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Question 1 ( 1.0 marks)
Write whether the rational number will have a terminating decimal expansion or a non-terminating repeating decimal expansion.
Solution:
The given rational number i.e., can be written as

This is of the form , where m = 2 and n = 3 are non-negative integers.
Thus, has a terminating decimal expansion.
Question 2 ( 1.0 marks)
Write the polynomial, the product and sum of whose zeroes are and respectively.
Solution:
Let α and β be the zeroes of the required polynomial.
It is given that and .
Therefore, the required polynomial is given by

For k = 2,
Thus, one of the polynomials which satisfy the given condition is .
Question 3 ( 1.0 marks)
Write whether the following pair of linear equations is consistent or not:
x + y = 14
x − y = 4
Solution:
The given pair of linear equations can be written as
x + y − 14 = 0
x − y − 4 = 0
Here, a1 = 1, b1 = 1, c1 = −14
a2 = 1, b2 = −1, c2 = −4
Now,

Thus, the given pair of linear equations is consistent.
Question 4 ( 1.0 marks)
Write the nature of roots of quadratic equation
Solution:
The given quadratic equation is .
This equation is of the form , where .

Thus, the given quadratic equation has two equal real roots.
Question 5 ( 1.0 marks)
For what value of k, are the numbers x, 2x + k and 3x + 6 three consecutive terms of an A.P.
Solution:
If a, b, and c are the three consecutive terms of an A.P., then we have 2b = a + c.
Now, will be the three consecutive terms of an A.P. if

Thus, the required value of k is 3.
Question 6 ( 1.0 marks)
In a ΔABC, DE||BC. IF DE = BC and area of ΔABC = 81 cm2, find the area of ΔADE.

Solution:
In ΔADE and ΔABC, we have
∠ADE = ∠ABC [Corresponding Angles]
∠DAE = ∠BAC [Common]
∴ΔADE ∼ ΔABC [By AA similarity criterion]
We know that the areas of two similar triangles are in the ratio of the squares of their corresponding sides.

It is given that and area of ΔABC = 81 cm2.


Thus, the area of ΔADE is 36 cm2.
Question 7 ( 1.0 marks)
If sec A = and A + B = 90°, find the value of cosec B.
Solution:
A + B = 90º
⇒ A = 90º − B
∴ sec A = sec (90º − B)
⇒ sec A = cosec B [sec(90º − θ) = cosec θ]
[sec A = ]
Thus, the value of cosec B is .
Question 8 ( 1.0 marks)
If the mid-point of the line segment joining the points P (6, b − 2) and Q (−2, 4) is (2, −3), find the value of b.
Solution:

The mid-point of the line segment joining the points P (6, b − 2) and Q (−2, 4) is (2, −3).
By mid-point formula, we obtain

Thus, the value of b is −8.
Question 9 ( 1.0 marks)
The length of the minute hand of a wall clock is 7 cm. How much area does it sweep in 20 minutes?
Solution:
Angle swept by the minute hand in 60 minutes = 360º
Angle swept by the minute hand in 20 minutes =
∴ Area swept by the minute hand in 20 minutes =

Question 10 ( 1.0 marks)
What is the lower limit of the modal class of the following frequency distribution?
Age in (years) 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60
Number of patients 16 13 6 11 27 18
Solution:
Modal class is defined as the class interval whose frequency is the maximum. In the given data, the class interval 40−50 has the maximum frequency. Therefore, this interval is the modal class.
Thus, the lower limit of the modal class of the given frequency distribution is 40.
Section B

Question number 11 to 15 carry 2 marks each.

Question 11 ( 2.0 marks)
Without drawing the graph, find out whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincident:

Solution:
A pair of linear equations in two variables i.e., a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 represents
(a) Intersecting lines if
(b) Coincident lines if
(c) Parallel lines if
The given pair of linear equations can be written as
9x − 10y − 21 = 0 and
Here, a1 = 9, b1 = −10, c1 = −21 and a2 = , b2 = , c2 =
∴

Thus, the given pair of linear equations represents coincident lines.
Question 12 ( 2.0 marks)
The 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.
Solution:
The nth term (an) of an A.P. is given by an = a + (n − 1)d, where a is the first term and d is the common difference.
∴a17 = a + (17 − 1)d = a + 16d
a10 = a + (10 − 1)d = a + 9d
According to the given information, a17 − a10 = 7.
∴ a + 16d − (a + 9d) = 7
⇒ 7d = 7
∴ d = 1
Thus, the required common difference of the A.P. is 1.
Question 13 ( 2.0 marks)
Without using trigonometric tables, evaluate:

Solution:
The given expression is .
We know that .

Thus, the value of the given expression is 5.
Question 14 ( 2.0 marks)
Show that the points (−2, 5); (3, −4) and (7, 10) are the vertices of a right angled isosceles triangle.
OR
The centre of a circle is (2α − 1, 7) and it passes through the point (−3, −1). If the diameter of the circle is 20 units, then find the values(s) of α.
Solution:
The given points are A (−2, 5), B (3, −4) and C (7, 10).
By distance formula, we obtain

Therefore, ΔABC is isosceles, where AB = CA.

Now,
Hence, by Pythagoras Theorem, ΔABC is right-angled at A.
Thus, the given points are the vertices of a right-angled isosceles triangle.
OR
The centre of the given circle is O (2α − 1, 7). This circle passes through the point P (−3, −1).

Now, by distance formula, we obtain the radius as

The diameter of the circle is given as 20 units.
∴2 OP = 20 units

If 2 + 2α = 6, then 2α = 4 ⇒ α = 2
If 2 + 2α = −6, then 2α = −8 ⇒ α = −4
Thus, the value of αis 2 or −4.
Question 15 ( 2.0 marks)
If C is a point lying on the line segment AB joining A (1, 1) and B (2, −3) such that 3 AC = CB, then find the coordinates of C.
Solution:
Let the coordinates of point C be (x, y).
It is given that C is a point lying on line segment AB, which is joining A (1, 1) and (2, −3) such that 3AC = CB.
⇒ AC: CB = 1:3
This means that C divides AB internally in the ratio 1: 3.
Now, by section formula, the coordinates of point C are given by

Thus, the coordinates of point C are .
Section C

Question number 16 to 25 carry 3 marks each.

Question 16 ( 3.0 marks)
Show that the square of any positive odd integer is of the form 8m + 1, for some integer m.
OR
Prove that is not a rational number.
Solution:
Let a be any positive integer and let b = 8.
Then, a = 8n + r where 0 ≤ r < 8 and n is an integer.
∴a = 8n or 8n + 1 or 8n + 2 or 8n + 3 or 8n + 4 or 8n + 5 or 8n + 6 or 8n + 7
So, every odd positive integer is of the form 8n + 1 or 8n + 3 or 8n + 5 or 8n + 7.
Now, consider these four cases.
Case 1: When a = 8n + 1
, where m = is an integer.
Case 2: When a = 8n + 3
, where m = is an integer.
Case 3: When a = 8n + 5
where m = is an integer.
Case 4: When a = 8n + 7
where m = is an integer.
Thus, it can be seen that square of any positive odd integer is of the form 8m + 1 for some integer m.
OR
If possible, suppose to be rational.
Therefore, we can find two integers a and b (b ≠ 0) such that

Since a and b are rational, should also be rational and thus, is rational as well. This contradicts the fact that is irrational. Therefore, our assumption is false and hence, is not a rational number.
Question 17 ( 3.0 marks)
If the polynomial 6x4 + 8x3 − 5x2 + ax + b is exactly divisible by the polynomial 2x2 − 5, the find the values of a and b.
Solution:
It is given that the polynomial p(x) = 6x4 + 8x3 − 5x2 + ax + b is exactly divisible by the polynomial q(x) = 2x2 − 5.
Now, p(x) can be divided by q(x) as follows:

Since p(x) is exactly divisible by q(x), the remainder should be zero.
∴ (a + 20)x + (b + 25) = 0
Comparing the coefficients of the term containing x and the constant term on both sides, we have
a + 20 = 0 ⇒ a = −20
b + 25 = 0 ⇒ b = −25
Thus, the values of a and b are −20 and −25 respectively.
Question 18 ( 3.0 marks)
If 9th term of an A.P. is zero, prove that its 29th term is double of its 19th term.
Solution:
It is given that the ninth term of an A.P. is zero.
Let a and d respectively be the first term and the common difference of the A.P.
The nth term of an A.P. is given by an = a + (n − 1)d
∴a9 = a + (9 − 1)d = a + 8d
a29 = a + (29 − 1)d = a + 28d
a19 = a + (19 − 1)d = a + 18d
Now, it is given that the ninth term is zero.
∴ a + 8d = 0
⇒ a = −8d
Now, a29 = a + 28d = −8d + 28d = 20d
a19 = a + 18d = −8d + 18d = 10d
∴ a29 = 20d = 2 × 10d = 2 × a19
Thus, the 29th term of the A.P. is double the 19th term.
Question 19 ( 3.0 marks)
Draw a circle of radius 3 cm. From a point P, 6 cm away from its centre, construct a pair of tangents to the circle. Measure the lengths of the tangents.
Solution:
Steps of construction:
i. First draw a circle with centre O and radius 3 cm. Take a point P such that OP = 6 cm and then join OP.
ii. Draw the perpendicular bisector of OP such that it intersects OP at M.
iii. With M as centre and OM as radius, draw a circle. Let it intersect the previously drawn circle at points A and B.
iv. Joint PA and PB. PA and PB are the required pair of tangents.

The length of the tangents PA and PB is 5.20 cm.
Question 20 ( 3.0 marks)
In figure 1, two triangles ABC and DBC lie on the same side of base BC. P is a point on BC such that PQ || BA and PR || BD. Prove that QR || AD.

Solution:

In ΔABC, PQ || BA, by basic proportionality theorem, we obtain
... (1)
In ΔBDC, PR || BD, by basic proportionality theorem, we obtain
... (2)
On comparing equations (1) and (2), we obtain

In ΔACD,
Therefore, by converse of basic proportionality theorem, it can be said that QR || AD.
Question 21 ( 3.0 marks)
In figure 2, a triangle ABC is right angled at B. Side BC is trisected at points D and E. Prove that 8 AE2 + 5 AD2.

OR
In figure 3, a circle is inscribed in a triangle ABC having side BC = 8 cm, AC = 10 cm and AB = 12 cm. Find AD, BE and CF.

Solution:

It is given that side BC is trisected at D and E.

In ΔABD, by Pythagoras theorem, we get

In ΔABE, by Pythagoras theorem, we get

In ΔABC, by Pythagoras theorem, we get

Now, 3AC2 + 5AD2

Thus, the given result is proved.
OR
Let AD = x, BE = y and CF = z.
The tangents drawn from an external point to a circle are equal.
∴AF = AD = x
BD = BE = y
CE = CF = z
∴AB = AD + BD = x + y = 12 cm … (1)
BC = BE + EC = y + z = 8 cm … (2)
AC = AF + FC = x + z = 10 cm … (3)
Now, AB + BC + CA = 12 cm + 8 cm + 10 cm = 30 cm
∴2 (x + y + z) = 30 cm
⇒ x + y + z = 15 cm … (4)
Subtracting equation (1) from equation (4), we get:
x + y + z − (x + y) = (15 − 12) cm
⇒ z = 3 cm
Similarly, subtracting equation (2) from equation (4), we get x = 7 cm.
Similarly, subtracting equation (3) from equation (4), we get y = 5 cm.
Thus, the lengths of AD, BE and CF are respectively 7 cm, 5 cm, and 3 cm.
Question 22 ( 3.0 marks)
Prove that

Solution:


Question 23 ( 3.0 marks)
Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
Solution:
It is given that the points (x, y), (1, 2) and (7, 0) are collinear. Thus, the area of the triangle formed with these points as its vertices will be zero.
We know that the area of triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is given by
Thus, we have
Area of triangle =

Now, the area of this triangle is zero.
∴ x + 3y − 7 = 0
Thus, x + 3y − 7 = 0 is the required relation between x and y for the given points to be collinear.
Question 24 ( 3.0 marks)
In figure 4, the shape of the top of a table in a restaurant is that of a sector of a circle with centre O and ∠BOD = 90°. If BO = OD = 60 cm, find
(i) the area of the top of the table.
(ii) the perimeter of the table top.
(Take π = 3.14)

OR
In figure 5, ABCD is a square of side 14 cm and APD and BPC are semicircles. Find the area of shaded region. (Take )

Solution:
(i) ∠BOD = 90º
⇒ Reflex ∠BOD = 360º − 90º = 270º
Now, the area of a sector of angle θ and radius r is given by
∴Area of the table top =

Thus, the area of the table top is 8478 cm2.
(ii) Perimeter of the table top = length of major arc BD + OB + OD
= × 2πr + OB + OD
=
= 402.6 cm
OR

From the figure it can be seen that
Area of shaded region = area of square ABCD − area of semi circles APD and BPC
Area of square ABCD = (14 cm)2 = 196 cm2
The diameter of each semicircle is equal to the side of square (14 cm).
∴ Radius of each semi-circle = 7 cm
∴Area of semicircles APD and BPC =
∴ Area of the shaded region = 196 cm2 − 154 cm2 = 42 cm2
Thus, the area of the shaded region is 42 cm2.
Question 25 ( 3.0 marks)
A box has cards numbered 14 to 99. Cards are mixed thoroughly and a card is drawn from the bag at random. Find the probability that the number on the card, drawn from the box is
(i) an odd number,
(ii) a perfect square number,
(iii) a number divisible by 7.
Solution:
It is given that the box has cards numbered from 14 to 99.
Thus, number of cards in the box = (99 − 14) + 1 = 86
The probability of an event is given by
i. From the numbers 14 to 99, there are 43 odd numbers.
Thus, the probability of getting an odd number is .
ii. From the numbers 14 to 99, there are 6 perfect squares, namely 16, 25, 36, 49, 64, and 81.
Thus, the probability of getting a perfect square is .
iii. From the numbers 14 to 99, there are 13 numbers that are divisible by 7 (14, 21, 28 … 98).
Thus, probability of getting a number divisible by 7 is .
Section D

Question number 26 to 30 carry 6 marks each.

Question 26 ( 6.0 marks)
A trader bought a number of articles for Rs 900. Five articles were found damaged. He sold each of the remaining articles at Rs. 2 more than what he paid for it. He got a profit of Rs. 80 on the whole transaction. Find the number of articles he bought.
OR
Two years ago the man’s age was three times the square of his son’s age. Three years hence his age will be four times his son’s age. Find their present ages.
Solution:
Let the number of articles be x.
It is given that the total cost of articles is Rs 900.
∴Cost of one article =
Five articles were found damaged. Therefore, the number of remaining articles is
(x − 5).
He sold the remaining articles at Rs 2 more than i.e., the selling price of each remaining article is .
It is given that the trader got a profit of Rs 80.
Therefore, according to the given information

Since the number of articles cannot be negative, x = 75.
Thus, the trader had bought 75 articles.
OR
Let the present age of the son be x.
Two years ago, son’s age = x − 2
⇒ Two years ago, man’s age = 3(x − 2)2 … (1)
After 3 years, son’s age = x +3
After 3 years, man’s age = 4(x +3) = 4x +12
⇒ Man’s present age = 4x +12 − 3 = 4x + 9 … (2)
Man’s age two years ago = 4x + 9 − 2 = 4x + 7
Using this in equation (1), we obtain
4x + 7 = 3(x − 2)2
⇒ 4x + 7 = 3(x2 + 4 − 4x)
⇒ 4x + 7 = 3x2 + 12 − 12x
⇒ 3x2 − 16x + 5 = 0
⇒ 3x2 − 15x − x + 5 = 0
⇒ 3x(x − 5) − 1(x − 5) = 0
⇒ (3x − 1) (x − 5) = 0
⇒ x = or 5
Since age cannot be a fraction, x = 5.
On substituting this in equation (2), we obtain
4x + 9 = 4(5) + 9 = 29
Thus, the present ages of the man and his son are 29 years and 5 years respectively.
Question 27 ( 6.0 marks)
Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
Using the above theorem prove the following:
The area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal.
Solution:
Let us consider two similar triangles ABC and PQR.
We have to prove .
To do this, let us draw AP ⊥ BC and DQ ⊥ EF.

We know that the area of a triangle is given by
× Base × Altitude
∴ ar (ΔABC) = × BC × AP … (1)
∴ ar (ΔDEF) = × EF × DQ … (2)
On dividing equation (1) by (2), we obtain

⇒ … (3)
Now, in ΔAPB and ΔDQE
∠APB = ∠DQE (90° each)
∠B = ∠E ( ΔABC ∼ ΔDEF)
∴ ΔAPB ∼ ΔDQE (AA similarity criterion)
⇒ … (4)
It is also given that ΔABC ∼ ΔDEF.
∴ … (5)
From equations (4) and (5), we obtain
… (6)
From equations (3) and (6), we obtain

⇒
On using equation (6), we obtain


Let ABCD be a square of side a.
Therefore, diagonal of the square
The two required equilateral triangles are formed as ΔABE and ΔDBF.
Side of equilateral triangle ABE, described on one of its sides = a
Side of equilateral triangle DBF, described on one of its diagonals
We know that all the angles of an equilateral triangle are equal to 60º. We also know that all the sides of an equilateral triangle are equal.
Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.
Question 28 ( 6.0 marks)
The angle of elevation of the top of a building from the foot of a tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Solution:

Let AB be the building and CD be the tower.
In ΔCDB,

In ΔABD,

Thus, the height of the building is .
Question 29 ( 6.0 marks)
A spherical copper shell, of external diameter 18 cm, is melted and recast into a solid cone of base radius 14 cm and height cm. Find the inner diameter of the shell.
OR
A bucket is in the form of a frustum of a cone with a capacity of 12308.8 cm3. The radii of the top and bottom circular ends of the bucket are 20 cm and 12 cm respectively. Find the height of the bucket and also the area of metal sheet used in making it.
Solution:
Let r and R be the inner and the external radii of the spherical copper shell respectively.
It is given that the external diameter of the shell is 18 cm.
∴2R = 18 cm ⇒ R = 9 cm
Volume of the spherical shell =

Since the shell is recast into the solid cone, volume of the shell = volume of the cone.
Radius of cone = 14 cm [Given]
Height of cone =
Volume of cone
Therefore, we now have

Thus, the inner diameter of the spherical copper shell is 2r = 2 × 8 cm = 16 cm.
OR
Let r1 and r2 be the radii of the top and the bottom circular ends of the bucket respectively.
Let h be the height of the bucket.
Accordingly, r1 = 20 cm and r2 = 12 cm
Since the capacity of the bucket is 12308.8 cm3, the volume of the bucket is also 12308.8 cm3.

Thus, the height of the bucket is 15 cm.
The area of the metal sheet used in making the bucket
= CSA of the bucket + area of the bottom circular end of the bucket


Thus, area of the metal sheet used in making the bucket

Question 30 ( 6.0 marks)
Find the mode, median and mean for the following data:
Marks obtained 25 − 35 35 − 45 45 − 55 55 − 65 65 − 75 75 − 85
Number of students 7 31 33 17 11 1
Solution:
(A). Calculation of mode
Marks obtained 25−35 35−45 45−55 55−65 65−75 75−85
Number of students 7 31 33 17 11 1
From the data given above, it can be observed that the maximum class frequency is 33 and the class corresponding to this frequency is 45−55.
Thus, the modal class is 45−55.
Lower class limit (l) of modal class = 45
Frequency (f1) of modal class = 33
Frequency (f0) of class preceding the modal class = 31
Frequency (f2) of class succeeding the modal class = 17
Class size (h) = 10


(B). Calculation of median
To find the median of the given data, cumulative frequency is calculated as follows.
Marks obtained Number of students Cumulative frequency
25 − 35 7 7
35 − 45 31 38
45 − 55 33 71
55 − 65 17 88
65 − 75 11 99
75 − 85 1 100
From the table, we obtain
n = 100
Cumulative frequency (cf) just greater than is 71
This belongs to interval 45−55. Thus, the median class is 45−55.
Lower limit (l) of median class = 45
Class size (h) = 10
Frequency (f) of median class = 33
Cumulative frequency (cf) of class preceding median class = 38

(C). Calculation of mean
The class marks for each interval can be calculated by using the formula

Taking 60 as assured mean (a), di, ui, and fiui can be calculated as follows:
Marks obtained Number of students (fi) xi di = xi − 60 , where h = 10 fiui
25¬−35 7 30 − 30 − 3 − 21
35−45 31 40 − 20 − 2 − 62
45−55 33 50 − 10 − 1 − 33
55−65 17 60 0 0 0
65−75 11 70 10 1 11
75−85 1 80 20 2 2
Total 100 − 103

Thus, the mode, median and mean of the given data are 46.11, 48.63 and 49.7 respectively
Question 1 ( 1.0 marks)
Find the [HCF × LCM] for the numbers 105 and 120.
Solution:
For two numbers a and b, HCF × LCM = a × b
∴ For the given numbers 105 and 120:
HCF × LCM = 105 × 120
HCF × LCM = 12600
Question 2 ( 1.0 marks)
Find the number of solutions of the following pair of linear equations:
x + 2y − 8 = 0
2x + 4y = 16
Solution:
The given pair of linear equations is
x + 2y − 8 = 0
2x + 4y − 16 = 0
On comparing with general equations

Hence, the given pair of linear equations has infinitely many solutions.
Question 3 ( 1.0 marks)
If , a, and 2 are three consecutive terms of an A.P., then find the value of a.
Solution:
If three terms a, b, and c are in A.P., then we have b − a = c − b
⇒ 2b = a + c
∴ If are three consecutive terms of an A.P., then

Thus, the value of a is .
Question 4 ( 1.0 marks)
Two coins are tossed simultaneously. Find the probability of getting exactly one head.
Solution:
If two coins are tossed simultaneously, then the possible outcomes are
S = {HT, TH, TT, HH}
Thus, the total number of possible outcomes is 4.
Out of all the four outcomes, {HT} and {TH} are cases of exactly one head.
∴ Required probability =
Question 5 ( 1.0 marks)
In figure 1, ΔABC is circumscribing a circle. Find the length of BC.

Solution:
BR = BP [Tangents drawn to a circle from a point outside the
circle are equal]
However, BR = 3 cm
∴BP = 3 cm … (1)
AR = AQ [Tangents drawn to a circle from a point outside the
circle are equal]
However, AR = 4 cm
∴AQ = 4 cm
AQ + QC = AC
⇒ QC = AC − AQ
Using the values AQ = 4 cm and AC = 11 cm,
QC = 11 cm − 4 cm
⇒ QC = 7 cm
CP = CQ [Tangents drawn to a circle from a point outside the
circle are equal]
∴CP = 7 cm … (2)
BC = BP + CP
On using equations (1) and (2), we obtain
BC = 3 cm + 7 cm
BC = 10 cm
Thus, the length of BC is 10 cm.
Question 6 ( 1.0 marks)
If the diameter of a semicircular protractor is 14 cm, then find its perimeter.
Solution:

Diameter = 14 cm
Radius =
Length of the semicircular part = πr
Total perimeter = Length of semicircular part + Diameter
= 22 cm + 14 cm
= 36 cm
Thus, the perimeter of the protractor is 36 cm.
Question 7 ( 1.0 marks)
If 1 is a zero of the polynomial p(x) = ax2 − 3(a − 1) x − 1, then find the value of a.
Solution:
If 1 is a zero of polynomial p(x), then p(1) = 0.
∴ p(1) = a(1)2 − 3(a − 1) (1) − 1 = 0
⇒ a − 3a + 3 − 1 = 0
⇒ −2a = −2
⇒ a = 1
Thus, the value of a is 1.
Question 8 ( 1.0 marks)
In ΔLMN, ∠L = 50° and ∠N = 60°. If ΔLMN ∼ ΔPQR, then find ∠Q.
Solution:

∠L + ∠M + ∠N = 180° (Angle sum property)
Substituting ∠L = 50° and ∠N = 60° in this equation:
50° + ∠M + 60° = 180°
∠M = 70°
It is given that ΔLMN ∼ ΔPQR.
We know that corresponding angles in similar triangles are of equal measures.
∴ ∠M = ∠Q = 70°
Thus, the measure of ∠Q is 70°.
Question 9 ( 1.0 marks)
If sec2θ (1 + sin θ) (1 − sin θ) = k, then find the value of k.
Solution:
sec2θ (1 + sinθ)(1 − sinθ) = k

Thus, the value of k is 1.
Question 10 ( 1.0 marks)
Find the discriminant of the quadratic equation
Solution:
For the quadratic equation ax2 + bx + c = 0,
Discriminant = D = b2 − 4ac
Hence, for the given equation, D =
= 100 − 36
= 64
Thus, the discriminant of the given equation is 64.
Section B

Question number 11 to 15 carry 2 marks each.

Question 11 ( 2.0 marks)
Find all the zeroes of the polynomial 2x3 + x2 − 6x − 3, if two of its zeroes are and .
Solution:
The given polynomial is .
It is given that and are two zeroes of p(x).
Thus, and are the factors of p(x).
This means, is also a factor of p(x).
We can divide by x2 − 3 as

∴p(x) = (x2 − 3) (2x + 1)
As (2x + 1) is a factor of the polynomial p(x), x = is a zero of the polynomial.
Thus, is the third zero of the given polynomial.
Thus, the three zeroes of are , , and .
Question 12 ( 2.0 marks)
If , then evaluate
OR
Find the value of tan 60°, geometrically.
Solution:
The given expression is .

Thus, the value of the given expression is .
OR
Consider an equilateral ΔABC.
Then, ∠ABC = ∠BCA = ∠CAB = 60° and AB = BC = CA = 2a (say)
Now, draw AD⊥ BC.

Comparing ΔABD and ΔACD, we have:
∠ADB = ∠ADC = 90° [By construction]
AB = AC [Sides of equilateral triangle]
AD = AD [Common]
∴ΔABD ≅ ΔACD [By RHS congruency axiom]
⇒ BD = DC [CPCT]
∴
On applying Pythagoras’ Theorem in ΔABD, we get:

∴tan 60º = tan B
Thus, the value of tan 60º is .
Question 13 ( 2.0 marks)
If the points A (4, 3) and B (x, 5) are on the circle with the centre O (2, 3), find the value of x.
Solution:
If the points A (4, 3) and B (x, 5) are on the circle with the centre O (2, 3), then OA and OB will be the radii of the circle.
∴OA = OB
⇒ OA2 = OB2
⇒ (4 − 2)2 + (3 − 3)2 = (x − 2)2 + (5 − 3)2 [Using distance formula]
⇒ 22 + 02 = (x − 2)2 + 22
⇒ 0 = (x − 2)2
⇒ x − 2 = 0
⇒ x = 2
Thus, the required value of x is 2.
Question 14 ( 2.0 marks)
Which term of the A.P. 3, 15, 27, 39… will be 120 more than its 21st term?
Solution:
First term, a = 3
Common difference, d = 15 − 3 = 12
The nth term of an A.P. is given by an= a + (n − 1) d.
∴ 21st term of the A.P. (a21) = 3 + (21 − 1) 12
⇒ a21 = 3 + (20 × 12) = 3 + 240 = 243
Let the rth term of the A.P. be 120 more than the 21st term of the A.P.
∴ ar = a21 + 120
⇒ ar = 243 + 120
⇒ ar = 363
Now, ar = a + (r −1) d
⇒ 363 = 3 + (r − 1) 12
⇒ 360 = (r − 1) 12
⇒ r − 1 = 30
⇒ r = 31
Thus, the 31st term of the given A.P. is 120 more than the 21st term of the A.P.
Question 15 ( 2.0 marks)
In Figure 2, ΔABD is a right triangle, right-angled at A and AC ⊥ BD. Prove that AB2 = BC . BD.

Solution:
In ΔACB and ΔDAB, we have
∠ACB = ∠DAB [Each is 90º]
∠ABC = ∠DBA [Common angle]
∴ΔACB ∼ ΔDAB [By AA similarity criterion]
It is known that if two triangles are similar, then the corresponding sides are proportional.

Hence, proved.
Section C

Question number 16 to 25 carry 3 marks each.

Question 16 ( 3.0 marks)
Prove that is an irrational number.
Solution:
If possible, suppose is rational.
Therefore, we can find two integers a, b (b ≠ 0) such that

Since a, b, 2, and 5 are integers, is a rational number.
Hence, should be rational.
This conclusion contradicts the fact that is irrational.
Therefore, our assumption is false.
Hence, is irrational.
Question 17 ( 3.0 marks)
In Figure, 3, AD ⊥ BC and BD CD. Prove that 2CA2 = 2AB2 + BC2.

OR
In Figure 4, M is mid-point of side CD of a parallelogram ABCD. The line BM is drawn intersecting AC at L and AD produced at E. Prove that EL = 2 BL.

Solution:

On applying Pythagoras theorem to ΔABD, we obtain
AB2 = BD2 + AD2
⇒ AD2 = AB2 − BD2 … (1)
On applying Pythagoras theorem to ΔACD, we obtain
AC2 = AD2 + DC2
⇒ AD2 = AC2 − DC2 … (2)
On comparing equations (1) and (2), we obtain
AB2 − BD2 = AC2 − DC2
⇒ AC2 − AB2 = DC2 − BD2 … (3)
It is given that BD CD
⇒ CD = 3BD
Now, BD + DC = BC
⇒ BD + 3BD = BC
⇒ BD BC … (4)
Also, CD = 3BD
⇒ DC BC … (5) [From (4)]
On substituting the values of BD and DC from (4) and (5) in (3), we obtain
AC2 − AB2

∴ 2(AC2 − AB2) = BC2
⇒ 2AC2 − 2AB2 = BC2
⇒ 2CA2 = 2AB2 + BC2
Hence, proved.
OR

Comparing ΔBCM and ΔEDM:
CM = MD [M is the mid-point of CD]
∠BMC = ∠DME [Vertically opposite angels]
∠BCM = ∠EDM [Alternate interior angles as AE||BC]
∴ ΔBCM ≅ ΔEDM [ASA congruence criterion]
⇒ BC = ED [C.P.C.T.]
⇒ BC = AD = ED [Opposite sides of parallelogram are equal]
Comparing ΔBLC and ΔELA:
∠LBC = ∠AEL [Alternate interior angles as AE||BC]
∠BLC = ∠ELA [Vertically opposite angels]
∴ Δ BLC ∼ ΔELA [AA similarity criterion]
It is known that the corresponding sides of similar triangles are proportional.

Hence, proved.
Question 18 ( 3.0 marks)
The area of an equilateral triangle is cm2. Taking each angular point as centre, circles are drawn with radius equal to half the length of the side of the triangle. Find the area of triangle not included in the circles. [Take = 1.73]
OR
Figure 5 shows a decorative block which is made of two solids − a cube and a hemisphere. The base of the block is a cube with edge 5 cm and the hemisphere, fixed on the top, has a diameter of 4.2 cm. Find the total surface area of the block. [Take π = ]

Solution:

Let ABC be the given equilateral triangle whose area is cm2.
Area of equilateral triangle
∴ cm2
⇒ (Side)2 cm2
⇒ Side = 2 × 7 cm = 14 cm
Thus, radius of each circle cm = 7 cm
We know, ∠A = ∠B = ∠C = 60° [ ΔABC is equilateral]
∴ Area of sector AEF

∴ Area of the three sectors cm2
= 77 cm2
Thus, area of the triangle not included in the circle
= (84.77 − 77) cm2
= 7.77 cm2
OR
TSA of the block = TSA of the cube − Base area of the hemisphere
+ CSA of the hemisphere …(1)
TSA of the cube = 6 (side)2 = 6 × 52 cm2 = 150 cm2
Base area of the hemisphere = π (radius)2

CSA of hemisphere = 2 × π(radius)2
= 2 × 13.86 cm2
From equation (1), we get:
T.S.A of the block = (150 − 13.86 + 2 × 13.86) cm2
= 150 + 13.86 cm2
= 163.86 cm2
Thus, the total surface area of the block is 163.86 cm2.
Question 19 ( 3.0 marks)
Two dice are thrown simultaneously. What is the probability that
(i) 5 will not come up on either of them?
(ii) 5 will come up on at least one?
(iii) 5 will come up at both dice?
Solution:
The sample space for the given experiment can be shown as:
1 2 3 4 5 6
1 (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
2 (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
3 (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
4 (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
5 (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
6 (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
∴Total number of possible outcomes = 6 × 6 = 36
(i) P (5 will come up on either of them)
∴ P (5 will not come up on either of them)
(ii) P (5 will come up on at least one)
(iii) P (5 will come up on both dice)
Question 20 ( 3.0 marks)
The sum of first six terms of an arithmetic progression is 42. The ratio of its 10th term to its 30th term is 1 : 3. Calculate the first and the thirteenth term of the A.P.
Solution:
Let a be the first term and d be the common difference of the given A. P.
It is known that the nth term of an A.P. is given by an = a + (n − 1)d and the sum of the first n terms is given by .
Sum of the first 6 terms = 42

⇒ 2a + 5d = 14 … (1)
10th term = a10 = a + (10− 1)d = a + 9d
30th term, a30 = a + (30− 1)d = a + 29d
It is given that

Substituting a = d in equation (1), we get:
2d + 5d = 14
⇒ 7d = 14
⇒ d = 2
∴a = d = 2
Now, 13th term = a13 = a + 12d = 2 + 12 × 2 = 2 + 24 = 26
Thus, the first term and the 13th term of the given A.P. are 2 and 26 respectively.
Question 21 ( 3.0 marks)
Evaluate:


Solution:

Thus, the value of the given expression is −1.
Question 22 ( 3.0 marks)
Find the ratio in which the point (x, 2) divides the line segment joining the points (−3, −4) and (3, 5). Also find the value of x.
Solution:
Let the point (x, 2) divide the line segment joining the points (−3, −4) and (3, 5) internally in the ratio k:1.
Then, by using the section formula, we have:

Comparing the y coordinates from equation (1), we get:

Therefore, the point (x, 2) divides the line segment joining the points (−3, −4) and (3, 5) in the ratio 2:1.
Now, comparing the x coordinates of equation (1), we get:

Hence, the value of x is 1.
Question 23 ( 3.0 marks)
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, − 1), (2, 1) and (0, 3).
Solution:
Let vertices of the triangle be A (0, − 1), B (2, 1), C (0, 3)

Let D, E, F are midpoints of the sides of this triangle.
The coordinates of D, E, and F are given by:

Question 24 ( 3.0 marks)
Solve for x and y:

ax − by = 2ab
OR
The sum of two numbers is 8. Determine the numbers if the sum of their reciprocals is
Solution:
The given pair of linear equations is:

ax − by = 2ab
These equations can also be written as:

Now, the given equations can be solved by cross-multiplication method as:

On taking the first and last terms, we get:

On taking the last two terms, we get:

Thus, the solution to the given pair of equations is x = b and y = − a.
OR
Let one of the two numbers be x. Then, the other number is 8 − x.
It is given that the sum of the reciprocals of the numbers =


If we take x = 5, then the other number is 8 − 5 = 3.
If we take x = 3, then the other number is 8 − 3 = 5.
In either case, the numbers are 3 and 5.
Thus, the required two numbers are 3 and 5.
Question 25 ( 3.0 marks)
Draw a right triangle in which sides (other than hypotenuse) are of lengths 8 cm and 6 cm. Then construct another triangle whose sides are times the corresponding sides of the first triangle.
Solution:
It is given that sides other than hypotenuse are of lengths 8 cm and 6 cm. Clearly, these will be perpendicular to each other.
The required triangle can be drawn as follows.
Step 1
Draw a line segment AB = 8 cm. Draw a ray SA making 90° with it.
Step 2
Draw an arc of 6 cm radius while taking A as its centre to intersect SA at C. Join BC. ΔABC is the required triangle.
Step 3
Draw a ray AX making an acute angle with AB, opposite to vertex C.
Step 4
Locate 4 points (as 4 is greater in 3 and 4), A1, A2, A3, A4 on line segment AX such that AA1 = A1A2 = A2A3 = A3A4.
Step 5
Join A4B. Draw a line through A3 parallel to A4B intersecting extended line segment AB at B\'.
Step 6
Through B\', draw a line parallel to BC intersecting extended line segment AC at C\'. ΔAB\'C\' is the required triangle.

Section D

Question number 26 to 30 carry 6 marks each.

Question 26 ( 6.0 marks)
A juice seller serves his customers using a glass as shown in Figure 6. The inner diameter of the cylindrical glass is 5 cm, but the bottom of the glass has a hemispherical portion raised which reduces the capacity of the glass. If the height of the glass is 10 cm, find the apparent capacity of the glass and its actual capacity. (Use π = 3.14)

OR
A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is full of water. A solid cone of base diameter 7 cm and height 6 cm is completely immersed in water. Find the volume of
(i) water displaced out of the cylindrical vessel.
(ii) water left in the cylindrical vessel.
[Take π ]
Solution:

Apparent capacity of the glass will be the same as the capacity of the cylindrical portion having its base diameter as 5 cm and height as 10 cm.
Base radius =
Apparent capacity = πr2h
= 3.14 × (2.5)2 × 10
= 196.25 cm3
Actual capacity of the glass will be the difference between the cylindrical portion and the hemispherical portion.
From the figure, it is clear that the base radius of the hemispherical portion is also cm or 2.5 cm.
Actual capacity of the glass =

= 196.25 cm3 − 32.71 cm3
= 163.54 cm3
Thus, the apparent capacity of the glass is 196.25 cm3 and the actual capacity of the glass is 163.54 cm3.
OR

(I)When the solid cone is completely immersed in water, some water will be displaced out of the cylindrical vessel.

The volume of water thus displaced out will be the same as the volume of the solid cone having its height as 6 cm and base diameter as 7 cm.
Base radius
∴Volume of the water displaced out

Thus, the water displaced out of the cylindrical vessel is .
(II) In the beginning, the volume of water in the completely-filled cylindrical vessel will be the same as the capacity of this cylindrical vessel.
Base radius
Volume of water in the cylindrical vessel = πr2h

Water left in the cylindrical vessel = Total water − Water displaced out
= 825 cm3 − 77 cm3
= 748 cm3
Thus, the water left in the cylindrical vessel is .
Question 27 ( 6.0 marks)
In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, prove that the angle opposite to the first side is a right angle.
Use the above theorem to find the measure of ∠PKR in the below figure.

Solution:
Given: ΔABC such that AC2 = AB2 + BC2
To prove: ΔABC is a right triangle, right angled at B
Construction: Draw a right ΔPQR, right angled at Q such that AB = PQ and BC = QR

Proof:
Applying Pythagoras theorem in right ΔPQR:
PR2 = PQ2 + QR2 … (i)
According to the construction, PQ = AB and QR = BC
Substituting PQ and QR in equation (i):
PR2 = AB2 + BC2 … (ii)
It is given that AC2 = AB2 + BC2 … (iii)
From equations (ii) and (iii):
PR2 = AC2
⇒ PR = AC
Now in ΔABC and ΔPQR:
AB = PQ (As constructed)
BC = QR (As constructed)
AC = PR (Proved above)
∴ ΔABC ΔPQR (SSS congruence criterion)
∴ ∠B = ∠Q (C.P.C.T.)
However, ΔPQR was constructed such that ∠Q = 90°
⇒ ∠B = ∠Q = 90°
∴ ∠B = 90°
Thus, ΔABC is a right triangle, right angled at B.
Hence proved

Applying Pythagoras theorem in right ΔPQR:
QR2 = PQ2 + PR2
⇒ (26 cm)2 = (24 cm)2 + PR2
⇒ PR2 = (676 − 576) cm2
⇒ PR2 = 100 cm2
Now, in ΔPKR:
PK2 = 82 = 64; KR2 = 62 = 36; PR2 = 100
It is seen that 100 = 64 + 36
⇒ PR2 = PK2 + KR2
In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle (Proved above)
In ΔPKR, PR2 = PK2 + KR2 and the angle opposite to PR is ∠PKR.
Thus, the measure of ∠PKR is 90°.
Question 28 ( 6.0 marks)
From the top of a building 60 m high, the angles of depression of the top and bottom of a vertical lamp post are observed to be 30° and 60° respectively. Find
(i) the horizontal distance between the building and the lamp post
(ii) The height of the lamp post.
[Take ]
Solution:
The given situation can be represented as:

In the figure, AC is the building and DE is the lamp post. Let x and y respectively be the horizontal distance between the building and the lamp post and the height of the lamp post.
Height of the building, AC = 60 m
CD = BE = x
ED = BC = y
AB = AC − BC = 60 m − y
Now in ΔACD,

In ΔABE,

Substitute the value of x from equation (1) to equation (2):

Thus, the horizontal distance between the building and the lamp post is 34.64 m and the height of the lamp post is 40 m.
Question 29 ( 6.0 marks)
During the medical check-up of 35 students of a class their weights were recorded as follows:
Weight (in kg) Number of students
38 − 40 3
40 − 42 2
42 − 44 4
44 − 46 5
46 − 48 14
48 − 50 4
50 − 52 3
Draw a less than type and a more than type ogive from the given data. Hence obtain the median weight from the graph.
Solution:
For the given data, the cumulative frequency distribution of the less than type can be computed as follows.
Weight (in kg) Number of students (Cumulative frequency)
Less than 40 3
Less than 42 3 + 2 = 5
Less than 44 5 + 4 = 9
Less than 46 9 + 5 = 14
Less than 48 14 + 14 = 28
Less than 50 28 + 4 = 32
Less than 52 32 + 3 = 35
To draw a less than ogive, we mark the upper class limits of the class intervals on the x-axis and their corresponding cumulative frequencies on the y-axis by taking a convenient scale.
Now, plot the points corresponding to the ordered pairs [(upper class limit, cumulative frequency) − i.e., (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35)] on the graph paper as follows:

Similarly, we can compute the cumulative frequency distribution of the more than type as follows:
Weight (in kg) Number of students (Cumulative frequency)
More than 38 35
More than 40 35 − 3 = 32
More than 42 32 − 2 = 30
More than 44 30 − 4 = 26
More than 46 26 − 5 = 21
More than 48 21 − 14 = 7
More than 50 7 − 4 = 3
Now, to draw a more than ogive, we mark the lower class limits of the class intervals on the x-axis and their corresponding cumulative frequencies on the y-axis by taking a convenient scale.
Now, plot the points corresponding to the ordered pairs [(lower class limit, cumulative frequency) − i.e., (38, 35), (40, 32), (42, 30), (44, 26), (46, 21), (48, 7), (50, 3)] on the graph paper as follows:

Now, to obtain the median weight from the graph, we draw both ogives on the same graph paper. They intersect at (46.5, 17.5). 46.5 kg is the median weight of the given data.

Question 30 ( 6.0 marks)
Solve the following equation for x:
9x2 − 9(a + b)x + (2a2 + 5ab + 2b2) = 0
OR
If (−5) is a root of the quadratic equation 2x2 + px − 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots, then find the values of p and k.
Solution:
9x2 − 9(a + b)x + (2a2 + 5ab + 2b2) = 0

OR
It is given that 2x2 + px − 15 = 0
If (−5) is a root of this quadratic equation, then
2(−5)2 + p(−5) − 15 = 0
⇒ 2(25) + p(−5) − 15 = 0
⇒ 50 − 5p − 15 = 0
⇒ 5p = 35
⇒ p = 7 … (1)
Again, it is given that p(x2 + x) + k = 0
⇒ px2 + px + k = 0
On putting p = 7, we get:
7x2 + 7x + k = 0
If this equation has real roots, then its discriminant must be 0.
(7)2 − 4 (7) (k) = 0
⇒ 49 − 28k = 0
⇒ k =
Thus, the values of p and k are 7 and respectively.
Question 1 ( 1.0 marks)
Find the discriminant of the quadratic equation
Solution:
For the quadratic equation ax2 + bx + c = 0,
Discriminant = D = b2 − 4ac
Hence, for the given equation, D =
= 100 − 36
= 64
Thus, the discriminant of the given equation is 64.
Question 2 ( 1.0 marks)
If , a, and 2 are three consecutive terms of an A.P., then find the value of a.
Solution:
If three terms a, b, and c are in A.P., then we have b − a = c − b
⇒ 2b = a + c
∴ If are three consecutive terms of an A.P., then

Thus, the value of a is .
Question 3 ( 1.0 marks)
If the areas of two similar triangles are in the ratio 25:64, write the ratio of their corresponding sides.
Solution:
We know thatthe ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
It is given that the areas of two similar triangles are in the ratio 25:64.

Thus, the ratio of the corresponding sides of the two similar triangles is 5:8.
Question 4 ( 1.0 marks)
In figure 1, ΔABC is circumscribing a circle. Find the length of BC.

Solution:
BR = BP [Tangents drawn to a circle from a point outside the
circle are equal]
However, BR = 3 cm
∴BP = 3 cm … (1)
AR = AQ [Tangents drawn to a circle from a point outside the
circle are equal]
However, AR = 4 cm
∴AQ = 4 cm
AQ + QC = AC
⇒ QC = AC − AQ
Using the values AQ = 4 cm and AC = 11 cm,
QC = 11 cm − 4 cm
⇒ QC = 7 cm
CP = CQ [Tangents drawn to a circle from a point outside the
circle are equal]
∴CP = 7 cm … (2)
BC = BP + CP
On using equations (1) and (2), we obtain
BC = 3 cm + 7 cm
BC = 10 cm
Thus, the length of BC is 10 cm.
Question 5 ( 1.0 marks)
Two coins are tossed simultaneously. Find the probability of getting exactly one head.
Solution:
If two coins are tossed simultaneously, then the possible outcomes are
S = {HT, TH, TT, HH}
Thus, the total number of possible outcomes is 4.
Out of all the four outcomes, {HT} and {TH} are cases of exactly one head.
∴ Required probability =
Question 6 ( 1.0 marks)
Find the [HCF × LCM] for the numbers 100 and 190.
Solution:
For two numbers a and b, HCF × LCM = a × b
∴ For the given numbers 100 and 190:
HCF × LCM = 100 × 190
HCF × LCM = 19000
Question 7 ( 1.0 marks)
If 1 is a zero of the polynomial p(x) = ax2 − 3(a − 1) x − 1, then find the value of a.
Solution:
If 1 is a zero of polynomial p(x), then p(1) = 0.
∴ p(1) = a(1)2 − 3(a − 1) (1) − 1 = 0
⇒ a − 3a + 3 − 1 = 0
⇒ −2a = −2
⇒ a = 1
Thus, the value of a is 1.
Question 8 ( 1.0 marks)
If sec2θ (1 + sin θ) (1 − sin θ) = k, then find the value of k.
Solution:
sec2θ (1 + sinθ)(1 − sinθ) = k

Thus, the value of k is 1.
Question 9 ( 1.0 marks)
If the diameter of a semicircular protractor is 14 cm, then find its perimeter.
Solution:

Diameter = 14 cm
Radius =
Length of the semicircular part = πr
Total perimeter = Length of semicircular part + Diameter
= 22 cm + 14 cm
= 36 cm
Thus, the perimeter of the protractor is 36 cm.
Question 10 ( 1.0 marks)
Find the number of solutions of the following pair of linear equations:
x + 2y − 8 = 0
2x + 4y = 16
Solution:
The given pair of linear equations is
x + 2y − 8 = 0
2x + 4y − 16 = 0
On comparing with general equations

Hence, the given pair of linear equations has infinitely many solutions.
Section B

Question number 11 to 15 carry 2 marks each.

Question 11 ( 2.0 marks)
If the points A (4, 3) and B (x, 5) are on the circle with the centre O (2, 3), find the value of x.
Solution:
If the points A (4, 3) and B (x, 5) are on the circle with the centre O (2, 3), then OA and OB will be the radii of the circle.
∴OA = OB
⇒ OA2 = OB2
⇒ (4 − 2)2 + (3 − 3)2 = (x − 2)2 + (5 − 3)2 [Using distance formula]
⇒ 22 + 02 = (x − 2)2 + 22
⇒ 0 = (x − 2)2
⇒ x − 2 = 0
⇒ x = 2
Thus, the required value of x is 2.
Question 12 ( 2.0 marks)
Which term of the A.P. 4, 12, 20, 28, … will be 120 more than its 21st term?
Solution:
The given A.P. is 4, 12, 20, 28, …
Here, first term = a = 4
Common difference = d = a2 − a = 12 − 4 = 8
a21 = a + (21 − 1) d [an = a + (n − 1) d]
= 4 + 20 × 8
= 4 + 160 = 164
120 more than the 21st term = 120 + 164 = 284
Let the nth term of the given A.P. be 284.
∴ an = a + (n − 1) d
⇒ 284 = 4 + (n − 1) 8
⇒ 280 = (n − 1) 8
⇒ n − 1 =
⇒ n = 35 + 1 = 36
Thus, the 36th term of the given A.P. is 120 more than its 21st term.
Question 13 ( 2.0 marks)
If , then evaluate
OR
Find the value of tan 60°, geometrically.
Solution:
The given expression is .

Thus, the value of the given expression is .
OR
Consider an equilateral ΔABC.
Then, ∠ABC = ∠BCA = ∠CAB = 60° and AB = BC = CA = 2a (say)
Now, draw AD⊥ BC.

Comparing ΔABD and ΔACD, we have:
∠ADB = ∠ADC = 90° [By construction]
AB = AC [Sides of equilateral triangle]
AD = AD [Common]
∴ΔABD ≅ ΔACD [By RHS congruency axiom]
⇒ BD = DC [CPCT]
∴
On applying Pythagoras’ Theorem in ΔABD, we get:

∴tan 60º = tan B
Thus, the value of tan 60º is .
Question 14 ( 2.0 marks)
Find all the zeroes of the polynomial x3 + 3x2 − 2x − 6, if two of its zeroes are and .
Solution:
The given polynomial is .
It is given that and are the two zeroes of p(x).
Thus, and are the factors of p(x).
This means, is also a factor of p(x).
We can divide by x2 − 2 as

∴p(x) = (x2 − 2) (x + 3)
As (x + 3) is a factor of the polynomial p(x), so x = −3 is a zero of the polynomial.
Thus, −3 is the third zero of the given polynomial.
Thus, the three zeroes of are −3, , and .
Question 15 ( 2.0 marks)
In Figure 2, ΔABD is a right triangle, right-angled at A and AC ⊥ BD. Prove that AB2 = BC . BD.

Solution:
In ΔACB and ΔDAB, we have
∠ACB = ∠DAB [Each is 90º]
∠ABC = ∠DBA [Common angle]
∴ΔACB ∼ ΔDAB [By AA similarity criterion]
It is known that if two triangles are similar, then the corresponding sides are proportional.

Hence, proved.
Section C

Question number 16 to 25 carry 3 marks each.

Question 16 ( 3.0 marks)
Prove that is an irrational number.
Solution:
If possible, suppose is a rational number.
Then, we can find two integers a, b (b ≠ 0) such that

Since a, b, 5 and 3 are integers, is rational and so is rational.
This conclusion contradicts the fact that is an irrational number.
Hence, our assumption that is rational is false.
Thus, is irrational.
Question 17 ( 3.0 marks)
Draw a right triangle in which sides (other than hypotenuse) are of lengths 8 cm and 6 cm. Then construct another triangle whose sides are times the corresponding sides of the first triangle.
Solution:
Let us assume that right ΔABC has base BC = 6 cm, side AB = 8 cm, and ∠B = 90°. Let ΔA\'BC\' have sides that are times those of ΔABC.
Now, ΔABC and ΔA\'BC\' can be drawn as follows:
(1) Draw a line segment BC = 6 cm. Draw a ray BX making 90° with BC.
(2) Draw an arc of 8 cm radius taking B as its centre to intersect BX at A. Join AC. ΔABC is the required triangle.
(3) Draw a ray BY making any acute angle with BC on the other side of line segment BC.
(4) Locate 4 points B1, B2, B3, and B4 on ray BY such that BB1 = B1B2 = B2B3 = B3B4
(5) Join B4C. Draw a line through B3 parallel to B4C intersecting BC at C\'.
(6) Through C\', draw a line parallel to AC intersecting AB at A\'. ΔA\'BC\' is the required triangle.

Question 18 ( 3.0 marks)
Two dice are thrown simultaneously. What is the probability that
(i) 5 will not come up on either of them?
(ii) 5 will come up on at least one?
(iii) 5 will come up at both dice?
Solution:
The sample space for the given experiment can be shown as:
1 2 3 4 5 6
1 (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
2 (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
3 (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
4 (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
5 (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
6 (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
∴Total number of possible outcomes = 6 × 6 = 36
(i) P (5 will come up on either of them)
∴ P (5 will not come up on either of them)
(ii) P (5 will come up on at least one)
(iii) P (5 will come up on both dice)
Question 19 ( 3.0 marks)
In Figure, 3, AD ⊥ BC and BD CD. Prove that 2CA2 = 2AB2 + BC2.

OR
In Figure 4, M is mid-point of side CD of a parallelogram ABCD. The line BM is drawn intersecting AC at L and AD produced at E. Prove that EL = 2 BL.

Solution:

On applying Pythagoras theorem to ΔABD, we obtain
AB2 = BD2 + AD2
⇒ AD2 = AB2 − BD2 … (1)
On applying Pythagoras theorem to ΔACD, we obtain
AC2 = AD2 + DC2
⇒ AD2 = AC2 − DC2 … (2)
On comparing equations (1) and (2), we obtain
AB2 − BD2 = AC2 − DC2
⇒ AC2 − AB2 = DC2 − BD2 … (3)
It is given that BD CD
⇒ CD = 3BD
Now, BD + DC = BC
⇒ BD + 3BD = BC
⇒ BD BC … (4)
Also, CD = 3BD
⇒ DC BC … (5) [From (4)]
On substituting the values of BD and DC from (4) and (5) in (3), we obtain
AC2 − AB2

∴ 2(AC2 − AB2) = BC2
⇒ 2AC2 − 2AB2 = BC2
⇒ 2CA2 = 2AB2 + BC2
Hence, proved.
OR

Comparing ΔBCM and ΔEDM:
CM = MD [M is the mid-point of CD]
∠BMC = ∠DME [Vertically opposite angels]
∠BCM = ∠EDM [Alternate interior angles as AE||BC]
∴ ΔBCM ≅ ΔEDM [ASA congruence criterion]
⇒ BC = ED [C.P.C.T.]
⇒ BC = AD = ED [Opposite sides of parallelogram are equal]
Comparing ΔBLC and ΔELA:
∠LBC = ∠AEL [Alternate interior angles as AE||BC]
∠BLC = ∠ELA [Vertically opposite angels]
∴ Δ BLC ∼ ΔELA [AA similarity criterion]
It is known that the corresponding sides of similar triangles are proportional.

Hence, proved.
Question 20 ( 3.0 marks)
The sum of first six terms of an arithmetic progression is 42. The ratio of its 10th term to its 30th term is 1 : 3. Calculate the first and the thirteenth term of the A.P.
Solution:
Let a be the first term and d be the common difference of the given A. P.
It is known that the nth term of an A.P. is given by an = a + (n − 1)d and the sum of the first n terms is given by .
Sum of the first 6 terms = 42

⇒ 2a + 5d = 14 … (1)
10th term = a10 = a + (10− 1)d = a + 9d
30th term, a30 = a + (30− 1)d = a + 29d
It is given that

Substituting a = d in equation (1), we get:
2d + 5d = 14
⇒ 7d = 14
⇒ d = 2
∴a = d = 2
Now, 13th term = a13 = a + 12d = 2 + 12 × 2 = 2 + 24 = 26
Thus, the first term and the 13th term of the given A.P. are 2 and 26 respectively.
Question 21 ( 3.0 marks)
The area of an equilateral triangle is cm2. Taking each angular point as centre, circles are drawn with radius equal to half the length of the side of the triangle. Find the area of triangle not included in the circles. [Take = 1.73]
OR
Figure 5 shows a decorative block which is made of two solids − a cube and a hemisphere. The base of the block is a cube with edge 5 cm and the hemisphere, fixed on the top, has a diameter of 4.2 cm. Find the total surface area of the block. [Take π = ]

Solution:

Let ABC be the given equilateral triangle whose area is cm2.
Area of equilateral triangle
∴ cm2
⇒ (Side)2 cm2
⇒ Side = 2 × 7 cm = 14 cm
Thus, radius of each circle cm = 7 cm
We know, ∠A = ∠B = ∠C = 60° [ ΔABC is equilateral]
∴ Area of sector AEF

∴ Area of the three sectors cm2
= 77 cm2
Thus, area of the triangle not included in the circle
= (84.77 − 77) cm2
= 7.77 cm2
OR
TSA of the block = TSA of the cube − Base area of the hemisphere
+ CSA of the hemisphere …(1)
TSA of the cube = 6 (side)2 = 6 × 52 cm2 = 150 cm2
Base area of the hemisphere = π (radius)2

CSA of hemisphere = 2 × π(radius)2
= 2 × 13.86 cm2
From equation (1), we get:
T.S.A of the block = (150 − 13.86 + 2 × 13.86) cm2
= 150 + 13.86 cm2
= 163.86 cm2
Thus, the total surface area of the block is 163.86 cm2.
Question 22 ( 3.0 marks)
Find the ratio in which the point (x, −1) divides the line segment joining the points (−3, 5) and (2, −5). Also find the value of x.
Solution:
Let the point (x, −1) divide the line segment joining the points (−3, 5) and
(2, −5) internally in the ratio k:1.
Then, by using the section formula, we have:

Comparing the y coordinates from equation (1), we get:

Therefore, the point (x, −1) divides the line segment joining the points (−3, 5) and (2, −5) in the ratio 3:2.
Now, comparing the x coordinates of equation (1), we get:

Hence, the value of x is 0.
Question 23 ( 3.0 marks)
Find the area of the quadrilateral ABCD whose vertices are A (1, 0), B (5, 3), C (2, 7) and D (¬−2, 4)
Solution:
The given quadrilateral ABCD can be drawn as:

We join AC to obtain ΔABC and ΔACD.
We know that the area of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) is given by the numerical value of the expression:


Question 24 ( 3.0 marks)
Evaluate:


Solution:


Thus, the value of the given expression is −1.
Question 25 ( 3.0 marks)
Solve for x and y:

ax − by = 2ab
OR
The sum of two numbers is 8. Determine the numbers if the sum of their reciprocals is
Solution:
The given pair of linear equations is:

ax − by = 2ab
These equations can also be written as:

Now, the given equations can be solved by cross-multiplication method as:

On taking the first and last terms, we get:

On taking the last two terms, we get:

Thus, the solution to the given pair of equations is x = b and y = − a.
OR
Let one of the two numbers be x. Then, the other number is 8 − x.
It is given that the sum of the reciprocals of the numbers =


If we take x = 5, then the other number is 8 − 5 = 3.
If we take x = 3, then the other number is 8 â