EXERCISE 1A

Question 1.
What do you mean by Euclid’s division lemma?
Solution:
For any two given positive integers a and b, there exist unique whole numbers q and r such that
a = bq + r, when 0 ≤ r < b.
Here, a is called dividend, b as divisor, q as quotient and r as remainder.
Dividend = (Divisor x Quotient) + Remainder.

Question 2.
A number when divided by 61 gives 27 as quotient and 32 as remainder. Find the number.
Solution:
Using Euclid’s divison Lemma
Dividend = (Divisor x Quotient) + Remainder
= (61 x 27) + 32
= 1647 + 32
= 1679
Required number = 1679

Question 3.
By what number should 1365 be divided to get 31 as quotient and 32 as remainder.
Solution:
Let the required divisor = x
Then by Euclid’s division Lemma,
Dividend = (Divisor x Quotient) + remainder
1365 = x x 31 + 32
=> 1365 = 31x + 32
=> 31x= 1365 – 32 = 1333
x = = 43
Divisor = 43

Question 4.
Using Euclid’s division algorithm, find the HCF of
(i) 405 and 2520
(ii) 504 and 1188
(iii) 960 and 1575
Solution:
(i) 405 and 2520
HCF of 405 and 2520 = 45

Question 5.
Show that every positive integer is either even or odd.
Solution:
Let n be an arbitrary positive integer.
On dividing n by 2, let m be the quotient and r be the remainder, then by Euclid’s division lemma
n = 2 x m + r = 2m + r, 0 ≤ r < 2
n = 2m or 2m + 1 for some integer m.
Case 1 : When n = 2m, then n is even
Case 2 : When n = 2m + 1, then n is odd.
Hence, every positive integer is either even or odd.

Question 6.
Show dial any positive odd integer is of the form (6m + 1) or (6m + 3) or (6m + 5), where m is some integer.
Solution:
Let n be a given positive odd integer.
On dividing n by 6, let m be the quotient and r be the remainder, then by Euclid’s division Lemma.
n = 6m + r, where 0 ≤ r < 6 => n = 6m + r, where r = 0, 1, 2, 3, 4, 5
=> n = 6m or (6m + 1) or (6m + 2) or (6m + 3) or (6m + 4) or (6m + 5)
But n = 6m, (6m + 2) and (6m + 4) are even.
Thus when n is odd, it will be in the form of (6m + 1) or (6m + 3) or (6m + 5) for some integer m.

Question 7.
Show that any positive odd integer is of the form (4m + 1) or (4m + 3), when m is some integer.
Solution:
Let n be an arbitrary odd positive integer.
On dividing by 4, let m be the quotient and r be the remainder.
So by Euclid’s division lemma,
n = 4m + r, where 0 ≤ r < 4
n = 4m or (4m + 1) or (4m + 2) or (4m + 3)
But 4m and (4m + 2) are even integers.
Since n is odd, so n ≠ 4m or n ≠ (4m + 2)
n = (4m + 1) or (4m + 3) for some integer m.
Hence any positive odd integer is of the form (4m + 1) or (4m + 3) for some integer m.

Question 8.
For any positive integer n, prove that n³ – n is divisible by 6.
Solution:
Let a = n³ – n
=> a = n (n² – 1)
=> a = n (n – 1) (n + 1) [(a² – b²) = (a – b) (a + b)]
=> a = (n – 1 ) n (n + 1)
We know that,
(i) If a number is completely divisible by 2 and 3, then it is also divisible by 6.
(ii) If the sum of digits of any number is divisible by 3, then it is also divisible by 3.
(iii) If one of the factor of any number is an even number, then it is also divisible by 2.
a = (n – 1) n (n + 1) [From Eq. (i)]
Now, sum of the digits
= n – 1 + n + n + 1 = 3n
= Multiple of 3, where n is any positive integer.
and (n – 1) n (n +1) will always be even, as one out of (n – 1) or n or (n + 1) must be even.
Since, conditions (ii) and (iii) is completely satisfy the Eq. (i).
Hence, by condition (i) the number n3 – n is always divisible by 6, where n is any positive integer.
Hence proved.

Question 9.
Prove that if x and y are both odd positive integers then x² + y² is even but not divisible by 4.
Solution:
Let x = 2m + 1 and y = 2m + 3 are odd positive integers, for every positive integer m.
Then, x² + y² = (2m + 1)² + (2m + 3)²
= 4m² + 1 + 4 m + 4m² + 9 + 12m [(a + b)² = a² + 2ab + b²]
= 8m² + 16m + 10 = even
= 2(4m² + 8m + 5) or 4(2m² + 4m + 2) + 1
Hence, x² + y² is even for every positive integer m but not divisible by 4.

Question 10.
Use Euclid’s algorithm to find HCF of 1190 and 1445. Express the HCF in the form 1190m + 1445n.
Solution:
We find HCF (1190, 1145) using the following steps:

(i) Since 1445 > 1190, we divide 1445 by 1190 to get 1 as quotient and 255 as remainder.
By Euclid’s division lemma, we get
1445 = 1190 x 1 + 255 …(i)
(ii) Since the remainder 255 ≠ 0, we divide 1190 by 255 to get 4 as a quotient and 170 as a remainder.
By Euclid’s division lemma, we get
1190 = 255 x 4 + 170 …(ii)
(iii) Since the remainder 170 ≠ 0, we divide 255 by 170 to get 1 as quotient and 85 as remainder.
By Euclid’s division lemma, we get
255 = 170 x 1 +85 …(iii)
(iv) Since the remainder 85 ≠ 0, we divide 170 by 85 to get 2 as quotient and 0 as remainder.
By Euclid’s division lemma, we get
170 = 85 x 2 + 0 …(iv)
The remainder is now 0, so our procedure steps
HCF (1190, 1445) = 85
Now, from (iii), we get
255 = 170 x 1 + 85
=> 85 = 255 – 170 x 1
= (1445 – 1190) – (1190 – 255) x 4
= (1445 – 1190) – (1190 – 255) x 4
= (1445 – 1190) x 2 + (1445 – 1190) x 4
= 1445 – 1190 x 2 + 1445 x 4 – 1190 x 4
= 1445 x 5 – 1190 x 6
= 1190 x (-6) + 1445 x 5
Hence, m = -6, n = 5

EXERCISE 1B

Question 1.
Using prime factorization, find the HCF and LCM of
(i) 36, 84
(ii) 23, 31
(iii) 96, 404
(iv) 144, 198
(v) 396, 1080
(vi) 1152, 1664
In each case, verily that:
HCF x LCM = product of given numbers.
Solution:
(i) 36, 84

36 = 2 x 2 x 3 x 3 = 2² x 3²
84 = 2 x 2 x 3 x 7 = 2² x 3 x 7
HCF = 2² x 3 = 2 x 2 x 3 = 12
LCM = 2² x 3² x 7 = 2 x 2 x 3 x 3 x 7 = 252
Now HCF x LCM = 12 x 252 = 3024
and product of number = 36 x 84 = 3024
HCF x LCM = Product of given two numbers.
(ii) 23, 31
23 = 1 x 23
31 = 1 x 31
HCF= 1
and LCM = 23 x 31 = 713
Now HCF x LCM = 1 x 713 = 713
and product of numbers = 23 x 31 = 713
HCF x LCM = Product of given two numbers
(iii) 96, 404

96 = 2 x 2 x 2 x 2 x 2 x 3 = 2⁵ x 3
404 = 2 x 2 x 101 = 2² x 101
HCF = 2² = 2 x 2 = 4
LCM = 2⁵ x 3 x 101 = 32 x 3 x 101 = 9696
Now HCF x LCM = 4 x 9696 = 38784
and product of two numbers = 96 x 404 = 38784
HCF x LCM = Product of given two numbers
(iv) 144, 198

144 = 2 x 2 x 2 x 2 x 3 x 3 = 2⁴ x 32
198 = 2 x 3 x 3 x 11 = 2 x 3² x 11
HCF = 2 x 3² = 2 x 3 x 3 = 18
LCM = 2⁴ x 3² x 11 = 16 x 9 x 11 = 1584
and product of given two numbers = 144 x 198 = 28512

and HCF x LCM = 18 x 1584 = 28512

HCF x LCM = Product of given two numbers
(v) 396, 1080

396 = 2 x 2 x 3 x 3 x 11 = 2² x 3² x 11
1080 = 2 x 2 x 2 x 3 x 3 x 3 x 5 = 2³ x 3³ x 5
HCF = 2² x 3² = 2 x 2 x 3 x 3 = 36
LCM = 2³ x 3³ x 11 x 5 = 2 x 2 x 2 x 3 x 3 x 3 x 5 x 11 = 11880
Now HCF x LCM = 36 x 11880 = 427680

Product of two numbers = 396 x 1080 = 427680

HCF x LCM = Product of two given numbers.
(vi) 1152, 1664

1152 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 = 2⁷ x 3²
1664 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 13 = 2⁷ x 13
HCF = 2⁷ = 2 x 2 x 2 x 2 x 2 x 2 x 2 = 128
LCM = 2⁷ x 3² x 13 = 128 x 9 x 13 = 128 x 117= 14976

Now HCF x LCM = 128 x 14976= 1916928

and product of given two numbers = 1152 x 1664= 1916928

HCF x LCM = Product of given two numbers.

Question 2.
Using prime factorization, find the HCF mid LCM of:
(i) 8, 9, 25
(ii) 12, 15, 21
(iii) 17, 23, 29
(iv) 24, 36, 40
(v) 30, 72, 432
(vi) 21, 28, 36, 45
Solution:

Question 3.
The HCF of two numbers is 23 and their LCM is 1449. If one of the number is 161, find die other.
Solution:
HCF of two numbers = 23
LCM =1449
One number = 161

Second number = 207

Question 4.
The HCF of two numbers is 145 and their LCM is 2175. If one of the numbers is 725, find die other.
Solution:
HCF of two numbers = 145
LCM = 2175
One number = 725

Second number = 435

Question 5.
The HCF of two numbers is 18 and their product is 12960. Find their LCM.
Solution:
HCF of two numbers = 18
and product of two numbers = 12960

LCM of two numbers = 720

Question 6.
Is it possible to have two numbers whose HCF is 18 and LCM is 760. Give reason.
Solution:
HCF= 18
LCM = 760
HCF always divides the LCM completely
760 – 18 = 42 and remainder 4
Hence, it is not possible.

Question 7.
Find the simplest form of
(a)
(b)
(c)
(d)
Solution:
(a)
HCF of 69 and 92 = 23

Question 8.
Find the largest number which divides 438 and 606, leaving remainder 6 in each case.
Solution:
Numbers are 428 and 606 and remainder in each case = 6
Now subtracting 6 from each number, we get 438 – 6 = 432
and 606 – 6 = 600
Required number = HCF of 432 and 600 = 24
The largest required number is 24

Question 9.
Find the largest number which divides 320 and 457, leaving remainders 5 and 7 respectively.
Solution:
The numbers are 320 and 457
and remainders are 5 and 7 respectively
320 – 5 = 315 and 457 – 7 = 450
Now the required greatest number of 315 and 450 is their HCF
Now HCF of 315 and 450 = 45

Question 10.
Find the least number which when divided by 35, 56 and 91 leaves the same remainder 7 in each case.
Solution:
The numbers are given = 35, 56, 91 and the remainder = 7 in each case,
Now the least number = LCM of 35, 56, 91 = 3640

LCM = 7 x 5 x 8 x 13 = 3640
Required least number = 3640 + 7 = 3647

Question 11.
Find the smallest number which when divided by 28 and 32 leaves remainders 8 and 12 respectively.
Solution:
Given numbers are 28 and 32
Remainders are 8 and 12 respectively
28 – 8 = 20
32 – 12 = 20
Now, LCM of 28 and 32 = 224

LCM = 2 x 2 x 7 x 8 = 224
Least required number = 224 – 20 = 204

Question 12.
Find the smallest number which when increased 17 is exactly divisible by both 468 and 520.
Solution:
The given numbers are 468 and 520
Now LCM of 468 and 520 = 4680

LCM = 2 x 2 x 13 x 9 x 10 = 4680
When number 17 is increase then required number = 4680 – 17 = 4663

Question 13.
Find the greatest number of four digits which is exactly divisible by 15, 24 and 36.
Solution:
LCM of 15, 24, 36 = 360

Required number = 9999 – 279 = 9720

Question 14.
Find the largest four-digits number which when divided by 4, 7 and 13 leaves a remainder of 3 in each case.
Solution:
Greatest number of 4 digits is 9999
LCM of 4, 7 and 13 = 364
On dividing 9999 by 364, remainder is 171
Greatest number of 4 digits divisible by 4, 7 and 13 = (9999 – 171) = 9828
Hence, required number = (9828 + 3) = 9831

Question 15.
Find the least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3.
Solution:
LCM of 5, 6, 4 and 3 = 60
On dividing 2497 by 60, the remainder is 37
Number to be added = (60 – 37) = 23

Question 16.
Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
Solution:
We can represent any integer number in the form of: pq + r, where ‘p is divisor, ‘q is quotient’, r is remainder*.
So, we can write given numbers from given in formation As :
43 = pq₁ + r …(i)
91 = pq₂ + r …(ii)
And 183 = pq₃ + r …(iii)
Here, we want to find greatest value of ‘p’ were r is same.
So, we subtract eq. (i) from eq. (ii), we get
Pq₂ – Pq₁ = 48
Also, subtract eq. (ii) from eq. (iii), we get
pq₃ – pq₂ = 92
Also, subtract eq. (i) from eq. (iii), we get
Pq₃ – Pq₁ = 140
Now, to find greatest value of ‘p’ we find HCF of 48, 92 and 140 as,
48 = 2 x 2 x 2 x 2 x 3
92 = 2 x 2 x 2 x 2 x 2 x 3
and 140 = 2 x 2 x 5 x 7
So, HCF (48, 92 and 140) = 2 x 2 = 4
Greatest number that will divide 43, 91 and 183 as to leave the same remainder in each case = 4.

Question 17.
Find the least number which when divided by 20, 25, 35 and 40 leaves remainder 14, 19, 29 and 34 respectively.
Solution:
Remainder in all the cases is 6, i.e.,
20 – 14 = 6
25 – 19 = 6
35 – 29 = 6
40 – 34 = 6
The difference between divisor and the corresponding remainder is 6.
Required number = (LCM of 20, 25, 35, 46) – 6 = 1400 – 6 = 1394

Question 18.
In a seminar, the number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively. Find the minimum number of rooms required, if in each room, the same number of participants are to be seated and all of them being in the same subject.
Solution:
Number of participants in Hindi = 60
Number of participants in English = 84
Number of participants in Mathematics =108
Minimum number of participants in one room = HCF of 60, 84 and 108 = 12

Question 19.
Three sets of English, mathematics and science books containing 336, 240 and 96 books respectively have to be stacked in such a way that all the books are stored subject wise and the height of each stack is the same. How many stacks will be there?
Solution:
Number of books in English = 336
Number of books in Mathematics = 240
Number of books in Science = 96
Minimum number of books of each topic in a stack = HCF of 336, 240 and 96 = 48

Question 20.
Three pieces of timber 42 m, 49 m and 63 m long have to be divided into planks of the same length. What is the greatest possible length of each plank ? How many planks are formed?
Solution:
Length of first piece of timber = 42 m
Length of second piece of timber = 49 m
and length of third piece of timber = 63 m

Question 21.
Find the greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm and 12 m 95 cm.
Solution:
Lengths are given as 7 m, 3 m 85 cm and 12 m 95 cm = 700 cm, 385 cm and 1295 cm
Greatest possible length that can be used to measure exactly = HCF of 700, 385, 1295 = 35 cm

Question 22.
Find the maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and the same number of pencils.
Solution:
Number of pens =1001
and number of pencils = 910
Maximum number of pens and pencils equally distributed to the students = HCF of 1001 and 910 = 91
Number of students = 91

Question 23.
Find the least number of square tiles required to pave the ceiling of a room 15 m 17 cm long and 9 m 2 cm broad.
Solution:
Length of the room = 15 m 17 cm = 1517 cm
and breadth = 9 m 2 cm = 902 cm
Maximum side of square tile used = HCF of 1517 and 902 = 41 cm

Question 24.
Three measuring rods are 64 cm, 80 cm and 96 cm in length. Find the least length of cloth that can be measured an exact number of times, using any of the rods.
Solution:
Measures of three rods = 64 cm, 80 cm and 96 cm
Least length of cloth that can be measured an exact number of times
= LCM of 64, 80, 96
= 960 cm
= 9 m 60 cm
= 9.6 m

LCM = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 3 = 960

Question 25.
An electronic device makes a beep after every 60 seconds. Another device makes a beep after every 62 seconds. They beeped together at 10 a.m. At what time will they beep together at the earliest?
Solution:
Beep made by first devices after every = 60 seconds
Second device after = 62 seconds
Period after next beep together = LCM of 60, 62

LCM = 2 x 30 x 31 = 1860 = 1860 seconds = 31 minutes
Time started beep together, first time together = 10 a.m.
Time beep together next time = 10 a.m. + 31 minutes = 10 : 31 a.m.

Question 26.
The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they all change simultaneously at 8 a.m., then at what time will they again change simultaneously?
Solution:
The traffic lights of three roads change after
48 sec., 72 sec. and 108 sec. simultaneously
They will change together after a period of = LCM of 48 sec., 72 sec. and 108 sec.

= 7 minutes, 12 seconds
First time they light together at 8 a.m. i.e., after 8 hr.
Next time they will light together = 8 a.m. + 7 min. 12 sec. = 8 : 07 : 12 hrs.

Question 27.
Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10, 12 minutes respectively. In 30 hours, how many times do they toll together?
Solution:
Tolling of 6 bells = 2, 4, 6, 8, 10, 12 minutes
They take time tolling together = LCM of 2, 4, 6, 8, 10, 12 = 120 minutes = 2 hours

LCM of 2 x 2 x 2 x 3 x 5 = 120 min. (2 hr)
They will toll together after every 2 hours Total time given = 30 hours
Number of times, there will toll together in 30 hours = = 15 times
Total numbers of times = 15 + 1 (of starting time) = 16 times

EXERCISE 1C

Question 1.
Without actual division, show that each of the following rational numbers is a terminating decimal. Express each in decimal form.

Solution:

Question 2.
Without actual division, show that each of the following rational numbers is a nonterminating repeating decimal.

Solution:

Question 3.
Express each of the following as a fraction in simplest form.

Solution:

EXERCISE 1D

Question 1.
Define:
(i) rational numbers
(ii) irrational numbers
(iii) real numbers.
Solution:
(i) Rational numbers: Numbers in the form of where p and q are integers and q ≠ 0, are called rational numbers.
(ii) Irrational numbers : The numbers which are not rationals, are called irrational numbers. Irrational numbers can be expressed in decimal form as non terminating non-repeating decimal.
(iii) Real numbers : The numbers which are rational or irrational, are called real numbers.

Question 2.
Classify the following numbers as rational or irrational:
(i)
(ii) 3.1416
(iii) π
(iv)
(v) 5.636363…
(vi) 2.040040004…
(vii) 1.535335333…
(viii) 3.121221222…
(ix) √21
(x)
Solution:
(i)
It is a rational number as it is in the form of
(ii) 3.1416
It is a rational number as it is a terminating decimal.
(iii) π
It is an irrational number as it is nonterminating non-repeating decimal.
(iv)
It is a rational number as it is nonterminating repeating decimal.
(v) 5.636363… = 5.63
It is a rational number as it is nonterminating repeating decimal.
(vi) 2.040040004…
It is an irrational number as it is nonterminating non-repeating decimal.
(vii) 1.535335333…
It is an irrational number as it is non terminating non-repeating decimal.
(viii) 3.121221222…
It is an irrational number as it is nonterminating non-repeating decimal.
(ix) √21
It is an irrational number aS it is not in the form of
(x)
It is an irrational number as it is not in the form of

Question 3.
Prove that each of the following numbers is irrational.
(i) √6
(ii) (2 – √3) [CBSE 2008]
(iii) (3 + √2 ) [CBSE 2009]
(iv) (2 + √5) [CBSE 2008C]
(v) (5 + 3√2 ) [CBSE 2008]
(vi) 3√7
(vii)
(viii) (2 – 3√5 )[CBSE2010]
(ix) (√3 + √5)
Solution:
(i) √6 is irrational.
Let √6 is not an irrational number, but it is a rational number in the simplest form of
√6 = (p and q have no common factors)
Squaring both sides,
6 =
p² = 6q²
p² is divisible by 6
=> p is divisible by 6
Let p = 6a for some integer a
6q² = 36a²
=> q² = 6a²
q² is also divisible by 6
=> q is divisible by 6
6 is common factors of p and q
But this contradicts the fact that p and q have no common factor
√6 is irrational
(ii) (2 – √3) is irrational
Let (2 – √3) is a rational and 2 is also rational, then
2 – (2 – √3 ) is rational (Difference two rationals is rational)
=> 2 – 2 + √3 is rational
=> √3 is rational
But it contradicts the fact
(2 – √3) is irrational
(iii) (3 + √2 ) is irrational
Let (3 + √2 ) is rational and 3 is also rational
(3 + √2 ) – 3 is rational (Difference of two rationals is rational)
=> 3 + √2 – 3 is rational
=> √2 is rational
But it contradicts the fact (3 + √2 ) is irrational
(iv) (2 + √5 ) is irrational
Let (2 + √5 ) is rational and 2 is also rational
(2 + √5) – 2 is rational (Difference of two rationals is rational)
=> 2 + √5 – 2 is rational
=> √5 is rational
But it contradicts the fact (2 + √5) is irrational
(v) (5 + 3√2 ) is irrational
Let (5 + 3√2 ) is rational and 5 is also rational
(5 + 3√2 ) – 5 is rational (Difference of two rationals is rational)
=>5 + 3√2 – 5 is rational
=> 3√2 is rational
Product of two rationals is rational
3 is rational and √2 is rational
√2 is rational
But it contradicts the fact
(5 + 3√2 ) is irrational
(vi) 3√7 is irrational
Let 3√7 is rational
3 is rational and √7 is rational (Product of two rationals is rational)
But √7 is rational, it contradicts the fact
3√7 is irrational
(vii) is irrational
Let is rational
is rational
is rational and √5 is rational
But √5 is a rational, it contradicts the fact
is irrational
(viii)(2 – 3√5) is irrational
Let 2 – 3√5 is rational, 2 is also rational
2 – (2 – 3√5) is rational (Difference of two rationals is rational)
2 – 2 + 3√5 is rational
=> 3√5 is rational
3 is rational and √5 is rational (Product of two rationals is rational)
√5 is rational
But it contradicts the fact
(2 – 3√5) is irrational
(ix) (√3 + √5) is irrational
Let √3 + √5 is rational
Squaring,
(√3 + √5)² is rational
=> 3 x 5 + 2√3 x √5 is rational
=> 8 + 2√15 is rational
=> 8 + 2√15 – 8 is rational (Difference of two rationals is rational)
=> 2√15 is rational
2 is rational and √15 is rational (Product of two rationals is rational)
√15 is rational
But it contradicts the fact
(√3 + √5) is irrational

Question 4.
Prove that is irrational.
Solution:
Let is rational
= is rational
is rational and √3 is rationals (Product of two rationals is rational)
√3 is rational But it contradicts the fact
is irrational

Question 5.
(i) Give an example of two irrationals whose sum is rational.
(ii) Give an example of two irrationals whose product is rational.
Solution:
(i) We can take two numbers 3 + √2 and 3 – √2 which are irrationals
Sum = 3 + √2 + 3 – √2 = 6 Which is rational
3 + √2 and 3 – √2 are required numbers
(ii) We take two. numbers
5 + √3 and 5 – √3 which are irrationals
Now product = (5 + √3) (5 – √3)
= (5)² – (√3 )² = 25 – 3 = 22 which is rational
5 + √3 and 5 – √3 are the required numbers

Question 6.
State whether the given statement is true or false.
(i) The sum of two rationals is always rational.
(ii) The product of two rationals is always rational.
(iii) The sum of two irrationals is always an irrational.
(iv) The product of two irrationals is always an irrational.
(v) The sum of a rational and an irrational is irrational.
(vi) The product of a rational and an irrational is irrational.
Solution:
(i) True.
(ii) True.
(iii) False, as sum of two irrational can be rational number also such as
(3 + √2) + (3 – √2) = 3 + √2 + 3 – √2 = 6 which is rational.
(iv) False, as product of two irrational numbers can be rational also such as
(3 + √2)(3 – √2 ) = (3)2 – (√2 )2 = 9 – 2 = 7
which is rational
(v) True.
(vi) True.

Question 7.
Prove that (2√3 – 1) is an irrational number. [CBSE 2010]
Solution:
Let (2√3 – 1) is a rational number and 1 is a rational number also.
Then sum = 2√3 – 1 + 1 = 2√3
In 2√3, 2 is rational and √3 is rational (Product of two rational numbers is rational)
But √3 is rational number which contradicts the fact
(2√3 – 1) is an irrational.

Question 8.
Prove that (4 – 5√2 ) is an irrational number.
Solution:
Let 4 – 5√2 is a rational number and 4 is also a rational number
Difference of two rational number is a rational numbers
4 – (4 – 5√2 ) is rational
=> 4 – 4 + 5√2 is rational
=> 5√2 is rational
Product of two rational number is rational
5 is rational and √2 is rational
But it contradicts the fact that √2 is rational √2 is irrational
Hence, 4 – 5√2 is irrational

Question 9.
Prove that (5 – 2√3) is an irrational number. [CBSE 2010]
Solution:
Let (5 – 2√3) is a rational number and 5 is also a rational number
Difference of two rational number is rational
=> 5 – (5 – 2√3) is rational
=> 5 – 5 + 2√3 or 2√3 is rational
Product of two rational number is rational
2 is rational and √3 is rational
But it contradicts the fact
(5 – 2√3) is an irrational number.

Question 10.
Prove that 5√2 is irrational.
Solution:
Let 5√2 is a rational
Product of two rationals is a rational
5 is rational and √2 is rational
But it contradicts the fact
5√2 is an irrational.

Question 11.
Prove that : is irrational.
Solution:

Let is a rational number, then
is rational and √7 is rational
But it contradicts the fact is an irrational number.

EXERCISE 1E

Question 1.
State Euclid’s division lemma.
Solution:
For any two given positive integers a and b there exist unique whole numbers q and r such that
a = bq + r, where 0 ≤ r < b.
Here, we call ‘a’ as dividend, b as divisor, q is quotient and r as remainder.
Dividend = (Divisor x Quotient) + Remainder

Question 2.
State fundamental theorem of Arithmetic.
Solution:
Every composite number can be uniquely expressed as a product of two primes, except for the order in which these prime factors occurs.

Question 3.
Express 360 as product of its prime factors.
Solution:
360 = 2 x 2 x 2 x 3 x 3 x 5 = 2³ x 3² x 5

Question 4.
If a and b are two prime numbers, then find HCF (a, b).
Solution:
We know that HCF of two primes is
HCF (a, b) = 1

Question 5.
If a and b are two prime numbers then find LCM (a, b).
Solution:
a and b are two prime numbers then their
LCM = Product of these two numbers
LCM(a, b) = a x b = ab.

Question 6.
If the product of two numbers is 1050 and their HCF is 25, find their LCM.
Solution:
We know that product of two numbers is equal to their HCF x LCM
LCM =
= = 42
LCM of two numbers = 42

Question 7.
What is a composite number?
Solution:
A composite number is a number which is not a prime. In other words, a composite number has more than two factors.

Question 8.
If a and b are relatively prime then what is their HCF?
Solution:
a and b are two primes, then their
HCF will be 1
HCF of a and b = 1

Question 9.
If the rational number has a terminating b decimal expansion, what is the condition to be satisfied by b. [CBSE 2008]
Solution:
is a rational number and it has terminating decimal
b will in the form 2^m x 5ⁿ where m and n are some non-negative integers.

Question 10.
Simplify : [CBSE 2010]
Solution:

Question 11.
Write the decimal expansion of [CBSE 2009]
Solution:

Question 12.
Show that there is no value of n for which (2ⁿ x 5ⁿ) ends in 5.
Solution:
2ⁿ x 5ⁿ = (2 x 5)ⁿ = (10)ⁿ
Which always ends in a zero
There is no value of n for which (2ⁿ x 5ⁿ) ends in 5

Question 13.
Is it possible to have two numbers whose HCF is 25 and LCM is 520?
Solution:
We know that HCF is always a factor is its LCM
But 25 is not a factor of 520
It is not possible to have two numbers having HCF = 25 and LCM = 520

Question 14.
Give an example of two irrationals whose sum is rational.
Solution:
Let two irrational number be (5 + √3) and (5 – √3).
Now their sum = (5 + √3) + (5 – √3) = 5 + √3 + 5 – √3 = 10
Which is a rational number.

Question 15.
Give an example of two irrationals whose product is rational.
Solution:
Let the two irrational number be (3 + √2) and (3 – √2)
Now, their product = (3 + √2) (3 – √2)
= (3)² – (√2)² {(a + b) (a – b) = a² – b²}
= 9 – 2 = 7
Which is a rational number.

Question 16.
If a and b are relatively prime, what is their LCM.
Solution:
a and b are relative primes
their HCF = 1

Question 17.
The LCM of two numbers is 1200. Show that the HCF of these numbers cannot be 500. Why?
Solution:
LCM of two numbers = 1200
and HCF = 500
But we know that HCF of two numbers divides their LCM.
But 500 does not divide 1200 exactly
Hence, 500 is not their HCF whose LCM is 1200.

Short-Answer Questions
Question 18.
Express as a rational number is simplest form.
Solution:
Let x = 0.4 = 0.444
Then 10x = 4.444….
Subtracting, we get
9x = 4 => x =
= which is in the simplest form.

Question 19.
Express as a rational number in simplest form.
Solution:

Let x = = 0.232323…….
and 100x = 23.232323……
Subtracting, we get
99x = 23 => x =
= which is in the simplest form.

Question 20.
Explain why 0.15015001500015…. is an irrational number.
Solution:
0.15015001500015
It is non-terminating non-repeating decimal.
It is an irrational number.

Question 21.
Show that is irrational
Solution:
= √2
Let √2 is a rational number
Product of two rational numbers is a rational
is rational and √2 is rational contradicts
But it contradicts the fact
or √2 is irrational.

Question 22.
Write a rational number between √3 and 2.
Solution:
√3 and 2.
√3 = 1.732 and 2.000
A rational number between 1.732 and 2.000 can be 1.8 or 1.9
Hence, 1.8 or 1.9 is a required rational.

Question 23.
Explain why is a rational number.
Solution:

It is non-terminating repeating decimal.
It is a rational number.

MULTIPLE CHOICE QUESTIONS

Choose the correct answer in each of the following questions.
Question 1.
Which of the following is a pair of co-primes.
(a) (14, 35)
(b) (18, 25)
(c) (31, 93)
(d) (32, 62)
Solution:
(b) We know that HCF of two co-prime number is 1
HCF of 14, 35 is 7
HCF of 18, 25 is 1
HCF of 31, 93 is 31
HCF of 32, 60 is 4
Required co-prime number is (18, 25)

Question 2.
If a = (2² x 3³ x 5⁴) and b = (2³ x 3² x 5) then HCF (a, b) = ?
(a) 90
(b) 180
(c) 360
(d) 540
Solution:
(b) a = (2² x 3³ x 5⁴), b = (2³ x 3² x 5)
HCF = 2² x 3² x 5 = 2 x 2 x 3 x 3 x 5 = 180

Question 3.
HCF of (2³ x 3² x 5), (2² x 3³ x 5²) and (2⁴ x 3 x 5³ x 7) is
(a) 30
(b) 48
(c) 60
(d) 105
Solution:
(c) HCF of 2³ x 3² x 5, 2² x 3³ x 5², 2⁴ x 3 x 5³ x 7
HCF = 2² x 3 x 5 = 2 x 2 x 3 x 5 = 60

Question 4.
LCM of (2³ x 3 x 5) and (2⁴ x 5 x 7) is
(a) 40
(b) 560
(c) 1120
(d) 1680
Solution:
(d) LCM of 2³ x 3 x 5, 2⁴ x 5 x 7 = 2⁴ x 3 x 5 x 7
=2 x 2 x 2 x 2 x 3 x 5 x 7
= 1680

Question 5.
The HCF of two numbers is 27 and their LCM is 162. If one of the number is 54, what is the other number:
(a) 36
(b) 45
(c) 9
(d) 81
Solution:
(d) HCF of two numbers = 27
LCM = 162
One number = 54

Question 6.
The product of two numbers is 1600 and their HCF is 5. The LCM of the numbers is
(a) 8000
(b) 1600
(c) 320
(d) 1605
Solution:
(c) Product of two numbers = 1600
HCF = 5

Question 7.
What is the largest number that divides each one of 1152 and 1664 exactly:
(a) 32
(b) 64
(c) 128
(d) 256
Solution:
(c) Largest number that divides each one of 1152 and 1664
HCF of 1152 and 1664 =128

Question 8.
What is the largest number that divides 70 and 125, leaving remainders 5 and 8 respectively
(a) 13
(b) 9
(c) 3
(d) 585
Solution:
(a) Largest number that divides 70 and 125 leaving remainders as 5 and 8 respectively.
Required number = 70 – 5 = 65
and 125 – 8= 117
HCF of 65, 117 = 13

Question 9.
What is the largest number that divides 245 and 1029, leaving remainder 5 in each case?
(a) 15
(b) 16
(c) 9
(d) 5
Solution:
(b) Largest number that divides 245 and 1029 leaving remainder as 5 in each case. .
Required number = 245 – 5 = 240 and 1029 – 5 = 1024
Now, HCF of 240 and 1020 = 16

Question 10.
The simplest form of
(a)
(b)
(c)
(d)
Solution:
(d)

Question 11.
Euclid’s division lemma states that for any positive integer a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy:
(a) 1 < r < b
(b) 0 < r ≤ b
(c) 0 ≤ r < b
(d) 0 < r < b
Solution:
(c) In a = bq + r
r must satisfy i.e. 0 ≤ r < b

Question 12.
A number when divided by 143 leaves 31 as remainder. What will be the remainder when the same number is divided by 13?
(a) 0
(b) 1
(c) 3
(d) 5
Solution:
(d) Let the given number when divided by 143 gives q as quotient and 31 as remainder.
Number = 143q + 31
= (13 x 11) q + 31
= 13 x 11 q+ 13 x 2 + 5
= 13 (110 + 2) + 5
The number where divided by 73, gives 5 as remainder.

Question 13.
Which of the following is an irrational number?
(a)
(b) 3.1416
(c)
(d) 3.141141114…
Solution:
(d) 3.141141114… is irrational because it is non terminating non-repeating.

Question 14.
π is
(a) an integer
(b) a rational number
(c) an irrational number
(d) none of these
Solution:
(c) π is an irrational number.

Question 15.
is
(a) an integer
(b) a rational number
(c) an irrational number
(d) none of these
Solution:
(b) is a rational number as it is non-terminating repeating decimal.

Question 16.
2.13113111311113… is
(a) an integer
(b) a rational number
(c) an irrational number
(d) none of these
Solution:
(c) 2.13113111311113… is an irrational number.
It is non-terminating non-repeating decimal.

Question 17.
The number 3.24636363… is
(a) an integer
(b) a rational number
(c) an irrational number
(d) none of these
Solution:
(b) 3.24636363…
=
It is non-terminating repeating decimal.
It is a rational number.

Question 18.
Which of the following rational numbers is expressible as a terminating decimal?
(a)
(b)
(c)
(d)
Solution:
(c) = is a rational because it has terminating decimal as q = 5⁴ which is in form of 2^m x 5ⁿ.

Question 19.
The decimal expansion of the rational number will terminate after
(a) one decimal place
(b) two decimal places
(c) three decimal places
(d) four decimal places
Solution:
(b)

Question 20.
The decimal expansion of the number will terminate after
(a) one decimal place
(b) two decimal places
(c) three decimal places
(d) four decimal places
Solution:
(d)

Question 21.
The number 1.732 is
(a) an irrational number
(b) a rational number
(c) an integer
(d) a whole number
Solution:
(b) 1.732 is a rational number.
As it is terminating decimal.

Question 22.
a and b are two positive integers such that the least prime factor of a is 3 and the least prime factor of b is 5. Then, the least prime factor of (a + b) is
(a) 2
(b) 3
(c) 5
(d) 8
Solution:
(a) Least prime factor of a positive integer a is 3 and b is 5
2 is neither a factor of a nor of b
a and b are odd
Then (a + b) = even
(Sum of two odd numbers is even)
(a + b) is divisible by 2
Which is the least prime factor.

Question 23.
√2 is
(a) a rational number
(b) an irrational number
(c) a terminating decimal
(d) a nonterminating repeating decimal
Solution:
(b) √2 is an irrational number.

Question 24.
is
(a) a fraction
(b) a rational number
(c) an irrational number
(d) none of these
Solution:
(c)

Question 25.
(2 + √2 ) is
(a) an integer
(b) a rational number
(c) an irrational number
(d) none of these
Solution:
(c) 2 + √2 is an irrational number as sum of a rational and an irrational is an irrational

Question 26.
What is the least number that is divisible by all the natural numbers from 1 to 10 (both inclusive)?
(a) 100
(b) 1260
(c) 2520
(d) 5040
Solution:
(c) LCM of 1 to 10 = 2 x 2 x 2 x 3 x 3 x 5 x 7 = 2520

TEST YOURSELF

Question 1.
The decimal representation of is
(a) a terminating decimal
(b) a nonterminating, repeating decimal
(c) a nonterminating and non repeating decimal
(d) none of these
Solution:
(b)

Its decimal will be nonterminating repeating decimal.

Question 2.
Which of the following has a terminating decimal expansion?
(a)
(b)
(c)
(d)
Solution:
(b) is terminating decimal if q = 2^m x 5ⁿ
Now, 91 = 7 x 13, 45 = 3² x 5
80 = 2⁴ x 5, 42 = 2 x 3 x 7
80 is of the form 2^m x 5ⁿ
is terminating decimal expansion,

Question 3.
On dividing a positive integer n by 9, we get 7 as remainder. What will be the remainder if (3n – 1) is divided by 9
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(b) Divisor = 9 and remainder = 7
Let b be the divisor, then
n = 9b + 7
Multiplying both sides by 3 and subtracting 1.
3n – 1 = 3(9b + 7) – 1
3n – 1 = 27b + 21 – 1
3n – 1 = 9(3b) + 9 x 2 + 2
3n – 1 = 9(3b + 2) + 2
Remainder = 2

Question 4.
+ = ?
(a)
(b)
(c)
(d) None of these
Solution:
(b) +
0.686868 ……… + 0.737373……
= 1.424241 =

Short-Answer Questions (2 marks)
Question 5.
Show that any number of the form 4ⁿ, n ∈ N can never end with the digit 0.
Solution:
4ⁿ, n ∈ N
4¹ = 4
4² = 4 x 4 = 16
4³ = 4 x 4 x 4 = 64
4⁴ = 4 x 4 x 4 x 4 = 256
4⁵ = 4 x 4 x 4 x 4 x 4 = 1024
We see that value of 4ⁿ, ends with 4 or 6 only.
Hence, the value of 4ⁿ, n ∈ N, never ends with 0.

Question 6.
The HCF of two numbers is 27 and their LCM is 162. If one of the number is 81, find the other.
Solution:
HCF of two numbers = 27 and LCM =162
One number = 81

Question 7.
Examine whether is a terminating decimal.
Solution:
=
Here, q is in the form of 2^m x 5ⁿ
It is not terminating decimal.

Question 8.
Find the simplest form of
Solution:

Question 9.
Which of the following numbers are irrational?

Solution:

Question 10.
Prove that 2 + √3 is irrational.
Solution:
Let (2 + √3) is rational and 2 is rational.
Difference of them is also rational.
=> (2 + √3) – 2 = 2 + √3 – 2
= √3 is rational
But it contradicts the fact.
(2 + √3) is irrational.

Short-Answer Questions (3 marks)
Question 11.
Find the HCF and LCM of 12, 15, 18, 27.
Solution:
HCF of 12, 15, 18, 27
12 = 2 x 2 x 3 = 2² x 3
15 = 3 x 5
18 = 2 x 3 x 3 = 2 x 3²
27 = 3 x 3 x 3 = 3³
Now, HCF = 3
and LCM = 2² x 3³ x 5 =2 x 2 x 3 x 3 x 3 x 5
= 4 x 27 x 5 = 540

Question 12.
Give an example of two irrationals whose sum is rational.
Solution:
Let 2 + √3 and 2 – √3 are two irrational number.
Sum = 2 + √3 + 2 – √3 = 4 which is a rational.

Question 13.
Give prime factorization of 4620.
Solution:
4620

Question 14.
Find the HCF of 1008 and 1080 by prime factorization method.
Solution:
1008

Question 15.
Find the HCF and LCM of , and
Solution:

Question 16.
Find the largest number which divides 546 and 764, leaving remainders 6 and 8 respectively.
Solution:
Give numbers are 546 and 764 and remainders are 6 and 8 respectively.
Remaining number 546 – 6 = 540
and 764 – 8 = 756
Now, required largest number = HCF of 540 and 756 = 108

Long-Answer Questions (4 marks)
Question 17.
Prove that √3 is an irrational number.
Solution:
Let √3 is a rational number.
Let √3 = where p and q are integers and have no common factor, other than 1 and q ≠ 0
Squaring both sides.
3 = => 3q² – p²
=> 3 divides p²
=> 3 divides p
Let p = 3c for some integer c
3q² = 9c² => q² – 3c²
=> 3 divides q² (3 divides 3c²)
=> 3 divides q
3 is common factors of p and q
But it contradicts the fact that p and q have
no common factors and also contradicts that √3 is a rational number.
Hence, √3 is irrational number.

Question 18.
Show that every positive odd integer is of the form (4q + 1) or (4q + 3) for some integer q.
Solution:
Let n be an arbitrary odd positive integer on dividing n by 4, let m be the quotient and r be the remainder.
By Euclid’s division lemma,
n = 4q + r where 0 ≤ r < 4
n = 4q or (4q + 1) or (4q + 2) or (4q + 3)
Clearly, 4q and (4q + 2) are even number
since n is odd.
n ≠ 4q and n ≠ (4q + 2)
n = (4 q + 1) or (4q + 3) for same integer n
Hence, any positive odd integer of the form 4q + 1 or 4q + 3 for some integer q.

Question 19.
Show that one and only one out of n, (n + 2) and (n + 4) is divisible by 3, where n is any positive integer.
Solution:
On dividing n by 3, let q be the quotient and r be the remainder, then
n = 3q + r where 0 ≤ r < 3 => n = 3q + r where r = 0, 1 or 2
n = 3q or n = 3q + 1 or n = 3q + 2
(i) Case (I)
If n = 3q then n is divisible by 3
(ii) Case (II)
If n = (3q + 1) then n + 2 = 3q + 3 = 3q (q + 1) which is divisible by 3
In this case, n + 2 is divisible by 3
(iii) Case (III)
If n = (3q + 2) then n + 1 (n + 1) = 3q + 3 = 3(q + 1) which also divisible by 3
In this case, (n + 1) is divisible by 3
Hence, one and only one out of n, (n + 1) and (n + 2) is divisible by 3.

Question 20.
Show that (4 + 3√2) is irrational.
Solution:
Let (4 + 3√2) is rational number and 4 is also a rational number.
Difference of two rational numbers is also a rational number.
4 + 3√2 – 4 = 3√2 is a rational number
Product of two rational numbers is rational
3 is rational and √2 is rational
But it contradicts the fact
√2 is irrational
Hence, (4 + 3√2 ) is irrational.