Class- X-CBSE-Mathematics Real Numbers
CBSE NCERT Solutions for Class 10 Mathematics Chapter 1
Back of Chapter Questions
1. Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Solution:
(i) 135 and 225
Step 1:
Since 225 is greater than 135, we can apply Euclid's division lemma to
a = 225 and b = 135 to find q and r such that
225 = 135q + r, 0 ≤ r < 135
So, dividing 225 by 135 we get 1 as the quotient and 90 as remainder.
i. e 225 = (135 × 1) + 90
Step 2:
Remainder r is 90 and is not equal to 0, we apply Euclid's division lemma
to b = 135 and r = 90 to find whole numbers q and r such that
135 = 90 × q + r, 0 ≤ r < 90
So, dividing 135 by 90 we get 1 as the quotient and 45 as remainder.
𝑖𝑖. 𝑒𝑒 135 = (90 × 1) + 45
Step 3:
Again, remainder r is 45 and is not equal to 0, so we apply Euclid's
division lemma to b = 90 and r = 45 to find q and r such that
90 = 45 × q + r, 0 ≤ r < 45
So, dividing 90 by 45 we get 2 as the quotientand 0 as remainder.
𝑖𝑖. 𝑒𝑒. 90 = (2 × 45) + 0
Step 4:
Since the remainder is zero, the divisor at this stage will be HCF of
(135, 225)
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Since the divisor at this stage is 45, therefore, the HCF of 135 and 225 is
45.
(ii) 196 and 38220
Step 1:
Since 38220 is greater than 196, we can apply Euclid's division lemma
to a = 38220 and b = 196 to find whole numbers q and r such that
38220 = 196 q + r, 0 ≤ r < 196
So dividing 38220 by 196, we get 195 as the quotient and 0 as remainder
r
𝑖𝑖. 𝑒𝑒 38220 = (196 × 195) + 0
Because the remainder is zero, divisor at this stage will be HCF
Since divisor at this stage is 196, therefore, HCF of 196 and 38220 is
196.
(iii) 867 and 255
Step 1:
Since 867 is greater than 255, we can apply Euclid's division lemma, to
a = 867 and b = 255 to find q and r such that 867 = 255q + r, 0 ≤ r <
255. So, dividing 867 by 255 we get 3 as the quotient and 102 as
remainder.𝑖𝑖. 𝑒𝑒 867 = 255 × 3 + 102
Step 2:
Since remainder is 102 and is not equal to 0, we can apply the division
lemma to a = 255 and b = 102 to find whole numbers q and r such that
255 = 102q + r where 0 ≤ r < 102. So, dividing 255 by 102 we get 2
as the quotient and 51 as remainder.𝑖𝑖. 𝑒𝑒 255 = 102 × 2 + 51
Step 3:
Again remainder 51 is not equal to zero, so we apply the division lemma
to a = 102 and b = 51 to find whole numbers q and r such that 102 =
51 q + r where 0 ≤ r < 51. So, dividing 102 by 51 we get 2 as the
quotient and 0 as remainder.𝑖𝑖. 𝑒𝑒 102 = 51 × 2 + 0. Since, the remainder
is zero, the divisor at this stage is the HCF. Since the divisor at this stage is
51, therefore, HCF of 867 and 255 is 51
2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5,
where q is some integer.
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Solution:
Let a be any odd positive integer, we need to prove that a is in the form of 6q +
1 , or 6q + 3 , or 6q + 5, where q is some integer.
Because a is an integer, we can consider b to be 6 as another integer. Applying
Euclid's division lemma, we get
a = 6q + r for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 Where 0 ≤ r < 6.
Therefore, a can be any of the form 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4
or 6q + 5
However, since a is odd, a cannot take the values 6q, 6q + 2 and 6q + 4
(since all these are divisible by 2)
Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k 2 + 1, where k 2 is an integer
6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k 3 + 1, where k 3 is an integer
Clearly, 6q + 1, 6q + 3, 6q + 5 are in the form of 2k + 1, where k is an integer.
Therefore, 6q + 1, 6q + 3 and 6q + 5 are odd numbers.
Therefore, any odd integer can be expressed in the form of
6q + 1, or 6q + 3, or 6q + 5 where q is some integer
3. An army contingent of 616 members is to march behind an army band of 32
members in a parade. The two groups are to march in the same number of
columns. What is the maximum number of columns in which they can march?
Solution:
Maximum number of columns in which the Army contingent and the band can
march is equal to the HCF of 616 and 32. Euclid's algorithm can be used here to
find the HCF.
Step 1:
Since 616 is greater than 32, so by applying Euclid's division lemma to a = 616
and b = 32 we get integers q and r as 19 and 8
𝑖𝑖. 𝑒𝑒 616 = 32 × 19 + 8
Step 2:
Since remainder r is 8 and is not equal to 0, we can again apply Euclid's lemma to
32 and 8 to get integers 4 and 0 as the quotient and remainder respectively.
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CBSE NCERT Solutions for Class 10 Mathematics Chapter 1
Back of Chapter Questions
1. Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Solution:
(i) 135 and 225
Step 1:
Since 225 is greater than 135, we can apply Euclid's division lemma to
a = 225 and b = 135 to find q and r such that
225 = 135q + r, 0 ≤ r < 135
So, dividing 225 by 135 we get 1 as the quotient and 90 as remainder.
i. e 225 = (135 × 1) + 90
Step 2:
Remainder r is 90 and is not equal to 0, we apply Euclid's division lemma
to b = 135 and r = 90 to find whole numbers q and r such that
135 = 90 × q + r, 0 ≤ r < 90
So, dividing 135 by 90 we get 1 as the quotient and 45 as remainder.
𝑖𝑖. 𝑒𝑒 135 = (90 × 1) + 45
Step 3:
Again, remainder r is 45 and is not equal to 0, so we apply Euclid's
division lemma to b = 90 and r = 45 to find q and r such that
90 = 45 × q + r, 0 ≤ r < 45
So, dividing 90 by 45 we get 2 as the quotientand 0 as remainder.
𝑖𝑖. 𝑒𝑒. 90 = (2 × 45) + 0
Step 4:
Since the remainder is zero, the divisor at this stage will be HCF of
(135, 225)
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Since the divisor at this stage is 45, therefore, the HCF of 135 and 225 is
45.
(ii) 196 and 38220
Step 1:
Since 38220 is greater than 196, we can apply Euclid's division lemma
to a = 38220 and b = 196 to find whole numbers q and r such that
38220 = 196 q + r, 0 ≤ r < 196
So dividing 38220 by 196, we get 195 as the quotient and 0 as remainder
r
𝑖𝑖. 𝑒𝑒 38220 = (196 × 195) + 0
Because the remainder is zero, divisor at this stage will be HCF
Since divisor at this stage is 196, therefore, HCF of 196 and 38220 is
196.
(iii) 867 and 255
Step 1:
Since 867 is greater than 255, we can apply Euclid's division lemma, to
a = 867 and b = 255 to find q and r such that 867 = 255q + r, 0 ≤ r <
255. So, dividing 867 by 255 we get 3 as the quotient and 102 as
remainder.𝑖𝑖. 𝑒𝑒 867 = 255 × 3 + 102
Step 2:
Since remainder is 102 and is not equal to 0, we can apply the division
lemma to a = 255 and b = 102 to find whole numbers q and r such that
255 = 102q + r where 0 ≤ r < 102. So, dividing 255 by 102 we get 2
as the quotient and 51 as remainder.𝑖𝑖. 𝑒𝑒 255 = 102 × 2 + 51
Step 3:
Again remainder 51 is not equal to zero, so we apply the division lemma
to a = 102 and b = 51 to find whole numbers q and r such that 102 =
51 q + r where 0 ≤ r < 51. So, dividing 102 by 51 we get 2 as the
quotient and 0 as remainder.𝑖𝑖. 𝑒𝑒 102 = 51 × 2 + 0. Since, the remainder
is zero, the divisor at this stage is the HCF. Since the divisor at this stage is
51, therefore, HCF of 867 and 255 is 51
2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5,
where q is some integer.
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Solution:
Let a be any odd positive integer, we need to prove that a is in the form of 6q +
1 , or 6q + 3 , or 6q + 5, where q is some integer.
Because a is an integer, we can consider b to be 6 as another integer. Applying
Euclid's division lemma, we get
a = 6q + r for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 Where 0 ≤ r < 6.
Therefore, a can be any of the form 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4
or 6q + 5
However, since a is odd, a cannot take the values 6q, 6q + 2 and 6q + 4
(since all these are divisible by 2)
Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k 2 + 1, where k 2 is an integer
6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k 3 + 1, where k 3 is an integer
Clearly, 6q + 1, 6q + 3, 6q + 5 are in the form of 2k + 1, where k is an integer.
Therefore, 6q + 1, 6q + 3 and 6q + 5 are odd numbers.
Therefore, any odd integer can be expressed in the form of
6q + 1, or 6q + 3, or 6q + 5 where q is some integer
3. An army contingent of 616 members is to march behind an army band of 32
members in a parade. The two groups are to march in the same number of
columns. What is the maximum number of columns in which they can march?
Solution:
Maximum number of columns in which the Army contingent and the band can
march is equal to the HCF of 616 and 32. Euclid's algorithm can be used here to
find the HCF.
Step 1:
Since 616 is greater than 32, so by applying Euclid's division lemma to a = 616
and b = 32 we get integers q and r as 19 and 8
𝑖𝑖. 𝑒𝑒 616 = 32 × 19 + 8
Step 2:
Since remainder r is 8 and is not equal to 0, we can again apply Euclid's lemma to
32 and 8 to get integers 4 and 0 as the quotient and remainder respectively.
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