KC Sinha Mathematics Solution Class 10 Chapter 1 Real numbers Exercise 1.2

Q1 A Q1 B Q1 C Q1 D Q1 E Q1 F Q1 G Q2 Q3 Q4 Q5 A Q5 B Q5 C Q6 A Q6 B Q6 C Q6 D Q7 A Q7 B Q7 C Q7 D Q7 E Q7 F Q7 G Q8 A Q8 B Q8 C Q8 D Q8 E Q8 F Q9 A Q9 B Q9 C Q9 D Q9 E Q9 F Q10 A Q10 B Q10 C Q10 D Q10 E Q10 F Q11 A Q11 B Q11 C Q11 D Q11 E Q11 F Q12 A Q12 B Q12 C Q13 Q14 Q15

Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4

 Exercise 1.2

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Question 1 A 

Express each of the following numbers as a product of its prime factors:
Sol :
Given number is 4320
Factorization of 4320 is

$\begin{array}{l|l}2 & 4320 \\\hline 2 & 2160 \\\hline 2 & 1080 \\\hline 2 & 540 \\\hline 2 & 270 \\\hline 3 & 135 \\\hline 3 & 45 \\\hline 3 & 15 \\\hline 5 & 5 \\\hline & 1\end{array}$

Hence, 4320 = 2⁵ × 3³ × 5 (Product of its prime factors)

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Question 1 B 

Express each of the following numbers as a product of its prime factors: 7560
Sol :
Given number is 7560
Factorization of 7560 is
$\begin{array}{l|l}2 & 7560 \\\hline 2 & 3780 \\\hline 2 & 1890 \\\hline 3 & 945 \\\hline 3 & 315 \\\hline 3 & 105 \\\hline 5 & 35 \\\hline 7 & 7 \\\hline & 1\end{array}$

Hence, 7560 = 2³×3³×5×7 (Product of its prime factors)

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Question 1 C 

Express each of the following numbers as a product of its prime factors: 140
Sol :
Given number is 140
Factorization of 140 is

$\begin{array}{l|l}
2 & 140 \\
\hline 2 & 70 \\
\hline 5 & 35 \\
\hline 7 & 7 \\
\hline & 1
\end{array}$
Hence, 140 = 2² × 5 × 7 (Product of its prime factors)

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Question 1 D 

Express each of the following numbers as a product of its prime factors: 5005
Sol :
Given number is 5005
Factorization of 5005 is

$\begin{array}{l|l}
5 & 5005 \\
\hline 7 & 1001 \\
\hline 11 & 143 \\
\hline 13 & 13 \\
\hline & 1
\end{array}$
Hence, 5005 = 5 × 7 × 11 × 13 (Product of its prime factors)

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Question 1 E 

Express each of the following numbers as a product of its prime factors: 32760
Sol :
Given number is 32760
Factorization of 32760 is

$\begin{array}{l|l}
2 & 32760 \\
\hline 2 & 16380 \\
\hline 2 & 8190 \\
\hline 3 & 4095 \\
\hline 3 & 1365 \\
\hline 5 & 455 \\
\hline 7 & 91 \\
\hline 13 & 13 \\
\hline & 1
\end{array}$
Hence, 32760 = 2³ × 3² × 5 × 7 × 13 (Product of its prime factors)

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Question 1 F 

Express each of the following numbers as a product of its prime factors: 156
Sol :
Given number is 156
Factorization of 156 is

$\begin{array}{l|l}
2 & 156 \\
\hline 2 & 78 \\
\hline 3 & 39 \\
\hline 13 & 13 \\
\hline & 1
\end{array}$
Hence, 156 = 2² × 3 × 13 (Product of its prime factors)

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Question 1 G 

Express each of the following numbers as a product of its prime factors: 729
Sol :
Given number is 729
Factorization of 729 is

$\begin{array}{l|l}
3 & 729 \\
\hline 3 & 243 \\
\hline 3 & 81 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}$
Hence, 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3⁶ (Product of its prime factors)

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Question 2 

Find the highest power of 5 in 23750.
Sol :
To find the highest power of 5 in 23750, we have to factorize 23570

$\begin{array}{l|l}
2 & 23750 \\
\hline 5 & 11875 \\
\hline 5 & 2375 \\
\hline 5 & 475 \\
\hline 5 & 95 \\
\hline 19 & 19 \\
\hline & 1
\end{array}$
Hence, 23750 = 2 × 5⁴ × 19
So, the highest power of 5 in 23750 is 4.

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Question 3 

Find the highest power of 2 in 1440.
Sol :
Given number is 1440
Factorization of 1440 is

$\begin{array}{l|l}
2 & 1440 \\
\hline 2 & 720 \\
\hline 2 & 360 \\
\hline 2 & 180 \\
\hline 2 & 90 \\
\hline 3 & 45 \\
\hline 3 & 15 \\
\hline 5 & 5 \\
\hline & 1
\end{array}$
Factors of 1440 = 2⁵ × 3² × 5
So, the highest power of 2 is 5.

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Question 4 

If 6370 = 2^m .5ⁿ.7^k.13^p, then find m+n+k+p.

Sol :
We have to factorize the 6370 to find the value of m, n, k and p
Factorization of 6370 is

$\begin{array}{l|l}
2 & 6370 \\
\hline 5 & 3185 \\
\hline 7 & 637 \\
\hline 7 & 91 \\
\hline 13 & 13 \\
\hline & 1
\end{array}$
6370 = 2 × 5 × 7² × 13
On Comparing, we get

6370 =2^m .5ⁿ.7^k.13^p

=2¹×5¹×7²×13¹

m = 1
n = 1
k = 2
p = 1
So, m+n+k+p=5

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Question 5 A 

Which of the following is a pair of co-primes?
(32,62)
Sol :
Given numbers are 32 and 62
For pairs to be co-primes there should be no common factor except 1
Factorization of 32 and 62 are

$\begin{array}{l|l}
2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}$
Factors of 32=2×2×2×2×2=2⁵

$\begin{array}{l|l}
2 & 62 \\
\hline 31 & 31 \\
\hline & 1
\end{array}$
Factors of 62=2×31

Here, we can see that 2 is the common factor. So, (32,62) is not co-prime.

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Question 5 B 

Which of the following is a pair of co-primes?
(18,25)
Sol :
Given numbers are 18 and 25
For pairs to be co-primes there should be no common factor except 1
Factorization of 18 and 25 are

$\begin{array}{l|l}
2 & 18 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}$

Factors of 32 = 2 × 3 × 3
$\begin{array}{l|l}5 & 25 \\\hline 5 & 5 \\\hline & 1\end{array}$
Factors of 62 = 5 × 5
Therefore, there is no common factor. So, (18,25) is co-prime.

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Question 5 C 

Which of the following is a pair of co-primes?
(31, 93)
Sol :
Given numbers are 31 and 93
For pairs to be co-primes there should be no common factor except 1
Factorization of 31 and 93 are

$\begin{array}{c|l}
3 & 93 \\
\hline 31 & 31 \\
\hline 2 & 1
\end{array}$

Factors of 93=31×3

$\begin{array}{l|l}
31 & 31 \\
\hline & 1
\end{array}$

Factors of 31=31×1
Here, we can see that 31 is the common factor. So, (31, 93) is not co-prime.

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Question 6 A 

Write down the missing numbers a, b, c, d, x, y in the following factor tree :

Sol :

Here, a=2520; b=2; c=315; d=3; x=3; y=5

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Question 6 B 

Write down the missing numbers a, b, c, d, x, y in the following factor tree :

Sol :

Here, a=15015; b=5005; c=5; d=143; x=13

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Question 6 C 

Write down the missing numbers a, b, c, d, x, y in the following factor tree :

Sol :

Here, a=18380; b=2; c=1365; d=3; x=5; y=13

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Question 6 D 

Write down the missing numbers a,b,c,d,x, y in the following factor tree :

Sol :

Here, a=3; b=147407; c=11339; d=667; x=29

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Question 7 A 

Find the LCM and HCF of the following integers by applying the prime factorization method :
96 and 404
Sol :
Given numbers are 96 and 404.
The prime factorization of 96 and 404 gives:

$\begin{array}{l|l}
2 & 96 \\
\hline 2 & 48 \\
\hline 2 & 24 \\
\hline 2 & 12 \\
\hline 2 & 6 \\
\hline 3 & 3 \\
\hline & 1
\end{array}$

96=2⁵×3

=2×2×2×2×2×3

$\begin{array}{l|l}
2 & 404 \\
\hline 2 & 202 \\
\hline 101 & 101 \\
\hline & 1
\end{array}$

404=2²×101

=2×2×101

Here, 2² is the smallest power of the common factor 2.
Therefore, the H.C.F of these two integers is 2×2 = 4
2⁵ × 3¹ × 101¹ are the greatest powers of the prime factors 2, 3 and 101 respectively involved in the given numbers.
Now, L.C.M of 96 and 404 is 2×2×2×2×2×3×101

= 9696

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Question 7 B 

Find the LCM and HCF of the following integers by applying the prime factorization method :
6 and 20
Sol :
Given numbers are 6 and 20
The prime factorization of 6 and 20 gives:

$\begin{array}{l|l}
2 & 6 \\
\hline 3 & 3 \\
\hline & 1
\end{array}$

6=2×3

$\begin{array}{l|l}
2 & 20 \\
\hline 2 & 10 \\
\hline 5 & 5 \\
\hline & 1
\end{array}$

20=2×2×5

Here, 2¹ is the smallest power of the common factor 2.
Therefore, the H.C.F of these two integers = 2
2² × 3¹ × 5¹ are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the given numbers.
Now, L.C.M of 6 and 20=2×2×3×5

= 60

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Question 7 C 

Find the LCM and HCF of the following integers by applying the prime factorization method :
26 and 91
Sol :
Given numbers are 26 and 91.
The prime factorization of 26 and 91 gives:

$\begin{array}{l|l}
2 & 26 \\
\hline 13 & 13 \\
\hline & 1
\end{array}$

26=2×3

$\begin{array}{l|l}
7 & 91 \\
\hline 13 & 13 \\
\hline & 1
\end{array}$

91=7×13

Here, 13¹ is the smallest power of the common factor 13.
Therefore, the H.C.F of these two integers = 13
2¹ × 7¹ × 13¹ are the greatest powers of the prime factors 2, 7 and 13 respectively involved in the given numbers.
Now, L.C.M of 6 and 21 is 2×7×13

= 182

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Question 7 D 

Find the LCM and HCF of the following integers by applying the prime factorization method :
87 and 145
Sol :
Given numbers are 87 and 145.
The prime factorization of 87 and 145 gives:

$\begin{array}{c|l}
3 & 87 \\
\hline 29 & 29 \\
\hline & 1
\end{array}$

87=3×29

$\begin{array}{c|l}
5 & 145 \\
\hline 29 & 29 \\
\hline & 1
\end{array}$

145=5×29

Here, 29¹ is the smallest power of the common factor 29.
Therefore, the H.C.F of these two integers = 29
3¹ × 5¹ × 29¹ are the greatest powers of the prime factors 3, 5 and 29 respectively involved in the given numbers.
Now, L.C.M of 87 and 145=3×5×29 

= 435

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Question 7 E 

Find the LCM and HCF of the following integers by applying the prime factorization method :
1485 and 4356
Sol :
Given numbers are 1485 and 4356.
The prime factorization of 1485 and 4356 gives:

$\begin{array}{l|l}
3 & 1485 \\
\hline 3 & 495 \\
\hline 3 & 165 \\
\hline 5 & 55 \\
\hline 11 & 11 \\
\hline & 1
\end{array}$

1485=3×3×3×5×11

$\begin{array}{l|l}
2 & 4356 \\
\hline 2 & 2178 \\
\hline 3 & 1089 \\
\hline 3 & 363 \\
\hline 11 & 121 \\
\hline 11 & 11 \\
\hline & 1
\end{array}$

4356=2×2×3×3×11×11

Here, 3²× 11 is the smallest power of the common factors 3 and 11.
Therefore, the H.C.F of these two integers = 3 × 3 × 11 = 99
2² × 3³× 5¹ × 11² are the greatest powers of the prime factors 2, 3 and 7 respectively involved in the given numbers.
Now, L.C.M of 1485 and 4356 =2×2×3×3×3×5×11×11 

= 65430

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Question 7 F 

Find the LCM and HCF of the following integers by applying the prime factorization method :
1095 and 1168
Sol :
Given numbers are 1095 and 1168.
The prime factorization of 1095 and 1168 gives:

$\begin{array}{c|l}
3 & 1095 \\
\hline 5 & 365 \\
\hline 73 & 73 \\
\hline & 1
\end{array}$

1485=3×5×73

$\begin{array}{l|l}
2 & 1168 \\
\hline 2 & 584 \\
\hline 2 & 292 \\
\hline 2 & 146 \\
\hline 73 & 73 \\
\hline & 1
\end{array}$

4356=2×2×2×2×73

Here, 73¹ is the smallest power of the common factor 73.
Therefore, the H.C.F of these two integers = 73
2⁴ × 3¹× 5¹ × 73¹ are the greatest powers of the prime factors 2, 3, 5 and 73 respectively involved in the given numbers.
Now, L.C.M of 1485 and 4356=2×2×2×2×3×5×73

= 17520

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Question 7 G 

Find the LCM and HCF of the following integers by applying the prime factorization method :
6 and 21
Sol :
Given numbers are 6 and 21.
The prime factorization of 6 and 21 gives:

$\begin{array}{l|l}
2 & 6 \\
\hline 3 & 3 \\
\hline & 1
\end{array}$

6=2×3

$\begin{array}{l|l}
3 & 21 \\
\hline 7 & 7 \\
\hline & 1
\end{array}$

21=3×7

Here, 3¹ is the smallest power of the common factor 3.
Therefore, the H.C.F of these two integers = 3
2¹ × 3¹ × 7¹ are the greatest powers of the prime factors 2, 3 and 7 respectively involved in the given numbers.
Now, L.C.M of 6 and 21 is 2×3×7

= 42

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Question 8 A 

Find the LCM and HCF of the following pair of integers and verify that LCM X HCF = Product of two numbers :
96 and 404
Sol :
Given numbers are 96 and 404
The prime factorization of 96 and 404 gives:
96 = 2⁵×3 and 404 =2²×101
Therefore, the H.C.F of these two integers = 2² = 4
Now, the L.C.M of 96 and 404 =2×2×2×2×2×3×101

= 9696

Now, we have to verify

L.C.M(a,b)×H.C.F(a,b)=Product of two numbers(a×b)

L.H.S = L.C.M × H.C.F = 9696 × 4 = 38784
R.H.S = Product of two numbers = 96 × 404 = 38784
Hence, L.H.S = R.H.S
So, the product of two numbers is equal to the product of their HCF and LCM.

---

Question 8 B 

Find the LCM and HCF of the following pair of integers and verify that LCM X HCF = Product of two numbers :
852 and 1491
Sol :
Given numbers are 852 and 1491
The prime factorization of 852 and 1491 gives:
852 = 2 × 2 × 3 × 71 and 1491 = 3 × 7 × 71
Therefore, the H.C.F of these two integers = 3 × 71 = 213
Now, the L.C.M of 96 and 404 = 2 × 2 × 3 × 7 × 71 = 5964
Now, we have to verify

L.C.M(a,b)×H.C.F(a,b)=Product of two numbers(a×b)

L.H.S = L.C.M × H.C.F 

= 5964 × 213

 = 1270332

R.H.S = Product of two numbers 

= 852 × 1491 

= 1270332

Hence, L.H.S = R.H.S
So, the product of two numbers is equal to the product of their HCF and LCM.

---

Question 8 C 

Find the LCM and HCF of the following pair of integers and verify that LCM X HCF = Product of two numbers :
777 and 1147
Sol :
Given numbers are 777 and 1147
The prime factorization of 777 and 1147 gives:
777 = 3 × 7 × 37 and 1147 = 31 × 37
Therefore, the H.C.F of these two integers = 37
Now, the L.C.M of 96 and 404 = 3 × 7 × 31 × 37

 = 24087

Now, we have to verify

L.C.M(a,b)×H.C.F(a,b)=Product of two numbers(a×b)

L.H.S = L.C.M × H.C.F

 = 24087 × 37 = 891219
R.H.S = Product of two numbers 

= 777 × 1147

 = 891219

Hence, L.H.S = R.H.S
So, the product of two numbers is equal to the product of their HCF and LCM.

---

Question 8 D 

Find the LCM and HCF of the following pair of integers and verify that LCM × HCF = Product of two numbers :
36 and 64
Sol :
Given numbers are 36 and 64
The prime factorization of 36 and 64 gives:
36 = 2 × 2 × 3 × 3 and 64 = 2⁶
Therefore, the H.C.F of these two integers = 2 × 2 = 4
Now, the L.C.M of 36 and 64 = 3 × 3 × 2⁶ = 576
Now, we have to verify

L.C.M(a,b)×H.C.F(a,b)=Product of two numbers(a×b)

L.H.S = L.C.M × H.C.F = 576 × 4 = 2304
R.H.S = Product of two numbers = 36 × 64 = 2304
Hence, L.H.S = R.H.S
So, the product of two numbers is equal to the product of their HCF and LCM.

---

Question 8 E 

Find the LCM and HCF of the following pair of integers and verify that LCM X HCF = Product of two numbers :
32 and 80
Sol :
Given numbers are 32 and 80
The prime factorization of 32 and 80 gives:
32 = 2⁵ and 80 = 2⁴ × 5
Therefore, the H.C.F of these two integers = 2⁴ = 16
Now, the L.C.M of 32 and 80 

= 5 × 2⁵ 

= 160

Now, we have to verify

L.C.M(a,b)×H.C.F(a,b)=Product of two numbers(a×b)

L.H.S = L.C.M × H.C.F

 = 160 × 16 

= 2560

R.H.S = Product of two numbers

 = 32 × 80 

= 2560

Hence, L.H.S = R.H.S
So, the product of two numbers is equal to the product of their HCF and LCM.

---

Question 8 F 

Find the LCM and HCF of the following pair of integers and verify that LCM X HCF = Product of two numbers :
902 and 1517
Sol :
Given numbers are 902 and 1517
The prime factorization of 902 and 1517 gives:
902 = 2 × 11 × 41 and 1517 = 37 × 41
Therefore, the H.C.F of these two integers = 41

Now, the L.C.M of 902 and 1517 = 2 ×
11 × 37 × 41 = 33374
Now, we have to verify

L.C.M(a,b)×H.C.F(a,b)=Product of two numbers(a×b)

L.H.S = L.C.M × H.C.F 

= 33374 × 41

 = 1368334

R.H.S = Product of two numbers 

= 902 × 1517 

= 1368334
Hence, L.H.S = R.H.S
So, the product of two numbers is equal to the product of their HCF and LCM.

---

Question 9 A 

Find LCM and HCF of the following integers by using prime factorization method:
6, 72 and 120
Sol :

Given numbers are 6, 72 and 120
Factorization of 6, 72 and 120

$\begin{array}{l|l}
2 & 6 \\
\hline 3 & 3 \\
\hline & 1
\end{array}$

$\begin{array}{c|c}
2 & 72 \\
\hline 2 & 36 \\
\hline 2 & 18 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}$

$\begin{array}{c|c}
2 & 120 \\
\hline 2 & 60 \\
\hline 2 & 30 \\
\hline 3 & 15 \\
\hline 5 & 5 \\
\hline & 1
\end{array}$

6=2×3=2¹×3¹

72=2×2×2×3×3=2³×3²

120=2×2×2×3×5=2³×3¹×5¹

Here, 2¹ × 3¹ are the smallest powers of the common factors 2 and 3, respectively.
So, HCF (6, 72, 120) = 2 × 3 = 6

2³×3²×5¹ are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the given three numbers .
LCM of these three integers =2×2×2×3×3×5 

= 360

---

Question 9 B 

Find LCM and HCF of the following integers by using prime factorization method:
8, 9, and 25
Sol :
Given numbers are 8, 9 and 25
Factorization of 8, 9 and 25

$\begin{array}{c|c}
2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}$

$\begin{array}{c|c}
3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}$


$\begin{array}{c|c}
5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}$

8=2×2×2×1=2³×1
9=3×3×1=3²×1
25=5×5×1=5²×1

Here, 1¹ is the smallest power of the common factor 1.
So, HCF (8, 9, 25) = 1
2³ × 3² × 5² are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the given three numbers .
LCM of these three integers = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800

---

Question 9 C 

Find LCM and HCF of the following integers by using prime factorization method:
12, 15, and 21
Sol :
Given numbers are 12, 15 and 21
Factorization of 12, 15 and 21

$\begin{array}{l|l}
2 & 12 \\
\hline 2 & 6 \\
\hline 3 & 3 \\
\hline & 1
\end{array}$


$\begin{array}{c|c}
3 & 15 \\
\hline 5 & 5 \\
\hline & 1
\end{array}$


$\begin{array}{c|c}
3 & 21 \\
\hline 7 & 7 \\
\hline & 1
\end{array}$

12=2×2×3=2²×3¹

15=3×5=3¹×5¹

21=3×7=3¹×7¹

Here, 3¹ is the smallest power of the common factor 3.

So, HCF (12, 15, 21) = 3
2² × 3¹ × 5¹ × 7¹ are the greatest powers of the prime factors 2, 3, 5 and 7 respectively involved in the given three numbers .

LCM of these three integers

=2×2×3×5×7

= 420

---

Question 9 D 

Find LCM and HCF of the following integers by using prime factorization method:
36, 45, and 72
Sol :
Given numbers are 36, 45 and 72
Factorization of 36, 45 and 72

$\begin{array}{c|c}
2 & 36 \\
\hline 2 & 18 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}$

$\begin{array}{c|c}
3 & 45 \\
\hline 3 & 15 \\
\hline 5 & 5 \\
\hline & 1
\end{array}$

$\begin{array}{c|c}
2 & 72 \\
\hline 2 & 36 \\
\hline 2 & 18 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}$

36=2×2×3×3=2²×3²

45=3×3×5=3²×5¹

72=2×2×2×3=2³×3²

Here, 3² is the smallest power of the common factor 3.
So, HCF (36, 45, 72) = 3 × 3 = 9
2³ × 3² × 5¹ are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the given three numbers .

LCM of these three integers 

= 2 × 2 × 2 × 3 × 3 × 5 

= 360

---

Question 9 E 

Find LCM and HCF of the following integers by using prime factorization method:
42, 63 and 140
Sol :
Given numbers are 42, 63 and 140
Factorization of 42, 63 and 140

$\begin{array}{l|l}
2 & 42 \\
\hline 3 & 21 \\
\hline 7 & 7 \\
\hline & 1
\end{array}$


$\begin{array}{c|c}
3 & 63 \\
\hline 3 & 21 \\
\hline 7 & 7 \\
\hline & 1
\end{array}$


$\begin{array}{c|c}
2 & 140 \\
\hline 2 & 70 \\
\hline 5 & 35 \\
\hline 7 & 7 \\
\hline & 1
\end{array}$

42=2×3×7=2¹×3¹×7¹

63=3×3×7=3²×7¹

140=2×2×5×7=2²×5¹×7¹

Here, 7¹ is the smallest power of the common factor 7.
So, HCF (42, 63, 140) = 7
2² × 3² × 5¹ × 7¹ are the greatest powers of the prime factors 2, 3, 5 and 7 respectively involved in the given three numbers .
LCM of these three integers = 2 × 2 × 3 × 3 × 5 × 7 = 1260

---

Question 9 F 

Find LCM and HCF of the following integers by using prime factorization method:
48, 72 and 108
Sol :
Given numbers are 48, 72 and 108
Factorization of 48, 72 and 108

$\begin{array}{l|l}
2 & 48 \\
\hline 2 & 24 \\
\hline 2 & 12 \\
\hline 2 & 6 \\
\hline 3 & 3 \\
\hline & 1 \\
\end{array}$


$\begin{array}{c|c}
2 & 72 \\
\hline 2 & 36 \\
\hline 2 & 18 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}$


$\begin{array}{c|c}
2 & 108 \\
\hline 2 & 54 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}$

48=2×2×2×2×3=2⁴×3

72=2×2×2×3×3=2³×3²

108=2×2×3×3×3=2²×3³

Here, 2² × 3¹ are the smallest powers of the common factors 2 and 3 respectively.
So, HCF (48, 72, 108) = 2 × 2 × 3 = 12
2⁴ × 3³ are the greatest powers of the prime factors 2 and 3 respectively involved in the given three numbers .
LCM of these three integers = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432

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Question 10 A 

If HCF (96, 404) and 4, then, find LCM (96, 404)
Sol :
Given: HCF (96 , 404) = 4
To Find: LCM (96, 404)
We use the formula
L.C.M (a,b) × H.C.F (a,b) = Product of two numbers (a×b)
LCM (96, 404) × HCF (96, 404) = 96 × 404
⇒ LCM (96, 404) × 4 = 96 × 404 [∴HCF (96, 404) = 4]
⇒ LCM (96, 404) = $\frac{96 \times 404}{4}$
⇒ LCM (96, 404) = 9696

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Question 10 B 

If LCM (72, 126) = 504, find HCF (72, 126)
Sol :
Given: LCM (72, 126) = 504
To Find: HCF (72, 126)
We use the formula
L.C.M (a,b) × H.C.F (a,b) = Product of two numbers (a×b)
LCM (72, 126) × HCF (72, 126) = 72 × 126
⇒ 504 × HCF (72, 126) = 72 × 126 [∵LCM(72,126)=504]
⇒ HCF (72, 126) = $\frac{72 \times 126}{504}$
⇒ HCF (72 , 126) = 18

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Question 10 C 

If HCF (18, 504) = 18, find LCM (18, 504)
Sol :
Given: HCF (18, 504) = 18
To Find: LCM (18, 504)
We use the formula
L.C.M (a,b) × H.C.F (a,b) = Product of two numbers (a×b)
LCM (18, 504) × HCF (18, 504) = 18 × 504
⇒ LCM (18, 504) × 18 = 18 × 504 [∵HCF(18, 504) = 18]
⇒ LCM (18, 504) = $\frac{18 \times 504}{18}$
⇒ LCM (18, 504) = 504

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Question 10 D 

If LCM (96, 168) = 672, find HCF (96, 168)
Sol :
Given: LCM (96, 168) = 672
To Find: HCF (96, 168)
We use the formula
L.C.M (a,b) × H.C.F (a,b) = Product of two numbers (a×b)
LCM (96, 168) × HCF (96, 168) = 96 × 168
⇒ 672 × HCF (96, 168) = 96 × 168 [∵LCM(96, 168)=672]

⇒ HCF (96, 168) = $\frac{96 \times 168}{672}$
⇒ HCF (96, 168) = 24

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Question 10 E 

If HCF (306, 657) = 9, find LCM (306, 657)
Sol :
Given: HCF (306, 657) = 9
To Find: LCM (306, 657)
We use the formula
L.C.M (a,b) × H.C.F (a,b) = Product of two numbers (a×b)
LCM (306, 657) × HCF (306, 657) = 306 × 657
⇒ LCM (306, 657) × 9 = 306 × 657 [∵HCF(306,657)= 9]

⇒ LCM (306, 657) = $\frac{306 \times 657}{9}$
⇒ LCM (306, 657) = 22338

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Question 10 F 

If HCF (36, 64) = 4, find LCM (36, 64)
Sol :
Given: HCF (36, 64) = 4
To Find: LCM (36, 64)
We use the formula
L.C.M (a,b) × H.C.F (a,b) = Product of two numbers (a×b)
LCM (36, 64) × HCF (36, 64) = 36 × 64
⇒ LCM (36, 64) × 4 = 36 × 64 [∵HCF (36, 64)= 4]

⇒ LCM (36, 64) = $\frac{36 \times 64}{4}$
⇒ LCM (36, 64) = 576

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Question 11 A 

Examine whether (15)ⁿ can end with the digit 0 for any n ϵ N.

Sol :
If (15)ⁿ end with the digit 0, then the number should be divisible by 2 and 5.
As 2 × 5 = 10
⇒This means the prime factorization of 15ⁿ should contain prime factors 2 and 5.
But (15)ⁿ = (3 × 5)ⁿ and it does not have the prime factor 2 but have 3 and 5.
∵, 2 is not present in the prime factorization, there is no natural number nor which 15ⁿ ends with digit zero.
So, 15ⁿ cannot end with digit zero.

---

Question 11 B 

Examine whether (24)ⁿ can end with the digit 5 for any n ϵ N.

Sol :
If (24)ⁿ end with the digit 5, then the number should be divisible by 5.
⇒This means the prime factorization of 24ⁿ should contain prime factors 5.
But (24)ⁿ = (2³ × 3)ⁿ and it does not have the prime factor 5 but have 3 and 2.
∵, 5 is not present in the prime factorization, there is no natural number nor which 24ⁿ ends with digit 5.
So, 24ⁿ cannot end with digit 5.

---

Question 11 C 

Examine whether (21)ⁿ can end with the digit 0 for any n ϵ N.

Sol :
If (21)ⁿ end with the digit 0, then the number should be divisible by 2 and 5.
As 2 × 5 = 10
⇒This means the prime factorization of 21ⁿ should contain prime factors 2 and 5.
But (21)ⁿ = (3 × 7)ⁿ and it does not have the prime factor 2 and 5 but have 3 and 7.
∵, 2 and 5 is not present in the prime factorization, there is no natural number nor which 21ⁿ ends with digit zero.
So, 21ⁿ cannot end with digit zero.

---

Question 11 D 

Examine whether (8)ⁿ can end with the digit 5 for any n ϵ N.

Sol :
If (8)ⁿ end with the digit 5, then the number should be divisible by 5.
⇒This means the prime factorization of 8ⁿ should contain prime factor 5.
But (8)ⁿ = (2³)ⁿ and it does not have the prime factor 5 but have 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of 8ⁿ.
∵, 5 is not present in the prime factorization, there is no natural number nor which 8ⁿ ends with digit 5.
So, 8ⁿ cannot end with digit 5.

---

Question 11 E 

Examine whether (4)ⁿ can end with the digit 0 for any n ϵ N.

Sol :
If (4)ⁿ end with the digit 0, then the number should be divisible by 5.
As 2 × 5 = 10
⇒This means the prime factorization of 4ⁿ should contain prime factor 5.
This is not possible because (4)ⁿ = (2²ⁿ), so the only prime in the factorization of 4ⁿ is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of 4ⁿ.
∵, 5 is not present in the prime factorization, there is no natural number nor which 4ⁿ ends with digit zero.
So, 4ⁿ cannot end with digit zero.

---

Question 11 F 

Examine whether (7)ⁿ can end with the digit 5 for any n ϵ N.

Sol :
If (7)ⁿ end with the digit 5, then the number should be divisible by 5.
⇒This means the prime factorization of 7ⁿ should contain prime factor 5.
But (7)ⁿ does not have the prime factor 5. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of 7ⁿ.
∵, 5 is not present in the prime factorization, there is no natural number nor which 7ⁿ ends with digit 5.
So, 7ⁿ cannot end with digit 5.

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Question 12 A 

Explain why 7 x 11 x 13 x 17 +17 is a composite number.
Sol :
Composite number: A whole number that can be divided exactly by numbers other than 1 or itself.
Given 7 x 11 x 13 x 17 +17
⇒17 (7 x 11 x 13 x 17 +1)
⇒17 (7 x 11 x 13 x 17 +1)
⇒17 (17017 + 1)
⇒17 (17018)
⇒17 (2 × 8509)
⇒17 × 2 × 8509
So, given number is the composite number because it is the product of more than one prime numbers.

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Question 12 B 

Explain why 5 x 7 x 13 + 5 is a composite number.
Sol :
Composite number: A whole number that can be divided exactly by numbers other than 1 or itself.
5 x 7 x 13 + 5
⇒5 (1 x 7 x 13 +1)
⇒5 (91 +1)
⇒5 (92)
⇒5 (2² × 23)
⇒5 × 2 × 2 × 23
So, given number is the composite number because it is the product of more than one prime numbers.

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Question 12 C 

Show that 5 x 7 x 11 x 13 + 55 is a composite number.

Sol :
Composite number: A whole number that can be divided exactly by numbers other than 1 or itself.
5 x 7 x 11 x 13 + 55
⇒5 (1 x 7 × 11 x 13 +11)
⇒5 × 11 (91 +1)
⇒5 × 11 (92)
⇒5 × 11 (2² × 23)
⇒5 × 11 × 2 × 2 × 23
So, given number is the composite number because it is the product of more than one prime numbers.

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Question 13 

Three measuring rods 64 cm, 80 cm and 96 cm in length. Find the least length of cloth that can be measured exact number of times using anyone of the above rods.
Sol :
Lengths of three measuring rods = 64cm, 80cm and 96cm
Least Length of cloth that can be measured = LCM (64, 80, 96)

$\begin{array}{l|l}
2 & 64 \\
\hline 2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}$


$\begin{array}{l|l}
2 & 80 \\
\hline 2 & 40 \\
\hline 2 & 20 \\
\hline 3 & 10 \\
\hline 5 & 5 \\
\hline & 1 \\
\end{array}$


$\begin{array}{l|r}
2 & 96 \\
\hline 2 & 48 \\
\hline 2 & 24 \\
\hline 2 & 12 \\
\hline 2 & 6 \\
\hline 3 & 3 \\
\hline & 1
\end{array}$

64 = 2⁶
80 = 2³ × 3 × 5
96 = 2⁵ × 3
So, 2⁶ × 3 × 5 are the greatest powers of the prime factors 2, 3 and 5
LCM (64, 80, 96) = 2⁶ × 3 × 5 = 960
Least Length of cloth that can be measured is 960 cm

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Question 14 

Three containers contain 27 litres, 36 litres and 72 litres of milk. What biggest measure can measure exactly the milk in the three containers?
Sol :
Milk in three containers = 27L, 36L, 72L
Biggest measure which can exactly measure the milk = HCF (27, 36, 72)

$\begin{array}{l|l}
2 & 36 \\
\hline 2 & 18 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}$

$\begin{array}{l|l}
2 & 72 \\
\hline 2 & 36 \\
\hline 2 & 18 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}$

$\begin{array}{c|c}
3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}$

27 = 3³
36 = 2² × 3²
72 = 2³ × 3²
Here, 3² is the smallest power of the common factor of the prime 3
HCF (27, 36, 72) = 9
So, biggest measure which can exactly measure the milk = 9L

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Question 15 

Three different containers contain different quantities of mixtures of milk and water, whose measurements are 403 kg, 434 kg and 465 kg, what biggest measure can measure all the different quantities exactly.
Sol :
Mixtures of milk and water in three containers = 403kg, 434kg, 465kg
Biggest measure which can exactly measure different quantities = HCF (403, 434, 465)

$\begin{array}{l|l}
13 & 403 \\
\hline 31 & 31 \\
\hline & 1
\end{array}$

$\begin{array}{c|c}
2 & 434 \\
\hline 7 & 217 \\
\hline 31 & 31 \\
\hline & 1
\end{array}$

$\begin{array}{c|c}
3 & 465 \\
\hline 5 & 155 \\
\hline 31 & 31 \\
\hline & 1
\end{array}$

403=13×31

434=2×7×31

465=3×5×31

Here, 31¹ is the smallest power of the common factor.
HCF (403, 434, 465) = 31
So, biggest measure which can exactly measure the milk = 31L

---

S.no	Chapters	Links
1	Real numbers	Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4
2	Polynomials	Exercise 2.1 Exercise 2.2 Exercise 2.3
3	Pairs of Linear Equations in Two Variables	Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5
4	Trigonometric Ratios and Identities	Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4
5	Triangles	Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5
6	Statistics	Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4
7	Quadratic Equations	Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5
8	Arithmetic Progressions (AP)	Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4
9	Some Applications of Trigonometry: Height and Distances	Exercise 9.1
10	Coordinates Geometry	Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4
11	Circles	Exercise 11.1 Exercise 11.2
12	Constructions	Exercise 12.1
13	Area related to Circles	Exercise 13.1
14	Surface Area and Volumes	Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4
15	Probability	Exercise 15.1