Date: 05/05/2022

Question Paper Code

430/2/1

Time: 2 Hrs.

Class-X

Max. Marks: 40

MATHEMATICS (Basic) Term-II (CBSE 2022)

GENERAL INSTRUCTIONS

1. This question paper contains 14 questions. All questions are compulsory.
2. This question paper is divided into 3 Sections – Section A, B and C.
3. Section-A comprises of 6 questions (Q. Nos. 1 to 6) of 2 marks each.

Internal choice has been provided in two questions.

1. Section-B comprises of 4 questions (Q. Nos. 7 to 10) of 3 marks each.

Internal choice has been provided in one question.

1. Section-C comprises of 4 questions (Q. Nos. 11 to 14) of 4 marks each. An internal choice has been provided in one question. It also contains two case study based questions.
2. Use of calculator is not permitted.

SECTION-A

Question Numbers 1 to 6 carry 2 marks each.

1. Find the nature of the roots of the quadratic equation: [2] <math>4x^2 - 5x - 1 = 0</math> Solution <math>4x^2 - 5x - 1 = 0</math>

<math>D = b^2 - 4ac</math>, where <math>a = 4</math>, <math>b = -5</math> and <math>c = -1</math> <math>[\frac{1}{2}]</math>

<math>\Rightarrow</math> D = 25 + 16 = 41 <math>[\frac{1}{2}]</math>

<math>\Rightarrow</math> D > 0 <math>[\frac{1}{2}]</math>

: . The given equation has real and distinct roots <math>[\frac{1}{2}]</math>

1. (a) Which term of the A.P. 3, 8, 13, 18, ... is 78? [2]

OR (b) Find the common difference of an A.P. whose <math>n^{th}</math> term is given by [2]

<math>a_n = 6n - 5</math>. Solution

1. Given A.P. is 3, 8, 13, 18, ....

Here, <math>a = 3</math> and <math>d = 8 - 3 = 5</math> <math>[\frac{1}{2}]</math>

<math>a_n = a + (n-1)d</math> [<math>n^{th}</math> term] <math>[\frac{1}{2}]</math>

<math>\Rightarrow</math> 78 = 3 + (n-1)5 <math>\Rightarrow \frac{75}{5} = n - 1</math> <math>[\frac{1}{2}]</math>

<math>\Rightarrow</math> n = 16 :. 78 is 16^th term of the given A.P. <math>[\frac{1}{2}]</math> OR

1. <math>n^{th}</math> term of A.P. is <math>a_n = 6n - 5</math> if <math>n = 1</math>,

<math>\Rightarrow</math> <math>a_1 = 6 - 5 = 1</math> <math>[\frac{1}{2}]</math>

if <math>n_2 = 1</math>, <math>a_2 = 6 \times 2 - 5 = 7</math> <math>[\frac{1}{2}]</math>

··. Common difference <math>(d) = 7 - 1</math> <math>[\frac{1}{2}]</math>

= 6 <math>[\frac{1}{2}]</math>

1. 3 cubes each of 8 cm edge are joined end to end. Find the total surface area of the cuboid so formed. [2] Solution Dimensions of cuboid are 24 cm, 8 cm, 8 cm

T.S.A of cuboid = <math>2(lb + bh + lh)</math> <math>[\frac{1}{2}]</math>

<math>= 2[24(8) + 8(8) + 24(8)]</math> <math>[\frac{1}{2}]</math>

<math>= 2[192 + 64 + 192]</math> <math>[\frac{1}{2}]</math>

<math>= 2[448] = 896 \text{ cm}^2</math> <math>[\frac{1}{2}]</math>