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Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 1 Number Systems MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 1 Number Systems MCQ Questions

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
The \(\frac{p}{q}\) form of \(0 . \overline{6}\) is ……………
A. \(\frac{3}{2}\)
B. \(\frac{2}{3}\)
C. \(\frac{9}{6}\)
D. \(\frac{6}{10}\)
Answer:
B. \(\frac{2}{3}\)

Question 2.
The \(\frac{p}{q}\) form of \(3.1\overline{23}\) is ……………..
A. \(\frac{3123}{999}\)
B. \(\frac{3123}{1000}\)
C. \(\frac{1546}{495}\)
D. \(\frac{3123}{990}\)
Answer:
C. \(\frac{1546}{495}\)

Question 3.
The \(\frac{p}{q}\) form of \(2.\overline{237}\) is ……………
A. \(\frac{2235}{999}\)
B. \(\frac{2235}{99}\)
C. \(\frac{2237}{990}\)
D. \(\frac{2237}{1000}\)
Answer:
A. \(\frac{2235}{999}\)

Question 4.
The decimal expression of \(\frac{5}{6}\) is ………………… .
A. non-terminating recurring
B. non-terminating non-recurring
C. un-determinate
D. terminating
Answer:
A. non-terminating recurring

Question 5.
(√3 + √2) (√3 – √2) = …………..
A. √3
B. √2
C. 5
D. 1
Answer:
D. 1

Question 6.
6√20 ÷ 2√5 = …………..
A. 6
B. 3
C. 3√5
D. 4√5
Answer:
A. 6

Question 7.
– \(\frac{\sqrt{48}}{\sqrt{27}}\) is a/an …………… .
A. Irrational number
B. negative Integer
C. positive Integer
D. rational number
Answer:
D. rational number

Question 8.
(2^{– 2})^{– 3} =
A. 8²
B. 8⁴
C. 15²
D. 15⁴
Answer:
A. 8²

Question 9.
\(5^{\frac{3}{4}} \times 5^{\frac{1}{4}}\) = ……….
A. 5
B. 5⁴
C. 5³
D. 5²
Answer:
A. 5

Question 10.
………….. are equivalent rational numbers.
A. \(\frac{26}{39}\) and \(\frac{51}{34}\)
B. \(\frac{33}{22}\) and \(\frac{65}{52}\)
C. \(\frac{14}{21}\) and \(\frac{27}{18}\)
D. \(\frac{63}{42}\) and \(\frac{69}{46}\)
Answer:
\(\frac{63}{42}\) and \(\frac{69}{46}\)

Question 11.
……………… is a rational number between 5 and 6.
A. \(\frac{17}{4}\)
B. \(\frac{17}{3}\)
C. \(\frac{17}{2}\)
D. \(\frac{13}{2}\)
Answer:
B. \(\frac{17}{3}\)

Question 12.
The \(\frac{p}{q}\) form of \(0.3 \overline{5}\) is …………… .
A. \(\frac{16}{45}\)
B. \(\frac{35}{9}\)
C. \(\frac{35}{99}\)
D. \(\frac{35}{90}\)
Answer:
A. \(\frac{16}{45}\)

Question 13.
Since \(\frac{2}{7}\) = \(0 . \overline{285714}\), \(\frac{6}{7}\) =
A. \(0 . \overline{571428}\)
B. \(0 . \overline{142857}\)
C. \(0 . \overline{857142}\)
D. \(0 . \overline{428571}\)
Answer:
C. \(0 . \overline{857142}\)

Question 14.
√1 + √4 is a/an …………… .
A. natural number
B. irrational number
C. negative number
D. fractional number
Answer:
A. natural number

Question 15.
√2 + √2 is a/an ……………. .
A. Integer
B. irrational number
C. rational number
D. whole number
Answer:
B. irrational number

Question 16.
2 √18 ÷ √50 is a / an …………………. .
A. Integer
B. rational number
C. whole number
D. Irrational number
Answer:
B. rational number

Question 17.
(√3 – √2)² is a/an ………………… number.
A. natural
B. irrational
C. rational
D. whole
Answer:
B. irrational

Question 18.
To rationalize the denominator of \(\frac{5}{2-\sqrt{3}}\), it should be multiplied by
A. \(\frac{5}{2-\sqrt{3}}\)
B. \(\frac{5}{\sqrt{3}-2}\)
C. \(\frac{2+\sqrt{3}}{2+\sqrt{3}}\)
D. \(\frac{2-\sqrt{3}}{5}\)
Answer:
C. \(\frac{2+\sqrt{3}}{2+\sqrt{3}}\)

Question 19.
\(\frac{3}{5+2 \sqrt{2}}\) will be expressed as …………………. with rational denominator.
A. \(\frac{15-6 \sqrt{2}}{17}\)
B. \(\frac{15+6 \sqrt{2}}{17}\)
C. \(\frac{15+6 \sqrt{2}}{33}\)
D. \(\frac{15-6 \sqrt{2}}{33}\)
Answer:
A. \(\frac{15-6 \sqrt{2}}{17}\)

Question 20.
If \(\sqrt[n]{a^{2}}\) = b, then b²ⁿ = …………………. ;
(a, b > 0, n is a natural number).
A. a
B. \(a^{\frac{n}{2}}\)
C. a²ⁿ
D. a⁴
Answer:
D. a⁴

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 1 Knowing Our Numbers Ex 1.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3

1. Estimate each of the following using general rule:

Question (a)
837 + 987
Solution:
While rounding off to hundreds place
837 + 987 = 800 + 1000
= 1800

Question (b)
783 – 427
Solution:
While rounding off to hundreds place
783 – 427 = 800 – 400
= 400

Question (c)
1391 + 2783
Solution:
(i) While rounding off to thousands place
1391 + 2783 = 1000 + 3000
= 4000
(ii) While rounding off to hundreds place
1391 + 2783 = 1400 + 2800
= 4000

Question (d)
28292 – 21496.
Solution:
While rounding off to ten thousands place.
28292 – 21496 = 30000 – 20000
= 10000

2. Estimate the product using general rule:

Question (a)
898 × 785
Solution:
898 rounds off to hundreds place = 900
785 rounds off to hundreds place = 800
Estimated product = 900 × 800
= 720000

Question (b)
9 × 795
Solution:
9 rounding off to tens place = 10
795 rounding off to tens place = 800
Estimated product = 10 × 800
= 8000

Question (c)
(c) 87 × 317
Solution:
87 rounded off to hundreds place = 100
317 rounded off to hundreds place = 300
Estimated product = 90 × 300
= 27000

Question (d)
9250 × 29
Solution:
9250 rounds off to thousands place = 9000
29 rounds off to tens place = 30
Estimated product = 9000 × 30
= 270000

3. Estimate by rounding off to nearest hundred:

Question (a)
439 + 334 + 4317
Solution:
439 rounds off to nearest hundreds = 400
334 rounds off to nearest hundreds = 300
4317 rounds off to nearest hundreds = 4300
Estimated sum = 400 + 300 + 4300 = 5000

Question (b)
108734 – 47599.
Solution:
108734 rounds off to nearest hundreds = 108700
47599 rounds off to nearest hundreds = – 47600
Estimated difference = 61100

4. Estimate by rounding off to nearest tens:

Question (a)
439 + 334 + 4317
Solution:
439 + 334 + 4317
439 rounds off to nearest tens = 440
334 rounds off to nearest tens = + 330
4317 rounds off to nearest tens = + 4320
Estimated sum = 5090

Question (b)
108734 – 47599
Solution:
108734 rounds off to nearest tens = 108730
47599 rounds off to nearest tens = – 47600
Estimated difference = 61130

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 1 Number Systems Ex 1.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6

Question 1.
Find:
(i) \(64^{\frac{1}{2}}\)
Answer:
\(64^{\frac{1}{2}}\) = \(\left(2^{6}\right)^{\frac{1}{2}}\)
= \(2^{6 \times \frac{1}{2}}\)
= 2³ = 8

(ii) \(32^{\frac{1}{5}}\)
Answer:
\(32^{\frac{1}{5}}\) = \(\left(2^{5}\right)^{\frac{1}{5}}\)
= \(2^{5 \times \frac{1}{5}}\)
= 2¹ = 2

(iii) \(125^{\frac{1}{3}}\)
Answer:
\(125^{\frac{1}{3}}\) = \(\left(5^{3}\right)^{\frac{1}{3}}\)
= \(5^{3} \times \frac{1}{3}\)
= 5¹
= 5

Question 2.
Find:
(i) \(9^{\frac{3}{2}}\)
Answer:
\(9^{\frac{3}{2}}\) = \(\left(3^{2}\right)^{\frac{3}{2}}\)
= \(3^{2 \times \frac{3}{2}}\)
= 3³
= 27

(ii) \(32^{\frac{2}{5}}\)
Answer:
\(32^{\frac{2}{5}}\) = \(\left(2^{5}\right)^{\frac{2}{5}}\)
= \(2^{5 \times \frac{2}{5}}\)
= 2² = 4

(iii) \(16^{\frac{3}{4}}\)
Answer:
\(16^{\frac{3}{4}}\) = \(\left(2^{4}\right)^{\frac{3}{4}}\)
= \(2^{4 \times \frac{3}{4}}\)
= 2³ = 8

(iv) \(125^{\frac{-1}{3}}\)
Answer:
\(125^{\frac{-1}{3}}\) = \(\left(5^{3}\right)^{-\frac{1}{3}}\)
= \(5^{3 \times \frac{-1}{3}}\)
= 5^-1 = \(\frac{1}{5}\)

Question 3.
Simplify
(i) \(2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}\)
Answer:
\(2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}\) = \(2^{\frac{2}{3}+\frac{1}{5}}\)
= \(2^{\frac{10+3}{15}}\)
= \(2^{\frac{13}{15}}\)

(ii) \(\left(\frac{1}{3^{3}}\right)^{7}\)
Answer:
\(\left(\frac{1}{3^{3}}\right)^{7}\) = (3^-3)⁷
= 3^{– 3 × 7}
= 3^{– 21}

(iii) \(\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}\)
Answer:
\(\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}\) = \(11^{\frac{1}{2}-\frac{1}{4}}\)
= \(11^{\frac{2-1}{4}}\)
= \(11^{\frac{1}{4}}\)

(iv) \(7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}\)
Answer:
\(7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}\) = \((7 \cdot 8)^{\frac{1}{2}}\)
= \(56^{\frac{1}{2}}\)

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