Given below are some sample questions for CBSE Class 10 Mathematics: Important 3 Marks Questions:
Q. Prove that √3 is irrational.
Sol.
Let √3 be a rational number
√3 = a/b (a and b are integers and co-primes and b ≠ 0)
On squaring both the sides, 3 = a2/b2
⟹ 3b2 = a2
⟹ a2 is divisible by 3
⟹ a is divisible by 3
We can write a = 3c for some integer c.
⟹ a2 = 9c2
⟹ 3b2 = 9c2
⟹ b2 = 3c2
⟹ b2 is divisible by 3
⟹ b is divisible by 3
From (i) and (ii), we get 3 as a factor of ‘a’ and ‘b’ which is contradicting the fact that a and b are co-primes. Hence our assumption that√3 is an rational number is false. So √3 is an irrational number.
Q. For any positive integer n, prove that n3 ‒ n is divisible by 6.
Sol.
Let x = n3‒ n
⟹ a = n(n2‒1)
⟹ x = n (n ‒ 1) ×(n+ 1) [∵ (a2 ‒ b2)=(a‒b)(a + b)]
x = (n ‒1) × n × (n+ 1) ... (i)
We know that, if a number is completely divisible by 2 and 3, then it is also divisible by 6.
Divisibility test for 3:
If the sum of digits of any number is divisible by 3, then it is divisible by 3:
Sum of the digits = (n ‒ 1) + (n) + (n + 1) = 3
⟹ Number is divisible by 3.
Divisibility test for 2:
If n is odd then (n ‒ 1) and (n + 1) will be even so, (n ‒1) × n × (n + 1) will be divisible by 2.
If n is even then, (n ‒1) × n × (n + 1) will be divisible by 2.
Therefore, for any positive integral value of n, n3 ‒ n is divisible by 6.
CBSE Class 10 Mathematics Previous Years' Solved Question Papers (2010-2019)
Q. At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.
Sol.
Let, Nisha’s present age be = x year
Therefore, according to the first condition, Asha's present age = x2 + 2
Nisha grows to her mother's present age after [(x2 + 2) – x] years.
Then, Asha's age will become (x2 + 2) + [(x2 + 2) –x] year.
According to the question,
(x2 + 2) + [(x2 + 2) – x] = 10x – 1
⟹ 2x2 – x + 4 = 10x – 1
⟹ 2x2 – 11x + 5 = 0
⟹ 2x2 – 10x – x + 5= 0
⟹ 2x (x – 5) –1(x – 5) = 0
⟹ (x – 5) (2x – 1) = 0
⟹ (x – 5) = 0 or (2x – 1) = 0
⟹ x = 5 or 1/2
Ignoring x = 1/2 because then Asha’s age = x2 + 2 =which is not possible.
Hence, present age of Nisha = 5year
And present age of Asha = x2 + 2 = (5)2 + 2 = 25 + 2 = 27 year.
Given below are some sample questions for CBSE Class 10 Mathematics: Important 3 Marks Questions:
Q. Prove that √3 is irrational.
Sol.
Let √3 be a rational number
√3 = a/b (a and b are integers and co-primes and b ≠ 0)
On squaring both the sides, 3 = a2/b2
⟹ 3b2 = a2
⟹ a2 is divisible by 3
⟹ a is divisible by 3
We can write a = 3c for some integer c.
⟹ a2 = 9c2
⟹ 3b2 = 9c2
⟹ b2 = 3c2
⟹ b2 is divisible by 3
⟹ b is divisible by 3
From (i) and (ii), we get 3 as a factor of ‘a’ and ‘b’ which is contradicting the fact that a and b are co-primes. Hence our assumption that√3 is an rational number is false. So √3 is an irrational number.
Q. For any positive integer n, prove that n3 ‒ n is divisible by 6.
Sol.
Let x = n3‒ n
⟹ a = n(n2‒1)
⟹ x = n (n ‒ 1) ×(n+ 1) [∵ (a2 ‒ b2)=(a‒b)(a + b)]
x = (n ‒1) × n × (n+ 1) ... (i)
We know that, if a number is completely divisible by 2 and 3, then it is also divisible by 6.
Divisibility test for 3:
If the sum of digits of any number is divisible by 3, then it is divisible by 3:
Sum of the digits = (n ‒ 1) + (n) + (n + 1) = 3
⟹ Number is divisible by 3.
Divisibility test for 2:
If n is odd then (n ‒ 1) and (n + 1) will be even so, (n ‒1) × n × (n + 1) will be divisible by 2.
If n is even then, (n ‒1) × n × (n + 1) will be divisible by 2.
Therefore, for any positive integral value of n, n3 ‒ n is divisible by 6.
CBSE Class 10 Mathematics Previous Years' Solved Question Papers (2010-2019)
Q. At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.
Sol.
Let, Nisha’s present age be = x year
Therefore, according to the first condition, Asha's present age = x2 + 2
Nisha grows to her mother's present age after [(x2 + 2) – x] years.
Then, Asha's age will become (x2 + 2) + [(x2 + 2) –x] year.
According to the question,
(x2 + 2) + [(x2 + 2) – x] = 10x – 1
⟹ 2x2 – x + 4 = 10x – 1
⟹ 2x2 – 11x + 5 = 0
⟹ 2x2 – 10x – x + 5= 0
⟹ 2x (x – 5) –1(x – 5) = 0
⟹ (x – 5) (2x – 1) = 0
⟹ (x – 5) = 0 or (2x – 1) = 0
⟹ x = 5 or 1/2
Ignoring x = 1/2 because then Asha’s age = x2 + 2 =which is not possible.
Hence, present age of Nisha = 5year
And present age of Asha = x2 + 2 = (5)2 + 2 = 25 + 2 = 27 year.