1. A cylindrical pencil sharpened at one edge is the combination of

(A) a cone and a cylinder

(B) frustum of a cone and a cylinder

(C) a hemisphere and a cylinder

(D) two cylinders.

Answer: (A)

Explanation: The shape of a sharpened pencil is:

2. A cone is cut through a plane parallel to its base and then the cone that is for medon one side of that plane is removed. The new part that is left over on the other side of the plane is called

(A) a frustum of a cone

(B) cone

(C) cylinder

(D) sphere

Answer:  (A)

Explanation: Observe figure

3. During conversion of a solid from one shape to another, the volume of the new shape will

(A) increase

(B) decrease

(C) remain unaltered

(D) be doubled

Answer:  (C)

Explanation: During conversion of one solid shape to another, the volume of the new shape will remain unaltered.

4. A right circular cylinder of radius r cm and height h cm (h>2r) just encloses a sphere of diameter

(A) r cm

(B) 2r cm

(C) h cm

(D) 2h cm

Answer: (B)

Explanation: Because the sphere is enclosed inside the cylinder, therefore the diameter of sphere is equal to the diameter of cylinder which is 2r cm.

5. A hollow cube of internal edge 22cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that 1/8th space of the cube remains unfilled. Then the number of marbles that the cube can accommodate is

(A) 142244

(B) 142396

(C) 142496

(D) 142596

Answer:  (A)

Explanation:

6. A metallic spherical shell of internal and external diameters 4 cm and 8 cm respectively, is melted and recast into the form of a cone with base diameter 8cm. The height of the cone is

(A) 12cm

(B) 14cm

(C) 15cm

(D) 18cm

Answer:  (B)

Explanation:Since volume will remain same, therefore,

7. A solid piece of iron in the form of a cuboid of dimensions 49cm × 33cm × 24cm, is moulded to form a solid sphere. The radius of the sphere is

(A) 21cm

(B) 23cm

(C) 25cm

(D) 19cm

Answer:  (A)

Explanation: Since volume will remain the same, therefore,

8. If two solid hemispheres of same base radii r, are joined together along their bases, then curved surface area of this new solid is

(A) 4πr²

(B) 6πr²

(C) 3πr²

(D) 8πr²

Answer:  (A)

Explanation: Because curved surface area of a hemisphere is and here we join two solid hemispheres along their bases of radii r, from which we get a solid sphere.

Hence the curved surface area of new solid = 2πr² + 2πr² = 4πr²

9. A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. The total surface area of the shape so formed is

(A) 4πrh + 4πr²

(B) 4πrh − 4πr²

(C) 4πrh + 2πr²           

(D) 4πrh − 2πr²

Answer:  (C)

Explanation: Since the total surface area of cylinder of radius r and height h = 2πrh + 2πr².

When one cylinder is placed over the other cylinder of same height and radius,

Then height of new cylinder = 2h

And radius of the new cylinder = r

Therefore total surface area of new cylinder

= 2πr (2h) + 2πr²

= 4πrh + 2πr²

10. The radii of the top and bottom of a bucket of slant height 45cm are 28cm and 7 cm respectively. The curved surface area of the bucket is:

(A) 4950 cm²

(B) 4951 cm²

(C) 4952 cm²

(D) 4953 cm²

Answer: (A)

Explanation:

Curved Surface area of the bucket = π (R + r) l

⇒ Curved surface area of the bucket = π (28 + 7) X 45

⇒ Curved surface area of the bucket = 4950 cm²

11. A medicine-capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of entire capsule is 2 cm. The capacity of the capsule is

(A) 0.36 cm³

(B) 0.35 cm³

(C) 0.34 cm³

(D) 0.33 cm³

Answer:  (A)

Explanation:

Since diameter of the cylinder = diameter of the hemisphere = 0.5cm

Radius of cylinder r = radius of hemisphere r = 0.5/2 = 0.25 cm

Observe the figure,

Total length of capsule = 2cm

Capacity of capsule is:

12. Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is

(A) 4 cm

(B) 3 cm

(C) 2 cm

(D) 6 cm

Answer:  (C)

Explanation:

Therefore diameter of each solid sphere = 2cm

13. The diameters of the two circular ends of the bucket are 44 cm and 24 cm. The height of the bucket is 35 cm. The capacity of the bucket is

(A) 32.7 litres

(B) 33.7 litres

(C) 34.7 litres

(D) 31.7 litres

Answer:  (A)

Explanation: Since shape of bucket is like Frustum,

Therefore, volume of bucket

14. Volumes of two spheres are in the ratio 64:27. The ratio of their surface areas is:

(A) 3 : 4

(B) 4 : 3

(C) 9 : 16

(D) 16 : 9

Answer: (D)

Explanation: According to question,

Therefore ratio of surface area is:

15. A mason constructs a wall of dimensions 270cm× 300cm × 350cm with the bricks each of size 22.5cm × 11.25cm × 8.75cm and it is assumed that 1/8 space is covered by the mortar. Then the number of bricks used to construct the wall is:

(A) 11100

(B) 11200

(C) 11000

(D) 11300

Answer:  (B)

Explanation: According to question,

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1. If the sum of the areas of two circles with radii R₁ and R₂ is equal to the area of a circle of radius R, then

(A) R₁ + R₂ = R

(B) R₁² + R₂² = R²

(C) R₁ + R₂ < R

(D) R₁² + R₂² < R²

Answer: (B)

Explanation: According to given condition,

Area of circle = Area of first circle + Area of second circle

2. If the circumference of a circle and the perimeter of a square are equal, then

(A) Area of the circle = Area of the square

(B) Area of the circle > Area of the square

(C) Area of the circle < Area of the square

(D) Nothing definite can be said about the relation between the areas of the circle and square.

Answer:  (B)

Explanation: According to given condition

Circumference of a circle = Perimeter of square

Hence Area of the circle > Area of the square

3. Area of the largest triangle that can be inscribed in a semi-circle of radius r units, in square units is:

(A) r²

(B) 1/2r²

(C) 2 r²

(D) √2r²

Answer:  (A)

Explanation: The triangle inscribed in a semi-circle will be the largest when the perpendicular height of the triangle is the same size as the radius of the semi-circle.

Consider the following figure:

4. If the perimeter of a circle is equal to that of a square, then the ratio of their areas is:

(A) 22:7

(B) 14:11

(C) 7:22

(D) 11:14

Answer: (B)

Explanation: Perimeter of circle = Perimeter of square

5. It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be

(A) 10 m

(B) 15 m

(C) 20 m

(D) 24 m

Answer: (A)

Explanation: Area of first circular park, whose diameter is 16m

= πr² = π (16/2)² = 64π m²

Area of second circular park, whose diameter is 12m

= πr² = π (12/2)² = 36π m²

According to question,

Area of new circular park =

6. The area of the circle that can be inscribed in a square of side 6 cm is

(A) 36 π cm²

(B) 18 π cm²

(C) 12 π cm²

(D) 9 π cm²

Answer:  (D)

Explanation: Given,

Side of square = 6 cm

Diameter of a circle = side of square = 6cm

Therefore, Radius of circle = 3cm

Area of circle

= πr²

= π (3)²

= 9π cm²

7. The area of the square that can be inscribed in a circle of radius 8 cm is

(A) 256 cm²

(B) 128 cm²

(C) 642 cm²

(D) 64 cm²

Answer:  (B)

Explanation: Radius of circle = 8 cm

Diameter of circle = 16 cm = diagonal of the square

Therefore side of square = diagonal/√2

= 16/√2

Therefore, are of square is = (side)² = (16/√2)²

= 256/2

= 128 cm²

8. The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36 cm and 20 cm is

(A) 56 cm

(B) 42 cm

(C) 28 cm

(D) 16 cm

Answer:  (C)

Explanation: According to question,

Circumference of circle = Circumference of first circle + Circumference of second circle

πD = πd₁ + πd₂

D = 36 + 20

D = 56cm

9. The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm respectively, is

(A) 31 cm

(B) 25 cm

(C) 62 cm

(D) 50 cm

Answer:  (D)

Explanation: According to question

Therefore diameter = 2 × 25 = 50cm

10. If the length of an arc of a circle of radius r is equal to that of an arc of a circle of radius 2r, then

(A) the angle of the corresponding sector of the first circle is double the angle of the corresponding sector of the other circle.

(B) the angle of the corresponding sector of the first circle is equal the angle of the corresponding sector of the other circle.

(C) the angle of the corresponding sector of the first circle is half the angle of the corresponding sector of the other circle.

(D) the angle of the corresponding sector of the first circle is 4 times the angle of the corresponding sector of the other circle.

Answer:  (A)

Explanation: According to Question,

11. The wheel of a motor cycle is of radius 35 cm. How many revolutions per minute must the wheel make so as to keep a speed of 66 km/h?

(A) 300

(B) 400

(C) 450

(D) 500

Answer: (D)

Explanation:

12. A cow is tied with a rope of length 14 m at the corner of a rectangular field of dimensions 20m × 16m, then the area of the field in which the cow can graze is:

(A) 154 m²

(B) 156 m²

(C) 158 m²

(D) 160 m²

Answer:  (A)

Explanation:Figure according to question is:

Area of the field in which cow can graze= Area of a sector AFEG

= (θ/360) X πr²

= (90/360) X π (14)²

= (1/4) X (22/7) X 196

= 154 m²

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13. The area of the shaded region in Fig., where arcs drawn with centres P, Q, Rand S intersect in pairs at mid-points A, B, C and D of the sides PQ, QR, RS and SP, respectively of a square PQRS, is:

(A) 25.25 cm²

(B) 27.45 cm²

(C) 29.65 cm²

(D) 30.96 cm²

Answer:  (D)

Explanation:

14. Area of a sector of central angle 120° of a circle is 3π cm². Then the length of the corresponding arc of this sector is:

(A) 5.8cm

(B) 6.1cm

(C) 6.3cm

(D) 6.8cm

Answer:  (C)

Explanation:

Given that

Area of a sector of central angle 120° of a circle is 3π cm²

15. A round table cover has six equal designs as shown in the figure. If the radius of thecover is 28 cm, then the cost of making the design at the rate of Rs. 0.35 per cm² is:

(A) Rs.146.50

(B) Rs.148.75

(C) Rs.152.25

(D) Rs.154.75

Answer:  (B)

Explanation: The area of the hexagon will be equal to six equilateral triangles with each side equal to the radius of circle.

Area of given hexagon = Area of 6 equilateral triangles.

= 6 X (√3/4) X (side)²

= 6 X (√3/4) X (28)²

= 1999.2 cm²                           (Taking √3 = 1.7)        

Area of circle = πr²

= π × 28²

= 2464 cm²

So, area of designed portion = 2464 – 1999.2 = 464.8 cm²

Cost of making design = 464.8 × 0.35

= Rs. 162.68

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1. To divide a line segment AB in the ratio 5:7, first a ray AX is drawn so that ∠BAX is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is:

(A) 8

(B) 10

(C) 11

D) 12

Answer: (D)

Explanation: We know that to divide a line segment in the ratio m : n, first draw a ray AX which makes an acute angle BAX , then marked m+n points at equal distances from each other.

Here m = 5, n = 7

So minimum number of these point = m + n = 5 + 7 = 12

2. To divide a line segment AB in the ratio 4:7, a ray AX is drawn first such that ∠BAX is an acute angle and then points A₁, A₂, A₃, ….are located at equal distances on the ray AX and the point B is joined to

(A) A₁₂                                                 

(B) A₁₁

(C) A₁₀                                               

(D) A₉

Answer: (B)

Explanation: Here minimum 4+7=11 points are located at equal distances on the ray AX and then B is joined to last point, i.e., A_11.

3. To divide a line segment AB in the ratio 5 : 6, draw a ray AX such that ∠BAX is an acute angle, then draw a ray BY parallel to AX and the points A₁, A₂, A₃, … and B₁, B₂, B₃, … are located at equal distances on ray AX and BY, respectively. Then the points joined are

(A) A₅ and B₆                                      

(B) A₆ and B₅

(C) A₄ and B₅                                      

(D) A₅ and B₄

Answer:  (A)

Explanation:   Observe the following figure:

4. To construct a triangle similar to a given ΔABC with its sides 3/7 of the corresponding sides of ΔABC, first draw a ray BX such that ∠CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then locate points B₁, B₂, B₃, … on BX at equal distances and next step is to join:

(A) B₁₀ to C                            

(B) B₃ to C

(C) B₇ to C              

(D) B₄ to C

Answer: (C)

Explanation: Here we locate points B₁,B₂,B₃,B₄,B₅,B₆ and B₇ on BX at equal distances and in next step join the last point B₇ to C

5. To construct a triangle similar to a given ΔABC with its sides 8/5 of the corresponding sides of ΔABC draw a ray BX such that ∠CBX is an acute angle and X is on the opposite side of A with respect to BC. The minimum number of points to be located at equal distances on ray BX is:

(A) 5                                                  

(B) 8

(C) 13                             

(D) 3

Answer: (B)

Explanation: To construct a triangle similar to a given triangle with its sides m/n of the corresponding sides of given triangle ,the minimum number of points to be located at equal distance is equal to the greater of m and n in m/n.

Here, m/n = 8/5

So the minimum number of points to be located at equal distance on ray BX is 8.

6. To draw a pair of tangents to a circle which are inclined to each other at an angle of 60°, it is required to draw tangents at end points of those two radii of the circle, the angle between them should be:

(A) 135° 

(B) 90°

(C) 60°

(D) 120⁰

Answer:  (D)

Explanation: The angle between them should be 120⁰ because the figure formed by the intersection point of pair of tangents, the two end points of those two radii (at which tangents are drawn) and the centre of circle, is a quadrilateral. Thus the sum of the opposite angles in this quadrilateral must be 180^o.

7. To divide a line segment AB in the ratio p : q (p, q are positive integers), draw a ray AX so that ∠BAX is an acute angle and then mark points on ray AX at equal distances such that the minimum number of these points is

(A) greater of p and q            

(B) p + q

(C) p + q – 1                            

(D) pq

Answer:  (B)

Explanation: We know that to divide a line segment in the ratio m : n, first draw a ray AX which makes an acute angle BAX , then mark m + n points at equal distances from each other.

Here m = p, n = q

So minimum number of these points = m + n = p + q

8. To draw a pair of tangents to a circle which are inclined to each other at an angle of 35°, it is required to draw tangents at the end points of those two radii of the circle, the angle between which is:

(A) 105°                                             

(B) 70°

(C) 140°                                     

(D) 145°

Answer:  (D)

Explanation: The angle between them should be 145⁰ because the figure formed by the intersection point of pair of tangents, the two end points of those two radii (at which tangents are drawn) and the centre of circle, is a quadrilateral. Thus the sum of the opposite angles in this quadrilateral must be 180^o.

9. By geometrical construction, it is possible to divide a line segment in the ratio:

Answer:  (A)

Explanation:

10. A pair of tangents can be constructed from a point P to a circle of radius 3.5 cm situated at a distance of ___________ from the centre.

(A) 5cm                                              

(B) 2cm

(C) 3cm                                            

(D) 3.5cm

Answer:  (A)

Explanation: The pair of tangents can be drawn from an external point only, so its distance from the centre must be greater than radius. Since only 5cm is greater than radius of 3.5cm. So the tangents can be drawn from the point situated at a distance of 5cm from the centre.

11. To divide a line segment AB in the ratio 5:6, draw a ray AX such that ∠BAX is an acute angle, then drawa ray BY parallel to AX and the points A₁, A₂, A₃,…. and B₁, B₂, B₃,…. are located to equal distances on ray AX and BY, respectively. Then, the points joined are

(A) A₅ and B₆                                             

(B) A₆ and B₅    

(C) A₄ and B₅                                              

(D) A₅ and B₄

Answer: (A)

Explanation:

To divide line segment AB in the ratio 5:6.

Steps of construction

1. Draw a ray AX making an acute ∠BAX.

2. Draw a ray BY parallel to AX by taking ∠ABY equal to ∠BAX.

3. Divide AX into five (m = 5) equal parts AA₁, A₁A₂, A₂A₃, A₃A₄ and A₄A₅

4. Divide BY into six (n = 6) equal parts and BB₁, B₁B₂, B₂B₃, B₃B₄, B₄B₅ and B₅B₆.

5. Join B₆ A₅. Let it intersect AB at a point C.

Then, AC : BC = 5 : 6

12. A rhombus ABCD in which AB = 4cm and ABC = 60^o, divides it into two triangles say, ABC and ADC. Construct the triangle AB’C’ similar to triangle ABC with scale factor 2/3. Select the correct figure.

Answer:  (A)

13. A triangle ABC is such that BC = 6 cm, AB = 4 cm and AC = 5 cm. For the triangle similar to this triangle with its sides equal to (3/4)th of the corresponding sides of ΔABC, correct figure is:

Answer:  (D)

14. For ∆ABC in which BC = 7.5 cm, ∠B =45⁰ and AB – AC = 4, select the correct figure.

(D) None of these

Answer:  (B)

15. Draw the line segment AB = 5 cm. From the point A draw a line segment AD = 6cm making an angle of 60⁰ with AB. Draw a perpendicular bisector of AD. Select the correct figure.

(D) None of these

Answer:  (A)

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1. If angle between two radii of a circle is 130º, then the angle between the tangents at the ends of the radii is: 

(A) 90º

(B) 50º

(C) 70º

(D) 40º

Answer: (B)

Explanation: If the angle between two radii of a circle is 130º, then the angle between tangents is 180º − 130º = 50º. (By the properties of circles and tangents)

2. The length of tangent from an external point P on a circle with centre O is

(A) always greater than OP

(B) equal to OP

(C) always less than OP

(D) information insufficient

Answer:  (C)

Explanation: Observe figure

The angle between tangent and radius is 90º. This implies OP is the hypotenuse for the right triangle OQP right angled at Q and hypotenuse is the longest side in aright triangle, therefore the length of tangent from an external point P on a circle with centre O is always less than OP.     

3. If angle between two tangents drawn from a point P to a circle of radius ‘a’and centre ‘O’ is 90°, then OP =

(A) 2a√2

(B) a√2

(C) a/√2

(D) 5a√2

Answer:  (B)

Explanation: From point P, two tangents are drawn

OT = a (given)

Also line OP bisects the RPT

Therefore,

 TPO =  RPO = 45º

Also

OT is perpendicular to PT

In right triangle OTP

sin 45° = OT/OP

⇒ 1/√2 = a/OP

⇒ OP = a√2

4. The length of tangent from an external point on a circle is

(A) always greater than the radius of the circle.

(B) always less than the radius of the circle.

(C) may or may not be greater than the radius of circle

(D) None of these.

Answer:  (C)

Explanation: Observe the figure,

In figure OQ is the radius and QP is the tangent.

For right triangle OQP, radius and tangents are two smaller sides, smaller than hypotenuse OP.

Also the length of tangent depends upon the distance of external point from circle. Thus,

The length of tangent from an external point on a circle may or may not be greater than the radius of circle.

5. In the following figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. Then AT is equal to

(A) 4 cm

(B) 2 cm

(C) 2√3 cm

(D) 4√3 cm

Answer: (C)

Explanation: Join OA

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Therefore,

∠OAT = 90°

In triangle OAT

cos30° = AT/OT

√3/2 = AT/4

AT = 2√3 cm

6. At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is

(A) 4 cm

(B) 5 cm

(C) 6 cm

(D) 8 cm

Answer:  (D)

Explanation: Observe the figure,

Since OC = OA= radius =5cm

Therefore

OE = AE – AO

OE = 8 – 5

OE = 3 cm

In right triangle OEC

OC² = OE² + CE²

⇒ 5² = 3² + CE²

⇒ CE² = 25 – 9

⇒ CE² = 16

⇒ CE = 4

Therefore length of chord CD = 2×4 = 8cm

7. In the following figure, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ, then ∠POQ is equal to

(A) 100°

(B) 80°

(C) 90°

(D) 75°

Answer:  (A)

Explanation: From figure using properties of circle and tangents

∠OPQ = 90° – 50°

∠OPQ = 40°

OP = OQ = radius

So (∠E) = ∠OQP = 40°

Now In ∆POQ

∠POQ = 180° – (∠OPQ + ∠OQP)

∠POQ = 180° – (40° + 40°)

∠POQ = 100°

8. In the following figure, PA and PB are tangents from a point P to a circle with centre O. Then the quadrilateral OAPB must be a

(A) Square

(B) Rhombus

(C) Cyclic quadrilateral

(D) Parallelogram

Answer:  (C)

Explanation: Since tangent and radius intersect at right angle,

So,

∠OAP = ∠OBP = 90°

⇒ ∠OAP + ∠OBP = 90° + 90° = 180°

Which are opposite angles of quadrilateral OAPB

So the sum of remaining two angles is also 180°

Therefore Quad OAPB is cyclic Quadrilateral.

9. If d₁, d₂ (d₂ > d₁) be the diameters of two concentric circles and c be the length of a chord of a circle which is tangent to the other circle, then

(A) d₂² = c² + d₁²

(B) d₂² = c² – d₁²

(C) d₁² = c² + d₂²

(D) d₁² = c² – d₂²

Answer:  (A)

Explanation:

Let AB be a chord of a circle which touches the other circle at C. Then ΔOCB is right triangle.

By Pythagoras theorem

10. If a chord AB subtends an angle of 60° at the centre of a circle, then angle between the tangents at A and B is:

(A) 60°

(B) 120°

(C) 80°

(D) 100°

Answer:  (A)

Explanation: Observe the figure

Using properties of circles and tangents, angle between tangents is:

= 180° – 60°

= 120°

11. If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm, then length of each tangent is equal to

(A) 2√3 cm

(B) 6√3 cm

(C) 3√3 cm

(D) 3 cm

Answer:  (C)

Explanation: Let P be an external point from which pair of tangents are drawn as shown in the figure given below:

Join OA and OP

Also OP is a bisector line of ∠APC

∠APO = ∠CPO = 30°

OA ⊥ AP

 Therefore, in triangle OAP

tan30° = OA/AP

1/√3 = 3/AP

AP = 3√3cm

12. In the following figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to

(A) 20°

(B) 40°

(C) 35°

(D) 45°

Answer:  (B)

Explanation: In the given figure

13. In the adjoining figure, Δ ABC is circumscribing a circle. Then, the length of BC is

(A) 7 cm

(B) 8 cm                               

(C) 9 cm

(D) 10 cm

Answer:  (D)

Explanation: Since lengths of tangents from same external point are equal.

Therefore,

BZ=BL=4cm

Also

AZ=AM=3cm

This gives

MC= 9 – 3 = 6 cm

Similarly LC = MC = 6cm

So length of BC = BL + LC = 4cm + 6cm = 10cm    

14. In the following figure, AB is a chord of the circle and AOC is its diameter such that ACB = 50°. If AT is the tangent to the circle at the point A, then BAT is equal to

(A) 65°

(B) 60°

(C) 50°

(D) 40°

Answer:  (C)

Explanation: Here

∠ABC = 90 (Angle in Semicircle)

In ∆ACB

∠A + ∠B + ∠C = 180°

∠A = 180° – (90° + 50°)

∠A = 40°

Or ∠OAB = 40°

Therefore, ∠BAT = 90° – 40° = 50°

15. In the following figure, tangents PQ and PR are drawn to a circle such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ, then ∠RQS.

(A) 30°

(B) 60°

(C) 90°

(D) 120°

Answer:  (A)

Explanation: Since PQ = PR

(Lengths of tangents from same external point are equal)

Therefore,

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1.If the length of the shadow of a tower is increasing, then the angle of elevation of the sun

(A) is also increasing                      

(B) is decreasing

(C) remains unaffected

(D) Don’t have any relation with length of shadow

Answer: (B)

Explanation: Observe the following figure, Let A represents sun, then as the length of shadow increases from DC to DB , the angle of elevation decreases from 60 to 30.

2. The angle of elevation of the top of a tower is 30°. If the height of the tower is doubled, then the angle of elevation of its top will

(A) also get doubled                                                      

(B) will get halved

(C) will be less than 60 degree                   

(D) None of these

Answer: (C)

Explanation: According to Question:

3. If the height of a tower and the distance of the point of observation from its foot,both, are increased by 10%, then the angle of elevation of its top

(A) increases                                                                                     

(B) decreases

(C) remains unchanged                                                

(D) have no relation.

Answer:  (C)

Explanation: Since          

tan θ = h/x

Where h is height and x is distance from tower,

If both are increased by 10%, then the angle will remain unchanged.

4. A ladder 15 metres long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, then the height of the wall will be

(A) 7.5m                                                                              

(B) 7.7m

(C) 8.5m                                                                              

(D) 8.8m

Answer:  (A)

Explanation: Given that the height of ladder is 15m

Let height of vertical be = h

And the ladder makes an angle of elevation 60° with the wall

In triangle QPR

5. An observer 1.5 metres tall is 20.5 metres away from a tower 22 metres high.Determine the angle of elevation of the top of the tower from the eye of the observer.

(A) 30°                                                                                                             

(B) 45°

(C) 60°                                                                                                 

(D) 90°

Answer:  B

Explanation: Let the angle of elevation of the tower from the eye of observer be θ.

Given that:

AB = 22m, PQ = 1.5m = MB

QB = PM = 20.5m

AM = AB − MB = 22 − 1.5 = 20.5m

Now in triangle APM

6. The angles of elevation of the top of a tower from two points distant s and t from its foot are complementary. Then the height of the tower is:

(A) st                                                                                    

(B) s²t²

(C) √st                                                                 

(D) s/t

Answer:  (C)

Explanation: Let the height of tower be h.

Construct figure according to given information as,

7. The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. Then the height of tower is:

(A) 20√3                                             

(B) 25√3

(C) 10√3                                             

(D) 30√3

Answer:  (B)

Explanation: Given condition can be represented as follows where SQ is the pole.

Let the height be h and RQ = x m

Then from figure:

8. If a man standing on a platform 3 metres above the surface of a lake observes a cloud and its reflection in the lake, then the angle of elevation of the cloud is

(A) equal to the angle of depression of its reflection.

(B) double to the angle of depression of its reflection

(C) not equal to the angle of depression of its reflection

(D) information insufficient

Answer:  (C)

Explanation: Observe the figure,

We know that if P is a point above the lake at a distance d, then the reflection of the point in the lake would be at the same distance d.

Also the angle of elevation and depression from the surface of the lake is same.

Here the man is standing on a platform 3m above the surface , so its angle of elevation to the cloud and angle of depression to the reflection of the cloud cannot be same.

9. If a pole 6m high casts a shadow 2√3 m long on the ground, then the sun’s elevation is

(A) 60°                                                                   

(B) 45°

(C) 30°                                                 

(D) 90°

Answer:  (A)

Explanation: According to Question:

Therefore,

tan θ = 6/2√3

⇒ tan θ = √3

⇒ tan θ = tan60°

⇒ θ = 60°

10. The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 metres towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower

(A) 10 (√3 + 1)                                

(B) 5√3

(C) 5 (√3 + 1)                                                   

 (D) 10√3

Answer:(A)

Explanation: Since after moving towards the tower the angle of elevation of the top increases by 15°.

Therefore angle becomes 30° + 15° = 45°

Observe the figure,

11. A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are α and β, respectively. Then the height of the tower is:

Answer:  (B)

Explanation: Observe the figure,

12. The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60°, then the distance between the two towers is:

(A) 10√3 m                                                                        

(B) 15√3 m

(C) 12√3 m                                                                        

(D) 36 m

Answer:  (A)

Explanation: Observe the figure,

Let the distance between two towers be x m.

From figure,

tan60° = 30/x

⇒ √3 = 30/x

⇒ x = 30/√3

⇒ x = 10√3m

13. The angle of elevation of the top of a vertical tower from a point on the ground is60°. From another point 10 m vertically above the first, its angle of elevation is45°. Find the height of the tower.

(A) 5 (√3 + 3) m                                                              

(B) (√3 +3) m

(C) 15 (√3 +3)                                                  

(D) 5√3

Answer:  (A)

Explanation: According to Question:

14. A window of a house is h metres above the ground. From the window, the angles of elevation and depression of the top and the bottom of another house situated on the opposite side of the lane are found to be A and B respectively. Then the height of the other house is:

Answer:  (B)

Explanation: Observe the figure,

Let the height of another house be H m and distance between two houses is x m.

From figure,

15. There are two windows in a house. A window of the house is at a height of 1.5 m above the ground and the other window is 3 m vertically above the lower window. Ram and Shyam are sitting inside the two windows. At an instant, the angle of elevation of a balloon from these windows are observed as 45° and 30° respectively. Find the height of the balloon from the ground.

(A) 7.598m                                                                         

(B) 8.269m

(C) 7.269m                                                                         

(D) 8.598 m

Answer:  (D)

Explanation: Let PQ be the ground level, Ram be sitting at A, Shyam be sitting at B and the balloon be at C from the ground.

Then

AP = 1.5m

And

AB = 3m

AP = DQ = 1.5m and BA = ED = 3m

Let the height of balloon from ground be h,

Then CE = (h − 4.5)m

In right triangle ADC

In right triangle CEB

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1. (sin30° + cos30°) – (sin 60° + cos60°)

(A) – 1

(B) 0

(C) 1

(D) 2

Answer: (B)

Explanation: According to question

2. Value of tan30°/cot60° is:

(A) 1/√2

(B) 1/√3

 (C) √3

(D) 1

Answer: (D)

Explanation:

3. sec²θ – 1 = ?

(A) tan²θ

(B) tan²θ + 1

(C) cot²θ – 1

(D) cos²θ

Answer:  (A)

Explanation: From trigonometric identity

1+ tan²θ = sec²θ

⇒sec²θ – 1 = tan²θ

4. The value of sin θ and cos (90° – θ)

(A) Are same

(B) Are different

(C) No relation

(D) Information insufficient

Answer: (A)

Explanation: Since from trigonometric identities,

cos(90° – θ) = sin θ

So, both represents the same value.

5. If cos A = 4/5, then tan A = ?

(A) 3/5

(B) 3/4

(C) 4/3

(D) 4/5

Answer: (B)

Explanation: From trigonometric identity

6. The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is

(A) 1

(B) −1

(C) 0

(D) 1/2

Answer: (C)

Explanation: Since

cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)

= cosec (75° + θ) – cosec [90° – (15° – θ)] – tan (55° + θ) + tan [90° – (35° – θ)]

= cosec (75° + θ) – cosec (75° + θ) – tan (55° + θ) + tan (55° + θ)

= 0

7. Given that: SinA = a/b, then cosA = ?

(C) b/a

(D) a/b

Answer:(B)

Explanation: We have

8. The value of (tan1° tan2° tan3° … tan89°) is

(A) 0

(B) 1

(C) 2

(D)1/2

Answer: (B)

Explanation: This can be written as,

(tan1° tan2° tan3° … tan89°)

(tan1° tan2° tan3° ……. tan44° tan45° tan46° ….. tan87°tan88°tan89°)

= [tan1° tan2° tan3° ……. tan44° tan45° tan (90 – 44)° ….. tan(90° – 3) tan (90° – 2) tan (90° – 1)]

= (tan1° tan2° tan3° ……. tan44° tan45° cot 44° ….. cot3° cot2° cot 1°)

= 1

Since tan and cot are reciprocals of each other, so they cancel each other.

9. If sin A + sin² A = 1, then cos² A + cos⁴ A = ?

(A) 1

(B) 0

(C) 2

(D) 4

Answer: (A)

Explanation: We have

sin A + sin ² A = 1

⇒ sin A = 1 – sin² A

⇒ sin A = cos² A               ……(i)

Squaring both sides

⇒sin²A = cos⁴A               ……(ii)

From equations (i) and (ii), we have

cos²A + cos⁴A = sin A + sin²A = 1

10. If sin A = 1/2 and cos B = 1/2, then A + B = ?

(A) 0⁰

(B) 30⁰

(C) 60⁰

(D) 90⁰

Answer: (D)

Explanation: Since

(A) 3

(B) 2

(C) 1

(D) 0

Answer: (B)

Explanation: Using trigonometric properties, we have:

12. If cos9α = sin α and 9α < 90°, then the value of tan 5α is

(A) √3

(B) 1/√3

(C) 0

(D) 1

Answer: (D)

Explanation: Since

cos9α = sinα

⇒ sin (90° – 9α) = sinα

⇒ (90° – 9α) = α

⇒ α = 9°

Therefore,

tan 5α = tan 5 (9°)

= tan45°

= 1

13. If a pole 6m high casts a shadow 2√3 m long on the ground, then the sun’s elevation is

(A) 60°

(B) 45°

(C) 30°

(D)90°

Answer: (A)

Explanation: Given condition can be represented as follows:

14. If cos (A + B) = 0, then sin (A – B) is reduced to:

(A) cos A

(B) cos 2B

(C) sin A

(D) sin 2B

Answer: (B)

Explanation: Since

cos (A + B) = 0

⇒ cos (A + B) = cos90°

⇒ (A + B) = 90°

⇒ A = 90° – B

This implies

sin (A – B) = sin (90° – B – B)

⇒ sin (A – B) = sin (90° – 2B)

sin (A – B) = cos 2B

(A) 2/3

(B) 1/3

(C) 1/2

(D) 3/4

Answer:(C)

Explanation: This can be solved as,

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1. In an AP, if d = –4, n = 7, a_n = 4, then a is

(A) 6                                                   

(B) 7

(C) 20                                                 

(D) 28

Answer:  (D)

Explanation:

For an A.P

a_n = a + (n – 1)d

4 = a + (7 – 1)( −4)

4 = a + 6(−4)

4 + 24 = a

a = 28

2. In an AP, if a = 3.5, d = 0, n = 101, then a_n will be

(A) 0                                       

(B) 3.5

(C) 103.5                                

(D) 104.5

Answer: (B)

Explanation:

For an A.P

a_n = a + (n – 1)d

    = 3.5 + (101 – 1) × 0

    = 3.5

3. The first four terms of an AP, whose first term is –2 and the common difference is –2, are

(A) – 2, 0, 2, 4

(B) – 2, 4, – 8, 16

(C) – 2, – 4, – 6, – 8

(D) – 2, – 4, – 8, –16

Answer:  (C)

Explanation:

Let the first four terms of an A.P are a, a+d, a+2d and a+3d

Given that the first termis −2 and difference is also −2, then the A.P would be:

– 2, (–2–2), [–2 + 2 (–2)], [–2 + 3(–2)]

= –2, –4, –6, –8

4. The famous mathematician associated with finding the sum of the first 100 natural numbers is

(A) Pythagoras                                               

(B) Newton

(C) Gauss                                                       

(D) Euclid

Answer:  (C)

Explanation:

Gauss is the famous mathematician associated with finding the sum of the first 100 natural Numbers.

(A) –20                                               

(B) 20

(C) –30                                               

(D) 30

Answer:  (B)

Explanation:

6. The 21st term of the AP whose first two terms are –3 and 4 is

(A) 17                                                 

(B) 137

(C) 143                                               

(D) –143

Answer:  (B)

Explanation:

First two terms are –3 and 4

Therefore,

a = −3

a + d = 4

⇒ d = 4 − a

⇒ d = 4 + 3

⇒ d = 7

Thus,

a₂₁ = a + (21 – 1)d

a₂₁ = –3 + (20)7

a₂₁ = 137

7. If the 2nd term of an AP is 13 and the 5th term is 25, what is its 7th term?

(A) 30                                                             

(B) 33

(C) 37                                                             

(D) 38

Answer:  (B)

Explanation:

Since

a₂ = 13

a₅ = 25

⇒ a + d = 13              ….(i)

⇒ a + 4d = 25           ….(ii)

Solving equations (i) and (ii), we get:

a = 9; d = 4

Therefore,

a₇ = 9 + 6 × 4

a₇ = 9 + 24

a₇ = 33

8. If the common difference of an AP is 5, then what is a₁₈ – a₁₃?

(A) 5                                       

(B) 20

(C) 25                                     

(D) 30

Answer:  (C)

Explanation:

Since, d = 5

a₁₈ – a₁₃ = a + 17d – a – 12d

                    = 5d

                    = 5 × 5

                    = 25

9. The sum of first 16 terms of the AP: 10, 6, 2,… is

(A) –320                                             

(B) 320

(C) –352                                             

(D) –400

Answer:  (A)

Given A.P. is 10, 6, 2,…

10. The sum of first five multiples of 3 is

(A) 45                                                 

(B) 55

(C) 65                                                 

(D) 75

Answer:  (A)

Explanation:

The first five multiples of 3 are 3, 6, 9, 12 and 15

11. The middle most term (s) of the AP:–11, –7, –3, …, 49 is:

(A) 18, 20                                           

(B) 19, 23

(C) 17, 21                                           

(D) 23, 25

Answer:  (C)

Explanation:

Here, a = −11

d = − 7 – (−11) = 4

And a_n = 49

We have,

a_n = a + (n – 1)d

⇒ 49 = −11 + (n – 1)4

⇒ 60 = (n – 1)4

⇒ n = 16

As n is an even number, there will be two middle terms which are16/2th and [(16/2)+1]th, i.e. the 8th term and the 9th term.

a₈ = a + 7d = – 11 + 7 × 4 = 17

a₉ = a + 8d = – 11 + 8 × 4 = 21

12. Two APs have the same common difference. The first term of one of these is –1 and that of the other is – 8. Then the difference between their 4th terms is

(A) –1                                                 

(B) – 8

(C) 7                                                   

(D) –9

Answer:  (C)

Explanation:

The 4th term of first series is

a₄ = a₁ + 3d

The 4th term of another series is

a`₄ = a₂ + 3d

Now,

As, a₁ = –1, a₂ = –8

Therefore,

a₄ – a`₄ = (–1 + 3d) – (–8 + 3d)

a₄ – a`₄ = 7

13. If 7 times the 7th term of an AP is equal to 11 times its 11th term, then its 18th term will be

(A) 7                                                               

(B) 11

(C) 18                                                             

(D) 0

Answer:  (D)

Explanation:

According to question

   7(a + 6d) = 11(a + 10d)

⇒ 7a + 42d = 11a + 110d

⇒ 4a + 68d = 0

⇒ 4(a + 17d) = 0

⇒ a + 17d = 0

Therefore,

a₁₈ = a + 17d

a₁₈ = 0

14. In an AP if a = 1, a_n = 20 and S_n = 399, then n is

(A) 19                                                 

(B) 21

(C) 38                                                 

(D) 42

Answer:  (C)

Explanation:

15. If the numbers n – 2, 4n – 1 and 5n +2 are in AP, then the value of n is:

(A) 1                                                   

(B) 2

(C) − 1                                                

(D) − 2

Answer:  (A)

Explanation:

Let

a = n – 2

b = 4n – 1

c = 5n + 2

Since the terms are in A.P,

Therefore,

2b = a + c

⇒ 2 (4n – 1) = n – 2 + 5n + 2

⇒ 8n – 2 = 6n

⇒ 2n = 2

⇒ n = 1

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1. The roots of quadratic equation 5x² – 4x + 5 = 0 are

(A) Real & Equal                                               

(B) Real & Unequal

(C) Not real                                                        

(D) Non-real and equal

Answer:  (C)

Explanation: To find the nature, let us calculate b² – 4ac

b² – 4ac = 4² – 4 x 5 x 5

= 16 – 100

= -84 < 0

2. Equation (x+1)² – x² = 0 has _____ real root(s).

(A) 1                                                     

(B) 2

(C) 3                                                      

(D) 4

Answer:  (A)

Explanation:

Since (x + 1)² – x² = 0

⟹ x² + 1 + 2x – x² = 0

⟹ 1 + 2x = 0

⟹ x= -1/2

This gives only 1 real value of x.

3. Which constant should be added and subtracted to solve the quadratic equation 4x² − √3x + 5 = 0 by the method of completing the square?

(A) 9/16                                               

(B) 3/16               

(C) 3/4                                                 

(D) √3/4

Answer:  (B)

Explanation:

This can be written as

Hence the given equation can be solved by adding and subtracting 3/16.

4. If 1/2 is a root of the equation x² + kx – (5/4) = 0 then the value of k is

(A) 2                                                                                                     

(B) – 2

(C) 3                                                                                                      

(D) –3

Answer:  (A)

Explanation:

As one root of the equation x² + kx – (5/4) = 0 is 1/2

5. A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.

(A) 3                                     

(B) 8      

(C) 4                                      

 (D) 7

Answer:  (B)

Explanation:

Let the number be x

Then according question,

x + 12 = 160/x

x² + 12x – 160 = 0

x² + 20x – 8x – 160 = 0

(x + 20) (x – 8) = 0

x = -20, 8

Since the number is natural, so we consider only positive value.

6. The product of two successive integral multiples of 5 is 300. Then the numbers are:

(A) 25, 30                                                                            

(B) 10, 15

(C) 30, 35                                                                            

(D) 15, 20

Answer:  (D)

Explanation:

Let the consecutive integral multiple be 5n and 5(n + 1) where n is a positive integer.

According to the question:

5n × 5(n + 1) = 300

⇒ n² + n – 12 = 0

⇒ (n – 3) (n + 4) = 0

⇒ n = 3 and n = – 4.

As n is a positive natural number so n = – 4 will be discarded.

Therefore the numbers are 15 and 20.

(A) 3.5

(B) 4

(C) 3                                                                      

(D) – 3

Answer: (C)

Explanation:

Since y cannot be negative as negative square root is not real so y = 3.

8. If p²x² – q² = 0, then x =?

(A) ± q/p                                             

(B) ±p/q

(C) p                                                     

(D) q

Answer:(A)

Explanation:

p²x² – q² = 0

⇒p²x² = q²

⇒x = ±p/q

(A) 3                                                                     

(B)5       

(C) 4                                                                      

(D) 7

Answer:(B)

Explanation:

10. If x² (a² + b²) + 2x (ac + bd) + c² +d² = 0 has no real roots, then

(A) ad≠bc                                                                           

(B) ad<bc

(C) ad>bc                                                                            

(D) all of these

Answer: (D)

Explanation:

If equation has no real roots then discriminant of the equation must be less than zero.

11. If the one root of the equation 4x² – 2x + p – 4 = 0 be the reciprocal of other. Then value of p is

(A) 8                                                                                     

(B) – 8

(C) – 4                                                                                  

(D) 4

Answer:A

Explanation:

If one root is reciprocal of other, then product of roots is:

12. Rohini had scored 10 more marks in her mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test?

(A) 14                                                   

(B) 16

(C) 15                                                                   

(D) 18

Answer:  (C)

Explanation:

Let her actual marks be x

Therefore,

9 (x + 10) = x²

⇒x² – 9x – 90 = 0

⇒x² – 15x + 6x – 90 = 0

⇒x(x – 15) + 6 (x – 15) = 0

⇒(x + 6) (x – 15) = 0

Therefore  x = – 6 or x =15

Since x is the marks obtained, x ≠ – 6. Therefore, x = 15.

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

(A) 42 km/hr                                                                                                      

(B) 44 km/hr

(C) 46 km/hr                                                                                                      

(D) 48 km/hr

Answer: (A)

Explanation:

Let the original speed be x,

Then according to question

This gives x = -3 and x = 42

Since speed cannot be negative, so we ignore –3,

Therefore original average speed is 42 km/hr.

14. Satvik observed that in a clock, the time needed by the minute hand of a clock to show 3 PM was found to be 3 min less than t²/4 minutes at t minutes past 2 PM. Then t is equal to

(a) 14                                                                    

(b) 15

(c) 16                                                                    

(d) None of these

Answer: (A)

Explanation: We know that the time between 2 PM to 3 PM = 1 hr = 60 min

Given that at t minutes past 2 PM, the time needed by the minute’s hand of a clock to show 3 PM was found to be 3 minutes less than t²/4minutes

Therefore,

15. A takes 6 days less than B to finish a piece of work. If both A and B together can finish the work in 4 days, find the time taken by B to finish the work.

(A)12 days                                                                                          

(B) 12 ½ Days

(C) 13 days                                                                                         

(D) 15days

Answer: (A)

Explanation: Let B alone finish the work in x days.

Therefore, A alone can finish the work in (x – 6) days

A’s one day work = 1/x-6

B’s one day work = 1/x

Given that (A + B) can finish the work in 4 days.

Therefore, A’s one day work + B’s one day work = (A + B)’s one day work

As, x ≠ 2 , because if x = 2 , then A alone can finish work in (2 – 6) = – 4 days which is not possible.

Therefore we consider x = 12.

This implies B alone can finish work in 12 days and A alone will finish the work in 12 – 6 = 6 days.

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Quick Revision Notes :Quadratic Equations

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1. Graphically, the pair of equations

6x – 3y + 10 = 0

2x – y + 9 = 0

Represents two lines which are:

(A) Intersecting at exactly one point.           

(B) Intersecting at exactly two points.

(C) Coincident.                                              

(D) Parallel

Answer:  (D)

Explanation:

Here

Therefore, lines are parallel.

2. The pair of equations x + 2y – 5 = 0 and −3x – 6y + 15 = 0 have:

(A) A unique solution                        

(B) Exactly two solutions

(C) Infinitely many solutions             

(D) No solution

Answer:  (C)

Explanation:

Here,

Therefore, the pair of equations has infinitely many solutions.

3. If a pair of linear equations is consistent, then the lines will be:

(A) Parallel                                         

(B) Always coincident

(C) Intersecting or coincident            

(D) Always intersecting

Answer:  (C)

Explanation: If a pair of linear equations is consistent the two lines represented by these equations definitely have a solution, this implies that either lines are intersecting or coincident.

4. The pair of equations y = 0 and y = –7 has

(A) One solution                                            

(B) Two solutions

(C) Infinitely many solutions                         

(D) No solution

Answer:  (D)

Explanation: The graph of equations will be parallel lines. So the equations have no solution.

5. If the lines given by

3x + 2ky = 2

2x + 5y + 1 = 0

are parallel, then the value of k is

(A) 5/4                                                            

(B) 2/5

(C) 15/4                                                          

(D) 3/2

Answer:  (C)

Explanation:

For parallel lines

6. The value of c for which the pair of equations cx – y = 2 and 6x – 2y = 3 will have infinitely many solutions is

(A) 3                                                   

(B) – 3

(C) –12                                               

(D) no value

Answer: (A)

Explanation: For infinitely many solutions:

7. One equation of a pair of dependent linear equations is –5x + 7y – 2 = 0. The second equation can be

(A) 10x + 14y + 4 = 0                                                

(B) –10x – 14y + 4 = 0

(C) –10x + 14y + 4 = 0                                  

(D) 10x – 14y = –4

Answer: (D)

Explanation: For dependent pair, the two lines must have

8. Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5. Then the numbers are:

(A) 40, 42                                           

(B) 42, 48

(C) 40, 48                                           

(D) 44, 50

Answer: (C)

Explanation:

According to given information

9. The solution of the equations x – y = 2 and x + y = 4 is:

(A) 3 and 5                             

(B) 5 and 3

(C) 3 and 1                             

(D) –1 and –3

Answer: (C)

Explanation: Adding both equations, we have:

10. For which values of a and b, will the following pair of linear equations have infinitely many solutions?

x + 2y = 1

(a – b)x + (a + b)y = a + b – 2

(A) a = 2 and b = 1                                                    

(B) a = 2 and b = 2

(C) a =  ̶ 3 and b = 1                                                   

(D) a = 3 and b = 1

Answer: (D)

Explanation: For infinitely many solutions:

Solving equation (i) and (ii), we get a = 3 and b = 1.

11. 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. The present ages, in years, of the son and the father are, respectively

(A) 4 and 24                                       

(B) 5 and 30

(C) 6 and 36                                       

(D) 3 and 24

Answer:  (C)

Explanation: Let the age of father be x and of son is y.

Then according to question,

x = 6y …..(i)

Four years hence age of son will be y + 4 and age of father will be x + 4

Then according to question,

x + 4 = 4 (y + 4)

x – 4y = 12      …..(ii)

Solving equations (i) and (ii) we get:

y = 6  and x = 36

12. Rakshita has only Rs. 1 and Rs. 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is Rs 75, then the number of Rs.1 andRs.2 coins is, respectively

(A) 35 and 15                                     

(B) 35 and 20

(C) 15 and 35                                     

(D) 25 and 25

Answer:  (D)

Explanation:

Let her number of Rs.1 coins are x

Let the number of Rs.2 coins are y

Then

By the given conditions

x + y = 50   …..(i)

1 × x + 2 × y = 75

⇒ x + 2y = 75    …..(ii)

Solving equations (i) and (ii) we get:

(x + 2y) – (x + y) = 75 – 50

⇒ y = 25

Therefore, x = 50 – 25 = 25

So the number of coins are 25, 25 each.

13. In a competitive examination, one mark is awarded for each correct answer while 1/2 mark is deducted for every wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly?

(A) 100                                                                       

(B) 95

(C) 90                                                             

(D) 60

Answer: (A)

Explanation: Let x be the number of correct answers of the questions in a competitive exam.

Then, 120 − x be the number of wrong answers

Then by given condition

14. The angles of a cyclic quadrilateral ABCD are:

Then value of x and y are:

(A) x = 20^o and y = 30^o                                              

(B) x = 40^o and y = 10^o

(C) x = 44^o and y = 15^o                                              

(D) x = 15^o and y = 15^o

Answer: (A)

Explanation: In cyclic quadrilateral, sum of opposite angles is 180⁰

Therefore

6x + 10 + x + y = 180

⇒ 7x + y = 170           …..(i)

5x + 3y – 10 = 180

⇒ 5x + 3y = 190         …..(ii)

Multiplying equations (i) and (ii), we get:

x = 20^o and y = 30^o

15. A shopkeeper gives books on rent for reading. She takes a fixed charge for the first two days, and an additional charge for each day thereafter. Reema paid Rs. 22 for a book kept for six days, while Ruchika paid Rs 16 for the book kept for four days, then the charge for each extra day is:

(A) Rs 5                                                         

(B) Rs 4

(C) Rs 3                                                          

(D) Rs.2

Answer: (C)

Explanation: Let Rs. x be the fixed charge and Rs. y be the charge for each extra day.

Then by the given conditions

x + 4y = 22                  …..(i)

x + 2y = 16                  …..(ii)

Subtracting equation (ii) from (i), we get:

y = Rs. 3

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1. The zeroes of the quadratic polynomial x² + 99x + 127are

(A) both positive                                            

(B) both negative

(C) one positive and one negative                 

(D) both equal

Answer: (B)

2. If the zeroes of the quadratic polynomial x² + bx + c , c ≠ 0are equal, then

(A) c and a have opposite signs

(B) c and b have opposite signs

(C) c and a have the same sign          

(D) c and b have the same sign

Answer: (C)

3. The number of polynomials having zeroes as –2 and 5 is

(A) 1                                                                     

(B) 2

(C) 3                                                                      

(D) more than 3

Answer:  (D)

4. The degree of the polynomial (x + 1)(x² – x – x⁴ +1) is:

(A)2                                                      

(B) 3

(C) 4                                                      

(D) 5

Answer:  (D)

Explanation: Since the highest degree variable in first bracket is x and in second bracket is x⁴ on multiplying x with x⁴.the highest power we obtain is 5.

5. If the zeroes of the quadratic polynomial x² + (a + 1)x + b are 2 and –3, then

(A) a = –7, b = –1                                                             

(B) a = 5, b = –1

(C) a = 2, b = – 6                                                               

(D) a = 0, b = – 6

Answer:  (D)

6. Given that one of the zeroes of the cubic polynomial ax³ + bx² + cx + d is zero, the product of the other two zeroes is

(A) –c/a                                                                               

(B) c/a

(C) 0                                                                                      

(D) 3

Answer:  (B)

7. If one of the zeroes of the cubic polynomial x³ + ax² + bx + c is –1, then the

Product of the other two zeroes is

(A) b – a + 1                                       

(B) b – a – 1

(C) a – b + 1                                       

(D) a – b –1

Answer: (A)

8. If one of the zeroes of the quadratic polynomial (k – 1)x² + kx + 1 is –3, then the value of k is

(A) 4/3                                                                 

(B) – 4/3

(C) 2/3                                                                 

(D) – 2/3

Answer:  (A)

10. The value of p for which the polynomial x³ + 4x² –px + 8 is exactly divisible by (x – 2) is: 

(A) 0                                                                                     

(B) 3

(C) 5                                                                                      

(D) 16 

Answer: (D)

11. If sum of the squares of zeroes of the quadratic polynomial 6x² + x + k is 25/36, the value of k is: 

(A) 4                                                                                     

(B) – 4

(C) 2                                                                                      

(D) – 2

Answer:  (D)

12. If α and β are zeroes of x² – 4x + 1, then 1/α + 1/β – αβ is 

(A) 3                                                                     

(B) 5

(C) –5

(D) –3 

Answer:  (A)

13. If (x + 1) is a factor of x²− 3ax +3a − 7, then the value of a is: 

(A) 1                                                                     

(B) –1

(C) 0                                                                      

(D) –2 

Answer: (A)

14. If one of the zeroes of a quadratic polynomial of the form x² + ax + b is the negative of the other, then it

(A) has no linear term and the constant term is negative.

(B) has no linear term and the constant term is positive.

(C) can have a linear term but the constant term is negative.

(D) can have a linear term but the constant term is positive

Answer: (A)

15. If α, β are zeroes of x² –6x + k, what is the value of k if 3α + 2β = 20? 

(A)–16                                                 

(B) 8

(C) 2                                                                      

(D) –8 

Answer: (A)

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Quick Revision Notes :Polynomials

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MCQs: Polynomials

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