Bhagya

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 2 Polynomials Ex 2.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.1

Question 1.
The graphs of y = p (x) are given in Fig. below, for some polynomials p(x). Find the number of zeroes of p (x), in each case.

Solution:
The graphs of y = p (x) are given in figure above, for some polynomials p (x). The number of zeroes of p (x) in each case are given below:
(i) From the graph, it is clear that it does not meet x-axis at any point.
Therefore, it has NIL; no. of zeroes.

(ii) From the graph, it is clear that it meets x- axis at only one point.
Therefore, it has only one no. of zeroes.

(iii) From the graph, it is clear that it meets x-axis at three points.
Therefore, it has three no. of zeroes.

(iv) From the graph, it is clear that it meets x- axis at two points.
Therefore, it has two no. of zeroes.

(v) From the graph, it is clear that it meets x- axis at four points.
Therefore, it has four no. of zeroes.

(vi) From the graph, it is clear that it meets pr-axis at three points.
Therefore, it has three no. of zeroes.

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) \(\frac {3}{4}\)