﻿ CBSE Class 10 Mathematics Solved Question Paper 2016 | Previous Year Papers | Zigya

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# CBSE Class 10 Mathematics Solved Question Paper 2016

#### Short Answer Type

1.

For what value of k will k+9, 2k-1 and 2k+7 are the consecutive terms of an A.P?

If k+9, 2k-1 and 2k+7 are the consecutive terms of A.P, then the common difference will be the same.
∴ (2k – 1) – (k + 9) = (2k + 7) – (2k – 1)

k – 10 = 8

∴ k = 18

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2.

If -5 is a root of the quadratic equation 2x2  + px – 15 = 0 and the quadratic equation p(x2 + x)k = 0 has equal roots, find the value of k.

We Given -5 is a root of the quadratic equation 2x2+px-15 =0
, -5 satisfies the given equation.
∴ 2 (-5)2 +p(-5)-15=0
50-5p-15 =0
35-5p=0
5p=35 ⇒ p=7

Substituting p=7 in (x2+x)+k =0, we get
7(x2+x)+k =0
7x2+7x+k=0
The roots of the equation are equal
∴ Discriminant =b2-4ac =0
Here, a= 7, b=7,c=k
b2-4ac=0
∴(7)2-4(7)(k) =0
49-28k =0
28k=49
k= 49/28
=7/4

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3.

In Fig, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If OP=2r, show that  ∠ OTS = ∠ OST = 30°. In the given figure, Given
OP= 2r
∠OTP = 90°  (radius drawn at the point of contact is perpendicular to the tangent)
Now ,
In ΔOTP,
Sin∠ OPT = OT/OP = 1/2 = Sin 30o
⇒ ∠ OPT = 30o
therefore,
∠ TOP=60o
∴ ΔOTP is a 30o-60o-90o, right triangle.

In ΔOTS,
OT = OS   … (Radii of the same circle)
therefore,
ΔOTS is an isosceles triangle.

∴∠OTS = ∠OST … (Angles opposite to equal sides of an isosceles triangle are equal)

In ΔOTQ and ΔOSQ

OS = OT … (Radii of the same circle)
OQ = OQ ...(side common to both triangles)
∠OTQ = ∠OSQ … (angles opposite to equal sides of an isosceles triangle are
equal)

∴ ΔOTQ ≅ ΔOSQ … (By S.A.S)
∴ ∠TOQ = ∠SOQ = 60° … (C.A.C.T)
∴ ∠TOS = 120° … (∠TOS = ∠TOQ + ∠SOQ = 60° + 60° = 120°)
∴ ∠OTS + ∠OST = 180° – 120° = 60°
∴ ∠OTS = ∠OST = 60° ÷ 2 = 30°

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4.

A card is drawn at random from a well -shuffled fled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen. 1.

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5.

Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right-angled isosceles triangle.

Given, the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right-angled isosceles triangle.
Therefore,
Let A (3,0), B (6,4) and C (-1,3) be the given points.
Now, 6944 Views

6.

A ladder leaning against a wall makes an angle of 60° with the horizontal. If the foot of the ladder is 2.5 m away from the wall, find the length of the ladder.

Let AB be the ladder and CA be the wall. The ladder makes an angle of 60o with the horizontal.
∴ ΔABC is a 30o-60o-90o, right triangle.

Given: BC = 2.5 m, ∠ABC = 60°
AB = 5 cm and ∠BAC = 30°

From Pythagoras Theorem, we have
AB2 = BC2 + CA2
52 = (2.5)2 + (CA)2
(CA)2 = 25 – 6.25 = 18.75 m
Hence, length of the ladder is 4627 Views

7.

In Fig. , PQ is tangent at point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°, find ∠PCA. In the given figure, In Δ ACO,
OA=OC (Radii of the same circle)
Therefore,
ΔACO is an isosceles triangle.
∠CAB = 30° (Given)
∠CAO = ∠ACO = 30° (angles opposite to equal sides of an isosceles triangle are equal)
∠PCO = 90° …(radius is drawn at the point of
contact is perpendicular to the tangent)
Now ∠PCA = ∠PCO – ∠CAO
Therefore,
∠PCA = 90° – 30° = 60°

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8.

The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.

We have given that
4th term of an A.P.= a4 = 0
∴ a + (4 – 1)d = 0
∴ a + 3d = 0
∴ a = –3d        ….(1)

25th term of an A.P. = a25
= a + (25 – 1)d
= –3d + 24d      ….[From the equation (1)]
= 21d

3 times 11th term of an A.P. = 3a11
= 3[a + (11 – 1)d]
= 3[a + 10d]
= 3[–3d + 10d]
= 3 × 7d
= 21d

∴ a25 = 3a11

i.e., the 25th term of the A.P. is three times its 11th term.

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9.

Let P and Q be the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4) such that P is nearer to A. Find the coordinates of P and Q.

P and Q are the points of trisection of AB,
therefore, AP=PQ=QB
Thus, P divides AB internally in the ratio 1:2 and Q divide AB internally in the ratio 2:1.  8136 Views

10.

In Fig.2, a quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively. Prove that AB + CD = BC + DA.  In the Given figure, since tangents are drawn from an exterior point to a circle are equal in length,
AP = AS ….(1)
BP = BQ ….(2)
CR = CQ ….(3)
DR = DS ….(4)
Adding equations (1), (2), (3) and (4), we get

AP + BP + CR + DS = AS + BQ + CQ + DS
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
AB + CD = BC + DA

Hence proved

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