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Page 1
Exercise - 1B
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 verify that HCF × LCM = product of given numbers.
Sol:
(i) Prime factorization:
36 = 2
2
× 3
84 = 2
2
× 3 × 7
HCF = product of smallest power of each common prime factor in the numbers
= 2
2
× 3 = 12
LCM = product of greatest power of each prime factor involved in the numbers
= 2
2
× 3
2
× 7 = 252
(ii) Prime factorization:
23 = 23
31 = 31
HCF = product of smallest power of each common prime factor in the numbers = 1
LCM = product of greatest power of each prime factor involved in the numbers
= 23 × 31 = 713
(iii) Prime factorization:
96 = 2
5
× 3
404 = 2
2
× 101
HCF = product of smallest power of each common prime factor in the numbers
= 2
2
= 4
LCM = product of greatest power of each prime factor involved in the numbers
= 2
5
× 3 × 101 = 9696
(iv)Prime factorization:
Page 2
Exercise - 1B
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 verify that HCF × LCM = product of given numbers.
Sol:
(i) Prime factorization:
36 = 2
2
× 3
84 = 2
2
× 3 × 7
HCF = product of smallest power of each common prime factor in the numbers
= 2
2
× 3 = 12
LCM = product of greatest power of each prime factor involved in the numbers
= 2
2
× 3
2
× 7 = 252
(ii) Prime factorization:
23 = 23
31 = 31
HCF = product of smallest power of each common prime factor in the numbers = 1
LCM = product of greatest power of each prime factor involved in the numbers
= 23 × 31 = 713
(iii) Prime factorization:
96 = 2
5
× 3
404 = 2
2
× 101
HCF = product of smallest power of each common prime factor in the numbers
= 2
2
= 4
LCM = product of greatest power of each prime factor involved in the numbers
= 2
5
× 3 × 101 = 9696
(iv)Prime factorization:
144 = 2
4
× 3
2
198 = 2 × 3
2
× 11
HCF = product of smallest power of each common prime factor in the numbers
= 2 × 3
2
= 18
LCM = product of greatest power of each prime factor involved in the numbers
= 2
4
× 3
2
× 11 = 1584
(v) Prime factorization:
396 = 2
2
× 3
2
× 11
1080 = 2
3
× 3
3
× 5
HCF = product of smallest power of each common prime factor in the numbers
= 2
2
× 3
2
= 36
LCM = product of greatest power of each prime factor involved in the numbers
= 2
3
× 3
3
× 5 × 11 = 11880
(vi) Prime factorization:
1152 = 2
7
× 3
2
1664 = 2
7
× 13
HCF = product of smallest power of each common prime factor in the numbers
= 2
7
= 128
LCM = product of greatest power of each prime factor involved in the numbers
= 2
7
× 3
2
× 13 = 14976
2. Using prime factorization, find the HCF and 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
Sol:
(i) 8 = 2 × 2 × 2 = 2
3
9 = 3 × 3 = 3
2
25 = 5 × 5 = 5
2
HCF = product of smallest power of each common prime factor in the numbers = 1
LCM = product of greatest power of each prime factor involved in the numbers
= 2
3
× 3
2
× 5
2
= 1800
(ii) 12 = 2 × 2 × 3 = 2
2
× 3
15 = 3 × 5
21 = 3 × 7
HCF = product of smallest power of each common prime factor in the numbers = 3
LCM = product of greatest power of each prime factor involved in the numbers
Page 3
Exercise - 1B
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 verify that HCF × LCM = product of given numbers.
Sol:
(i) Prime factorization:
36 = 2
2
× 3
84 = 2
2
× 3 × 7
HCF = product of smallest power of each common prime factor in the numbers
= 2
2
× 3 = 12
LCM = product of greatest power of each prime factor involved in the numbers
= 2
2
× 3
2
× 7 = 252
(ii) Prime factorization:
23 = 23
31 = 31
HCF = product of smallest power of each common prime factor in the numbers = 1
LCM = product of greatest power of each prime factor involved in the numbers
= 23 × 31 = 713
(iii) Prime factorization:
96 = 2
5
× 3
404 = 2
2
× 101
HCF = product of smallest power of each common prime factor in the numbers
= 2
2
= 4
LCM = product of greatest power of each prime factor involved in the numbers
= 2
5
× 3 × 101 = 9696
(iv)Prime factorization:
144 = 2
4
× 3
2
198 = 2 × 3
2
× 11
HCF = product of smallest power of each common prime factor in the numbers
= 2 × 3
2
= 18
LCM = product of greatest power of each prime factor involved in the numbers
= 2
4
× 3
2
× 11 = 1584
(v) Prime factorization:
396 = 2
2
× 3
2
× 11
1080 = 2
3
× 3
3
× 5
HCF = product of smallest power of each common prime factor in the numbers
= 2
2
× 3
2
= 36
LCM = product of greatest power of each prime factor involved in the numbers
= 2
3
× 3
3
× 5 × 11 = 11880
(vi) Prime factorization:
1152 = 2
7
× 3
2
1664 = 2
7
× 13
HCF = product of smallest power of each common prime factor in the numbers
= 2
7
= 128
LCM = product of greatest power of each prime factor involved in the numbers
= 2
7
× 3
2
× 13 = 14976
2. Using prime factorization, find the HCF and 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
Sol:
(i) 8 = 2 × 2 × 2 = 2
3
9 = 3 × 3 = 3
2
25 = 5 × 5 = 5
2
HCF = product of smallest power of each common prime factor in the numbers = 1
LCM = product of greatest power of each prime factor involved in the numbers
= 2
3
× 3
2
× 5
2
= 1800
(ii) 12 = 2 × 2 × 3 = 2
2
× 3
15 = 3 × 5
21 = 3 × 7
HCF = product of smallest power of each common prime factor in the numbers = 3
LCM = product of greatest power of each prime factor involved in the numbers
= 2
2
× 3 × 5 × 7 = 420
(iii) 17 = 17
23 = 23
29 = 29
HCF = product of smallest power of each common prime factor in the numbers = 1
LCM = product of greatest power of each prime factor involved in the numbers
= 17 × 23 × 29 = 11339
(iv) 24 = 2 × 2 × 2 × 3 = 2
3
× 3
36 = 2 × 2 × 3 × 3 = 2
2
× 3
2
40 = 2 × 2 × 2 × 5 = 2
3
× 5
HCF = product of smallest power of each common prime factor in the numbers
= 2
2
= 4
LCM = product of greatest power of each prime factor involved in the numbers
= 2
3
× 3
2
× 5 = 360
(v) 30 = 2 × 3 × 5
72 = 2 × 2 × 2 × 3 × 3 = 2
3
× 3
2
432 = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 2
4
× 3
3
HCF = product of smallest power of each common prime factor in the numbers
= 2 × 3 = 6
LCM = product of greatest power of each prime factor involved in the numbers
= 2
4
× 3
3
× 5 = 2160
(vi) 21 = 3 × 7
28 = 2 × 2 × 7 = 2
2
× 7
36 = 2 × 2 × 3 × 3 = 2
2
× 3
2
45 = 5 × 3 × 3 = 5 × 3
2
HCF = product of smallest power of each common prime factor in the numbers = 1
LCM = product of greatest power of each prime factor involved in the numbers
= 2
2
× 3
2
× 5 × 7 = 1260
3. The HCF of two numbers is 23 and their LCM is 1449. If one of the numbers is 161, find the
other.
Sol:
Let the two numbers be a and b.
Let the value of a be 161.
Given: HCF = 23 and LCM = 1449
We know, a × b = HCF × LCM
161 × b = 23 × 1449
b = = = 207
Hence, the other number b is 207.
Page 4
Exercise - 1B
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 verify that HCF × LCM = product of given numbers.
Sol:
(i) Prime factorization:
36 = 2
2
× 3
84 = 2
2
× 3 × 7
HCF = product of smallest power of each common prime factor in the numbers
= 2
2
× 3 = 12
LCM = product of greatest power of each prime factor involved in the numbers
= 2
2
× 3
2
× 7 = 252
(ii) Prime factorization:
23 = 23
31 = 31
HCF = product of smallest power of each common prime factor in the numbers = 1
LCM = product of greatest power of each prime factor involved in the numbers
= 23 × 31 = 713
(iii) Prime factorization:
96 = 2
5
× 3
404 = 2
2
× 101
HCF = product of smallest power of each common prime factor in the numbers
= 2
2
= 4
LCM = product of greatest power of each prime factor involved in the numbers
= 2
5
× 3 × 101 = 9696
(iv)Prime factorization:
144 = 2
4
× 3
2
198 = 2 × 3
2
× 11
HCF = product of smallest power of each common prime factor in the numbers
= 2 × 3
2
= 18
LCM = product of greatest power of each prime factor involved in the numbers
= 2
4
× 3
2
× 11 = 1584
(v) Prime factorization:
396 = 2
2
× 3
2
× 11
1080 = 2
3
× 3
3
× 5
HCF = product of smallest power of each common prime factor in the numbers
= 2
2
× 3
2
= 36
LCM = product of greatest power of each prime factor involved in the numbers
= 2
3
× 3
3
× 5 × 11 = 11880
(vi) Prime factorization:
1152 = 2
7
× 3
2
1664 = 2
7
× 13
HCF = product of smallest power of each common prime factor in the numbers
= 2
7
= 128
LCM = product of greatest power of each prime factor involved in the numbers
= 2
7
× 3
2
× 13 = 14976
2. Using prime factorization, find the HCF and 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
Sol:
(i) 8 = 2 × 2 × 2 = 2
3
9 = 3 × 3 = 3
2
25 = 5 × 5 = 5
2
HCF = product of smallest power of each common prime factor in the numbers = 1
LCM = product of greatest power of each prime factor involved in the numbers
= 2
3
× 3
2
× 5
2
= 1800
(ii) 12 = 2 × 2 × 3 = 2
2
× 3
15 = 3 × 5
21 = 3 × 7
HCF = product of smallest power of each common prime factor in the numbers = 3
LCM = product of greatest power of each prime factor involved in the numbers
= 2
2
× 3 × 5 × 7 = 420
(iii) 17 = 17
23 = 23
29 = 29
HCF = product of smallest power of each common prime factor in the numbers = 1
LCM = product of greatest power of each prime factor involved in the numbers
= 17 × 23 × 29 = 11339
(iv) 24 = 2 × 2 × 2 × 3 = 2
3
× 3
36 = 2 × 2 × 3 × 3 = 2
2
× 3
2
40 = 2 × 2 × 2 × 5 = 2
3
× 5
HCF = product of smallest power of each common prime factor in the numbers
= 2
2
= 4
LCM = product of greatest power of each prime factor involved in the numbers
= 2
3
× 3
2
× 5 = 360
(v) 30 = 2 × 3 × 5
72 = 2 × 2 × 2 × 3 × 3 = 2
3
× 3
2
432 = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 2
4
× 3
3
HCF = product of smallest power of each common prime factor in the numbers
= 2 × 3 = 6
LCM = product of greatest power of each prime factor involved in the numbers
= 2
4
× 3
3
× 5 = 2160
(vi) 21 = 3 × 7
28 = 2 × 2 × 7 = 2
2
× 7
36 = 2 × 2 × 3 × 3 = 2
2
× 3
2
45 = 5 × 3 × 3 = 5 × 3
2
HCF = product of smallest power of each common prime factor in the numbers = 1
LCM = product of greatest power of each prime factor involved in the numbers
= 2
2
× 3
2
× 5 × 7 = 1260
3. The HCF of two numbers is 23 and their LCM is 1449. If one of the numbers is 161, find the
other.
Sol:
Let the two numbers be a and b.
Let the value of a be 161.
Given: HCF = 23 and LCM = 1449
We know, a × b = HCF × LCM
161 × b = 23 × 1449
b = = = 207
Hence, the other number b is 207.
4. The HCF of two numbers is 145 and their LCM is 2175. If one of the numbers is 725, find
the other.
Sol:
HCF of two numbers = 145
LCM of two numbers = 2175
Let one of the two numbers be 725 and other be x .
Using the formula, product of two numbers = HCF × LCM
we conclude that
725 × x = 145 × 2175
x =
= 435
Hence, the other number is 435.
5. The HCF of two numbers is 18 and their product is 12960. Find their LCM.
Sol:
HCF of two numbers = 18
Product of two numbers = 12960
Let their LCM be x .
Using the formula, product of two numbers = HCF × LCM
we conclude that
12960 = 18 × x
x =
= 720
Hence, their LCM is 720.
6. Is it possible to have two numbers whose HCF is 18 and LCM is 760?
Give reason.
Sol:
No, it is not possible to have two numbers whose HCF is 18 and LCM is 760.
Since, HCF must be a factor of LCM, but 18 is not factor of 760.
7. Find the simplest form of
(i)
69
92
(ii)
473
645
(iii)
1095
1168
(iv)
368
496
Sol:
(i) Prime factorization of 69 and 92 is:
69 = 3 × 23
92 = 2
2
× 23
Therefore, = = =
Page 5
Exercise - 1B
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 verify that HCF × LCM = product of given numbers.
Sol:
(i) Prime factorization:
36 = 2
2
× 3
84 = 2
2
× 3 × 7
HCF = product of smallest power of each common prime factor in the numbers
= 2
2
× 3 = 12
LCM = product of greatest power of each prime factor involved in the numbers
= 2
2
× 3
2
× 7 = 252
(ii) Prime factorization:
23 = 23
31 = 31
HCF = product of smallest power of each common prime factor in the numbers = 1
LCM = product of greatest power of each prime factor involved in the numbers
= 23 × 31 = 713
(iii) Prime factorization:
96 = 2
5
× 3
404 = 2
2
× 101
HCF = product of smallest power of each common prime factor in the numbers
= 2
2
= 4
LCM = product of greatest power of each prime factor involved in the numbers
= 2
5
× 3 × 101 = 9696
(iv)Prime factorization:
144 = 2
4
× 3
2
198 = 2 × 3
2
× 11
HCF = product of smallest power of each common prime factor in the numbers
= 2 × 3
2
= 18
LCM = product of greatest power of each prime factor involved in the numbers
= 2
4
× 3
2
× 11 = 1584
(v) Prime factorization:
396 = 2
2
× 3
2
× 11
1080 = 2
3
× 3
3
× 5
HCF = product of smallest power of each common prime factor in the numbers
= 2
2
× 3
2
= 36
LCM = product of greatest power of each prime factor involved in the numbers
= 2
3
× 3
3
× 5 × 11 = 11880
(vi) Prime factorization:
1152 = 2
7
× 3
2
1664 = 2
7
× 13
HCF = product of smallest power of each common prime factor in the numbers
= 2
7
= 128
LCM = product of greatest power of each prime factor involved in the numbers
= 2
7
× 3
2
× 13 = 14976
2. Using prime factorization, find the HCF and 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
Sol:
(i) 8 = 2 × 2 × 2 = 2
3
9 = 3 × 3 = 3
2
25 = 5 × 5 = 5
2
HCF = product of smallest power of each common prime factor in the numbers = 1
LCM = product of greatest power of each prime factor involved in the numbers
= 2
3
× 3
2
× 5
2
= 1800
(ii) 12 = 2 × 2 × 3 = 2
2
× 3
15 = 3 × 5
21 = 3 × 7
HCF = product of smallest power of each common prime factor in the numbers = 3
LCM = product of greatest power of each prime factor involved in the numbers
= 2
2
× 3 × 5 × 7 = 420
(iii) 17 = 17
23 = 23
29 = 29
HCF = product of smallest power of each common prime factor in the numbers = 1
LCM = product of greatest power of each prime factor involved in the numbers
= 17 × 23 × 29 = 11339
(iv) 24 = 2 × 2 × 2 × 3 = 2
3
× 3
36 = 2 × 2 × 3 × 3 = 2
2
× 3
2
40 = 2 × 2 × 2 × 5 = 2
3
× 5
HCF = product of smallest power of each common prime factor in the numbers
= 2
2
= 4
LCM = product of greatest power of each prime factor involved in the numbers
= 2
3
× 3
2
× 5 = 360
(v) 30 = 2 × 3 × 5
72 = 2 × 2 × 2 × 3 × 3 = 2
3
× 3
2
432 = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 2
4
× 3
3
HCF = product of smallest power of each common prime factor in the numbers
= 2 × 3 = 6
LCM = product of greatest power of each prime factor involved in the numbers
= 2
4
× 3
3
× 5 = 2160
(vi) 21 = 3 × 7
28 = 2 × 2 × 7 = 2
2
× 7
36 = 2 × 2 × 3 × 3 = 2
2
× 3
2
45 = 5 × 3 × 3 = 5 × 3
2
HCF = product of smallest power of each common prime factor in the numbers = 1
LCM = product of greatest power of each prime factor involved in the numbers
= 2
2
× 3
2
× 5 × 7 = 1260
3. The HCF of two numbers is 23 and their LCM is 1449. If one of the numbers is 161, find the
other.
Sol:
Let the two numbers be a and b.
Let the value of a be 161.
Given: HCF = 23 and LCM = 1449
We know, a × b = HCF × LCM
161 × b = 23 × 1449
b = = = 207
Hence, the other number b is 207.
4. The HCF of two numbers is 145 and their LCM is 2175. If one of the numbers is 725, find
the other.
Sol:
HCF of two numbers = 145
LCM of two numbers = 2175
Let one of the two numbers be 725 and other be x .
Using the formula, product of two numbers = HCF × LCM
we conclude that
725 × x = 145 × 2175
x =
= 435
Hence, the other number is 435.
5. The HCF of two numbers is 18 and their product is 12960. Find their LCM.
Sol:
HCF of two numbers = 18
Product of two numbers = 12960
Let their LCM be x .
Using the formula, product of two numbers = HCF × LCM
we conclude that
12960 = 18 × x
x =
= 720
Hence, their LCM is 720.
6. Is it possible to have two numbers whose HCF is 18 and LCM is 760?
Give reason.
Sol:
No, it is not possible to have two numbers whose HCF is 18 and LCM is 760.
Since, HCF must be a factor of LCM, but 18 is not factor of 760.
7. Find the simplest form of
(i)
69
92
(ii)
473
645
(iii)
1095
1168
(iv)
368
496
Sol:
(i) Prime factorization of 69 and 92 is:
69 = 3 × 23
92 = 2
2
× 23
Therefore, = = =
Thus, simplest form of is .
(ii) Prime factorization of 473 and 645 is:
473 = 11 × 43
645 = 3 × 5 × 43
Therefore, = =
Thus, simplest form of is .
(iii) Prime factorization of 1095 and 1168 is:
1095 = 3 × 5 × 73
1168 = 2
4
× 73
Therefore, = =
Thus, simplest form of is .
(iv) Prime factorization of 368 and 496 is:
368 = 2
4
× 23
496 = 2
4
× 31
Therefore, = =
Thus, simplest form of is .
8. Find the largest number which divides 438 and 606 leaving remainder 6 in each case.
Answer:
Largest number which divides 438 and 606, leaving remainder 6 is actually the largest
number which divides 438 6 = 432 and 606 6 = 600, leaving remainder 0.
Therefore, HCF of 432 and 600 gives the largest number.
Now, prime factors of 432 and 600 are:
432 = 2
4
× 3
3
600 = 2
3
× 3 × 5
2
HCF = product of smallest power of each common prime factor in the numbers = 2
3
× 3 =
24
Thus, the largest number which divides 438 and 606, leaving remainder 6 is 24.
9. Find the largest number which divides 320 and 457 leaving remainders 5 and 7 respectively.
Answer:
We know that the required number divides 315 (320 5) and 450 (457 7).
Required number = HCF (315, 450)
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FAQs on RS Aggarwal Solutions: Real Numbers - 2 - Mathematics (Maths) Class 10
| 1. What are real numbers? |  |
Ans. Real numbers are numbers that can be represented on the number line and include both rational numbers (which can be expressed as a fraction) and irrational numbers (which cannot be expressed as a fraction and have an infinite decimal representation).
| 2. How do you determine if a number is rational or irrational? | |
Ans. A number is rational if it can be expressed as a fraction, where the numerator and denominator are both integers. On the other hand, a number is irrational if it cannot be expressed as a fraction and has a non-repeating, non-terminating decimal representation.
| 3. What are some examples of irrational numbers? | |
Ans. Some examples of irrational numbers include √2, π (pi), and e (Euler's number). These numbers cannot be expressed as fractions and have decimal representations that go on forever without repeating.
| 4. What is the difference between a real number and a complex number? | |
Ans. A real number is a number that can be represented on the number line, while a complex number is a number that consists of both a real part and an imaginary part. The imaginary part is a multiple of the imaginary unit "i", which is equal to the square root of -1.
| 5. How are real numbers used in everyday life? | |
Ans. Real numbers are used in various aspects of everyday life, such as measuring distances, calculating time, determining temperature, and making financial transactions. They are essential for representing and quantifying quantities in the physical world.
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Additional Information about RS Aggarwal Solutions: Real Numbers - 2 for Class 10 Preparation
RS Aggarwal Solutions: Real Numbers - 2 Free PDF Download
The RS Aggarwal Solutions: Real Numbers - 2 is an invaluable resource that delves deep into the core of the Class 10 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
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Importance of RS Aggarwal Solutions: Real Numbers - 2
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RS Aggarwal Solutions: Real Numbers - 2 Notes
RS Aggarwal Solutions: Real Numbers - 2 Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to RS Aggarwal Solutions: Real Numbers - 2.
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RS Aggarwal Solutions: Real Numbers - 2 Class 10 Questions
The "RS Aggarwal Solutions: Real Numbers - 2 Class 10 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 10 exam. It focuses on providing a wide range of practice questions to help students gauge
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The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
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The content of RS Aggarwal Solutions: Real Numbers - 2 is prepared as per the latest Class 10 syllabus.
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