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

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

PSEB Solutions for Class 10 Math's 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) 133125

(ii) 178

(iii) 64455

(iv) 151600

(v) 29343

(vi) 232352

(vii) 129255775

(viii) 615

(ix) 3550

(x) 77210
Solution:
(i)Let x = 133125 ………..(1)
Compare (1) with x = pq
Here p = 13 and q = 3125
Prime factors of q = 3125 = 5 × 5 × 5 × 5 × 5 = 55 × 20
which are of the form 2n × 5m here n = 0, m = 5
which are non negative integers.
∴ x = 133125 have a terminating decimal expansion.

(ii) Let x = 178 …………………(1)
Compare (1) with x = pq
Here p = 17 and q = 8
Prime factors of q = 8 = 2 × 2 × 2 = 23 = 23 × 50

which are of the form 2n × 5m here n = 3, m = 0
and these are non negative integers.
∴ x = 178 have a terminating decimal expansion.

(iii) Let x = 64455 …………(1)
Compare (1) with x = pq
Here p = 64, q = 455
Prime factors of q = 455 = 5 × 7 × 13 which are not of the form 2n × 5m
∴ x = 64455 has a non – terminating decimal expansion.

(iv) Let x = 151600 ……….(1)
Compare (1) with x = pq
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 × 5m, here n = 6, m = 2 and these are non negative integers.
∴ x = having terminating decimal expansion.

(v) Let x = 29343 ………… (1)
Compare (1) with x = pq
Here p = 29 and q = 343
prime factors of q = 343
= 7 × 7 × 7 = 73
which are not of the form 2n × 5m, here n = 0, m = 0
∴ x = 29343 will have a non – terminating decimal expansion.

(vi) Let x = 232352 ……….(1)
Compare (1) with x = pq
Here p = 23, q = 2352
Prime factors of q = 2352
which are of the form 2n × 5m, here n = 3, m = 2 and these are non negative integers.
∴ x = 232352 will have a terminating decimal expansion.

(vii) Let x = 129255775 ……………….(1)
Compare (1) with x = pq
Here p = 129 and q = 25 57 75
Prime factors of q = 25 57 75
which are not of the form 2n × 5m,
∴ x = 129255775 have a non – terminating decimal expansion.

(ix) Let x = 3550=710 …………. (1)
Compare (1) with x = pq
Here p = 7, q = 10
Prime factors of q = 10 = 2 × 5 = 21 × 51
Which is of the form 2n × 5m here n = 1, m = 1
both n and m are non negative integer.
∴ x = 3550 have a terminating decimal expansion.

(x) Let x = 77210=1130 ……………..(1)
Compare (1) with x = pq
Here p = 11, q = 30
Prime factors of q = 30 = 2 × 5 × 3
which are not of the form 2n × 5m,
∴ x = 77210 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 = 133125
Compare (1) with x = pq
Here p = 13,q = 3125
Prime factors of q = 3125 = 5 × 5 × 5 × 5 × 5 = 55 × 20
Which are of the form 2n × 5m, where n = 0, m = 5 and these are non negative integers
∴ x = have a terminating decimal expansion.
To Express in Decimal form
x = 133125=1355×20

x = 13×2555×25
[∵ we are to make 10 in the denominator so multiply and divide by 25]

x = 13×32(2×5)5

x = 416(10)5=416100000

x = 0.00416

(ii) Let x = 178 ………………..(1)
Compare (1) with x = pq
Here p = 17, q = 8
Prime factors of q = 8 = 2 × 2 × 2 = 23 × 50
Which are of the form 2n × 5m, where n = 3, m = 0 and these are non negative integers
∴ x = 178 can be expressed as a terminating decimal expansion.
To Express in Decimal form

x = 178=1723×50

x = 17×5323×53
[Multiply and divide with 53 to make the denominator 10]

x = 17×125(2×5)3

x = 2125(10)3=21251000

x = 2.125

⇒ 178 = 2.125

(iii) Let x = 151600 …………….(1)
Compare (1) with x = pq
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 × 5m, where n = 6, m = 2 and these are non negative integers
∴ x = 151600 can be expressed as a terminating decimal expansion.
To Express in Decimal form

x = 151600

x = 15×5426×52×54
[To make denominator a power of 10 multiply and divide by 54]

x = 15×62526×56

x = 9375(2×5)6

x = 9375(10)6=93751000000 = 0.009375

In Decimal form, x = 151600 = 0.009375

(iv) Let x = 232352 …………….(1)
Compare (1) with x = pq
Here p = 15, q = 23 52
Prime factors of q = 23 52
Which are of the form 2n × 5m, where n = 3, m = 2 and these are non negative integers
∴ x = 232352 have a terminating decimal expansion.
To Express in Decimal form

x = 232352=23×523×52×5=11523×53

x = 115(2×5)3=1151000 = 0.115

In Decimal form,
x = 232352 = 0.115

(v) Let x = 615=25
Compare (1) with x = pq
Here p = 2, q = 5
Prime factors of q = 5 = 20 × 51
Which are of the form 2n × 5m, where n = 0, m = 1 and these are non negative integers
∴ x = 615 have a terminating decimal expansion.
To Express in Decimal form

x = 615=25

x = 2×2121×51=410 = 0.4

In Decimal form,
x = 615 = 0.4

(vi) Let x = 3550=710
Compare (1) with x = pq
Here p = 7, q = 10
Prime factors of q = 5 = 21 × 51
Which are of the form 2n × 5m, where n = 1, m = 1 and these are non negative integers
∴ x = 710 have a terminating decimal expansion.
To Express in Decimal form

x = 3550

x = 710

x = 721×51

x = 7(2×5)1=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 = 431234567891000000000

= 43123456789109 …………….(2)
From (2) x is a rational number and of the pq.

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 pq
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 pq
x = 4791495194111111111
Here p = 4791495194, q = 111111111
x = 479149519432(12345679)
Hence, prime factors of q are 32 (123456789)