Multiple Choice Questions Q1: A bucket is in the form of a frustum of a cone ad holds 28.490 liters of water. The radii of the top and bottom are 28cm and 21cm respectively. Find the height of the bucket. (a) 20 cm (b) 15 cm (c) 10 cm (d) None of the above
Q2: Lead spheres of diameter 6cm are dropped into a cylindrical beaker containing some water and are fully submerged. If the diameter of the beaker is 18cm and water rises by 40cm. find the no. of lead sphere dropped in the water. (a) 90 (b) 150 (c) 100 (d) 80
Short Answer Questions Q3: A cone, a hemisphere and a cylinder stand on equal bases and have the same height. Show that their volumes are in the ratio 1: 2: 3.
Q4: Prove that the surface area of a sphere is equal to the curved surface area of the circumscribed cylinder.
Q5: 500 persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is 0.04m3?
Q6: The radii of the basses of two right circular solid cones of same height are r1 and r2 respectively. The cones are melted and recast into a solid sphere of radius R. Show that the height of each cone is given by
Q7: The slant height of the frustum of a cone is 4cm, and the perimeter of its circular bases is 18cm and 6cm respectively. Find the curved surface area of the frustum ?
Q8: A bucket of height 8cm and made up of copper sheet is in the form of frustum of a right circular cone with radii of its lower and upper ends as 3cm and 9m respectively. Calculate – (a) the height of cone of which the bucket is a part. (b) the volume of water which can be filled in the bucket. (c) the area of copper shut required to make the bucket.
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Multiple Choice Questions Q1: The part of the circular region enclosed by a chord and the corresponding arc of a circle is called (a) a segment (b) a diameter (c) a radius (d) a sector
Q2: If ‘r’ is the radius of a circle, then it’s circumference is given by (a) 2πr (b) None of these (c) πr (d) 2πd
Q3: The angle described by the minute hand between 4.00 pm and 4.25 pm is (a) 900 (b) 1500 (c) 1250 (d) 1000
Q4: If a line meets the circle in two distinct points, it is called (a) a chord (b) a radius (c) secant (d) a tangent
Q5: The perimeter of a protractor is (a) πr (b) πr+2r (c) π+r (d) π+2r
Q6: The perimeter of a circle having radius 5cm is equal to: (a) 30 cm (b) 3.14 cm (c) 31.4 cm (d) 40 cm
Q7: Area of the circle with radius 5cm is equal to: (a) 60 sq.cm (b) 75.5 sq.cm (c) 78.5 sq.cm (d) 10.5 sq.cm
Q8: The largest triangle inscribed in a semi-circle of radius r, then the area of that triangle is; (a) r2 (b) 1/2r2 (c) 2r2 (d) √2r2
Q9: If the perimeter of the circle and square are equal, then the ratio of their areas will be equal to: (a) 14:11 (b) 22:7 (c) 7:22 (d) 11:14
Q10: The area of the circle that can be inscribed in a square of side 8 cm is (a) 36 π cm2 (b) 16 π cm2 (c) 12 π cm2 (d) 9 π cm2
Very Short Answer Question
Q1: A bicycle wheel makes 5000 revolutions in moving 11km.Find the diameter of the wheel.
Q2: A chord AB of a circle of radius 10 cm makes a right angle at the centre of the circle. Find the area of the minor and major segments.
Q3: Find the difference of the areas of a sector of angle 1200 and its corresponding major sector of a circle of radius 21 cm.
Q4: A boy is cycling such that the wheels of the cycle are making 140 revolutions per minute. If the diameter of the wheel is 60cm, calculate the speed per hour with which the boy is cycling
Long Question Answer
Q1: On a square cardboard sheet of area 784 cm2, four circular plates of maximum size are placed such that each circular plate touches the other two plates and each side of the square sheet is tangent to circular plates. Find the area of the square sheet not covered by the circular plates.
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Multiple Choice Question Q1: If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is:
(a) 6 cm (b) 8 cm (c) 10 cm (d) 12 cm
Q2: In Fig, if ∠DAB = 60º, ∠ABD = 50º, then ∠ACB is equal to:
(a) 60º (b) 50º (c) 70º (d) 80º
Q3: AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is:
(a) 17 cm (b) 15 cm (c) 4 cm (d) 8 cm
Q4: In Fig, if AOB is a diameter of the circle and AC = BC, then ∠CAB is equal to:
(a) 30º (b) 60º (c) 90º (d) 45º
Q5: In Fig, ∠AOB = 90º and ∠ABC = 30º, then ∠CAO is equal to:
(a) 30º (b) 45º (c) 90º (d) 60º
True or False
Q6: Through three collinear points a circle can be drawn.
Q7: If A, B, C, D are four points such that ∠BAC = 30° and ∠BDC = 60°, then D is the centre of the circle through A, B and C.
Q8: Two chords AB and AC of a circle with centre O are on the opposite sides of OA. Then ∠OAB = ∠OAC.
Q9. If AOB is a diameter of a circle and C is a point on the circle, then AC2 + BC2 = AB2
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Q1: If the length of the shadow of a tree is decreasing then the angle of elevation is: (a) Increasing (b) Decreasing (c) Remains the same (d) None of the above Q2. The angle of elevation of the top of a building from a point on the ground, which is 30 m away from the foot of the building, is 30°. The height of the building is: (a) 10 m (b) 30/√3 m (c) √3/10 m (d) 30 m
Q3: If the height of the building and distance from the building foot’s to a point is increased by 20%, then the angle of elevation on the top of the building: (a) Increases (b) Decreases (c) Do not change (d) None of the above
Q4: If a tower 6m high casts a shadow of 2√3 m long on the ground, then the sun’s elevation is: (a) 60° (b) 45° (c) 30° (d) 90° Q5: The angle of elevation of the top of a building 30 m high from the foot of another building in the same plane is 60°, and also the angle of elevation of the top of the second tower from the foot of the first tower is 30°, then the distance between the two buildings is: (a) 10√3 m (b) 15√3 m (c) 12√3 m (d) 36 m Q6: The angle formed by the line of sight with the horizontal when the point is below the horizontal level is called: (a) Angle of elevation (b) Angle of depression (c) No such angle is formed (d) None of the above
Q7: The angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level is called: (a) Angle of elevation (b) Angle of depression (c) No such angle is formed (d) None of the above
Q8: From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. The height of the tower (in m) standing straight is: (a) 15√3 (b) 10√3 (c) 12√3 (d) 20√3 Q9: The line drawn from the eye of an observer to the point in the object viewed by the observer is said to be (a) Angle of elevation (b) Angle of depression (c) Line of sight (d) None of the above
Q10: The height or length of an object or the distance between two distant objects can be determined with the help of: (a) Trigonometry angles (b) Trigonometry ratios (c) Trigonometry identities (d) None of the above
Solve the following Questions
Q1: Two poles of equal heights are standing opposite to each other on either side of the road which is 80m wide. From a point between them on the road the angles of elevation of the top of the poles are 60°and 30°.find the height of the poles and the distances of the point from the poles.
Q2: A tree standing on a horizontal plane leaning towards east. At two points situated at distances a and b exactly due west on it, the angles of elevation of the top are respectively α and β .Prove that the height of the top from the ground is .
Q3: A man sitting at a height of 20m on a tall tree on a small island in the middle of the river observes two poles directly opposite to each other on the two banks of the river and in line with the foot of tree. If the angles depression of the feet of the poles from a point at which the man is sitting on the tree on either side of the river are 60° and 30° respectively. Find the width of the river.
Q4: Consider right triangle ABC, right angled at B. If AC = 17 units and BC = 8 units determine all the trigonometric ratios of angle C.
Q5: If C and Z are acute angles and that cos C = cos Z prove that ∠C = ∠Z.
Q6: In triangle ABC, right angled at B if sin A = 1/2 . Find the value of 1. sin C cos A – cos C sin A 2. cos A cos C + sin A sin C
Q7: In triangle ABC right angled at B, AB = 12cm and ∠CAB = 60°. Determine the lengths of the other two sides.
Q8: If θ is an acute angle and find θ.
Q9: Find the value of x in each of the following.
(i) cosec 3x = (ii) cos x = 2 sin 45° cos 45° – sin 30°
Q10: Given sin A = 12/37, find cos A and tan A.
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Multiple choice Questions Q1: Find the centroid of the triangle XYZ whose vertices are X (3, – 5) Y (- 3, 4) and Z (9, – 2). (a) (0, 0) (b) (3, 1) (c) (2, 3) (d) (3,-1)
Q2: If the mid-point of the line segment joining the points A (3, 4) and B (a, 4) is P (x, y) and x + y – 20 = 0, then find the value of a (a) 0 (b) 1 (c) 40 (d) 45
Q3: Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A (2, – 2) and B (3, 7). (a) 2:9 (b) 1:9 (c)1:2 (d) 2:3
Short Answer type Q4: The coordinates of one end point of a diameter of a circle are (4, -1) and the co-ordinates of the centre of the circle are (1, -3). Find the co- ordinates of the other end of the diameter.
Q5: If the mid- point of the line joining (3, 4) and (z, 7) is (x, y) and 2x + 2y + 1 = 0 find the value of z.
Long Answer Type Q6: Find the centre of the circle passing through (6, -6), (3, -7) and (3, 3)
Q7: Prove that the points (3, 0), (4, 5), (-1, 4) and (-2, -1), taken in order, form a rhombus. Also, find its area.
Q8: Two opposite vertices of a square are (-1, 2) and (3, 2). Find thee co-ordinates of other two vertices.
Q9: Find a point on y – axis which is equidistant from the points (5, -2) and (-3, 2).
Q10: If the coordinates of the mid points of the sides of a triangle are (1, 2), (0, -1) and (2, -1). Find the coordinates of its vertices.
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Q1: Which of the following triangles have the same side lengths? (a) Scalene (b) Isosceles (c) Equilateral (d) None of these
Q2: Area of an equilateral triangle with side length a is equal to: (a) (√3/2)a (b) (√3/2)a2 (c) (√3/4) a2 (d) (√3/4) a
Q3: D and E are the midpoints of side AB and AC of a triangle ABC, respectively, and BC = 6 cm. If DE || BC, then the length (in cm) of DE is: (a) 2.5 (b) 3 (c) 5 (d) 6 Q4: The diagonals of a rhombus are 16 cm and 12 cm, in length. The side of the rhombus in length is: (a) 20 cm (b) 8 cm (c) 10 cm (d) 9 cm
Q5: Corresponding sides of two similar triangles are in the ratio of 2:3. If the area of the small triangle is 48 sq.cm, then the area of the large triangle is: (a) 230 sq.cm. (b) 106 sq.cm (c) 107 sq.cm. (d) 108 sq.cm
Q6: If the perimeter of a triangle is 100 cm and the length of two sides are 30 cm and 40 cm, the length of the third side will be: (a) 30 cm (b) 40 cm (c) 50 cm (d) 60 cm
Q7: If triangles ABC and DEF are similar and AB = 4 cm, DE = 6 cm, EF = 9 cm, and FD = 12 cm, the perimeter of triangle ABC is: (a) 22 cm (b) 20 cm (c) 21 cm (d) 18 cm
Q8: The height of an equilateral triangle of side 5 cm is: (a) 4.33 cm (b) 3.9 cm (c) 5 cm (d) 4 cm
Q9: If ABC and DEF are two triangles and AB/DE = BC/FD, then the two triangles are similar if (a) ∠A = ∠F (b) ∠B = ∠D (c) ∠A = ∠D (d) ∠B = ∠E
Q10: Sides of two similar triangles are in the ratio 4: 9. Areas of these triangles are in the ratio (a) 2: 3 (b) 4: 9 (c) 81: 16 (d) 16: 81
Solve the following Questions
Q1: The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR.
Q2: Determine whether the triangle having sides (b − 1) cm, 2√b cm and (b + 1) cm is a right angled triangle.
Q3: Sides of triangles are given below. Determine which of them are right triangles.In case of a right triangle, write the length of its hypotenuse. (i) 7 cm, 24 cm, 25 cm (ii) 3 cm,4 cm,5 cm (iii) 40 cm, 80 cm, 100 cm (iv) 13 cm, 12 cm, 5 cm
Q4: In ΔABC, AD is perpendicular to BC. Prove that: a. AB2 + CD2 = AC2 + BD2 b. AB2 − BD2 = AC2 − CD2
Q5: Triangle ABC is right- angled at B and D is the mid – point of BC. Prove that: AC2 = 4AD2 − 3AB2
Q6: ABC is an isosceles triangle, right -angled at C. Prove that AB2 = 2BC2 .
Q7: A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Q8: The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its tip reach?
Q9: The areas of two similar triangles are 121 cm2 and 64 cm2 respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other.
Q10: DEF is an equilateral triangle of side 2b. Find each of its altitudes.
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Q.1.A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.
Ans: Given, a natural number increased by 12 equals to 160 times its reciprocal. We have to find the natural number. Let the natural number be x. So, x + 12 = 160(1/x) On transposing 1/x to L.H.S ,
we get, x2 + 12x = 160 x2 + 12x – 160 = 0 On factoring, x2 + 20x – 8x – 160 = 0 x(x + 20) – 8(x + 20) = 0 (x – 8)(x + 20) = 0 Now, x – 8 = 0 x = 8 Also, x + 20 = 0 x = -20 Since a negative number cannot be negative, x = -20 is neglected. Therefore, the natural number is x = 8.
Q.2. By increasing the list price of a book by ₹ 10, a person can buy 10 books less for ₹ 1200. Find the original list price of the book.
Ans: Let original book price = X Number of books that can be bought in 1200 =1200/x Suppose price increases by 10. New price = x + 10 No of books that can be bought in new price = 1200/(x + 10) Now, this number is 10 less than the no of books that can be bought in 1200 at ‘x’ price.We are given that, 1200/x−(1200/x+10)=10 ⇒1200×10=10x(x+10) ⇒x2+10x−1200=0 ⇒x2+40x−30x−1200=0 ⇒x(x+40)−30(x+40)=0 ⇒(x+40)(x−30)=0 ⇒x=−40,30 This is a quadratic equation having two solutions, 30 & -40. Ignoring negative value the answer is x = 30 So original book price is Rs 30
Q.3.The hypotenuse of a right-angled triangle is 1 cm more than twice the shortest side. If the third side is 2 cm less than the hypotenuse, find the sides of the triangle.
Ans:Let the shortest side of the right triangle be x m Then, Hypotenuse = (2x – 1) m And Third side = (x + 1) m Using Pythagoras Theorem Hypotenuse² = (Side 1)² + (Side2)² (2x – 1)² = x² + (x + 1)² 4x² – 4x + 1 = x² + x² + 2x + 1 4x² – 4x + 1 = 2x² + 2x + 1 ⇒ 4x² – 2x² – 4x – 2x + 1 – 1 = 0 ⇒ 2x² – 6x = 0 ⇒ 2x (x – 3) = 0 ⇒ 2x = 0 or x – 3 = 0 ⇒ x = 0 or x = 3 ⇒ x = 3 [ By rejecting x = 0 Since , Side cannot be 0 ] Thus, x = 3 ⇒ 2x – 1 = 2(3) – 1 = 5 { Hypotenuse } ⇒ x + 1 = 3 + 1 = 4 { Third side } Obviously x = 0 cm is not a realistic answer in this case. The only acceptable answer is x = 8 cm The two shorter sides are 8, 15 and the hypotenuse is 17 cm.
Q.4. A passenger train takes 2 hours less for a journey of 300 km, if its speed is increased by 5 km/hr from its usual speed. Find its usual speed.
Ans: Given: A train takes two hours less for a journey of 300 km Formula Used: Distance = Speed × Time Calculation: Suppose the usual speed of the train = x km/hr According to the question 300/x – 300/(x + 5) = 2 ⇒ 300x + 1500 – 300x = 2x2 + 10x ⇒ 2x2 + 10x – 1500 = 0 ⇒ 2x2 + 60x – 50x – 1500 = 0 ⇒ 2x (x + 30) – 50(x + 30) = 0 ⇒ (x + 30) (2x – 50) = 0 ⇒ x = 25 (As negative value of x is not possible) ∴ Usual speed of train is 25 km/hr.
Q.5. The numerator of a fraction is one less than its denominator. If three is added to each of the numerator and denominator, the fraction is increased by 3/28. Find the fraction.
Ans: Let the denominator = x Given that numerator is one less than the denominator ⇒ numerator = x – 1 So, the fraction = (x – 1)/x According to the question,
Q.6.The difference of squares of two natural numbers is 45. The square of the smaller number is four times the larger number. Find the numbers
Ans: Let the smaller natural number be x and larger natural number be y Hence x2 = 4y → (1) Given y2 – x2 = 45 ⇒ y2 – 4y = 45 ⇒ y2 – 4y – 45 = 0 ⇒ y2 – 9y + 5y – 45 = 0 ⇒ y ( y – 9 ) + 5 ( y – 9 ) = 0 ⇒ ( y – 9 ) ( y + 5 ) = 0 ⇒ ( y – 9 ) = 0 o r ( y + 5 ) = 0 ∴ y = 9 o r y = − 5 But y is natural number, hence y ≠ – 5 Therefore, y = 9 Equation (1) becomes, x2 = 4 (9) = 36 ∴ x = 6 Thus the two natural numbers are 6 and 9.
Q.7.Solve: x2 + 5√5x – 70 = 0
Ans: Given, the equation is x2 + 5√5x – 70 = 0 We have to find whether the equation has real roots or not. A quadratic equation ax2 + bx + c = 0 has 2 distinct real roots when the discriminant of the equation is greater than zero. Discriminant = b2 – 4ac Here, a = 1, b = 5√5 and c = -70 So, b2 – 4ac = (5√5)2 – 4(1)(-70) = 125 + 280 = 405 > 0 So, the equation has 2 distinct real roots. By using the quadratic formula, x = [-b ± √b2 – 4ac]/2a x = (-5√5 ± √405)/2(1) = (-5√5 ± 9√5)/2 Now, x = (-5√5 + 9√5)/2 = 4√5/2 = 2√5 x = (-5√5 – 9√5)/2 = -14√5/2 = -7√5 Therefore, the roots of the equation are –7√5 and 2√5.
Q.8.A train travels a distance of 300 km at a uniform speed. If the speed of the train is increased by 5 km an hour, the journey would have taken two hours less. Find the original speed of the train.
Ans: Total distance travelled = 300 km Let the speed of train = x km/hr We know that,
Hence, time taken by train = 300/x According to the question, Speed of the train is increased by 5km an hour ∴ the new speed of the train = (x + 5)km/hr Time taken to cover 300km = 300/(x + 5) Given that time taken is 2hrs less from the previous time
⇒ 300x + 1500 – 300x = 2x (x + 5) ⇒ 1500 = 2x2 + 10x ⇒ 750 = x2 + 5x ⇒ x2 + 5x – 750 = 0 ⇒ x2 + 30x – 25x – 750 = 0 ⇒ x (x + 30) – 25 (x + 30) = 0 ⇒ (x – 25) (x + 30) = 0 ⇒ (x – 25) = 0 or (x + 30) = 0 ∴ x = 25 or x = – 30 Since, speed can’t be negative. Hence, the speed of the train is 25km/hr
Q.9.The speed of a boat in still water is 11 km/ hr. It can go 12 km upstream and returns downstream to the original point in 2 hours 45 minutes. Find the speed of the stream.
Ans: Let the speed of the stream be x km/hr Given that, the speed boat in still water is 11 km/hr. ⇒ speed of the boat upstream = (11 – x) km/hr ⇒ speed of the boat downstream = (11 + x) km/hr It is mentioned that the boat can go 12 km upstream and return downstream to its original point in 2 hr 45 min. ⇒ One-way Distance traveled by boat (d) = 12 km ⇒ Tupstream + Tdownstream = 2 hr 45 min = (2 + 3/4) hr = 11/4 hr ⇒ [distance / upstream speed ] + [distance / downstream speed] = 11/4 ⇒ [ 12/ (11-x) ] + [ 12/ (11+x) ] = 11/4 ⇒ 12 [ 1/ (11-x) + 1/(11+x) ] = 11/4 ⇒ 12 [ {11 – x + 11 + x} / {121 – x2} ] = 11/4 ⇒ 12 [ {22} / {121 – x2} ] = 11/4 ⇒ 12 [ 2 / {121 – x2} ] = 1/4 ⇒ 24 / {121 – x2} = 1/4 ⇒ 24 (4) = {121 – x2} ⇒ 96 = 121 – x2 ⇒ x2 = 121 – 96 ⇒ x2 = 25 ⇒ x = + 5 or -5 As speed to stream can never be negative, we consider the speed of the stream(x) as 5 km/hr. Thus, the speed of the stream is 5 km/hr.
Q.10. Determine the value of k for which the quadratic equation 4x2 – 3kx + 1 = 0 has equal roots.
Ans: The given equation 4x2 – 3kx + 1 = 0 is in the form of ax2 + bx + c = 0 Where a = 4, b = -3k, c = 1 For the equation to have real and equal roots, the condition is D = b2 – 4ac = 0 ⇒ (-3k)2 – 4(4)(1) = 0 ⇒ 9k2 – 16 = 0
Q.11. Using quadratic formula, solve the following equation for ‘x’: ab x2 + (b2 – ac) x – bc = 0
⇒ x = −(b² − ac) ± (b² + ac)2ab or −(b² − ac) ± (b² − ac)2ab
⇒ x = 2ac2ab or −2b²2ab
⇒ x = cb or −ba
Q.12. The sum of the numerator and the denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes 1/2. Find the fraction.
Ans: Let’s assume the numerator of the fraction to be x and the denominator of the fraction to be y. So, the required fraction is x/y. From the question it’s given as, The sum of the numerator and denominator of the fraction is 12. Thus, the equation so formed is, x + y = 12 ⇒ x + y – 12 = 0 And also it’s given in the question as, If the denominator is increased by 3, the fraction becomes 1/2. Putting this as an equation, we get x/ (y + 3) = 1/2 ⇒ 2x = (y + 3) ⇒ 2x – y – 3 = 0 The two equations are, x + y – 12 = 0…… (i) 2x – y – 3 = 0…….. (ii) Adding (i) and (ii), we get x + y – 12 + (2x – y – 3) = 0 ⇒ 3x -15 = 0 ⇒ x = 5 Using x = 5 in (i), we find y 5 + y – 12 = 0 ⇒ y = 7 Therefore, the required fraction is 5/7.
Q.13. Rewrite the following as a quadratic equation in x and then solve for x:
Ans: Given expression is
Solve the above expression
Cross multiplying the above equation (4 – 3x) (2x + 3) = 5x 8x + 12 – 6x2 – 9x = 5x – 6x2 + 8x – 9x – 5x + 12 = 0 – 6x2 – 6x + 12 = 0 Divide the above equation by -6 we get x2 + x – 2 = 0 By following factorization method x2 + 2x – x – 2 = 0 x (x + 2) -1(x + 2) = 0 (x + 2)(x – 1) = 0 x + 2 = 0 or x – 1= 0 x = –2, 1 The solution of the given expression is x = -2 and x = 1
Q.14. A two digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.
Ans: Let us consider, one’s digit of a two digit number = x and Ten’s digit = y The number is x + 10y After interchanging the digits One’s digit = y Ten’s digit = x The number is y + 10x As per the statement, xy = 18 ………(1) And, x + 10y -63 = y + 10x 9y – 9x – y – 10x = 63 y – x = 7 …..(2) using algebraic identity: (x + y)2 = (x – y)2 + 4xy (x + y)2 = (-7)2 + 4(18) = 121 (Using (1)) (x + y)2 = 112 or x + y = 11 …..(3) Add (1) and (2) 2y = 18 or y = 9 From (1): xy = 18 9x = 18 x = 2 Answer: The number is: x + 10y = 2 + 10(9) = 92
Q.15.A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/ hr more, it would have taken 30 minutes less for the journey. Find the original speed of the train.
Ans: Let the original speed of the train be x km/hr. According to the question:
90x − 90x + 15 = 12
⇒ 90(x + 15) − 90xx(x + 15) = 12
⇒ 90x + 1350 − 90xx² + 15x = 12
⇒ 1350x² + 15x = 12
⇒ 2700 = x² + 15x
⇒ x² + 15x − 2700 = 0
⇒ x² + (60 − 45)x − 2700 = 0
⇒ x(x + 60) − 45(x + 60) = 0
⇒ (x + 60)(x − 45) = 0
⇒ x = −60 or x = 45
x cannot be negative; therefore, the original speed of train is 45 km/hr.
Substituting values, we get: x = 9(a + b) ± √[9(a − b)²]2 × 9
⇒ x = 3(a + b) ± 3(a − b)6
Simplifying, we get: x = (2a + 4b)6 or (a + 2b)3
Hence, x = 2a + b3 and x = a + 2b3 are the required solutions.
Q.18. A number consisting of two digits is seven times the sum of its digits. When 27 is subtracted from the number, the digits are reversed. Find the number.
Ans: Let the tens and the units digits of the required number be x and y, respectively. Required number = (10x + y) 10x + y = 7(x + y) 10x + 7y = 7x + 7y or 3x – 6y = 0 ……….(i) Number obtained on reversing its digits = (10y + x) (10x + y) – 27 = (10y + x) ⇒ 10x – x + y – 10y = 27 ⇒ 9x – 9y = 27 ⇒9(x – y) = 27 ⇒ x – y = 3 ……..(ii) On multiplying (ii) by 6, we get: 6x – 6y = 18 ………(iii) On subtracting (i) from (ii), we get: 3x = 18 ⇒ x = 6 On substituting x = 6 in (i) we get 3 × 6 – 6y = 0 ⇒ 18 – 6y = 0 ⇒ 6y = 18 ⇒ y = 3 Number = (10x + y) = 10 × 6 + 3 = 60 + 3 = 63 Hence, the required number is 63.
Q.19.The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
Ans: Let the two consecutive odd numbers be x and x + 2. ∴ x2 + (x + 2)2 = 394 ⇒ x2 + x2 + 4x + 4 = 394 ⇒ 2x2 + 4x – 390 = 0 ⇒ x2 + 2x – 195 = 0 ⇒ x2 + 15x – 13x – 195 = 0 or x(x + 15) – 13(x + 15) = 0 ⇒ x= 13 or x = – 15 ∴ For x = 13, x + 2 = 13 + 2 = 15 Thus, the required numbers are 13 and 15.
Q.20.Find the number which exceeds its positive square root by 20.
Ans: Let the Number be x According to the given Question √x + 20 = x √x = x – 20 Squaring both the sides x = x2 + 400 – 40x [(a – b)2 = a2 – 2ab + b2] x2 – 41x + 400 = 0 x2 – 16x – 25x + 400 = 0 x(x – 16) – 25(x – 16) = 0 (x – 16)(x – 25) = 0 Ans = x is equal to 25 or 16 both are correct answers
Q.21.A two digit number is such that the product of its digits is 14. If 45 is added to the number, the digits interchange their places. Find the number.
Ans: Let the no be ‘xy’ with x × y = 14 As xy + 45 = yx(x, y > 0) ⇒ (10x + y) + 45 = 10y + x ⇒ 9x − 9y + 45 = 0 ⇒ x − y + 5 = 0 & x × y = 14 ⇒ x− 14/x + 5 = 0 ⇒ x − 14/x + 5 = 0 ⇒ x2 + 5x − 14 = 0 ⇒ x = 2,−7 & y = 7, −2 Ad x, y > 0 ⇒ no is 27
Q.22. A two digit number is such that the product of its digits is 20. If 9 is added to the number, the digits interchange their places. Find the number.
Ans: Let the one’s digit be ‘a’ and ten’s digit be ‘b’. Given, two – digit number is such that the product of its digits is 20. ⇒ ab = 20 —– (1) Also, If 9 is added to the number, the digits interchange their places. ⇒ 10b + a + 9 = 10a + b ⇒ a – b = 1 —– (2) Substituting value of a from eq1 in to eq2 ⇒ 20/b – b = 1 ⇒ b2 + b – 20 = 0 ⇒ (b + 5)(b – 4) = 0 Thus, b = 4 and a = b + 1 = 5 Number is 45.
Q.23.A two digit number is such that the product of its digits is 15. If 8 is added to the number, the digits interchange their places. Find the number.
Ans: Let the ten’s digit be x and one’s digit be y. The number will be 10x + y. Given product of its digits is 15 xy = 15 y = 15/x — (1). Given that when 18 is added to the number the digits get interchanged. 10x + y + 18 = 10y + x 9x – 9y + 18 = 0 x – y + 2 = 0 ——- (2) Substitute (1) in (2), we get x – (15/x) + 2 = 0 x2 + 2x – 15 = 0 x2 + 5x – 3x – 15 = 0 x(x + 5) – 3(x + 5) = 0 (x + 5)(x – 3) = 0 x = -5, 3. Since the digits cannot be negative, x = 3. Substitute x = 3 in (1) y = 15/3 = 5. The number = 10x + y = 10 * 3 + 5 = 30 + 5 = 35.
Q.24.Solve for x: abx2 + (b2 − ac) x − bc = 0
Ans: We have: abx2 + (b2 − ac)x − bc = 0
Here, A = ab, B = b2 − ac, C = −bc
Using the quadratic formula: x = −B ± √(B2 − 4AC)2A
Q.25.The sum of two numbers is 18. The sum of their reciprocals is 1/4. Find the numbers.
Ans: Let The numbers be x and 18 – x
According to the given hypothesis: 1x + 118 − x = 14
⇒ 18 − x + xx(18 − x) = 14
⇒ 72 = 18x − x2
⇒ x2 − 18x − 72 = 0
Factorizing: x2 − 6x − 12x − 72 = 0
⇒ x(x − 6) − 12(x − 6) = 0
⇒ (x − 6)(x − 12) = 0
⇒ x = 6 or x = 12
∴ The two numbers are 6 and 12.
Q.26.The sum of two numbers ‘a’ and ‘b’ is 15, and sum of their reciprocals 1/a and 1/b is 3/10. Find the numbers ‘a’ and ‘b’.
Ans: The two numbers are a and b According to the question, ⇒ a + b = 15 ⇒ a = 15 − b —- ( 1 ) ⇒ 1/a + 1/b = 3/10 ⇒ 1/15−b + 1/b = 3/10 ⇒ b + 15 − b/b(15 − b) = 3/10 ⇒ 150 = 3b(15−b) ⇒ 50 = b(15 − b) ⇒ 50 = 15b − b2 ⇒ b2 − 15b + 50 = 0 ⇒ b2 − 10b − 5b + 50 = 0 ⇒ b(b − 10) − 5(b − 10) = 0 ⇒ (b − 10)(b − 5) = 0 ∴ b = 5 or b = 10 Putting both values of b in equation ( 1), ⇒ a = 15 − 5 or a = 15 − 10 ∴ a = 10 or a = 5
Q.27.Find the roots of the following quadratic equation:
Ans: Given, 2/5 x2 – x – 3/5 = 0 ⇒ 2x2 – 5x – 3 = 0 By splitting the middle term, ⇒ 2x2 – 6x + x – 3 = 0 Taking common in the expression, ⇒ 2x (x – 3) + 1 (x – 3) = 0 ⇒ (2x + 1) (x – 3) = 0 ∴ 2x + 1 = 0 and ∴ x – 3 = 0 ∴ x = -1/2 and ∴ x = 3
Q.28. A natural number when subtracted from 28, becomes equal to 160 times its reciprocal.
Find the number.
Ans: Let the number be x Now, (28 – x) = 160 / x 28x – x2 = 160 x2 – 28x + 160 = 0 x2 – 20x – 8x + 160 = 0 x(x – 20) – 8(x – 20) = 0 (x – 8)(x – 20) = 0 x = 8 , 20
Q.29. Find two consecutive odd positive integers, sum of whose squares is 290.
Ans: Let one of the odd positive integer be x then the other odd positive integer is x + 2 their sum of squares = x2 + (x + 2)2 = x2 + x2 + 4x + 4 = 2x2 + 4x + 4 Given that their sum of squares = 290 2x2 + 4x + 4 = 290 2x2 + 4x = 286 2x2 + 4x − 286 = 0 x2 + 2x − 143 = 0 x2 + 13x − 11x − 143 = 0 x(x + 13) − 11(x + 13) = 0 (x − 11) = 0,(x + 13) = 0 Therefore, x = 11 or −13 We always take positive value of x So, x = 11 and (x + 2) = 11 + 2 = 13 Therefore , the odd positive integers are 11 and 13
Q.30. Find the values of k for which the quadratic equation (k + 4) x2 + (k + 1) x + 1 = 0 has equal roots.
Ans: In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.
Q.33.Find the value of k, for which one root of quadratic equation kx2 – 14x + 8 = 0 is six times the other.
Ans: Let one root = α Other root = 6α
Sum of roots = α + 6α = 14k
Or, 7α = 14k ⇒ α = 2k (i)
Product of roots: α × 6α = 8k
Or, 6α2 = 8k (ii)
Solving (i) and (ii):
6 × (2/k)28/k
⇒ 6 × 4k2 = 8k
⇒ 3k2 = 1k
Or, 3k = k2
Or, k(3 − k) = 0
⇒ k = 0 or k = 3
k = 0 is not possible.
Hence, k = 3
Q.34.If x = 2/3 and x = –3 are roots of the quadratic equation ax2 + 7x + b = 0, find the value of a and b.
Ans: Let us assume the quadratic equation be, Ax2 + Bx + C = 0. Sum of the roots = -B/A
Given:
Sum of roots = -7a = 23
Multiply both sides by a, we get:
-7 = 2a3
Solving for a:
a = 3
Product of roots = CA
Given:
ba = 23 × (-3)
Simplify:
b = -6
Q.35.If –5 is a root of the quadratic equation 2x2 + px – 15 = 0 and the quadratic equation p (x2 + x) + k = 0 has equal roots, find the value of k.
Ans: Given: -5 is a root of the quadratic equation 2x2 + px – 15 = 0 Substitute the value of x = -5 2(-5)2 + p(-5) – 15 = 0 50 – 5p – 15 = 0 35 – 5p = 0 p = 7 Again, In quadratic equation p(x2 + x) + k = 0 7 (x2 + x) + k = 0 (put value of p = 7) 7x2 + 7x + k = 0 Compare given equation with the general form of quadratic equation, which is ax2 + bx + c = 0 a = 7, b = 7, c = k Find Discriminant: D = b2 – 4ac = (7)2 – 4 x 7 x k = 49 – 28k Since roots are real and equal, put D = 0 49 – 28k = 0 28k = 49 k = 7 / 4 The value of k is 7/4.
A. Multiple Choice Questions Q1: A pair of linear equations which have a unique solution x = 2, y = – 3 is: (a) 2x – 3y = – 5, x + y = – 1 (b) 2x + 5y + 11 = 0, 4x + 10y + 22 = 0 (c) x – 4y – 14 = 0, 5x – y – 13 = 0 (d) 2x – y = 1, 3x + 2y = 0
Q2: If a system of a pair of linear equations in two unknowns is consistent, then the lines representing the system will be (a) Parallel (b) Always coincident (c) Always intersecting (d) Intersecting or coincident
Q3: The pair of equations x = 0 and y = 0 has (a) One solution (b) Two solutions (c) Infinitely many solutions (d) No solution
Q4: A pair of system of equations x = 2, y = -2; x = 3, y = – 3 when represented graphically enclose (a) Square (b) Trapezium (c) Rectangle (d) Triangle
Q5: If two lines are parallel to each other then the system of equations is (a) Consistent (b) Inconsistent (c) Consistent dependent (d) (a) and (c) both B. Fill in the blanks
Q6:If in a system of equations corresponding to coefficients of member, equations are proportional then the system has ______________ solution (s).
Q7: A pair of linear equations is said to be inconsistent if its graph lines are ____________.
Q8:A pair of linear equations is said to be ____________ if its graph lines intersect or coincide.
Q9: A consistent system of equations where straight lines fall on each other is also called _____________ system of equations.
Q10:Solution of linear equations representing 2x – y = 0, 8x + y = 25 is ____________ .
C. Very Short Answer Questions
Q11:In Fig., ABCD is a rectangle. Find the values of x and y.
Q12: Graphically, determine whether the following pair of equations has no solution, a unique solution, or infinitely many solutions: (1) 2x – 3y + 4 = 0 (2) 4x – 6y + 8 = 0
Q13: If 51x + 23y = 116 and 23x + 51y = 106, then find the value of (x – y).
Q14: For what value of V does the point (3, a) lie on the line represented by 2x – 3y = 5?
Q15: Determine whether the following system of linear equations is inconsistent or not. 3x – 5y = 20 6x – 10y = -40
D. Short Answer Type Questions
Q16: The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. Find the present ages of the son and the father.
Q17: If the lines x + 2y + 7 = 0 and 2x + ky + 18 = 0 intersect at a point, then find the value of k.
Q18: Find the value of k for which the system of equations x + 2y -3 = 0 and ky + 5x + 7 = 0 has a unique solution.
Q19: Find the values of a and b for which the following system of linear equations has an infinite number of solutions: 2x + 3 y = 7 2αx + (a + b) y = 28
Q20: In a cyclic quadrilateral ABCD, Find the four angles. a. ∠A = (2 x + 4), ∠B = (y + 3), ∠C = (2y + 10) , ∠D = (4x − 5) . b. ∠A = (2 x − 1) , ∠ B = (y + 5) , ∠C = (2 y + 15) and ∠D = (4 x − 7)
E. Long Answer Type Questions
Q21:Draw the graph of 2x + y = 6 and 2x – y + 2 = 0. Shade the region bounded by these lines and the x-axis. Find the area of the shaded region.
Q22: Draw the graphs of the following equations: 2x – y = 1, x + 2y = 13 (i) Find the solution of the equations from the graph. (ii) Shade the triangular region formed by lines and the y-axis.
Q23: A man travels 370 km partly by train and partly by car. If he covers 250 km by train and the rest by car, it takes him 4 hours. But, if he travels 130 km by train and the rest by car, he takes 18 minutes longer. Find the speed of the train and that of the car.
Q24: The taxi charges in a city comprise a fixed charge together with the charge for the distance covered. For a journey of 10 km, the charge paid is ₹75 and for a journey of 15 km, the charge paid is ₹110. What will a person have to pay for traveling a distance of 25 km?
Q25: Solve the following system by drawing their graph: (3/2)x – (5/4)y = 6, 6x – 6y = 20. Determine whether these are consistent, inconsistent, or dependent.
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