Category: Maths

Maths

SET-I

Section – A

Q.1. If the quadratic equation px²– 2√5px + 15 = 0 has two equal roots, then find the value of p.

Answer. The given quadratic equation can be written as px²– 2√5px + 15 = 0

a = p, b = -2√5p, c = 15

For equal roots, D = 0

D = b² – 4ac

0 = (– 2√5p)² – 4 ×p × 15

0 = 4 ×5p² – 60p

0 = 20p² – 60p

p = 60p / 20p = 3 

∴ p = 3

 

Q.2. In Figure 1, a tower AB is 20 m high and BC, its shadow on the ground, is 20-√3 m long. Find the Sun’s altitude.

Answer.

 

Q.3. Two different dices are tossed together. Find the probability that the product of the two numbers on the top of the dice is 6.

Answer. Total outcomes = 6ⁿ = 6² = 36

Possible outcomes having the product of the two numbers on the top of the dice as 6 are (3 × 2, 2 × 3, 6 × 1, 1 × 6), i.e., 4

P(Product of two numbers is 6) = 4/36 = 1/9 

Q.4. In Figure 2, PQ is a chord of a circle with centre O and PT is a tangent. If ∠QPT = 60°, find ∠PRQ.

Answer.

Section – B

Q.5. In Figure 3, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ = 120°, then prove that OR = PR + RQ.

Answer.

 

Q.6. In Figure 4, a triangle ABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are respectively of lengths 6 cm and 9 cm. If the area of ∆ABC is 54 cm2, then find the lengths of sides AB and AC.

Answer.

 

Q.7. Solve the following quadratic equation for x: 4x² + 4bx -(a² – b²) = 0

Answer.

 

Q.8. In an AP, if S₅+ S7 = 167 and S₁₀ = 235, then find the AP, where S_n denotes the sum of its first n terms.

Answer.

 

Q.9. The points A(4,7), B(p,3) and C(7,3) are the vertices of a right triangle, right-angled at B. Find the value of p.

Answer.

 

Q.10. Find the relation between x and y if the points A(x, y), B(-5, 7) and C(-4, 5) are collinear.

Answer.

Section – C

Q.11. The 14th term of an AP is twice its 8th term. If its 6th term is -8, then find the sum of its first 20 terms.

Answer.

 

Q.12. Solve for x:

Answer.

 

Q.13. The angle of elevation of an aeroplane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the aeroplane is flying at a constant height of 1500 √3 m, find the speed of the plane in km/hr.

Answer.

 

Q.14. If the coordinates of points A and B are (-2, -2) and (2, -4) respectively, find the coordinates of P such that AP = 3/5 AB, where P lies on the line segment AB.

Answer.

 

Q.15. The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is 1/4. The probability of selecting a blue ball at random from the same jar is 1/3 . If the jar contains 10 orange balls, find the total number of balls in the jar.

Answer.

 

Q.16. Find the area of the minor segment of a circle of radius 14 cm, when its central angle is 60°. Also find the area of the corresponding major segment. [Use π = 22/7 ]

Answer.

 

Q.17. Due to sudden floods, some welfare associations jointly requested the government to get 100 tents fixed immediately and offered to contribute 50% of the cost. If the lower part of each tent is of the form of a cylinder of diameter 4.2 m and height 4 m with the conical upper part of same diameter but of height 2.8 m, and the canvas to be used costs Rs 100 per sq. m, find the amount, the associations will have to pay. [Use π = 22/7 ] What values are shown by these associations?

Answer.

 

Q.18. A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of each bottle, if 10% liquid is wasted in this transfer.

Answer.

 

Q.19. A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of Rs 5 per 100 sq. cm. [Use π = 3.14]

Answer. Let the side of cuboidal block (a) = 10cm

Let the  radius of hemisphere be r

Side of cube = Diameter of hemisphere

Largest possible diameter of hemisphere = 10cm

 

Q.20. 504 cones, each of diameter 3.5 cm and height 3 cm, are melted and recast into a metallic sphere. Find the diameter of the sphere and hence find its surface area. [Use π = 22/7 ]

Answer.

Section – D

Q.21. The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the lengths of the sides of the field.

Answer.

Q.22. Find the 60th term of the AP 8,10,12,…, if it has a total of 60 terms and hence find the sum of its last 10 terms.

Answer.

Q.23. A train travels at a certain average speed for a distance of 54 km and then travels a . distance of 63 km at an average speed of 6 km/h more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed?

Answer.

Q.24. Prove that the lengths of the tangents drawn from an external point to a circle are equal.

Answer. Given : Let circle be with centre O and P be a point outside circle PQ and PR are two tangents to circle intersecting at point Q and R respectively

To Prove : Lengths of tangents are equal i.e. PQ = PR

Construction:  Join OQ, OR and OP

Proof: As PQ is a tangent OQ⊥PQ [Tangent at any point of circle is perpendicular to the radius through point of contact]

So, ∠OQP = 90°

Hence ΔOQP is right triangle.

 

Q.25. Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

Answer.

 

Q.26. Construct a ∆ABC in which AB = 6 cm, ∠A = 30° and ∠B = 60°. Construct another ∆AB’C’ similar to ∆ABC with base AB’ = 8 cm.

Answer.

 

Q.27. At a point A, 20 metres above the level of water in a lake, the angle of elevation of a cloud is 30°. The angle of depression of the reflection of the cloud in the lake, at A is 60°. Find the distance of the cloud from A.

Answer.

 

Q.28. A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is

(i) a card of spade or an ace. (ii) a black king.

(iii) neither a jack nor a king. (iv) either a king or a queen.

Answer.

 

Q.29. Find the values of k so. that the area of the triangle with vertices (1, -1), (-4, 2k) and (-k, -5) is 24 sq. units.

Answer.

 

Q.30. In Figure 5, PQRS is a square lawn with side PQ = 42 metres. Two circular flower beds are there on the sides PS and QR with centre at O, the intersection of its diagonals. Find the total area of the two flower beds (shaded parts).

Answer.

Q.31. From each end of a solid metal cylinder, metal was scooped out in hemispherical form of same diameter. The height of the cylinder is 10 cm and its base is of radius 4.2 cm. The rest of the cylinder is melted and converted into a cylindrical wire of 1.4 cm thickness. Find the length of the wire. [Use π = 22/7 ]

Answer.

SET II

Q.10. If A(4, 3), B(-l, y) and C(3, 4) are the vertices of a right triangle ABC, right-angled at A, then find the value of y.

Answer.

Q.18. All the vertices of a rhombus lie on a circle. Find the area of the rhombus, if the area of the circle is 1256 cm2. [Use π= 3.14]

Answer.

Q.19. Solve for x:

Answer.

Q.20. The 16th term of an AP is five times its third term. If its 10th term is 41, then find the sum of its first fifteen terms.

Answer.

Q.28. A bus travels at a certain average speed for a distance of 75 km and then travels a distance of 90 km at an average speed of 10 km/h more than the first speed. If it takes 3 hours to complete the total journey, find its first speed.

Answer.

Q.29. Prove that the tangent at any point of a circle is perpendicular to the radius through the I point of contact.

Answer. Given : Let circle be with centre O and P be a point outside circle PQ and PR are two tangents to circle intersecting at point Q and R respectively

To Prove : Lengths of tangents are equal i.e. PQ = PR

Construction:  Join OQ, OR and OP

Proof: As PQ is a tangent OQ⊥PQ [Tangent at any point of circle is perpendicular to the radius through point of contact]

So, ∠OQP = 90°

Hence ΔOQP is right triangle

Q.30. Construct a right triangle ABC with AB = 6 cm, BC = 8 cm and ∠B = 90°. Draw BD, the perpendicular from B on AC. Draw the circle through B, C and D and construct the tangents from A to this circle.

Answer. 

Q.31. Find the values of k so that the area of the triangle with vertices (k + 1, 1), (4, -3) and (7, -k) is 6 sq. units.

Answer.A(k + 1, 1), B(4, -3) and C(7, -k)

Area of ΔABC = ½ [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]

6 = ½ [(k+1)(-3 + k) + 4(-k-1) + 7(1+3)]

12 = [-3k + k² -3 + k-4k-4 + 28]

12 = [ k² -6k + 21]

⇒  k² -6k + 21-12    ⇒  k² -6k + 9

⇒ k² -3k -3k + 9      ⇒ k(k-3)-3(k-3) = 0

⇒ k-3 = 0 ⇒ k-3 = 0

⇒ k = 3 ⇒ k = 3

Solving get k = 3

SET III

Q.10. Solve the following quadratic equation for x:  x² – 2ax – (4b² – a²) = 0

Answer.

Q.18. The 13th term of an AP is four times its 3rd term. If its fifth term is 16, then find the sum of its first ten terms.

Answer.

Q.19. Find the coordinates of a point P on the line segment joining A(1, 2) and B(6, 7) such that AP=2/5 AB.

Answer.

Q.20. A bag contains, white, black and red balls only. A ball is drawn at random from the bag. If the probability of getting a white ball is 3/10 and that of a black ball is 2/5, then find the probability of getting a red ball. If the bag contains 20 black balls, then find the total number of balls in the bag.

Answer.

Q.28. A truck covers a distance of 150 km at a certain average speed and then covers another 200 km at an average speed which is 20 km per hour more than the first speed. If the truck covers the total distance in 5 hours, find the first speed of the truck.

Answer.

Q .29. Arithmetic Progressions, 12,19,… has 50 terms. Find its last term. Hence find the sum of its last 15 terms.

Answer.

Q.30. Construct a triangle ABC in which AB = 5 cm, BC = 6 cm and ∠ABC = 60°. Now construct another triangle whose sides are 5/7 times the corresponding sides of ∆ABC.

Answer. 

Q.31. Find the values of k for which the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear.

Answer.

Maths

SET-I

Section – A

Q.1. In Fig. 1, PQ is a tangent at a point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°. Find ∠PCA.

Answer.

∠ACB = 90°            …………..[Angle in the semi-circle

In ΔABC, ∠CAB + ∠ACB + ∠CBA = 180°

30° + 90° + ∠CBA = 180°

∠CBA = 180° – 30° – 90° = 60° [Angle-sum-property of a Δ]

∠PCA = ∠CBA       ………….[Angle in the alternate Segment]

∴ ∠PCA = 60°

 

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

Answer. As we know, a₂ – a₁ = a₃ -a₂

2k -1-(k+9) = 2k +7 – (2k -1)

2k -1- k – 9 = 2k +7 – 2k + 1

k – 10 = 8 

∴ k = 8 + 10 = 18

 

Q 3. 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.

Answer.

Let AC be the ladder

Cos60° = AB/AC

½ = 2.5/AC

∴  Length of ladder, AC = 5cm

 

Q. 4. A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen.

Answer.

S = 52

P (neither a red card nor a queen)

= 1 – P(red card or a queen)

= 1- [(26+4-2)/52]  [red cards = 26, Queen = 4, Red queen = 2]

= 1 – 28/52 = 24/52 = 6/13

 

Section-B

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

Answer. 2x² + px – 15 = 0

Since (-5) is a root of the given equation

∴ 2(-5)² + p(-5) – 15 = 0

=  2(25) – 5p – 15 = 0

=  50 – 15 = 5p 

=  35 = 5p

=  p = 7 ——(i)

     p(x²+x) + k   ⇨ px² + px + k = 0

Here, a = p, b = p, c = k

D = 0                (Roots are equal)

       b² – 4ac = 0      , (p)² – 4(p)k = 0

(7)² – 4(7)k = 0

49 – 28k = 0

∴ k = 49/28 = 7/4

 

Q. 6. 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.

Answer.

 

Q. 7. 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.

Answer.

AP = AS

BP = BQ

CR = CQ

DR = DS

[∴ Tangents drawn from an external point are equal in length]

By adding (i) to (iv)

(AP + BP) + (CR + DR) = AS + BQ + CQ + DS

AB + CD = (BQ + CQ) + (AS + DS)

∴ AB + CD = BC + AD (Hence Proved)

 

Q. 8. Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right angled isosceles triangle.

Answer.

 

Q. 9. The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.

Answer. Let 1st term = a,  Common difference = d

a₄ = 0 ⇒ a + 3d ⇒ a = -3d          …………(i)

a₂₅ = a + 24d ⇒ -3d + 24d = 21d ….[From (i)

3(a₁₁) = 3(a + 10d) ⇒ 3(-3d + 10d) =21d ….[From (i)

From above, a₂₅ = 3(a₁₁) (Hence proved)

 

Q. 10. In Fig. 3, 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°.

Answer.

Section – C

Q 11. In Fig. 4, O is the centre of a circle such that diameter AB = 13 cm and AC = 12 cm. BC is joined. Find the area of the shaded region. (Take π = 3.14)

Answer.

 

Q 12. In Fig. 5, a tent is in the shape of a cylinder surmounted by a conical top of same diameter. If the height and diameter of cylindrical part are 2.1 m and 3 m respectively and the slant height of conical part is 2.8 m, find the cost of canvas needed to make the tent if the canvas is available at the rate of Rs 500/sq. metre. (Use π = 22/7 )

Answer.

 

Q 13. If the point P(x, y) is equidistant from the points A (a + b,b – a) and B(a -b,a + b), prove that bx = ay.

Answer.

 

Q 14. In Fig. 6, find the area of the shaded region, enclosed between two concentric circles of radii 7 cm and 14 cm where ∠AOC = 40°. (Use π= 22/7 )

Answer.

 

Q 15. If the ratio of the sum of first n terms of two A.P’s is (7n + 1) : (4n + 27), find the ratio of their mth terms.

Answer.

 

Q 16. Solve for x:

Answer.

 

Q 17. A conical vessel, with bash radius 5 cm and height 24 cm, is full of water. This water is emptied into a cylindrical vessel of base radius 10 cm. Find the height to which the water will rise in the cylindrical vessel. (Use π= 22/7)

Answer.

 

Q 18. A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by 3 (5/9) cm. Find the diameter of the cylindrical vessel.

Answer.

 

Q 19. A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Find the distance of the hill from the ship and the height of the hill.

Answer.

 

Q 20. Three different coins are tossed together. Find the probability of getting (i) exactly two heads (ii) at least two heads (ii) at least two tails.

Answer.

Section  – D

Q 21. Due to heavy floods in a State, thousands were rendered homeless. 50 schools collectively offered to the State Government to provide place and the canvas for 1,500 tents to be fixed by the Government and decided to share the whole expenditure equally. The lower part of each tent is cylindrical of base radius 2.8 m and height 3.5 m, with conical upper part of same base radius but of height 2.1 m. If the canvas used to make the tents costs Rs 120 per sq. m, find the amount shared by each school to set up the tents. What value is generated by the above problem? (Use π= 22/7)

Answer.

 

Q 22. Prove that the lengths of the tangents drawn from an external point to a circle are equal.

Answer. Given : Let circle be with centre O and P be a point outside circle PQ and PR are two tangents to circle intersecting at point Q and R respectively

To Prove : Lengths of tangents are equal i.e. PQ = PR

Construction:  Join OQ, OR and OP

Proof: As PQ is a tangent OQ⊥PQ [Tangent at any point of circle is perpendicular to the radius through point of contact]

So, ∠OQP = 90°

Hence ΔOQP is right triangle

 

Q. 23. Draw a circle of radius 4 cm. Draw two tangents to the circle inclined at an angle of 60° to each other.

Answer. 

 

Q 24. In Fig. 7, two equal circles, with centres O and O’, touch each other at X. OO’ produced meets the circle with centre O’ at A. AC is tangent to the circle with centre O, at the point C. O’D is perpendicular to AC. Find the value of DO’/CO.

Answer. Given: two equal circles, with centres O and O’, touch each other at point X. OO’ is produced to meet the circle with centre O’ at A. AC is tangent to the circle with centre O, at the point C. O’D is perpendicular to AC.

 

Q 25. Solve for x:

Answer.

 

Q 26. The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower is 45°. Find the height of the tower PQ and the distance PX. (Use √3= 1.73)

Answer.

 

Q 27. The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses preceding the house numbered X is equal to sum of the numbers of houses following X.

Answer.

 

Q 28. In Fig. 8, the vertices of ∆ABC are A(4, 6), B(l, 5) and C(7, 2). A line segment DE is drawn to intersect the sides AB and AC at D and E respectively such that AD/AB= AE/AC= 1/3 .Calculate the area of ∆ADE and Calculate the area compare it with area of ∆ABC.

Answer.

 

Q 29. A number x is selected at random from the numbers 1, 2, 3 and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that the product of x and y is less than 16.

Answer.

 

Q 30. In Fig. 9, is shown a sector OAP of a circle with centre O, containing ∠θ. AB is perpendicular to the radius OA and meets OP produced at B. Prove that the perimeter of shaded region is r :

Answer.

 

Q 31. A motor boat whose speed is 24 km/h in still water takes 1 hr more to go 32 km upstream than to return downstream to the same spot. Find the speed of the stream.

Answer.

 

SET II

Q 10.Solve for x:

Answer.

 

Q 18. The digits of a positive number of three digits are in A.P. and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.

Answer.

 

Q 19. If the roots of the quadratic equation (a – b)x² + (b – c)x + (c – a) = 0 are equal, prove that 2a = b + c.

Answer.

 

Q 20. From a pack of 52 playing cards, Jacks, Queens and Kings of red colour are removed. From the remaining, a card is drawn at random. Find the probability that drawn card is: (i) a black King (ii) a card of red colour (iii) a card of black colour

Answer.

 

Q 28. Draw an isosceles ∆ABC in which BC = 5.5 cm and altitude AL = 3 cm. Then construct another triangle whose sides are 3/4 of the corresponding sides of ∆ABC.

Answer.

 

Q 29. Prove that the tangent drawn at any point of a circle is perpendicular to the radius through the point of contact.

Answer. Given: XY is a tangent at point P to the circle with centre O.

To prove: OP⏊XY

Construction: Take a point Q on XY other than P and join OQ.

Proof: If point Q lies inside the circle, then XY will become a secant and not a tangent to the circle.

∴ OQ > OP

This happen with every point on the line XY except the point P.

OP is the shortest of all the distances of the point O to the points of XY

∴ OP⏊XY

 

Q 30. As observed from the top of a lighthouse, 100 m high above sea level, the angles of depression of a ship, sailing directly towards it, changes from 30° to 60°. Find the distance travelled by the ship during the period of observation. (Use √3 = 1.73)

Answer.

 

Q 31. A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m. Find the length and breadth of the rectangular park.

Answer.

SET III

Q 10. Solve for x:

Answer.

 

Q 18. There are 100 cards in a bag on which numbers from 1 to 100 are written. A card is taken out from the bag at random. Find the probability that the number on the selected card (i) is divisible by 9 and is a perfect square (ii) is a prime number greater than 80.

Answer.

 

Q 19. Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Find the numbers.

Answer. Let three consecutive natural numbers are x, x+1,x+2

According to the question, (x+1)² -[(x+2)² –x² ] = 60

⇒ x²+1+2x -[x²+4+4x –x² ] = 60

⇒ x² + 1 + 2x – x²– 4 – 4x + x² = 60

⇒ x² – 2x – 63 = 0

⇒ x² – 9x +7x – 63 = 0

⇒ x(x –9) + 7(x – 9) = 0

⇒ (x –9)(x + 7) = 0

⇒ x –9 = 0 , x = 9

⇒ x + 7 = 0 , x = -7

Natural No’s can not be -ve, ∴ x = 9

∴ Numbers are 9, 10, 11

 

Q 20. The sums of first n terms of three arithmetic progressions are S₁, S₂ and S₃ respectively. The first term of each A.P. is 1 and their common differences are 1, 2 and 3 respectively. Prove that S₁+ S₃ = 2S₂.

Answer.

 

Q 28. Two pipes running together can fill a tank in 11 (1/9) minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately.

Answer.

 

Q 29. From a point on the ground, the angle of elevation of the top of a tower is observed to be 60°. From a point 40 m vertically above the first point of observation, the angle of elevation of the top of the tower is 30°. Find the height of the tower and its horizontal distance from the point of observation.

Answer.

 

Q 30. Draw a triangle with sides 5 cm, 6 cm and 7 cm. Then draw another triangle whose sides are 4/5 of the corresponding sides of first triangle.

Answer.

 

Q 31. A number x is selected at random from the numbers 1, 4, 9, 16 and another number y is selected at random from the numbers 1, 2, 3, 4. Find the probability that the value of xy is more than 16.

Answer. x can be any one of 1,4,9, or 16, i.e. 4 ways y can be any one of 1,2,3 or 4 ways

 Total number of cases of xy = 4×4 = 16 ways

Number of cases, where product is more than 16 

(9,2)(9,3)(9,4)(16,2)(16,3)(16,4) i.e. 6 ways

9×2 = 18 9×3 = 27

9×4 = 36 16×2 = 32

16×3 = 48 16×4 = 64

{18,27,36,32,48,64}

∴ Required Probability = 6/16 = 3/8

Maths

Section – A

Q.1. What is the common difference of an A.P. in which a₂₁ – a₇ = 84 ?

Solution: Given, a₂₁ – a₇ = 84

⇒ (a + 20d) – (a + 6d) = 84

⇒ a + 20d – a – 6d = 84

⇒ 20d – 6d = 84

⇒ 14d = 84

Hence common difference = 6

 

Q. 2.If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is 60°, then find the length of OP.

Solution: Given, ∠APB = 60°

∠APO = 30°

In right angle ΔOAP,

OP/OA = cosec 30°

⇒ OP/a = 2

⇒ OP = 2a.

Q. 3.If a tower 30 m high, casts a shadow 10√3 m long on the ground, then what is the angle of elevation of the sun?]

Solution: In ΔABC,

tan θ = AB/ BC

⇒ tan θ = 30/10√3 = √3

⇒ tan θ = tan 60°

⇒ θ = 60°

Hence angle of elevation is 60°.

 

Q. 4. The probability of selecting a rotten apple randomly from a heap of 900 apples is 0-18. What is the number of rotten apples in the heap? 

Solution: Total apples = 900

P(E) = 0.18

No. of rotten apples / Total No. of apples = 0.18

No. of rotten apples / 900 = 0.18

No. of rotten apples = 900 × 0.18 = 162

Section – B

Q. 5. Find the value of p, for which one root of the quadratic equation px² – 14x + 8 = 0 is 6 times the other. 

Solution: Given equation is px2 – 14x + 8 = 0

Let one root = α

then other root = 6α

Sum of roots = -b/a

α+6α=-(-14)/p

7α=14/p or α= 2/p   ……….(1)

Product of roots = c/a

(α)(6α)=8/p

6α²=8/p  ……….(2)

Putting value of α from eq. (i),

⇒6×(2/p)2 = 8/p

⇒6×4/p2 = 8/p

⇒24p = 8p2

⇒8p2-24p = 0

⇒8p(p-3) = 0

⇒ Either 8p = 0

p = 0

or        (p-3) = 0

p = 3

For p=0, given condition is not satisfied

ஃ p=3

 

Q 6.Which term of the progression 20, 19¼  , 18½ , 17 ¾, … is the first negative term ? 

Solution: Given, A.P. is 20, 19¼  , 18½ , 17 ¾, …

 

Q. 7. Prove that the tangents drawn at the endpoints of a chord of a circle make equal angles with the chord.

Solution: Given, a circle of radius OA and centred at O with chord AB and tangents PQ & RS are drawn from point A and B respectively.

Draw OM ⊥ AB, and join OA and OB.

In ∆OAM and ∆OMB,

OA = OB (Radii)

OM = OM (Common)

∠OMA = ∠OMB (Each 90°)

∆OAM = ∆OMB (By R.H.S. Congurency)

∠OAM = ∠OBM (C.PC.T.)

Also, ∠OAP = ∠OBR = 90° (Line joining point of contact of tangent to centre is perpendicular on it)

On addition,

∠OAM + ∠OAP = ∠OBM + ∠OBR

⇒ ∠PAB = ∠RBA

⇒ ∠PAQ – ∠PAB = ∠RBS – ∠RBA

⇒ ∠QAB = ∠SBA

Hence Proved

 

Q. 8. A circle touches all the four sides of a quadrilateral ABCD. Prove that AB + CD = BC + DA 

Solution: Given, a quad. ABCD and a circle touch its all four sides at P, Q, R, and S respectively.

To prove: AB + CD = BC + DA

Now, L.H.S. = AB + CD

= AP + PB + CR + RD

= AS + BQ + CQ + DS (Tangents from same external point are always equal)

= (AS + SD) + (BQ + QC)

= AD + BC

= R.H.S.

Hence Proved.

 

Q 9.A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, -5) is the mid-point of PQ, then find the coordinates of P and Q. 

Solution: Let co-ordinate of P (0, y)

Co-ordinate of Q (x, 0)

 

Q. 10.If the distances of P(x, y), from A(5, 1) and B(-1, 5) are equal, then prove that 3x = 2y. 

Solution: Given, PA = PB

⇒ x² + 25 – 10x + y² + 1 – 2y = x² + 1 + 2x + y² + 25 – 10y

⇒ -10x – 2y = 2x – 10y

⇒ -10x – 2x = -10y + 2y

⇒ 12x = 8y

⇒ 3x = 2y

Hence Proved.

Section – C

Q 11. If ad ≠ bc, then prove that the equation (a2 +b2) x2 + 2 (ac + bd) x +  (c2 + d2) = 0 has no real roots. 

Solution: Given, ad ≠ bc

(a2 + b2) x2 + 2(ac + bd)x + (c2 + d2) = 0

D = b2 – 4ac

= [2(ac + bd)]2 – 4 (a2 + b2) (c2 + d2)]

= 4[a2c2 + b2d2 + 2abcd] – 4(a2c2 + a2d2 + b2c2 + b2d2)

= 4[a2c2 + b2d2 + 2abcd – a2c2 – a2d2 – b2c2 – b2d2]

= 4[-a2d2 – b2c2 + 2abcd]

= -4[a2d2 + b2c2 – 2abcd]

= -4[ad – bc]2

D is negative

Hence given equation has no real roots.

 

Q 12.The first term of an A.E is 5, the last term is 45 and the sum of all its terms is 400. Find the number of terms and the common difference of the A.P. 

Solution: Given, a = 5, an = 45, Sn = 400

We have, Sn = ⇒ 400 = n/2 [5 + 45]

⇒ 400 = n/2 [50]

⇒ 25n = 400

⇒ n = 16

Now, an = a + (n – 1) d

⇒ 45 = 5 + (16 – 1)d

⇒ 45 – 5 = 15d

⇒ 15d = 40

⇒ d = 8/3

So n = 16 and d = 8/3

 

Q 13.On a straight line passing through the foot of a tower, two points C and D are at distances of 4 m and 16 m from the foot respectively. If the angles of elevation from C and D of the top of the tower are complementary, then find the height of the tower. 

Solution:

Let height AB of tower = h  m.

 

Q14. A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag. 

Solution: Given, no. of white balls = 15

Let no. of black balls = x

Total balls = (15 + x)

According to the question,

P(Blackball) = 3 × P(White ball)

⇒ x/15+x = 3 × 15/15+x

⇒ x = 45

No. of black balls in bag = 45

 

Q 15.In what ratio does the point (2411, y) the line segment joining the points P(2, -2) and Q(3, 7) ? Also, find the value of y. 

Solution: Let point R divides PQ in the ratio k : 1

Q 16. Three semicircles each of diameter 3 cm, a circle of diameter 4.5 cm and a semi-circle of radius 4.5 cm are drawn in the given figure. Find the area of the shaded region. 

Solution: Given, radius of large semi-circle = 4.5 cm

Q17. In the given figure, two concentric circles with centre O have radii 21 cm and 42 cm. If ∠AOB = 60°, find the area of the shaded region. [Use π = 227]

Solution: Angle for shaded region = 360° – 60° = 300°

Area of shaded region

 

Q 18.Water in a canal, 5-4 m wide and 1.8 m deep, is flowing with a speed of 25 km/hour. How much area can it irrigate in 40 minutes, if 10 cm of standing water is required for irrigation ? 

Solution: Width of canal = 5.4 m

Depth of canal = 1.8 m

Length of water in canal for 1 hr = 25 km = 25000 m

Volume of water flown out from canal in 1 hr = l × b × h = 5.4 × 1.8 × 25000 = 243000 m3

Volume of water for 40 min = 243000 × 40 60 = 162000 m3

Area to be irrigated with 10 cm standing water in field = Volume/ Height

= (162000×100)/10  m2

= 1620000 m2

= 162 hectare

 

Q 19.The slant height of a frustum of a cone is 4 cm and the perimeters of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum. 

Solution: Slant height of frustum ‘l’ = 4 cm

Perimeter of upper top = 18 cm

Q 20. The dimensions of a solid iron cuboid are 4.4 m × 2.6 m × 1.0 m. It is melted and recast into hollow cylindrical pipe of 30 cm inner radius and thickness 5 cm. Find the length of the pipe. 

Solution:  Inner radius of pipe ‘r’ = 30 cm 

The thickness of pipe = 5 cm

Outer radius ‘R’ = 30 + 5 = 35 cm

Now, Volume of hollow pipe = Volume of Cuboid

Section – D

Q. 21.Solve for x:

Solution:

Q 22.Two taps running together can fill a tank in 3 ¹/13  hours. If one tap takes 3 hours more than the other to fill the tank, then how much time will each tap take to fill the tank ? 

Solution: Let tank fill by one tap = x hrs

other tap = (x + 3) hrs

Together they fill by (3) 1/13 = 40/13 hrs

Either x – 5 = 0 or 13x + 24 = 0

x = 5, x = -24/13 (Rejected)

One tap fill the tank in 5 hrs

So other tap fill the tank in 5 + 3 = 8 hrs

 

Q 23.If the ratio of the sum of the first n terms of two A.P.S is (7n + 1) : (4n + 27), then find the ratio of their 9th terms. 

Solution:

Ratio of the sum of first n terms of two A.P.s are

Hence ratio of 9th terms of two A.P.s is 24 : 19

 

Q 24.Prove that the lengths of two tangents drawn from an external point to a circle are equal. 

Solution: Given, a circle with centre O and external point P. |

Two tangents PA and PB are drawn.

To Prove: PA = PB

Construction: Join radius OA and OB also join O to P.

Proof: In ∆OAP and ∆OBP,

OA = OB (Radii)

∠A = ∠B (Each 90°)

OP = OP (Common)

∆AOP = ∆BOP (RHS cong.)

PA = PB [By C.PC.T.]

Hence Proved.

 

Q 25.In the given figure, XY and XY are two parallel tangents to a circle with centre O and another tangent AB with a point of contact C, is intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°. 

Solution: Given, XX’ & YY’ are parallel.

Tangent AB is another tangent which touches the circle at C.

To prove: ∠AOB = 90°

Construction: Join OC.

Proof: In ∆OPA and ∆OCA,

OP = OC (Radii)

∠OPA = ∠OCA (Radius ⊥ Tangent)

OA = OA (Common)

∆OPA = ∆OCA (CPCT)

∠1 = ∠2 …(i)

Similarly, ∆OQB = ∆OCB

∠3 = ∠4 …(ii)

Also, POQ is a diameter of circle

∠POQ = 180° (Straight angle)

∠1 + ∠2 + ∠3 + ∠4 = 180°

From eq. (i) and (ii),

∠2 + ∠2 + ∠3 + ∠3 = 180°

⇒ 2(∠2 + ∠3) = 180°

⇒ ∠2 + ∠3 = 90°

Hence, ∠AOB = 90°

Hence Proved.

 

Q 26.Construct a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then construct another triangle whose sides are 3 4  times the corresponding sides of the ∆ABC. 

Solution: BC = 7 cm, ∠B = 45°, ∠A = 105°

∠C = 180 ° – (∠B + ∠A) = 180° – (45° + 105°) = 180° – 150° = 30°

Steps of construction:

1. Draw a line segment BC = 7 cm.
2. Draw an angle 45° at B and 30° at C. They intersect at A.
3. Draw an acute angle at B.
4. Divide angle ray in 4 equal parts as B1, B2, B3 and B4.
5. Join B4 to C.
6. From By draw a line parallel to B4C intersecting BC at C’.
7. Draw another line parallel to CA from C’ intersecting AB ray at A.
   Hence, ∆A’BC’ is required triangle such that ∆A’BC’ ~ ∆ABC with A’B = ¾ AB

 

Q 27. An aeroplane is flying at a height of 300 m above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are 45° and 60° respectively. Find the width of the river. [Use √3 = 1.732] 

Solution: Let aeroplane is at A, 300 m high from a river. C and D are opposite banks of river.

Q 28. If the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear, then find the value of k. 

Solution: Since A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear points, so area of triangle = 0.

 

Q. 29. Two different dice are thrown together. Find the probability that the numbers obtained have

(i) even sum, and

(ii) even product. 

Solution: When two different dice are thrown together

Total outcomes = 6 × 6 = 36

(i) For even sum: Favourable outcomes are

(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6),

(3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6),

(5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)

No. of favourable outcomes = 18

(ii) For even product: Favourable outcomes are

(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6).

No. of favourable outcomes = 27

Q. 30. In the given figure, ABCD is a rectangle of dimensions 21 cm × 14 cm. A semicircle is drawn with BC as diameter. Find the area and the perimeter of the shaded region in the figure.

Solution: Area of Shaded region = Area of a rectangle – Area of a semi-circle

 

Q. 31.In a rain-water harvesting system, the rainwater from a roof of 22 m × 20 m drains into a cylindrical tank having a diameter of base 2 m and height 35 m. If the tank is full, find the rainfall in cm. Write your views on water conservation.

Solution: Volume of water collected in system = Volume of a cylindrical tank

Set II

Note: Except for the following questions, all the remaining questions have been asked in previous sets.

Section – B

Q. 10.Which term of the A.P. 8, 14, 20, 26,… will be 72 more than its 41st term? 

Solution: A.P. is 8, 14, 20, 26,….

a = 8, d = 14 – 8 = 6

Let a_n = a₄₁ + 72

a + (n – 1)d = a + 40d + 72

⇒ (n – 1) 6 = 40 × 6 + 72 = 240 + 72 = 312

⇒ n – 1 = 52

⇒ n = 52 + 1 = 53rd term

Section – C

Q. 18.From a solid right circular cylinder of height 24 cm and radius 0.7 cm, a right circular cone of the same height and same radius is cut out. Find the total surface area of the remaining solid.

Solution: Given, Height of cylinder ‘h’ = 2.4 cm,

Radius of base ‘r’ = 0.7 cm

 

Q. 19.If the 10th term of an A.E is 52 and the 17th term is 20 more than the 13th term, find the A.P. 

Solution: Given, a₁₀ = 52;

a₁₇ = a₁₃ + 20

⇒ a + 16d = a + 12d + 20

⇒ 16d = 12d + 20

⇒ 4d = 20

⇒ d = 5

Also, a + 9d = 52

⇒ a + 9 × 5 = 52

⇒ a + 45 = 52

⇒ a = 7

Therefore A.E = 7, 12, 17, 22, 27,….

 

Q. 20. If the roots of the equation (c^2 – ab) x^2 – 2(a^2 – bc) x + b^2 – ac = 0 in x are equal, then show that either a = 0 or a^3 + b^3 + c^3 = 3abc.

Solution:

Section – D

Q. 28. Solve for x:

Solution:

Q 29.A train covers a distance of 300 km at a uniform speed. If the speed of the train is increased by 5 km/hour, it takes 2 hours less on the journey. Find the original speed of the train.

Solution: Let original speed of train = x km/hr

Increased speed of train = (x + 5) km/hr

Distance = 300 km

According to the question,

Q 30.A man observes a car from the top of a tower, which is moving towards the tower with a uniform speed. If the angle of depression of the car changes from 30° to 45° in 12 minutes, find the time taken by the car now to reach the tower.

Solution: Let AB is a tower, the car is at point D at 30° and goes to C at 45° in 12 minutes.

Q. 31.In the given figure, ΔABC is a right-angled triangle in which ∠A is 90°. Semi-circles are drawn on AB, AC and BC as diameters. Find the area of the shaded region.

Solution: In right ΔBAC, by Pythagoras theorem,

Set III

Note: Except for the following questions, all the remaining questions have been asked in previous sets.

Section – B

Q. 10.For what value of n, are the terms of two A.Ps 63, 65, 67,…. and 3, 10, 17,…. equal ? 

Solution:1st A.P. is 63, 65, 67,…

a = 63, d = 65 – 63 = 2

a_n = a + (n – 1 )d = 63 + (n – 1) 2 = 63 + 2n – 2 = 61 + 2n

2nd A.E is 3, 10, 17,…

a = 3, d = 10 – 3 = 7

an = a + (n – 1 )d = 3 + (n – 1) 7 = 3 + 7n – 7 = 7n – 4

According to question,

61 + 2n = 7n – 4

⇒ 61 + 4 = 7n – 2n

⇒ 65 = 5n

⇒ n = 13

Hence, 13th term of both A.P. is equal.

Section – C

Q. 18.A toy is in the form of a cone of radius 3-5 cm mounted on a hemisphere of the same radius on its circular face. The total height of the toy is 15*5 cm. Find the total surface area of the toy. 

Solution: Given, radius of base ‘r’ = 3.5 cm

Total height of toy = 15.5 cm

Height of cone ‘h’ = 15.5 – 3.5 = 12 cm

Q. 19.How many terms of an A.E 9, 17, 25,… must be taken to give a sum of 636? 

Solution: A.P. is 9, 17, 25,….,

S_n = 636

a = 9, d = 17 – 9 = 8

Q. 20. If the roots of the equation (a² + b²) x² – 2 (ac + bd) x + (c² + d²) = 0 are equal, prove that a/b = c/d

Solution:

Section – D

Q. 28.Solve for x:

Solution:

Q. 29. A takes 6 days less than B to do a work. If both A and B working together can do it in 4 days, how many days will B take to finish it? 

Solution: Let B can finish a work in x days

so, A can finish work in (x – 6) days

Together they finish work in 4 days

Now,

⇒ 4 (2x – 6) = x² – 6x

⇒ 8x – 24 = x² – 6x

⇒ x² – 14x + 24 = 0

⇒ x² – 12x – 2x + 24 = 0

⇒ x(x – 12) – 2(x – 12) = 0

⇒ (x – 12) (x – 2) = 0

Either x – 12 = 0 or x – 2 = 0

x = 12 or x = 2 (Rejected)

B can finish work in 12 days

A can finish work in 6 days.

 

Q. 30.From the top of a tower, 100 m high, a man observes two cars on the opposite sides of the tower and in a same straight line with its base, with angles of depression 30° and 45°. Find the distance between cars.

[Take √3 = 1.732]

Solution: Let AB is a tower.

Cars are at point C and D respectively

Distance between two cars = x + y = 173.2 + 100 = 273.2 m

 

Q. 31.In the given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of the shaded region.

Solution: Given, C (O, OB) with AC = 24 cm AB = 7 cm and ∠BOD = 90°

∠CAB = 90° (Angle in semi-circle)

Using pythagoras theorem in ∆CAB,

BC² = AC² + AB² = (24)² + (7)² = 576 + 49 = 625

⇒ BC = 25 cm

Radius of circle = OB = OD = OC = 25/2cm

Area of shaded region = Area of semi-circle with diamieter BC – Area of ∆CAB + Area of sector BOD

Maths

Section – A

Q.1. If x = 3 is one root of the quadratic equation x² – 2kx – 6 = 0, then find the value of k. 

Solution:

Given quadratic equation is, x² – 2kx – 6 = 0

x = 3 is a root of above equation, then

(3)² – 2k (3) – 6 = 0

 ⇒ 9 – 6k – 6 = 0

⇒ 3 – 6k = 0

 ⇒ 3 = 6k

⇒ k = ³/₆

⇒ k =½

 

Q.2. What is the HCF of the smallest prime number and the smallest composite number? 

Solution:

Smallest prime number = 2

Smallest composite number = 4

Prime factorisation of 2 is 1 × 2

Prime factorisation of 4 is 1 × 2²

HCF (2, 4) = 2

 

Q.3.Find the distance of a point P(x, y) from the origin. 

Solution:

The given point is P (x, y).

The origin is O (0, 0)

The distance of point P from the origin,

 

Q.4. In an AP if the common difference (d) = -4 and the seventh term (a⁷) is 4, then find the first term. 

Solution:

Given,

d = -4, a⁷ = 4

a + 6d = 4

⇒ a + 6(-4) = 4

⇒ a – 24 = 4

⇒ a = 4 + 24

⇒ a = 28

 

Q.5. What is the value of (cos² 67° – sin² 23°) ? 

Solution:

We have, cos² 67° – sin² 23°

= cos² 67° – cos² (90° – 23°)          [∵ sin (90° – θ) = cos θ]

= cos² 67° – cos² 67°

= 0

 

Q.6. Given ΔABC ~ ΔPQR, if ^AB/_PQ=¹/₃  then find ^arΔABC/_arΔPQR

Solution:

 

Section – B

Q.7.Given that √2 is irrational, prove that (5 + 3√2) is an irrational number. 

Solution:

Given, √2 is an irrational number.

Let √2 = m

Suppose, 5 + 3√2 is a rational number.

 But ^a-5b/_3b is rational number, so m is rational number which contradicts the fact that m = √2 is irrational number.

So, our supposition is wrong.

Hence, 5 + 3√2 is also irrational.

Hence Proved.

Q.8.In fig. 1, ABCD is a rectangle. Find the values of x and y.

Solution:

Given, ABCD is a rectangle.

AB = CD

⇒ 30 = x + y

or 

⇒ x + y = 30 …(i)

Similarly, AD = BC

⇒ 14 = x – y

or 

⇒ x – y = 14 …(ii)

On adding eq. (i) and (ii), we get

2x = 44

⇒ x = 22

Putting the value of x in eq. (i), we get

 22 + y = 30

 ⇒ y = 30 – 22

 ⇒ y = 8

 So, x = 22, y = 8.

 

Q.9.Find the sum of the first 8 multiples of 3. 

Solution:First 8 multiples of 3 are 3, 6, 9,….. up to 8 terms

 We can observe that the above series is an AP with

 a = 3, d = 6 – 3 = 3, n = 8

 Sum of n terms of an A.P is given by,

 

Q.10.Find the ratio in which P(4, m) divides the line segment joining the points A(2, 3) and B(6, -3). Hence find m. 

Solution:Let P divides line segment AB in the ratio k : 1

 Coordinates of P

 

Q.11. Two different dice are tossed together. Find the probability:

 (i) of getting a doublet.

 (ii) of getting a sum 10, of the numbers on the two dice. 

Solution:

 Total outcomes on tossing two different dice = 36

 (i) A: getting a doublet

 A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

 Number of favourable outcomes of A = 6

 (ii) B: getting a sum 10.

 B = {(4, 6), (5, 5), (6, 4)}

 Number of favourable outcomes of B = 3

 

Q.12. An integer is chosen at random between 1 and 100. Find the probability that it is:

 (i) divisible by 8.

 (ii) not divisible by 8. 

Solution: The total number are 2, 3, 4, …….. 99

 (i) Let E be the event of getting a number divisible by 8.

 E = {8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96} = 12

 (ii) Let E’ be the event of getting a number not divisible by 8.

Then, P(E’) = 1 – P(E) = 1 – 0.1224 = 0.8756

 

Section – C

Q.13. Find HCF and LCM of 404 and 96 and verify that HCF × LCM = Product of the two given numbers. 

Solution:

 Prime factorization of 404 = 2 × 2 × 101

 Prime factorization of 96 = 2 × 2 × 2 × 2 × 2 × 3

 HCF = 2 × 2 = 4

 And LCM = 2 × 2 × 2 × 2 × 2 × 3 × 101 = 9696

 HCF = 4, LCM = 9696

 Verification:

 HCF × LCM = Product of the two given numbers

 4 × 9696 = 404 × 96

 38784 = 38784

 Hence Verified.

 

Q.14. Find all zeroes of the polynomial (2x⁴ – 9x³ + 5x² + 3x – 1) if two of its zeroes are (2 + √3) and (2 – √3). 

Solution:

Here, p(x) = 2x⁴ – 9x³ + 5x² + 3x – 1

And two of its zeroes are (2 + √3) and (2 – √3).

Quadratic polynomial with zeroes is given by,

 {x – (2 + √3)}. {x – (2 – √3)}

 ⇒ (x – 2 – √3) (x – 2 + √3)

 ⇒ (x – 2)² – (√3)²

 ⇒ x² – 4x + 4 – 3

 ⇒ x² – 4x + 1 = g(x) (say)

 Now, g(x) will be a factor of p(x) so g(x) will be divisible by p(x)

 For other zeroes,

 2x² – x – 1 = 0

 2x² – 2x + x – 1 = 0

 or 

2x (x – 1) + 1 (x – 1) = 0

(x – 1) (2a + 1) = 0

x – 1 = 0 and 2x + 1 = 0

x = 1, x = -½

Zeroes of p(x) are

 1, -½, 2 + √3 and 2 – √3.

 

Q.15. If A(-2, 1) and B(a, 0), C(4, b) and D( 1, 2) are the vertices of a parallelogram ABCD, find the values of a and b. Hence find the lengths of its sides. 

 OR

Q.15. If A(-5, 7), B(-4, -5), C(-1, -6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.

Solution:

Given ABCD is a parallelogram.

 

Q.16. A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed. 

Solution: Let the usual speed of plane be x km/h.

 Increased speed = (x + 100) km/h.

 Distance to cover = 1500 km.

 Time taken by plane with usual speed = ¹⁵⁰⁰ /x hr

 Time taken by plane with increased speed = ¹⁵⁰⁰ /_{100 + x}

 According to the question,

 x² + 100x = 300000

 x² + 100x – 300000 = 0

 x² + 600x – 500x – 300000 = 0

 x(x + 600) – 500(x + 600) = 0

 (x + 600) (x – 500) = 0

 Either x + 600 = 0 ⇒ x = -600 (Rejected)

 or 

 x – 500 = 0 ⇒ x = 500

 Usual speed of plane = 500 km/hr.

 

Q.17. Prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonal. 

 OR

Q.17. If the area of two similar triangles is equal, prove that they are congruent.

Solution: Let ABCD be a square with side ‘a’.

 

Q.18. Prove that the lengths of tangents drawn from an external point of a circle are equal. 

Solution:

Given: A circle with centre O on which two tangents PM and PN are drawn from an external point P.

 To Prove: PM = PN

 Construction: Join OM, ON and OP

 Proof: Since tangent and radius are perpendicular at point of contact,

 ∠OMP = ∠ONP = 90°

 In ΔPOM and ΔPON,

 OM = ON (Radii)

 ∠OMP = ∠ONP

 PO = OP (Common)

 ΔOMP = ΔONP (RHS cong.)

 PM = PN (C.P.C.T)

 Hence Proved.

 

Q.19. If 4 tanθ = 3, evaluate [(4sinθ – cosθ + 1)/ (4sinθ + cosθ – 1)]

 OR

Q.19. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

Solution:

 Given, 4 tan θ = 3

 ⇒ tan θ = ¾ (=P/B)

 OR

 Solution: Given, tan 2A = cot (A – 18°)

 ⇒ cot (90° – 2A) = cot (A – 18°)

 [∵ tan θ = cot (90° – θ)]

 ⇒ 90° – 2A = A – 18°

 ⇒ 90° + 18° = A + 2A

 ⇒ 108° = 3A

 ⇒ A = 36°

 

Q.20. Find the area of the shaded region in Fig. 2, where arcs are drawn with centres A, B, C and D intersect in pairs at mid-points P, Q, R and S of the sides AB, BC, CD and DA respectively of a square ABCD of side 12 cm. [Use π = 3.14] 

 Solution:

 Given, ABCD is a square of side 12 cm.

 P, Q, R and S are the midpoints of sides AB, BC, CD and AD respectively.

 Area of shaded region = Area of square – 4 × Area of quadrant

 = a² – 4 × ¼πr² 

 = (12)² – 3.14 × (6)²

 = 144 – 3.14 × 36

 = 144 – 113.04

 = 30.96 cm²

 

Q.21. A wooden article was made by scooping out a hemisphere form each end of a solid cylinder, as shown in Fig. 3. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm. Find the total surface area of the article. 

 OR

Q.21. A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap?

Solution:

 Given, Radius (r) of cylinder = Radius of hemisphere = 3.5 cm.

 Total SA of article = CSA of cylinder + 2 × CSA of hemisphere

 Height of cylinder, h = 10 cm

 TSA = 2πrh + 2 × 2πr²

 = 2πrh + 4πr²

 = 2πrh (h + 2r)

 = 2 × ²²⁄7× 3.5 (10 + 2 × 3.5)

 = 2 × 22 × 0.5 × (10 + 7)

 = 2 × 11 × 17

 = 374 cm²

 OR

 Solution: Base diameter of cone = 24 m.

 Radius r = 12 m

 Height of cone, h = 3.5 m

 Volume of rice in conical heap = ⅓πr²h

 = ⅓ × ²²⁄7 × 12 × 12 × 3.5 = 528 cm³

 

Q.22. The table below shows the salaries of 280 persons:

Calculate the median salary of the data.

Solution:

 ^N∕₂= ²⁸⁰∕₂  = 140

The cumulative frequency just greater than 140 is 182.

Median class is 10 -15.

 l = 10, h = 5, N = 280, c.f. = 49 and f = 133

 

Section – D

Q.23. A motorboat whose speed is 18 km/hr in still water takes 1 hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. 

 OR

Q.23. A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete the total journey, what is the original average speed?

Solution:

Given, speed of motorboat instil

 water = 18 km/hr.

 Let speed of stream = x km/hr.

 Speed of boat downstream = (18 + x) km/hr.

 And speed of boat upstream = (18 – x) km/hr.

 Time of the upstream journey = ²⁴∕_18-x 

 Time of the downstream journey = ²⁴∕_18+x

 According to the question,

 ⇒ x² + 48x – 324 = 0

 ⇒ x² + 54x – 6x – 324 = 0

 ⇒ x(x + 54) – 6(x + 54) = 0

 ⇒ (x + 54)(x – 6) = 0

 Either x + 54 = 0 ⇒ x = -54

 Rejected, as speed cannot be negative

 Or

 x – 6 = 0 ⇒ x = 6

 Thus, the speed of the stream is 6 km/hr.

 OR

 Solution: Let the original average speed of train be x km/hr.

 Increased speed of train = (x + 6) km/hr.

 Time taken to cover 63 km with average speed = ⁶⁴∕_x hr.

 Time taken to cover 72 km with increased speed = ⁷²∕_x+6 hr.

 According to the question,

⇒ 135x + 378 = 3(x² + 6x)

 ⇒ 135x + 378 = 3x² + 18x

 ⇒ 3x² + 18x – 135x – 378 = 0

 ⇒ 3x² – 117x – 378 = 0

 ⇒ 3(x² – 39x – 126) = 0

 ⇒ x² – 39x – 126 = 0

 ⇒ x² – 42x + 3x – 126 – 0

 ⇒ x(x – 42) + 3(x – 42) = 0

 ⇒ (x – 42) (x + 3) = 0

 Either x – 42 = 0 ⇒ x = 42

 or 

x + 3 = 0 ⇒ x = -3

 Rejected (as speed cannot be negative)

 Thus, average speed of train is 42 km/hr.

 

Q.24. The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7:15. Find the numbers. 

Solution: Let the first term of AP be a and d be a common difference.

 Let your consecutive term of an AP be a – 3d, a – d, a + d and a + 3d

 According to the question,

 a – 3d + a – d + a + d + a + 3d = 32

 ⇒ 4a = 32

 ⇒ a = 8 …(i)

 Also,

 (a – 3d) (a + 3d) : (a – d) (a + d) = 7 : 15

 For d = 2, four terms of AP are,

 a – 3d = 8 – 3 (2) = 2

 a – d = 8 – 2 = 6

 a + d = 8 + 2 = 10

 a + 3d = 8 + 3(2) = 14

 For d = -2, four term are

 a – 3d = 8 – 3(-2) = 14

 a – d = 8 – (-2) = 10

 a + d = 8 + (-2) = 6

 a + 3d = 8 + 3 (-2) = 2

 Thus, the four terms of AP series are 2, 6, 10, 14 or 14, 10, 6, 2.

 

Q.25. In an equilateral ∆ABC, D is a point on side BC such that BD = ⅓ BC. Prove that 9(AD)² = 7(AB)². 

OR

Q.25. Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

Solution:

 Given, ABC is an equilateral triangle and D is a point on BC such that BD = ⅓  BC.

To prove: 9AD² = 7AB²

 Construction : Draw AE ⊥ BC

 Proof: BD = ⅓ BC …(i) (Given)

 AE ⊥ BC

 We know that perpendicular from a vertex of equilateral triangle to the base divides base in two equal parts.

 BE = EC = ½ BC …(ii)

 In ∆AEB,

 AD² = AE² + DE² (Pythagoras theorem)

 or 

AE² = AD² – DE² …(iii)

 Similarly, In ∆AEB,

 AB² = AE² + BE²

 OR

Solution: Given: ∆ABC is a right angle triangle, right-angled at A.

 To prove : BC² = AB² + AC²

 Construction : Draw AD ⊥ BC.

 Proof: In ∆ADB and ∆BAC,

 ∠B = ∠B (Common)

 ∠ADB = ∠BAC (Each 90°)

 ∆ADB ~ ∆BAC (By AA similarity axiom)

 ^AB∕_BC= ^BD∕_AB (CPCT)

 AB² = BC × BD

 Similarly,

 ∆ADC ~ ∆CAB

 ^AC∕_BC= ^DC∕_AC

 AC² = BC × DC …(ii)

 On adding equation (i) and (ii)

 AB² + AC² = BC × BD + BC × CD = BC (BD + CD) = BC × BC

 AB² + AC² = BC²

 BC² = AB² + AC²

Hence Proved.

 

Q.26. Draw a triangle ABC with BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are ¾ of the corresponding sides of the ∆ABC. 

Solution:

Draw a line segment BC = 6 cm.

Construct ∠XBC = 60°.

With B as centre and radius equal to 5 cm, draw an arc intersecting XB at A.

Join AC. Thus, ∆ABC is obtained.

Draw an acute angle ∠CBY below of B.

Mark 4-equal parts on BY as B₁, B₂, B₃ and B₄

Join B₄ to C.

From By draw a line parallel to B₄C intersecting BC at C’.

Draw another line parallel to CA from C’, intersecting AB at A’.

∆A’BC’ is required triangle which is similar to ∆ABC such that BC’ = ¾  BC.

 

Q.27.

 Solution:

 

Q.28. The diameters of the lower and upper ends of a bucket in the form of a frustum of a cone are 10 cm and 30 cm respectively. If its height is 24 cm, find:

 (i) The area of the metal sheet used to make the bucket.

 (ii) Why we should avoid the bucket made by ordinary plastic? [Use π = 3.14] 

Solution: Given, Height of frustum, h = 24 cm.

 Diameter of lower end = 10 cm

 Radius of lower end, r = 5 cm.

 Diameter of upper end = 30 cm

 Radius of upper end, R = 15 cm.

(i) Area of metal sheet used to make the bucket = CSA of frustum + Area of base

 = πl(R + r) + πr²

 = π[26 (15 + 5) + (5)²]

 = 3.14 (26 × 20 + 25)

 = 3.14 (520 + 25)

 = 3.14 × 545

 = 1711.3 cm²

(ii) We should avoid the bucket made by ordinary plastic because plastic is harmful to the environment and to protect the environment its use should be avoided.

 

Q.29. As observed from the top of a 100 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. [Use √3 = 1.732] 

Solution: Let AB be the lighthouse and two ships are at C and D.

 Distance between two ships = y – x

 = 100√3 – 100 [from equation (i) and (ii)]

 = 100 (√3 – 1)

 = 100(1.732 – 1)

 = 100 (0.732)

 = 73.2 m

 

Q.30. The mean of the following distribution is 18. Find the frequency f of the class 19-21. 

 OR

Q.30. The following distribution gave the daily income of 50 workers of a factory:

Convert the distribution above to a less than type cumulative frequency distribution and draw its give.

Solution:

Maths

Section -A

Q.1. If HCF (336,54) = 6, find LCM (336, 54),

Answer :  Given, HCF (336, 54) = 6 

We know HCF × LCM = one number × other number 

6 × LCM = 336 × 54 

LCM = 336×54/6 = 336 × 9 = 3024 

 

Q.2. Find the nature of roots of the quadratic equation 2x² – 4x + 3 = 0. 

Answer: Given, 2x² – 4x + 3 = 0

Comparing it with quadratic equation ax² + bx + c = 0

Here, a = 2, b = -4 and c = 3

D = b² – 4ac = (-4)² – 4 × (2)(3) = 16 – 24 = -8 < 0

Hence, D < 0 this shows that roots will be imaginary.

 

Q.3. Find the common difference of the Arithmetic Progression (A.P.).

Answer :

Q.4. Evaluate: sin² 60° + 2 tan 45° – cos² 30° 

OR

Q.4. If sin A = ¾, Calculate sec A.

Answer :

We know,

 

Q.5. Write the coordinates of a point P on the x-axis which is equidistant from point A(-2, 0) and B(6, 0).

Answer: Let coordinates of P on x-axis is (x, 0) 

Given, A(-2, 0) and B(6, 0)

Here, PA = PB 

On squaring both sides, we get

(x + 2)² = (x – 6)²

⇒ x² + 4 + 4x = x² + 36 – 12x

⇒ 4 + 4x = 36 – 12x

⇒ 16x = 32

⇒ x = 2

Coordinates of P are (2, 0)

 

Q.6. Find the 21st term of the A.P. -4½, – 3, -1½ ………

Answer : 

Section- B

Q.7. For what value of k, will the following pair of equations have infinitely many Answers: 

2x + 3y = 7 and (k + 2)x – 3(1 – k)y = 5k + 1 [2]

Answer: Given, The system of equations is

2x + 3y = 7 and (k + 2) x – 3 (1 – k) y = 5k +1

These equations are of the form a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

where, a₁ = 2, b₁ = 3, c₁ = -7

a₂ = (k + 2), b₂ = -3(1 – k), c₂ = -(5k + 1)

Since, the given system of equations have infinitely many Answers. 

Hence, the given system of equations has infinitely many Answers when k = 4.

 

Q.8. Find a relation between x and y if the points A(x, y), B(-4, 6) and C(-2, 3) are collinear. 

OR

Q.8. Find the area of a triangle whose vertices are given as (1, -1) (-4, 6) and (-3, -5).

Answer: Given, A(x, y), B(-4, 6), C(-2, 3)

x¹ = x, y¹ = y, x² = -4, y² = 6, x³ = -2, y³ = 3

If these points are collinear, then area of triangle made by these points is 0.

 

Q.9. The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is ⅕. The probability of selecting a black marble at random from the same jar is ¼. If the jar contains 11 green marbles, find the total number of marbles in the jar. 

Answer: Let the probability of selecting a blue marble, black marble and green marbles are P(x), P(y), P(z) respectively.

P(x) = ⅕ , P (y) = ¼ Given

We Know,

P(x) + P(y) + P(z) = 1

 

Q.10. Find the value(s) of k so that the pair of equations x + 2y = 5 and 3x + ky + 15 = 0 has a unique Answer. 

Answer: Given, x + 2y = 5, 3x + ky + 15 = 0

Comparing above equations with

a¹x + b¹y + c¹ = 0 and a²x + b²y + c² = 0,

We get,

a¹ = 1, b¹ = 2, c¹ = -5

a² = 3, b² = k, c² = 15

Condition for the pair of equations to have unique Answer is

k can have any value except 6.

 

Q.11. The larger of the two supplementary angles exceeds the smaller by 18°. Find the angles. 

OR

Q.11. Sumit is 3 times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present?

Answer: Let two angles A and B are supplementary.

A + B = 180° …(i)

Given, A = B + 18°

On putting A = B + 18° in equation (i),

we get B + 18° + B = 180°

⇒ 2B + 18° = 180°

⇒ 2B = 162°

⇒ B = 81°

A = B + 18°

⇒ A = 99°

OR

Answer: Let age of Sumit be x years and age of his son be y years.Then, according to question we have, x = 3y …… (i)

Five years later, x + 5 = 2½ (y + 5)……….(ii) 

On putting x = 3y in equation (ii)

 

Q.12. Find the mode of the following frequency distribution:

Answer :

Here, the maximum frequency is 50.

So, 35 – 40 will be the modal class.

l = 35, f₀ = 34, f₁ = 50, f₂ = 42 and h = 5

Section – C

Q.13. Point A lies on the line segment XY joining X(6, -6) and Y (-4, -1) in such a way that XA/XY = ⅖. 

If Point A also lies on the line 3x + k (y + 1) = 0, find the value of k. 

Answer: Given,

Since, point A(2, -4) lies on line 3x + k (y + 1) = 0.

Therefore it will satisfy the equation.

On putting x = 2 and y = -4 in the equation, we get

3 × 2 + k(-4 + 1) = 0

⇒ 6 – 3k = 0

⇒ 3k = 6

⇒ k = 2

 

Q.14. Solve for x: x² + 5x – (a² + a – 6) = 0 [3]

Answer: Taking (a² + a – 6)

= a² + 3a – 2a – 6

= a(a + 3) – 2 (a + 3)

= (a + 3) (a – 2)

x² + 5x – (a + 3) (a – 2) = 0

⇒ x² + (a + 3)x – (a – 2)x – (a + 3)(a – 2) = 0

⇒ x[x + (a + 3)] – (a – 2) [x + (a + 3)] = 0

⇒ (x – a + 2)(x + a + 3) = 0

Hence, x – a + 2 = 0 and x + a + 3 = 0

x = a – 2 and x = -(a + 3)

Required values of x are (a – 2), -(a + 3).

 

Q.15. Find A and B if sin (A + 2B) = √3/2 and cos (A + 4B) = 0, where A and B are acute angles. 

Answer: Given 

Sin (A + 2B) = √3/2 and cos (A + 4B) = 0 

⇒ sin (A + 2B) = 60° (∵ sin 60° = √3/2)

A + 2B =60 …(i)

cos (A + 4B) = cos 90° (∵ cos 90° = 0)

⇒ A + 4B = 90° …(ii)

On solving equation (i) and (ii), we get

B = 15° and A = 30°

 

Q.16. Find the ratio in which the line x – 3y = 0 divides the line segment joining the points (-2, -5) and (6, 3). Find the coordinates of the point of intersection.

Answer: Let the required ratio be k : 1

By section formula, we have

 

Q.17.  Evaluate:

Answer : 

 

Q.18. In Figure 4, a square OABC is inscribed in a quadrant OPBQ. If OA = 15 cm, find the area of the shaded region. (Use π = 3.14)

OR

Q.18. In Figure 5, ABCD is a square with side 2√2 cm and inscribed in a circle. Find the area of the shaded region. (Use π = 3.14)

Answer: Given, OABC is a square with OA = 15 cm

OB = radius = r

 Let side of square be a then,

a² + a² = r²

⇒ 2a² = r²

⇒ r = √2 a

⇒ r = 15√2 cm (∵ a = 15 cm)

Area of square = Side × Side = 15 × 15 = 225 cm²

Area of shaded region = Area of quadrant OPBQ – Area of square

= 353.25 – 225 = 128.25 cm

OR

Answer: Given, ABCD is a square with side 2√2 cm

BD = 2r

In ∆BDC,

BD² = DC² + BC²

⇒ 4r² = 2(DC)² (∵ DC = CB = Side = 2√2 )

⇒ 4r² = 2 × 2√2 × 2√2

⇒ 4r² = 8 × 2

⇒ 4r² = 16

⇒ r² = 4

⇒ r = 2 cm

Area of square BCDA = Side × Side = DC × BC = 2√2 × 2√2 = 8 cm²

Area of circle = πr² = 3.14 × 2 × 2 = 12.56 cm²

Area of shaded region = Area of circle – Area of square. = 12.56 – 8 = 4.56 cm²

 

Q.19. A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 20 cm and the diameter of the cylinder is 7 cm. Find the total volume of the solid. (Use π = 22/7).

Answer: ABCD is a cylinder and BFC and AED are two hemispheres which has radius (r) = 7/2 cm

Total volume of solid = Volume of two hemisphere + Volume of cylinder

= 179.67 + 500.5 = 680.17 cm

 

Q.20. The marks obtained by 100 students in an examination are given below:

Find the mean marks of the students.

Answer :

 

Q.21. For what value of k, is the polynomial f(x) = 3x⁴ – 9x³ + x² + 15x + k completely divisible by 3x² – 5? 

OR

Q.21. Find the zeroes of the quadratic polynomial 7y2 – 11/2y – ⅔ and verify the relationship between the zeroes and the coefficients.

Answer: Given, f(x) = 3x⁴ – 9x³ + x² + 15x + k

It is completely divisible by 3x² – 5

 

Q.22. Write all the values of p for which the quadratic equation x² + px + 16 = 0 has equal roots. Find the roots of the equation so obtained. 

Answer: Given, equation is x² + px + 16 = 0

This is of the form ax² + bx + c = 0

where, a = 1, b = p and c = 16

D = b² – 4ac = p² – 4 × 1 × 16 = p² – 64

for equal roots, we have D = 0

p² – 64 = 0

⇒ p² = 64

⇒ p = ±8 Putting p = 8 in given equation we have,

x2 + 8x + 16 = 0

⇒ (x + 4)² = 0

⇒ x + 4 = 0⇒ x = -4

Now, putting p = -8 in the given equation, we get

x²  > – 8x + 16 = 0

⇒ (x – 4)² = 0

⇒ x = 4

Required roots are -4 and -4 or 4 and 4.

 

Section – D

Q.23. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares on their corresponding sides.

Answer: Given, ΔABC ~ ΔDEF

 

Q.24. Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point P between them on the road, the angle of elevation of the top of a pole is 60° and the angle of depression from the top of the other pole of point P is 30°. Find the heights of the poles and the distance of the point P from the poles.

Answer: Let AC is the road of 80 m width. P is the point on road AC and height of poles AB and CD is h m.

h———–(i)

Equating the values of h from equation (i) and (ii) we get

⇒ x√3 =

⇒ 3x = 80 – x

⇒ 4x = 80

⇒ x = 20m

On putting x = 20 in equation (i), we get

h = √3 × 20 = 20√3

h = 20√3 m

Thus, height of poles is 20√3 m and point P is at a distance of 20 m from left pole and (80 – 20) i.e., 60 m from right pole.

 

Q.25. The total cost of a certain length of a piece of cloth is ₹ 200. If the piece was 5 m longer and each metre of cloth costs ₹ 2 less, the cost of the piece would have remained unchanged. How long is the piece and what is its original rate per metre? 

Answer: Let the original length of the piece of cloth is x m and rate of cloth is ₹ y per metre.

Then according to question, we have

x × y = 200 …(i)

and if length be 5 m longer and each meter of cloth be ₹ 2 less than

(x + 5) (y – 2) = 200

⇒ (x + 5) (y – 2) = 200

⇒ xy – 2x + 5y – 10 = 200 …(ii)

On equating equation (i) and (ii), we have

xy = xy – 2x + 5y – 10

⇒ 2x – 5y = -10 …… (iii)

⇒ y = 200/x from equation (i)

⇒ 2x – 5 × 200/x = -10

= 2x – 1000/x -10

⇒ 2x² – 1000 = -10x

⇒ 2x² + 10x – 1000 = 0

⇒ x² + 5x – 500 = 0

⇒ x² + 25x – 20x – 500 = 0

⇒ x(x + 25) – 20 (x + 25) = 0

⇒ (x + 25) (x – 20) = 0

⇒ x = 20 (x ≠ -25 length of cloth can never be negative)

∴ x × y = 200

20 × y = 200

y = 10

Thus, the length of the piece of cloth is 20 m and original price per metre is ₹ 10.

 

Q.26. Construct an equilateral ∆ABC with each side 5 cm. Then construct another triangle whose sides are ⅔ times the corresponding sides. Draw two concentric circles of radii 2 cm and 5 cm. Take a point P on the outer circle and construct a pair of tangents PA and PB to the smaller circle. Measure PA.

Answer: Steps for construction are as follows:

1. Draw a line segment BC = 5 cm
2. At B and C construct ∠CBX = 60° and ∠BCX = 60°
3. With B as centre and radius 5 cm, draw an arc cutting ray BX at A. On graph paper, we take the scale.
4. Join AC. Thus an equilateral ∆ABC is obtained.

1. Below BC, make an acute angle ∠CBY
2. Along BY, mark off 3 points B₁, B₂, B₃ Such that BB₁, B₁B₂, B₂B₃ are equal.
3. Join B³C
4. From B₂ draw B₂D || B₃C, meeting BC at D
5. From D, draw DE || CA, meeting AB at E.

Then ∆EBD is the required triangle, each of whose sides is ⅔ of the corresponding side of ∆ABC.

 

Q.27. Change the following data into ‘less than type’ distribution and draw its olive:

Answer :

 

Q.28. Prove that:

Answer :

 

Q.29. Which term of the Arithmetic Progression -7, -12, -17, -22,…..will be -82? Is -100 any term of the A.P.? Give a reason for your answer. 

OR

Q.29. How many terms of the Arithmetic Progression 45, 39, 33, … must be taken so that their sum is 180? Explain the double answer.

Answer:-7, -12, -17, -22, …….

Here a = -7, d = -12 – (-7) = -12 + 7 = -5

Let T_n = -82T_n = a + (n – 1) d

⇒ -82 = -7 + (n – 1) (-5)

⇒ -82 = -7 – 5n + 5

⇒ -82 = -2 – 5n

⇒ -82 + 2 = -5n

⇒ -80 = -5n

⇒ n = 16

Therefore, 16th term will be -82.

Let T_n = -100

Again, T_n = a + (n -1) d

⇒ -100 = -7 + (n – 1) (-5)

⇒ -100 = -7 – 5n + 5

⇒ -100 = – 2 – 5n

⇒ -100 + 2 = -5n

⇒ -98 = -5n

⇒ n = 98/5 

But the number of terms can not be in fraction.

So, -100 can not be the term of this A.P.

OR

Answer: 45, 39, 33, …..

Here a = 45, d = 39 – 45 = -6

Let S_n = 180

⇒ n/2 [ 2a + (n – 1) d] = 180

⇒ n/2 [2 × 45 + (n – 1) (-6)] = 180

⇒ n/2 [90 – 6n + 6] = 180

⇒ n/2 [96 – 6n] = 180

⇒ n(96 – 6n) = 360

⇒ 96n – 6n² = 360

⇒ 6n² – 96n + 360 = 0

On dividing the above equation by 6

⇒ n² – 16n + 60 = 0

⇒ n² – 10n – 6n + 60 = 0

⇒ n(n – 10) – 6 (n – 10) = 0

⇒ (n – 10) (n – 6) = 0

⇒ n = 10, 6

Sum of first 10 terms = Sum of first 6 terms = 180

This means that the sum of all terms from 7th to 10th is zero.

 

Q.30. In a class test, the sum of Aran’s marks in Hindi and English is 30. Had he got 2 marks more in Hindi and 3 marks less in English, the product of the marks would have been 210. Find his marks in the two subjects.

Answer: Let Aran marks in Hindi be x and marks in English be y.

Then, according to question, we have

x + y = 30 …(i)(x + 2)(y – 3) = 210 …(ii)

from equation (i) put x = 30 – y in equation (ii)

(30 – y + 2) (y – 3) = 210

⇒ (32 – y) (y – 3) = 210

⇒ 32y – 96 – y² + 3y = 210

⇒ y² – 35y + 306 = 0

⇒ y² – 18y – 17y + 306 = 0

⇒ y(y – 18) – 17(y – 18) = 0

⇒ (y – 18) (y – 17) = o

⇒ y = 18, 17

Put y = 18 and 17 in equation (i), we get x = 12, 13

Hence his marks in hindi can be 12 and 13 and in english his marks can be 18 and 17.

 

 

Maths

Section -A

Q.1. If HCF (336,54) = 6, find LCM (336, 54),

Answer :  Given, HCF (336, 54) = 6 

We know HCF × LCM = one number × other number 

6 × LCM = 336 × 54 

LCM = 336×54/6 = 336 × 9 = 3024 

 

Q.2. Find the nature of roots of the quadratic equation 2x² – 4x + 3 = 0. 

Answer: Given, 2x² – 4x + 3 = 0

Comparing it with quadratic equation ax² + bx + c = 0

Here, a = 2, b = -4 and c = 3

D = b² – 4ac = (-4)² – 4 × (2)(3) = 16 – 24 = -8 < 0

Hence, D < 0 this shows that roots will be imaginary.

 

Q.3. Find the common difference of the Arithmetic Progression (A.P.).

Answer :

Q.4. Evaluate: sin² 60° + 2 tan 45° – cos² 30° 

OR

Q.4. If sin A = ¾, Calculate sec A.

Answer :

We know,

 

Q.5. Write the coordinates of a point P on the x-axis which is equidistant from point A(-2, 0) and B(6, 0).

Answer: Let coordinates of P on x-axis is (x, 0) 

Given, A(-2, 0) and B(6, 0)

Here, PA = PB 

On squaring both sides, we get

(x + 2)² = (x – 6)²

⇒ x² + 4 + 4x = x² + 36 – 12x

⇒ 4 + 4x = 36 – 12x

⇒ 16x = 32

⇒ x = 2

Coordinates of P are (2, 0)

 

Q.6. In Figure 1, ABC is an isosceles triangle right angled at C with AC = 4 cm. Find the length of AB.

OR

Q.6. In Figure 2, DE || BC. Find the length of side AD, given that AE = 1.8 cm, BD = 7.2 cm and CE = 5.4 cm.

Answer:

Given, ∠C = 90° and AC = 4 cm, AB = ?

 ∆ABC is an isosceles triangle so, BC = AC = 4 cm

 On applying Pythagoras theorem, we have

AB² = AC² + BC²

⇒ AB² = AC² + AC² (∵ BC = AC)

⇒ AB² = 42 + 42 = 16 + 16 = 32

⇒ AB = √32 = 4√2 cm

OR

Answer: Given, DE || BCOn applying, Thales theorem, we have

Section – B

 

Q.7. A die is thrown twice. Find the probability that

(i) 5 will come up at least once. 

(ii) 5 will not come up either time.

Answer: When two dice are thrown simultaneously, all possible outcomes are

(1.1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

Total number of outcomes = 36Total outcomes where 5 comes up at least once = 11

 

Q.8. Find a relation between x and y if the points A(x, y), B(-4, 6) and C(-2, 3) are collinear. 

OR

Q.8. Find the area of a triangle whose vertices are given as (1, -1) (-4, 6) and (-3, -5).

Answer: Given, A(x, y), B(-4, 6), C(-2, 3)

x¹ = x, y¹ = y, x² = -4, y² = 6, x³ = -2, y³ = 3

If these points are collinear, then area of triangle made by these points is 0.

 

Q.9. The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is ⅕. The probability of selecting a black marble at random from the same jar is ¼. If the jar contains 11 green marbles, find the total number of marbles in the jar. 

Answer: Let the probability of selecting a blue marble, black marble and green marbles are P(x), P(y), P(z) respectively.

P(x) = ⅕ , P (y) = ¼ Given

We Know,

P(x) + P(y) + P(z) = 1

 

Q.10. Find the value(s) of k so that the pair of equations x + 2y = 5 and 3x + ky + 15 = 0 has a unique Answer. 

Answer: Given, x + 2y = 5, 3x + ky + 15 = 0

Comparing above equations with

a¹x + b¹y + c¹ = 0 and a²x + b²y + c² = 0,

We get,

a¹ = 1, b¹ = 2, c¹ = -5

a² = 3, b² = k, c² = 15

Condition for the pair of equations to have unique Answer is

k can have any value except 6.

 

Q.11. The larger of the two supplementary angles exceeds the smaller by 18°. Find the angles. 

OR

Q.11. Sumit is 3 times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present?

Answer: Let two angles A and B are supplementary.

A + B = 180° …(i)

Given, A = B + 18°

On putting A = B + 18° in equation (i),

we get B + 18° + B = 180°

⇒ 2B + 18° = 180°

⇒ 2B = 162°

⇒ B = 81°

A = B + 18°

⇒ A = 99°

OR

Answer: Let age of Sumit be x years and age of his son be y years.Then, according to question we have, x = 3y …… (i)

Five years later, x + 5 = 2½ (y + 5)……….(ii) 

On putting x = 3y in equation (ii)

 

Q.12. Find the mode of the following frequency distribution:

Answer :

Here, the maximum frequency is 50.

So, 35 – 40 will be the modal class.

l = 35, f₀ = 34, f₁ = 50, f₂ = 42 and h = 5

Section – C

 

Q.13. Find the ratio in which the y-axis divides the line segment joining the points (-1, -4) and (5, -6). Also, find the coordinates of the point of intersection. 

Answer: Let the y-axis cut the line joining point A(-1, -4) and point B(5, -6) in the ratio k : 1 at the point P(0, y)

Then, by section formula, we have

 

Q.14. Find the value of:

Answer :

 

Q.15. Two spheres of same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the new sphere. 

Answer: Given, a radius of small sphere be r = 3 cm

Both spheres are made by the same metal, then their densities will be same.

Let radius of bigger sphere = r’ then,

Then according to question, we have,

Volume of bigger sphere + Volume of smaller sphere = Volume of new sphere.

4/3(r)³  + 4/3 (r)³ = 4/3 (R)³

⇒ r’³ + r³ = R³

⇒ 189 + 27 = R³

⇒ 216 = R³

⇒ R = 6

D = 6 × 2 = 12

Radius of new sphere is 6 cm.

So, the diameter is 12 cm.

 

Q.16. Find the ratio in which the line x – 3y = 0 divides the line segment joining the points (-2, -5) and (6, 3). Find the coordinates of the point of intersection.

Answer: Let the required ratio be k : 1

By section formula, we have

 

Q.17.  Evaluate:

Answer : 

 

Q.18. In Figure 4, a square OABC is inscribed in a quadrant OPBQ. If OA = 15 cm, find the area of the shaded region. (Use π = 3.14)

OR

Q.18. In Figure 5, ABCD is a square with side 2√2 cm and inscribed in a circle. Find the area of the shaded region. (Use π = 3.14)

Answer: Given, OABC is a square with OA = 15 cm

OB = radius = r

 Let side of square be a then,

a² + a² = r²

⇒ 2a² = r²

⇒ r = √2 a

⇒ r = 15√2 cm (∵ a = 15 cm)

Area of square = Side × Side = 15 × 15 = 225 cm²

Area of shaded region = Area of quadrant OPBQ – Area of square

= 353.25 – 225 = 128.25 cm

OR

Answer: Given, ABCD is a square with side 2√2 cm

BD = 2r

In ∆BDC,

BD² = DC² + BC²

⇒ 4r² = 2(DC)² (∵ DC = CB = Side = 2√2 )

⇒ 4r² = 2 × 2√2 × 2√2

⇒ 4r² = 8 × 2

⇒ 4r² = 16

⇒ r² = 4

⇒ r = 2 cm

Area of square BCDA = Side × Side = DC × BC = 2√2 × 2√2 = 8 cm²

Area of circle = πr² = 3.14 × 2 × 2 = 12.56 cm²

Area of shaded region = Area of circle – Area of square. = 12.56 – 8 = 4.56 cm²

 

Q.19. A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 20 cm and the diameter of the cylinder is 7 cm. Find the total volume of the solid. (Use π = 22/7).

Answer: ABCD is a cylinder and BFC and AED are two hemispheres which has radius (r) = 7/2 cm

Total volume of solid = Volume of two hemisphere + Volume of cylinder

= 179.67 + 500.5 = 680.17 cm

 

Q.20. The marks obtained by 100 students in an examination are given below:

Find the mean marks of the students.

Answer :

 

Q.21. For what value of k, is the polynomial f(x) = 3x⁴ – 9x³ + x² + 15x + k completely divisible by 3x² – 5? 

OR

Q.21. Find the zeroes of the quadratic polynomial 7y2 – 11/2y – ⅔ and verify the relationship between the zeroes and the coefficients.

Answer: Given, f(x) = 3x⁴ – 9x³ + x² + 15x + k

It is completely divisible by 3x² – 5

 

Q.22. Write all the values of p for which the quadratic equation x² + px + 16 = 0 has equal roots. Find the roots of the equation so obtained. 

Answer: Given, equation is x² + px + 16 = 0

This is of the form ax² + bx + c = 0

where, a = 1, b = p and c = 16

D = b² – 4ac = p² – 4 × 1 × 16 = p² – 64

for equal roots, we have D = 0

p² – 64 = 0

⇒ p² = 64

⇒ p = ±8 Putting p = 8 in given equation we have,

x2 + 8x + 16 = 0

⇒ (x + 4)² = 0

⇒ x + 4 = 0⇒ x = -4

Now, putting p = -8 in the given equation, we get

x²  > – 8x + 16 = 0

⇒ (x – 4)² = 0

⇒ x = 4

Required roots are -4 and -4 or 4 and 4.

 

Section – D

Q.23. In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then prove that the angle opposite the first side is a right angle. 

Answer: Given, ∆ABC in which AC² = AB² + BC²

To prove: ∠B = 90°Construction : Draw a ∆DEF such that

DE = AB, EF = BC and ∠E = 90°.

Proof: In ∆DEF we have ∠E = 90°

So, by Pythagoras theorem, we have

DF² = DE² + EF²

⇒ DF² = AB² + BC² …(i)

(∵ DE = AB and EF = BC)

AC² = AB² + BC² …(ii) (Given)

From equation (i) and (ii), we get

AC² = DF² ⇒ AC = DF.

Now, in ∆ABC and ∆DEF, we have

AB = DE, BC = EF and AC = DF.

∆ABC = ∆DEF.

Hence, ∠B = ∠E = 90°.

Hence Proved.

 

Q.24. From a point P on the ground, the angle of elevation of the top of the tower is 30° and that of the top of the flag-staff fixed on the top of the tower is 45°. If the length of the flag-staff is 5 m, find the height of the tower. (Use √3 = 1.732) [4]

Answer: Let AB be the tower and BC be the flag-staff.

Let P be a point on the ground such that

∠APB = 30° and ∠APC = 45°, BC = 5 m

Let AB = h m and PA = x metres

From right ∆PAB, we have

Hence, the height of the tower is 6.83 m

 

Q.25. A right cylindrical container of radius 6 cm and height 15 cm is full of ice-cream, which has to be distributed to 10 children in equal cones having a hemispherical shape on the top. If the height of the conical portion is four times its base radius, find the radius of the ice-cream cone.

Answer: Let R and H be the radius and height of the cylinder.

Given, R = 6 cm, H = 15 cm.

Volume of ice-cream in the cylinder = πR²H = π × 36 × 15 = 540π cm³

Let the radius of cone be r cm

Height of the cone (h) = 4r

Radius of hemispherical portion = r cm.

Volume of ice-cream in cone = Volume of cone + Volume of the hemisphere

Number of ice cream cones distributed to the children = 10

⇒ 10 × Volume of ice-cream in each cone = Volume of ice-cream in cylindrical container

⇒ 10 × 2πr =540π

⇒ 20r³ = 540 

⇒ r³ = 27

⇒ r = 3

Thus, the radius of the ice-cream cone is 3 cm.

 

Q.26. Construct an equilateral ∆ABC with each side 5 cm. Then construct another triangle whose sides are ⅔ times the corresponding sides. Draw two concentric circles of radii 2 cm and 5 cm. Take a point P on the outer circle and construct a pair of tangents PA and PB to the smaller circle. Measure PA.

Answer: Steps for construction are as follows:

1. Draw a line segment BC = 5 cm
2. At B and C construct ∠CBX = 60° and ∠BCX = 60°
3. With B as centre and radius 5 cm, draw an arc cutting ray BX at A. On graph paper, we take the scale.
4. Join AC. Thus an equilateral ∆ABC is obtained.

1. Below BC, make an acute angle ∠CBY
2. Along BY, mark off 3 points B₁, B₂, B₃ Such that BB₁, B₁B₂, B₂B₃ are equal.
3. Join B³C
4. From B₂ draw B₂D || B₃C, meeting BC at D
5. From D, draw DE || CA, meeting AB at E.

Then ∆EBD is the required triangle, each of whose sides is ⅔ of the corresponding side of ∆ABC.

 

Q.27. Change the following data into ‘less than type’ distribution and draw its olive:

Answer :

 

Q.28. Prove that:

Answer :

 

Q.29. Which term of the Arithmetic Progression -7, -12, -17, -22,…..will be -82? Is -100 any term of the A.P.? Give a reason for your answer. 

OR

Q.29. How many terms of the Arithmetic Progression 45, 39, 33, … must be taken so that their sum is 180? Explain the double answer.

Answer:-7, -12, -17, -22, …….

Here a = -7, d = -12 – (-7) = -12 + 7 = -5

Let T_n = -82T_n = a + (n – 1) d

⇒ -82 = -7 + (n – 1) (-5)

⇒ -82 = -7 – 5n + 5

⇒ -82 = -2 – 5n

⇒ -82 + 2 = -5n

⇒ -80 = -5n

⇒ n = 16

Therefore, 16th term will be -82.

Let T_n = -100

Again, T_n = a + (n -1) d

⇒ -100 = -7 + (n – 1) (-5)

⇒ -100 = -7 – 5n + 5

⇒ -100 = – 2 – 5n

⇒ -100 + 2 = -5n

⇒ -98 = -5n

⇒ n = 98/5 

But the number of terms can not be in fraction.

So, -100 can not be the term of this A.P.

OR

Answer: 45, 39, 33, …..

Here a = 45, d = 39 – 45 = -6

Let S_n = 180

⇒ n/2 [ 2a + (n – 1) d] = 180

⇒ n/2 [2 × 45 + (n – 1) (-6)] = 180

⇒ n/2 [90 – 6n + 6] = 180

⇒ n/2 [96 – 6n] = 180

⇒ n(96 – 6n) = 360

⇒ 96n – 6n² = 360

⇒ 6n² – 96n + 360 = 0

On dividing the above equation by 6

⇒ n² – 16n + 60 = 0

⇒ n² – 10n – 6n + 60 = 0

⇒ n(n – 10) – 6 (n – 10) = 0

⇒ (n – 10) (n – 6) = 0

⇒ n = 10, 6

Sum of first 10 terms = Sum of first 6 terms = 180

This means that the sum of all terms from 7th to 10th is zero.

 

Q.30. In a class test, the sum of Aran’s marks in Hindi and English is 30. Had he got 2 marks more in Hindi and 3 marks less in English, the product of the marks would have been 210. Find his marks in the two subjects.

Answer: Let Aran marks in Hindi be x and marks in English be y.

Then, according to question, we have

x + y = 30 …(i)(x + 2)(y – 3) = 210 …(ii)

from equation (i) put x = 30 – y in equation (ii)

(30 – y + 2) (y – 3) = 210

⇒ (32 – y) (y – 3) = 210

⇒ 32y – 96 – y² + 3y = 210

⇒ y² – 35y + 306 = 0

⇒ y² – 18y – 17y + 306 = 0

⇒ y(y – 18) – 17(y – 18) = 0

⇒ (y – 18) (y – 17) = o

⇒ y = 18, 17

Put y = 18 and 17 in equation (i), we get x = 12, 13

Hence his marks in hindi can be 12 and 13 and in english his marks can be 18 and 17.