# CBSE Class 10 Mathematics Exam 2018: Important 3 Marks Questions

Feb 14, 2018 13:38 IST
CBSE Class 10 Mathematics Important 3 Marks Questions

Get here the CBSE Class 10 Mathematics: Important 3 Marks Questions to prepare for the board exam 2018. Each question has been provided with a proper solution.

In CBSE Class 10 Mathematics Board Exam 2018, Section - C will comprise 10 questions of 3 marks each.

Our subject experts have prepared these questions by analysing trend followed in previous years’ board exams. Students must practice these important questions to acquaint with the important topics of class 10 Maths and track their preparedness for the board exams 2018. Moreover, the solutions provided here give students an idea about what how to write proper solutions to each question in the board exam so as to score optimum marks.

CBSE Class 10 Mathematics Exam 2018: Important 1 Mark Questions

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.

CBSE Class 10 Mathematics Exam 2018: Important 2 Marks Questions

Q. For any positive integer n, prove that n3n is divisible by 6.

Sol.

Let x = n3n

a = n(n2‒1)

x = n (n ‒ 1) ×(n+ 1)                                              [∵ (a2b2)=(ab)(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.

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

⟹  2x2x + 4 = 10x – 1

⟹  2x2 – 11x + 5 = 0

⟹  2x2 – 10xx + 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.

To get the complete set of questions, click on the following link:

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