Chapter 3 – HCF and LCM
Practice Set 10
1. Which number is neither a prime number nor a composite number?
Ans: 1 is neither a prime number nor a composite number.
2. Which of the following are pairs of co-primes?
(i) 8, 14
Solution:
Factors of 8: 1, 2, 4, 8
Factors of 14: 1, 2, 7, 14
Common factors of 8 and 14: 1, 2
∴ 8 and 14 are not a pair of co-prime numbers.
(ii) 4, 5
Solution:
Factors of 4: 1, 4
Factors of 5: 1, 5
Common factors of 4 and 5: 1
∴ 4 and 5 are a pair of co-prime numbers.
(iii) 17, 19
Solution:
Factors of 17: 1, 17
Factors of 19: 1, 19
Common factors of 17 and 19: 1
∴ 17 and 19 are a pair of co-prime numbers.
(iv) 27, 15
Solution:
Factors of 27: 1, 3, 9, 27
Factors of 15: 1, 3, 5, 15
Common factors of 27 and 15 : 1, 3
∴ 27 and 15 are not a pair of co-prime numbers.
3. List the prime numbers from 25 to 100 and say how many they are.
Ans: 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
There are altogether 16 prime numbers from 25 to 100.
4. Write all the twin prime numbers from 51 to 100.
Ans:
(i) 59 and 61
(ii) 71 and 73
5. Write 5 pairs of twin prime numbers from 1 to 50.
Ans:
(i) 3, 5
(ii) 5, 7
(iii) 11, 13
(iv) 17, 19
(v) 29, 31
(vi) 41, 43
6. Which are the even prime numbers?
Ans: 2 is the only even prime number.
Practice set 11
Factorise the following numbers into primes.
(i) 32
Solution:
∴ 32 = 2 × 2 × 2 × 2 × 2
(ii) 57
Solution:
∴ 57 = 3 × 19
(iii) 23
Solution:
∴ 23 = 23 × 1
(iv) 150
Solution:
∴ 150 = 2 × 3 × 5 × 5
(v) 216
Solution:
∴ 216 = 2 × 2 × 2 × 3 × 3 × 3
(vi) 208
Solution:
∴ 208 = 2 × 2 × 2 × 2 × 13
(vii) 765
Solution:
∴ 765 = 3 × 3 × 5 × 17
(viii) 342
Solution:
∴ 342 = 2 × 3 × 3 × 19
(ix) 377
Solution:
∴ 377 = 13 × 29
(x) 559
Solution:
∴ 559 = 13 × 43
Practice set 12
1. Find the HCF.
(i) 25, 40
Solution:
25 = 5 × 5
40 = 2 × 2 × 2 × 5
∴ HCF of 25 and 40 = 5
(ii) 56, 32
Solution:
56 = 2 × 2 × 2 × 7
32 = 2 × 2 × 2 × 2 × 2
∴ HCF of 56 and 32 = 2 × 2 × 2
∴ HCF of 56 and 32 = 8
(iii) 40, 60, 75
Solution:
40 = 2 × 2 × 2 × 5
60 = 2 × 2 × 3 × 5
75 = 3 × 5 × 5
∴ HCF of 40, 60 and 75 = 5
(iv) 16, 27
Solution:
16 = 2 × 2 × 2 × 2
27 = 3 × 3 × 3
∴ HCF of 16 and 27 = 1
(v) 18, 32, 48
Solution:
18 = 2 × 3 × 3
32 = 2 × 2 × 2 × 2 × 2
48 = 2 × 2 × 2 × 2 × 3
∴ HCF of 18, 32 and 48 = 2
(vi) 105, 154
Solution:
777 = 3 × 7 × 37
315 = 3 × 3 × 5 × 7
588 = 2 × 2 × 3 × 7 × 7
∴ HCF of 777, 315 and 588 = 3 × 7
∴ HCF of 777, 315 and 588 = 21
2. Find the HCF by the division method and reduce to the simplest form.
(i) \(\large \frac {275}{525}\)
Solution:
HCF of 275 and 525 = 25
∴ \(\large \frac {275}{525}\) = \(\large \frac {275\,÷\,25}{525\,÷\,25}\) = \(\large \frac {11}{21}\)
(ii) \(\large \frac {76}{133}\)
Solution:
HCF of 76 and 133 = 19
∴ \(\large \frac {76}{133}\) = \(\large \frac {76\,÷\,19}{133\,÷\,19}\) = \(\large \frac {4}{7}\)
(iii) \(\large \frac {161}{69}\)
Solution:
HCF of 161 and 69 = 23
∴ \(\large \frac {161}{69}\) = \(\large \frac {161\,÷\,23}{69\,÷\,23}\) = \(\large \frac {7}{3}\)
Practice set 13
1. Find the LCM.
(i) 12, 15
Solution:
12 = 2 × 2 × 3
15 = 3 × 5
∴ LCM of 12 and 15 = 2 × 2 × 3 × 5
∴ LCM of 12 and 15 = 60
(ii) 6, 8, 10
Solution:
6 = 2 × 3
8 = 2 × 2 × 2
10 = 2 × 5
∴ LCM of 6, 8 and 10 = 2 × 3 × 2 × 2 × 5
∴ LCM of 6, 8 and 10 = 120
(iii) 18, 32
Solution:
18 = 2 × 3 × 3
32 = 2 × 2 × 2 × 2 × 2
∴ LCM of 18 and 32 = 2 × 3 × 3 × 2 × 2 × 2 × 2
∴ LCM of 18 and 32 = 288
(iv) 10, 15, 20
Solution:
10 = 2 × 5
15 = 3 × 5
20 = 2 × 2 × 5
∴ LCM of 10, 15 and 20 = 2 × 5 × 3 × 2
∴ LCM of 10, 15 and 20 = 60
(v) 45, 86
Solution:
45 = 5 × 3 × 3
86 = 2 × 43
∴ LCM of 45 and 86 = 5 × 3 × 3 × 2 × 43
∴ LCM of 45 and 86 = 3870
(vi) 15, 30, 90
Solution:
15 = 3 × 5
30 = 2 × 3 × 5
90 = 2 × 5 × 3 × 3
∴ LCM of 15, 30 and 90 = 3 × 5 × 2 × 3
∴ LCM of 15, 30 and 90 = 90
(vii) 105, 195
Solution:
105 = 3 × 5 × 7
195 = 3 × 5 × 13
∴ LCM of 105 and 95 = 3 × 5 × 7 × 13
∴ LCM of 105 and 95 = 1365
(viii) 12, 15, 45
Solution:
12 = 2 × 2 × 3
15 = 3 × 5
45 = 5 × 3 × 3
∴ LCM of 12, 15 and 45 = 3 × 5 × 2 × 2 × 3
∴ LCM of 12, 15 and 45 = 180
(ix) 63, 81
Solution:
63 = 3 × 3 × 7
81 = 3 × 3 × 3 × 3
∴ LCM of 63 and 81 = 3 × 3 × 7 × 3 × 3
∴ LCM of 63 and 81 = 567
(x) 18, 36, 27
Solution:
18 = 2 × 3 × 3
36 = 2 × 2 × 3 × 3
27 = 3 × 3 × 3
∴ LCM of 18, 36 and 27 = 2 × 3 × 3 × 2 × 3
∴ LCM of 18, 36 and 27 = 108
2. Find the HCF and LCM of the numbers given below. Verify that their product is equal to the product of the given numbers.
(i) 32, 37
Solution:
32 = 2 × 2 × 2 × 2 × 2
37 = 37
HCF of 32 and 37 = 1
LCM of 32 and 37 = 2 × 2 × 2 × 2 × 2 × 37
∴ LCM of 32 and 37 = 1184
HCF × LCM = 1 × 1184
∴ HCF × LCM = 1184
Product of the given numbers = 32 × 37
∴ Product of the given numbers = 1184
∴ HCF × LCM = Product of the given numbers
Hence verified.
(ii) 46, 51
Solution:
46 = 2 × 23
51 = 3 × 17
HCF of 46 and 51 = 1
LCM of 46 and 51 = 2 × 23 × 3 × 17
∴ LCM of 46 and 51 = 2346
HCF × LCM = 1 × 2346
∴ HCF × LCM = 2346
Product of the given numbers = 46 × 51
∴ Product of the given numbers = 2346
∴ HCF × LCM = Product of the given numbers
Hence verified.
(iii) 15, 60
Solution:
15 = 3 × 5
60 = 2 × 2 × 3 × 5
HCF of 15 and 60 = 3 × 5
∴ HCF of 15 and 60 = 15
LCM of 15 and 60 = 3 × 5 × 2 × 2
∴ LCM of 15 and 60 = 60
HCF × LCM = 15 × 60
∴ HCF × LCM = 900
Product of the given numbers = 15 × 60
∴ Product of the given numbers = 900
∴ HCF × LCM = Product of the given numbers
Hence verified.
(iv) 18, 63
Solution:
18 = 2 × 3 × 3
63 = 3 × 3 × 7
HCF of 18 and 63 = 3 × 3
∴ HCF of 18 and 63 = 9
LCM of 18 and 63 = 3 × 3 × 2 × 7
∴ LCM of 18 and 63 = 126
HCF × LCM = 9 × 126
∴ HCF × LCM = 1134
Product of the given numbers = 18 × 63
∴ Product of the given numbers = 1134
∴ HCF × LCM = Product of the given numbers
Hence verified.
(v) 78, 104
Solution:
78 = 2 × 3 × 13
104 = 2 × 2 × 2 × 13
HCF of 78 and 104 = 2 × 13
∴ HCF of 78 and 104 = 26
LCM of 78 and 104 = 2 × 13 × 3 × 2 × 2
∴ LCM of 78 and 104 = 312
HCF × LCM = 26 × 312
∴ HCF × LCM = 8112
Product of the given numbers = 78 × 104
∴ Product of the given numbers = 8112
∴ HCF × LCM = Product of the given numbers
Hence verified.
Practice set 14
1. Choose the right option.
(i) The HCF of 120 and 150 is ________.
(1) 30
(2) 45
(3) 20
(4) 120
Ans: Option (1) – 30
Solution:
120 = 2 × 2 × 2 × 3 × 5
150 = 2 × 3 × 5 × 5
HCF of 120 and 150 = 2 × 3 × 5
∴ HCF of 120 and 150 = 30
(ii) The HCF of this pair of numbers is not 1.
(1) 13, 17
(2) 29, 20
(3) 40, 20
(4) 14, 15
Ans: Option (3) – 40, 20
Solution:
40 = 2 × 2 × 2 × 5
60 = 2 × 2 × 5
HCF of 120 and 150 = 2 × 2 × 5
∴ HCF of 120 and 150 = 20 ≠ 1
2. Find the HCF and LCM.
(i) 14, 28
Solution:
14 = 2 × 7
28 = 2 × 2 × 7
HCF of 14 and 28 = 2 × 7
∴ HCF of 14 and 28 = 14
LCM of 14 and 28 = 2 × 7 × 2
∴ LCM of 14 and 28 = 28
(ii) 32, 16
Solution:
32 = 2 × 2 × 2 × 2 × 2
16 = 2 × 2 × 2 × 2
HCF of 32 and 16 = 2 × 2 × 2 × 2
∴ HCF of 32 and 16 = 16
LCM of 32 and 16 = 2 × 2 × 2 × 2 × 2
∴ LCM of 32 and 16 = 32
(iii) 17, 102, 170
Solution:
17 = 17
102 = 2 × 3 × 17
170 = 2 × 5 × 17
HCF of 17, 102 and 170 = 17
LCM of 17, 102 and 170 = 17 × 2 × 3 × 5
∴ LCM of 17, 102 and 170 = 510
(iv) 23, 69
Solution:
23 = 23
69 = 3 × 23
HCF of 23 and 69 = 23
LCM of 23 and 69 = 2 × 23
∴ LCM of 23 and 69 = 69
(v) 21, 49, 84
Solution:
21 = 3 × 7
49 = 7 × 7
84 = 2 × 2 × 3 × 7
HCF of 21, 49 and 84 = 7
LCM of 21, 49 and 84 = 7 × 3 × 7 × 2 × 2
∴ LCM of 21, 49 and 84 = 588
3. Find the LCM.
(i) 36, 42
Solution:
36 = 2 × 2 × 3 × 3
42 = 2 × 3 × 7
∴ LCM of 36 and 42 = 2 × 3 × 2 × 3 × 7
∴ LCM of 36 and 42 = 252
(ii) 15, 25, 30
Solution:
15 = 3 × 5
25 = 5 × 5
30 = 2 × 3 × 5
∴ LCM of 15, 25 and 30 = 3 × 5 × 5 × 2
∴ LCM of 15, 25 and 30 = 150
(iii) 18, 42, 48
Solution:
18 = 2 × 3 × 3
42 = 2 × 3 × 7
48 = 2 × 2 × 2 × 2 × 3
∴ LCM of 18, 42 and 48 = 2 × 3 × 3 × 7 × 2 × 2 × 2
∴ LCM of 18, 42 and 48 = 1008
(iv) 4, 12, 20
Solution:
4 = 2 × 2
12 = 2 × 2 × 3
20 = 2 × 2 × 5
∴ LCM of 4, 12 and 20 = 2 × 2 × 3 × 5
∴ LCM of 18, 42 and 48 = 60
(v) 24, 40, 80, 120
Solution:
24 = 2 × 2 × 2 × 3
40 = 2 × 2 × 2 × 5
80 = 2 × 2 × 2 × 2 × 5
120 = 2 × 2 × 2 × 3 × 5
∴ LCM of 24, 40, 80 and 120 = 2 × 2 × 2 × 3 × 5 × 2
∴ LCM of 24, 40, 80 and 120 = 240
4. Find the smallest number which when divided by 8, 9, 10, 15, 20 gives a remainder of 5 every time.
Solution:
The smallest number for division will be the LCM of 8, 9, 10,15 and 20.
8 = 2 × 2 × 2
9 = 3 × 3
10 = 2 × 5
15 = 3 × 5
20 = 2 × 2 × 5
∴ LCM of 8, 9, 10,15 and 20 = 2 × 2 × 2 × 3 × 3 × 5
∴ LCM of 8, 9, 10,15 and 20 = 360
∴ Smallest number = LCM + Remainder
∴ Smallest number = 360 + 5
∴ Smallest number = 365
Ans: The required smallest number is 365.
5. Reduce the fractions \(\large \frac {348}{319}\), \(\large \frac {221}{247}\), \(\large \frac {437}{551}\) to the lowest terms.
Solution:
HCF of 348 and 319 = 29
\(\large \frac {348}{319}\) = \(\large \frac {348\,÷\,29}{319\,÷\,29}\) = \(\large \frac {12}{11}\)
HCF of 221 and 247 = 13
∴ \(\large \frac {221}{247}\) = \(\large \frac {221\,÷\,13}{247\,÷\,13}\) = \(\large \frac {17}{19}\)
HCF of 437 and 551 = 19
∴ \(\large \frac {437}{551}\) = \(\large \frac {437\,÷\,19}{551\,÷\,19}\) = \(\large \frac {23}{29}\)