Bhagya

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 1 Knowing Our Numbers Ex 1.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.2

1. Convert the following measurements as directed:

Question (a)
5 km into metre
Solution:
1 km= 1000 m
∴ 5 km = 5 × 1000 m
= 5000 m

Question (b)
35 kilometre into metre
Solution:
1 km = 1000 m
∴ 35 km = 35 × 1000 m
= 35000 m

Question (c)
2000 milligram into gram
Solution:
1000 mg = 1 gm
∴ 2000 mg = \(\frac {1}{1000}\) × 2000 gm
= 2 gm

Question (d)
500 decigram into gram
Solution:
10 decigram = 1 gm
∴ 500 decigram = \(\frac {1}{10}\) × 500 gm
= 50 gm

Question (e)
2000 millilitre into litre
Solution:
1000 ml = 1 litre
∴ 2000 ml = \(\frac {1}{1000}\) × 2000 litre
= 2 litre

Question (f)
12 kilolitre into litre
Solution:
1 kilolitre = 1000 litre
∴ 12 kilolitre = 12 × 1000 litres
= 12000 litres

2. In an election, the successful candidate registered 6317 votes whereas his nearest rival could attain only 3761 votes. By what margin did the successful candidates defeat his rival?
Solution:
Votes attained by successful candidate = 6317
Votes attained by nearest rival = 3761
Difference between their votes = 6317 – 3761 = 2556
Successful candidate defeat his rival by 2556 votes.

3. A monthly magazine having 37 pages is published on 20th day of each month. This month 23791 copies were printed. Tell us how many pages were printed in all?
Solution:
Number of pages in one copy = 37
Number of pages in 23791 copies
= 23791 × 37
= 880267

4. A shopkeeper has 37 reams. One ream contain 480 pages and he wants to make quires of all these sheets to sell in retail. One quire of sheets contain 24 sheets. How many quires will be made?
Solution:
Number of pages in one ream = 480
Number of pages in 37 reams = 37 × 480
= 17760

Number of quires in 24 sheets = 1
Number of quires in 17760 sheets
= \(\frac {1}{24}\) × 17760
= 740 quires

5. Veerpal serves milk to the guests in glasses of capacity 250 ml each. Suppose that the glasses are filled to capacity and there was 5 litre milk that got consumed. How many guests were served with milk?
Solution:
Total quantity of milk consumed = 5 litre
= 5 × 1000 ml
= 5000 ml
Capacity of the glass = 250 ml
Number of glasses served = 5000 ÷ 250 = 20
Now

The milk is served in 20 glasses

6. A box of medicine contain 2,00,000 tablets each weighing 20 mg. What is the total weight of tablets inbox?
Solution:
Weight of each tablet = 20 mg
Weight of 2,00,000 tablets
= 2,00,000 × 20 mg
= 40,00,000 mg
= \(\frac {40,00,000}{1000}\) g
= 4000 g
= \(\frac {4000}{1000}\) = 4 kg
Hence, total weight of tablets is 4 kg

7. A bookstore sold books worth Rupees Two lakh eighty-five thousand eight hundred ninety-one in the first week of June. They sold books worth Rupees Four lakh seven hundred sixty-eight in the second week of June. How much was the total sale for two weeks together?
Solution:
Worth of books sold in first week = Rupees Two lakh eighty-five thousand eight hundred ninety one only.
= ₹ 2,85,891
Worth of books sold in second week = Rupees Four lakh seven hundred sixty-eight = ₹ 4,00,768.
Total sale for two weeks

8. A famous cricket player has so far scored 6978 runs in test matches. He wishes to complete 10,000 runs. How many more runs he need?
Solution:
The number of runs player wishes to complete = 10,000
The number of runs he scored = – 6,978
The number of more runs he needed = 3,022.
He needed 3,022 more runs

9. Surinder has ₹ 78592 with him. He placed an order for purchasing 39 radio sets at ₹ 1234 each. How much money will remain with him after the purchase?
Solution:
Cost of one radio set = ₹ 1234
Cost of 39 radio sets = 39 × ₹ 1234
= ₹ 48126

Total money Surinder has = ₹ 78592
Cost of 39 radio sets = – ₹ 48126
Money remained with him = ₹ 30466

10. A vessel has 3 litre 650 ml of curd. In how many glasses each of 25 ml capacity can it be distributed?
Solution:
Total quantity of curd = 3 l 650 ml
= 3 × 1000 ml + 650 ml
= 3000 ml + 650 ml
= 3650 ml
Capacity of one glass = 25 ml
Number of glasses distributed
= 3650 ÷ 25 = 146

∴ The curd can be distributed in 146 glasses.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 1 Number Systems Ex 1.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5

Question 1.
Classify the following numbers as rational or irrational:
(i) 2 – √5
Answer:
2 – √5 is an irrational number as it is the difference of rational number (2) and irrational number (√5).

(ii) (3 + √23) – √23
Answer:
Thus, (3 + √23) – √23 is a rational number even if it is the difference of two irrational numbers (3 + √23) and √23.

(iii) \(\frac{2 \sqrt{7}}{7 \sqrt{7}}\)
Answer:
\(\frac{2 \sqrt{7}}{7 \sqrt{7}}\) = \(\frac{2}{7}\), Thus, \(\frac{2 \sqrt{7}}{7 \sqrt{7}}\) is a rational number even if it is a quotient of two irrational numbers 2√7 and 7√7.

(iv) \(\frac{1}{\sqrt{2}}\)
Answer:
\(\frac{1}{\sqrt{2}}\) is an irrational number as it is the quotient of rational number (1) and irrational number (√2).

(v) 2π
Answer:
2π is an irrational number as it is the product of rational number 2 and irrational number π.

Question 2.
Simplify each of the following expressions:
(i) (3 + √3) (2 + √2)
Answer:
(3 + √3) (2 + √2) = 6 + 3√2 + 2√3 = √6

(ii) (3 + √3) (3 – √3)
Answer:
(3 + √3) (3 – √3) = (3)² – (√3)² = 9 – 3 = 6

(iii) (√5 + √2)²
Answer:
(√5)² + 2(√5) (√2) + (√2)²
= 5 + 2√10 + 2
= 7 + 2√10

(iv) (√5 – √2) (√5 + √2)
Answer:
(√5 – √2) (√5 + √2) = (√5)² – (√2)²
= 5 – 2 = 3

Question 3.
Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = \(\frac{c}{d}\). This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
Answer:
π is defined as the ratio of circumference (C) of a circle to its diameter (d). As seen in the process of successive magnification used to represent real numbers on the number line, we see that more and more accuracy can be obtained by successive magnification. But, since the real numbers exhibit gaps. we can never measure the exact length of the circumference and the diameter. Any one or both may be having length represented by an irrational number. Hence, there is no contradiction that π being the ratio of c and d is still an irrational number.

Note: in the study of mathematics at higher level, you may study that π is a transcendental number and also the proof of π being an irrational number.

Question 4.
Represent √9.3 on the number line.
Answer:

Steps of construction:

• Draw ray AX.
• Mark B on ray AX such that AB = 9.3 cm.
• Mark C on ray BX such that BC = 1 cm.
• By drawing perpendicular bisector of seg AC, obtain its midpoint P.
• Draw a semicircle with centre P and radius AP.
• Draw perpendicular to seg AC at B intersecting the semicircle at D.
• Then, BD = √9.3.
• Let line AX be the number line on which B corresponds to 0 and BC = 1 unit.
• Draw an arc with centre B and radius BD to intersect ray BX at K.
• Point K represents √9.3 on the number line.

Question 5.
Rationalise the denominators of the following:
(i) \(\frac{1}{\sqrt{7}}\)
Answer:
\(\frac{1}{\sqrt{7}}\) = \(\frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}\)
= \(\frac{\sqrt{7}}{7}\)

(ii) \(\frac{1}{\sqrt{7}-\sqrt{6}}\)
Answer:
\(\frac{1}{\sqrt{7}-\sqrt{6}}\) = \(\frac{1}{\sqrt{7}-\sqrt{6}} \times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}\)
= \(\frac{\sqrt{7}+\sqrt{6}}{7-6}\)
= √7 + √8

(iii) \(\frac{1}{\sqrt{5}+\sqrt{2}}\)
Answer:
\(\frac{1}{\sqrt{5}+\sqrt{2}}\) = \(\frac{1}{\sqrt{5}+\sqrt{2}} \times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}\)
= \(\frac{\sqrt{5}-\sqrt{2}}{5-2}\)
= \(\frac{\sqrt{5}-\sqrt{2}}{3}\)

(iv) \(\frac{1}{\sqrt{7-2}}\)
Answer:
\(\frac{1}{\sqrt{7-2}}\) = \(\frac{1}{\sqrt{7}-2} \times \frac{\sqrt{7}+2}{\sqrt{7}+2}\)
= \(\frac{\sqrt{7}+2}{7-4}\)
= \(\frac{\sqrt{7}+2}{3}\)

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.