{"id":623894,"date":"2023-08-31T15:05:34","date_gmt":"2023-08-31T15:05:34","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=623894"},"modified":"2023-09-01T02:49:00","modified_gmt":"2023-09-01T02:49:00","slug":"kc-sinha-exercise-1-2-mathematics-solution-class-10-chapter-1-real-numbers","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-1-2-mathematics-solution-class-10-chapter-1-real-numbers\/","title":{"rendered":"KC Sinha: Exercise 1.2 – Mathematics Solution Class 10 Chapter 1 Real numbers"},"content":{"rendered":"\n\n\n\n\n

Question 1 A <\/h4>\n\n\n\n

Express each of the following numbers as a product of its prime factors:<\/strong>
Sol :
Given number is 4320
Factorization of 4320 is<\/p>\n\n\n\n

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

Hence, 4320 = 25<\/sup> \u00d7 33<\/sup> \u00d7 5 (Product of its prime factors)<\/p>\n\n\n\n

\n\n\n\n

Question 1 B <\/h4>\n\n\n\n

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

Hence, 7560 = 23<\/sup>\u00d733<\/sup>\u00d75\u00d77 (Product of its prime factors)<\/p>\n\n\n\n

\n\n\n\n

Question 1 C <\/h4>\n\n\n\n

Express each of the following numbers as a product of its prime factors: 140<\/strong>
Sol :
Given number is 140
Factorization of 140 is<\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 140 \\\\
\\hline 2 & 70 \\\\
\\hline 5 & 35 \\\\
\\hline 7 & 7 \\\\
\\hline & 1
\\end{array}$
Hence, 140 = 22<\/sup> \u00d7 5 \u00d7 7 (Product of its prime factors)<\/p>\n\n\n\n

\n\n\n\n

Question 1 D <\/h4>\n\n\n\n

Express each of the following numbers as a product of its prime factors: 5005<\/strong>
Sol :
Given number is 5005
Factorization of 5005 is<\/p>\n\n\n\n

$\\begin{array}{l|l}
5 & 5005 \\\\
\\hline 7 & 1001 \\\\
\\hline 11 & 143 \\\\
\\hline 13 & 13 \\\\
\\hline & 1
\\end{array}$
Hence, 5005 = 5 \u00d7 7 \u00d7 11 \u00d7 13 (Product of its prime factors)<\/p>\n\n\n\n

\n\n\n\n

Question 1 E <\/h4>\n\n\n\n

Express each of the following numbers as a product of its prime factors: 32760<\/strong>
Sol :
Given number is 32760
Factorization of 32760 is<\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 32760 \\\\
\\hline 2 & 16380 \\\\
\\hline 2 & 8190 \\\\
\\hline 3 & 4095 \\\\
\\hline 3 & 1365 \\\\
\\hline 5 & 455 \\\\
\\hline 7 & 91 \\\\
\\hline 13 & 13 \\\\
\\hline & 1
\\end{array}$
Hence, 32760 = 23<\/sup> \u00d7 32<\/sup> \u00d7 5 \u00d7 7 \u00d7 13 (Product of its prime factors)<\/p>\n\n\n\n

\n\n\n\n

Question 1 F <\/h4>\n\n\n\n

Express each of the following numbers as a product of its prime factors: 156<\/strong>
Sol :
Given number is 156
Factorization of 156 is<\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 156 \\\\
\\hline 2 & 78 \\\\
\\hline 3 & 39 \\\\
\\hline 13 & 13 \\\\
\\hline & 1
\\end{array}$
Hence, 156 = 22<\/sup> \u00d7 3 \u00d7 13 (Product of its prime factors)<\/p>\n\n\n\n

\n\n\n\n

Question 1 G <\/h4>\n\n\n\n

Express each of the following numbers as a product of its prime factors: 729<\/strong>
Sol :
Given number is 729
Factorization of 729 is<\/p>\n\n\n\n

$\\begin{array}{l|l}
3 & 729 \\\\
\\hline 3 & 243 \\\\
\\hline 3 & 81 \\\\
\\hline 3 & 27 \\\\
\\hline 3 & 9 \\\\
\\hline 3 & 3 \\\\
\\hline & 1
\\end{array}$
Hence, 729 = 3 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3 = 36<\/sup> (Product of its prime factors)<\/p>\n\n\n\n

\n\n\n\n

Question 2 <\/h4>\n\n\n\n

Find the highest power of 5 in 23750.<\/strong>
Sol :
To find the highest power of 5 in 23750, we have to factorize 23570<\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 23750 \\\\
\\hline 5 & 11875 \\\\
\\hline 5 & 2375 \\\\
\\hline 5 & 475 \\\\
\\hline 5 & 95 \\\\
\\hline 19 & 19 \\\\
\\hline & 1
\\end{array}$
Hence, 23750 = 2 \u00d7 54<\/sup> \u00d7 19
So, the highest power of 5 in 23750 is 4.<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 3 <\/h4>\n\n\n\n

Find the highest power of 2 in 1440.<\/strong>
Sol :
Given number is 1440
Factorization of 1440 is<\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 1440 \\\\
\\hline 2 & 720 \\\\
\\hline 2 & 360 \\\\
\\hline 2 & 180 \\\\
\\hline 2 & 90 \\\\
\\hline 3 & 45 \\\\
\\hline 3 & 15 \\\\
\\hline 5 & 5 \\\\
\\hline & 1
\\end{array}$
Factors of 1440 = 25<\/sup> \u00d7 32<\/sup> \u00d7 5
So, the highest power of 2 is 5.<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 4 <\/h4>\n\n\n\n

If 6370 = 2m<\/sup> .5n<\/sup>.7k<\/sup>.13p<\/sup>, then find m+n+k+p.<\/strong><\/p>\n\n\n\n

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

$\\begin{array}{l|l}
2 & 6370 \\\\
\\hline 5 & 3185 \\\\
\\hline 7 & 637 \\\\
\\hline 7 & 91 \\\\
\\hline 13 & 13 \\\\
\\hline & 1
\\end{array}$
6370 = 2 \u00d7 5 \u00d7 72<\/sup> \u00d7 13
On Comparing, we get<\/p>\n\n\n\n

6370 =2m<\/sup> .5n<\/sup>.7k<\/sup>.13p<\/sup><\/p>\n\n\n\n

=21<\/sup>\u00d751<\/sup>\u00d772<\/sup>\u00d7131<\/sup><\/p>\n\n\n\n

m = 1
n = 1
k = 2
p = 1
So, m+n+k+p=5<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 5 A <\/h4>\n\n\n\n

Which of the following is a pair of co-primes?<\/strong>
(32,62)<\/strong>
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<\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 32 \\\\
\\hline 2 & 16 \\\\
\\hline 2 & 8 \\\\
\\hline 2 & 4 \\\\
\\hline 2 & 2 \\\\
\\hline & 1
\\end{array}$
Factors of 32=2\u00d72\u00d72\u00d72\u00d72=25<\/sup><\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 62 \\\\
\\hline 31 & 31 \\\\
\\hline & 1
\\end{array}$
Factors of 62=2\u00d731<\/p>\n\n\n\n

Here, we can see that 2 is the common factor. So, (32,62) is not co-prime.<\/p>\n\n\n\n

\n\n\n\n

Question 5 B <\/h4>\n\n\n\n

Which of the following is a pair of co-primes?<\/strong>
(18,25)<\/strong>
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<\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 18 \\\\
\\hline 3 & 9 \\\\
\\hline 3 & 3 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

Factors of 32 = 2 \u00d7 3 \u00d7 3
$\\begin{array}{l|l}5 & 25 \\\\\\hline 5 & 5 \\\\\\hline & 1\\end{array}$
Factors of 62 = 5 \u00d7 5
Therefore, there is no common factor. So, (18,25) is co-prime.<\/p>\n\n\n\n

\n\n\n\n

Question 5 C <\/h4>\n\n\n\n

Which of the following is a pair of co-primes?<\/strong>
(31, 93)<\/strong>
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<\/p>\n\n\n\n

$\\begin{array}{c|l}
3 & 93 \\\\
\\hline 31 & 31 \\\\
\\hline 2 & 1
\\end{array}$<\/p>\n\n\n\n

$\\begin{array}{l|l}
31 & 31 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

Factors of 31=31\u00d71
Here, we can see that 31 is the common factor. So, (31, 93) is not co-prime.<\/p>\n\n\n\n

\n\n\n\n

Question 6 A <\/h4>\n\n\n\n

Write down the missing numbers a, b, c, d, x, y in the following factor tree :<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Here, a=2520; b=2; c=315; d=3; x=3; y=5<\/p>\n\n\n\n

\n\n\n\n

Question 6 B <\/h4>\n\n\n\n

Write down the missing numbers a, b, c, d, x, y in the following factor tree :<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Here, a=15015; b=5005; c=5; d=143; x=13<\/p>\n\n\n\n

\n\n\n\n

Question 6 C <\/h4>\n\n\n\n

Write down the missing numbers a, b, c, d, x, y in the following factor tree :<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Here, a=18380; b=2; c=1365; d=3; x=5; y=13<\/p>\n\n\n\n

\n\n\n\n

Question 6 D <\/h4>\n\n\n\n

Write down the missing numbers a,b,c,d,x, y in the following factor tree :<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Here, a=3; b=147407; c=11339; d=667; x=29<\/p>\n\n\n\n

\n\n\n\n

Question 7 A <\/h4>\n\n\n\n

Find the LCM and HCF of the following integers by applying the prime factorization method :<\/strong>
96 and 404<\/strong>
Sol :
Given numbers are 96 and 404.
The prime factorization of 96 and 404 gives:<\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 96 \\\\
\\hline 2 & 48 \\\\
\\hline 2 & 24 \\\\
\\hline 2 & 12 \\\\
\\hline 2 & 6 \\\\
\\hline 3 & 3 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

96=25<\/sup>\u00d73<\/p>\n\n\n\n

=2\u00d72\u00d72\u00d72\u00d72\u00d73<\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 404 \\\\
\\hline 2 & 202 \\\\
\\hline 101 & 101 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

404=22<\/sup>\u00d7101<\/p>\n\n\n\n

=2\u00d72\u00d7101<\/p>\n\n\n\n

Here, 22<\/sup> is the smallest power of the common factor 2.
Therefore, the H.C.F<\/strong> of these two integers is 2\u00d72 = 4<\/strong>
25<\/sup> \u00d7 31<\/sup> \u00d7 1011<\/sup> are the greatest powers of the prime factors 2, 3 and 101 respectively involved in the given numbers.
Now, L.C.M<\/strong> of 96 and 404 is 2\u00d72\u00d72\u00d72\u00d72\u00d73\u00d7101<\/p>\n\n\n\n

= 9696<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 7 B <\/h4>\n\n\n\n

Find the LCM and HCF of the following integers by applying the prime factorization method :<\/strong>
6 and 20<\/strong>
Sol :
Given numbers are 6 and 20
The prime factorization of 6 and 20 gives:<\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 6 \\\\
\\hline 3 & 3 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

6=2\u00d73<\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 20 \\\\
\\hline 2 & 10 \\\\
\\hline 5 & 5 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

20=2\u00d72\u00d75<\/p>\n\n\n\n

Here, 21<\/sup> is the smallest power of the common factor 2.
Therefore, the H.C.F<\/strong> of these two integers = 2<\/strong>
22<\/sup> \u00d7 31<\/sup> \u00d7 51<\/sup> are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the given numbers.
Now, L.C.M<\/strong> of 6 and 20=2\u00d72\u00d73\u00d75<\/p>\n\n\n\n

= 60<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 7 C <\/h4>\n\n\n\n

Find the LCM and HCF of the following integers by applying the prime factorization method :<\/strong>
26 and 91<\/strong>
Sol :
Given numbers are 26 and 91.
The prime factorization of 26 and 91 gives:<\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 26 \\\\
\\hline 13 & 13 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

$\\begin{array}{l|l}
7 & 91 \\\\
\\hline 13 & 13 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

91=7\u00d713<\/p>\n\n\n\n

Here, 131<\/sup> is the smallest power of the common factor 13.
Therefore, the H.C.F<\/strong> of these two integers = 13<\/strong>
21<\/sup> \u00d7 71<\/sup> \u00d7 131<\/sup> are the greatest powers of the prime factors 2, 7 and 13 respectively involved in the given numbers.
Now, L.C.M<\/strong> of 6 and 21 is 2\u00d77\u00d713<\/p>\n\n\n\n

= 182<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 7 D <\/h4>\n\n\n\n

Find the LCM and HCF of the following integers by applying the prime factorization method :<\/strong>
87 and 145<\/strong>
Sol :
Given numbers are 87 and 145.
The prime factorization of 87 and 145 gives:<\/p>\n\n\n\n

$\\begin{array}{c|l}
3 & 87 \\\\
\\hline 29 & 29 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

$\\begin{array}{c|l}
5 & 145 \\\\
\\hline 29 & 29 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

145=5\u00d729<\/p>\n\n\n\n

Here, 291<\/sup> is the smallest power of the common factor 29.
Therefore, the H.C.F<\/strong> of these two integers = 29<\/strong>
31<\/sup> \u00d7 51<\/sup> \u00d7 291<\/sup> are the greatest powers of the prime factors 3, 5 and 29 respectively involved in the given numbers.
Now, L.C.M<\/strong> of 87 and 145=3\u00d75\u00d729 <\/p>\n\n\n\n

= 435<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 7 E <\/h4>\n\n\n\n

Find the LCM and HCF of the following integers by applying the prime factorization method :<\/strong>
1485 and 4356<\/strong>
Sol :
Given numbers are 1485 and 4356.
The prime factorization of 1485 and 4356 gives:<\/p>\n\n\n\n

$\\begin{array}{l|l}
3 & 1485 \\\\
\\hline 3 & 495 \\\\
\\hline 3 & 165 \\\\
\\hline 5 & 55 \\\\
\\hline 11 & 11 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

1485=3\u00d73\u00d73\u00d75\u00d711<\/p>\n\n\n\n

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

4356=2\u00d72\u00d73\u00d73\u00d711\u00d711<\/p>\n\n\n\n

Here, 32<\/sup>\u00d7 11 is the smallest power of the common factors 3 and 11.
Therefore, the H.C.F<\/strong> of these two integers = 3 \u00d7 3 \u00d7 11 = 99<\/strong>
22<\/sup> \u00d7 33<\/sup>\u00d7 51<\/sup> \u00d7 112<\/sup> are the greatest powers of the prime factors 2, 3 and 7 respectively involved in the given numbers.
Now, L.C.M<\/strong> of 1485 and 4356 =2\u00d72\u00d73\u00d73\u00d73\u00d75\u00d711\u00d711 <\/p>\n\n\n\n

= 65430<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 7 F <\/h4>\n\n\n\n

Find the LCM and HCF of the following integers by applying the prime factorization method :<\/strong>
1095 and 1168<\/strong>
Sol :
Given numbers are 1095 and 1168.
The prime factorization of 1095 and 1168 gives:<\/p>\n\n\n\n

$\\begin{array}{c|l}
3 & 1095 \\\\
\\hline 5 & 365 \\\\
\\hline 73 & 73 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

1485=3\u00d75\u00d773<\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 1168 \\\\
\\hline 2 & 584 \\\\
\\hline 2 & 292 \\\\
\\hline 2 & 146 \\\\
\\hline 73 & 73 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

4356=2\u00d72\u00d72\u00d72\u00d773<\/p>\n\n\n\n

Here, 731<\/sup> is the smallest power of the common factor 73.
Therefore, the H.C.F<\/strong> of these two integers = 73<\/strong>
24<\/sup> \u00d7 31<\/sup>\u00d7 51<\/sup> \u00d7 731<\/sup> are the greatest powers of the prime factors 2, 3, 5 and 73 respectively involved in the given numbers.
Now, L.C.M<\/strong> of 1485 and 4356=2\u00d72\u00d72\u00d72\u00d73\u00d75\u00d773<\/p>\n\n\n\n

= 17520<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 7 G <\/h4>\n\n\n\n

Find the LCM and HCF of the following integers by applying the prime factorization method :<\/strong>
6 and 21<\/strong>
Sol :
Given numbers are 6 and 21.
The prime factorization of 6 and 21 gives:<\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 6 \\\\
\\hline 3 & 3 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

6=2\u00d73<\/p>\n\n\n\n

$\\begin{array}{l|l}
3 & 21 \\\\
\\hline 7 & 7 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

21=3\u00d77<\/p>\n\n\n\n

Here, 31<\/sup> is the smallest power of the common factor 3.
Therefore, the H.C.F<\/strong> of these two integers = 3<\/strong>
21<\/sup> \u00d7 31<\/sup> \u00d7 71<\/sup> are the greatest powers of the prime factors 2, 3 and 7 respectively involved in the given numbers.
Now, L.C.M<\/strong> of 6 and 21 is 2\u00d73\u00d77<\/p>\n\n\n\n

= 42<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 8 A <\/h4>\n\n\n\n

Find the LCM and HCF of the following pair of integers and verify that LCM X HCF = Product of two numbers :<\/strong>
96 and 404<\/strong>
Sol :
Given numbers are 96 and 404
The prime factorization of 96 and 404 gives:
96 = 25<\/sup>\u00d73 and 404 =22<\/sup>\u00d7101
Therefore, the H.C.F of these two integers = 22<\/sup> = 4<\/strong>
Now, the L.C.M of 96 and 404 =2\u00d72\u00d72\u00d72\u00d72\u00d73\u00d7101<\/p>\n\n\n\n

= 9696<\/strong><\/p>\n\n\n\n

Now, we have to verify<\/p>\n\n\n\n

L.C.M(a,b)\u00d7H.C.F(a,b)=Product of two numbers(a\u00d7b)<\/p>\n\n\n\n

L.H.S = L.C.M \u00d7 H.C.F = 9696 \u00d7 4 = 38784
R.H.S = Product of two numbers = 96 \u00d7 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.<\/p>\n\n\n\n

\n\n\n\n

Question 8 B <\/h4>\n\n\n\n

Find the LCM and HCF of the following pair of integers and verify that LCM X HCF = Product of two numbers :<\/strong>
852 and 1491<\/strong>
Sol :
Given numbers are 852 and 1491
The prime factorization of 852 and 1491 gives:
852 = 2 \u00d7 2 \u00d7 3 \u00d7 71 and 1491 = 3 \u00d7 7 \u00d7 71
Therefore, the H.C.F of these two integers = 3 \u00d7 71 = 213<\/strong>
Now, the L.C.M of 96 and 404 = 2 \u00d7 2 \u00d7 3 \u00d7 7 \u00d7 71 = 5964<\/strong>
Now, we have to verify<\/p>\n\n\n\n

L.C.M(a,b)\u00d7H.C.F(a,b)=Product of two numbers(a\u00d7b)<\/p>\n\n\n\n

L.H.S = L.C.M \u00d7 H.C.F <\/p>\n\n\n\n

= 5964 \u00d7 213<\/p>\n\n\n\n

 = 1270332<\/p>\n\n\n\n

R.H.S = Product of two numbers <\/p>\n\n\n\n

= 852 \u00d7 1491 <\/p>\n\n\n\n

= 1270332<\/p>\n\n\n\n

Hence, L.H.S = R.H.S
So, the product of two numbers is equal to the product of their HCF and LCM.<\/p>\n\n\n\n

\n\n\n\n

Question 8 C <\/h4>\n\n\n\n

Find the LCM and HCF of the following pair of integers and verify that LCM X HCF = Product of two numbers :<\/strong>
777 and 1147<\/strong>
Sol :
Given numbers are 777 and 1147
The prime factorization of 777 and 1147 gives:
777 = 3 \u00d7 7 \u00d7 37 and 1147 = 31 \u00d7 37
Therefore, the H.C.F of these two integers = 37<\/strong>
Now, the L.C.M of 96 and 404 = 3 \u00d7 7 \u00d7 31 \u00d7 37<\/p>\n\n\n\n

 = 24087<\/strong><\/p>\n\n\n\n

Now, we have to verify<\/p>\n\n\n\n

L.C.M(a,b)\u00d7H.C.F(a,b)=Product of two numbers(a\u00d7b)<\/p>\n\n\n\n

L.H.S = L.C.M \u00d7 H.C.F<\/p>\n\n\n\n

 = 24087 \u00d7 37 = 891219
R.H.S = Product of two numbers <\/p>\n\n\n\n

= 777 \u00d7 1147<\/p>\n\n\n\n

 = 891219<\/p>\n\n\n\n

Hence, L.H.S = R.H.S
So, the product of two numbers is equal to the product of their HCF and LCM.<\/p>\n\n\n\n

\n\n\n\n

Question 8 D <\/h4>\n\n\n\n

Find the LCM and HCF of the following pair of integers and verify that LCM <\/strong>\u00d7 HCF = Product of two numbers :<\/strong>
36 and 64<\/strong>
Sol :
Given numbers are 36 and 64
The prime factorization of 36 and 64 gives:
36 = 2 \u00d7 2 \u00d7 3 \u00d7 3 and 64 = 26<\/sup>
Therefore, the H.C.F of these two integers = 2 \u00d7 2 = 4<\/strong>
Now, the L.C.M of 36 and 64 = 3 \u00d7 3 \u00d7 26<\/sup> = 576<\/strong>
Now, we have to verify<\/p>\n\n\n\n

L.C.M(a,b)\u00d7H.C.F(a,b)=Product of two numbers(a\u00d7b)<\/p>\n\n\n\n

L.H.S = L.C.M \u00d7 H.C.F = 576 \u00d7 4 = 2304
R.H.S = Product of two numbers = 36 \u00d7 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.<\/p>\n\n\n\n

\n\n\n\n

Question 8 E <\/h4>\n\n\n\n

Find the LCM and HCF of the following pair of integers and verify that LCM X HCF = Product of two numbers :<\/strong>
32 and 80<\/strong>
Sol :
Given numbers are 32 and 80
The prime factorization of 32 and 80 gives:
32 = 25<\/sup> and 80 = 24<\/sup> \u00d7 5
Therefore, the H.C.F of these two integers = 24<\/sup> = 16<\/strong>
Now, the L.C.M of 32 and 80 <\/p>\n\n\n\n

= 5 \u00d7 25<\/sup> <\/p>\n\n\n\n

= 160<\/strong>
<\/p>\n\n\n\n

Now, we have to verify<\/p>\n\n\n\n

L.C.M(a,b)\u00d7H.C.F(a,b)=Product of two numbers(a\u00d7b)<\/p>\n\n\n\n

L.H.S = L.C.M \u00d7 H.C.F<\/p>\n\n\n\n

 = 160 \u00d7 16 <\/p>\n\n\n\n

= 2560<\/p>\n\n\n\n

R.H.S = Product of two numbers<\/p>\n\n\n\n

 = 32 \u00d7 80 <\/p>\n\n\n\n

= 2560<\/p>\n\n\n\n

Hence, L.H.S = R.H.S
So, the product of two numbers is equal to the product of their HCF and LCM.<\/p>\n\n\n\n

\n\n\n\n

Question 8 F <\/h4>\n\n\n\n

Find the LCM and HCF of the following pair of integers and verify that LCM X HCF = Product of two numbers :<\/strong>
902 and 1517<\/strong>
Sol :
Given numbers are 902 and 1517
The prime factorization of 902 and 1517 gives:
902 = 2 \u00d7 11 \u00d7 41 and 1517 = 37 \u00d7 41
Therefore, the H.C.F of these two integers = 41<\/strong>
<\/p>\n\n\n\n

Now, the L.C.M of 902 and 1517 = 2 \u00d7
11 \u00d7 37 \u00d7 41 = 33374<\/strong>
Now, we have to verify<\/p>\n\n\n\n

L.C.M(a,b)\u00d7H.C.F(a,b)=Product of two numbers(a\u00d7b)<\/p>\n\n\n\n

L.H.S = L.C.M \u00d7 H.C.F <\/p>\n\n\n\n

= 33374 \u00d7 41<\/p>\n\n\n\n

 = 1368334<\/p>\n\n\n\n

R.H.S = Product of two numbers <\/p>\n\n\n\n

= 902 \u00d7 1517 <\/p>\n\n\n\n

= 1368334
Hence, L.H.S = R.H.S
So, the product of two numbers is equal to the product of their HCF and LCM.<\/p>\n\n\n\n

\n\n\n\n

Question 9 A <\/h4>\n\n\n\n

Find LCM and HCF of the following integers by using prime factorization method:<\/strong>
6, 72 and 120<\/strong>
Sol :<\/p>\n\n\n\n

Given numbers are 6, 72 and 120
Factorization of 6, 72 and 120<\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 6 \\\\
\\hline 3 & 3 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

$\\begin{array}{c|c}
2 & 72 \\\\
\\hline 2 & 36 \\\\
\\hline 2 & 18 \\\\
\\hline 3 & 9 \\\\
\\hline 3 & 3 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

$\\begin{array}{c|c}
2 & 120 \\\\
\\hline 2 & 60 \\\\
\\hline 2 & 30 \\\\
\\hline 3 & 15 \\\\
\\hline 5 & 5 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

6=2\u00d73=21<\/sup>\u00d731<\/sup><\/p>\n\n\n\n

72=2\u00d72\u00d72\u00d73\u00d73=23<\/sup>\u00d732<\/sup><\/p>\n\n\n\n

120=2\u00d72\u00d72\u00d73\u00d75=23<\/sup>\u00d731<\/sup>\u00d751<\/sup><\/p>\n\n\n\n

Here, 21<\/sup> \u00d7 31<\/sup> are the smallest powers of the common factors 2 and 3, respectively.
So, HCF (6, 72, 120)<\/strong> = 2 \u00d7 3 = 6<\/strong><\/p>\n\n\n\n

23<\/sup>\u00d732<\/sup>\u00d751<\/sup> are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the given three numbers .
LCM<\/strong> of these three integers =2\u00d72\u00d72\u00d73\u00d73\u00d75 <\/p>\n\n\n\n

= 360<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 9 B <\/h4>\n\n\n\n

Find LCM and HCF of the following integers by using prime factorization method:<\/strong>
8, 9, and 25<\/strong>
Sol :
Given numbers are 8, 9 and 25
Factorization of 8, 9 and 25<\/p>\n\n\n\n

$\\begin{array}{c|c}
2 & 8 \\\\
\\hline 2 & 4 \\\\
\\hline 2 & 2 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

$\\begin{array}{c|c}
3 & 9 \\\\
\\hline 3 & 3 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

$\\begin{array}{c|c}
5 & 25 \\\\
\\hline 5 & 5 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

8=2\u00d72\u00d72\u00d71=23<\/sup>\u00d71
9=3\u00d73\u00d71=32<\/sup>\u00d71
25=5\u00d75\u00d71=52<\/sup>\u00d71
<\/p>\n\n\n\n

Here, 11<\/sup> is the smallest power of the common factor 1.
So, HCF (8, 9, 25) = 1<\/strong>
23<\/sup> \u00d7 32<\/sup> \u00d7 52<\/sup> are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the given three numbers .
LCM<\/strong> of these three integers = 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 5 \u00d7 5 = 1800<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 9 C <\/h4>\n\n\n\n

Find LCM and HCF of the following integers by using prime factorization method:<\/strong>
12, 15, and 21<\/strong>
Sol :
Given numbers are 12, 15 and 21
Factorization of 12, 15 and 21<\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 12 \\\\
\\hline 2 & 6 \\\\
\\hline 3 & 3 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

$\\begin{array}{c|c}
3 & 15 \\\\
\\hline 5 & 5 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

$\\begin{array}{c|c}
3 & 21 \\\\
\\hline 7 & 7 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

12=2\u00d72\u00d73=22<\/sup>\u00d731<\/sup><\/p>\n\n\n\n

15=3\u00d75=31<\/sup>\u00d751<\/sup><\/p>\n\n\n\n

21=3\u00d77=31<\/sup>\u00d771<\/sup><\/p>\n\n\n\n

Here, 31<\/sup> is the smallest power of the common factor 3.
<\/p>\n\n\n\n

So, HCF (12, 15, 21) = 3<\/strong>
22<\/sup> \u00d7 31<\/sup> \u00d7 51<\/sup> \u00d7 71<\/sup> are the greatest powers of the prime factors 2, 3, 5 and 7 respectively involved in the given three numbers .
<\/strong><\/p>\n\n\n\n

LCM<\/strong> of these three integers<\/p>\n\n\n\n

=2\u00d72\u00d73\u00d75\u00d77<\/p>\n\n\n\n

= 420<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 9 D <\/h4>\n\n\n\n

Find LCM and HCF of the following integers by using prime factorization method:<\/strong>
36, 45, and 72<\/strong>
Sol :
Given numbers are 36, 45 and 72
Factorization of 36, 45 and 72<\/p>\n\n\n\n

$\\begin{array}{c|c}
2 & 36 \\\\
\\hline 2 & 18 \\\\
\\hline 3 & 9 \\\\
\\hline 3 & 3 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

$\\begin{array}{c|c}
3 & 45 \\\\
\\hline 3 & 15 \\\\
\\hline 5 & 5 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

$\\begin{array}{c|c}
2 & 72 \\\\
\\hline 2 & 36 \\\\
\\hline 2 & 18 \\\\
\\hline 3 & 9 \\\\
\\hline 3 & 3 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

36=2\u00d72\u00d73\u00d73=22<\/sup>\u00d732<\/sup><\/p>\n\n\n\n

45=3\u00d73\u00d75=32<\/sup>\u00d751<\/sup><\/p>\n\n\n\n

72=2\u00d72\u00d72\u00d73=23<\/sup>\u00d732<\/sup><\/p>\n\n\n\n

Here, 32<\/sup> is the smallest power of the common factor 3.
So, HCF (36, 45, 72)<\/strong> = 3 \u00d7 3 = 9<\/strong>
23<\/sup> \u00d7 32<\/sup> \u00d7 51<\/sup> are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the given three numbers .<\/p>\n\n\n\n

LCM<\/strong> of these three integers <\/p>\n\n\n\n

= 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 5 <\/p>\n\n\n\n

= 360<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 9 E <\/h4>\n\n\n\n

Find LCM and HCF of the following integers by using prime factorization method:<\/strong>
42, 63 and 140<\/strong>
Sol :
Given numbers are 42, 63 and 140
Factorization of 42, 63 and 140<\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 42 \\\\
\\hline 3 & 21 \\\\
\\hline 7 & 7 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

$\\begin{array}{c|c}
3 & 63 \\\\
\\hline 3 & 21 \\\\
\\hline 7 & 7 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

$\\begin{array}{c|c}
2 & 140 \\\\
\\hline 2 & 70 \\\\
\\hline 5 & 35 \\\\
\\hline 7 & 7 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

42=2\u00d73\u00d77=21<\/sup>\u00d731<\/sup>\u00d771<\/sup><\/p>\n\n\n\n

63=3\u00d73\u00d77=32<\/sup>\u00d771<\/sup><\/p>\n\n\n\n

140=2\u00d72\u00d75\u00d77=22<\/sup>\u00d751<\/sup>\u00d771<\/sup><\/p>\n\n\n\n

Here, 71<\/sup> is the smallest power of the common factor 7.
So, HCF (42, 63, 140) = 7<\/strong>
22<\/sup> \u00d7 32<\/sup> \u00d7 51<\/sup> \u00d7 71<\/sup> are the greatest powers of the prime factors 2, 3, 5 and 7 respectively involved in the given three numbers .
LCM<\/strong> of these three integers = 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 5 \u00d7 7 = 1260<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 9 F <\/h4>\n\n\n\n

Find LCM and HCF of the following integers by using prime factorization method:<\/strong>
48, 72 and 108<\/strong>
Sol :
Given numbers are 48, 72 and 108
Factorization of 48, 72 and 108<\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 48 \\\\
\\hline 2 & 24 \\\\
\\hline 2 & 12 \\\\
\\hline 2 & 6 \\\\
\\hline 3 & 3 \\\\
\\hline & 1 \\\\
\\end{array}$<\/p>\n\n\n\n

$\\begin{array}{c|c}
2 & 72 \\\\
\\hline 2 & 36 \\\\
\\hline 2 & 18 \\\\
\\hline 3 & 9 \\\\
\\hline 3 & 3 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

$\\begin{array}{c|c}
2 & 108 \\\\
\\hline 2 & 54 \\\\
\\hline 3 & 27 \\\\
\\hline 3 & 9 \\\\
\\hline 3 & 3 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

48=2\u00d72\u00d72\u00d72\u00d73=24<\/sup>\u00d73<\/p>\n\n\n\n

72=2\u00d72\u00d72\u00d73\u00d73=23<\/sup>\u00d732<\/sup><\/p>\n\n\n\n

108=2\u00d72\u00d73\u00d73\u00d73=22<\/sup>\u00d733<\/sup><\/p>\n\n\n\n

Here, 22<\/sup> \u00d7 31<\/sup> are the smallest powers of the common factors 2 and 3 respectively.
So, HCF (48, 72, 108)<\/strong> = 2 \u00d7 2 \u00d7 3 = 12<\/strong>
24<\/sup> \u00d7 33<\/sup> are the greatest powers of the prime factors 2 and 3 respectively involved in the given three numbers .
LCM<\/strong> of these three integers = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 3 = 432<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 10 A <\/h4>\n\n\n\n

If HCF (96, 404) and 4, then, find LCM (96, 404)<\/strong>
Sol :
Given: HCF (96 , 404) = 4
To Find: LCM (96, 404)
We use the formula
L.C.M (a,b) \u00d7 H.C.F (a,b) = Product of two numbers (a\u00d7b)<\/strong>
LCM (96, 404) \u00d7 HCF (96, 404) = 96 \u00d7 404
\u21d2 LCM (96, 404) \u00d7 4 = 96 \u00d7 404 [\u2234HCF (96, 404) = 4]
\u21d2 LCM (96, 404) = $\\frac{96 \\times 404}{4}$
\u21d2 LCM (96, 404) = 9696<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 10 B <\/h4>\n\n\n\n

If LCM (72, 126) = 504, find HCF (72, 126)<\/strong>
Sol :
Given: LCM (72, 126) = 504
To Find: HCF (72, 126)
We use the formula
L.C.M (a,b) \u00d7 H.C.F (a,b) = Product of two numbers (a\u00d7b)<\/strong>
LCM (72, 126) \u00d7 HCF (72, 126) = 72 \u00d7 126
\u21d2 504 \u00d7 HCF (72, 126) = 72 \u00d7 126 [\u2235LCM(72,126)=504]
\u21d2 HCF (72, 126) = $\\frac{72 \\times 126}{504}$
\u21d2 HCF (72 , 126) = 18<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 10 C <\/h4>\n\n\n\n

If HCF (18, 504) = 18, find LCM (18, 504)<\/strong>
Sol :
Given: HCF (18, 504) = 18
To Find: LCM (18, 504)
We use the formula
L.C.M (a,b) \u00d7 H.C.F (a,b) = Product of two numbers (a\u00d7b)<\/strong>
LCM (18, 504) \u00d7 HCF (18, 504) = 18 \u00d7 504
\u21d2 LCM (18, 504) \u00d7 18 = 18 \u00d7 504 [\u2235HCF(18, 504) = 18]
\u21d2 LCM (18, 504) = $\\frac{18 \\times 504}{18}$
\u21d2 LCM (18, 504) = 504<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 10 D <\/h4>\n\n\n\n

If LCM (96, 168) = 672, find HCF (96, 168)<\/strong>
Sol :
Given: LCM (96, 168) = 672
To Find: HCF (96, 168)
We use the formula
L.C.M (a,b) \u00d7 H.C.F (a,b) = Product of two numbers (a\u00d7b)<\/strong>
LCM (96, 168) \u00d7 HCF (96, 168) = 96 \u00d7 168
\u21d2 672 \u00d7 HCF (96, 168) = 96 \u00d7 168 [\u2235LCM(96, 168)=672]<\/p>\n\n\n\n

\u21d2 HCF (96, 168) = $\\frac{96 \\times 168}{672}$
\u21d2 HCF (96, 168) = 24<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 10 E <\/h4>\n\n\n\n

If HCF (306, 657) = 9, find LCM (306, 657)<\/strong>
Sol :
Given: HCF (306, 657) = 9
To Find: LCM (306, 657)
We use the formula
L.C.M (a,b) \u00d7 H.C.F (a,b) = Product of two numbers (a\u00d7b)<\/strong>
LCM (306, 657) \u00d7 HCF (306, 657) = 306 \u00d7 657
\u21d2 LCM (306, 657) \u00d7 9 = 306 \u00d7 657 [\u2235HCF(306,657)= 9]<\/p>\n\n\n\n

\u21d2 LCM (306, 657) = $\\frac{306 \\times 657}{9}$
\u21d2 LCM (306, 657) = 22338<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 10 F <\/h4>\n\n\n\n

If HCF (36, 64) = 4, find LCM (36, 64)<\/strong>
Sol :
Given: HCF (36, 64) = 4
To Find: LCM (36, 64)
We use the formula
L.C.M (a,b) \u00d7 H.C.F (a,b) = Product of two numbers (a\u00d7b)<\/strong>
LCM (36, 64) \u00d7 HCF (36, 64) = 36 \u00d7 64
\u21d2 LCM (36, 64) \u00d7 4 = 36 \u00d7 64 [\u2235HCF (36, 64)= 4]<\/p>\n\n\n\n

\u21d2 LCM (36, 64) = $\\frac{36 \\times 64}{4}$
\u21d2 LCM (36, 64) = 576<\/strong><\/p>\n\n\n\n

\n\n\n\n

Question 11 A <\/h4>\n\n\n\n

Examine whether (15)n<\/sup> can end with the digit 0 for any n \u03f5 N.<\/strong><\/p>\n\n\n\n

Sol :
If (15)n<\/sup> end with the digit 0, then the number should be divisible by 2 and 5.
As 2 \u00d7 5 = 10
\u21d2This means the prime factorization of 15n<\/sup> should contain prime factors 2 and 5.
But (15)n<\/sup> = (3 \u00d7 5)n<\/sup> and it does not have the prime factor 2 but have 3 and 5.
\u2235, 2 is not present in the prime factorization, there is no natural number nor which 15n<\/sup> ends with digit zero.
So, 15n<\/sup> cannot end with digit zero.<\/p>\n\n\n\n

\n\n\n\n

Question 11 B <\/h4>\n\n\n\n

Examine whether (24)n<\/sup> can end with the digit 5 for any n \u03f5 N.<\/strong><\/p>\n\n\n\n

Sol :
If (24)n<\/sup> end with the digit 5, then the number should be divisible by 5.
\u21d2This means the prime factorization of 24n<\/sup> should contain prime factors 5.
But (24)n<\/sup> = (23<\/sup> \u00d7 3)n<\/sup> and it does not have the prime factor 5 but have 3 and 2.
\u2235, 5 is not present in the prime factorization, there is no natural number nor which 24n<\/sup> ends with digit 5.
So, 24n<\/sup> cannot end with digit 5.<\/p>\n\n\n\n

\n\n\n\n

Question 11 C <\/h4>\n\n\n\n

Examine whether (21)n<\/sup> can end with the digit 0 for any n \u03f5 N.<\/strong><\/p>\n\n\n\n

Sol :
If (21)n<\/sup> end with the digit 0, then the number should be divisible by 2 and 5.
As 2 \u00d7 5 = 10
\u21d2This means the prime factorization of 21n<\/sup> should contain prime factors 2 and 5.
But (21)n<\/sup> = (3 \u00d7 7)n<\/sup> and it does not have the prime factor 2 and 5 but have 3 and 7.
\u2235, 2 and 5 is not present in the prime factorization, there is no natural number nor which 21n<\/sup> ends with digit zero.
So, 21n<\/sup> cannot end with digit zero.<\/p>\n\n\n\n

\n\n\n\n

Question 11 D <\/h4>\n\n\n\n

Examine whether (8)n<\/sup> can end with the digit 5 for any n \u03f5 N.<\/strong><\/p>\n\n\n\n

Sol :
If (8)n<\/sup> end with the digit 5, then the number should be divisible by 5.
\u21d2This means the prime factorization of 8n<\/sup> should contain prime factor 5.
But (8)n<\/sup> = (23<\/sup>)n<\/sup> 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<\/sup>.
\u2235, 5 is not present in the prime factorization, there is no natural number nor which 8n<\/sup> ends with digit 5.
So, 8n<\/sup> cannot end with digit 5.<\/p>\n\n\n\n

\n\n\n\n

Question 11 E <\/h4>\n\n\n\n

Examine whether (4)n<\/sup> can end with the digit 0 for any n \u03f5 N.<\/strong><\/p>\n\n\n\n

Sol :
If (4)n<\/sup> end with the digit 0, then the number should be divisible by 5.
As 2 \u00d7 5 = 10
\u21d2This means the prime factorization of 4n<\/sup> should contain prime factor 5.
This is not possible because (4)n<\/sup> = (22n<\/sup>), so the only prime in the factorization of 4n<\/sup> is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of 4n<\/sup>.
\u2235, 5 is not present in the prime factorization, there is no natural number nor which 4n<\/sup> ends with digit zero.
So, 4n<\/sup> cannot end with digit zero.<\/p>\n\n\n\n

\n\n\n\n

Question 11 F <\/h4>\n\n\n\n

Examine whether (7)n<\/sup> can end with the digit 5 for any n \u03f5 N.<\/strong><\/p>\n\n\n\n

Sol :
If (7)n<\/sup> end with the digit 5, then the number should be divisible by 5.
\u21d2This means the prime factorization of 7n<\/sup> should contain prime factor 5.
But (7)n<\/sup> 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<\/sup>.
\u2235, 5 is not present in the prime factorization, there is no natural number nor which 7n<\/sup> ends with digit 5.
So, 7n<\/sup> cannot end with digit 5.<\/p>\n\n\n\n

\n\n\n\n

Question 12 A <\/h4>\n\n\n\n

Explain why 7 x 11 x 13 x 17 +17 is a composite number.<\/strong>
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
\u21d217 (7 x 11 x 13 x 17 +1)
\u21d217 (7 x 11 x 13 x 17 +1)
\u21d217 (17017 + 1)
\u21d217 (17018)
\u21d217 (2 \u00d7 8509)
\u21d217 \u00d7 2 \u00d7 8509
So, given number is the composite number because it is the product of more than one prime numbers.<\/p>\n\n\n\n

\n\n\n\n

Question 12 B <\/h4>\n\n\n\n

Explain why 5 x 7 x 13 + 5 is a composite number.<\/strong>
Sol :
Composite number: A whole number that can be divided exactly by numbers other than 1 or itself.
5 x 7 x 13 + 5
\u21d25 (1 x 7 x 13 +1)
\u21d25 (91 +1)
\u21d25 (92)
\u21d25 (22<\/sup> \u00d7 23)
\u21d25 \u00d7 2 \u00d7 2 \u00d7 23
So, given number is the composite number because it is the product of more than one prime numbers.<\/p>\n\n\n\n

\n\n\n\n

Question 12 C <\/h4>\n\n\n\n

Show that 5 x 7 x 11 x 13 + 55 is a composite number.<\/strong><\/p>\n\n\n\n

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
\u21d25 (1 x 7 \u00d7 11 x 13 +11)
\u21d25 \u00d7 11 (91 +1)
\u21d25 \u00d7 11 (92)
\u21d25 \u00d7 11 (22<\/sup> \u00d7 23)
\u21d25 \u00d7 11 \u00d7 2 \u00d7 2 \u00d7 23
So, given number is the composite number because it is the product of more than one prime numbers.<\/p>\n\n\n\n

\n\n\n\n

Question 13 <\/h4>\n\n\n\n

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.<\/strong>
Sol :
Lengths of three measuring rods = 64cm, 80cm and 96cm
Least Length of cloth that can be measured = LCM (64, 80, 96)<\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 64 \\\\
\\hline 2 & 32 \\\\
\\hline 2 & 16 \\\\
\\hline 2 & 8 \\\\
\\hline 2 & 4 \\\\
\\hline 2 & 2 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 80 \\\\
\\hline 2 & 40 \\\\
\\hline 2 & 20 \\\\
\\hline 3 & 10 \\\\
\\hline 5 & 5 \\\\
\\hline & 1 \\\\
\\end{array}$<\/p>\n\n\n\n

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

64 = 26<\/sup>
80 = 23<\/sup> \u00d7 3 \u00d7 5
96 = 25<\/sup> \u00d7 3
So, 26<\/sup> \u00d7 3 \u00d7 5 are the greatest powers of the prime factors 2, 3 and 5
LCM (64, 80, 96) = 26<\/sup> \u00d7 3 \u00d7 5 = 960
Least Length of cloth that can be measured is 960 cm<\/p>\n\n\n\n

\n\n\n\n

Question 14 <\/h4>\n\n\n\n

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

$\\begin{array}{l|l}
2 & 36 \\\\
\\hline 2 & 18 \\\\
\\hline 3 & 9 \\\\
\\hline 3 & 3 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

$\\begin{array}{l|l}
2 & 72 \\\\
\\hline 2 & 36 \\\\
\\hline 2 & 18 \\\\
\\hline 3 & 9 \\\\
\\hline 3 & 3 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

$\\begin{array}{c|c}
3 & 27 \\\\
\\hline 3 & 9 \\\\
\\hline 3 & 3 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

27 = 33<\/sup>
36 = 22<\/sup> \u00d7 32<\/sup>
72 = 23<\/sup> \u00d7 32<\/sup>
Here, 32<\/sup> 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<\/p>\n\n\n\n

\n\n\n\n

Question 15 <\/h4>\n\n\n\n

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.<\/strong>
Sol :
Mixtures of milk and water in three containers = 403kg, 434kg, 465kg
Biggest measure which can exactly measure different quantities = HCF (403, 434, 465)<\/p>\n\n\n\n

$\\begin{array}{l|l}
13 & 403 \\\\
\\hline 31 & 31 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

$\\begin{array}{c|c}
2 & 434 \\\\
\\hline 7 & 217 \\\\
\\hline 31 & 31 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

$\\begin{array}{c|c}
3 & 465 \\\\
\\hline 5 & 155 \\\\
\\hline 31 & 31 \\\\
\\hline & 1
\\end{array}$<\/p>\n\n\n\n

403=13\u00d731<\/p>\n\n\n\n

434=2\u00d77\u00d731<\/p>\n\n\n\n

465=3\u00d75\u00d731<\/p>\n\n\n\n

Here, 311<\/sup> is the smallest power of the common factor.
HCF (403, 434, 465) = 31
So, biggest measure which can exactly measure the milk = 31L<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 10 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Question 1 A  Express each of the following numbers as a product of its prime factors:Sol :Given number is 4320Factorization 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 […]<\/p>\n","protected":false},"author":302,"featured_media":623903,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[24],"tags":[],"boards":[],"class_list":["post-623894","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-10","entry"],"yoast_head":"\nKC Sinha: Exercise 1.2 - Mathematics Solution Class 10 Chapter 1 Real numbers - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Question 1 A  Express each of the following numbers as a product of its prime factors:Sol :Given number is 4320Factorization of 4320 is\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-1-2-mathematics-solution-class-10-chapter-1-real-numbers\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"KC Sinha: Exercise 1.2 - 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