Class: 10
Subject: Mathematics (Standard) – Theory
Set: 1
Code No: 30/5/1
Time allowed : 3 Hours
Maximum Marks: 80 Marks
Class-10 CBSE Board Math Paper Solution-2020
General instructions:
Read the following instructions very carefully and strictly
follow them:
(i) This question paper comprises four sections – A, B, C, and D. This question paper carries 40 questions All questions are compulsory.
(ii) Section A: Question Numbers 1 to 20 comprises 20 questions of one mark each.
(iii) Section B: Question Numbers 21 to 26 comprises 6 questions of two marks each.
(iv) Section C: Question Numbers 27 to 34 comprise 8 questions of three marks each.
(v) Section D: Question Numbers 35 to 40 comprise 6 questions of four marks each.
(vi) There is no overall choice in the question paper. However, an internal choice has been provided in 2 questions of the mark, 2 questions of one mark, and 2 questions of two marks. 3 questions of three marks and 3 questions of four marks. You have to attempt
only one of the choices in such questions. (vii) In addition to this. Separate instructions are given with each section and question, wherever necessary.
(viii) Use of calculations is not permitted.
Read Also – CBSE CLASS 10 2022 QUESTION PAPERS
Class-10 CBSE Board Math Paper Solution-2020
1.Section A
Question numbers 1 to 20 carry 1 mark each. Question numbers 1 to 10 are multiple-choice questions. Choose the correct option.
1. On dividing a polynomial p(x) by x2 – 4, quotient and remainder are found to be x and 3 respectively. The polynomial p(x) is
(A) 3x² + x – 12
(B) x³ -4x + 3
(C) x² + 3x – 4
(D) x² – 4x – 3
Answer:
Correct Answer: (B) x³ –4x+3
Explanation:
P(x) = (divisor)×(quotient) + Remainder
=(x² – 4)x+3
= x³ – 4x+3
2) In Figure-1, ABC is an isosceles triangle, right angled at C. Therefore
(A) AB² = 2AC²
(B) BC² = 2AB²
(C) AC² = 2AB²
(D) AB² = 4AC²
Answer:
Correct Answer: (A) AB² = 2AC²
Explanation
Given that ACB is an isosceles triangle right angled at
C.
Therefore, AC = BC
Using Pythagoras theorem in the given triangle,
we have
AB² = AC² + BC²
= AC² + AC²
= 2AC²
3) The point on the x-axis which is equidistant
from (─ 4, 0) and 10, 0) is
(A) (7, 0)
(B) (5, 0)
(C) (0, 0)
(D) (3, 0)
Answer:
Correct Answer: (D) (3, 0)
Explanation:
The required point and the given points as well lie on
the x-axis.
The required point (x, 0) is the mid-point of the line
joining points (–4, 0) and (10, 0).
So, x = (– 4+10)/2
= 6/2
= 3
Required point = (x, 0)
= (3, 0)
4) The value(s) of k for which the quadratic
equation 2×2 + kx + 2 = 0 has equal roots, is
(A) 4
(B)
4
(C) ─ 4
(D) 0
Answer:
Correct Answer: (B) ±4
Explanation:
The given equation is:
2×2 + kx + 2 =0
Discriminant = b2–4ac
Here, b =k, a =2, and c =2
So, Discriminant = k2–4×2×2
= k2–16
A quadratic equation has equal roots if its
discriminant is zero.
k
2–16 = 0
www.cbsesolution.com
k
2 =16
k = ±4