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CBSE Board Paper Solution-
Class :^ X Subject :^ Mathematics (Standard) - Theory Set :^1 Code No :^ 30/5/ Time allowed :^ 3 Hours Maximum Marks :^80 Marks
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 question All questions are compulsory (ii) Section A : Question Numbers 1 to 20 comprises of 20 question of one mark each. (iii) Section B : Question Numbers 21 to 26 comprises of 6 question of two marks each. (iv) Section C : Question Numbers 27 to 34 comprises of 8 question of three marks each. (v) Section D : Question Numbers 35 to 40 comprises of 6 question of four marks each. (vi) There is no overall choice in the question paper. However, an internal choice has been provided in 2 question of the mark, 2 question of one mark, 2 questions of two marks. 3 question of three marks
and 3 question 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.
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 x^2 – 4, quotient and remainder are found to be x and 3 respectively. The polynomial p(x) is
(A) 3x^2 + x - 12 (B) x^3 -4x + 3 (C) x^2 + 3x - 4 (D) x^2 - 4x - 3 Answer: Correct Answer: (B) x^3 – 4x+ Explanation: P(x) = (divisor)×(quotient) + Remainder =(x^2 – 4)x+ = x^3 – 4x+
2) In Figure-1, ABC is an isosceles triangle, right- angled at C. Therefore
(B) (5, 0)
(C) (0, 0)
(D) (3, 0)
OR
The centre of a circle whose end points of a diameter are (─ 6, 3) and 6, 4) is
(A) (8, ─ 1)
(B) (4, 7)
(C)^ ^ ,
(D)^ ^ ,
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)/ = 6/ = 3
Required point = (x, 0) = (3, 0)
OR
Correct Answer: (C) (0, 7/2) Explanation: The centre of a circle is the mid-point of its diameter. End points of the diameter are: (–6, 3) and (6, 4) Coordinates of the centre = ((– 6+6)/2, (3+4)/2) = (0, 7/2)
4) The value(s) of k for which the quadratic equation 2x^2 + kx + 2 = 0 has equal roots, is (A) 4 (B) 4
(C) ─ 4 (D) 0
Answer: Correct Answer: (B) ± Explanation: The given equation is: 2x^2 + kx + 2 = Discriminant = b^2 – 4ac Here, b =k, a =2, and c = So, Discriminant = k^2 – 4×2× = k^2 – 16 A quadratic equation has equal roots if its discriminant is zero. k^2 – 16 = 0
Difference between consecutive terms is not same. So, this is not an A.P.
6) The pair of linear equations
3x (^) + 5y= 7 and 9x + 10y = 14 is 2 3 (A) consistent (B) inconsistent (C) consistent with one solution (D) consistent with many solutions
Answer: Correct Answer: (B) Inconsistent Explanation:
3x 5y (^7) 2 3 9x 10y 7 6 9x 10y 42 ...(1) 9x 10y 14 ...(2) Ratios of coefficients of x and that of y are 9 10 1 9 10 1 Ratio of constants= 42 =^3 14 1 1 Ratios of coefficients of x and y are equal but they are not equal to the ratio of constants. So, the given equations represent a pair of parallel lines and so they do not have a common solution.
7) In Figure-2 PQ is tangent to the circle with centre at O, at the point B. If AOB = 100°, then ABP is equal to (A) 50° (B) 40° (C) 60° (D) 80°
Answer: Correct Answer: (C)32/ Explanation:
3
3
3 2 2 /
Volume of sphere = r 3 4 12 r 3 r 3 r 3
9) The distance between the points (m,–n) and (–m, n) is
(A) m^2 + n^2 (B) m + n
(C) 2 m^2 + n^2
(D) 2m^2 + 2n^2
Answer: Correct Answer: (C) 2 m^2 n^2 Explanation: (^2 )
2 2 2 2
Distance = m ( m) (–n – n)
(m m) (–2n) 2 m n
10) In Figure-3. From an external point P, two tangents PQ and PR are drawn to a circle of radius 4 cm with centre O. If QPR = 90°, then length of PQ is (A) 3cm (B) 4cm (C) 2cm
(D) 2 2 cm
Answer: Correct Answer: (B) 4 cm Explanation: Tangents are drawn from an external point P.
So, line joining centre O and point P bisects ∠PQR. OP bisects ∠QPR = 90°. In ∆ OQP, ∠Q = 90° (radius meets tangent at 90°) ∠QPO = 45° = ∠QOP
14) i^ i i i
fu
In the formula x a h, u = _____.
f
Answer: xi a h
15) All concentric circles are ______ to each other.
Answer: similar
Answer the following question numbers 16 to 20.
16) Find the sum of the first 100 natural numbers.
Answer:
1 2 3 ....100 is an A. P. Here first term a 1 Common difference d =
Sum of n terms of an A.P. = n 2a + n 1 d 2 The sum of first 100 natural numbers
= 100 2×1 + 100 1 1 2 100 (101) = 2 50 101 5050
17) In Figure-4 the angle of elevation of the top of a tower from a point C on the ground, which is
30 m away from the foot of the tower, is 30°. Find the height of the tower.
Answer:
tan30 AB 30 1 AB 3 30
AB 30 10 3 3 So, the height of the tower is 10 3 m.
18) The LCM of two numbers is 182 and their HCF is
13. If one of the numbers is 26, Find the other.
Answer:
LCM HCF Pr oduct of the two numbers 182 × 13 = 26 × x
x= 182 × 13 91 26 So, the other number is 91.
Answer:
2 tan45° ×cos60° sin 2 1 1 2 1 2 2 SECTION B
Question number 21 to 26 carry 2 marks each.
21) In the given Figure-5, DE ||AC and DF||AE.
Prove that
BF BE
FE EC
Answer:
In ABC, DE AC So, using basic proportionality theorem, we get BD BE (^) ...(1) DA EC In BAE, DF AE So, using basic proportionality theorem, we get BD BF DA FE
...(2) From (1) and (2), we get BE BF EC FE
22) Show that 5 +2 7 is an irrational number, where 7 is given to be an irrational number. OR Check whether 12n^ can end with the digit 0 for any natural number n.
Answer:
^
Let us assume, to the contrary, that 5 2 7 is rational. That is, we can find coprime a and b (b 0) such that
5 2 7
Rearranging this equation, we get 7 5 2 2
a b a b a a b b b
^ ^
^
Given that A, B and C are interior angles of a triangle ABC. A B C 180 or A 180 B C Now,
cos B^ C^ sin 90 B^ C 2 2
sin 180 B^ C 2
sin A 2
24) In Figure 6, a quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = BC +AD.
OR
In Figure-7, find the perimeter of ABC, if AP = 12 cm.
Answer:
