Bhagya

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 1 Knowing Our Numbers Ex 1.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.1

1. Write the smallest and the greatest number:

Question (a)
30900, 30594, 30945, 30495
(b) 10092, 10029, 10209, 10920.
Solution:
(a) All the given numbers are: 30900, 30594, 30945, 30495 are five-digit numbers. Let us examine digits on extreme left side of each number. First digit and second digit of all the numbers are same.

Then by observing the third and fourth digits from left side we conclude that
Smallest number = 30495
Greatest number = 30945

Question (b)
10092, 10029, 10209, 10920.
Solution:
All the given numbers are: 10092, 10029, 10209, 10920 are five digit numbers. Let us examine digits on extreme left side of each number. First digit and second digit from left of all the numbers are same.

Then by observing third and fourth digits from left we conclude that
Smallest number = 10029
Greatest number = 10920

2. Arrange the numbers in ascending order:

Question (a)
6089, 6098, 5231, 3953
Solution:
Ascending order is:
3953, 5231, 6089, 6098

Question (b)
49905, 6073, 58904, 7392
Solution:
Ascending order is:
6073, 7392, 49905, 58904

Question (c)
9801, 25751, 36501, 38802.
Solution:
Ascending order is:
9801, 25751, 36501, 38802

3. Arrange the numbers in descending order:

Question (a)
75003, 20051, 7600, 60632
Solution:
Descending order is:
75003, 60632, 20051, 7600

Question (b)
2934, 2834, 667, 3289
Solution:
Descending order is:
3289, 2934, 2834, 667

Question (c)
1971, 45321, 88715, 92547.
Solution:
Descending order is:
92547, 88715, 45321, 1971.

4. Use the given digits without repetition and make the greatest and smallest 4 digit number:

Question (a)
6, 4, 3, 2
Solution:
6432, 2346

Question (b)
9, 7, 0, 3
Solution:
9730, 3079

Question (c)
5, 4, 0, 3
Solution:
5430, 3045

Question (d)
3, 2, 7, 1.
Solution:
1321, 1237.

5. Using any one digit twice make the greatest and the smallest 4 digit number:

Question (i)
(a) 2, 3,7
(b) 5,0,3
(c) 2, 3, 0
(d) 1, 3, 4
(e) 2, 5, 8
(f) 1, 2, 3
Solution:
(a) 7732, 2237
(b) 5530, 3005
(c) 3320, 2003
(d) 4431, 1134
(e) 8852, 2258
(f) 3321, 1123

6. Read the following numbers using place value chart:

Question (i)
(a) 638975
(b) 84321
(c) 29061058
(d) 60003608.
Solution:
Place Value Chart:

	C	TL	L	TTh	Th	H	T	O
(a)			6	3	8	9	7	5
(b)				8	4	3	2	1
(c)	2	9	0	6	1	0	5	8
(d)	6	0	0	0	3	6	0	8

(a) Six lakh thirty-eight thousand nine hundred seventy-five
(b) Eighty-four thousand three hundred twenty-one
(c) Two crore ninety lakh sixty one thousand fifty-eight
(d) Six crore three thousand six hundred eight.

7. Insert commas suitably and write the names according to Indian System of Numeration:

Question (a)
98606873
Solution:
9,86,06,873
Nine crore eighty-six lakh six thousand eight hundred seventy-three.

Question (b)
7635172
Solution:
76,35,172
Seventy-six lakh thirty-five thousand one hundred seventy-two.

Question (c)
89700057
Solution:
8,97,00,057
Eight crore ninety-seven lakh fifty-seven.

Question (d)
89322602
Solution:
8,93,22,602
Eight crore ninety-three lakh twenty-two thousand six hundred two.

Question (e)
4503217
Solution:
45,03,217
Forty-five lakh three thousand two hundred seventeen.

Question (f)
90032045.
Solution:
9,00,32,045
Nine crore thirty-two thousand forty-five.

8. Insert commas suitably and write the names according to International System of Numeration:

Question (a)
89832081
Solution:
89,832,081
Eighty-nine million eight hundred thirty-two thousand eighty-one.

Question (b)
6543374
Solution:
6,543,374
Six million five hundred fourty three thousand three hundred seventy-four.

Question (c)
88976306
Solution:
88,976,306
Eighty-eight million nine hundred seventy-six thousand three hundred six.

Question (d)
9860001
Solution:
9,860,001
Nine million eight hundred sixty thousand one.

Question (e)
90032045
Solution:
90,032,045
Ninety million thirty-two thousand forty-five.

Question (f)
4503217
Solution:
4,503,217
Four million five hundred three thousand two hundred seventeen.

9. Write the number names as numerals:

Question (a)
Seven lakh fifty-four thousand
Solution:
7,54,000

Question (b)
Nine crore fifty-three lakh seventy-four thousand five hundred twenty-three.
Solution:
9,53,74,523

Question (c)
Six hundred forty-seven thousand five hundred twenty-five.
Solution:
647,525

Question (d)
Seventy-two million three hundred thirty-two thousand one hundred twelve.
Solution:
72,332,112

Question (e)
Fifty-eight million four hundred twenty-three thousand two hundred two.
Solution:
58,423,202

Question (f)
Twenty-three lakh thirty thousand ten.
Solution:
23,30,010.

10. How many eight-digit numbers are there in all?
Solution:
Largest eight-digit number is 99999999.
Largest seven-digit number is 9999999.
Total number of eight digit numbers = Largest eight digit – Largest seven digit number
= 99999999 – 9999999
= 90000000

11. Fill in the blanks:

Question (i)
(a) 1 Lakh = ten thousand
(b) 1 Million = hundred thousand
(c) 1 Crore = ten lakh
(d) 1 Crore = million
(e) 1 Million = lakh.
Solution:
(a) Ten
(b) Ten
(c) Ten
(d) Ten
(e) Ten

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 1 Number Systems Ex 1.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.4

Question 1.
Visualise 3.765 on the number line, using successive magnification.
Answer:

Point P in fig. 4 represents 3.765 on the number line.

Question 2.
Visualise \(4 . \overline{26}\) on the number line, up to 4 decimal places.
Answer:
\(4 . \overline{26}\) = 4.2626 up to 4 decImal places.

Point P in fig 5 represents 4.2626. i.e.. \(4 . \overline{26}\) up to 4 decimal places on the number line.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 7 Congruence of Triangles Ex 7.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.1

1. Identify the pairs of congruent figures and write the congruence in symbolic form.

Question (i).

Answer:
In figure
Radius of circle C₁ = 2 cm
Radius of circle C₂ = 1.5 cm
As radius of circle C₁ ≠ Radius of circle C₂
∴ Circle C₁ is not congruent to circle C₂

Question (ii).

Answer:
In figure
Length of line segment AB = 6 cm
Length of line segment MN = 7 cm
As length of line segment AB ≠ Length of line segment MN
∴ AB is not congruent to MN

Question (iii).

Answer:
In ΔXYZ and ΔPQR
XY = PQ, YZ = PR, XZ = QR
So, ΔXYZ and ΔPQR have the same size and shape
∴ ΔXYZ ≅ ΔQPR

Question (iv).

Answer:
In figure ΔABC and ΔDEF do not have the same size and shape
∴ ΔABC and ΔDEF are do not have the same size and shape.
∴ ΔABC and ΔDEF are not-congruent.

Question (v).

Answer:

Question (vi).

Answer:

2. If ΔPQR as ΔOMN under the correspondence PQR ↔ OMN, write all the corresponding congruent parts of the triangle.
Solution:
For better understanding of the correspondence, let us draw a diagram of given correspondence.

The correspondence is PQR → OMN
This means vertices P ↔ O, Q ↔ M, R ↔ N
Sides : PQ ↔ OM, QR ↔ MN, RP ↔ NO
and Angles: ∠PQR ↔ ∠OMN, ∠QRP ↔ ∠MNO, ∠RPQ ↔ ∠NOM

3. Draw any two pairs of congruent triangles.
Solution:
Two pairs of congruent triangles are :
(i) Draw a ΔABC in which AB = 5 cm, BC = 4 cm and CA = 6 cm.
Draw another ΔPQR in which PQ = 6 cm, QR = 5 cm and RP = 4 cm shown in the following figure.


Make a trace copy of ΔABC using a tracig paper and superinpose it on ΔPQR, where C falls on P, A falls on Q and B falls on R. We observe that ΔABC will corr. ΔPQR.
∴ ΔABC ≅ ΔQRP

(ii) Draw a ΔXYZ in which XY = 5 cm, YZ = 6 cm and ZX = 3 cm. Draw another ΔLMN in which LM = 5 cm, MN = 6 cm and NL = 3 cm. Since both ΔXYZ and ΔLMN have the same size and shape.

4. If ΔABC ≅ ΔZYX, write the parts of ΔZYX that correspond to.
(i) ∠B
(ii) CA
(iii) AB
(iv) ∠C
Solution:
First of all we draw a diagram of given correspondence.

The correspondence is ABC ↔ ZYX.
This means A ↔ Z, B ↔ Y, C ↔ X
Therefore
(i) ∠B = ∠Y
(ii) CA = XZ
(iii) AB = ZY
(iv) ∠C = ∠X

5. Multiple Choice Questions :

Question (i).
If ΔABC as ΔXYZ under the correspondence ABC ↔ XYZ. Then
(a) ∠A = ∠Z
(b) ∠X = ∠B
(c) ∠A = ∠X
(d) ∠C = ∠X.
Answer:
(c) ∠A = ∠X

Question (ii).
Two line segments are congruent if,
(a) They are parallel
(b) They intersect each other
(c) They are part of same line
(d) They are of equal length.
Answer:
(d) They are of equal length.

Question (iii).
Two triangles ΔABC and ΔLMN are congruent AB = LM, BC = MN. If AC = 5 cm then LN is :
(a) 3 cm
(b) 15 cm
(c) 5 cm
(d) Can’t find.
Answer:
(c) 5 cm

6. Two right angles are always congruent. (True/False)
Answer:
True

7. Two opposite sides of a rectangle are always congruent. (True/False)
Answer:
True

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 1 Number Systems Ex 1.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3

Question 1.
Write the following in decimal form and say
what kind of decimal expansion each has:

(i) \(\frac{36}{100}\)
Answer:
The decimal expansion of \(\frac{36}{100}\) is terminating.

(ii) \(\frac{1}{11}\)
Answer:

Thus, \(\frac{1}{11}\) = 0.0909…. = \(0.\overline{09}\)
The decimal expansion of \(\frac{1}{11}\) is non- terminating.

(iii) 4\(\frac{1}{8}\)
Answer:
4\(\frac{1}{8}\) = \(\frac{33}{8}\)

Thus, 4\(\frac{1}{8}\) = 4.125
The decimal expansion of 4\(\frac{1}{8}\) is terminating.

(iv) \(\frac{3}{13}\)
Answer:

Thus, \(\frac{3}{13}\) = 0.230769230769…. = \(0.\overline{230769}\)
The decimal expansion of 4\(\frac{2}{11}\) is non-terminating recurring.

(v) \(\frac{2}{11}\)
Answer:

Thus, \(\frac{2}{11}\) = 0.1818….. = \(0 . \overline{18}\)
The decimal expansion of \(\frac{2}{11}\) is non-terminating recurring.

(vi) \(\frac{329}{400}\)
Answer:

Thus, \(\frac{329}{400}\) = 0.8225
The decimal expansion of \(\frac{329}{400}\) is terminating recurring.

Question 2.
You know that \(\frac{1}{7}\) = \(0 . \overline{142857}\). Can you predict what the decimal expansions of \(\frac{2}{7}\), \(\frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}\) are, without actually doing the long division? if so, how? [Hint: Study the remainders while finding the value of \(\frac{1}{7}\) carefully.]
Answer:
While finding the decimal expansion of \(\frac{1}{7}\) by long division, we observe that the remainder 3, 2, 6, 4, 5 and 1 repeat in that order and the digits 1, 4, 2, 8, 5 and 7 repeat in that order in the quotient. So, in the decimal expansion of \(\frac{2}{7}\), the same digits recur beginning with 2. Thus,

Question 3.
Express the following in the form \(\frac{p}{q}\), where p and q are integers and q ≠ 0:
(i) 0.\(\overline{6}\)
Answer:
Let x = 0.\(\overline{6}\)
∴ x = 0.6666…
∴ 10x = 6.6666…
∴ 10x = 6 + x
∴ 10x – x = 6
∴ 9x = 6
∴ x = \(\frac{6}{9}\)
∴ x = \(\)
Thus. 0.\(\overline{6}\) = \(\frac{2}{3}\)

(ii) 0.4\(\overline{7}\)
Answer:
Let x = 0.4\(\overline{7}\)
∴ x = 0.4777…..
∴ 10x = 4.7777…..
∴ 10x = 4.3 + 0.4777…..
∴ 10x = 4.3 + x
∴ 9x = 4.3
∴ 9x = \(\frac{43}{10}\)
∴ x = \(\frac{43}{90}\)
Thus, 0.4\(\overline{7}\) = \(\frac{43}{90}\)

(iii) 0.\(\overline{001}\)
Answer:
Let x = 0.\(0 . \overline{001}\)
∴ x = 0.001001001 ……
∴ 1000 x = 1.001001001 ……
∴ 1000 x = 1 + 0.001001001 ……
∴ 1000 x = 1 + x
∴ 999 x = 1
∴ x = \(\frac{1}{999}\)
Thus, \(0 . \overline{001}\) = \(\frac{1}{999}\)

Question 4.
Express 0.99999 ….. in the form \(\frac{p}{q}\). Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Answer:
Let x = 0.99999 …..
∴ 10x = 9.99999 ……
∴ 10x = 9 + 0.99999 ……
∴ 10x = 9+x
∴9x = 9
∴ x = 1
Thus, 0.99999 ….. = 1
in other words, 0.\(\overline{9}\) = 1.
This also follows the rule mentioned in ‘Remember’ section that 0.\(\overline{m}\) = \(\frac{m}{9}\). As per rule, 0.\(\overline{9}\) = \(\frac{9}{9}\) = 1.

The answer seems to be tricky. Actually, there is no situation that gives the decimal expansion of a rational number \(\left[\frac{p}{q}\right]\) as 0.\(\overline{9}\).
This is an imaginary illustration of recurring decimal number.
You may say that 0.\(\overline{3}\) + 0.\(\overline{6}\) = 0.\(\overline{9}\) But in simple fraction form 0.\(\overline{3}\) and 0.\(\overline{6}\) add up to give 1. Thus, 0.\(\overline{9}\) = 1.

Question 5.
What can the maximum number of digits be in the repeating block of digits in the decimal expansion of \(\frac{1}{17}\)? Perform the division to check your answer.
Answer:
The maximum number of digits in the repeating block of digits in the decimal expansion of \(\frac{1}{17}\) can be 16(17 – 1).

Thus, \(\frac{1}{17}\) = \(0.\overline{0588235294117647}\)
Here, 17 – 1 16 gives the maximum possible number of digits in the repeating block of digits in the decimal expansion of \(\frac{1}{17}\). In some other cases, the number of digits in the repeating block of digits in the decimal expansion of \(\frac{1}{m}\) can be smaller than (m – 1), where m is a natural number.

For example, \(\frac{1}{13}\) = \(0 . \overline{076923}\) has only 6 digits (not 12) in the repeating block of digits in the decimal expansion.

Question 6.
Look at several examples of rational numbers in the form \(\frac{p}{q}\) (q ≠ 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?
Answer:
\(\frac{1}{2}\) = 0.5, \(\frac{1}{5}\) = 0.2, \(\frac{1}{10}\) = 0.1,
\(\frac{1}{4}\) = 0.25, \(\frac{1}{8}\) = 0.125, \(\frac{1}{16}\) = 0.0625,
\(\frac{1}{25}\) = 0.04, \(\frac{1}{125}\) = 0.008, \(\frac{1}{625}\) = 0.0016,
\(\frac{1}{20}\) = 0.05, \(\frac{1}{50}\) = 0.02 etc.
But, \(\frac{1}{3}\) = 0.\(\overline{3}\), \(\frac{1}{7}\) = \(0 . \overline{142857}\), \(\frac{1}{6}\) = 0.1\(\overline{6}\) etc.

This suggest that the decimal expansion of a rational number \(\frac{\boldsymbol{P}}{\boldsymbol{q}}\) is terminating if and only if q has no prime factors other than 2 and 5. in other words, q = 2^m5ⁿ, where m and n are whole numbers.

Question 7.
Write three numbers whose decimal expansions are non-terminating non-recurring.
Answer:
We know that the decimal expansion of an irrational number is non-terminating non recurring. There are infinitely many irrational numbers. We can state few of them as below:
o.o1001000100001 …,
0.02002000200002…., 0.50500500050000… are a few required numbers in the decimal form. The decimal expansions of numbers like √2, √3, √5, \(\sqrt[3]{10}\), etc. are also non- terminating non-recurring.

Question 8.
Find three different irrational numbers between the rational numbers \(\frac{5}{7}\) and \(\frac{9}{11}\)
Answer:
We know that \(\frac{5}{7}\) = \(0 . \overline{714285}\) and \(\frac{9}{11}\) = \(0 . \overline{81}\)
There are actually infinitely many irrational numbers between these recurring decimals. Three irrational numbers between them can be stated as below:
0.720720072000 …, 0.750750075000 …
0.780780078000….
As you see, the above numbers are non-terminating non-recurring. Hence, they are irrational numbers.

Question 9.
Classify the following numbers as rational or irrational:
(i) √23
Answer:
√23 is an irrational number.

(ii) √225
Answer:
√225 = 15 is a rational number

(iii) 0.3796
Answer:
0.3796 is a rational number.

(iv) 7.478478……..
Answer:
7.478478…… = \(7 . \overline{478}\) is a rational number.

(v) 1.101001000100001…….
Answer:
1.101001000100001……. is an irrational number.