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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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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1. O is the point of intersection of two equal chords ABand CD such that OB = OD, then triangles OAC and ODB are

(A) Equilateral but not similar

(B) Isosceles but not similar

(C) Equilateral and similar

(D) Isosceles and similar

Answer: (D)

Explanation:

Since O is the point of intersection of two equal chords AB and CD such that OB = OD,

As chords are equal and OB = OD, so AO will also be equal to OC

Also ∠AOC = ∠DOB = 45⁰

Now in triangles OAC and ODB

AO/OB = CO/OD

And ∠AOC = ∠DOB = 45⁰

So triangles are isosceles and similar

2. D and E are respectively the midpoints on the sides AB and AC of a triangle ABC and BC = 6 cm. If DE || BC, then the length of DE (in cm) is

(A) 2.5                                                  

(B) 3

(C) 5                                                                      

(D) 6

Answer:  B

Explanation:

By midpoint theorem,

If D and E are respectively the midpoints on the sides AB and AC of a triangle ABC, DE||BC and BC = 6 cm

So, DE will be half of BC i.e. 3cm

3. In triangle PQR, if PQ = 6 cm, PR = 8 cm, QS = 3 cm, and PS is the bisector of angle QPR, what is the length of SR?

(A) 2                                                                     

(B) 4

(C) 6                                                                      

(D) 8

Answer:  (B)

Explanation:

Since, PS is the angle bisector of angle QPR

So, by angle bisector theorem,

QS/SR = PQ/PR

⇒ 3/SR = 6/8

⇒ SR = (3 X 8)/6 cm = 4 cm

4. The lengths of the diagonals of a rhombus are 16 cm and 12cm. Then, thelength of the side of the rhombus is

(A) 9 cm

(B) 10 cm

(C) 8 cm                                                               

(D) 20 cm

Answer:(B)

Explanation:

The diagonals of rhombus bisect each other at right angle, so side of rhombus is the hypotenuse for the triangles formed.

Therefore,

By Pythagoras theorem

(16/2)² + (12/2)² = Side²

⇒ 8² + 6² = Side²

⇒ 64 + 36 = Side²

⇒ Side = 10 cm

5. A flag pole 18 m high casts a shadow 9.6 m long. Find the distance of the top of the pole from the far end of the shadow.

(A) 25.6                                                                               

(B) 20.4

(C) 23.7                                                                

(D) 32.5

Answer:(B)

Explanation:

According to given question

The far end of shadow is represented by point A,

Therefore we need to Find AC

By Pythagoras theorem,

(18)² + (9.6)² = (AC)²

⇒ AC² = 416.16

⇒ AC = 20.4 m (approx)

6. Diagonals of a trapezium PQRS intersect each other at the point O, PQ || RS and PQ = 3 RS, Then the ratio of areas of triangles POQ and ROS is:

(A) 1:9                                                                  

(B) 9:1

(C) 3:1                                                                  

(D) 1:3

Answer:(B)

Explanation:According to given Question

Since

SR || PQ,

So, ∠OSR= ∠OQP (alternate interior angles)

Also ∠SOR= ∠POQ (vertically opposite angles)

So triangles SOR and POQ are similar,

Therefore,

ar(POQ)/ar(SOR) = (PQ/SR)²

ar(POQ)/ar(SOR) = (3 SR/SR)²

ar(POQ)/ar(SOR) = 9/1

7. ABCD is a trapezium in which AB|| DC and P, Q are points on ADand BC respectively such that PQ || DC. If PD = 18 cm, BQ = 35 cm andQC = 15 cm, find AD.

(A) 55cm                                                             

(B) 57cm

(C) 60cm                                                             

(D) 62cm

Answer:(C)

Explanation:

According to question

ABCD is a trapezium in which AB || DC and P and Q are points on AD and BC, respectively such that PQ || DC. If PD = 18 cm, BQ = 35 cm and QC = 15 cm,

In triangle ABD

DP/AP = OD/OB

In triangle BDC

BQ/QC = OB/OD

This implies

DP/AP = QC/BQ

18/AP = 15/35

AP = (18 x 35)/15

AP = 42

Therefore, AD = AP + DP = 42 + 18 = 60cm

8. Areas of two similar triangles are 36 cm² and 100 cm². If the length of a side of the larger triangle is 20 cm, then the length of the corresponding side of the smaller triangle is:

(A) 12cm                                                                             

(B) 13cm

(C) 14cm                                                                             

(D) 15cm

Answer:(A)

Explanation:

Let the side of smaller triangle be x cm.

ar(Larger Triangle)/ar(Smaller Triangle) = (side of larger triangle/side of smaller triangle)²

100/36 = (20/x)²

x = √144

X = 12 cm

9. In the figure if ∠ACB = ∠CDA, AC = 8 cm and AD = 3 cm, find BD.

(A) 53/3 cm                                                        

(B) 55/3 cm

(C) 64/3 cm                                                        

(D) 35/7 cm

Answer:(B)

Explanation:

In triangle ACB and ADC

∠A=∠A

∠ACB = ∠CDA

Therefore triangle ACB and ADC are similar,

Hence

AC/AD = AB/AC

AC² = AD X AB

8² = 3 x AB

⇒ AB = 64/3

This implies,

BD = 64/3 – AD

⇒ BD = 55/3

10. If ABCD is parallelogram, P is a point on side BC and DP when produced meets AB produced at L, then select the correct option

(A) DP/BL = DC/PL

(B) DP/PL = DC/BL

(C) DP/PL = BL/DC

(D) DP/PL = AB/DC

Answer: (B)

Explanation:

In ΔALD, we have

BP || AD

∴ LB/BA = LP/PD

⇒ BL/AB = PL/DP

⇒ BL/DC = PL/DP [∵ AB = DC

⇒ DP/PL = DC/BL

11. In the figure given below DE || BC. If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, the value of x is:

(A) 4                                                                                                     

(B) 8

(C) 16                                                                                                   

(D) 32

Answer: (A)

Explanation:

In triangle ABC, we have DE || BC

∴ AD/DB = AE/EC (By Thale’s Theorem)

⇒ x/x – 2 = (x + 2)/(x – 1)

⇒ x (x – 1) = (x – 2)(x + 2)

⇒ x² – x = x² – 4

⇒ x = 4

12. The length of altitude of an equilateral triangle of side 8cm is

(A) √3 cm                                                            

(B) 2√3 cm

(C) 3√3 cm                                                          

(D) 4√3 cm

Answer:(D)

Explanation:

The altitude divides the opposite side into two equal parts,

Therefore, BD = DC = 4 cm

In triangle ABD

AB² = AD² + BD²

8² = AD² + 4²

AD² = 64 – 16

AD² = 48

AD = 4√3 cm

13. If ΔABC ~ ΔDEF, AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm, find the perimeter of ABC.

(A) 18    cm                                                         

(B) 20 cm

(C) 21    cm                                                         

(D) 22 cm

Answer:(A)

Explanation:

According to question,

ΔABC ~ ΔDEF,

 AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm,

Therefore,

AB/DE = BC/EF = AC/DF

4/6 = BC/9 = AC/12

⇒ 4/6 = BC/9

⇒ BC = 6 cm

And

4/6 = AC/12

⇒ AC = 8 cm

Perimeter = AB + BC + CA

= 4 + 6 + 8

= 18 cm

14. A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, then the distance by which the top of the ladder would slide upwards on the wall is:

(A) 2 m                                                                                

(B) 1.2 m

(C) 0.8 m                                                                             

(D) 0.5 m

Answer:(C)

Explanation:

Let AC be the ladder of length 5m and BC = 4m be the height of the wall where ladder is placed. If the foot of the ladder is moved 1.6m towards the wall i.e. AD = 1.6 m, then the ladder is slided upward to position E i.e. CE = x m.

In right triangle ABC

AC² = AB² + BC²

⇒5² = AB² + 4²

⇒ AB = 3m

⇒ DB = AB – AD = 3 – 1.6 = 1.4m

In right angled ΔEBD

ED² = EB² + BD²

⇒ 5² = EB² + (1.4)²

⇒ EB = 4.8m

EC = EB – BC = 4.8 – 4 = 0.8m

Hence the top of the ladder would slide upwards on the wall at distance 0.8 m.

15. Corresponding sides of two similar triangles are in the ratio of 2 : 3. If the area of the smaller triangle is 48 cm², then the area of the larger triangle is:

(A) 108 m²                                                                          

(B) 107 m²

(C) 106 m²                                                                          

(D) 230 m²

Answer:(A)

Explanation:

According to given Question

ar(Larger Triangle)/ar(Smaller Triangle) = (side of larger triangle/side of larger triangle)²

ar(Larger Triangle)/48 = (3/2)²

ar(Larger Triangle) = (9 x 48 )/4

ar(Larger Triangle) = 108 cm²

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1. If the distance between the points (2, –2) and (–1, x) is 5, one of the values of x is

(A) –2

(B) 2

(C) –1                                                 

(D) 1

Answer: (B)

Explanation: According to question

2. The mid-point of the line segment joining the points A (–2, 8) and B (– 6, – 4) is

(A) (– 4, – 6)                                       

(B) (2, 6)

(C) (– 4, 2)                                          

(D) (4, 2)

Answer:  (C)

Explanation: Let the coordinates of midpoint be (x, y) then

Therefore the coordinates are

3. The points A (9, 0), B (9, 6), C (–9, 6) and D (–9, 0) are the vertices of a

(A) Square                                          

(B) Rectangle

(C) Rhombus                                      

(D) Trapezium

Answer:  (B)

Explanation: Here we will calculate the measure of all four sides of the quadrilateral fromed by given points A, B, C and D.

Since, AB = CD and BC = AD

Therefore given points A,B,C and D are the vertices of a rectangle.

4. The distance of the point P (2, 3) from the x-axis is

(A) 2                                                               

(B) 3

(C) 1                                                               

(D) 5

Answer:  (B)

Explanation: Distance of the point P (2, 3) from the x-axis =Ordinate of the point (2, 3) i.e.3.

5. The distance between the points A (0, 6) and B (0, –2) is

(A) 6                                                   

(B) 8

(C) 4                                                   

(D) 2

Answer:  (B)

Explanation: Here, x₁ = 0, y₁ = 6, x₂ = 0, y₂ = –2

6. AOBC is a rectangle whose three vertices are vertices A (0, 3), O (0, 0) and B (5, 0). The length of its diagonal is

(A) 5                                                   

(B) 3

(C) √34                                               

(D) 4

Answer:  (C)

Explanation:

The length of the diagonal is distance between the points AB.

 The distance is calculated as,

7. If P (a/3, 4) is the mid-point of the line segment joining the points Q (– 6, 5) and R (– 2, 3), then the value of a is

(A) – 4                                                

(B) – 12

(C) 12                                                 

(D) – 6

Answer:  (B)

Explanation: As (a/3, 4) is the mid – point of the line segment joining the points Q (– 6, 5) and R (– 2, 3). Therefore

8. The coordinates of the point which is equidistant from the three vertices of the Δ AOB as shown in the figure is:

(A) (x, y)                                                         

(B) (y, x)

(C) (x/2, y/2)                                                  

(D) (y/2, x/2)  

Answer:  (A)

Explanation: As we have to find the coordinates which are equidistant from A and B, Let the points be (a, b).

Then (a, b) will be the midpoint of AB.

Therefore,

Hence the coordinates are (x, y)

9. A circle drawn with origin as the centre passes through The point which does not lie in the interior of the circle is

(C) (5, –1/2)                                       

(D) (–6, 5/2)

Answer:  (D)

Explanation:If the point lies in the interior of circle, the distance of the point from the centre should be less than radius of circle.

The radius of circle is the distance between origin and the point

Distance between origin and (-3/4, 1) is

Similarly the distance of points (2, 7/3) and (5, –1/2) is also less than 6.5

But the distance of (–6, 5/2) is equal to 6.5.

So the point (–6, 5/2) does not lie in the interior of circle.

10. If the distance between the points (4, p) and (1, 0) is 5, then the value of p is

(A) 4 only                                                        

(B) ± 4

(C) – 4 only                                                     

(D) 0

Answer:B

Explanation:   According to question:

11. The area of a triangle with vertices A (3, 0), B (7, 0) and C (8, 4) is:

(A) 14                                                 

(B) 28

(C) 8                                                   

(D) 6

Answer:  (C)

Explanation: Area of triangle is calculated as,

12. The point which divides the line segment joining the points (7, –6) and (3, 4) in ratio 1 : 2 internally lies in the

(A) I quadrant                                                 

(B) II quadrant

(C) III quadrant                                               

(D) IV quadrant

Answer: (D)

Explanation: Let the point be (x, y)

Then, by using section formula

Therefore, the point is (17/3, -8/3) which lies in fourth quadrant.

13. One of the two points of trisection of the line segment joining the points A (7, – 2) and B (1, – 5) which divides the line in the ratio 1:2 are:

(A) (5, –3)                                                       

(B) (5, 3)

(C) (–5, –3)                                                     

(D) (13, 0)

Answer:  (A)

Explanation: Required point of trisection that divides the given line in the ratio 1: 2 is

14. 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 the coordinates of P and Q are, respectively.

(A) (0, – 5) and (2, 0)                                     

(B) (0, 10) and (– 4, 0)

(C) (0, 4) and (– 10, 0)                                   

(D) (0, – 10) and (4, 0)

Answer:  (D)

Explanation: As the line intersects the y and x axis, let the coordinates be (0, b) and (a, 0) respectively. Since (2, –5) is the midpoint of the axis. Therefore,

Therefore, the coordinates are (0, –10) and (4, 0).

15. The ratio in which the point P (3/4, 5/12)divides the line segment joining the Points A (1/2, 3/2) and B (2, –5) is:

(A) 1:5                                                            

(B) 5:1

(C) 1:3                                                            

(D) 3:1

Answer:  (A)

Explanation: Let the ratio be m : n then, according to the question:

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