)
= 27\(\frac {1}{5}\)

3. Solve :

Question (i).
\(\frac {1}{2}\) of 46
Answer:
\(\frac {1}{2}\) of 46
= \(\frac {1}{2}\) × 46
= 23

Question (ii).
\(\frac {2}{3}\) of 27
Answer:
\(\frac {2}{3}\) of 27
= \(\frac {2}{3}\) × 27
= 18

Question (iii).
\(\frac {1}{3}\) of 36
Answer:
\(\frac {1}{3}\) of 36
= \(\frac {1}{3}\) × 36
= 12

Question (iv).
\(\frac {3}{4}\) of 16
Answer:
\(\frac {3}{4}\) of 16
= \(\frac {3}{4}\) × 16
= 12

Question (v).
\(\frac {5}{7}\) of 35
Answer:
\(\frac {5}{7}\) of 35
= \(\frac {5}{7}\) × 35
= 25

4. Shade :

(i) \(\frac {1}{3}\) of the rectangles in box (a)
(ii) \(\frac {2}{3}\) of the circles in box (b)
(iii) \(\frac {1}{2}\) of the triangles in box (c)

Answer:

5. Rahul earns ₹ 44,000 per month. He spends \(\frac {3}{4}\) th of his income every month and saves the rest of his earning. Find his monthly savings ?
Answer:
Rahul earns per month = ₹ 44000
He spends = \(\frac {3}{4}\)th of ₹ 44000
= ₹ \(\frac {3}{4}\) × 44000
= ₹ 33000
His savings = ₹ 44,000 – ₹ 33000
= ₹ 11000

6. The cost of a book is ₹ 117\(\frac {1}{2}\). Find the cost of 8 books ?
Answer:
Cost of one book = ₹ 117 \(\frac {1}{2}\)
= ₹ \(\frac {235}{2}\)
Cost of 8 books = 8 × ₹ \(\frac {235}{2}\)
= ₹ 4 × 235
= ₹ 940

7. Multiple choice questions :

Question (i).
\(\frac {1}{2}\) × 8 is equal to
(a) 8
(6) 2
(c) 4
(d) 1
Answer:
(c) 4

Question (ii).
\(\frac {3}{2}\) of 16 is :
(a) 48
(b) 8
(c) 3
(d) 24
Answer:
(d) 24

Question (iii).
What fraction of an hour is 40 minutes ?
(a) \(\frac {2}{3}\)
(b) 40
(c) \(\frac {1}{4}\)
(d) \(\frac {1}{2}\)
Answer:
(a) \(\frac {2}{3}\)

Question (iv).
What fraction does the shaded part of figure represent ?

(a) \(\frac {1}{3}\)
(b) \(\frac {2}{3}\)
(c) \(\frac {3}{4}\)
(d) \(\frac {1}{2}\)
Answer:
(c) \(\frac {3}{4}\)

Punjab State Board PSEB 10th Class Hindi Book Solutions Hindi Grammar vilom shabd विलोम शब्द Exercise Questions and Answers, Notes.

PSEB 10th Class Hindi Grammar विलोम शब्द

निम्नलिखित शब्दों के विलोम शब्द लिखिए

उत्तर:

निम्नलिखित बहुविकल्पी प्रश्नों के उत्तर एक सही विकल्प चुनकर लिखें.

प्रश्न 1.
असली का विलोम शब्द है
(क) निष्ठुर
(ख) अनुचित
(ग) नवीन
(घ) नकली।
उत्तर:
(घ) नकली

प्रश्न 2.
चेतन का विलोम शब्द है
(क) सुप्त
(ख) जड़
(ग) जागृत
(घ) निद्रा।
उत्तर:
(ख) जड़

प्रश्न 3.
कोमल का विलोम शब्द है
(क) निष्ठुर
(ख) निकृष्ट
(ग) कर्कश
(घ) अधम।
उत्तर:
(ग) कर्कश

प्रश्न 4.
सार्थक का विलोम शब्द है,
(क) व्यर्थ
(ख) निरर्थक
(ग) निर्गुण
(घ) क्षीण।
उत्तर:
(ख) निरर्थक

प्रश्न 5.
पूर्व का विलोम है, पश्चिम (हाँ या नहीं में उत्तर लिखें)
उत्तर:
हाँ

प्रश्न 6.
दानव का विलोम है, मानव (हाँ या नहीं में उत्तर लिखें)
उत्तर:
नहीं

प्रश्न 7.
भय का विलोम है, साहस (हाँ या नहीं में उत्तर लिखें)
उत्तर:
हाँ

प्रश्न 8.
ताप का विलोम है, शीत (सही या गलत लिखकर उत्तर दें)
उत्तर:
सही

प्रश्न 9.
पतन का विलोम है, गिरावट (सही या गलत लिखकर उत्तर दें)
उत्तर:
गलत

प्रश्न 10.
मिलन का विलोम है, विरह (सही या गलत लिख कर उत्तर दें)
उत्तर:
सही।

निम्नलिखित में से किसी एक शब्द का विलोम शब्द लिखिए

वर्ष
1. कंजूस, जिंदाबाद।
उत्तर:
कंजूस = दानी
जिंदाबाद = मुर्दाबाद।

2. हानि, ईमानदार।
उत्तर:
हानि = लाभ
ईमानदार = बेईमान।

3. निंदा, हार।
उत्तर:
निंदा = स्तुति
हार = जीत।

वर्ष
1. करीब, मित्र।
उत्तर:
करीब = दूर
मित्र = शत्रु।

2. आशा, मान
उत्तर:
आशा = निराशा
मान = अपमान।

3. विधवा, स्वस्थ।
उत्तर:
विधवा = सधवा
स्वस्थ = अस्वस्थ।

वर्ष
कंजूस, विस्तार।
उत्तर:
कंजूस = खर्चीला
विस्तार = संक्षेप।

प्रश्न 1.
विलोम शब्द किसे कहते हैं? उदाहरण सहित लिखिए।
उत्तर:
परस्पर विपरीत अर्थ का ज्ञान कराने वाले किसी शब्द को विलोम शब्द कहते हैं। इसे विपरीत या विलोमार्थी शब्द भी कहते हैं। उदाहरण-
(I) रीना सदा न्याय का पक्ष लेती है।
विलोम अर्थ-रीना सदा अन्याय का पक्ष लेती है।

(II) नीरज साधारण परिवार से संबंधित है।
विलोम अर्थ-नीरज असाधारण परिवार से संबंधित है।

(III) गली में एक व्यक्ति था।
विलोम अर्थ-गली में अनेक व्यक्ति थे।

(IV) गीताजंली का व्यवहार उचित था।
विलोम अर्थ-गीतांजली का व्यवहार अनुचित था।

(V) आप की यह हरकत पाक नहीं कहला सकती।
विलोम अर्थ-आप की यह हरकत नापाक नहीं कहला सकती।

प्रश्न 2.
विलोम शब्द किस-किस प्रकार बनाए जाते हैं? उदाहरण सहित लिखिए।
उत्तर:
विलोम शब्द प्रायः तीन प्रकार से बनाए जाते हैं-
(क) उपसर्ग के योग से
(ख) उपसर्ग बदलने से
(ग) विलोम शब्द मूल रूप से।
उदाहरण-
(क) उपसर्ग के योग सेइच्छा-अनिच्छा, साधारण-असाधारण, यश-अपयश, गुण-अवगुण, सुगंध-दुर्गंध, पसंद-नापसंद।
(ख) उपसर्ग बदलने से-साक्षर-निरक्षर, ईमानदार-बेईमानदार, स्वदेश-विदेश, संपन्न-विपन्न, अधुनातन-पुरातन, आयात-निर्यात।
(ग) विलोम शब्द मूल से-उदय-अस्त, आधुनिक-प्राचीन, उजड़ा-बसा, झगड़ा-समझौता, निकास-प्रवेश, दिवस-रात।

(क) उपसर्ग के योग से बने विलोम शब्द
(i) ‘अ’ उपसर्ग के योग से बने विलोम शब्द

(ii) ‘अन्’ उपसर्ग के योग से बने विलोम शब्द

(iii) ‘अप’ उपसर्ग के योग से बने विलोम शब्द

(iv) ‘अव’ उपसर्ग के योग से बने विलोम शब्द

(v) ‘कु’ उपसर्ग के योग से बने विलोम शब्द

(vi) ‘दुः/दुर्’ उपसर्ग के योग से बने विलोम शब्द

(vii) ‘ना’ उपसर्ग के योग से बने विलोम शब्द

(viii) ‘निः/निर्’ उपसर्ग के योग से बने विलोम शब्द

(ix) ‘पर’ उपसर्ग के योग से बने विलोम शब्द

(x) ‘प्रति’ उपसर्ग के योग से बने विलोम शब्द

(xi) ‘वि’ उपसर्ग के योग से बने विलोम शब्द

(xii) ‘बे’ उपसर्ग के योग से बने विलोम शब्द

कुछ विपरीत शब्द मूल रूप में ही प्रयुक्त होते हैं। जैसे-

Punjab State Board PSEB 9th Class Science Book Solutions Chapter 9 Force and Laws of Motion Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Science Chapter 9 Force and Laws of Motion

PSEB 9th Class Science Guide Force and Laws of Motion Textbook Questions and Answers

Question 1.
An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.
Answer:
Yes, it is possible for an object to move with non-zero velocity even when net zero unbalanced force is experienced by it. In this situation the magnitude of velocity and direction will be same. As for example, in case of a rain drop falling freely with constant velocity, the weight of the drop is balanced by upthrust so to say the net unbalanced force on drop is zero.

Question 2.
When a carpet is beaten with a stick, the dust comes out of it? Explain.
Answer:
When a carpet is beaten with a stick, the carpet is set into motion while the dust particles due to inertia tend to remain at rest. In this way dust particles get detached from the carpet and come out of it.

Question 3.
Why is it advised to tie any luggage kept on the roof of a bus with rope?
Answer:
When a fast-moving bus suddenly takes a turn round a sharp bend then the luggage placed on the roof of a bus gets displaced. The reason for this is that the luggage tends to remain with linear motion while an unbalanced force is applied by the engine to change the direction of the bus so that the luggage kept at the roof of the bus gets displaced. So it is advised to tie the luggage with a rope on the roof of bus.

Question 4.
A batsman hits a cricket ball when then rolls on a level ground. After covering short distance, the ball comes to rest. The ball comes to a stop because
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball so that ball would want to come to rest.
Answer:
(c) is correct. There is a force of friction on the ball in direction opposite to that of motion.

Question 5.
A truck starts from rest and rolls down a hill with constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force on it if its mass is 7 metric tonnes. (1 tonne = 100 kg)
Solution:
Here initial velocity (u) = 0
Time (t) = 20 s
Distance (s) = 400 m
S = ut + \(\frac{1}{2}\)at²
400 = 0 × 30 + \(\frac{1}{2}\) × a × (20)²
400 = 0 + \(\frac{1}{2}\) × a × 20 × 20
400 = \(\frac{1}{2}\) × 20 × 20 × a
400 = 200 × a
or a = \(\frac{400}{200}\)
∴ a = 2ms^-2
Now mass of the truck(m) = 7 tonne
= 7 × 1000 kg
Acceleration (a) = 2ms^-2
But Force, F = m × a
= 7000 kg × 2 ms^-2
= 14000 kg – ms^-2
= 14000 N

Question 6.
A stone of 1 kg is thrown with a velocity of 20 m s^-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Solution:
Here, mass of stone (m) = 1 kg
Initial velocity of stone (u) = 20 ms^-1
Distance travelled by the stone (S) = 50 m
Final velocity of stone (υ) = 0 (at rest)
Force of friction between the stone and ice (F) = ?
Using υ² – u² = 2aS, we have
(0)² – (20)² = 2 × (a) × 50
– 20 × 20 = 100 × a
a = \(\frac{-20 \times 20}{100}\)
a = – 4 ms^-2
F = ma = 1 × (- 4)
F = – 4 N
Minus sign shows the force of friction is in direction opposite to direction of motion of stone.

Question 7.
A 8,000 kg engine pulls a train of 5 wagons, each of 2,000 kg along a horizontal track. If the engine exerts a force of 40,000 N and track offers a force of friction of 35,000 N, then calculate the
(a) net accelerating force;
(b) acceleration of the train; and
(c) force of wagon 1 on wagon 2.
Solution:
Mass of the engine = 8000 kg
Mass of 5 wagons = 5 × 2000 kg = 10,000 kg
∴ Total mass of engine and 5 wagons = 8000 kg + 10,000 kg = 18,000 kg
Total force of engine = 40,000 N
Frictional force offered by the track= 5000 N
(a) ∴ Net Accelerating Force (F) = Total force of engine – Frictional force of track
= 40,000 N – 5000 N = 35,000 N

(b) Acceleration of the train (a) = \(\frac{Accelerating force on rail(F)}{Mass of the train(m)}\)
= \(\frac{35000}{18000}\)
= \(\frac{35}{18}\)m^-2
= 1.94ms^-2

(c) Force exerted by wagon 1 on wagon 2 = Net Accelerating Force – Mass of wagon × Acceleration

= 35000 – 2000 × \(\frac{35}{18}\)
= 35000 – 3888.8
= 31111.2 N

Question 8.
An automobile vehicle has mass of 1,500 kg. What must be the force between vehicle and the road if vehicle is to be stopped with negative acceleration of 1.7 ms^-2?
Solution:
Here the mass of automobile (m) = 1500 kg
Acceleration of vehicle (a) = -1.7ms^-2
Frictional Force between road and vehicle (F) = ?
We know, F = m x a
= 15000 x (- 1.7)
= – 2550 N
∴ Backward frictional force (F) = 2550 N

Question 9.
What is the momentum of an object of mass m moving with velocity υ?
(a) (mυ)²; (b) mυ²; (c) \(\frac{1}{2}\)mυ², (d) mυ.
Answer:
(d) is correct. Momentum = mυ.

Question 10.
Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor with constant velocity. What is the friction force that will be exerted on the cabinet?
Answer:
When no acceleration is to be produced (i.e. body is to be moved with constant velocity), the net force has to be zero. Force of friction should be equal and opposite to the force applied i.e., force of friction has to be 200 N.

Question 11.
Two objects, each of mass 1.5 kg, are moving in the same straight line but in the opposite directions. The velocity of each object is 2.5 ms-1 before the collision during which they stick together. What will be the velocity of combined object after collision?
Solution:
Given, mass of the lirst object (m₁) – 1.5 kg
, and mass of the second object (m₂) = m1 = 1.5 kg
Initial velocity of the first object (u₁) = 2.5 ms^-1
Initial velocity of the second object (u₂) = – 2.5 ms^-1
(Since both the objects move in the direction opposite to each other therefore velocity of first object will be taken as positive and that of the other as negative.)
Suppose after collision the velocity of the combination of two objects is ‘υ’
∴ According to the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = m₁υ + m₂υ
1.5 × 2.5 + 1.5 × (-2.5) = (1.5 × υ + 1.5 × υ)
1.5 [2.5 + (-2.5)] = (1.5 + 1.5) × υ
1.5 [2.5 – 2.5] = 3 × υ
1.5 × 0 = 3 × υ
0 = 3 × υ
∴ υ = 0 ms^-1
i. e. Both the objects will come to rest after collision.

Question 12.
According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the road side, it will probably not move. A student justifies by answering that the two forces cancel each other. Comment on the logic and explain why the truck does not move.
Answer:
Student is justified. Friction is equal and opposite to force applied till the force applied crosses the force of limiting friction. When he applies a force slightly more than force of limiting friction, the truck will move. Till the truck moves uniformly, the force applied is exactly equal to force of friction at that instant.

Question 13.
A hockey ball of mass 200 g travelling at 10 ms^-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms^-1. Calculate the change in momentum occurred in the motion of hockey ball by the force applied by hockey stick.
Sol. Mass of the ball (m) = 200 g
= \(\frac{200}{1000}\)kg = 0.2 kg
Initial velocity of the ball (u) = 10 ms^-1
Final velocity of the ball (v) = – 5ms^-1
[∵ the direction of the ball is opposite to the first direction]
Change in momentum of the ball = Final momentum – Initial momentum.
= mυ – mu
= m (υ – u)
= 0.2 (- 5 – 10)
= 0.2 × (- 15)
= – 3.0 kg – ms^-1

Question 14.
A bullet of mass 10 kg travelling horizontally with a velocity of 150 ms^-1 strikes a stationary wooden block and come to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Calculate the magnitude of force exerted by the wood in block in the bullet.
Solution:
Here, mass of the bullet (m) = 10 g = 0.01 kg
Initial velocity of the bullet (u) = 150 ms^-1
Final velocity of the bullet (υ) = 0
Time (t) = 0.03 s
We know, acceleration of the bullet (a) = \(\frac{v-u}{t}\)
= \(\frac{0-150}{0.03}\)
= – 5000 ms^-2

(a) Net force exerted by the wooden block on the bullet (F) = m × a
= 0.01 × (- 5000)
= 1 × (-50)
= -50N
∴ Magnitude of force = 50 N

(b) Distance covered by the bullet after penetration in the wooden block (S) = ?
using S = ut + \(\frac{1}{2}\)at²
= 15 × 0.03 + \(\frac{1}{2}\) × (- 5000) × (0.03)²
= 4.5 + (- 2.25)
= 4.5 – 2.25
S = 2.25 m

Question 15.
An object of mass 1 kg travelling in straight line with a velocity of 10 ms^-1 collides with it and sticks to a stationary wooden block of mass 5 kg. Then both move off together in the same straight line. Calculate the total momentum before the impact and just after the impact. Also calculate the velocity of combined object.
Solution:
Mass of the object (m₁) = 1 kg
Initial velocity of the object (u₁) = 10 ms^-1
Mass of the wooden block (m₂) = 5 kg
Initial velocity of the wooden block (u₂) = 0 [Wooden block at rest]
Suppose ‘υ’ is the velocity of the combination of the object and wooden block after collision
∴ Before collision total momentum of the object and block
= m₁u₁ + m₂u₂
= 1 × 10 + 5 × 0
= 10 + 0
= 10kg ms^-1 ……………… (i)
After collision. Total momentum of the object and block = m₁υ + m₂υ
= (m₁ + m₂) × υ
= (1 + 5) × υ
= 6υ kg m – ms ……… (ii)
According to the law of conservation of momentum,
Total momentum of the combinations before collision = Total momentum of the combination after collision 10 = 6υ
υ = \(\frac{10}{6}\)
∴ υ = \(\frac{5}{3}\) ms^-1
= 1.67 ms^-1
Substituting the value of υ in (ii) above
∴ Total momentum of the combination (after collision) = 6υ
= 6 × \(\frac{5}{3}\)
= 10kg – ms^-1

Question 16.
An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms^-1 to 8 ms^-1 in 6s. Calculate the initial and final momentum of the object. Also find the force exerted on the object.
Solution:
Here, mass of the object (m) = 100kg
Initial velocity of the object (u) = 5ms^-1
Final velocity of the object (v) = 8ms^-1
Time interval (t) = 6s
Initial momentum of the object (p₁) = m × u
= 100 × 5 = 500 kg – ms^-1
Final momentum of the object (p₂) = m × υ
= 100 × 8 = 800 kg – ms^-1
Force acting on the object (F) = \(\frac{p_{2}-p_{1}}{t}\)
= \(\frac{(800-500) \mathrm{kg}-\mathrm{ms}^{-1}}{6 \mathrm{~s}}\)
= 50kg – ms^-2 = 50N

Question 17.
Akhtar, Kiran and Rahul were riding in a motor car that was moving with a high velocity on an express-way when an insect hit the windshield and got struck on wind-screen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of motor car (because change in the velocity of insect was much more than that of motor car). Akhtar said that since the motor car was moving with a larger velocity, it exerted a larger force on the insect. As a result, the insect died. Rahul while putting in entirely new explanation said that both the motor car and the insect experienced the same force and same change in their momentum. Comment on these suggestions.
Answer:
I agree with Rahul’s explanation. According to law of conservation of momentum, during collision, the momentum of the system (insect and motor car) remains conserved. Therefore, both insect and motor car experience the same force and hence same change in momentum. The insect having smaller mass would suffer greater change in velocity as a result of this, it will crush the insect while the motor car does not suffer any noticeable change in velocity.

Question 18.
How much momentum will a dumbbell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 ms^-2.
Solution:
Here, momentum of the dumb-bell (m) = 10 kg
Initial velocity of the bell (u) = 0 (at rest)
Distance covered by the bell (S) = (h) = 80 cm
= 0.80 m
Acceleration of the ball (a) = 10 ms^-2 (downward direction)
Let υ be the final velocity of the bell on reaching the ground.
Using υ² – u² = 2aS
υ² – (0)² = 2 × 10 × 0.80
υ² = 2 × 10 × 0.80
or υ² = 16
∴ Final velocity of the bell (υ) = \(\sqrt{16}\) = 4ms
Momentum transferred by the bell to the floor (p) = m × υ
= 10 × 4 = 40 kg – ms^-1

Science Guide for Class 9 PSEB Force and Laws of Motion InText Questions and Answers

Question 1.
Which of the following has more inertia:
(a) rubber ball and a stone of the same size?
(b) a bicycle and a train?
(c) a five rupees coin and a one-rupee coin?
Answer:
We know that mass of an object is the measure of its inertia. The more is the mass of an object, the more is its inertia hence.
(a) a stone of the same size has more inertia than a rubber ball.
(b) a train has more inertia than a bicycle.
(c) a ₹ 5 coin has more inertia than ₹ 1 coin because a ₹ 5 coin has more mass than a ₹ 1 coin.

Question 2.
In the following example, try to identify the number of times the velocity of the ball changes:
“A football player kicks a football to another player of his team who kicks the football towards the goal-keeper. The goal-keeper of opposite team collects the football and kicks it towards a player of his own team”. Also identify the agent supplying the force in each case.
Answer:
To push, to strike or to pull, all these activities act as a force for changing the velocity or for changing the direction of motion of the object. Therefore, in the above given example the velocity of ball changes four times.

1. First time the football player of first-team kicks the football to another player of his team and thus changes the velocity of the football.
2. In second time velocity changes when the second player kicks the football towards the goal-keeper of the opposite team and applies force on the ball.
3. Third time the goal-keeper pushes the ball and reduces its velocity to zero by applying force.
4. The goal-keeper now applies a force by kicking the football towards player of his team. In this case the force increases the velocity of the football.

Question 3.
Explain why some of the leaves may get detached from the tree if we vigorously shake its branch.
Answer:
Before shaking, the branch of the tree, both the branch and leaves were at rest. When we shake the branch of the tree, branch moves but the leaves remain at rest due to inertia of rest and get detached from the branch.

Question 4.
Why do you fall in the forward direction when a moving bus brakes to a stop and falls backward when it accelerates from rest?
Answer:
When the bus is moving, whole of our body is moving forward. When brakes are applied, the lower part of our body touching the bus (e.g., feet etc.) comes to rest and upper part of the body not touching the bus continue move forward due to inertia of motion and fall in forward direction.

When the bus suddenly starts and accelerates from rest, the lower part of our body starts moving forward (accelerating) along with the bus while upper part of our body tends to remain at rest due to inertia of rest and we fall backward.

Question 5.
If action is always equal to reaction, explain how a horse can pull a cart?
Answer:
According to Newton’s third law of motion “Action and Reaction are equal and opposite.”
The horse pulls (Action) the cart with some force in the forward direction and the cart applies equal force on the cart in the backward direction (Reaction). These two forces balance each other. When the horse pushes the ground with its feet in the backward direction with force P along OP it gets reaction R due to ground along OR in the upward direction.

This force of reaction can be resolved into two rectangular components.
1. Vertical component ‘V’ which balances the weight mg of the horse and cart in the downward direction.
The horizontal component ‘H’ which helps to move the cart in the forward direction. The force of friction between wheels and ground acts in the backward direction but the horizontal component ‘H’ acts in the forward direction is more than the backward force of friction, it succeeds to move the cart forward.

Question 6.
Explain why is it difficult for a fireman to hold a hose, which ejects large amount of water at a high velocity?
Answer:
Water is ejected from rubber hose in forward direction with a force (action), it exerts an equal reaction on the hose in backward direction. Due to backward reaction, fire man finds it difficult to hold the hose.

Question 7.
From a rifle of mass 4 kg, a bullet df> mass 50 g is fired with an initial velocity of 35 m s^-1. Calculate the recoil velocity of the rifle.
Solution:
Mass of the bullet (m₁) = 50 g = 0.05 kg
Mass of the rifle (m₂) = 4 kg
Initial velocity of the bullet (u₁) = 0
Initial velocity of the rifle (u₂) = 0
Final velocity of the bullet (υ₁) = 35 m s^-1
Final velocity of the rifle (υ₂) = ?
According to the law of conservation of momentum,
Total initial momentum of bullet and rifle = Total final momentum of the bullet and rifle.

∴ Negative sign indicates that the rifle moves in a direction opposite to the direction of motion of the bullet.

Question 8.
Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 ms^-1 and 1ms^-1 respectively. They collide and after the collision, the first object moves with a velocity of 1.67 m s^-1. Determine the velocity of the second object.
Solution: