## MATHEMATICS

### Previous year question paper with solutions for MATHEMATICS Mar-2017

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### Question paper 1

(i) For which value of p does the pair of equation given below has unique solution

4x + 2y + 1 = 0

3px - 3y + 4 = 0

Answer:

4x + 2y + 1 = 0

3px - 3y + 4 = 0

Here a

_{1}=4 , b_{1}=2 , c_{1}=1 , a_{2}=3p , b_{2}=-3 , c_{2}=4For Unique Solution

\({a_1 \over a_2} = {b_1 \over b_2} = {4 \over 3p} = {2\over -3}\)

P Not equal to 2

given equation have unique solution for all value of P other then 2

(ii) For AP: -5,- 1.3,7 ..........write first term 'a'and comrnon difference 'd'.

Answer:

AP: -5,- 1.3,7 ......

first term is -5

and common difference is = -1-(-5) = 4

(iii) Write the definition of pythagoras theorem.

Answer:

If Trangle ABC is a Right angle triangle Having right angle at B then AC

^{2}= BC^{2}+ AB^{2}(iv) Fill in thc blank :

sect

^{2}0 ........... = 1Answer:

Tan

_{2}Q(v) Find the surface area of sphere radius is 3 crn.

Answer:

Surface area of sphere with radius r is \(4 \pi r^2\)

here r = 3 cm

surface area is = \(36 \pi\)

2. Find H.C.F. and L.C.M . of 26 and 91 .

Answer:

HCF of 26 and 91

26 = 2 * 13 , 91 = 13 * 7

HCF = 13

And LCM = 2 * 13 * 7

LCM is = 182

3. Find the Zeroes of the quadratic polynomial 4x

^{2}+ 8xAnswer:

4x

^{2}+ 8x = 04x(x + 2) = 0

x= 0, x=-2

4. Find the nature ofthe roots of euadratic equation 2x

^{2}- 6 x + 3 = 0.Answer:

Given equation is 2x

^{2}- 6 x + 3 = 0.Discreminent D = b

^{2}- 4acHere a= 2 b = -6 c = 3

D = (-6)

^{2}-4(2)(3)36 - 24 = 12 > 0

Roots are Real

5. For AP :21,18, 15, ........ check which term is -27 ?

or

Find the sumof 12 terms of the AP: - 37,-33,-29,............

Answer:

- 37,-33,-29,............

n = 12

a = -37, d = -33 - (-37) = -33 + -37

d = 4

sum of AP S

_{n}= \({n \over 2} [2a + (n-1) d]\)= \({12 \over 2} [2(-37) + (12-1) 4] \)

= 6 [-74 + 44] = 6 [-30] = -180

6. Prove that the length of a tangent drawn from an external point to a circle are equal.

or

The length of tangent from point T at distance 10 cm from the centre ofthe circle is 6 cm. Find the radius ofthe circle.

Answer:

Given a circle with center O and point A laying outside Circle and two Tangents AB, AC

we have to pove AB =AC

For this join OA ,OB and OC

angle ABO and angle ACO are of 90

^{0}['.' The tangents at any point of circle is perpendicular to the radius through the point of contact ]

OB = OC

OA = OA

trangle OBA = trangle OCA

AB = AC

7. Draw a line segment of length 7 cm and divide it in the ratio 2:3 .

or

Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at 60

^{0}.Answer:

8. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting (i) a king of red colour

(ii) the queen of diamonds.

Answer:

Total cards = 52

(i) No of King in Red color = 2

Probability of king = 2/52 = 1/26

(ii) No of Queen in Diamond = 1

Probability of Queen = 1/52

9. Solve by cross multiplication method ofthe following linear equation :

8x+5y:9

3x+2y:4

Answer:

8x+5y = 9

3x+2y = 4

8x+5y - 9 = 0

3x+2y - 4 = 0

a

_{1}=8 , b_{1}=5 , c_{1}=-9 , a_{2}=3 , b_{2}=2 , c_{2}= -4Cross multiplication methord

\({x \over (5 * 4) - (2* (-9))} = {y \over (-9 * 3) - (-8 *4)} {1 \over (8*2) - (5 * 3)}\)

\({x \over 38} = {y \over -59} ={ 1 \over 1}\)

\({x \over 38 } = 1 , {y \over -59 } =1\)

x = 38 , y = -59

10. Find the ratio in which the line segment joining the points (- 3, 10) and (6, - 8) is divided by (- 1,6)

Answer:

11. Prove that:

\({cosA - cosA\over cosA + cosA} = {cosesA - 1\over cosesA + 1}\)

Answer:

\({cosA - cosA\over cosA + cosA} \) = \({{cosA\over sin A }- cosA }\over{ {cosA\over sin A } + cosA} \)

\({cos A - cos A sin A\over sin A} \over {cos A + cos A sin A\over sin A} \) = \({cos A - cos A sin A\over cos A + cos A sinA}\)

\(cos A (1-sin A)\over cos A (1+sin A)\) = \(1 - sin A\over 1+ sinA\)

Divide through out by sinA

\({1 \over sinA} - {sinA \over sin A} \over {1 \over sinA} +{sinA \over sin A}\) = \(cosec A - 1 \over cosec A+ 1 \) = RHS

12. The angle of elevation of the top of a tower from a point on the ground which is 30 m away fror the foot ofthe tower is 30

^{0}. Find the height of the tower.Answer:

Let BC be a lower of height h . A be a point which is 30m away from foot of tower B

angle BAC = 30

^{0}Tan Q = \({p \over b} = { h \over 30}\)

Tan 30 = \({ h \over 30}\)

\({1 \over \sqrt 3} = { h \over 30}\)

h = \({30 \over \sqrt 3} * {\sqrt 3 \over \sqrt 3} = 10 \sqrt 3\)

Height of tower is \(10 \sqrt 3\)

13. In the figure given below a quadrilateral PQRS is drawn to circumscribe a circle. Prove that : PQ +RS:PS + QR

Answer:

14. In a circle ofradius 14 cm, an arc subtends an angle of 60

^{0}at the centre. Find(i) The length ofthe arc

(ii) Area of sector.

Answer:

Area of sector = \({\pi \over 360} r^2Q\)

= \({\pi \over 360} * 14^2 * 60 = {\pi \over 6} *196 = {98 \pi \over 3}\)

or

Area of shaded reagin is given by (Area of AOB) + (Area of circle) - (Area of sector OCD) --------(i)

Area of Equlaterial Triangle AOB with side 10 cm

= \({\sqrt 3 \over 4 }(side)^2 = {\sqrt 3 \over 4 } * 10^2 = {\sqrt 3 \over 4 } *100 = {50\sqrt 3 \over 2 }\)

Area of circle with Radius 5 = \(\pi r^2 = \pi 5^2 = 25 \pi\)

Area of sector = \({\pi \over 360} r^2 Q = {\pi \over 360} r^2 Q \)

here r = 5 , Q = 60 [Equlatrial trangle]

Area of sector = \({\pi \over 360} 5^2 60^0 = {25 \pi \over 6}\)

put all these values in equation (i)

Area of shaded position = \({50 \sqrt 3 \over 2 } - {125 \pi \over 6} \)

16. In a triangle if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

or

BL and CM are the medians of a triangle ABC right angled at A Prove that:

4(BL

^{2}+CM^{2})=5 BC^{2}Answer:

Given that in a trangle ABC

\(AC^2 = AB^2 + BC^2\) ------(i)

we have to prove that angle B = 90

^{0}For this , we construct a Trangle PQR which is right angle at Q such that PQ = AB and QR = BC

In PQR

\(PR^2 = RQ^2 + PQ^2\) [by pythagours theorm]

\(PR^2 = AB^2 + BC^2\)

\(BC^2 = AB^2 + AC^2\) from (1)

\(PR^2 = AC^2\)

PQ = AB , QR = BC , and PR = AC

LB = LQ

Lb = 90

^{0 }A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of Ice cream. The Ice cream is to be filled into cones of height 12 cm and diameter 6 cm having hemispherical shape on the top. Find the number of such cones which can be filled with lce cream.

or

A solid Iron pole consists of a cylinder of height 220 cm and base diameter 24 cm which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 oflronhas 8 gmrnass. (Use n = 3.14)

Answer:

Diameter of cylinder = 12 cm

Radius = 6cm

height 15cm

Volume of Cylinder = \(\pi r^2 h = \pi(6)^2 *15 = \pi * 36 * 35 = 540\pi\)

Now Height of cone = 12 cm

Diameter = 6cm

Radius = 3cm

Volume of Cone = \({1 \over 3 }\pi r^2 h = {1 \over 3} \pi (3)^2 *12 \)

= \(\pi * 3* 12 = 36 \pi\)

No of cones = \(Volume of Clyinder\over Volume of Cone\) \({540 \pi \over 36 \pi} ={ 540\over 36}\) = 15 Cones

### Question paper 2

(i) For which value of k does the pair of equation given below has unique solution ? 2kx+3y+3=0

4x+y+4 = 0

Answer:

2kx+3y+3=0

4x+y+4= 0

Here a

_{1}=2k , b_{1}=3 , c_{1}=-3 , a_{2}=4 , b_{2}=1 , c_{2}=4Condition from unique solution

\({a_1 \over a_2} = {b_1 \over b_2} = {2k \over 4} = {3 \over 1} = k = 6\)

K not equal to 6

(ii) For AP :6, 9,12.15......... Write the first term 'a' and common difference

Answer:

AP :6, 9,12.15........

Here a =6, d = 9-6 = 3

(iii) Write the definition of Thalis theorm

Answer:

Thalis theorm : If ABC are distrinct points on circle where AC is diameter then LABC is Right angle

(iv) Fill inthe blank:

sin

^{2}0+........... =1Answer:

Cos

^{2}Q(v) Find the volume of sphere whose radius is 21 cm

Answer:

Volume of Sphere is = \({4 \over 3} \pi r^3\)

= \({4 \over 3} \pi 21^3\)

= \({4 \over 3}*{ 22 \over 7} * 21 *21 *21\)

38808 cm

^{3}2. If H.C.F. of 306 and 657 is 9 then find L.C.M. of 306 and 657

Answer:

HCF * LCM = Product of Number

9 * LCM = 306 * 657

LCM = \(306 * 657 \over 9 \)

LCM = 34 * 657 = 22338

3. Divide the polynomial p(x) by g(x) and find quotient and remainder

p(x):x

^{4}-5x+6 and g(x)=-x^{2}+2Answer:

4. Find such value of p for quadratic equation p x

^{2}- 3 x + 5 = 0 that has two equal rootsAnswer:

5. For AP:11,8,5,2,......check -151 is which term of the AP?

or

Find the sum of T6 terms of AP -5,+(-8)+(-11)+........+(-230)

Answer:

6. The length of a tangent from point P at distance 13 cm from the centre of the circle is 12 cm. Find the radius ofthe circle.

or

Prove that in two concentric circles the chord of the larger circle, which touches the smaller circle is bisected at the point of contact.

Answer:

7. Draw a line segment of length 4.5 cm and divide it in the ratio 1:2

or

Draw a pair of tangents to a circle of radius 4 cm which are inclined to each other at an angle 60

^{0}.Answer:

8. A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from a lot. What is the probabilitythatthis bulb is defective ?

Answer:

Total Bulb 20

defective = 4

Let Probability of Defective bulb = 4/20 = 1/5

9. A fraction becomes 1/3 when 1 is subtracted frorn numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.

Answer:

Let numerator r = x

denominator = y

fraction = x/y

Acc to Question,

\({ x - 1 \over y} = {1 \over 3} \)

3x - 3 = y

3x - y = 3

again

\({ x \over y +8 } = {1 \over 4}\)

4x = y + 8

4x -y = 8

Solve x and y

-x = -5

3(5) - y = 3

15 -y = 3

y =12

Fraction is 5/12

10. Find the area of quadrilateral whose vertices, taken in order are (- 4,-2), (- 3, - 5), (3,-2) and (2,3).

Answer:

Let ABC,

Area of triangle ABC = \({ 1 \over 2 }[x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2)]\)

= \({ 1 \over 2 }[-4 (-5 + 2) + (-3) (-2 + 2) + 3 (-2 + 5)]\)

= \({1 \over 2} [12 + 0 + 9] = {21 \over 2}\)

Area of triangle ACB = \({ 1 \over 2 }[-4 (-2 - 3) + 3 (3 + 2) + 2 (-2 - 2)] = {35 \over 2}\)

Total Area of Quadilaterial = \({21 \over 2} + {35 \over 2}\) = 28

= 28 Sq unit

12. The angle of elevation of the top of a building from the foot ofthe tower is 30' and the angle of elevation of the top of the tower from the foot ofthe building is 60". Ifthe tower is 50 m high, 2 2 find the height of the building.

Answer:

Let AC = Bulding = h

BD be Tower with hight 50 m

In Triangle ABD

Tan

^{0}60 =\({BD \over AB} = { 50 \over AB}\)\(\sqrt 3 = {50 \over AB}\)

\(AB = {50 \over \sqrt 3} * {\sqrt 3 \over \sqrt 3} = { 50 \sqrt 3 \over 3}\)

In Trangle ABC

Tan 30

^{0 }= \({AC \over AB} = {h \over {50\sqrt 3 \over \sqrt 3 }} ={3h \over 50 \sqrt 3}\)\({1 \over \sqrt 3} = {3h \over 50 \sqrt 3}\)

h = \(50 \over 3\)

14. In a circle ofradius 7 cm,afiarc subtends an angle of60o atthe centre. Find

(i) The length of the arc

(ii) Area of sector.

Answer:

r = 7 cm , Q = 60

^{0}(i) arc length = \({ Q \over 360} * 2 \pi r = {60 \over 360} * 2 * \pi * 7\)

\({14 \over 6} * \pi = {7 \pi \over 3}\)

(ii) Area of Sector = \({ Q \over 360} * \pi r^2 \)

= \({60 \over 360} * \pi * 7 *7 \)

16. Prove that if a line drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the othertwo sidps are divided in the same ratio.

or

The ratio of areas oftwo similar triangles is equal to the squares of the ratio of their corresponding sides.

Answer:

We Given trangle ABC In which a line parllel to BC Intersect two Sides AB and AC at D, E Respectively

To Prove : \({AD \over DB } = {AE \over EC}\)

Construction: Join BE and CD

EN I AB , BM I AC

Now Area (ADE) = 1/2 * AE * DM

Area (DEC) = 1/2 * EC * DM

\({area (ADE) \over area (DEC)} = {{1\over 2} AD * EN \over {1 \over 2} DB * EN } = {AD \over DB } ---(i)\)

\({area (ADE) \over area (DEC)} = {AE \over EC } --- (ii)\)

Now Area (BDE) = Area (DEC) ----(iii)

From i ii and iii

\({AD\over DB } ={AE\over EC}\)

17. A 20 m deep well with diameter 7 m is dug and the soil from digging is evenly spread out to form a platform 22m x 14m.Find the height of the platform.

or

A toy is in the form of a cone of radius 7 cm mounted on a hemisphere of same radius. The total height of the toy is 1 5.5 cm. Find the total surtace area of the toy

Answer:

Depth of well = h = 20m

diameter = 7m

volume of well = \(\pi r^2 h\) = \(\pi * (3.5)^2 * 20\)

Now Length of Platform = 22m

Breadth = 14m

Let hight =H

vol = 12 * 14 * H

\(\pi * (3.5)^2 * 20\) = 12 * 14 * H

H = \({22 \over 7} * 3.5 * 3.5 * 20 \over 14 * 12\)

H = \({22} * 3.5 * 3.5 * 20 \over 7 * 14 * 12\) = \({5390 \over 1092} = 4.94 m\)

Hight of platform = 4.94 m

### Question paper 3

(i) For which value of p does the pair of equation given below has unique solution

4x + py + 8 = 0

2 x + 2y + 2 = 0

Answer:

4x + py + 8 = 0

2 x + 2y + 2 = 0

Here a

_{1 }= 4 b_{1 }= p c_{1 }= 8a

_{2}_{ }= 4 b_{2}_{ }= p c_{2}_{ }= 8condition for unique solution is:

\({a_1 \over a_2 } \notin { 4 \over 2 } \notin {p \over 2} = 2 \notin {p \over 2 } P \notin 2\)

(ii) For AP : \({3\over 2},{1\over 2},{-1\over 2},{-3\over 2}\) ........ write the first term 'a' and cofilmon difference 'd'.

Answer:

Here is first term is 3/2

Common Difference is 1/2 -3/2 = -1

(iii) Write the definition of Pythagoras theorem.

Answer:

Pythagoras theorem : If Triangle ABC is a right angled have 90

^{0 }at B thenAC

^{2 }= BC^{2 }+ AB^{2 }(iv) Fill in the blank :

..........-cot

^{2 }0 = 1Answer:

Cosec

^{2}Q - cot^{2 }0 = 1(v) Find the volume of sphere rvhose radius is 3 cm.

Answer:

Radius of Sphere = 3cm

Volume = \({4 \over 3} \pi r_2\)

= \({4 \over 3} \pi 3\)

=\(4 \pi * 9 = 36\pi \)

2. Find the L.C.M and H.C.F of 6 and 20 by prime factorisation method .

Answer:

6 = 2 * 3

20 = 2 * 5 * 2

HCF = 2

LCM = 2 * 3 * 5 * 2 = 60

3. Check whether the first polynomial is a factor of second polynomial by dividing the second polynomial by first polynomial

t

^{2}- 3,2t + 3 t^{3}-2t^{2}- 9t - 12Answer:

4. Find such value of k for quadratic equation kx

^{2}- 2 kx + 6 = 0 so that they have two equal rootsAnswer:

kx

^{2}- 2 kx + 6 = 0Here a=k, b= -2k, c= 6

D = B

^{2}-4ac = (2k)^{2}- 4.k.6= 4k

^{2}-24kFor Equal Roots D =0

4k

^{2}-24k = 04k(k-6) = 0

k=0 , or k =6

but k not equal to Zero

K = 6

5. Show that a

_{1 },a_{2 , }a_{3}.........a_{n}form AP where a_{n }: 3+4_{n}_{or}For AP : 7, 11 , 15 , 19 , ...... Find the sum of first 15 terms.

Answer:

a

_{n}= 3 + 4n Put n = 1 , 2 ,3 , 4 ...........we get , a

_{1 }= 3 + 4(1) = 7a2

_{ }= 3 + 4(2) = 11a3 = 3 + 4(3) = 15

7 , 11, 15, 19 ............ is an AP

a = 7, d = 4 given n = 15

S

_{n }= \({n \over 2} [ 2a + (n-1)d ] = {15\over 2} [ 2(2) + (15-1)4 ] = {15\over 2} [(14 + 56) ] = {15\over 2} [70] ] = 15 * 35 = 525 \)6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius ofthe circle.

or

Two concentric circles are of radii 5 cm and 3 cm. Find the length of chord of the larger circle which touches the smaller circle.

Answer:

7. Draw a line segment of length 6.5 cm and divide it in the ratio3:4

or

Draw a circle of radius 4 cm. From a point 7 cm away from a centre construct the pair of Tangent to the circle and measure its length

Answer:

8. A bag contains 3 red and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is

(i) red (ii) not red.

Answer:

Red Ball = 3

Black Ball = 5

Total = 8

PRobability of Red Ball = 3/8

Probability of not Red = 1 - 3/8 = 5/8 Answer

9. The cost of I pencil and 3 erasers is rupess 10 and the cost of 4 pencils and 6 erasers is rupess 28. Find the cost of 5 Pencils and 4 erasers.

Answer:

Let cost of 1 Pencil = x

cost of 1 Eraser = y

x + 3y = 10

4x + 6y = 28

6y = 12

y = 2

x + 3(2) = 10

x = 10 - 6 = x= 4

Cost of 1 pencil = 4 Rupees

cost of 1 eraser = 2 Rupees

Cost of 5 pencils = 4 * 5 = 20

10. Find the co-ordinates of the points of trisection (i.e. point dividing the three equal parts) of the segment joining the points A(2,-2) and B(-7,4)

Answer:

If tan (a+n) = \({\sqrt{3}}\) and tan (A-B) = \({1\over\sqrt{3}}\); 0

^{0}< A + B <= 90^{0}; A > B then find the value of A and B.or

Prove that:

\({\sqrt{1 + sin A \over 1 - sin A}}\) = sec A + tan A; A< 90

^{0}Answer:

Tan (A + B) = \(\sqrt 3\)

Tan (A - B) = 1/\(\sqrt 3\)

Tan (A + B) = Tan 60

^{0}Tan (A - B) = Tan 30

^{0}A + B = 60

A - B = 30

2A = 90

A = 45

A + B = 60

B = 60 - A

= 60 - 45

B = 15

^{0}or

\( \sqrt{ 1 + Sin A \over 1 + Sin A } = \sqrt{ {1 + Sin A \over 1 - Sin A} * {1 + Sin A \over 1 + Sin A} } \)

\({1 + SinA \over \sqrt {Cosec^2A}} = {1+ SinA \over cos A} = {1 \over cos A} + {SinA \over cos A }\)

= SecA + TanA

12. Two poles of equal heights are standing opposite each other on either side of the road which is g0 m rvide. From a point betrveen them on the road the angles of elevation of top of the poles are 60o and 30o. Find the height of the poles and distance of the point from the pole

Answer:

14. In a circle of radius 2l cm,an arc subtends an angle of 60

^{0}at the centre. Find(i) The length of the arc

(ii) Area of sector

Answer:

Radius of circle = 21 cm

Q = 60

^{0}r = 21cm

(i) Q = l/r l =rQ = 21 * 60

^{0}ar length = 21 * \({\pi \over 3} = 7 \pi\)

(ii) Area of Sector = \({Q \over 360} * \pi r^2 = { 60 \over 360} * \pi * 21* 21 = {441\pi \over 6}\)

Prove that the ratio of the areas of two similar triangle is equal to the squares of the ratio of their corresponding sides.

or

Prove that in a right triangle, the square ofthe hypotenuse is equal to the sum of the squares of the othertwo sides.

Answer: