CLASS X : MATH SOLUTIONS

Q1. It is given that x = - 1 2

is the solution of the quadratic equation 3x2+2kx-3 = 0.

∴ 3�−12

2� + 2k (-

1 2

) - 3 =0

⇒ 3 4

- k - 3=0

k = 3 4

– 3 = −94

Hence, the value of k is −94

Q2. Let AB and CD be the two towers of heights x and y, respectively.

Suppose E is the centre of the line joining the feet of the two towers i.e. BD.

Now, in ∆ ABE,

𝐴𝐴𝐴𝐴𝐴𝐴𝐵𝐵

= tan30°

⇒ 𝑥𝑥𝐴𝐴𝐵𝐵

= √3

⇒BE = √3x ……… 1

Also,

In ∆ CDE,

𝐶𝐶𝐶𝐶𝐶𝐶𝐵𝐵

= tan60°

⇒DE = 𝑦𝑦√3

………. (2)

NOW BE = DE …… (3) (E is mid – point of BD)

So, from (1) , (2) and (3) we get

√3x = 𝑦𝑦√3

⇒ 𝑥𝑥𝑦𝑦

= 13

Hence, the ratio of x and y is 1: 3.

Q3. There are 26 letters in English alphabets.

∴ Total number of outcomes = 26

We know that there are 5 vowels and 21 consonants in English alphabets.

∴ Total number of favourable outcomes = 21

∴ Probability that the chosen letter is a consonant = 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑇𝑇𝑜𝑜 𝑜𝑜𝑇𝑇𝑓𝑓𝑇𝑇𝑛𝑛𝑛𝑛𝑇𝑇𝑛𝑛𝑇𝑇𝑛𝑛 𝑇𝑇𝑛𝑛𝑇𝑇 𝑐𝑐𝑇𝑇𝑛𝑛𝑛𝑛𝑐𝑐𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑇𝑇𝑜𝑜 𝑇𝑇𝑛𝑛𝑇𝑇𝑐𝑐𝑇𝑇𝑛𝑛𝑛𝑛𝑐𝑐

=2126

Q4. PA and PB are tangents drawn from an external point P to the circle.

∴ PA = PB (Length of tangents drawn from an external point to the circle are equal)

In ∆ PAB,

PA = PB

⇒ ∠ PBA = ∠PAB ……… (1) (Angles opposite to equal sides are equal.)

Now,

∠APB + ∠PBA + ∠PAB = 180°

⇒ 50° + ∠PAB + ∠PAB = 180° [Using (1)]

⇒ 2∠PAB = 130°

⇒ ∠PAB = 130°2

= 65°

We know that radius is perpendicular to the tangent at the point of contact

∴ ∠ OAP = 90° (OA ⊥ PA)

⇒ ∠PAB + ∠OAB = 90°

⇒ 65° + ∠OAB = 90°

⇒ ∠OAB = 90° - 65° = 25°

Hence the measure of ∠OAB is 25°

SECTION – B

Q5. We know that the angle subtended by an arc of a circle at the centre is twice the angle subtended by it at any point on the remaining part of the circle.

∴ ∠AOQ = 2 ∠ ABQ

⇒ ∠ABQ = 12 ∠ AOQ

⇒ ∠ABQ = 12 × 58° = 29°

or ∠ABT = 29°

We know that the radius is perpendicular to the tangent at the point of contact

∴ ∠OAT = 90° (OA ⊥ AT)

Or ∠BAT = 90°

Now, in ∆BAT,

∠BAT + ABT + ∠ATB = 180°

⇒90° + 29°+∠ATB= 180°

⇒∠ATB = 61° ∠ATQ = 61°

Q6. We have

4x2 – 4a2x + (a4 – b4) =0

⇒ (4x2 -4a2x +a4) –b4 =0

⇒[(2x) 2 -2 (2x) (a) + (a2)2] – b4 = 0

⇒(2x – a2)2 – (b2)2 = 0

⇒ (2x – a2 +b2) (2x – a2 –b2) =0

⇒ 2x – a2 +b2 = 0 or 2x – a2 – b2 = 0

⇒ 2x = a2 – b2 or 2x = a2 + b2

⇒ x = 𝑇𝑇2−𝑛𝑛2

2 or 𝑇𝑇

2+𝑛𝑛2

2

Q7.

TP and TQ are tangents drawn from an external point T to the circle. O is the centre of the circle.

Suppose OT intersect PQ at point R.

In ∆OPT AND ∆OQT,

OP= OQ (Radii of the circle)

TP = TQ (Lengths of tangents drawn from an external point to a circle are equal )

OT = OT (Common sides)

∴ ∆ OPT ≅ ∆ OQT (BY SSS congruence rule)

So, ∠PTO = ∠QTO (BY CPCT) .……… (1)

Now, in ∆PRT and ∆QRT,

TP =TQ (Lengths of tangents drawn from an external point to a circle are equal)

∠PTO = ∠QTO [From (1)]

RT = RT (Common sides)

∴ ∆PRT ≅ ∆QRT (By SAS congruence rule)

So, PR = QR ………. (2) (By CPCT)

And, ∠PRT = ∠QRT (BY CPCT)

Now,

∠PRT +∠QRT = 180° (Linear Pair)

⇒2∠PRT = 90°

⇒ 2 ∠PRT = 90°

∴ ∠PRT = ∠QRT = 90° ……(3)

From (2) and (3) we can conclude that

OT is the right bisector of the line segment PQ.

Q8. The given anthmetic progression is

6, 13, 20…., 216

Let 216 be the nth term of the given AP. So, a = 6 d =7 an = 216 Now, an = a + (n - 1)d

⇒216 = 6 + (n - 1) × 7

⇒7(n-1) = 210

⇒n - 1 = 30

⇒n = 31, which is odd

∴ Middle term of the AP

= (31+12

)P

th term of the AP

= 16th term of the AP

∴ a16 =6+16-1×7=6+15×7=6+105=111

Thus, the middle term of the given AP is 111.

Q9. Using distance formula, we have

AB = �(2 − 5)2 + (−2 − 2)2 = √9 + 16 = 5 ……. (1)

BC =�(−2 − 2)2 + (𝑇𝑇 + 2)2 =√𝑇𝑇2 + 4𝑇𝑇 + 20 ……… (2)

AC = �(−2 − 5)2 + (𝑇𝑇 − 2)2 = = √𝑇𝑇2 − 4𝑇𝑇 + 53 …. (3)

Now, it is given that ∆ABC is right angled at B.

Using the Pythagorean theorem, we have

AB2 +BC2 = AC2

∴ 25 + t2 + 4t + 20 = t2 - 4t + 53 [From (1) (2) and (3)]

⇒45 + 4t = -4t + 53

⇒ 8t = 8

⇒ t = 1

Hence, the value of t is 1.

Q10. Let P ( 34

, 512

) divide AB in the ratio m1:m2.

Using the section formula, we have

( 34

, 512

) = ��2𝑛𝑛1+

12𝑛𝑛2

𝑛𝑛1+𝑛𝑛2� , �

−5𝑛𝑛1+32𝑛𝑛2

𝑛𝑛1+𝑛𝑛2��

34 =

2𝑛𝑛1+12𝑛𝑛2

𝑛𝑛1+𝑛𝑛2 …….. (1)

512

=−5𝑛𝑛1+3

2𝑛𝑛2

𝑛𝑛1+𝑛𝑛2 …… (2)

From (1) we have

34 =

2𝑛𝑛1+12𝑛𝑛2

𝑛𝑛1+𝑛𝑛2

3(𝑛𝑛1 + 𝑛𝑛2) = 4(2𝑛𝑛1 + 1

2𝑛𝑛2)

3m1 +3m2 =8m1 + 2m2

-5 m1 =- m2

m1: m2 = 1: 5

SECTION - C

Q11. Let the coordinates of points B and C of the ∆ABC be (a1, b1) and (a2, b2), respectively.

Q is the midpoint of AB.

Using midpoint formula, we have

(0, -1)= 𝑇𝑇1+12

,𝑛𝑛1−42

⇒ 𝑇𝑇1+12

=0 and 𝑛𝑛1−42

=-1

⇒ a1 =-1 and b1 =2

Therefore, the coordinates of B are (-1,2).

P is the midpoint of AC.

Now,

(2, -1) = 𝑇𝑇2+12

, 𝑛𝑛2−42

⇒ 𝑇𝑇2+12

= 2 and 𝑛𝑛2−42

= -1

⇒ a2= 3 and b2=2

Therefore, the coordinates of C are (3,2).

Thus, the vertices of ∆ABC are A(1, -4), B (-1,2) and C(3,2).

Now,

Area of the triangle having vertices (x1, y1),(x2, y2),(x3 ,y3) = 12x1y2-y3+2y3-y1+x3y1-y2

∴ Area of ∆ABC = 12 [x1(y2 - y3) + x2 (y3 - y1) + 3 (y1 - y2)]

=12

× [1 (2 - 2) + -1(2 + 4) + 3(-4 - 2)]

=-12

Since area is a measure that cannot be negative, we will take the numerical value of – 12, that is 12.

Thus , the area of ∆ABC is 12 square units.

Q12

We have

kx2 + 1 - 2(k - 1) x + x2 = 0

This equation can be rearranged as

(k + 1) x2 - 2(k - 1)x + 1 = 0

Here, a = k+1, b = -2(k-1) and c =1

∴ D = b2 – 4ac

= [-2(k-1)2] – 4 × (k + 1) × 1

=4[(k-1)2 – (k+1)]

= 4[k2 – 3k]

= 4[k (k -3)]

The given equations will have equal roots, if D = 0.

⇒ 4[k (k-3)] = 0

⇒ k = 0 or k -3 = 0

⇒ k= 3

Putting k= 3 in the given equation , we get

4x2 - 4x + 1 =0

⇒(2x-1)2 = 0

⇒ 2x -1 =0

⇒ x =1/2

Hence, the roots of the given equation are 1/2 and 1/2.

Q13

Let AB the building and CD be the tower.

CD/BD=tan 45°

⇒30/BD =1

⇒ BD =30m

In ∆ABD,

𝐴𝐴𝐴𝐴𝐴𝐴𝐶𝐶

= tan 30°

⇒ AB = BD × 1√3

⇒ AB= 10√3 m

Therefore, the height of the building is 10√3 m.

Q14.

When two dice are thrown simultaneously, the possible outcomes are

∴ Total number of possible outcomes = 36

We know

Probability of an event =𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑇𝑇𝑜𝑜 𝑜𝑜𝑇𝑇𝑓𝑓𝑇𝑇𝑛𝑛𝑛𝑛𝑇𝑇𝑛𝑛𝑇𝑇𝑛𝑛 𝑇𝑇𝑛𝑛𝑇𝑇𝑐𝑐𝑇𝑇𝑛𝑛𝑛𝑛𝑐𝑐𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑇𝑇𝑜𝑜 𝑇𝑇𝑛𝑛𝑇𝑇𝑐𝑐𝑇𝑇𝑛𝑛𝑛𝑛𝑐𝑐

(1) The outcomes favorable of the event’ the sum of the numbers on two dice to be 5’ denoted by E are (1, 4), (2, 3), (4, 1) and (3, 2).

Total number of favorable outcomes = 4

∴ PE =4/36 =1/9

(ii) The outcomes favourable to the event ‘even numbers on both dice’ denoted by F are

(2, 2), ( 2,4),(2,6), (4,2),(4,4),(4,6), (6,2), (6,4) and (6,6).

Total number of favouable outcomes = 9

∴PF =9/36 = 1/4

Q15

Let a and d be the first term and the common difference of AP, respectively,

We know

Sn = 𝑛𝑛2

[2a+ (n-1) d]

∴S8 = 82 [2a+ (8-1) d and S4 = 8

2 [2a+ (4-1) d]

⇒ S8 = 4(2a + 7d) and S4 = 2(2a + 3d)

⇒S8 = 8a + 28d and S4=4a + 6d

Now,

3(S8 – S4) =3(8a+28d-4a-6d)

= 3(4a+22d)

=6(2a+11d)

= 122

[2a+ (12-1) d]

=S12

∴ S12=3(S8-S4)

16.

Let the radius of the semi-circle APB be r.

⇒ The radius of the semi- circle AQO = 𝑛𝑛2

Now,

Perimeter of the given figure = Length of arc AQO + Length of arc APB + OB

= r (3 2

× 227

+ 1)

= r 8014

= r(40 7

) cm

⇒ r(40 7

) = 40

⇒ r = 7cm

∴Area of the shaded region = Area of semi – circle AQO + Area of semi- circle APB

= 𝜋𝜋(7

2)2

2+ 𝜋𝜋72

2

= 49 𝜋𝜋 (18

+ 12)

= 96.25 cm2

17. The remaining solid is a frustum of the given cone.

Total surface area of the frustum

Where

h = Height of the frustum = 12 - 4 = 8 cm

r1 = Larger radius of the frustum = 6 m

r2 = Smaller radius of the frustum

l = Slant height of the frustum

In the given figure, ∆ABC ~∆ ADE by AA similarity criterion.

∴BC/DE = AB/AD

𝑛𝑛26

= 412

⇒ r2 = 2 cm

We know

⇒ l = 4√5 cm.

∴Total surface area of the frustum = 𝜋𝜋𝑇𝑇(r1 +r2) +πr12 + πr2

2

= 350.592 cm2

Hence, the total surface area of the remaining solid is 350.592 cm2.

18. Let h be the height of the cone and r be the radius of the base of cone.

The volume of the wooden toy

= 𝜋𝜋𝑛𝑛 2ℎ3

+ 23πr3

According to the question,

776

(h + 7) = 10016

⇒h= 6 cm The height of the wooden toy = 6 cm + 3.5 cm

= 9.5 cm Now,

Curved surface area of the hemispherical part = 2X(22/7)X3.5X3.5 = 2 ×227

× (3.5)2

=77 cm2

Hence, the cost painting the hemispherical part of the toy = 77 X 10 = Rs 770

Q19. Surface area of the remaining block = Surface area of the cuboid + Curved surface area of cylinder -2 × Area of base of cylinder

= 2( 15 × 10 + 10 × 5 +15 × 5) + (2 × 227

× 72

× 5 ) –( 2 × 227

× 72

× 72

)

= 583 cm2

Thus, the surface area of the remaining block is 583 cm2. Q20. From the symmetry of the given figure, we have

∴ Area of the shaded region = Area of the square with side 14 cm – (Area of square with side 4 cm + 4 × Area of each semicircle with radius 2 cm )

= 14 × 14 –[ 4 × 4 + 4 × 12 × 3.14 ×(2)2]

= 196 –( 16 + 25.12) = 196 – 41.12 = 154.88 cm2

Section - D

21. Let the numerator and the denominator of the fraction be x + 3, respectively.

∴ Original fraction = 𝑥𝑥𝑥𝑥+3

Now, 2 is added to both the numerator and the denominator.

∴ New fraction = (𝑥𝑥+2)(𝑥𝑥+3)

According to the question,

𝑥𝑥𝑥𝑥 + 3

+𝑥𝑥 + 2𝑥𝑥 + 5

=2920

𝑥𝑥(𝑥𝑥 + 5) + (𝑥𝑥 + 2)(𝑥𝑥 + 3) (𝑥𝑥 + 3)(𝑥𝑥 + 5)

=2920

Solving we get

𝑥𝑥 = 7

10

Or𝑥𝑥 = −4511

Now, 𝑥𝑥 ≠ (−4511

)as it is a fraction.

So, the original fraction becomes (7/10).

22. Since Ramkali increased her weekly savings uniformly every week by a fixed number, her savings will form an AP.

Let Sn be the sum of savings in all 12 weeks.

∴ Sn=𝑛𝑛2

(2𝑇𝑇 + (𝑛𝑛 − 1)𝑑𝑑) (Here a is the money saved in the first week and d is the fixed

increase in the weekly savings.)

⇒ Sn=122

(2 × 100 + (12 − 1)20)=Rs 2,520

Ramkali required Rs2,500 after 12 weeks, but she saved Rs 2,520. So, she will be able to send her daughter to school after 12 weeks.

It shows that Ramkali is aware of the importance of girl child education.

23. 2𝑥𝑥+1

+ 32(𝑥𝑥−2)

= 235𝑥𝑥

4(𝑥𝑥 − 2) + 3(𝑥𝑥 + 1)2(𝑥𝑥 + 1)(𝑥𝑥 − 2)

=235𝑥𝑥

5𝑥𝑥�4(𝑥𝑥 − 2) + 3(𝑥𝑥 + 1)� = 46(𝑥𝑥 + 1)(𝑥𝑥 − 2)

11𝑥𝑥2 − 21𝑥𝑥 − 92 = 0

11𝑥𝑥2 − 44𝑥𝑥 + 23𝑥𝑥 − 92 = 0

11𝑥𝑥(𝑥𝑥 − 4) − 23(𝑥𝑥 − 4) = 0

Solving we get x =(-23/11) or 4

24.

Given: A circle C (o,r) and a tangent l at point A

To prove: ⊥OA /

Construction:

Take a point B, other than A, on the tangent l. Join OB. Suppose OB meets the circle in C.

Proof: We know that among all line segments joining the point O to a point on /, the perpendicular is shortest to /.

OA = OC (Radius of the same circle)

Now,

OB=OC+BC

⇒OB > OC

⇒OB > OA

⇒ OA < OB

B is an arbitrary point on the tangent l . Thus, OA is shorter than any other line segment joining O to any point on l .

Here,

⊥OA l

Hence, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

25.

We know that tangents from an external point are equal in length.

∴PQ = PR

In ∆PQR,

PQ = PR

PRQPQR ∠=∠∴ (Angles opposite to equal sides are equal).

Now in ∆ PQR,

∠ PQR + ∠ PRQ + ∠ RQP =1800

2∠ RQP = 1800-300

∠ RQP = 750

Also, radius is perpendicular to the tangent at the point of contact.

090=∠=∠∴ ORPOQP

Now, in PQOR,

∠ ROQ + ∠OQP +∠QPR +∠ PRO = 3600

900 + 900+300 +∠ ROQ = 3600

∠ ROQ = 1500

Since, angle subtended by an arc at any point on the circle is half the angle subtended at the centre by the same arc.

angle QSR = 750

Also, SQTQSR ∠=∠ (Alternate interior angles)

SQT∠∴ =750

Now,

∠ SQT +∠ PQR + ∠ SQR = 1800

750 + 750 + ∠ SQR = 1800

∠ SQR =300

26.

Given: A ABC∆ with BC = 7 cm, 060=∠B = and AB = 6 cm

Steps of Construction:

1. Draw a line segment AB = 6 cm. 2. With B as centre, 060=∠B . 3. With B as centre and radius BC=7 cm, draw an arc. 4. Join AC to obtain ∆ABC. 5. Below AB, make an acute angle BAX∠ . 6. Along AX, mark off four points A1, A2, A3 and A4 such that AA1=A1A2=A2A3=A3A4. 7. Join A4B. 8. From A3, draw A3B’//A4B 9. From B’ draw B’C’//BC

Thus, ∆AB’C’ is the required triangle with each of its side (3/4) times the size of the corresponding sides of ∆ABC.

27.

Let RQ be the tower and SR be the flag staff.

In PQR

Tan300 = (RQ/PQ)

1√3

= ℎ𝑥𝑥---------------------------(i)

x = h√3

In PQS,

Tan600=𝑆𝑆𝑆𝑆𝑃𝑃𝑆𝑆

√3 = ℎ+5𝑥𝑥

----------------(ii)

From (i) and (ii), we get

3h = h+5

2h = 5

H = 2.5 m

Hence, the height of the tower is 2.5metres.

28. Total number of cards = 20

(i) Numbers divisible by 2 or 3 are 2,3,4,6,8,9,10,12,14,15,16,18and 20.

Total favorable number of cards =13

Probability that the number on the drawn card is divisible by 2 or 3=(Total favorable number of cards /Total number of cards)

= (13/20)

(ii) Prime numbers are 2, 3, 5, 7, 11, 13,17and19

Total favorable number of cards =8

Probability that the number on the drawn card is a prime number=

(Total favorable number of cards/ Total number of cards)=(8/20)=(2/5)

29.

Let the vertices of the quadrilateral be A(-4,8), B(-3,-4), C(0,-5)and D(5,6).

Join AC to form two triangles, namely, ∆ABC and ∆ACD.

Area of quadrilateral ABCD = Area of ∆ABC+ Area of ∆ACD

We know

Area of triangle having vertices (x1,y1),(x2,y2),(x3,y3) = (1/2)(x1(y2-y3)+x2(y3-y1)+x3(y1-y2))

Now,

Area of ∆ABC = (1/2)((-4)(-4+5) + (-3)(-5-8) + 0 (8+4))

= (1/2)(-4+39)

= (35/2)

Area of ∆ACD =(1/2)((-4)(-6-5) +5(-5-8) +0 (8-6))

= (1/2)(-44-65)

= (109/2)

∴Area of quadrilateral ABCD=Area of ∆ABC + Area of ∆ACD = (35/2)+(109/2)=72 square units

Thus, the area of quadrilateral ABCD is 72 square units.

30.

Depth (h1) of the well = 14 m

Radius (r1) of the circular end of the well = (4/2) m =2 m

Height (h2) of embankment = 40 cm = 0.4 m

Let the width of embankment be x.

From the figure, it can be observed that the embankment will be cylindrical in shape having outer radius (r2) as (2 +x) m and inner radius (r1) as 2 m.

Volume of earth dug from the well = Volume of earth used to form embankment

Πr12h1 = π(r2

2 – r12)h2

Π(2)214 = π((2+x)2 – 22)

X2+4x-40 =0

(x-10)(x+14) =0

X =10 m

Therefore, the width of the embankment will be 10 m.

31. Increase in the level of water in half an hour, h=3.15 m=315 cm

Radius of the water tank, r=40 cm

Volume of water that falls in the tank in half an hour

= πr2h

= π(40)2(315)

= 5,04,000π cm3

Rate of flow of water = 2.52 km/h

Length of water column in half an hour = 2.522 = 1.26km = 1,26,000 cm

Let the internal diameter of the cylindrical pipe be d.

Volume of the water that flows through the pipe in half an hour =π (d/2)2(126000) cm3

We know

Volume of the water that flows through the pipe in half an hour = Volume of water that falls in the tank in half an hour

Π(d/2)2(126000) =5,04,000π

(d/2)2 = 4

d2 =16

d =4

Thus, the internal diameter of the pipe is 4 cm.