CBSE Class 10 Previous Question Paper (2015) > Maths and Solution
CBSE Class 10 Maths Previous Year Question Paper 2015: All Set Previous Year Question Paper PDF Class 10 Maths 2015 with Solution (also name as Marking Scheme) Download from here.
(1) If x = – 1/2 , is a solution of the quadratic equation 3ϰ2 + 2kϰ – 3 = 0, find the value of k.
Solution: -9/4
(2) The tops of two towers of height x and y, standing on level ground, subtend angles of 30° and 60° respectively at the centre of the line joining their feet, then find x:y
Solution: 1:3
(3) A letter of English alphabet is chosen at random. Determine the Probability that the chosen letter is a consonant.
Solution: 21/26
(4) In Fig. 1, PA and PB are tangents to the circle with center O such that <ABP = 50° . write the measure of <OAB.
Solution: 25°
(5) In Fig. 2, AB is the diameter of a circle with center O and AT is a tangent. If <AOQ = 58° , find <ATQ.
Solution: <ABQ = 1/2 <AOQ = 29°
<ATQ = 180° – (<ABQ+<BAT) = 180° – 119° = 61°
(6) Solve the following quadratic equation for x:
Solution:The given quadratic equation can be written as
(4x2 – 4a2 x + a2) – b4 = 0
Or(2x – a2)2 – (b2)2 = 0
\ (2x – a2 + b2) (2x – a2 – b2)2 = 0
Þ x = a2 – b2/2, a2 + b2/2
(7) From a point T outside a circle of center o, tangents TP and TQ are drawn to the circle. prove that OT is the right bisector of line Segment PQ.
Solution:
In ∆s’ TPC and TQC
TP= TQ
TC=TC
<1= <2 (TP and TQ are equally
inclined to OT)
∴ Δ TPC ≅ Δ TQC
∴ PC = QC and <3 = <4
But <3 + <4 = 180° = <3 = <4 = 90°
∴ OT is the right bisector of PQ
(8) Find the middle term of the A.P. 6, 13, 20, — , 216.
Solution:
The given A.P. is 6, 13, 20, —, 216
Let n be the number of terms, d = 7, a = 6
∴ 216 = 6 + (n – 1) .7 n = 31
∴ Middle term is 16th
∴ a 16= 6 + 15 × 7 = 111
(9) If A (5,2), B (2,-2) and C (-2.t) are the vertices of a right angle triangle with <B = 90°,
then find the value of t.
Solution:
∴ ABC is right triangle
∴ AC2 = BC2 + AB2
AB = (5 – 2)2 + (2 + 2)2 = 25 = AB = 5
Bc = (2+2)2 + (t+2)2 = 16 + (t+2)2
AC2 = (5+2)2 + (2 – t)2 = 49+ (2-t)2
∴ 49+ (2-t)2 = 41 + (t +2)2
(t+2) – (2-t) = 8
4 x 2t = 8 ⇒ t = 1
(10) Find the ratio in which the point p (3/4, 5/12) divides the line segment joining the points A (1/2, 3/2) and B (2, -5).
Solution:
Let P divide AB in the ratio of k : 1
∴ 2k+1/2/k+1 = 3/4 ⇒ 8k + 2 = 3k + 3
⇒ k = 1/5
∴ Required ratio = 1 : 5
(11) Find the area of area of triangle ABC with A (1,-4) and mid points of sides through a being (2,-1) and(0,-1)
Solution:
P is the mid-point of AB
∴ x + 1 = 4 ⇒ x = 3
similarly y = 2 ⇒ B (3, 2)
similarly finding C (–1, 2)
∴ Area Δ ABC = 1/2 [1(2-2) +3 (2+4) -1 (-4 -2) ] = 1/2 = 24 = 12 sq. u.
(12) Find that non – zero value of k, for which the quadratic equation kx +1 -2(k-1) x+x = 0 has equal roots. Hence find the roots o0f the equation.
Solution: The given quadratic eqn. can be written as
(k + 1)x2 – 2(k – 1)x + 1 = 0
For qual roots 4(k-1) -4 (k + 1) = 0 k 2 – 3k 0
∴ Non-zero value of k = 3 : Roots are 1/2, 1/2
(13) The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 45° . If the tower is 30 m high, find the height of the building.
Solution:
Fiqure
(i) 30/y = tan 45° = 1 ⇒ y = 30
(ii) x/y tan = 30° = 1/√3 ⇒ x = √y/3 = 30/√3 = 10 √3
∴ Height of building is √10 3 m
(14) Two different dice are rolled together. Find the probability of getting:
Solution: Total possible out comes = 36
(i) The possible outcomes are (2, 3), (3, 2), (1, 4), (4, 1) : Number : 4
∴ Required Probability = 4/36 = 1/9
(ii) The possible outcomes are
(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)
their number is 9
∴ Required Probability = 9/36 = 1/
(15) If S denotes the sum of first n terms of an A. P., prove that s = 3 (S – S).
Solution: Let a be the first term and d the common difference
S12 = 6 [2a + 11d] 12a + 66d
S8 = 4 [2a + 7d] = 8a + 28d
S4 = 2 [2a + 3d] = 4a + 6d
3 (S8 – 34) = 3 (4a + 22d) = 12a + 66d = S12
(16) In Fig. 3, APB and AQO are semicircles, and Ao = OB. If the perimeter of the figure is 40 cm, find the area of the shaded region. [Use π = 22/7]
Solution: Let OA = OB = r
∴ 22/7 x r/2 + 22/7 x r + r ⇒ 280 = 40r
r = 7
∴ shaded area = (1/2 X 22/7 X 7/2 X 7/2 X + 1/2 X 22/7 X 7 X 7) cm2
= (77 x 5/4) or 385 Cm2 = 96 ¼ cm2
(17) In Fig. 4, from the top of a solid cone of height 12 cm and base radius 6 cm, a cone of height 4 cm is removed by a plane parallel to the base. Find the total surface area of the remaining solid. (Use π = 22/7 and √5 = 2.236)
Solution:
D ADC ~ D ARQ
\ x/6 = 4/12 Þ x = 2
QC = √82 + 42 = 4 √5
Total surface area of frustum PQCB
= π [(6+2) x 4 √5 + (6) 2+ (2) 2 ]
= 22/7 [32x 2.236+40] = 22/7 (111.552) = 22x 15.936
=350.592
(18) A solid wooden toy is in the form of a hemisphere surmounted by a cone of same radius. The radius of hemisphere is 3.5 cm and the total wood used in the making of toy is 166 5/6 cm3. Find the height of the toy. Also, find the cost of painting the hemispherical part of the toy at the rate of ₹ 10 per cm2. [Use π =22/7]
Solution:
Volume of solid wooden toy
166 5/6 = 2/3 x 22/7 x 7/2 x 7/2 + 1/3 x 22/7 x 7/2 x 7/2 x h
Or 1001/6 = 22/7 x 7/2 x 7/2x [7+h]
Þ 7 + h = 1000×1/22×7 = 13 Þ h= 6 cm
Area of hemispherical part of toy = (2x 22/7 x 7/2 x 7/2) cm2
=77 cm2
Cost of Paenting = Rs\ (77 x 10) = Rs. 770
(19) In Fig. 5, from a cuboidal solid metallic block, of dimensions 15 cm x 10 cm x 5 cm, a cylindrical hole of diameter 7 cm is drilled out. Find the surface area of the remaining block.[ Use π =22/7]
Solution:
Total surfacearea of solid cuboidal block
= 2 ( 15 x 10 + 10 x 5 + 15 x 5) cm2
Area of curved surface of cylinder = 2 hrp = 2 x 22/7 x 7/2 x 5 = 110 cm2
Reqd – area = (550+110 – 77) cm2 = 583 cm2
(20) In Fig. 6, find the area of shaded region [Use π = 3.14]
Solution:
Area of Sq. ABCD = 142 or 196 cm2
Area of small Sq. = 42 or 16 cm2
Area of 4 semi circles = [4.1/23.14(2)2] cm2
=25.12 cm2
(21) The numerator of a fraction is 3 less then its denominator. If 2 is added to both the numerator and the denominator, thyen the sum of the new fraction and orginal fraction is 29/20. Find the original fraction.
Solution: Let the fraction be x-3/x
By the given condition, new fraction x-3+2/x+2 = x-1/x+2
\ x-3/x + x-1/x+2 = 29/20
Þ 20 [(x-3) (x-2) +x (x-1)] = 29 (x2 +2x)
= 20(x2 – x – 6 + x2 –x) = 29x2 + 58x
Or 11x2 – 98x – 120 = 0
Or 11x2 – 110x – 12x – 120 = 0
(11x + 12) (x-10) = 0 Þ x = 10
\ The Fraction is 10/ 7
(22) Ramkali required ₹ 2500 after 12 weeks to send her daughter to school. She saved ₹ 100 in the first week and increased her weekly saving by ₹ 20 every week. Find weather she will be able to send her daughter to school after 12 weeks. What value is generated in the above situation?
Solution: Money required for Ramkate for admission of daughter = Rs. 2500
A.P. formed by saving
(i) = 100, 120, 140, — upto 12 terms
Sum of AP (i) = 12/2 [ 2x 100+ 11×20] =6[420]
= Rs. 2520
\ She can get her doughter admitied
Value : Small saving can fulfill your big desires or any else
(23) Solve for x:
2/x+1 + 3/2(x-2)= 23/5x, x ≠ 0, – 1,2
Solution:
2/x+1 + 3/2(x-2) = 23/5x
Or 5x [4(x-2)+ 3x + 3]= 46(x+1) (x-2)
5x(7x-5)=46(x2 – x -2) Þ 11x2 -21x – 92 = 0
Þ x = 21±√441 + 4048/22 = 21± 67/22
= 4, -23/11
(24) Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Solution:
Correctly stated
Given, to Prove, Construction and correct figure
correct Proof
(25) In Fig. 7, tangents PQ and PR are drawn from an external point p to a circle with centre O, such that <RPQ = 30°. A chord RS is drawn parallel to the tangent PQ. Find <RQS.
Solution:
PQ =PR Þ <PRQ = <PQR = (180-30) °/ 2 = 75°
SR | | QP and QR is a transversal Þ <SRQ = 75°
\ <ORQ = <RQO =90° – 75° = 15°
\ <QOR =(180 – 2 X 15) ° = 150 ° Þ <QSR = 75°
< RQS = 180° – (<SRQ + <SQR) = 30°
(26) Construct a triangle ABC with BC = 7 cm, <B = 60° and AB = 6 cm. Construct another triangle whose sides are ¾ times the corresponding sides of D ABC.
Solution:
Correctly drawn D ABC
Correctly drawn a triangle similar to D ABC of given scale factor
(27) From a point P on the ground the angle of elevation of the top of a tower is 30° and that of the top of a fixed on the top of the tower, is 60°. If the length of the flag staff is 5 m, find the height of the tower.
Solution:
Writing the trigonometric equations
(i) x/y = tan 30° = 1/√3 Þ y= √3 x
(ii) x+5/y = tan 60° = √3 or x+5/√3x = √3
Þ 3x = x + 5
Or x = 2.5
\ Height of tower = 2.5 m
(28) A box contains 20 cards numbered from 1 to 20. A card is drawn at random from the box. Find the probability that the number on the drawn card is
(i) divisible by 2 or 3
(ii) a prime number
Solution:
(i) Numbers divisible by 2 or 3 from 1 to 20 are
2, 4, 6, 8, 10, 12, 14, 16, 18, 3, 9, 15 Their number is 13
\ Required Probability = 13/20
(ii) Prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19 : 8 in number
\ Required Probability = 8/20 or 2/5
(29) If A(-4, 8), B(-3, -4), C(0,-5) and D(5,6) are the vertices of a quadrilateral ABCD, find its area.
Solution:
ABC DArea
= ½ [-4(-4 + 5) – 3 (-5 – 8) +0 (8+4)]
=1/2 | – 4+39 | = 35/2
= Area of D ACD
=1/2[-4(-5 -6)+0(6-8)+5(8+5)]
=109/2
\ Area of Qurd. ABCD = 35/2+ 109/2 = 72sq.u.
(30) A well of diameter 4 m is dug 14 m deep. The earth taken out is spread evenly all around the well to from a 40 cm high embankment. Find the width of the embankment.
Solution: Volume of earth taken out after digging the well
=(22/7x2x2x14) cu.m = 176 cu.m……….(i)
Let x be the width of embankment formed by using (i)
Volume of embankment = 22/7[(2+x)2 – (2)2 x40/100 = 176
Þ x2 + 4x – 140 = 0 Þ (x + 14) (x – 10) = 0
Þ x = 10
\ Width of embankment = 10 m
(31) Water is flowing at the rate of 2.52 Km/h through a cylindrical pipe into a cylindrical tank, the radius of whose base is 40 cm, If the increase in the level of water in the tank, in half an hour is 3.15 m, find the internal diameter of the pipe.
Solution:
Let x m be the internal radius of the pipe
Radius of base of tank = 40 cm = 2/5 m
Level of water raised in the tank = 3.15 or 315/100
2.52 km/hour Þ 1.26 km in half hour = 1260 m
\ Getting the equation
π . X2 .1260 = π . . 2/5.2/5 x 315/100
Þ X2 = 4/25. 315/100 x 1/1260 = 1/2500
Þ x= 1/50 m = 2 cm
\ Internal diameter of pipe = 4 cm
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