RD Sharma 2017 Solutions for Class 10 Math Chapter 18 Summative Assessment 2 are provided here with simple step-by-step explanations. These solutions for Summative Assessment 2 are extremely popular among class 10 students for Math Summative Assessment 2 Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2017 Book of class 10 Math Chapter 18 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2017 Solutions. All RD Sharma 2017 Solutions for class 10 Math are prepared by experts and are 100% accurate.

#### Page No SA 2.12:

#### Question 31:

question

#### Answer:

Number x when selected can take any value among 1, 2 and 3.

Corresponding to each such selection of x there can be three ways of selecting y i.e 1, 4, 9.

So, the two numbers x and y can be selected in following ways:

(1, 1), (1, 4), (1, 9)

(2, 1), (2, 4), (2, 9)

(3, 1), (3, 4), (3, 9)

There are thus 9 ways. So, total elementary events = 9

Now the product xy will be less than 9 if the numbers are chosen in the following ways:

(1, 1), (1, 4), (2, 1), (2, 4), (3, 1)

So, the number of favorable outcomes = 5

Hence the probability that the product xy of the two numbers will be less than 9 will be = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favorable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}$

= $\frac{5}{9}$

#### Page No SA2.10:

#### Question 1:

If the *n*^{th} term of an A.P. is (2*n* + 1), then the sum, of its first three terms is

(a) 6*n* + 3

(b) 15

(c) 12

(d) 21

#### Answer:

First term(*n *= 1) = 2(1) + 1 = 3

Second term(*n *= 2) = 2(2) + 1 = 5

Third term(*n *= 3) = 2(3) + 1 = 7

Sum of first three terms = 3 + 5 + 7 = 15

Hence, the correct answer is option (b).

#### Page No SA2.10:

#### Question 2:

In figure 1, AP, AQ and BC are tangents to the circle. If AB = 5 cm, AC = 6 cm and BC = 4 cm, then the length of AP (in cm) is

(a) 7.5

(b) 15

(c) 10

(d) 9

#### Answer:

We know that tangent segments to a circle from the same external point are congruent.

Therefore, we have

AP = AQ

BP = BD

CQ = CD

Now, AB + BC + AC = 5 + 4 + 6 = 15

⇒AB + BD + DC + AC = 15 cm

⇒AB + BP + CQ + AC = 15 cm

⇒AP + AQ= 15 cm

⇒2AP = 15 cm

⇒AP = 7.5 cm

Hence, the correct answer is option (a).

#### Page No SA2.10:

#### Question 3:

The circumference of a circle is 22 cm. The area of its quadrant (in cm^{2}) is

(a) $\frac{77}{2}$

(b)$\frac{77}{4}$

(c)$\frac{77}{8}$

(d)$\frac{77}{16}$

#### Answer:

Let the radius of the circle be *r *cm.

Now, circumference of circle = 22

$\Rightarrow 2\mathrm{\pi}r=22\phantom{\rule{0ex}{0ex}}\Rightarrow 2\times \frac{22}{7}\times r=22\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{7}{2}\mathrm{cm}$

Area of quadrant = $\frac{1}{4}\mathrm{\pi}{r}^{2}=\frac{1}{4}\times \frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}=\frac{77}{8}{\mathrm{cm}}^{2}$

Hence, the correct answer is option (c).

#### Page No SA2.10:

#### Question 4:

A solid right circular cone is cut into two parts at the middle of its height by a planeparallel to its base. The ratio of the volume of the smaller cone to the whole cone is

(a) 1 : 2

(b) 1 : 4

(c) 1 : 6

(d) 1 : 8

#### Answer:

Let the height of the cone be *H* and radius of the cone be *R*.

Now, the cone is divided into two parts by the parallel plane

∴ OC = CA = $\frac{H}{2}$

Now, In ∆OCD and OAB

∠OCD = OAB (Corresponding angles)

∠ODC = OBA (Corresponding angles)

By AA-similarity criterion ∆OCD ∼ ∆OAB

$\therefore \frac{\mathrm{CD}}{\mathrm{AB}}=\frac{\mathrm{OC}}{\mathrm{OA}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{CD}}{R}=\frac{H}{2\times H}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{CD}=\frac{R}{2}$

$\frac{\mathrm{Volume}\mathrm{of}\mathrm{smaller}\mathrm{cone}}{\mathrm{Volume}\mathrm{of}\mathrm{whole}\mathrm{cone}}=\frac{{\displaystyle \frac{1}{3}}\mathrm{\pi}{\left(\mathrm{CD}\right)}^{2}\mathrm{OC}}{\frac{1}{3}\mathrm{\pi}{\left(\mathrm{AB}\right)}^{2}\mathrm{OA}}\phantom{\rule{0ex}{0ex}}=\frac{{\left({\displaystyle \frac{R}{2}}\right)}^{2}{\displaystyle \frac{H}{2}}}{{R}^{2}H}\phantom{\rule{0ex}{0ex}}=\frac{1}{8}\frac{{R}^{2}H}{{R}^{2}H}\phantom{\rule{0ex}{0ex}}=\frac{1}{8}$

Hence, the correct answer is option (d).

#### Page No SA2.10:

#### Question 5:

Find the value of *p* for which the roots of the equation *px* (*x *− 2) + 6 = 0 are equal.

#### Answer:

*px* (*x *− 2) + 6 = 0

⇒*px*^{2} − 2*px* + 6 = 0

Since, we know that for the equal roots in a quadratic equation, we have

$D=0\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}-4ac=0\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(-2p\right)}^{2}-4p\times 6=0\phantom{\rule{0ex}{0ex}}\Rightarrow 4{p}^{2}-24p=0\phantom{\rule{0ex}{0ex}}\Rightarrow 4p\left(p-6\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow p-6=0\left[\because p\ne 0\right]\phantom{\rule{0ex}{0ex}}\Rightarrow p=6\phantom{\rule{0ex}{0ex}}$

Hence, the value of *p* is 6.

#### Page No SA2.10:

#### Question 6:

How many two-digit numbers are divisible by 3?

#### Answer:

The first two digit number divisible by 3 is 12, then 15, 18........99.

Here, *a* = 12, *d* = 3 and *a*_{n}* *= 99

*a*_{n }= *a* + (*n* − 1)*d*

⇒ 99 = 12 + (*n* − 1)3

⇒ 99 = 12 + 3*n* − 3

⇒ 3*n* = 90

⇒ *n* = 30

Hence, there are 30 two-digit numbers-are divisible by 3

#### Page No SA2.10:

#### Question 7:

In Figure 2. a right triangle ABC. circumscribes a circle of radius *r*. If AB and BC are of length 8 cm and 6 cm respectively. find the value of *r*.

#### Answer:

In right triangle ABC

By using Pythagoras theorem we have

AC

^{2}= AB

^{2}+ BC

^{2}

= 8

^{2}+ 6

^{2}

= 64 + 36

= 100

∴ AC

^{2}= 100

⇒ AC = 10 cm

Now,

#### Page No SA2.11:

#### Question 8:

Prove that the tangents drawn at the ends of a diameter ofa circle are parallel.

#### Answer:

We know that the radius and tangent are perperpendular at their point of contact.

∠OQC = ∠ORA = 90

^{∘}

Now, ∠OQC + ∠ORA = 90

^{∘ }+ 90

^{∘ }= 180

^{∘}

Since, the sum of cointerior angles on the same side of a transversal line is 180

^{∘}

Hence, CD || AB

#### Page No SA2.11:

#### Question 9:

In Figure 3. ABCD is a square of side 4 cm. A quadrant of a circle of radius 1 cm is drawn at each vertex of the square and a circle of diameter 2 cm is also drawn. Find the area of the shaded region. (Use π = 3.14)

#### Answer:

Area of the shaded region = Area of square − (4 ⨯ Area of quarter having radius 1 cm + Area of circle having diameter 2 cm)

$={\left(4\right)}^{2}-\left[4\times \frac{1}{4}\times \mathrm{\pi}{\left(1\right)}^{2}+\mathrm{\pi}{\left(\frac{2}{2}\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=16-\left[3.14+3.14\right]\phantom{\rule{0ex}{0ex}}=16-6.28\phantom{\rule{0ex}{0ex}}=9.72{\mathrm{cm}}^{2}$

Hence, the area of shaded region is 9.72 cm^{2}

#### Page No SA2.11:

#### Question 10:

A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting

(i) a red king

(ii) a queen or a jack.

#### Answer:

(i) There are two red kings in a pack of 52 cards.

Therefore, the probability is given by

$\frac{\mathrm{Number}\mathrm{of}\mathrm{red}\mathrm{king}\mathrm{cards}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{cards}}\phantom{\rule{0ex}{0ex}}=\frac{2}{52}\phantom{\rule{0ex}{0ex}}=\frac{1}{26}$

(ii) There are two 4 queen and 4 jacks in a pack of 52 cards

Therefore, the probability is given by

$\frac{\mathrm{Number}\mathrm{of}\mathrm{queen}\mathrm{or}\mathrm{jack}\mathrm{cards}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{cards}}\phantom{\rule{0ex}{0ex}}=\frac{4+4}{52}\phantom{\rule{0ex}{0ex}}=\frac{8}{52}\phantom{\rule{0ex}{0ex}}=\frac{2}{13}$

#### Page No SA2.11:

#### Question 11:

From a rectangular sheet of paper ABCD with AB = 40 cm and AD = 28 cm, a semicircular portion with BC as diameter is cut off. Find

the area of the remaining paper.

#### Answer:

We know that the opposite sides of rectangle are equal

Now, Radius of semicircular portion =

∴ Area of remaining paper = Area of rectangular sheet − Area of semicircular portion

$=40\times 28-\frac{1}{2}\left(\frac{22}{7}\times 14\times 14\right)\phantom{\rule{0ex}{0ex}}=1120-308\phantom{\rule{0ex}{0ex}}=812{\mathrm{cm}}^{2}$

Hence, the area of the remaining paper is 812 cm

^{2}

#### Page No SA2.11:

#### Question 12:

A solid sphere of radius 10.5 cm is melted and recast into smaller solid cones, each of radius 3.5 cm and height 3 cm. Find the number of cones so formed. (Use $\mathrm{\pi}=\frac{22}{7}$ )

#### Answer:

Let there are *n* number of small solid cones.

Now, Volume of solid sphere = *n* ⨯ volume of one small cone

$\Rightarrow \frac{4}{3}\mathrm{\pi}{R}^{3}=n\times \frac{1}{3}\mathrm{\pi}{r}^{2}h\phantom{\rule{0ex}{0ex}}\Rightarrow 4{\left(10.5\right)}^{3}=n\times {\left(3.5\right)}^{2}\times 3\phantom{\rule{0ex}{0ex}}\Rightarrow 4{\left(10.5\right)}^{3}=n\times {\left(3.5\right)}^{2}\times 3\phantom{\rule{0ex}{0ex}}\Rightarrow n=126$

Hence, there are 126 number of small solid cones.

#### Page No SA2.11:

#### Question 13:

Find the value of *k*, if the point P (2, 4) is equidistant from the points A (5, *k*) and B (*k*, 7).

#### Answer:

P (2, 4) is equidistant from the points A (5, *k*) and B (*k*, 7)

$\mathrm{AP}=\mathrm{BP}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{{\left(2-5\right)}^{2}+{\left(4-k\right)}^{2}}=\sqrt{{\left(2-k\right)}^{2}+{\left(4-7\right)}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(2-5\right)}^{2}+{\left(4-k\right)}^{2}={\left(k-2\right)}^{2}+{\left(4-7\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(4-k\right)}^{2}={\left(2-k\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 16-8k+{k}^{2}=4-4k+{k}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 4k=12\phantom{\rule{0ex}{0ex}}\Rightarrow k=3$

Hence, the value of *k* is 3

#### Page No SA2.11:

#### Question 14:

Solve the following quadratic equation for *x*:

${x}^{2}-4ax-{b}^{2}+4{a}^{2}=0$

#### Answer:

${x}^{2}-4ax-{b}^{2}+4{a}^{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-4ax+4{a}^{2}-{b}^{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(x\right)}^{2}-2\left(2a\right)x+{\left(2a\right)}^{2}-{\left(b\right)}^{2}=0$

$\Rightarrow {\left(x-2a\right)}^{2}-{\left(b\right)}^{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x-2a-b\right)\left(x-2a+b\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-2a-b=0\mathrm{or}x-2a+b=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=2a+b\mathrm{or}x=2a-b$

#### Page No SA2.11:

#### Question 15:

Find the sum of all multiples of 7 lying between 500 and 900.

#### Answer:

The first multiple of 7 between 500 and 900 is 504 and the last multiple between 500 and 900 is 896.

So, we are to find the value of 504 + 511 + ..............+ 896

Here, *a* = 504, *d* = 7 and *a*_{n}* *= 896

*a*_{n }= *a* + (*n* − 1)*d*

⇒ 896 = 504 + (*n* − 1)7

⇒ 896 = 504 + 7*n* − 7

⇒ 7*n* = 399

⇒ *n* = 57

Now,

${S}_{n}=\frac{n}{2}\left[a+{a}_{n}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow {S}_{57}=\frac{57}{2}\left[504+896\right]\phantom{\rule{0ex}{0ex}}=39900$

Hence, sum of all multiples of 7 lying between 500 and 900 is 39900.

#### Page No SA2.11:

#### Question 16:

Draw a triangle ABC with BC = 7 cm, ∠B = 45° and ∠C = 60°. Then construct the another triangle, whose sides are $\frac{3}{5}$ times the corresponding sides of △ABC.

#### Answer:

Steps of Construction:

__Step I:__ BC = 7 cm is drawn.

__Step II__: At B, a ray is drawn making an angle of 45° with BC.

__Step III:__ At C, a ray making an angle of 60° with BC is drawn intersecting the previous ray at A.

__Step IV:__ A ray BX is drawn making an acute angle with BC opposite to vertex A.

__Step V:__ Five points B_{1}, B_{2}, B_{3}, B_{4} and B_{5} at equal distance is marked on BX.

__Step VI:__ B_{5}C is joined and B_{3}C' is made parallel to B_{5}C.

__Step VII:__ C'A' is made parallel CA.

Thus, A'BC' is the required triangle.

#### Page No SA2.11:

#### Question 17:

In Figure 4, a circle is inscribed in a triangle PQR with PQ = 10 cm, QR = 8 cm and PR = 12 cm. Find the lengths QM, RN and PL.

#### Answer:

We know that tangent segments to a circle from the same external point are congruent.

Now, we have

PL = PN, QL = QM and RM = RN

Now, PL + LQ = 10 cm .....(1)

PN + NR = 12 cm

⇒ PL + NR = 12 cm .....(2)

QM + MR = 8 cm

⇒ LQ + NR = 8 cm .....(3)

Adding all these we get

PL + LQ + PL + NR + LQ + NR = 30

⇒2(PL + LQ + NR) = 30

⇒PL + LQ + NR = 15 cm .....(4)

Solving (1) and (4), we get

NR = 5 cm

Solving (2) and (4), we get

LQ = 3 cm

Solving (3) and (4), we get

and PL = 7 cm

∴ PL = PN = 7 cm, QL = QM = 3 cm and RM = RN = 5 cm

#### Page No SA2.11:

#### Question 18:

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

[Use π = 3.14]

#### Answer:

In right triangle ABC

BC^{2} = AB^{2} + AC^{2}

= (7)^{2} + (24)^{2}

= 49 + 576

= 625

∴ BC^{2} = 625

⇒ BC = 25

Now, ∠COD + ∠BOD = 180° (Linear pair angles)

⇒∠COD = 180° − 90° = 90°

Now, Area of the shaded region = Area of sector having central angle (360° − 90°) − Area of triangle ABC

$=\frac{270\xb0}{360\xb0}\mathrm{\pi}{\left(\frac{\mathrm{BC}}{2}\right)}^{2}-\frac{1}{2}\mathrm{AB}\times \mathrm{AC}\phantom{\rule{0ex}{0ex}}=\frac{3}{4}\times 3.14{\left(\frac{25}{2}\right)}^{2}-\frac{1}{2}\times 7\times 24\phantom{\rule{0ex}{0ex}}=367.97-84\phantom{\rule{0ex}{0ex}}=283.91{\mathrm{cm}}^{2}$

Hence, the area of shaded region is 283.91 cm^{2}

#### Page No SA2.12:

#### Question 19:

#### Answer:

$={\left(14\right)}^{2}-\left[\frac{1}{2}\mathrm{\pi}{\left(\frac{14}{2}\right)}^{2}+\frac{1}{2}\mathrm{\pi}{\left(\frac{14}{2}\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=196-154\phantom{\rule{0ex}{0ex}}=42{\mathrm{cm}}^{2}$

Hence, the area of the shaded region is 42 cm

^{2}.

#### Page No SA2.12:

#### Question 20:

A hemispherical bowl of internal radius 9 cm is full of water. Its contents are emptied in a cylindrical vessel of internal radius 6 cm. Find the height of water in the cylindrical vessel.

#### Answer:

Volume of hemispherical bowl = Volume of cylindrical vessel

$\Rightarrow \frac{2}{3}\mathrm{\pi}{R}^{3}=\mathrm{\pi}{r}^{2}h\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{3}{\left(9\right)}^{3}={\left(6\right)}^{2}h\phantom{\rule{0ex}{0ex}}\Rightarrow h=13.5\mathrm{cm}$

#### Page No SA2.12:

#### Question 21:

The angles of depression of the top and bottom of a tower as seen from the top of a $60\sqrt{3}$ m high cliff are 45° and 60° respectively. Find the height of the tower.

#### Answer:

Let AE be the tower and BD be the cliff of $60\sqrt{3}$ height

In right triangle ABC

$\mathrm{tan}45\xb0=\frac{\mathrm{AC}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow 1=\frac{h}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=h$

Now, In right triangle BCE

$\mathrm{tan}60\xb0=\frac{EC}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}=\frac{60\sqrt{3}}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=60\mathrm{m}\phantom{\rule{0ex}{0ex}}\Rightarrow x=h=60\mathrm{m}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{AE}=\mathrm{AC}+\mathrm{CE}=60+60=120\mathrm{m}$

Hence, the height of the tower is 120 m.

#### Page No SA2.12:

#### Question 22:

Find the coordinates of a point P, which lies on the line segment joining the points A (− 2, − 2) and B (2, − 4) such that $\mathrm{AP}=\frac{3}{7}\mathrm{AB}$

#### Answer:

$\mathrm{AP}=\frac{3}{7}\mathrm{AB}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{AB}}{\mathrm{AP}}=\frac{7}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{AB}}{\mathrm{AP}}-1=\frac{7}{3}-1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{AB}-\mathrm{AP}}{\mathrm{AP}}=\frac{7-3}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{PB}}{\mathrm{AP}}=\frac{4}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{AP}}{\mathrm{PB}}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}$

So, the point P divides the line segment AB in the ratio 3 : 4.

Let the coordinates of P be (*x*, *y*)

$\therefore x=\frac{3\times 2+4\left(-2\right)}{3+4}=\frac{-2}{7}\mathrm{and}y=\frac{3\left(-4\right)+4\left(-2\right)}{3+4}=\frac{-20}{7}$

Hence, the coordinates of the point P is $\left(\frac{-2}{7},\frac{-20}{7}\right)$.

#### Page No SA2.12:

#### Question 23:

If the points A(*x*,*y*), B(3, 6) and C(−3,4) are collinear,show that *x* − 3*y* + 15 = 0

#### Answer:

Since, the points A(*x*,*y*), B(3, 6) and C(−3,4) are collinear

Hence, the area of traiangle is zero.

$\Rightarrow \frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right]=0\phantom{\rule{0ex}{0ex}}\Rightarrow x\left(6-4\right)+3\left(4-y\right)-3\left(y-6\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2x+12-3y-3y+18=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2x-6y+30=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-3y+15=0$

#### Page No SA2.12:

#### Question 24:

All kings, queens and aces are removed from a pack of 52 cards. The remaining cards are well shuffled and then a card is drawn from it. Find the probability that the drawn card is

(i) a black face card.

(ii) a red card.

#### Answer:

Total number of cards = 52

Number of cards after removing kings, queens and aces = 52 − (4 + 4 + 4) = 40

(i) There are 8 black face cards out of which 2 black kings, 2 black queens and 2 black aces are already removed

Hence, the required number of cards are 8 − 6 = 2

Therefore, the probability is given by

$\frac{\mathrm{Number}\mathrm{of}\mathrm{black}\mathrm{face}\mathrm{cards}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{cards}}\phantom{\rule{0ex}{0ex}}=\frac{2}{40}\phantom{\rule{0ex}{0ex}}=\frac{1}{20}$

(ii) There are 26 red cards out of which 2 red kings, 2 red queens and 2 red aces are already removed

Hence, the required number of cards are 26 − 6 = 20

Therefore, the probability is given by

$\frac{\mathrm{Number}\mathrm{of}\mathrm{red}\mathrm{cards}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{cards}}\phantom{\rule{0ex}{0ex}}=\frac{20}{40}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$

#### Page No SA2.12:

#### Question 25:

In a flight of 2800 km, an aircraft was slowed down due to bad weather. Its average speed is reduced by 100 km/h and time increased by 30 minutes. Find the original duration of the flight.

#### Answer:

Let the speed be *x*

Now, time taken at the normal speed is given by

$\frac{2800}{x}\mathrm{hr}$

As the speed is reduced by 100 km/h

Hence, time taken after the reduction in speed is given by

$\frac{2800}{x-100}\mathrm{hr}$

A.T.Q

$\frac{2800}{x-100}=\frac{2800}{x}+\frac{30}{60}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2800}{x-100}-\frac{2800}{x}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2800x-2800x+280000}{x\left(x-100\right)}=\frac{1}{2}$

$\Rightarrow x\left(x-100\right)=560000\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-100x-560000=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-800x+700x-560000=0\phantom{\rule{0ex}{0ex}}\Rightarrow x\left(x-800\right)+700\left(x-800\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x-800\right)\left(x+700\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=800\mathrm{km}/\mathrm{h}\left[\because x\ne -700\right]$

Thus, the original duration of the flight is given by

$\frac{2800}{800}=\frac{7}{2}=3\frac{1}{2}\mathrm{or}3\mathrm{hours}30\mathrm{minutes}$

#### Page No SA2.12:

#### Question 26:

Find the common difference of an A.P. whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.

#### Answer:

Let common difference = *x*

First term = 5

Hence next term will be 5 , (5 + *x*) (5 + 2*x*) (5 + 3*x*) (5 + 4*x*) (5 + 5*x*) (5 + 6*x*) (5 + 7*x*) (5 + 8*x*)

Sum of first four terms = 5 + 5 + *x* + 5 + 2*x* + 5 + 3*x* = 20 + 6*x*

Sum of next four terms = 5 + 4*x* + 5 + 5*x* + 5 + 6*x* + 5 + 7*x* + 5 + 8*x* = 20 + 22*x*

Now, Sum of first four terms is 1/2 of next four terms sum

⇒(20 + 6*x*)2 = (20 + 22*x*)

⇒ 40 + 12*x* = 20 + 22*x*

⇒ 12*x *− 22*x* = 40 − 20

⇒ 10*x* = 20

⇒ *x* = 2

#### Page No SA2.12:

#### Question 27:

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

#### Answer:

We know that the radius and tangent are perperpendular at their point of contact

∴ ∠OAC = ∠OBC = 90^{∘}

Now, In △OCA and △OCB

∠OAC = ∠OBC = 90^{∘}

OA = OB (Radii of the circle)

OC = OC (Common)

By RHS congruency

△OCA ≅ △OCB

∴ CA = CB

#### Page No SA2.12:

#### Question 28:

A drinking glass is in the shape of the frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

[Use $\mathrm{\pi}=\frac{22}{7}$]

#### Answer:

Capacity of the glass = $\frac{\mathrm{\pi}h}{3}\left({R}^{2}+Rr+{r}^{2}\right)$

$=\frac{22\times 14}{7\times 3}\left({2}^{2}+2\times 1+{1}^{2}\right)\left[\because R=\frac{4}{2}\mathrm{and}r=\frac{2}{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{22\times 14}{7\times 3}\times 7\phantom{\rule{0ex}{0ex}}=\frac{308}{3}{\mathrm{cm}}^{3}$

Hence, the capacity of the glass is $\frac{308}{3}{\mathrm{cm}}^{3}$

#### Page No SA2.12:

#### Question 29:

A military tent of height 8.25 m is in the form of a right circular cylinder of base diameter 30 m and height 5.5 m surmounted by a right circular cone of same base radius. Find the length of the canvas use in making the tent, if the breadth of the canvas is 1.5 m.

#### Answer:

Let the radius of the cylinder be *r*.

Then, $r=\frac{30}{2}=15\mathrm{m}$ which will be equal to the radius of the cone.

Now, Height of the circular cone = Height of tent − Height of cylinder = 8.25 − 5.5 = 2.75 m

Lateral height of cone (*l*)= $\sqrt{{15}^{2}+2.{75}^{2}}$

$=\sqrt{225+7.5625}\phantom{\rule{0ex}{0ex}}=\sqrt{232.5625}\phantom{\rule{0ex}{0ex}}=15.25\mathrm{m}$

Curved surface area of tent = curved surface area of cylinder + curved surface area of cone

$=2\mathrm{\pi}rh+\mathrm{\pi}rl\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}r\left(2h+l\right)\phantom{\rule{0ex}{0ex}}=\frac{22}{7}\times 15\left(2\times 5.5+15.25\right)\phantom{\rule{0ex}{0ex}}=\frac{22}{7}\times 15\times 26.25\phantom{\rule{0ex}{0ex}}=1237.5{\mathrm{m}}^{2}$

Let the length of canvas be *x*

Now, curved surface area of tent = rectangular piece of canvas

$\Rightarrow 1237.5=x\times 1.5\phantom{\rule{0ex}{0ex}}\Rightarrow x=825\mathrm{m}\phantom{\rule{0ex}{0ex}}$

Hence, the length of the canvas is 825 m

#### Page No SA2.12:

#### Question 30:

The angles of elevation and depression of the top and bottom of a light-house from the top of a 60 m high building are 30° and 60° respectively. Find

(i) the difference between the heights of the light-house and the building.

(ii) the distance between the light-house and the building .

#### Answer:

Let AE be the light house and BD be the building of 60 m height

In right triangle ABC

$\mathrm{tan}30\xb0=\frac{\mathrm{AC}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\sqrt{3}h$

Now, In right triangle BDE

$\mathrm{tan}60\xb0=\frac{\mathrm{BD}}{\mathrm{DE}}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}=\frac{60}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{60}{\sqrt{3}}=20\sqrt{3}\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}x=\sqrt{3}h\phantom{\rule{0ex}{0ex}}\Rightarrow 20\sqrt{3}=\sqrt{3}h\phantom{\rule{0ex}{0ex}}\Rightarrow h=20\mathrm{m}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

(i) The difference between the heights of the light-house and the building is *h* = 20 m

(ii) The distance between the light-house and the building is *x* = $20\sqrt{3}$ m

#### Page No SA2.2:

#### Question 1:

Which of the following equations has the sum of its roots as 3?

(*a*) *x*^{2} + 3*x* − 5 = 0

(*b*) −*x*^{2} + 3*x* + 3 = 0

(*c*)

(*d*) 3*x*^{2} − 3*x* − 3 = 0

#### Answer:

Given the following quadratic equations

(*a*) *x*^{2} + 3*x* − 5 = 0 (*b*) −*x*^{2} + 3*x* + 3 = 0

(*c*) (*d*) 3*x*^{2} − 3*x* − 3 = 0

We are to find out which of the above equations has sum of roots = 3.

The sum of the roots of the quadratic equation* ax*^{2} + *bx* + *c* = 0 is given by .

**For given equation**** (****a****)**

*a* = 1,* b* = 3,* c* = −5

**For given equation**** (****b****)**

*a* = −1*, b* = 3,* c* = 3

**For given equation**** (****c****)**

**For given equation**** (****d****)**

*a *= 3*, b *= −3*, c* = −3

Hence option (*b*) is correct.

#### Page No SA2.2:

#### Question 2:

In Fig. 1, *PQ* and *PR* are tangents to the circle with centre *O* such that ∠*QPR* = 50°,

(a) 25°

(b) 30°

(c) 40°

(d) 50°

#### Answer:

We are given the below figure in which

*PQ* and *PR* are tangents to the circle with centre *O* and

We have to find

*PQ *is the tangent to circle

Therefore [Since Radius of a circle is perpendicular to tangent]

Similarly

We know that sum of angles of a quadrilateral

Therefore in Quadrilateral PQOR

Hence option (*a*) is correct.

#### Page No SA2.2:

#### Question 3:

question

#### Answer:

Height of original cone = 60 cm.

Let the radius be *r*_{1}

Let the_{ }height of the new cone formed be *h*_{2} and the radius be *r*_{2}.

B and D are the centers of the bases of these cones

Given, *h* _{1} = AB = 60 cm

In ∆ADE and ∆ABC

∠DAE = ∠BAC (Common)

∠ADE = ∠ABC (Corresponding angles as DE || BC)

∴ ∆ADE ∼ ∆ABC (AA similarity)

$\Rightarrow \frac{AD}{AB}=\frac{DE}{BC}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{h}_{2}}{{r}_{2}}=\frac{{r}_{2}}{{r}_{1}}...\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}$

Given that

$\frac{\mathrm{Volume}\mathrm{of}\mathrm{cone}AFE}{\mathrm{Volume}\mathrm{of}\mathrm{cone}AGC}=\frac{1}{64}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\mathrm{\pi r}}_{2}^{2}{\mathrm{h}}_{2}}{{\mathrm{\pi r}}_{1}^{2}{\mathrm{h}}_{1}}=\frac{1}{64}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{{r}_{2}}{{r}_{1}}\right)}^{2}\times \left(\frac{{h}_{2}}{{h}_{1}}\right)=\frac{1}{64}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{{h}_{2}}{{h}_{1}}\right)}^{3}=\frac{1}{64}\left(From\left(\mathrm{i}\right)\right)$

$\Rightarrow {\left(\frac{{h}_{2}}{60}\right)}^{3}=\frac{1}{64}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{h}_{2}}{60}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow {h}_{2}=15\mathrm{cm}$

Hence, height from the base from which the cut is made = 60 − 15 = 45 cm

#### Page No SA2.2:

#### Question 4:

If the circumference of a circle is equal to the perimeter of a square then the ratio their areas is:

(a) 22 : 7

(b) 14 :11

(c) 7 :22

(d) 7 :11

#### Answer:

It is given that circumference of a circle = perimeter of a square

We have to find the ratio of their areas

Let the radius of circle = *r*

Let the circumference of circle =

Let the area of circle =

Let the side of the square = *b*

Let the perimeter of square =

Let the area of square =

Therefore

…… (1)

Area of Circle

[From Equation (1)]

Hence Option (*b*) is correct.

#### Page No SA2.2:

#### Question 5:

Find the roots of the following quadratic equation:

$\frac{2}{5}{x}^{2}-x-\frac{3}{5}=0.$

#### Answer:

It is given that:

We have to find the roots of above equation.

Multiplying both sides by 5

2*x*^{2} − 5*x* − 3 = 20

2*x*^{2} − 6*x* + *x* − 3 = 0

2*x*(*x* − 3) + 1 (*x* − 3) = 0

(*x* − 3) (2*x* + 1) = 0

*x* = 3, *x* =

#### Page No SA2.3:

#### Question 6:

If the numbers *x* − 2, 4*x* − 1 and 5*x* + 2 are in A.P. find the value of *x*.

#### Answer:

It is given that the numbers

We have to find the value of *x*

We know that if *x, y* and *z* are in A.P, then

Therefore for the given numbers

Hence

#### Page No SA2.3:

#### Question 7:

Two tangents *PA* and *PB* are drawn from an external point *P* to a circle with centre *O*. Prove that *AOBP* is a cyclic quadrilateral.

#### Answer:

We are given two tangents *PA *and *PB *drawn to a circle with centre *O *from external point *P*

We are to prove that quadrilateral *AOBP* is cyclic

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

Therefore from figure

That is

In quadrilateral *AOBP*,

[Sum of angles of a quadrilateral = 360°]

We know that the sum of opposite angles of cyclic quadrilateral = 180°

Therefore from (1) and (2)

Quadrilateral *AOBP* is a cyclic quadrilateral.

#### Page No SA2.3:

#### Question 8:

In Fig. 2, a circle of radius 7 cm is inscribed in a square. Find the area of the shaded region

$\left(Use\mathrm{\pi}=\frac{22}{7}\right).$

#### Answer:

It is given that a circle of radius 7 cm is inscribed in a square

We have to find the area of shaded region shown in figure

We are given the following figure

Let the side of the square = *a* cm

Since the circle in inscribed in the square

Diameter of the circle = *a* cm

Radius of circle * *cm

Given that radius of circle = 7 cm

Therefore

#### Page No SA2.3:

#### Question 9:

How many spherical lead shots each having diameter 3 cm can be made from a cuboidal lead solid of dimensions 9 cm × 11 cm × 12 cm?

#### Answer:

Given a cuboidal lead solid with dimensions 9 cm × 11 cm × 12 cm

We have to find the number of spherical lead shots each having a diameter

of 3cm which can be made from the cuboidal lead solid.

Let the length of cuboidal lead solid L* *=* *9 cm

Let the breadth of cuboidal lead solid B* *=* *11 cm

Let the length of cuboidal lead solid H =* *9 cm

Let the number of spherical lead shots = *x*

Volume of a spherical lead shot =

Volume of the cubical lead shot =

#### Page No SA2.3:

#### Question 10:

Point *P*(5, −3) is one of the two points of trisection of the line segment joining the points *A* (7, −2) and *B* (1, − 5) near to *A*. Find the coordinates of the other point of trisection.

#### Answer:

We are given a line segment joining points *A *(7, −2) and *B *(1, −5)

*P *(5, −3) is one of the two points of trisection of line segment *AB*

*P *is near to *A*

We are to find the coordinates of other points of trisection

Let the other point of trisection is *Q*

Therefore

*AP* = *PQ* = *QB*

That is *Q* is the mid point of line segment *PB*

We know that the coordinates of mid point of line segment with coordinates of end points

#### Page No SA2.3:

#### Question 11:

Show that the point P (−4, 2) lies on the line segment joining the points A (−4, 6) and B (−4, −6).

#### Answer:

We have to show that point *P *(−4, 2) lies on line segment *AB *with points *A *(−4, 6) and

*B *(−4, −6)

If *P* (−4, 2) lies on the line segment joining *A *(−4, 6) and *B *(−4, −6), then the three points

must be collinear.

Let the three points be not collinear and form a triangle *PAB*

We know that area of a triangle with coordinates of vertices $({x}_{1},{y}_{1}),({x}_{2},{y}_{2}),({x}_{3},{y}_{3})$

Since area of the triangle is 0, no triangle exists.

Therefore points *P* (−4, 2),* A *(−4, 6) and *B *(−4, −6) are collinear.

#### Page No SA2.3:

#### Question 12:

Two dice are thrown at the same time. Find the probability of getting different numbers on both dice.

#### Answer:

It is given that two dice are thrown at the same time.

We have to find the probability of getting different numbers on both dice.

Total number of possible choices in rolling a dice = 6

Total number of possible choices in rolling two dice [Using multiplication rule]

Probability of getting same number if two dice are thrown

Probability of getting different number if two dice are thrown

OR

It is given that a coin is tossed two times.

Therefore, sample space is given by,

{HH, HT, TH, TT}

Let E be the event of getting two heads. I.e., E = {HH}

Then, probability of getting atmost one head is given by,

P (E′) = P (HT or TH or TT) = 1 − P (E) = 1 − P (HH) =

#### Page No SA2.3:

#### Question 13:

A natural number, when increased by 12, becomes equal to 160 times its reciprocal. Find the number.

#### Answer:

It is given that a number when increased by 12 becomes 160 times its reciprocal.

We have to find the number.

Let the number be *x*

According to the question

#### Page No SA2.3:

#### Question 14:

Find the sum of the integers between 100 and 200 that are divisible by 9.

#### Answer:

We have to find the sum of integers between 100 and 200 that are divisible by 9.

Integers divisible by 9 between 100 and 200 are

108, 117, 126, … 198

The above equation forms an Arithmetic Progression (A.P)

Let there be n such integers in the above A.P

We know that the nth term of an A.P

Where = First term of A.P

= Common difference of successive members

Therefore total number of integers in the A.P = 11

We know that the sum of the *n* terms of an arithmetic progression

#### Page No SA2.3:

#### Question 15:

In Fig. 3, two tangents *PQ* are *PR* are drawn to a circle with centre *O* from an external point *P*. Prove that ∠*QPR* = 2 ∠*OQR*.

#### Answer:

Given a figure as shown. We have to prove that

Join *OR*

We know that sum of opposite angles of a cyclic quadrilateral

Therefore in quadrilateral *PQOR*,

…… (1)

In

[Since *OQ, OR* are radii of the circle]

Therefore is an isosceles triangle.

Hence [Angles opposite to equal side of isosceles triangle]

Now, [since,]

[From (1)]

Hence proved.

#### Page No SA2.3:

#### Question 16:

Draw a triangle *ABC *with side *BC* = 6 cm, *AB* = 5 cm and ∠*ABC** *= 60°. Then construct a triangle whose sides are $\frac{3}{4}$ time the corresponding sides of Δ*ABC*.

#### Answer:

We have to draw a triangle

Then we have to construct a similar triangle with side of

Steps of construction

1. Draw a line segment

2. At *B* draw so that a ray *BX* is made

3. Keep compass at *B* and mark an arc of 5 cm at ray *BX* and name that point as *A*

4. Join *AC* to make

5. Draw a ray making an acute angle with BC.

6. Mark 4 points, B_{1}, B_{2}, B_{3}, and B_{4} along BY such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}.

7. Join CB_{4}

8. Through the point B_{3}, draw a line parallel to B_{4}C by making an angle equal to ∠BB_{4}C, intersecting BC at C′.

9. Through the point C′, draw a line parallel to AC, intersecting BA at A′.

Thus, ΔA′BC′ is the required triangle.

#### Page No SA2.3:

#### Question 17:

In Fig. 4, *OABC* is a square inscribed in a quadrant *OPBQ*. If *OA* = 20 cm, find the area of shaded region [Use π = 3.14]

#### Answer:

It is given that OABC is a square in the quadrant OPBQ.OA is 20 cm.

We have to find the area of the shaded region.

*OABC* is a square , therefore sides of the *OABC* must be equal

Hence *OA, AB, BC, OC* = 20 cm.

Join the points *O* and *B *to form a line segment *OB.*

Since *OB* is the diagonal of *OABC*, is a right angled triangle.

Applying Pythagoras Theorem in

It can be seen from the figure that Quadrant *OPBQ* is of a circle with radius *OB( r )*

Therefore area of *OPBQ *

Now, area of the shaded region

#### Page No SA2.4:

#### Question 18:

A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter 'l' of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

#### Answer:

It is given that a hemisphere is cut from a cubical box with edge *l *such that

diameter of hemisphere is also *l.*

We have to find the surface area of the remaining solid.

Surface area of the cubical box with side *l*

Let *r* be the radius of hemisphere

Surface area of the hemisphere

( since )

#### Page No SA2.4:

#### Question 19:

A tower stands vertically on the ground. From a point on the ground which is 20 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.

#### Answer:

Given a tower at the ground such that at a distance of 20 m away from foot of tower,

the top of tower makes an angle of with the ground.

We have to find the height of the tower.

Let the tower be *AB*

Height of the tower = *h*

Distance of the tower from point *C* where top of the tower makes an angle of = 20 m

Since B is the foot of the tower, *CB* = 20 m

Height of the tower

#### Page No SA2.4:

#### Question 20:

Prove that the points *A* (4, 3), *B *(6, 4), *C* (5, −6) and *D* (3, −7) in that order are the vertices of a parallelogram.

#### Answer:

Given points *A (*4, 3), *B* (6, 4), *C* (5, −6) and *D *(3, −7).

We have to prove that these points form vertices of a parallelogram.

The above four points will form 4 line segments.

*AB, BC, CD, AD*

We know that length of a line segment having coordinates

Therefore

It can be seen that *AB = CD , BC = AD*

Since opposite sides are equal, *ABCD* is a parallelogram

Hence proved.

#### Page No SA2.4:

#### Question 21:

The points *A* (2, 9), *B* (*a*, 5), *C* (5, 5) are the vertices of a triangle *ABC* right angled at *B*. Find the value of '*a*' and hence the area of Δ*ABC*.

#### Answer:

It is given that ABC is a right angled triangle..Vertices of triangle are

*A*( 2,9 ), *B*( *a,*5 ), *C*( 5, 5 ).

We have to find the value of *a* and area of the triangle.

is a right angled triangle.

We know that length of a line with coordinates of end points

Hence in

…… (1)

…… (2)

Therefore, applying Pythagoras Theorem to right

Putting *a *= 2 in (1) and (2)

#### Page No SA2.4:

#### Question 22:

Cards with numbers 2 to 101 are placed in a box. A card is selected at random from the box. Find the probability that the card which is selected has a number which is a perfect square.

#### Answer:

It is given that cards with numbers 2 to 101 are placed in a box. A card is picked at random.

We have to find the probability that the card selected has a number which is a perfect square.

Perfect squares between 2 and 101 are 4, 9, 16, 25, 36, 48, 64, 81, 100

Total number of perfect squares

Total number of cards

Probability (perfect square)

Hence probability for the selected card number to be a perfect square

#### Page No SA2.4:

#### Question 23:

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/hr more than than its original speed. If it takes 3 hours to complete the total journey, what is the original average speed?

#### Answer:

It is given that a train travels a distance of 63 km with a certain average speed and 72 km with a speed which is 6 km/hr more than the original average speed.

Time taken for the whole journey is 3hr.

We have to find out the original average speed of the train.

Let the original average speed of the train = *x* km/hr

While covering the distance of 72 km, the speed of train

Total time taken by the train

Therefore

Dividing both sides by3

Speed cannot be negative

#### Page No SA2.4:

#### Question 24:

Find two consecutive odd positive integers, sum of whose squares is 290.

#### Answer:

We have to find two consecutive integers sum of whose squares is 290.

Let the two consecutive integers be *x* and *x*+2

According to the question

Dividing both sides by 2

Therefore two consecutive integers are

#### Page No SA2.4:

#### Question 25:

A sum of Rs. 1400 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs. 40 less than the preceding price, find the value of each of the prizes.

#### Answer:

It is given that total prize money is Rs 1400 /-. There are a total of 7 prizes distributed in a way that each prize is less than the previous prize by Rs 40/-

We have to find the value of the prizes.

Let *a* is the value of a prize

The difference between the consecutive prizes *d*

Total number of prizes n

Now it can be seen that the value of prizes forms an Arithmetic Progression (A.P)

Therefore

We know that for an A.P

Substituting the values

Therefore the value of prizes

#### Page No SA2.4:

#### Question 26:

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

#### Answer:

We have to prove that the lengths of tangents drawn from an external point to a circle are equal.

Draw a circle with centre O and tangents *PA* and *PB*, where *P* is the external point and *A *and *B* are the points of contact of the tangents.

Join *OA*, *OB* and *OP.*

Now

Hence proved.

#### Page No SA2.4:

#### Question 27:

A well of diameter 3 m ad 14 m deep is dug. The earth, taken out of it, has been evenly spread all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

#### Answer:

Given a well with diameter 3 m and height 14 m. The earth dug out from well is used to make a circular embankment of 4m width.

We have to find the height of the embankment.

Let *R *be the radius of well

Let *H *be the height of well

Let *r *be the radius of embankment

Let *h* be the height of embankment

*H*

Width of the circular embankment

According to the question

Volume of the earth dug = Volume of the circular embankment

Therefore,

=

#### Page No SA2.4:

#### Question 28:

21 glass spheres, each of radius 2 cm are packed in a cuboidal box of internal dimensions 16 cm × 8 cm × 8 cm and, then the box is filled with water. Find the volume of water filled in the box.

#### Answer:

It is given that 21 glass spheres each of radius 2 cm are packed in a cuboidal box of inner dimensions and then filled with water.

We have to find the volume of water.

Radius of glass sphere *r*

Volume of a glass sphere

Length of cuboidal box

Breadth of cuboidal box

Height of cuboidal box

Volume of water = Volume of cuboidal box − Volume of glass spheres

#### Page No SA2.4:

#### Question 29:

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

#### Answer:

It is given that slant height of frustum of a cone is 4 cm. Circumferences of its ends are

18 cm and 6 cm.

We have to find the curved surface area of the frustum.

Let *l* be the slant height

Let be the radii of two circular ends of the cone

Circumference of one end

Circumference of other end

Now,

Curved surface area of frustum

(since )

#### Page No SA2.4:

#### Question 30:

From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find height of the tower.

#### Answer:

It is given that tower is placed at a 20 m high building. The top and bottom of the tower makes an angle of respectively with the ground.

We have to find the height of the tower.

Let *DB* is the tower

Let *AD* is the building

Height of the building = 20 m

Height of the tower = *x *m

According to the figure,

…… (1)

…… (2)

Since (1) = (2)

#### Page No SA2.4:

#### Question 31:

question

#### Answer:

After spinning the arrow, it may stop in any of the six sectors.

a is the number of sector where the arrow stops after first spin.

b is the number of sector where the arrow stops after second spin.

So, the possible values of a can be 6 and so is the case for b.

So, (a,b) can have the following values:

$\left[\begin{array}{cccccc}\left(1,1\right)& \left(1,2\right)& \left(1,3\right)& \left(1,4\right)& \left(1,5\right)& \left(1,6\right)\\ \left(2,1\right)& \left(2,2\right)& \left(2,3\right)& \left(2,4\right)& \left(2,5\right)& \left(2,6\right)\\ \left(3,1\right)& \left(3,3\right)& \left(3,3\right)& \left(3,4\right)& \left(3,5\right)& \left(3,6\right)\\ \left(4,1\right)& \left(4,2\right)& \left(4,3\right)& \left(4,4\right)& \left(4,5\right)& \left(4,6\right)\\ \left(5,1\right)& \left(5,2\right)& \left(5,3\right)& \left(5,4\right)& \left(5,5\right)& \left(5,6\right)\\ \left(6,1\right)& \left(6,2\right)& \left(6,3\right)& \left(6,4\right)& \left(6,5\right)& \left(6,6\right)\end{array}\right]$

We have 36 elementary events out of which there are 6 elementary events where a = b.

We have 15 ordered pairs where a > b

$\Rightarrow \frac{a}{b}>1$ ...(i)

Also, there are 15 ordered pairs where a < b

$\Rightarrow \frac{a}{b}<1$ ...(ii)

So, the favorable events are 15 (From (i))

Hence the required probability = $\frac{\mathrm{Favorable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{15}{36}=\frac{5}{12}$

View NCERT Solutions for all chapters of Class 10