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#### Question 1:

Mark (✓) against the correct answer in each of the following:
Which of the following equations represents a line parallel to the x-axis?
(a) x + y = 5
(b) 2x + 5 = 8
(c) 2y − 4 = 1
(d) x + 4 = 0

(c) 2y − 4 = 1

$2y-4=1\phantom{\rule{0ex}{0ex}}⇒2y=4+1\phantom{\rule{0ex}{0ex}}⇒2y=5\phantom{\rule{0ex}{0ex}}⇒y=2.5$
A linear equation that contains only the y variable is parallel to the x-axis.

#### Question 2:

Mark (✓) against the correct answer in each of the following:
If the mode of the data 9, 7, x, 10, 12, 11, 14, 15, 13, 12, 10, 15 is 15, then the value of x is
(a) 10
(b) 10.5
(c) 12
(d) 15

(d) 15
The mode is 15 and we know that mode is the value which occurs most frequently.

#### Question 3:

Mark (✓) against the correct answer in each of the following:
The surface area of a solid hemisphere is
(a) 2πr2
(b) 3πr2
(c) 4πr2
(d) $\frac{2}{3}{\mathrm{\pi r}}^{3}$

(b) 3πr2

#### Question 4:

Mark (✓) against the correct answer in each of the following:
The height of a cone of diameter 10 cm and slant height 13 cm is
(a) 12 cm
(b) 13 cm
(c)
(d)

(a) 12 cm

#### Question 5:

Mark (✓) against the correct answer in each of the following:
Each side of a rhombus is 10 cm long and one of its diagonals measures 16 cm. The length of the other diagonal is
(a) 6 cm
(b) 12 cm
(c) 13 cm
(d)

(b) 12 cm We know that the diagonals of a rhombus bisect each other at right angles.
So, BO = OD = 8 cm

#### Question 6:

Mark (✓) against the correct answer in each of the following:
In the given figure, ​∠B < ​∠A and ​∠D > ​C. Then In $△$AOB,
$\angle$B < $\angle$A
So, AO < BO .........(i)
(Side opposite the larger angle is larger)
In $△$DOC,
$\angle$C < $\angle$D
So, DO  < CO .........(i)
(Side opposite the larger angle is larger)
Adding equations (i) and (ii), we get:
AO + DO < BO + CO

#### Question 7:

Mark (✓) against the correct answer in each of the following:
In the given figure, ABDC and ABFE are cyclic quadrilaterals. If ACD = 110°, then ∠AEF = ?
(a) 55°
(b) 70°
(c) 90°
(d) 110° (d) 110°

Since ACDE is a cylic quadrilateral, so

$\angle B+\angle C=180°\phantom{\rule{0ex}{0ex}}⇒\angle B=180°-\angle C=180°-110°=70°\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Again, AEFB is a cylic quadrilateral.
So,
$\angle B+\angle E=180°\phantom{\rule{0ex}{0ex}}⇒\angle E=180°-\angle B=180°-70°=110°\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Question 8:

Mark (✓) against the correct answer in each of the following:
In the given figure, the measure of BCD is
(a) 100°
(b) 70°
(c) 80°
(d) 30° (c) 80°

#### Question 9:

A dice was rolled 100 times and the number of times that 6 appeared was noted. If the probability of getting a 6 is $\frac{2}{5}$, how many times did 6 come up?

So, 6 appeared 40 times.

#### Question 10:

In a circle of radius 10 cm, find the length of a chord which is at a distance of 5 cm from its centre. We know that the perpendicular drawn from the centre of a circle bisects the chord of the circle.
So, AC = BC

#### Question 11:

The radius and slant height of a cone are in the ratio 4 : 7. If its curved surface area is 792 cm2, find its radius.

Suppose, the radius and slant height of the cone are 4x cm and 7x cm.
Curved surface are of the cone  = 792 cm2

Hence, radius of the cone  = 4x = 4$×$3 = 12 cm

#### Question 12:

Prove that a cyclic parallelogram is a rectangle. Given : ABCD is a cyclic parallelogram.
To prove : ABCD is a rectangle.
Proof :
ABCD is a paralleogram.

So, $\angle$ABC = $\angle$CDA = 90$°$
Similarly, $\angle$DAB =$\angle$BCD = 90$°$
Since, all the angles of the parallelogram are 90$°$, ABCD is a rectangle.

#### Question 13:

In the given figure, O is the centre of circle, ABO = 20° and ∠ACO = 30°. If ∠BOC = x°, find the value of x. In $△$AOB,
So, $\angle$OBA = $\angle$BAO = 20°  (Angles opposite equal sides are equal)
In $△$AOC,
So, $\angle$OCA = $\angle$CAO = 30°  (Angles opposite equal sides are equal)
$\angle$BAC = $\angle$CAO + $\angle$BAO
= 30°+ 20°
= 50°

$\angle$BOC = 100°
$⇒$x° = 100°
$⇒$x = 100

#### Question 14:

In the given figure, ABCD is a cyclic quadrilateral with O as the centre of the circle. If BOD = 136°, find ∠BCD. #### Question 15:

The diameter of a sphere is decreased by 20%. By what per cent does its curved surface area decrease?

Let the diameter of the sphere be d.
Curved surface area of the sphere = $4\mathrm{\pi }{r}^{2}$ = $\frac{4\mathrm{\pi }{d}^{\mathit{2}}}{4}=\mathrm{\pi }d$2
The diameter of the sphere is decreased by 20%.

New curved surface area = $4\mathrm{\pi }{{r}_{1}}^{2}$
$=4\mathrm{\pi }×\frac{4{d}^{2}}{25}\phantom{\rule{0ex}{0ex}}=\frac{16\mathrm{\pi }{d}^{2}}{25}$
Decrease in curved surface area =
Percentage decrease = $\frac{\frac{9\mathrm{\pi }{d}^{2}}{25}}{\mathrm{\pi }{d}^{2}}×100=\frac{9×100}{25}=36%$

#### Question 16:

A conical tent is 10 m high and its base radius is 24 m.
Find (i) the slant height of the tent and (ii) the cost of canvas required to make the tent, if the cost of 1 m2 of canvas is Rs 70. (i)

So, slant height of the cone = 26 m

(ii) Curved surface area of the cone = $\mathrm{\pi }rl$

=
Cost of 1 m2 area of the canvas = Rs 70
Cost of =

#### Question 17:

A cuboid has total surface area of 40 m2 and lateral surface area of 26 m2. Find the area of it base.

Surface area of top and bottom face of cuboid = Total surface area of cuboid $-$ lateral surface area of cuboid
= 40$-$26 = 14 m2
We know that the area of top and bottom face of a cuboid are equal.
So, area of base =

#### Question 18:

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in square metres.

Radius of the cylindrical roller = $\frac{84}{2}$ = 42 cm
Length of the roller = 120 cm
So, curved surface area of the roller = $2\mathrm{\pi }rh$

Area covered by the roller in 1 revolution = 31680 cm2
Area covered by the roller in 500 revolution = $500×$31680 = 15840000 cm2

= 1584 m2

#### Question 19:

Prove that an angle in a semicircle is a right angle.

Given: AB is a diameter of a circle C(O,r) and ∠ACB is an angle in the semicircle.
To Prove:ACB = 90° Proof: We know that the angle subtended by an arc at the centre of a circle is
twice the angle formed by it at any point on the remaining part of the circle.
∴ ∠AOB = 2∠ACB          [Since angle subtended by arc AB = 2 × angle formed by it at C]
⇒ 2∠ACB = 180°             [Since ∠AOB is a straight angle]
⇒ ∠ACB = 90°

#### Question 20:

Prove that two angles in the same segment of a circle are equal. Given: C(O,r) is a circle and ∠ACB and ∠ADB are angles in the same segment.
Proof: We know that the angle subtended by an arc at the centre of a circle is
twice the angle formed by it at any point on the remaining part of the circle.
∴ ∠AOB = 2∠ACB.................(i)
[Since angle subtended by arc AB = 2 × angle formed by it at C]
[Since angle subtended by arc AB = 2 × angle formed by it at D]
From equations (i) and (ii):

#### Question 21:

In ABC, if E is the mid-point of median AD, prove that $\mathrm{ar}\left(∆BED\right)=\frac{1}{4}\mathrm{ar}\left(∆ABC\right)$. Given:  D is the mid-point of BC and E is the mid-point of AD.
To prove: ar(BED) =$\frac{1}{4}$ ar(ABC)

Proof:
Since, D is the mid-point of BC, AD is the median of  ABC and BE is the median of  ABD.
We know that a median of a triangle divides it into two triangles of equal area.
So, ar ( ABD ) = $\frac{1}{2}$ ar ( ABC).................(i)

So, ar ( BED ) =$\frac{1}{2}$ar ( ABD).................(ii)

From (i) and (ii), we have:
ar ( BED  = $\frac{1}{2}$​​ ⨯ $\frac{1}{2}$⨯​ ar (∆ ABC
∴​     ar (∆ BED )​ =$\frac{1}{4}$  ⨯ ar(∆ABC)

#### Question 22:

Diagonals AC and BD of a trapezium ABCD, with AB || DC, intersect at O. Prove that ar(AOD) = ar(BOC). CDA and ​∆ CBD lie on the same base and between the same parallel lines.
So, ar( ​∆ CDA) = ar( CDB)...(i)
Subtracting ar (​​∆ OCD) from both sides of equation(i), we get:
ar(​​∆ CDA) $-$ ar (​​∆ OCD) = ​ ar(​​∆ CDB) $-$ar (​​∆ OCD)
⇒ ​ ar(​​∆ AOD)  =  ar (​​∆ BOC)

#### Question 23:

Draw a cumulative frequency curve for the following frequency distribution.

 Class Interval 0−9 10−19 20−29 30−39 40−49 50−59 60−69 Frequency 5 15 20 23 17 11 9

 Class Marks Frequency Cumulative Frequency 4.5 5 5 14.5 15 20 24.5 20 40 34.5 23 63 44.5 17 80 54.5 11 91 64.5 9 100

We plot the points A(4.5,5),B(14.5,20),C(24.5,40),D(34.5,63),E(44.5,80),F(54.5,91) and G(64.5,100). #### Question 24:

The arithmetic mean of 35 values is 45. What will be the new mean if each of these values is
(a) increased by 7?
(b) multiplied by 5?
(c) divided by 5?

Arithmetic mean = 45
(a) If each value is increased by 7, the mean will also be increased by 7.
New mean = 45+7 =52
(b) If each value is multiplied by 5, the mean will also be multiplied by 5.
New mean = 45$×$5 =225
(c) If each value is divided by 5, the mean will also be divided by 5.
New mean = 45$÷$5 = 9

#### Question 25:

The mean of 16 items was found to be 30. On rechecking, it was found that two items were wrongly taken as 22 and 18 instead of 32 and 28, respectively. Find the correct mean.

Given mean of 16 items = 30
Sum of 16 items = $16×30=480$
Correct sum of 16 items  = $480-22-18+32+28$ = 500
So, correct mean of 16 items = $\frac{500}{16}=31.25$

#### Question 26:

Find the median for the following frequency distribution.

 Variate 17 20 22 15 30 25 Frequency 5 9 4 3 10 6

First, we will arrange the following data in ascending order.

 Variate Frequency 15 3 17 5 20 9 22 4 25 6 30 10

Now, based on this data, we will make a cumulative frequency table

 Variate Frequency Cumulative Frequency 15 3 3 17 5 8 20 9 17 22 4 21 25 6 27 30 10 37

Here, n =37, which is odd.

#### Question 27:

The volume of a right circular cone of height 24 cm is 1232 cm3. Find its curved surface area.

Height of the cone = 24 cm

#### Question 28:

The diameter of a copper sphere is 6 cm. It is melted and drawn into a long wire of uniform circular cross-section. If the length of the wire is 36 cm, find its radius.

Diameter of the sphere = 6 cm
So, radius of the sphere = 3 cm
Volume of the sphere = $\frac{4}{3}\mathrm{\pi }{r}^{3}$

Length of the wire = 36 cm
Since, the sphere is melted and drawn into a wire, the volume of the sphere is equal to the wire.
Volume of the wire =

#### Question 29:

The curved surface area and volume of a pillar are 264 m2 and 396 m3, respectively. Find the diameter and height of the pillar.

Curved surface area of the pillar = 264 m2
Volume of  the pillar = 396 m3
Dividing the curved surface area of the pillar by its volume, we get:

Diameter = 2r = 6 m
So, diameter of the pillar is 6 m.
Curved surface area = 264 m2

Height of the pillar is 14 m.

#### Question 30:

In the adjoining figure, ABCD is a parallelogram and X, Y are the points on diagonal BD, such that DX = BY. Prove that CXAY is a parallelogram.  Given : ABCD is a parallelogram.
To prove: CXAY is a parallelogram.
Construction : Join AC, meeting BD at O.
Proof: We know that the diagonals of a parallelogram bisect each other.
So, OA = OC and OD = OB.
OD = OB...(i)
DX = BY.........(ii)
Subtracting (ii) from (i),we get:
$⇒$OD$-$DX = OB$-$BY
$⇒$OX = OY

Since, CXAY is a quadrilateral whose diagonals bisect each other, it is a parallelogram.

#### Question 31:

Construct a ABC in which BC = 5.6 cm, ∠B = 30° and the difference between the other sides is 3 cm. Steps of construction:

1. Draw a line segment BC = 5.6 cm.
2. Construct $\angle$CBX = 30$°$.
3. Set off BP = 3 cm.
4. Join PC.
5. Draw the right bisector of PC, meeting BP produced to A.
6. Join AC.
$△$ABC is thus constructed.

#### Question 32:

In the given figure, calculate the area of quadrilateral ABCD. (i) ​
ar(quadrilateral ABCD) = ar(∆ ABD) + ar(∆ BDC)
ABD and ∆ BDC are right-angled triangles. So,

BD
and AB
Now,
ar(∆ ABD) = $\frac{1}{2}$AD ⨯ AB
$\frac{1}{2}$⨯9 ⨯ 12
= 54 cm2
ar(∆ BDC) = $\frac{1}{2}$BD ⨯ BC
$\frac{1}{2}$ ⨯15 ⨯ 8
= 60 cm2

Hence, area of quadrilateral ABCD  =  54 + 60 = 114 cm2.

#### Question 33:

In the given figure, AB is a chord of a circle with centre O and AB is produced to C, such that BC = OB. Also, CO is joined and produced to meet the circle at D. If ACD = y° and ∠AOD = x°, prove that x = 3y. OB BC (Given)
So,
BOC∠BCO = y°   (Angles opposite equal sides are equal)
In $△$OBC,
OBA = BOC + BCO  (Exterior angle property)
= (
2y
Again,
So, OABOBA =(2y (Angles opposite equal sides are equal )
In $△$AOB,
AOD = ∠OAC + ∠ACO  (Exterior angle property)
= ∠OAB + ∠BCO
= (2y)° + y°
= 3y°
Hence, x = 3y.

#### Question 34:

In the given figure, AB and CD are straight lines through the centre O of a circle. If AOC = 80° and ∠CDE = 40°, find (i) ∠DCE, (ii) ∠OBC. (i)
CED = 90° (Angle in a semi-circle)
In ΔCED,
CED +EDC + DCE = 180°  (Angle sum property of a triangle)
90° + 40° + DCE = 180°
DCE = (180° – 130°) = 50° ............(i)
DCE = 50°
(ii)
AOC and BOC form a linear pair.
BOC = (180° – 80°) = 100° .............(ii)
In ΔBOC,
OBC + OCB + BOC = 180° (Angle sum property of a triangle)
∠OBC + DCE + BOC = 180°     [Since OCB = DCE]
OBC = 180° – (BOC + DCE)
= 180° – (100° + 50°)          [from (i) and (ii)]
= (180° - 150°) = 30°

#### Question 1:

Mark (✓) against the correct answer in each of the following:
The equation 2x + 3y = 6 has
(a) a unique solution
(b) only two solutions
(c) no solution
(d) infinitely many solutions

(d) infinitely many solutions

If the number of equations are less than the number of variables to be solved, then we get infinitely many solutions.
Here, we have two variables and only one equation. So, we will get infinitely many solutions.

#### Question 2:

Mark (✓) against the correct answer in each of the following:
In a ||gm ABCD, DL ​⊥ AB and BM ​⊥ AD. If AB = 10 cm, DL = 6 cm and BM = 8 cm, then AD = ?
(a) 8 cm
(b) 9 cm
(c) 7 cm
(d) 7.5 cm (d) 7.5 cm
Area of a parallelogram  = Base $×$Height
Area of parallelogram ABCD = AB $×$DL
$⇒$ AD$×$MB = AB $×$DL
$⇒$ AD$×$8 = 10$×$6
$⇒$ AD$×$8 = 60
$⇒$AD = 7.5 cm

#### Question 3:

Mark (✓) against the correct answer in each of the following:
How many bricks will be required to construct a wall 13.5 m long, 6 m high and 22.5 cm thick if each brick measures (27 cm × 12.5 cm × 9 cm)?
(a) 6000
(b) 7500
(c) 5000
(d) 3750

(a) 6000

#### Question 4:

Mark (✓) against the correct answer in each of the following:
The radius of a sphere is 21 cm. What is the surface area of the sphere?
(a) 9702 cm2
(b) 5544 cm2
(c) 12932 cm2
(d) 4312 cm2

(b) 5544 cm2

#### Question 5:

Mark (✓) against the correct answer in each of the following:
In the given figure, ABCD is rhombus in which A = 70°. Then, ∠CDB = ?
(a) 35°
(b) 20°
(c) 55°
(d) 50° (c) 55°

#### Question 6:

Mark (✓) against the correct answer in each of the following:
In the given figure, O is the centre of a circle and BAC = 40°. Then, ∠OBC = ?
(a) 40°
(b) 50°
(c) 80°
(d) 20° (b) 50°

#### Question 7:

Mark (✓) against the correct answer in each of the following:
In the given figure, ABC = 70° and ∠ACB = 30°. Then, ∠BDC = ?
(a) 70°
(b) 60°
(c) 80°
(d) 90° (c) 80°

#### Question 8:

Mark (✓) against the correct answer in each of the following:
In a medical examination of students of a class, the following blood groups were recorded.

 Blood group A AB B O No. of students 10 12 13 5
A student is selected at random. The probability that the chosen student's blood group is AB is
(a) $\frac{1}{4}$

(b) $\frac{1}{8}$

(c) $\frac{3}{10}$

(d) $\frac{13}{40}$

(c) $\frac{3}{10}$

Number of students with blood group AB = 12
Total number of students = 10 + 12 + 13 + 5 = 40

#### Question 9:

Find the area formed by the line x + 3y = 12 with the coordinate axes.

We can draw the graph of the given line by using the table below .

 x 0 3 y 4 3 Area of triangle AOB =

#### Question 10:

How many metres of cloth, 5 m wide, will be required to make a conical tent of base radius 7 m and height 24 m?

Radius of conical tent, r  = 7 m
Height of conical tent, h = 24 m

Length of cloth =

#### Question 11:

Find the curved surface area of a cylinder whose base radius is 14 m and height is 30 cm.

Radius of cylinder, r = 14 m =1400 cm
Height of cylinder, h  = 30 cm

#### Question 12:

The angles of a quadrilateral are in the ratio 2 : 4 : 5 : 7. Find these angles.

Suppose the angles of the quadrilateral are 2x, 4x, 5x and 7x.

#### Question 13:

In the given figure, O is the centre of a circle and ADC = 130°. Find ∠BAC. We know that ABCD is a cyclic quadrilateral.

#### Question 14:

In the given figure, O is the centre of a circle, such that OA = 5 cm, AB = 8 cm and OD AB, meeting AB at C. Find the length of CD. #### Question 15:

Fill in the blanks.
(i) For the line 4x + 3y = 12, x-intercept = ...... and y-intercept = ......
(ii) If x = 4, y = 3 is a solution of 2x + ky = 14, then k = ......
(iii) If the point P(p, 4) lies on the line 3x + y = 10, then p = ......

(i) For the line 4x + 3y = 12, x-intercept = 3 and y-intercept= 4.

We will draw the graph of the number line by using the values of x and y axes given below.

 x 0 3 y 4 0 So, the line intersects the x-axis at 3 and y-axis at 4.
(ii) If x = 4, y = 3 is a solution of 2x + ky = 14, then k = 2.

By putting the values of x and y we can find the value of .
$2x+ky=14\phantom{\rule{0ex}{0ex}}⇒2\left(4\right)+k\left(3\right)=14\phantom{\rule{0ex}{0ex}}⇒8+3k=14\phantom{\rule{0ex}{0ex}}⇒3k=14-8\phantom{\rule{0ex}{0ex}}⇒3k=6\phantom{\rule{0ex}{0ex}}⇒k=2$
(iii) If the point P(p, 4) lies on the line 3x + y = 10, then p = 2.

Since the point P(p,4) lies on the given line, it will satisfy the given equation of the line.
$3x+y=10\phantom{\rule{0ex}{0ex}}⇒3\left(p\right)+4=10\phantom{\rule{0ex}{0ex}}⇒3p=10-4\phantom{\rule{0ex}{0ex}}⇒3p=6\phantom{\rule{0ex}{0ex}}⇒p=2$

#### Question 16:

Which of the following statements are true?
I. If two non-parallel sides of a trapezium are equal, then it is cyclic.
II. A cyclic parallelogram is a rectangle.
III. The opposite angles of a cyclic quadrilateral area equal.
(a) I only
(b) II only
(c) I and II
(d) II and III

(c) I and II
(i) Given : ABCD is an isosceles trapezium.
To prove : ABCD is cyclic.
Construction : Draw CE parallel to DA.
Proof :
.
Hence, ABCD is a parallelogram.
So,
AD = EC  (Opposite sides of a parallelogram are equal)
and AD = BC (ABCD is an isosceles trapezium)
So, BC = CE
and

Hence, ABCD is cyclic.

(ii) Given : ABCD is a cyclic parallelogram.
To prove : ABCD is a rectangle.
Proof :
ABCD is a parallelogram.

So, $\angle$ABC = $\angle$CDA = 90$°$
Similarly, $\angle$DAB =$\angle$BCD = 90$°$
Since, all the angles of the parallelogram ABCD are 90$°$, it is a rectangle.
(iii) We know that the opposite angles of a cyclic quadrilateral are supplementary.
They are not necessarily equal.

#### Question 17:

Prove that an isosceles trapezium is always cyclic. Given : ABCD is an isosceles trapezium.
To prove : ABCD is cyclic.
Construction : Draw CE parallel to DA.
Proof :
.
Hence, ABCD is a paralleogram.
So,
AD = EC  (Opposite sides of a parallelogram are equal)
and AD = BC (ABCD is an isosceles trapezium)
So, BC = CE.
And

Also,

Hence, $\square$ABCD is cyclic.

#### Question 18:

If the diagonals of a quadrilateral bisect each other, prove that the quadrilateral is a parallelogram. Since, the opposite sides of the quadrilateral are equal, $\square$ABCD is a parallelogram.

#### Question 19:

Heights of two cones are in the ratio 1 : 3 and the radii of their bases are in the ratio 3 : 1. Show that their volumes are in the ratio 3 : 1.

Let the volume of the two cones be V1 and V2 , their radii be 3r and  and their heights be h and 3h.

#### Question 20:

How many spherical bullets can be made out of a solid cube of lead whose edge measures 44 cm, each bullet being 4 cm in diameter?

Side of cube  = 44 cm

#### Question 21:

The mean of 40 numbers was found to be 38. Later on, it was detected that the number 56 was misread as 36. Find the correct mean of the given numbers.

Given mean of 40 numbers = 38
Sum of 40 numbers = $38×40=1520$
Correct sum of 40 numbers = $1520-36+56$ =1540
So, correct mean of 40 numbers = $\frac{1540}{40}=38.5$

#### Question 22:

In the given figure, PQ is the diameter of a circle with centre O. If PQR = 65°, ∠SPR = 40° and ∠PQM = 50°, find (i) ∠QPR (ii) ∠QPM and (iii) ∠PRS. #### Question 23:

In a quadrilateral ABCD, M is the midpoint of AC. Prove that . Since, M is the mid-point of AC, AM is the median of $△$ADC.
We know that a median divides a triangle in two triangles of equal area.
Hence,ar($△$AMD) = ar($△$CMD) ......(i)
Also, BM is the median of $△$ABC.
Hence,ar($△$AMB) = ar($△$CMB).....(ii)
By adding equations (i) and (ii), we get:
ar($△$AMD)+ar($△$AMB) = ar($△$CMD) +ar($△$CMB)

#### Question 24:

The curved surface area of a cylinder is 4400 cm2 and the circumference of the base is 110 cm. Find the volume of the cylinder.

#### Question 25:

Calculate the mode for the following data:

 Marks 10 15 20 25 30 35 40 No. of students 8 12 36 35 28 18 9

 Marks (xi) No. of students (fi) Cumulative frequency (xi )(fi) 10 8 8 80 15 12 20 180 20 36 56 720 25 35 91 875 30 28 119 840 35 18 137 630 40 9 146 360 N = 146 3685

Mode = 3(Median) $-$2(Mean)
or, Mode  = 3(25) $-$ 2(25.23)
= 75$-$50.46
= 24.54

#### Question 26:

Prove that the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Given : A circle C(O,r) in which arc AB subtends $\angle$AOB at the centre and $\angle$ACB at any point C on the remaining part of the circle.   To prove: $\angle$AOB = 2$\angle$ACB when AB is a minor arc and or a semi-circle.
Reflex $\angle$AOB = $\angle$ACB when AB is a major arc.
Construction: Join AB and CO and produce CO to a point D outside the circle.
Proof: There are three cases:
Case-I  AB is a minor arc (Figure -a)
Case-II  AB is a semicircle (Figure -b)
Case-III  AB is a major arc (Figure -c)
Now, in $△$AOC,
OA = OC (Radii of the circle)
$\angle$1 = $\angle$3  (Angles opposite equal sides)
$\angle$5 = $\angle$1 + $\angle$3 (Exterior angle property)
$⇒$$\angle$5 = $\angle$1 + $\angle$1
$⇒$$\angle$5 = 2$\angle$1...................(i)
Now, in $△$BOC,
OB = OC (Radii of the circle)
$\angle$2 = $\angle$4  (Angles opposite equal sides)
$\angle$6 = $\angle$2 + $\angle$4   (Exterior angle property)
$⇒$$\angle$6 = $\angle$2 + $\angle$2
$⇒$$\angle$6 = 2$\angle$2...................(ii)
Adding equations (i) and (ii), we get:
In figures (a) and (b),
$\angle$5 + $\angle$6 = 2$\angle$1 + 2$\angle$2
$⇒$$\angle$5 + $\angle$6 = 2($\angle$1 + $\angle$2)
$⇒$$\angle$AOB = 2$\angle$ACB
In figure (c),
$\angle$5 + $\angle$6 = 2($\angle$1 + $\angle$2)
$⇒$Reflex $\angle$AOB = 2$\angle$ACB

#### Question 27:

The mean of 100 items was found to be 64. Later on, it was discovered that two items were misread as 26 and 9 instead of 36 and 90, respectively. Find the correct mean.

Given mean of 100 items = 64
Sum of 100 items = $64×100=6400$
Correct sum of 100 items  = $6400-26-9+36+90$ =6491
So, correct mean of 100 items = $\frac{6491}{100}=64.91$

#### Question 28:

Construct a ABC whose perimeter is 14 cm and the sides are in the ratio 2 : 3 : 4.

Steps of construction:
1. Draw a line segment XY = 14 cm.
2. Construct an acute angle on XY and drawn in the downward direction
3. Starting from  X , mark (2+3+4) = 9 equal distances along XZ.
4. Mark points L,M and N, such that XL = 2 units, LM = 3 units and MN = 4 units.
5. Join YN.
6. Through L and M, draw LB$\parallel$YN and MC$\parallel$YN, cutting XY at B and C, respectively.
7. With B as centre and radius XB, draw an arc.
8. With C as centre and radius CY, draw an arc, cutting the previously drawn arcs at point A.
9. Join AB  and AC.
$△$ABC, the required triangle, is thus drawn. #### Question 29:

Prove that there is one and only one circle passing through three given points.

Given: Three non-collinear points P, Q and R.
To prove: There is one and only one circle passing through the points P, Q and R.
Construction: Join PQ and QR.
Draw perpendicular bisectors AB of PQ and CD of QR.
Let the perpendicular bisectors intersect at the point O.
Now join OP, OQ and OR. A circle is obtained passing through the points P, Q and R. Proof:
We know that each and every point on the perpendicular bisector of a line segment is equidistant from its ends points. Thus,
OP = OQ           (Since O lies on the perpendicular bisector of PQ)
OQ = OR           (Since O lies on the perpendicular bisector of QR)
OP = OQ = OR
Let OP = OQ = OR = r      (Radius of a circle)
Now, draw a circle C (O, r) with O as centre and r as radius.
Then, circle C(O, r) passes through the points P, Q and R.

We have to show that this circle is the only circle passing through the points P, Q and R.
If possible, suppose there is another circle C(O′, t) which passes through the points P, Q and R.
Then, O will lie on the perpendicular bisectors AB and CD.
But O was the intersection point of the perpendicular bisectors AB and CD.
O must coincide with the point O.  (Since, two lines can not intersect at more than one point)
As, O′P = t , OP = r  and O coincides with O,
t = r
C(O, r) and C(O', t) are congruent.
Hence, there is one and only one circle passing through the given non-collinear points.

#### Question 30:

In the adjoining figure, O is the centre of a circle. Chord CD is parallel to diameter AB. If ABC = 25°, calculate ∠CED. and
$\angle CED=\frac{1}{2}\angle COD\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×80°\phantom{\rule{0ex}{0ex}}=40°$

#### Question 31:

If O is the centre of the given circle, such that AOC = 110° and side AB is produced to D, find ∠CBD. #### Question 32:

In the given figure, O is the centre of a circle and BOD = 150°. Find the values of x and y. #### Question 33:

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of their volumes and the ratio of their curved surfaces.

Let there be two cylinder whose volumes are V1 and V2 and curved
surfaces are S1and S2. Let their radii be 2r and 3r and their heights be 5h and 3h.

#### Question 34:

How many spherical bullets can be made out of a solid cube of lead whose edge measures 44 cm, each bullet being 4 cm in diameter?