Q 1. The point P on x-axis equidistant from the points A(–1, 0) and B(5, 0) is
A. (2, 0)
B. (0, 2)
C. (3, 0)
D. (2, 2)
Solution: Correct option : A
Let P(x, 0) … (Since, the point P is on x – axis)
The points A(–1, 0) and B(5, 0) are also lying on x – axis and P is equidistant from
the points A and B.
⇒ P is a midpoint of AB.
By using mid – point formula, we get

Therefore, the coordinates of the point P is (2, 0).
Q 2. The co-ordinates of the point which is reflection of point (–3, 5) in x-axis are
A. (3, 5)
B. (3, -5)
C. (-3, -5)
D. (-3, 5)
Solution: Correct option : C
The coordinates of the point which is the reflection of point (–3, 5) is (–3, –5).

Q 3. If the point P(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3:1, then the value of y is
A. 4
B. 3
C. 2
D. 1
Solution: Correct option : D
The point P(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3:1.
By using section formula, we get

Q 4. The sum of exponents of prime factors in the prime-factorisation of 196 is
A. 3
B. 4
C. 5
D. 2
Solution: Correct option : B
Prime factorisation of 196 = 2 × 2 × 7 × 7 = 22 × 72
The sum of the exponents of the prime factors in the prime factorisation of 196 is 2 + 2 = 4
Q 5. Euclid’s division Lemma states that for two positive integers a and b, there exists unique integer q and r satisfying a = bq + r, and
A. 0 < r < b
B. 0 < r < b
C. 0 ≤ r < b
D. 0 ≤ r ≤ b
Solution: Correct option : C
Euclid’s Division Lemma states that for two positive integers a and b, there exists unique integer q and r satisfying a = bq + r, and 0 ≤ r < b.
Q 6. The zeroes of the polynomial x2 – 3x – m(m+3) are
A. m, m+3
B. –m, m+3
C. m, –(m+3)
D. –m, –(m+3)
Solution: Correct option: B
p(x) = x2 – 3x – m(m + 3)
Substitute x = –m
p(–m) = (–m)2 – 3(–m) – m(m + 3)= m2 + 3m – m2 – 3m = 0 ...(i)
Substitute x = m + 3
p(m + 3) = (m + 3)2 – 3(m + 3) – m(m + 3)
= (m + 3)(m + 3 – 3 – m) ...(ii)
From (i) and (ii),
–m and m + 3 are the zeroes of the given polynomial.
Q 7. The value of k for which the system of linear equations x + 2y = 3, 5x + ky + 7 = 0 is inconsistent is
A. 
B. 
C. 5
D. 10
Solution: Correct option: D
Q 8. The roots of the quadratic equation x2 – 0.04 = 0 are
A. ±0.2
B. ±0.02
C. 0.4
D. 2
Solution: Correct option: A

Taking square root on both the side, we get
Q 9. The common difference of the A.P.

A. 1
B. 
C. -1
D. 2
Solution: Correct option: C

Q 10. The nth term of the A.P. a, 3a, 5a, …. Is
A. na
B. (2n – 1)a
C. (2n + 1)a
D. 2na
Solution: Correct option: B
Here a1 = a, d = 3a – a = 2a
We know that,
an = a1 + (n – 1)d
= a + (n – 1)2a
= a + 2an – 2a
= 2an – a = (2n – 1)a
In Q. Nos. 11 to 15, fill in the blanks. Each question carries 1 mark:
Q 11. In fig. 1, the angles of depressions from the observing positions O1 and O2 respectively of the object A are ______, ______.

Solution: In fig. 1, the angles of depressions from the observing positions O1 and O2 respectively of the object A are 30° and 45°.
In ΔO1CA,
⇒ ∠O1CA + ∠CAO1 + ∠CO1A = 180°
⇒ 90° + ∠CAO1 + 60° = 180°
⇒ ∠CAO1 = 30°
So, the angle of depression from the observing position O1 of the object A is 30°
And the angle of depression from the observing position O2 of the object A is 45°.
Q 12. In ΔABC, AB=6 cm, AC = 12 cm and BC = 6 cm, then ∠B=______.
OR
Two triangles are similar if their corresponding sides are ______.
Solution: In ∆ABC, AB =
, AC = 12 cm and BC = 6 cm, then ∠B = 90°.
AB2 =
= 108, BC2 = (6)2 = 36 and AC2 = (12)2 = 144
Here, AC2 = AB2 + BC2
By the converse of Pythagoras theorem, we get
m∠B = 90°.
OR
Two triangles are similar if their corresponding sides are in the same ratio.
Q 13. In given Fig. 2, the length PB=______cm.

Solution: In given Fig 2, the length PB = 4 cm.
AB is a tangent to the internal circle at P and OP is its radius.
⇒ OP ⊥ AB
In ∆APO,
OA2 = OP2 + AP2 ... (OP ⊥ AB)
⇒ AP2 = 52 – 32 = 16
⇒ AP = 4 cm
We know that,
In two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.
⇒ AP = PB = 4 cm
Q 14. In fig. 3, MN || BC and AM : MB =1 : 2, then 

Solution: In fig. 3 MN || BC and AM: MB = 1: 2, then
.
AM: MB = 1: 2
Let AM = x and MB = 2x
⇒ AB = AM + MB = 3x
Here, ∆AMN ~ ∆ABC ... (AA test)

Q 15. The value of sin 32° cos 58° +cos 32° sin 58° is ______.
OR
The value of
is ______.
Solution: The value of sin 32° cos 58° + cos 32° sin 58° is 1.
sin 32° cos 58° + cos 32° sin 58°
= sin 32° cos(90° – 32°) + cos 32° sin(90° – 32°)
= sin 32° sin 32° + cos 32° cos 32°
= sin232° + cos232°
= 1 …. (sin2θ + cos2θ = 1)
OR
The value of
is 2.
Explanation:

Q Nos. 16 to 20 are short answer type questions of 1 mark each.
Q 16. A die is thrown once. What is the probability of getting a prime number?
Solution: A die is thrown once.
S = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6
Let A be the event of getting a prime number.
A = {2, 3, 5} ⇒ n(A) = 3
⇒ P(A) =
.
Q 17. If a number x is chosen at random from the numbers –3, –2, –1, 0, 1, 2, 3, then find the probability of x2 < 4.
OR
What is the probability that a randomly taken leap year has 52 Sundays?
Solution: Here, S = {–3, –2, –1, 0, 1, 2, 3} ⇒ n(S) = 7
Let A be the event that the square of the chosen number is less than 4.
A = {–1, 0, 1}
⇒ P(A) =
.
OR
In the leap year, total number of days = 366 = 52 × 7 + 2
Total 52 weeks ⇒ 52 Sundays, now the 2 remaining days will be as follows.
S = {(Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday), (Sunday, Monday)}
⇒ n(S) = 7
So, the probability of not getting Sunday in the remaining 2 days = 
⇒ The probability that a randomly taken leap year has 52 Sundays is
.
Q 18. If sin A + sin2 A = 1, then find the value of the expression (cos2 A + cos4 A).
Solution: sinA + sin2A = 1
⇒ sinA = 1 – sin2A = cos2A … (i)
⇒ sin2A = cos4A … (ii)
cos2A + cos4 A
= sinA + sin2A … from (i) and (ii)
= 1 (sinA + sin2A = 1)
Q 19. Find the area of the sector of a circle of radius 6 cm whose central angle is 30°. (Take π =3.14).
Solution: Here, r = 6cm, θ = 30°
Area of the sector of a circle =
× 3.14 × 6 × 6 = 9.42 cm2.
Q 20. Find the class marks of the classes 20 – 50 and 35 – 60.
Solution: Class mark of the class 20 – 50 = 
Class mark of the class 35 – 60 = 
Q 21. A teacher asked 10 of his students to write a polynomial in one variable on a paper and then to handover the paper.
The following were the answers given by the students:
2x + 3, 3x2 + 7x + 2, 4x3 + 3x2 + 2, x3 +
+ 7, 7x +
, 5x3 – 7x + 2, 2x2 + 3 –
, 5x –
, ax3 + bx2 + cx + d, x +
.
Answer the following questions:
- How many of the above ten, are not polynomials?
- How many of the above ten, are quadratic polynomials?
Solution:
- From the above ten, only three are not polynomials.
They are
because in all three polynomials, the degree of x is not a positive integer.
- From the above ten, only one is a quadratic polynomial which is 3x2 + 7x + 2 because highest degree of the variable x is 2.
Q 22 .A child has a die whose six faces show the letters as shown below:

The die is thrown once. What is the probability of getting (i) A, (ii) D?
Solution: From the given question, n(S) = 6
- Let E be the event of getting A on the die.
As there are 2 A’s, n(E) = 2
Hence, the required probability is 
- Let F be the event of getting D on the die.
As there is only one D, n(F) = 1
Hence, the required probability is 
Q 23. In fig. 4, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that
.

OR
In fig.5, if AD ⊥ BC, then prove that AB2 + CD2 = BD2 + AC2.
Solution: Given: Triangle ABC and triangle DBC are on the same base BC.
To prove:

Construction:
Draw AM perpendicular to BC from point A and draw DN perpendicular to BC from point D.
Prove:
In Δ AMO and Δ DNO
∠AOM = ∠DON vertically opposite angles
∠AMO = ∠DNO 90Ëš angles
⇒ ΔAMO ≈ ΔDNO AA test of similarity
Ratio of the areas of two similar triangles is equal to the square of ratio of their corresponding sides
OR
In ΔADB,
AB2 = BD2 + AD2 Pythagoras theorem
⇒ AD2 = AB2 – BD2 … (i)
Also in ΔADC,
AC2 = CD2 + AD2 Pythagoras theorem
⇒ AD2 = AC2 – CD2 … (ii)
AB2 – BD2 = AC2 – CD2 from (i) and (ii)
⇒ AB2 + CD2 = BD2 + AC2
Q 24. Prove that 
OR
Show that tan4θ + tan2θ = sec4θ - sec2θ
Solution:
OR
LHS = tan4 θ + tan2 θ
= (sec2 θ - 1)2 + sec2 θ - 1
= sec4 θ - 2sec2 θ + 1 + sec2 θ - 1
= sec4 θ - sec2 θ
Hence, tan4 θ + tan2 θ = sec4 θ - sec2 θ.
Q 25. Find the mode of the following frequency distribution:

Solution: Here, the maximum frequency is 10, and the class corresponding to this is 30 – 35.
So the modal class is 30 – 35.
Therefore, l = 30, h = 5, f0 = 9, f1 = 10 and f2 = 3
Mode = 
Hence, the mode of the above data is 30.625.
Q 26. From a solid right circular cylinder of height 14 cm and base radius 6 cm, a right circular cone of same height and same base radius is removed. Find the volume of the remaining solid.
Solution: h = 14 cm and r = 6 cm
As the circular cone of same height and same radius is removed from the cylinder, therefore we have
Volume of the remaining solid
= Volume of cylinder – volume of cone
= πr2h -
πr2h
=
πr2h
= 
= 1056 cm3
Q. Nos. 27 to 34 carry 3 marks each.
Q 27. If a circle touches the side BC of a triangle ABC at P and extended sides AB and AC at Q and R, respectively, prove that
.
Solution:

Lengths of the tangents from an external point are equal.
Therefore, AQ = AR, BQ = BP and CP = CR
Perimeter of ΔABC = AB + BC + CA
= AB + BP + PC + AR – CR
= AB + BQ + PC + AQ – PC
= AQ + AQ
⇒ AB + BC + CA = 2AQ
⇒
(AB + BC + CA) = AQ
Q 28. The area of a circular play ground is 22176 cm2. Find the cost of fencing this ground at the rate of Rs. 50 per metre.
Solution: Area of a circular playground = 22176 cm2
⇒ πr2 = 22176
⇒
r2 = 22176
⇒ r2 = 22176 × 
⇒ r = 84 cm
Circumference of a playground = 2πr2 = 2 ×
× 84 = 528 cm = 5.28 m
The cost of fencing the playground at Rs. 50 per metre
= 5.28 × 50 = Rs. 264
Q 29. If the mid – point of the line segment joining the points A (3, 4) and B (k, 6) is P(x, y) and x + y – 10 = 0, find the value of k.
OR
Find the area of triangle ABC with A (1, –4) and the mid-points of sides through A being (2, –1) and (0, –1).
Solution: The coordinates of the midpoint of the line segment joining A(3, 4) and B(k, 6) are

It is given that x + y – 10 = 0

⇒ 3 + k = 10
⇒ k = 7
Hence, the value of k is 7.
OR
In ∆ABC, coordinates of A are (1, –4)
Let (a, b) and (x, y) be the coordinates of points B and C respectively.
Midpoint of AB is (2, –1) and of AC is (0, –1)
Midpoint of AB = 
⇒ a = 3, b = 2
Midpoint of AC = 
⇒ x = -1, y = 2
Therefore, B(3, 2) and C(-1, 2)
Area of Δ ABC =
∣x1(y2 - y3)+ x2(y3 - y1) + x3(y1 - y2)∣
=
∣1(2 - 2 ) + 3(2 + 4) - 1(-4 - 2)∣
=
∣0 + 18 + 6∣ = 12
Hence the area of ΔABC is 12 square units.
Q 30. If Fig.6, if ΔABC ~ ΔDEF and their sides of lengths (in cm) are marked along them, then find the lengths of sides of each triangle.
Solution: Given that ΔABC ~ ΔDEF
⇒
⇒ 4x – 2 = 18
⇒ 4x = 20
⇒ x = 5
Then the sides of triangles are
AB = 2x – 1 = 9, BC = 2x + 2 = 12, AC = 3x = 15
DE = 18, EF = 3x + 9 = 24, DF = 6x = 30
Q 31. If 2x + y = 23 and 4x – y = 19, find the value of (5y – 2x) and
.
OR
Solve for 
Solution: 2x + y = 23 …(i)
4x – y = 19 …(ii)
Adding (i) and (ii) we get
⇒ 6x = 42
⇒ x = 7
Put it in (i) we get
⇒ 14 + y = 23
⇒ y = 9
5y – 2x = 45 – 14 = 31

OR
According to the paper, condition given is x ≠ -4, 7 which implies that the equation should be
.

⇒ -30 = x2 - 3x -28
⇒ x2 - 3x + 2 = 0
⇒ (x - 2)(x - 1) = 0
⇒ x = 2 or x = 1x
Q 32. Which term of A.P. 20, 19
, 18
, 17
,.... is the first negative term.
OR
Find the middle term of the A.P. 7, 13, 19, …., 247.
Solution: 20, 19
, 18
, 17
,...
The above sequence is an A.P.
⇒ a1 = 20 and d = 
Consider, an < 0
⇒ a1 + (n – 1)d < 0
⇒ 20 + (n - 1) ×
< 0
⇒ 80 - 3(n - 1) < 0
⇒ 80 – 3n + 3 < 0
⇒ 83 < 3n
⇒ n >
Hence, 28th term of the given sequence is first negative term.
OR
7, 13, 19, …, 247
Given sequence is in A.P.
a1 = 7, d = 6 and an = 247
⇒ a1 + (n – 1)d = 247
⇒ 7 + (n – 1)6 = 247
⇒ 6(n – 1) = 240
⇒ n – 1 = 40
⇒ n = 41
The middle term should be a21.
a21 = a1 + 20d = 7 + 20 × 6 = 127
The middle term of the given sequence is 127.
Q 33. Water in a canal, 6m wide and 1.5m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8cm standing water is required?
Solution: Area of the face of canal = 6 × 1.5 m2
Speed of water flow = 10 km/h = 10000 m/h
So the water flow through the canal in 1 hour
= 6 × 1.5 × 10000
= 90000 m3
Water flow through the canal in 30 min = 90000/2 = 45000 m3
The height of standing water = 8 cm = 0.08 m
Area can irrigate = 45000 ÷ 0.08 = 562500 m2
Q 34. Show that: 
Solution:

Q 35. The mean of the following frequency distribution is 18. The frequency f in the class interval 19 – 21 is missing. Determine f.

OR
The following table gives production yield per hectare of wheat of 100 farms of a village:

Change the distribution to a ‘more than’ type distribution and draw its ogive.
Solution:


⇒ 704 + 20f = 720 + 18f
⇒ 2f = 16
⇒ f = 8
OR


Q 36. From a point on the ground, the angles of elevation of the bottom and the top of a tower fixed at the top of a 20m high building are 45° and 60° respectively. Find the height of the tower.
Solution:

Let BC be the building, AB be the tower and D be the point on ground from where elevation angles are to be measured.
In ΔBCD,
CD = 20 m …(i)
In ΔACD,
Hence, the height of the tower is 
Q 37. It can take 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for four hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. How long would it take for each pipe to fill the pool separately?
Solution: Let the time taken to fill the pool by pipe A be x hours and by pipe B be y hours.
Thus, the quantity of water filled by the pipe A in one hour is
and by the pipe is
.
In 1 hour the quantity of water filled by both the pipes is 
It takes 12 hours to fill the pool for both the pipes.
…(i)
Let A be the pipe with larger diameter and B be the pipe with smaller diameter.
In 4 hours the quantity of water filled by pipe A is
.
In 9 hours the quantity of water filled by pipe B is
.
According to the question,
Multiplying (i) by 4 we get
4m + 4n =
…(v)
Subtracting (v) from (iv) we get
n = 
Put it in (iii) we get
Hence, x = 20 and y = 30
Hence, the pipe A and pipe B fills the pool in 20 hours and 30 hours respectively.
Q 38. Prove that
is an irrational number.
Solution: Let us assume that
is a rational number
We can write it in the form of
where q ≠ 0.
Also p and q are co-prime numbers.
Here p is divisible by 5
p = 5k where k is a positive integer
Hence, p2 = 25k2
Substituting 5q2 = p2
⇒ 5q2 = 25k2
⇒ q2 = 5k2
q is divisible by 5
From this we can say that p and q have a common factor 5
It contradicts to our assumption p and q are co-prime.
Hence,
is an irrational number.
Q 39. Draw a circle of radius 3.5cm. From a point P, 6cm from its centre, draw two tangents to the circle.
OR
Construct a ΔABC with AB=6cm, BC=5cm and ∠B=60°. Now construct another triangle whose sides are
times the corresponding sides of ΔABC.
Solution: Steps of construction:
- Draw a line segment OP = 6 cm
- With centre O and radius 3.5 cm, draw a circle.
- Draw the perpendicular bisector of OP intersecting it at M.
- With centre M and radius OM, draw a circle which intersects the circle at T and S.
- Join PT and PS which are the required tangents.