1. society
2. crime
3. property crime
4. usury

iBOOKS SAMPLE PAPER – 01
[FOR XTH CBSE BOARD]
QUESTION
ANSWER
SECTION – A
SECTION – A
1.
2.
[1×10]
State the Fundamental Theorem of Arithmetic.
1.
Every composite number can be expressed as the
product of primes, and this factorization is unique, apart
from the order in which they occur.
[1]
2.
-1 and 3 are the zeroes of the polynomial.
3.
For infinitely many solution,
In figure, the graph of the polynomial P(x) is
given. Find the zeroes of the polynomial.
[1]
Y
4
3
2
1
X'
-3
-2
-1
0
-1
1
2
3
4
X
-2
a1 b1 c 1
=
=
a2 b2 c 2
Here
a1
2 b1
3 c1 k 1
=
,
=
,
=
= .
a 2 k − 1 b 2 k + 2 c 2 3k 3
Now
2
3
1
2
1
=
=
⇒
=
k −1 k + 2 3
k −1 3
-3
⇒ k – 1 = 6, ⇒ k = 7.
-4
[1]
Y'
3.
For what values of k, the following pair of linear
equations has infinitely many solutions ?
4.
2x + 3y = k
=
(k – 1)x + (k + 2) y = 3k.
4.
The Arc of length 'l' =
θ
× 2πr
360°
30
22
× 2 × × 21 = 11 cm
360°
7
∴ Perimeter = l + 2r = 11 + 2 × 21 = 53 cm.
If the following figure is a sector of a circle of
radius 21 cm. Find the perimeter of this sector.
B
5.
l
[1]
Total number of outcomes, throwing two dice are = 36.
The favourable outcomes to the even.
The total of numbers on the dice is 13 are 0.
[1]
30°
A
O
5.
Two dice are thrown simultaneously. Find the
probability that the total of the numbers on the
dice is 13.
6.
Here total frequency, Σfi = 5 + 8 + 20 + 15 + 7 + 5 = 60.
∴ n/2 = 30
Marks obtained: 0–10 10–20 20–30 30–40 40–50 50–60
Frequency
6.
:
5
8
20
15
7
5
5
13
33
48
45
60
Find the median class of the following data :
Cumulative
Marks obtained
Frequency
Frequency
0 – 10
5
10 – 20
8
Now 20 – 30 is the median class since its cumulative
frequency is 33 which is greater than 30.
[1]
20 – 30
20
30 – 40
15
40 – 50
7
50 – 60
5
:
QUESTION
7.
ANSWER
In figure, P and Q are points on the sides AB
and AC respectively of ∆ ABC such that AP =
3.5 cm, PB = 7 cm, AQ =3 cm and QC=6 cm. If
PQ = 4.5 cm, Find BC.
7.
In ∆ ABC, we have,
AP 3.5 1
AQ 3 1
=
= and
= =
PB
7
2
QC 6 2
∴ PQ || BC
∴ ∆ AQP ~ ∆ ABC.
2
⇒
2
Ar( ∆AQP)  AQ 
 AP 
 PQ 
=
=
 =  BC 
Ar( ∆ABC)  AC 
 AB 


2
2
2
2
1
 3 
 PQ 
 PQ 
⇒
 =  BC  ⇒  BC  = 3


3 + 6


⇒ BC = 3 PQ = 3 (4.5) = 13.5 cm.
8.
9.
If tan A = 3/4 and A + B = 90°, then what is the
value of cot B ?
Determine the 10th term from the end of the
AP., 4, 9, 14, …… , 254.
10. The length of the tangent from a point A at a
distance of 26 cm from the centre of the circle
is 10 cm. What will be the radius of the circle ?
[1]
8.
Given A + B = 90° ⇒ B = 90° - A.
[1]
9.
⇒ cot B = cot (90° - A)
⇒ cot B = tan A = 3/4.
Here a = 254, d = -5
∴ a10 = a + 9d = 254 + 9(-5) = 254 – 45 = 209.
[1]
10. Since the tangent to a circle is perpendicular to the
radius through the point of contact.
∴ ∠OTA = 90°
2
2
2
In rt. ∆ OAT, we have, OA = OT + AT
T
2
2
2
2
⇒ 26 = OT + 10 ⇒ 676 = OT + 100
10 cm
2
⇒ OT = 676 – 100 =576 ⇒ OT = 24 cm.
O
[1]
A
26 cm
SECTION – B
11. If the ten, jack, queen, king and ace of diamonds are
removed, then the total of cards left is 47.
Now the queens left in the deck is 3.
[1]
SECTION – B
[2 × 5]
11. The ten, jack, queen, king and ace of
diamonds are removed from the deck of cards
and then shuffled. One card is then picked up
at random (i) What is the probability that the
card is the queen ? (ii) What is the probability
that the card is a face card ?
OR
A game consists of tossing a one rupee coin 3
times and noting its outcomes each time. Anjali
wins if all the tosses give the same result and
losses otherwise. Calculate the probability that
Anjali will lose the game.
So P (getting a queen) = 3/47.
If the jack, queen and king of diamonds are removed,
then the face cards left = 9.
So P (getting a face card) = 9/47.
[1]
OR
When a coin is tossed 3 times, the possible outcomes
[1]
are : HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
The favourable outcomes that Anjali will lose the game if
all the tosses do not give the same results i.e. all heads
or all tails are HHT, HTH, THH, HTT, THT, TTH.
So P (lose the game) = 6/8 = 3/4.
[1]
12. Any point on x – axis is P (x, 0).
Given P is at a distance of 5 units from the point A.
So PA = 5
2
2
2
12. Find the points on the x – axis, each of which
is at a distance of 5 units from the point A (5, 3).
2
⇒ PA = 25. ⇒ (x – 5) + (0 + 3) = 25.
⇒ (x – 5) = 25 – 9 = 16. ⇒ x – 5 = ± 4.
[1]
⇒ x = 5 ± 4 = 9 or 1.
Therefore, there are two points (1, 0) and (9, 0) which are
at a distance of 5 units from the given point.
[1]
QUESTION
ANSWER
13. PQR is a right angled triangle having
2
2
∠Q = 90°, If QS = SR, Show that PR = 4PS –
2
3PQ .
13. Given PQR is a right angled triangle having ∠Q = 90°
and QS = SR.
P
2
2
2
To show PR = 4PS – 3PQ .
2
2
2
In ∆ PQR, PR = PQ + QR
(by Pythagoras theorem)
2
2
2
2
⇒ PR = PQ + (2QS)
( ∵ QS = SR)
2
Q
S
2
⇒ PR = PQ + 4QS ……….. (1)
2
2
R
[1]
2
Again in ∆ PQS, PS = PQ + QS
2
2
2
⇒ QS = PS – PQ .
Putting in equation (1) we get
2
2
2
2
2
2
2
PR = PQ + 4(PS – PQ ) = PQ + 4PS – 4PQ
2
2
= 4 PS – 3PQ .
14. Prove the identity : tan
2
2
A – tan
2
B =.
sin 2 A − sin 2 B
.
cos 2 A.cos 2 B
2
14. L.H.S. = tan A – tan B =
=
sin 2
sin 2 A
cos 2 A
A.cos 2 B − sin 2 B.cos 2
−
[1]
sin 2 B
cos 2 B
A
cos 2 A.cos 2 B
=
sin 2 A(1 − sin 2 B) − sin 2 B(1 − sin 2 A)
cos 2 A.cos 2 B
=
sin 2 A − sin 2 A.sin 2 B − sin 2 B + sin 2 A.sin 2 B
cos 2 A.cos 2 B
[1]
[1]
sin 2 A − sin 2 B
cos 2 A.cos 2 B
2
2
15. We have f(x) = abx + (b – ac) x – bc.
2
2
= abx + b x – acx – bc
= bx (ax + b) – c (ax + b) = (ax + b) (bx – c).
The value of f(x) is zero when ax + b = 0 or bx – c = 0.
−b
c
⇒x=
or x = .
a
b
Therefore zeroes are –b/a or c/b.
[1]
−b c −b 2 + ac −(b 2 − ac)
Sum of the zeroes =
+ =
=
a b
ab
ab
coefficient of x
=−
[½]
coefficient of x 2
Product of zeroes
c −bc
cons tan t term
 −b  c
=
[½]
=   =− =
a ab coefficient of x 2
 a b
=
15. Find the zeroes of the quadratic polynomial f(x)
2
2
= abx + (b – ac) x – bc and verify the
relationship between the zeroes and their
coefficients.
SECTION – C
[3×10]
16. Prove that for any prime positive integer p, √p
is an irrational number.
SECTION – C
16. Let us assume that √p is rational.
Then there exists positive co-primes a and b such that
√p = a/b
[1]
⇒ p = a2 /b2
⇒ pb2 = a2
⇒ p divides a2
⇒ p divides a ......(1)
⇒ Let a = pc for some positive integer c
2
2 2
Now a = p c
⇒ p c = pb2
⇒ b2 = pc2
2 2
[1]
QUESTION
ANSWER
⇒ p divides b2
⇒ p divides b .......(2)
17. Solve the following system of linear equations
graphically :
From (1) and (2), p is a common factor of a and b
This contradicts that a and b are co-prime.
Hence √p is irrational.
5−x
17. x + 2y = 5, ⇒ y = 2
x
y
x + 2y = 5
1
2
5
0
x
y
Also find the points where the lines meet the x
– axis.
1
2
-2
0
(-3, 4)
[1]
-3
4
Again 2x – 3y = -4, ⇒ y =
2x - 3y = -4
[1]
2x + 4
3
[1]
4
4
Y
2x–3y =- 4
(4, 4)
A(1, 2)
x+2y=5
X'
O
(5, 6)B
C(-2, 0)
Y'
The Lines meet x – axis at (5, 0) and (-2, 0)
18.
L.H.S. =
18. Prove the identity :
cos2 θ
sin3 θ
+
= 1 + sin θ.cos θ
1 − tan θ sin θ − cos θ
OR
Prove the identity :
tan θ + sec θ − 1 1 + sin θ
=
tan θ − sec θ + 1
cos θ
X
[½]
[½]
cos2 θ
sin3 θ
+
1 − tan θ sin θ − cos θ
cos2 θ
sin3 θ
+
sin θ sin θ − cos θ
1−
cos θ
[1]
=
cos3 θ
sin3 θ
cos3 θ − sin3 θ
−
=
cos θ − sin θ cos θ − sin θ
cos θ − sin θ
[1]
=
(cos θ − sin θ)(cos2 θ + sin2 θ + cos θ.sin θ)
(cos θ − sin θ)
[1]
=
= 1 + cos θ . sin θ = R. H. S.
OR
tan θ + sec θ − 1
L.H.S. = tan θ − sec θ + 1
(sec θ + tan θ) − (sec 2 θ − tan2 θ)
(1 − sec θ + tan θ)
(sec θ + tan θ) − (sec θ + tan θ)(sec θ − tan θ)
=
(1 − sec θ + tan θ)
(sec θ + tan θ){1 − (sec θ − tan θ)}
=
(1 − sec θ + tan θ)
(sec θ + tan θ)(1 − sec θ + tan θ)
=
(1 − sec θ + tan θ)
= sec θ + tan θ
=
[1]
[1]
QUESTION
ANSWER
2
19. In an A.P, the sum of first n terms is
3n
5n
+
.
2
2
th
Find its 25 term.
1
sin θ 1 + sin θ
+
=
[1]
cos θ cos θ
cos θ = R.H.S.
19. Let Sn denote the sum of n terms of an AP whose nth
term is an.
=
3n2 5n
+
2
2
3
5
⇒ Sn −1 = (n − 1)2 + (n − 1)
2
2
We have Sn =
[1]
 3n
5n   3
5

∴ Sn − Sn −1 = 
+  −  (n − 1)2 + (n − 1)
[1]
2   2
2

 2
3
5
⇒ an = [n2 − (n − 1)2 ] + [n − (n − 1)]
2
2
3
5
⇒ an = (2n − 1) +
2
2
3
5
⇒ a25 = × 49 + = 76
[1]
2
2
20. Given well is of diameter 3 m and 14 m deep. Hence its
radius = 3/2 m
2
Volume of the earth taken out from the well = πr h
3 3
= π × × × 14
2 2
22 3 3
=
× × × 14 = 99 m3
[1]
7 2 2
The earth taken out of the well has been spread evenly
all around it in the shape of a circular ring of width 4m to
form an embakment.
2
2
Area over which the earth is spread out = π (r1 – r2 )
22
2
2
= 7 (5.5 − 1.5 )
[1]
2
20. A well of diameter 3 m is dug 14 m deep. The
earth taken out of it has been spread evenly all
around it in the shape of a circular ring of width
4m to form an embankment. Find the height of
the embankment.
OR
Find the area of the shaded region in figure, if
PQ = 24 cm, PR = 7 cm and O is the centre of
the circle.
Q
=
O
22  3025 225  22
=
× 28 = 88m2
−
7
7  100 100 
∴ Height of embankment = 99/88 = 9/8 = 1.125 m.
[1]
OR
Q
R
P
O
R
P
In figure, given PQ = 24 cm, PR = 7 cm.
Here, ∠P is a right angle. [∵ Angle in a semicircle is
right angle]
∴ By Pythagoras Theorem, we have
2
2
2
2
2
2
QR = PQ + PR = (24) + (7) = 576 + 49 = 625 = (25)
⇒ QR = 25 cm.
[1]
Since, QR is a diameter of a circle (as passes through
the centre O of the circle),
Therefore, radius of a circle (r) = 25/2 cm.
Now, Area of the shaded region = Area of semicircle –
Area of right ∆ QPR.
[1]
1 2 1
= πr − (Base × Height)
2
2
QUESTION
ANSWER
2
1 22  25 
1
× ×
− × 7 × 24 [ ∵ Base = PR = 7 cm,
2 7  2 
2
Height = PQ = 24 cm]
11× 25 × 25

=
− 7 × 12 cm 2
7
×
4


=
21. Construct a triangle of sides 4 cm, 5 cm and 6
cm and then a triangle similar to it whose sides
are 2/3 of the corresponding sides of the first
triangle.
 6875 2352 
2  6875 − 2352  cm 2
=
−
 cm = 

28
28
28




4523
=
cm 2
[1]
28
21. Steps of Construction :
1.
Draw a line segment BC = 6 cm.
2.
With B as centre and radius = 4 cm, draw an arc.
3.
Again, C as centre and radius = 5 cm, draw another
are cutting the arc in step 2 at A. Then, ABC is the
required triangle.
4.
Draw any ray BX making an acute angle with BC on
the side opposite to the vertex A.
5.
Locate 3 points (the greater of 2 and 3 in 2/3) B1, B2
and B3 on BX so that BB1 = B1B2 = B2B3.
6.
Join B3C and draw a line through B2 parallel to B3C,
to intersect BC at C'.
7.
Draw a line through C' parallel to the line CA to
intersect BA at A'.
Then, ∆ A'BC' is the required triangle.
A
A'
5 cm
4 cm
B
6 cm
C
C'
B1
B2
X
B3
22. If the diagonals of a quadrilateral divide each
other proportionally, prove that it is a
trapezium.
[3]
22. Given : A quadrilateral ABCD whose diagonals AC and
AO BO
BD intersect each other at O such that
=
.
OC OD
To prove : Quadrilateral ABCD is a trapezium, i.e., AB ||
DC.
Construction : Draw OE || BA, meeting AD in E.
[1]
D
C
E
O
A
Proof : In ∆ ABD, we have OE || BA.
AE BO
So,
=
………….. (1) [By BPT]
ED OD
B
QUESTION
ANSWER
AO BO
=
………….. (2) [Given]
OC OD
AE AO
From (1) and (2), we get
=
[1]
ED OC
Thus, in ∆ DCA, O and E are points on AC and AD
AE AO
respectively such that
=
ED OC
Therefore, by the converse of BPT, we have EO || DC.
But OE || BA
∴ DC || BA ⇒ AB || DC.
Hence, ABCD is trapezium.
[1]
23. Let the speed of the steam be x km/h
Speed of the boat in still water = 5 km/h
But
23. Swati row her boat at a speed of 5 km/h in still
water. If it takes her 1 hour more to row the
boat 5.25 km upstream than to return
downstream, find the speed of the stream.
∴ Speed of the boat upstream = (5 – x) km/h.
Speed of the boat down stream = (5 + x) km/h
5.25 5.25
−
= 1 (given)
5−x 5+x
[1]
1 
21  5 + x − 5 + x 
 1
⇒ (5.25) 
−
 = 1, ⇒

 =1
4  (5 − x)(5 + x) 
5 − x 5 + x 
⇒ 21 × 2x = 4(25 – x2), ⇒ 21 x = 50 – 2x2
⇒ 2x2 + 21 x – 50 = 0, ⇒ 2x2 + 25 x – 4 x – 50 = 0 [1]
⇒ (2x + 5) (x – 2) = 0, ⇒ Either 2x + 5 = 0 or x – 2 = 0
⇒ x = -5/2 or x = 2, We reject x = -5/2 ⇒ x = 2
∴The speed of the stream = 2 km/h.
[1]
24. Find the lengths of the medians of the triangles
whose vertices (1, -1), (0, 4) and (-5, 3).
24. Let the three vertices of the triangle ABC be A(1, -1), B
(0, 4) and C(-5, 3), respectively. Let D, E, F be the midpoints of sides BC, CA and AB respectively.
Then, coordinates of points D, E and F are
 0 − 5 4 + 3   −5 + 1 3 − 1
 1 + 0 −1 + 4 
 2 , 2  ,  2 , 2  and  2 , 2 

 



 5 7
 1 3
i.e.  − ,  ,( −2,1) and  , 
 2 2
 2 2
 5 7
1 3
i.e., D  − ,  ,E( −2,1) and F  ,  , respectively.
 2 2
2 2
A (1, -1)
∴ Length of the median AD
2
 5 
7 
=  − − 1 +  + 1
 2 
2 
2
 7
9
= −  +  
 2
 2
2
F
2
49 81
130
130
+
=
=
4 4
4
2
Length of the median BE
=
E
B (0, 4)
D
= ( −2 − 0) 2 + (1 − 4) 2 = 4 + 9 = 13
and length of the median CF
2
2
[1]
2
1

3

 11
 3
=  + 5 +  − 3 =   +  − 
2

2

2
 2
C (-5, 3)
[1]
2
121 9
121 + 9
130
130
+ =
=
=
4 4
4
4
2
Hence, the lengths of the medians of the triangle whose
QUESTION
ANSWER
25. The coordinates of the centroid of a triangle
are (1, 4) and two of its vertices are (-8, 6) and
(9, 5). Find the third vertex. Also, find the
coordinates of the centroid of the triangle when
the third vertex is (2, 4).
OR
Find the point on the x – axis which is
equidistant from (2, -5) and (-2, 9).
SECTION – D
vertices are (1, -1), (0, 4) and (-5, 3) are
130
130
, 13 and
.
[1]
2
2
25. Let the two vertices of the triangle be A (-8, 6) and B (9,
5) and let third vertex of the triangle ABP be P(x, y), the
centroid of the triangle is given as G(1, 4).
[1]
x + x2 + x3
y + y2 + y3
∴ 1
=1
and 1
=4
3
3
−8 + 9 + x
6+5+y
⇒
=1
and
=4
3
3
⇒1+x=3
and 11 + y = 12
⇒x=3–1
and y = 12 – 11
⇒x=2
and y = 1
Thus, the coordinates of the third vertex are (2, 1). [1]
Let the centroid of the triangle ABC be G1 (x, y). Also, let
the three vertices of the triangle ABC are A (-8, 6),
B (9, 5), C (2, 4) respectively.
( −8) + 9 + 2
6+5+4
Then, x =
and y =
3
3
⇒ x = 3/3
and y = 15/3
⇒x=1
and y = 5
Thus, the coordinates of the centroid are (1, 5).
[1]
OR
Let the required point P be on the x – axis, then its
ordinate is zero. Let the abscissa of the point P be x.
∴ The coordinates of the point P are (x, 0).
Let the given points be A(2, -5) and B(-2, 9).
It is given that the point P(x, 0) on x – axis is equidistant
from A(2, -5) and B (-2, 9).
[1]
∴ AP = BP (given)
2
2
⇒ AP = BP
2
2
2
2
⇒ (x – 2) + (0 + 5) = (x + 2) + (0 – 9)
2
2
⇒ (x – 4x + 4) + 25 = (x + 4x + 4) + 81
⇒ -4x + 29 = 4x + 85
[1]
⇒ - 4x - 4x = 85 – 29
⇒ -8x = 56, ⇒ x = -7.
Thus, the required point on the x – axis is (-7, 0).
[1]
[6×5]
26. Prove that : The ratio of areas of similar
triangle is equal to the ratio of the squares of
the corresponding sides.
Using the above theorem, prove that the area
of equilateral triangle described on the side of
a square is half the area of the equilateral
triangle described its diagonal.
SECTION – D
26. Given: Triangles ABC and PQR such that ∆ ABC ~ ∆
PQR
( ar ( ABC ) ) =  AB 2 =  BC 2 =  CA 2
 QR 
 RP 
To prove: ( ar (PQR ) )  PQ 




Construction: Draw altitudes AM and PN of the
triangles.
P
A
B
M
C
Q
N
1
Proof: Now, ar(ABC) = 2 BC × AM
1
and ar(PQR) = 2 QR × PN
R
[1]
QUESTION
ANSWER
 1
So,

( ar ( ABC ) ) =   2  BC × AM  = (BC × AM)
( ar (PQR ) )   1  QR × PN  ( QR × PN) (1)
 
 2 

Now, in ∆ ABM and ∆ PQN,
∠B = ∠Q (As ∆ ABC ~ ∆ PQR)
[1]
∠M = ∠N (Each is of 90°)
(AA similarity criterion)
So, ∆ ABM ~ ∆ PQN
AM AB
Therefore, PN = PQ
(2)
Also, ∆ ABC ~ ∆ PQR
(Given)
AB BC CA
So, PQ = QR = RP
(3)
( ar ( ABC ) ) AB AM
Therefore, ( ar (PQR ) ) = PQ × PN [From (1) and (3)]
AB AB
= PQ × PQ
 AB 
=

 PQ 
[1]
[From (2)]
2
Using (3),
( ar ( ABC ) ) =  AB 2 =  BC 2 =  CA 2
( ar (PQR ) )  PQ   QR   RP 
nd
2 Part :
Given: ABCD is a square and ∆ABE is an equilateral
triangle on side AB and ∆BDF is also an equilateral
triangle on side BD i.e. diagonal of square ABCD.
A
D
E
C
A
F
1
To prove: ar (∆ABE) = 2 ar (∆BDF)
[1]
Proof: In a right ∆ABD.
2
2
2
AB + AD = BD
2
2
2
Or AB + AB = BD
( ‹ ABCD is a square)
2
2
⇒ 2AB = BD
or BD = √2AB
…(i)
∆ABE and ∆DBF are equilateral triangles.
∴ ∆ABE ~ ∆DBF
We know that areas of two similar triangles are in the
ratio of the squares of their corresponding sides.
area ( ∆ABE ) AB2
∴ area ( ∆DBF ) = BD2
[1]
From (i) BD = √2AB
ar ( ∆ABE )
AB2
AB2
1
=
=
=
2
2
∴ area ( ∆DBF )
2
2AB
2AB
(
)
1
or ar (∆ABE) = 2 ar(∆DBF).
QUESTION
ANSWER
27. An open metal bucket is in the shape of a
frustum of a cone, mounted on a hollow
cylindrical base made of the same metallic
sheet. The diameters of the two circular ends
of the bucket are 45 cm and 25 cm, the total
vertical height of the bucket is 40 cm and that
of the cylindrical base is 6 cm. Find the area of
the metallic sheet used to make the bucket,
where we do not take into account the handle
of the bucket. Also, find the volume of water
27. The total height of the bucket = 40 cm, which includes
the height of the base. So, the height of the frustum of
the cone = (40 – 6) cm = 34 cm.
Therefore, the slant height of the frustum,

22 
the bucket can hold.  Take π = 7 


l = h2 + (r1 − r2 )2 ,
where r1 = 22.5 cm, r2 = 12.5 cm and h = 34 cm
[1]
2
2
So, l = 34 + (22.5 − 12.5) cm
= 342 + 102 = 35.44 cm
[1]
The area of metallic sheet used = curved surface area of
frustum of cone
+ area of circular base
[1]
+ curved surface area of cylinder
2
= [π × 35.44 (22.5 + 12.5) + π × (12.5)
2
+ 2π × 12.5 × 6] cm
22
2
=
7 (1240.4 + 156.25 + 150) cm
2
28. Two poles of equal heights are standing
opposite each other on either side of the road,
which is 80m wide. From a point between them
on the road, the angles of elevation of the top
of the poles are 60° and 30° respectively. Find
the height of the poles and the distance of the
point from the poles.
OR
A man standing on the deck of a ship, which is
10 m above the water level, observes the
angle of elevation of the top of a hill as 60° and
angle of depression of the base of the hill as
30°. Find the distance of the hill from the ship
and height of the hill.
= 4860.9 cm
[1]
Now, the volume of water that the bucket can hold (also,
known as the capacity of the bucket)
π×h
2
2
=
3 × (r1 + r2 + r1r2)
22 34
2
2
3
=
×
7
3 ×[(22.5) + (12.5) + 22.5 × 12.5] cm
22 34
3
=
×
7
3 ×943.75 = 33615.48 cm
= 33.62 litres (approx.)
[1]
28. Let AB = DE = Poles
Point C is on the road such that ∠BCA = 60°, and ∠DCE
= 30°
To find AB = DE & BC, CD
[1]
From ∆ ABC, tan 60° = AB/BC
AB
⇒ 3=
BC
⇒ AB = BC √3 = x √3 ………. (i)
[1]
DE
AB
From ∆ CDE, tan 30° = CD = 80 − x
1
AB
⇒
=
[1]
3 80 − x …….. (ii)
[1]
Putting the value of AB from (i) in (ii), we get
1
x 3
=
3 80 − x
⇒ 3x = 80 – x ⇒ 4x = 80
⇒ x = 20
∴ BC = 20 cm, CD = 60 cm
Point C is 20 m from B and 60 m from D.
[1]
QUESTION
ANSWER
Height of poles AB = x√3
= 20 × 1.732 = 34.64 m
OR
Let the man is standing on deck of a ship AB = 10 m at a
point A above the water level BC.
Let CD be the hill. It is given that the angle of elevation of
the top D of a hill CD observed from the point A is 60°
and the angle of depression of the base C of the hill is
30° observed from A.
[1]
Draw AE ⊥ CD. Let AE = x m and DE = y m.
Then, ∠DAE = 60° and ∠EAC = ∠BCA = 30°.
In right ∆ EAD, we have tan 60° = DE/EA.
y
⇒ 3 = ⇒ y = √3 x ……….. (1)
[1]
x
Again, in right ∆ ABC, we have tan 30° = AB/AC.
1 10
⇒
=
[ ∵ BC = AE = x m]
3 x
29. A train travels 360 km at a uniform speed. If
the speed had been 5 km/hr more, it would
have taken 1 hour less for the same journey.
Find the speed of the train.
OR
⇒ x = 10√3 ……………. (2)
[1]
From (1) and (2), we get
y = √3 × 10√3 = 30 cm……………. (3)
[1]
Also, CD = DE + EC
⇒ CD = y + 10 [ ∵ DE = y m and EC = 10 m]
⇒ CD = 30 + 10 = 40 m. [using (3)]
Hence, the distance of the hill from the ship is AE = x =
10√3 (10 × 1.732 m) = 17.32 m and the height of the hill
CD = 40 m.
[1]
29. Let original speed = x km/hr.
Increased speed = (x + 5) km/hr
[1]
Distance traveled = 360 km
The according to question,
360 360
360x + 1800 − 360x
−
= 1⇒
=1
[2]
x
x+5
x(x + 5)
2
A two digit number is such that the product of
its digits is 15. If 18 is added to the number,
the digits interchange their places. Find the
number.
⇒ 1800 = x(x + 5) = x + 5x
2
⇒ x + 5x – 1800 = 0
2
⇒ x + 45 x – 40x – 1800 = 0
⇒ x(x + 45) – 40 (x + 45) = 0
[1]
⇒ (x + 45) (x – 40) = 0
Either x + 45 = 0 ⇒ x = -45 Rejecting negative values
or x – 40 = 0 ⇒ x = 40
∴ Original speed = 40 km/hr.
[1]
OR
Let the unit place digit be x and ten's place digit be y,
then according to the given condition xy = 15.
⇒ y = 15/x ………………. (1)
 15 
Original number = 10y + x = 10   + x
[1]
 x
 150 
=
 + x [using (1)]
 x 
When digits interchange their places, then New number =
10x + y = 10x + 15/x [using (1)]
According to the given condition, we have original
number + 18 = New number
QUESTION
ANSWER
150
15
+ x + 18 = 10x +
x
x
 15 150 
⇒ (10x − x) +  −
− 18 = 0
x 
 x
⇒
[1]
135
− 18 = 0
x
2
⇒ 9x – 135 – 18x = 0
2
⇒ 9(x – 2x – 15) = 0
2
⇒ x – 2x – 15 = 0
[1]
2
⇒ x – 5x + 3x – 15 = 0
⇒ x(x – 5) + 3(x – 5) = 0
⇒ (x – 5) (x +3) = 0
⇒ Either x – 5 = 0 or x + 3 = 0
⇒ Either x = 5 or x = -3
[1]
⇒ x = 5[ ∵ Digits can't be negative]
Thus, unit place digit = 5 and ten's place digit =
15 15
=
=3
x
5
Hence, the required number is 35.
[1]
⇒ 9x −
30. Draw the two types of cumulative frequency
curves and determine the median.
Wages in Rs.
No. of Workers
80 - 90
9
90 - 100
17
100 - 110
19
110 – 120
45
120 – 130
33
130 – 140
15
140–150
12
30.
C.I.(wages)
80 – 90
90 – 100
100 – 110
110 – 120
120 – 130
130 – 140
140 – 150
No. of
Cum.
workers Frequency <
frequency
type
9
9
17
26
19
45
45
90
33
123
15
138
12
150
Cum.
Frequency >
type
150
141
124
105
60
27
12
We plot both these types :
The two curves shown in the figure give more than type
and less than type.
They must at P where frequency is 75 and class interval
is at 117.
∴ Median is Rs. 117.
[1
QUESTION
ANSWER
iBOOKS SAMPLE PAPER – 02
[FOR XTH CBSE BOARD]
QUESTION
ANSWER
1.
SECTION – A
2
2
3825 = 3 × 5 × 17.
[1]
2.
1.
[1]
3.
Here
SECTION – A
[1×10]
1.
Write 3825 as a product of prime factors.
2.
In figures, the graph of some polynomial p(x) is
given. Find the number of zeroes of the
polynomial.
Y
c 1 −5 5
=
=
c 2 −k k
4
3
For a pair of linear equations to have no solution, if
2
a1 b1 c 1
3 1 5
=
≠
⇒ = ≠ . ⇒ k ≠ 10.
6 2 k
a2 b2 c 2
1
X'
-4 -3 -2 -1 O
1
1 2 3
4
5
X
2
4.
3
Y'
4.
5.
Find the value of k, so that the following
system of equations has no solution : 3x – y –
5 = 0; 6x – 2y = k.
5.
X
P
Y
Given, 5 tan α = 4. ⇒ tan α = 4/5.
=
5 sin α − 3 cos α
If 5 tan α = 4, Find the value of
.
5 sin α + 2 cos α
In the given figure, PQ || YZ. Find the length of
QX.
4
5 −3 4−3 1
4
=
= .
4
4+2 6
5 +2
5
XY XZ
=
XP XQ
Now in ∆ XYZ and ∆XPQ, we have
6 cm
Q
XY XZ
=
and ∠X = ∠X.
XP XQ
Z
9 cm
[1]
In ∆ XYZ, we have PQ || YZ.
⇒
cm
33 cm
[1]
5 sin α
5 sin α − 3 cos α cos α −3 5 tan α − 3
Now
=
=
5 sin α + 2cos α
5 sin α
5 tan α + 2
cos α +2
4
3.
a1 3 1 b1 −1 1
= = ,
=
= .
a 2 6 2 b 2 −2 2
6.
For what values of k, the quadratic equation
2
12 x + 4kx + 3 = 0 has equal roots ?
7.
In the given figure, PQ and PR are tangents to
the circle drawn from an external point P. LM is
a third tangent touching the circle at M. If PQ =
15 cm, KM = 4 cm, Find the length PL.
[1]
∴ ∆ XYZ ~ ∆ XPQ
⇒
6.
YZ XZ
9
6
=
⇒ =
⇒ XQ = 2 cm.
PQ XQ
3 XQ
[1]
2
For equal roots D = b – 4ac = 0
2
⇒ (4k) – 4.12.3 = 0
[1]
2
⇒ 16k – 144 = 0
L
P
k
M
Q
2
⇒ k = 144/16, ⇒ k = ± 3.
O
R
7.
[1]
We have PQ = PR = 15 cm. and KL = KM = 4 cm.
Now length of PL = PQ – LQ = 15 – 4 = 11 cm.
[1]
QUESTION
Cumulative frequency (No. of students)
8.
9.
ANSWER
A student draws a cumulative frequency curve
for the marks obtained by 40 students of a
class as shown below. Find the median
marks obtained by the student of the class.
Y
40
30
20
10
O
X
10
20
30
40 50 60 70 80
Upper limit of class intervals (Marks)
Two dice are thrown at the same time. What is
the probability that the sum of two numbers
appearing on the top of the dice is 13 ?
10. A cone, a cylinder and a hemisphere are of
equal base and have the same height. What is
the ratio of their volumes ?
SECTION – B
[2×5]
11. Without drawing the graphs, state whether the
following pair of linear equations will represent
intersecting lines, coincident lines or parallel
lines :
6x – 3y = - 10
2x – y + 9 = 0
OR
For which value of k will the following system
of linear equations have no solution ?
3x + y = 1
(2k – 1) x + (k – 1) y = 2k + 1.
12. If A (-2, -1), B (x, 0), C (4, y) and D (1, 2) are
the vertices of a parallelogram taken in order,
find x and y.
n 40
=
= 20
2 2
Locate 20 on the y – axis. From this pint, draw a line
parallel to the x – axis cutting the curve at a point. From
this point, draw a perpendicular to the x – axis. The point
of intersection of this perpendicular with the x – axis
gives the median of the data.
Hence, the median marks obtained by the student of the
class = 54 marks.
[1]
9.
0.
[1]
10. Let r be the radius of cone, cylinder and the hemisphere
respectively and 'h' be the height of the cone. Cylinder
and the hemisphere respectively.
Then w = r.
2
V1 = Volume of the cone = 1/3 πr h.
2
V2 = Volume of the cylinder = πr h.
3
V3 = Volume of the hemisphere = 2/3 πr .
2
2
3
∴ V1 : V2 : V3 = 1/3 πr h : πr h : 2/3 πr
= h : 3h : 2r = 1 : 3 : 2.
[1]
SECTION – B
a
6
11. Here 1 = = 3 .
a2 2
8.
Here n = 40 ⇒
b 1 −3
=
=3
b 2 −1
[1]
c 1 −10 10
=
=
c2
−9
9
a1 b1 c 1
=
≠
, the given pair of linear equations
a2 b2 c 2
represents parallel lines.
[1]
OR
a
b
c
The system of equation has no solution, if 1 = 1 ≠ 1
a2 b2 c 2
Since
3
1
1
=
≠
2k − 1 k − 1 2k + 1
3
1
1
1
⇒
=
and
≠
[1]
2k − 1 k − 1
k − 1 2k + 1
3
1
Now
=
2k − 1 k − 1
⇒ 3k – 3 = 2k – 1
⇒ k = 2.
Hence, the given system of linear equations will have no
solution, if k = 2.
[1]
12. We know that the diagonals of a parallelogram bisect
each other. Therefore, the coordinates of the mid-point of
the diagonal AC = the coordinates of the mid-point of the
diagonal BD.
 −2 + 4 −1 + y   x + 1 0 + 2 
⇒
,
=
,
2   2
2 
 2
⇒
 y − 1  x + 1 
⇒  1,
,1
=
 2   2

x +1
y −1
⇒
= 1and
=1
2
2
⇒ x + 1 = 2 and y – 1 = 2
⇒ x = 1 and y = 3
Hence, x = 1 and y = 3.
[1]
[1]
QUESTION
13. Prove that (1 + cot A – cosec A) (1+tan A +
sec A) = 2.
ANSWER
13. (1 + cot A – cosec A) (1 + tan A + sec A)
1 
sin A
1 
 cos A
1+
= 1 +
−
+
sin A sin A   cos A cos A 

 sin A + cos A − 1  cos A + sin A + 1
= 


sin A
cos A



(sin A + cos A − 1)(cos A + sin A + 1)
=
sin A.cos A
[1]
(sin A + cos A) 2 − 1 1 + 2 sin A.cos A − 1
=
sin A.cos A
sin A.cos A
2 sin A.cos A
=
= 2.
[1]
sin A.cos A
14. Given : ∆ ABC ~ DEF and AP and DQ are the medians.
=
14. Prove that the ratio of the areas of two similar
triangles is equal to the square of the ratio of
their corresponding medians.
To prove:
area ( ∆ABC )
area ( ∆DEF )
=
AP2
DQ2
D
A
B
P
C
E
Proof: ‹ ∆ABC ~ ∆DEF
AB BC AC
⇒
=
=
DE EF DF
AB BC
⇒
=
……….. (1)
DE EF
AB BP + PC
2 BP BP
=
⇒
=
=
DE EQ + QF 2 EQ EQ
Q
F
[1]
Now, in ∆ABP and ∆DEQ
AB BP
=
DE EQ
∠ABP = ∠DEQ
⇒ ∆ABP ~ ∆DEQ
Their corresponding sides must be proportional
AB BP AP
∴
=
=
……….. (2)
[1]
DE EQ DQ
AP BC
From (1) and (2),
=
DQ EF
We know that the areas of two similar triangles are in the
ratio of the square of their corresponding sides.
area ( ∆ABC ) BC2 AP2
∴
=
=
.
area ( ∆DEF ) EF2 DQ2
15. One card is drawn from a well shuffled deck of
52 playing cards. Find the probability of getting
(i) a non-face card. (ii) a black king or a red
queen.
15.
No. of face cards in a deck of playing cards = 12.
No. of non-face cards = 40.
40 10
∴ Probability of getting a non-face card =
=
. [1]
52 13
Since there are 2 black kings and 2 red queens in 52
playing cards ⇒ P (getting a black king or a red queen)
4
1
=
=
.
[1]
52 13
QUESTION
ANSWER
SECTION – C
[3×10]
16. Prove that 5 + √2 is irrational.
17. For what values of a and b, the following
system of linear equations have an infinite
number of solution ?
2x + 3y – 7 = 0.
(a – b) x + (a + b) y = 3a + b – 2.
18. Represent the following pair of equations
graphically and write the coordinates of points
where the lines intersect y – axis :
x + 3y = 6
2x – 3y = 12
SECTION – C
16. Let 5 + √2 be rational.
a
⇒ 5 + 2 = , [a, b are coprimes and b ≠ 0]
[1]
b
a
⇒ 2 = −5
b
a − 5b
⇒ 2=
.
b
a
Since a and b are integers, we get − 5 is rational.
b
So √2 is rational.
[1]
But this contradicts the fact that √2 is irrational.
This contradiction has arisen due to our wrong
assumption.
So 5 + √2 is irrational.
[1]
17. Given 2x + 3y – 7 = 0
(a – b) x + (a + b) y = 3a + b – 2
2
3
7
For infinitely many solutions
=
=
[1]
a − b a + b 3a + b − 2
2
3
3
7
=
Again
=
a −b a +b
a + b 3a + b − 2
⇒ a = 5b …… (1)
⇒ 2a – 4b = 6 ………..(2) [1]
Solving (1) and (2), we get
b=1
⇒a=5
[1]
18. Two solutions of each of the equations :
x + 3y = 6 …………. (1)
2x – 3y = 12 ………. (2)
are given in tables below :
[1]
x + 3y = 6
2x – 3y = 12
x
6
0
x
6
0
y
0
2
y
0
-4
A
B
A
C
4
Y
3
X'
B(0,2)
1
-6 -5 -4 -3 -2 -1 O
-1
x+3y=6
X
1
-2
-3
B(0,-4)
19. Determine an AP whose 3rd term is 16 and
when 5th term is subtracted from 7th term, we
get 12.
2 3
4 5 A(6,0)
2x-3y=12
Y'
[1.5]
The first line x + 3y = 6 intersects the y – axis at B (0, 2).
The second line 2x – 3y = 12 intersects the y – axis at C
(0, -4).
[0.5]
19. Let the first term and common difference of AP be a and
d, respectively.
Let a3, a5 and a7 denote the 3rd term, 5th term and 7th
term of an AP, then
a3 = a + (3 – 1)d
[∵ n = 3]
⇒ 16 = a + 2d ………. (1) [∵ a3 = 16 (given)]
It is given that the difference of 5th term from the 7th
term is 12.
[1]
QUESTION
20. Prove that :
1 + cos A
sin A
+
= 2 cos ec A .
sin A
1 + cos A
OR
Prove that : (cosec A – sin A) (sec A – cos A) =
1
.
tan A + cot A
ANSWER
∴ a7 – a5 = 12 (given)
⇒ [a + (7 – 1)d] – [a+(5 – 1)d] = 12
⇒ (a + 6d) – (a + 4d) = 12
[1]
⇒ 2d = 12
⇒ d = 6 ………. (4)
From (1) and (4), we get a + 2 × 6 = 16.
⇒ a = 16 – 12 = 4
Thus, the AP is 4, 10, 16, 22, 28, 34, 40, …….. [1]
1 + cos A
sin A
20. L.H.S. =
+
sin A
1 + cos A
=
(1 + cos A) 2 + sin 2 A
sin A(1 + cos A)
[1]
=
1 + 2 cos A + cos 2 A + sin 2 A
1 + 2 cos A + 1
=
sin A(1 + cos A)
sin A(1 + cos A)
[1]
2 + 2cos A
2 + (1 + cos A)
2
=
=
sin A(1 + cos A) sin A(1 + cos A) sin A
= 2 cosec A = R.H.S.
OR
We have
L.H.S. = (cosec A – sin A) (sec A – cos A)
 1
 1

=
− sin A  
− cos A 
 sin A
  cos A

=
[1]
[1]
 1 − sin 2 A   1 − cos 2 A 
=


 sin A   cos A 
21. Find the area of the quadrilateral whose
vertices taken in order are A(-5, -3), B(-4, -6),
C(2, -1) and D(1, 2).
 cos 2 A   sin 2 A 
2
2
=
[1]
  cos A  [ ∵ sin A + cos A = 1]
sin
A



= cos A sin A
1
Now, R.H.S. =
tan A + cot A
1
1
=
=
sin A cos A sin 2 A + cos 2 A
+
cos A sin A
sin A cos A
2
2
= sin A cos A
[∵ sin A + cos A = 1]
Hence, L.H.S. = R.H.S.
[1]
21. By joining A to C, we will get two triangles ABC and ACD.
Now, the area of ∆ ABC
1
= [x1(y2 – y3) + x2(y3 – y1) + x3 (y1 – y2)]
2
1
= [(-5)(-6 + 1) + (-4) (-1 + 3) + 2 (-3 + 6)]
[1]
2
D(1,2)
1
C(2,-1)
= [25 – 8 + 6]
2
23
=
sq. units.
2
Also, the area of ∆ ACD A(-5,-3)
B(-4,-6)
1
= [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
[1]
2
1
= [(-5)(-1 -2) + 2(2 + 3) + 1 (-3 + 1)]
2
1
= [15 + 10 – 2]
2
23
=
sq. units.
2
QUESTION
22. Show that the points A(2, -2), B (14, 10), C(11,
13) and D(-1, 1) are the vertices of a rectangle.
OR
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). Find the
ratio of this area to the area of the given
triangle.
ANSWER
So, the area of the quadrilateral ABCD
= area of ∆ ABC + area of ∆ ACD
23 23
=
+
sq. units = 23 sq. units.
[1]
2
2
22. Coordinates of the given points are A(2, -2), B (14, 10),
C(11, 13) and D(-1, 1) respectively.
(14 − 2) 2 + (10 + 2) 2
Then AB =
= (12) 2 + (12) 2 = 144 + 144
= 288 = 144 × 2 = 12 2
BC =
(11 − 14) 2 + (13 − 10) 2
[1]
= ( −3) 2 + (3) 2 = 9 + 9
= 9 + 9 = 18 = 9 × 2 = 3 2
CD =
( −1 − 11) 2 + (1 − 13) 2
= ( −12) 2 + ( −12) 2
D(-1, 1)
C(11, 13)
A(2, -2)
B(14, 10)
= 144 + 144 = 288
= 144 × 2 = 12 2
DA =
(2 + 1) 2 + ( −2 − 1) 2
= (3) 2 + ( −3) 2 = 9 + 9
= 18 = 9 × 2 = 3 2
Thus, AB = CD and BC = DA.
⇒ Opposite sides are equal in length.
Now, AC =
[1]
(11 − 2) 2 + (13 + 2) 2
= (9) 2 + (15) 2 = 81 + 225 = 306
BD =
( −1 − 14) 2 + (1 − 10) 2
= ( −15) 2 + ( −9) 2 = 225 + 81 = 306
Thus, Length of diagonal AC = Length of diagonal BD.
Since, opposite sides are equal and diagonals are also
equal.
Hence, the points A(2, -2), B (14, 10), C(11, 13) and
D (-1, 1) are the vertices of a rectangle.
[1]
OR
Vertices of ∆ABC are (0, -1), (2, 1) and (0, 3)
respectively.
A (0, -1)
F
B (2, 1)
E
D
C (0, 3)
[1]
Their mid-points of sides are :
 2 + 0 1+ 3 
 0 + 0 3 − 1
D=
,
,
 = (1,2) , E = 
 = (0,1)
2 
2 
 2
 2
 0 + 2 −1 + 1 
F=
,
 = (1,0)
2 
 2
Area of ∆ DEF = ½ [1(1-0)+0(0-2)+1(2-1)]
[1]
QUESTION
ANSWER
23. O is any point inside a rectangle ABCD. Prove
2
2
2
2
that OB + OD = OA + OC .
= ½ [1+1] = 1 sq. units.
Area of ∆ABC = ½ [0(1 -3)+2(3 +1) +0(-1-1)]
= ½ (0 + 8+ 0) = 8/2 = 4 sq. units.
Ratio of areas = 1 : 4.
[1]
23. Through O, draw a line segment PQ || BC so that P lies
on AB and Q lies on DC.
D
A
D
A
P
Q
O
O
B
B
C
Now, PQ || BC
Therefore, PQ ⊥ AB and PQ ⊥ DC [∠B = 90° and ∠C =
[1]
90°]
So, ∠BPQ = 90° and ∠CQP = 90°
Therefore, BPQC and APQD are both rectangles.
2
2
2
In ∆ OPA, we have OA = OP + PA …… (1) [By
Pythagoras Theorem]
[1]
C
2
2
2
In ∆ OCQ, we have OC = OQ + CQ
Adding (1) and (2), we get
2
2
2
2
2
2
OA + OC = OP + PA + OQ + CQ
2
2
2
2
= OP + DQ + OQ + PB [∵ AP = DQ and QC = PB]
2
2
2
2
= (OP + PB ) + (OQ + DQ )
2
2
= OB + OD
2
[In ∆ OPB and ∆ OQD by Pythagoras Theorem, OP +
2
2
2
2
2
PB = OB and OQ + DQ = OD ]
2
2
2
2
Hence, OA + OC = OB + OD .
[1]
24. Draw a pair of tangents to a circle of radius 5
cm which are inclined to each other at an
angle of 60°.
24.
[1]
Steps of construction :
1.
Take a point O on the plane of the paper and draw
a circle of radius OA = 5 cm.
2.
Extend OA to B such that OA = AB = 5 cm.
3.
With A as centre draw a circle of radius OA = AB =
5 cm. Suppose it intersect the circle drawn in step 1
at the points P and Q.
4.
Join BP and BQ.
[1]
Then BP and BQ are the required tangents which are
inclined to each other at an angle of 60° (see figure).
For justification of the construction :
In ∆ OAP, we have
QUESTION
ANSWER
OA = OP = 5 cm (= Radius)
Also, AP = 5 cm (= Radius of circle with centre A).
∆ OAP is equilateral
⇒ ∠PAO = 60° ⇒ ∠BAP = 120°
In ∆ BAP, we have AB = AP and ∠BAP = 120°
∴ ∠ABP = ∠APB = 30°
Similarly we can prove that
∠ABQ = ∠AQB =30°
⇒ PBQ = 60°.
25. Find the area of the shaded region from figure,
if the diameter of the circle with centre O is 28
cm and AQ =
1
AB.
4
[1]
25. We have
AB = Diameter of a circle = 28 cm
∴ OA = OB = 14 cm [Each = radius of a circle]
1
It is given that AQ = AB
4
1
⇒ AQ =
× 28 = 7 cm
4
∴ AO = AQ + QO
⇒ 14 cm = 7 cm + QO
⇒ QO = (14 – 7) cm = 7 cm
[1]
Now, QB = QO + OB
⇒ QB = 7cm + 14 cm = 21 cm
Area of the shaded region
= Area of the semi-circle on diameter AQ (= 7 cm)
+ Area of the semi-circle on diameter QB (= 21 cm)
SECTION – D
[6×5]
26. Some students arranged a picnic. The budget
for food was Rs. 240. Because four students of
the group failed to go, the cost of food to each
student got increased by Rs. 5. How many
students went for the picnic ?
OR
A plane left 30 minutes late than its scheduled
time and in order to reach the destination 1500
km away in time, it had to increase the speed
 1  AQ  2 1  QB  2   π × AQ 2 π × QB 2 
=  π×
+

 + 2π× 2   = 
8
8
 2
 2 

  

π
2
2
2
= [(7) + (21) ] cm
[1]
8
π
2
= [49 + 441] cm
8
22 1
11
=
× × 490cm 2 = × 70cm 2
7 8
4
11
385
2
2
= × 35cm =
cm 2 = 192.5 cm .
[1]
2
2
SECTION – D
26. Let the original number of students be x.
Total cost of food for x students = Rs. 240.
[1]
∴ The cost of food for each student = Rs. 240/x.
It is given that 4 students of the group failed to go.
∴ The number of students go for a picnic = (x – 4)
[1]
240
Now, the cost of food for each student = Rs.
x−4
It is given that the four students of a group failed to go,
the cost of food to each student got increased by Rs. 5.
QUESTION
by 250 km/h from the usual speed. Find its
usual speed.
ANSWER
240 240
−
= 5(given)
[1]
x−4
x
240x − 240x + 960
⇒
=5
x(x − 4)
2
⇒ 960 = 5(x – 4x)
[1]
2
⇒ 192 = x – 4x
2
⇒ x – 4x – 192 =0
[1]
2
⇒ x – 16x + 12x – 192 = 0
⇒ x(x – 16) + 12(x -16) = 0
⇒ (x – 16) (x + 12) = 0
[1]
⇒ Either x = 16 or x = -12
⇒ x = 16, as x cannot be negative.
Hence, the number of students go for the picnic = x – 4,
i.e., 16 – 4 = 12.
[1]
OR
Let the usual speed of the plane be x km/h.
Then, time taken to cover 1500 km with usual speed =
1500
h
[1]
x
When the speed of a plane is increased by 250 km/h,
then the time taken to cover 1500 km with the speed of (x
1500
h.
[1]
+ 250) km/h =
x + 250
1500 1500
1
∴
−
= hour (or 30 minutes)
x
x + 250 2
1500 + 1500 × 250 − 1500x 1
⇒
=
[1]
x(x + 250)
2
∴
1500 × 250 1
=
x 2 + 250x 2
2
⇒ 3000 × 250 = x + 250x
[1]
2
⇒ x + 250x – 750000 = 0
2
⇒ x + 1000x – 750x – 750000 = 0
⇒ x(x + 1000) – 750(x + 1000) = 0
[1]
⇒ (x + 1000) (x – 750) = 0
⇒ Either x = 750 or x = -1000
[1]
⇒ x = 750, as x cannot be negative.
[1]
Hence, the usual speed of the plane is 750 km/h.
27. Let AB be the height of lighthouse = 75 m. Let C and D
be the position of the two ships approaching to a light
house such that the distance CD between the two ships =
x m. [1]
It is given that as observed from the top A of the light
house the angles of depression of the two ships at C and
D be 30° and respectively.
[1]
∴ ∠BCA = angle of depression = 30° and ∠BDA = angle
of depression = 45°.
⇒
27. As observed from the top of a 75 m high
lighthouse (from sea-level), the angles of
depression of two ships are 30° and 45°. If one
ship is exactly behind the other on the same
side of the light house, find the distance
between the two ships.
[1]
In right triangle ADB, we have tan 45° =
AB
DB
QUESTION
ANSWER
75
⇒ DB = 75 m.
DB
⇒ DB = 75 m …………. (1)
⇒1=
In right triangle ACB, we have tan 30° =
Using the above theorem, do the following : If
ABC is an equilateral triangle with AD ⊥ BC,
2
2
then AD = 3DC .
AB
CB
75
=
[ ∵ CB = CD + DB]
3 CD + DB
⇒ CD + DB = 75 × √3 [Using (1)]
[1]
⇒ x = [75√3 – 75]m
⇒ x = 75(√3 – 1) m
⇒ x = 75(1.73 – 1) m
[1]
⇒ x = 75 × 0.73 m [∵ √3 = 1.73]
⇒ x = 54.75 m
Thus, the distance between the two ships = 54.75 m. [1]
28. Given : A right triangle ABC, right angled at B.
2
2
2
To prove : (Hypotenuse) = (Base) + (Perpendicular)
2
2
2
i.e., AC = AB + BC
[1]
Construction : Draw BD ⊥ AC
⇒
28. Prove that in a right triangle, the square of the
hypotenuse is equal to the sum of the squares
of the other two sides.
[1]
1
[1]
Proof : ∆ ADB ~ ∆ ABC.
[If a perpendicular is drawn from the vertex of the right
angle of a right triangle to the hypotenuse then triangles
on both sides of the perpendicular are similar to the
whole triangle and to each other].
[1]
AD AB
So,
=
[Sides are proportional]
AB AC
2
⇒ AD. AC = AB . ……………. (1)
[1]
Also, ∆ BDC ~ ∆ ABC
CD BC
So,
=
[Sides are proportional]
BC AC
2
⇒ CD. AC = BC ………….. (2)
[1]
Adding (1) and (2), we have
2
2
AD. AC + CD. AC = AB + BC
2
2
⇒ (AD + CD). AC = AB + BC
2
2
⇒ AC . AC = AB + BC
2
2
2
Hence, AC = AB + BC
[1]
For the Second Part :
Let ABC be an equilateral triangle and AD ⊥ BC.
In ∆ ADB and ∆ ADC, we have
AB = AC
[Given]
QUESTION
29. An iron pillar has lower part in the form of a
right circular cylinder and the upper part in the
form of a right circular cone. The radius of the
base of each of the cone and a cylinder is 8
cm. The cylindrical part is 240 cm high and
conical part is 36 cm high. Find the weight of
3
the pillar if 1cm of iron weighs 7.5 grams.
[Take π =
ANSWER
∠B = ∠C
[Each = 60°]
[Each 90°]
and ∠ADB = ∠ADC
∴ ∆ ADB ≅ ∆ ADC
⇒ BD = DC ……….. (1)
∴ BC = BD + DC = DC + DC = 2. DC …(2) [Using (1)]
In right angled ∆ ADC, we have
2
2
2
AC = AD + DC
2
2
2
⇒ BC =AD + DC [ ∵ AC=BC, sides of an equilateral ∆]
2
2
2
⇒ (2DC) = AD + DC
[using (2)]
2
2
2
⇒ AD = 4DC – DC
2
2
⇒ AD = 3DC .
29. Let r1 and r2 be the radii of the cone and cylinder
respectively, then r1 and r2 = 8 cm
Let h1 and h2 be the heights of the cone and cylinder
[1]
respectively, then h1 = 36 cm and h2 = 240 cm.
Total volume of the iron pillar = Volume of the cone +
Volume of the cylinder.
22
]
7
30. Find the mean, median and mode of the
following data
Class
Frequency
0 – 20
6
20 – 40
8
40 – 60
10
60 – 80
12
80 – 100
6
100 – 120
5
[1]
1 2
2
= πr1 h1 + πr2 h2
3
π
2
2
× (8) × 36 + π × (8) × 240
[1]
=
3
= π × 64 × 12 + π × 64 × 240
= 64π [12 + 240]
[1]
3
3
22
= 64 × × 252 = 64 × 22 × 36 cm = 50688 cm
7
Hence, total weight of the iron pillar = 50688 × 7.5 g
3
[ ∵ Weight of 1 cm = 7.5 grams (given)]
= 50688 × 15 g
[1]
2
= 25344 × 15 g = 380160 g = 380160 = 380.16 kg. [1]
1000
30. The cumulative frequency distribution table with the given
frequency becomes :
Class Frequency Cumulative Class di=xi-70
x − 70
fiui
u = i
(fi)
0 – 20
6
20– 40
8
40 – 60
10f0
60 – 80
12 f1
80 – 100
6 f2
100– 120
5
120– 140
3
Total N=Σfi=50
frequency mark
(cf)
(xi)
6
10
14
30
24cf
50
36
70=a
42
90
47
110
50
130
i
-60
-40
-20
0
20
40
60
20
-3
-2
-1
0
1
2
3
-18
-16
-10
0
6
10
9
Σfiui=-19
From the table, n = Σfi = 50 ⇒ n/2 = 25, a = 70, h = 20
Using the formula for calculating the mean :
[1]
QUESTION
120 – 140
3
OR
The mean of the following frequency table is
53. But the frequencies f1 and f2 in the classes
20
–
40
and
60 – 80 are missing. Find the missing
frequencies.
Age
(in years)
No. of
People
0 – 20
15
20 – 40
f1
40 – 60
21
60 – 80
f2
80 – 100
17
Total
100
ANSWER
Mean = a + ∑ f iu i × h = 70 + ( −19) × 20
∑ fi
50
38 = 70 – 7.6 – 62.4,
[1]
= 70 −
5
Now, 60 – 80 is the class whose cumulative frequency 36
is greater than n/2 = 25.
Therefore, 60 – 80 is the median class. Thus, the lower
[1]
limit (l) of the median class is 60.
Using the formula for calculating the median :
n

 − cf 
25 − 24
Median = l +  2
× h = 60 +
× 20
f 
12




20
5
= 60 +
= 60 + = 60 + 1.67 = 61.67.
[1]
12
3
Since the maximum frequency is 12, therefore, the modal
class is 60 – 80. Thus, the lower limit (l) of the modal
class = 60.
Using the formula for calculating the mode :
 fi − f0 
Mode = l + 
×h
 2f i − f 0 − f 2 
12 − 10
× 20
[1]
2 × 12 − 10 − 6
2 × 20
40
= 60 +
= 6+
= 60 + 5 = 65.
[1]
24 − 16
8
OR
The missing frequencies are f1 and f2.
Calculation of Mean
fixi
Age
Mid-point Frequency
(xi)
(fi)
(in years)
0 – 20
10
15
150
20 – 40
30
f1
30 f1
40 – 60
50
21
1050
60 – 80
70
F2
70 f2
80 – 100
90
17
1530
Total
n=53+f1+f2 Σfixi=2730+30f1+70f2
We have, n = 100 = Total frequency = Σfi
[2]
⇒ 53 + f1 + f2 = 100
⇒ f1 + f2 = 100 – 53
⇒ f1 + f2 = 47
⇒ f1 = 47 – f2
[1]
Also, Mean = 53 (given)
2730 + 30f1 + 70f 2
⇒ 53 =
100
273 + 3f1 + 7f 2
⇒ 53 =
10
⇒ 530 = 273 + 3f1 + 7f2
[1]
⇒ 530 – 273 – 7f2 = 3f1
⇒ 257 – 7f2 = 3f1
⇒ 257 – 7f2 = 3(47 – f2)
⇒ 257 – 7f2 = 141 – 3f2
[1]
⇒ 257 – 141 = 7f2 – 3f2
⇒ 116 = 4f2
⇒ 29 = f2
Now, from (1), f1 = 47 – f2 = 47 – 29 = 18.
Hence, f1 = 18 and f2 = 29. [1
= 60 +
QUESTION
ANSWER
iBOOKS SAMPLE PAPER – 03
[FOR XTH CBSE BOARD]
QUESTION
ANSWER
SECTION – A
[1×10]
1.
What is the maximum no. of factors of a prime
number?
2.
In the figure, the graph of some polynomial
p(x) is given. Find the number of zeroes of the
polynomial.
1.
2.
3.
2 3 4 5
Y
1
y = f(x)
X'
-5 -4 -3 -2 -1
O
X
-11
-2
-3
-4
-5
2
3
4
5
Y'
3.
If the co-ordinates of the midpoints of the sides
of a triangle are (1, 1), (2, -3) and (3, 4). Find
its centroid.
4.
Which measures of central tendency is given
by the x-co-ordinate of the point of intersection
of the 'more than' ogive and 'less than' ogive ?
5.
6.
7.
8.
9.
4.
5.
Cards bearing of numbers 3 to 20 are placed
in a bag and mixed thoroughly. A card is taken
out from the bag at random. What is the
probability that the number on the card taken
out is an even number ?
6.
A chord of a circle of radius 14 cm subtends an
angle 60° at the centre. Find the area of its
major sector.
7.
From a point P, the length of the tangent to a
circle is 15 cm and distance of P from the
centre of the circle is 17 cm. Then what is the
radius of the circle?
For what value of k, are the roots of the
2
quadratic equation 3x + 2kx + 27 = 0 real and
equal ?
Given that tan θ =
2
2
cos ec θ − sec θ
cos ec 2θ + sec 2 θ
?
1
3
2
2
x1 + x 2
=1
2
⇒ x1 + x2 = 2
…….. (1)
x 2 + x3
= 2 ⇒x2 + x3 = 4
…….. (2)
2
x1 + x3 = 6
…….. (3)
Again
y1 + y1
=1
2
⇒y1 + y2 = 2
…….. (4)
…….. (5)
y2 + y3 = -6
……..(6)
y1 + y3 = 8
Adding (1) (2) and (3)
⇒ (x1 + x2 + x3) = 12
(x1, y1)
⇒ x1 + x2 + x3 = 6
Now x3 = 6 – 2 = 4
x2 = 0
(3,4)
(1,1)
x1 = 2
Adding (4), (5) and (6)
⇒ (y1 + y2 + y3) = 4
y1 + y2 + y3 = 2
(x2, y2)
(2,-3)
Now y3 = 0
y2 = -6
y1 = 8
y1 + y 2 + y 3   2 
 x + x 2 + x3
So G =  1
 =  2, 
3
3

  3
Median
9
1
or
18
2
Here θ = 300°
O
θ
So Area =
× πr 2
60°
14
360°
Here
=
300° 22
×
× 14 × 14 = 513.3 cm2
360° 7
A
Here OA = OP2 − AP2
15 cm
= 289 − 225
17 cm
P
= 64 = 8 cm
For real and equal roots D = 0
2
⇒ (2k) – 4×3×27 = 0
2
2
2
⇒ 4k – 324 = 0 ⇒ 4k = 81⇒ k = 81⇒ k = ± 9
9.
tan θ =
1
3
2
Now
cos ec θ − sec 2 θ
cos ec 2θ + sec 2 θ
2
Now multiply sin θ
, what is the value of
=
1 − tan2 θ
1 + tan2 θ
−
B
A
= 172 − 152
8.
(x3, y3)
1 − 31
1 + 31
=
2
3
4
3
=
2 1
=
4 2
O
QUESTION
ANSWER
10. In the given figure, determine whether LM ||
QR when PL =5.2 cm, LQ = 7.8 cm, PM = 6
cm and PR = 15 cm.
10.
11.
P
Product of zeroes = 4(-3) = -12 =
−12
cons tan t term
=
1
coefficient of x 2
M
L
12.
Q
R
SECTION – B
[2×5]
11. Find a quadratic polynomial whose zeroes are
4
and
-3. Verify the relationship between the coefficient and zeroes of the polynomial.
13.
12. Without using trigonometric tables, find the
value of cos 70° + cos 57°. cosec 33° - 2 cos
sin 20°
60°.
13. A card is drawn at random from a well shuffled
deck of playing cards. Find the probability of
drawing a (i) face card (ii) card which is neither
a king nor a red card.
14.
OR
Five cards – ten, jack, queen, king and ace,
are well shuffled with their face downwards.
One card is then picked up random.
(i)
What is the probability that the card is a
queen?
(ii)
If the queen is drawn and put aside, what
is the probability that the second card
picked up is (a) an ace (b) a queen.
15.
14. Determine the ratio in which the line 2x + y – 4
= 0 divides the line segment joining the points
A(2, -2) and B(3, 7). Also find the co-ordinates
of the point of division.
15. In the figure, DE || BC. If AD = 2.4 cm, DB =
3.6 cm and AC = 5 cm find AE.
A
D
B
Yes
Given the zeroes are 4 and -3
2
So the quadratic equation is x – x – 12
−1 −coefficient of x
So sum of zeroes = 4+(-3) = 1 =
=
1 coefficient of x 2
16.
E
C
SECTION – C
[3×10]
16. Show that 3 + 5√2 is an irrational number.
cos 70°
+ cos 57° · cosec 33° - 2cos60°
sin 20°
cos(90° − 20°)
=
+ cos 57°icos ec(90° − 57°) − 2 cos 60°
sin 20°
sin 20°
=
+ cos 57°isin 57° − 2 cos 60°
sin 20°
1
= 1 + 1 – 2 · = 2- 1 = 1
2
12 3
(i)
P(face card) =
=
52 13
24 6
(ii)
P(neither king nor a red card) =
=
52 13
OR
1
(i)
P(a queen) =
5
1
(ii)
(a) P(ace) = (b) P(queen) = 0
4
Suppose the line 2x + y – 4 = 0 divides the line
segment joining A(2, -2) and B(3, 7) in the ratio k : f at
point C. Then the coordinates of C are
 3k + 2 7k − 2 
,

.
 k +1 k +1 
But C lies on the line 2x + ky – 4 = 0
 3k + 2  7k − 2
⇒ 2
− 4 = 0 ⇒ 6k+4+7k – 2–4k – 4 = 0
+
 k +1  k +1
⇒ 9k – 2 = 0 ⇒ k = 2/9
So the required ratio is 2 : 9
4
 24
Now the coordinate of C are  , − 
 11 11 
Given DE || BC
⇒ ∠ADE = ∠ABC (corresponding)
A
In ∆ ADE and ∆ABC
∠A = ∠A (common)
∠ADE = ∠ABC
D
E
⇒ ∆ ADE = ∆ABC
AD AE
2.4 AE
B
C
⇒
=
⇒
=
AB AC
6
AC
⇒ 6AE = 12 ⇒ AE = 2 cm.
Let 3 + 5√2 is an irrational number.
a
⇒ 3 + 5√2 = , [a, b are co-primes and b≠0]
b
a
a − 3b
a − 3b
⇒ 5 2 = −3 ⇒ 5 2 =
⇒ 2=
b
b
5b
a − 3b
Since a and b are integers ⇒
is rational
5b
a − 3b
But √2 is an irrational number ⇒
is irrational.
5b
QUESTION
ANSWER
17. Solve the following system of linear equations
graphically : 2x – y = 4, x – y = 1.
17.
Also shade the region bounded by the lines
and
x-axis.
Which is a contradiction. This contradiction has arisen
due to our wrong assumption.
A 3 + 5√2 is an irrational number
2x – y = 4 y = 2x – 4
x = 0 ⇒ y = -4
Y
x = 1 ⇒ y = -2
6
x=2⇒y=0
x
0
1
2
y
-4
-2
0
x – y = 1⇒ y = x – 1
x = 0 ⇒ y = -1
x=1⇒y=0
x=2⇒y=1
18. Find the roots of the equation
1
1
11
,
−
=
x + 4 x − 7 30
18.
x ≠ -4, 7.
OR
Find the value of k, so that the quadratic
equation
2
2x – (k – 2)x + 1 = 0 has equal roots.
19. The first and the last term of an AP are 4 and
19.
81 respectively. If the common difference is 7,
how many terms are there in the A. P. and
what is their sum ?
20. Prove that : sinθ(1 + tan θ)+cos θ(1 + cot θ) =
secθ + cosec θ
20.
x
0
1
2
y
-1
0
1
5
4
3
2
1
X'
-5 -4
-3 -2 -1
O
1
2
3
4
5
-1
-2
-3
-4
-5
Y'
Here x = 4, y = 4
1
1
11
x − 7 − x − 4 11
−
=
⇒
=
x + 4 x − 7 30
(x + 4)(x − 7) 30
11
11
2
2
⇒
=
⇒ x –3x – 28=-30 x – 3x + 2 = 0
2
x − 3x − 28 30
2
⇒ x – 2x – x + 2 = 0⇒ x(x – 2) – 1 (x – 2) = 0
⇒ (x – 2) (x – 1) = 0⇒ x –2 = 0 x – 1 = 0⇒ x=2 or x = 1
OR
2
2x – (k – 2) x + 1 = 0
The quadratic equation has equal roots when D = 0
2
2
⇒ [-(k – 2)] – 4 × 2 × 1 = 0⇒ k + 4 – 4k – 8 = 0
2
2
2
⇒ k – 4k – 4 = 0⇒ k – 2.k.(2) + 4 = 8⇒ (k – 2) = 8
⇒ k . 2 = ± 2√2⇒ k = 2 ±2√2
So k = 2 + 2√2 or 2 – 2√2
st
Let ‘a’ be the 1 term of the given A.P.
Given a=4, d=7
Let an be the last term
⇒ a +(n-1)d = 81⇒ 4 + (n-1)7 = 81
⇒ 7n = 84⇒ n = 12
So there are 12 term in the given AP
n st
Now S12 = (1 term + last term)
2
12
=
( 4+81)=6(85)=510
2
L.H.S = sinθ(1+tanθ) + cosθ(1+cotθ)
1 

= sinθ(1+tanθ) + cosθ  1 +

tan θ 

 tan θ + 1 
= sinθ(1+tanθ) + cosθ 

 tan θ 
OR
2
Prove that : (sin θ + cosec θ) + (cos θ + sec
2
2
2
θ) = 7 + tan θ + cot θ.

1 
cos2 θ 

= (1+tanθ)  sin θ + cos θ
 = (1+tanθ)  sin θ + sin θ 
tan θ 



sin θ 1 
 1   1
= (1+tanθ) 
+
.
 =

 sin θ   sin θ cos θ sin θ 
=cosecθ + secθ = R.H.S
2
2
LHS = (sinθ+cosecθ) + (cosθ+secθ)
2
2
2
2
=sin θ + cosec θ + 2 + cos θ+sec θ+2
2
2
2
2
= sin θ + cos θ + cosec θ + sec θ+4
2
2
2
2
= 1 + cosec θ +sec θ+4= 5 + cosec θ+ sec θ
2
2
2
2
=5 + 1+cot θ+1 + tan θ=7+cot θ+ tan θ
6 X
QUESTION
21.
ANSWER
D and E are points on the sides CA and CB
respectively of ∆ ABC right angled at C. Prove
2
2
2
2
that AE +BD =AB + DE .
21.
OR
In the given figure, DB ⊥ BC, DE ⊥ AB and
AC ⊥ BC. Prove that BE = AC .
DE
BC
A
D
E
B
C
22. Draw a triangle ABC with side BC = 6 cm, AB
= 5 cm and ∠ABC = 60°. Construct a ∆ AB'C'
similar to ∆ ABC such that sides of ∆ AB'C' are
th
3/4 of the corresponding sides of ∆ ABC.
22.
23.
23. If A(4, -8), B(3, 6) and C(5, -4) are the vertices
of a ∆ ABC, D is the midpoint of BC and P is a
point on AD joined such that AP = 2 , find the
PD
co-ordinate of P.
Given in ∆ ABC, ∠C = 90°. D and E are points on the
sides CA and CB respectively.
2
2
2
In ∆ ACE, AE = AC + CE ………..(1)
2
2
2
In ∆ BCD, BD = BC + CD ………..(2)
B
2
2
Adding (1) and (2), We get AE + BD
2
2
2
2
= AC + CE + BC + CD
2
2
2
2
E
=AC + BC + CE + CD
2
2.
=AB + DE
D
C
OR
In the given figure, DB ⊥ BC, DE ⊥ AB
and AC ⊥ BC
D
BE AC
To prove :
=
DE BC
Proof : Since DB ⊥ BC
⇒ ∠DBC = 90°
E
⇒∠DBE + ∠CBE = 90° …….(1)
Again in ∆ BDE, DE ⊥ AB
E
⇒ ∠DEB = 90°
⇒∠BDE + ∠DBE = 90° ………(2)
From (1) and (2), ∠DBE + ∠CBE = ∠BDE + ∠DBE
⇒ ∠CBE = ∠BDE
Now in ∆ BDE and ∆ABC
∠DEB = ∠ACB (=90°)
∠BDE = ∠CBE ⇒ ∆BDE ~ ∆ ABC (By A – A criterion)
BE DE
BE AC
⇒
=
⇒
=
AC BC
DE AC
Try yourself
Given A (4, -8), B(3, 6) and C(5, -4) are the vertices of
a ∆ ABC
D is the midpoint of BC
 3 + 5 6 + ( −4) 
So co-ordinate of D is 
,

2
 2

= (4, 1)
AP
P is a point on AD. Such that
=2
PD
Let the co-ordinates of P be (x, y)
2 × 4 + 1× 4
2 × 1 + 1× ( −8)
So, x =
=4, y=
= −2
3
3
Hence the co-ordinates of P are (4, 2)
A
2
P
1
24. If the point P(x, y) is equidistant from the points
24.
A(3, 6) and B(-3, 4), prove that 3x + y – 5 = 0.
25. OACB is a quadrant of a circle with centre O
and radius 3.5 cm. If OD = 2 cm, find the area
of the (i) quadrant OACB, (ii) shaded region.
25.
B (3,6)
(5,4) C
D
Since the point (x, y) is equidistant from the points A(3,
6) and B(-3, 4)
⇒ PA = PB
2
2
2
2
2
2
⇒ PA = PB ⇒ (x – 3) + (y – 6) = (x + 3) + (y – 4)
2
2
2
2
⇒ x + 9-6x + y + 36 – 12 y =x + 9 + 6x + y +16 – 8y
⇒ 12x + 4y – 20 = 0⇒ 3x + y – 5 = 0
Here radius = 3.5 cm ⇒ OA = OB = 3.5 cm
OD = 2cm
2
Area of the quadrant OACB = ¼ πr
1
7 1 22 35 35 77
2
2
(3.5 × 2) = =
×
×
=
cm = 9.625 cm
2
2 4 7 10 10
8
Area of ∆ BOD = ½ Base × Height
A
A
C
QUESTION
ANSWER
1
7
2
(3.5 × 2) = = 3.5 cm
2
2
Area of the shaded region = 9.625 – 3.5
2
= 6.125 cm
A
C
D
= ½ OB × OD =
SECTION – D
[6×5]
26. Prove that the ratio of areas of two similar
triangles is equal to the ratio of the squares on
their corresponding sides. Using this do the
following :
The diagonals of a trapezium ABCD, with AB ||
DC, intersect each other at the point O. If AB =
2 CD, find the ratio of the area of the ∆ AOB to
the area of the ∆ COD.
OR
Prove that the lengths of tangents drawn from
an external point to a circle are equal. Using
this do the following :
P
O•
Q
A
R
T
B
In the figure TP and TQ are tangents from T to
the circle with centre O and R is any point on
the circle. If AB is a tangent to the circle at R,
prove
that
TA + AR = TB + BR.
26.
B
Given - ∆ ABC ~ ∆ DEF
A
B
C
D
To prove -
O
D
E
M
ar ( ABC )
2
F
2
 AB 
 BC 
 AC 
=
=
 =  DF 
ar (DEF )  DE 
 EP 


2
Construction – Draw AD ⊥ BC and DM ⊥ EF
1
Proof – Here area of ∆ ABC = × BC × AD
2
1
area of ∆ DEF = × EF × DM
2
∴
ar ABC
=
ar DEF
1
2
× BC × AD
1
2
× EF × DM
=
BC × AD
……..(1)
EF × DM
In ∆ ABD and ∆ DEM
LB = LE ∵ (∆ ABC ~ ∆ DEF)
∠ADB = ∠DME (Each 90°)
∴ By AA condition ∆ ABD ~ ∆ DEM
AB AD
∴
=
………(2)
DE DM
∴ (sides are proportional)
Since ∆ ABC ~ ∆ DEF
AB BC AC
∴
=
=
……..(3)
DE EF DF
AB BC
∴
=
………(4)
DE EF
ar ABC BC × AD  AB AB   AB 
=
=
×
=
ar DEF EF × DM  DE DE   DF 
using (2) and (4)
2
From (1)
2
2
2
ar ABC  AB 
 BC 
 AC 
=
=
=


 using (3)
ar DEF  DE 
 EP 
 DF 
Hence, proved
2nd Part
A
B
4
2
Now
O
1
D
Since AB || CD
∴ ∠1 = ∠2
3
C
QUESTION
ANSWER
∠3 = ∠4 (all interior angles are equal) ……….(1)
In ∆ ABO and ∆ COD
∠1 = ∠2
∠3 = ∠4 Using (1)
∴ By AA condition ∆ ABO ~ ∆COD
As we know that ratio of the area of 2 similar triangles
are equal to ratio of the square of their corresponding
sides
ar AOB  AB 
=
ar COD  CD 
∴
2
Since AB = 2CD ∴
ar AOB  2 CD
=
ar COD  CD
2

4
 = Hence 4 : 1
1

OR
A
B
O
C
Given : AB and BC are tangent.
To prove : AB = BC
Construction : O to A, O to C and O to B
Proof : Since tangent at the point of contact with circle
makes right angle with radius.
∴ ∠OAB = ∠OCB …… (1)
In ∆ OAB acid ∆ OBC
OA = OC (each radius)
∠OAB = ∠OCB (each 90°) flow
Hence by RHS configuration condition
∆OAB ≅ ∆ OBC
∴ by CPCT AB = AC.
2
nd
part :P
A
R
O
Q
T
B
given : TP = TQ
to prove : TA + AR = TB + BR
since tangent drawn flows are external point to a circle
are equal in length
∴ TP = TQ = ... (1)
AP = AR = …. (2)
BQ = BR ….. (3)
Using (1) twice TP = TQ
∴ AT + AP = BT + BQ
AT + AR = BT + BR [using (k) and 3]
Hence TA + AR = TB + BR
Hence verified.
2
⇒ 8x – 70x – 750 = 0
70 ± 4900 + 24000
⇒ x=
16
70 ± 170 240 100
=
=
=
rejected= 15 hr.
16
16
10
QUESTION
ANSWER
27. A motor boat whose speed is 18 km/h in still
27.
Let the speed of the stream be x km/hr
Given speed of the boat in still water is 18 km/hr
∴ speed in up stream = (18-x) km/hr
∴ speed in down stream = (18+x) km/hr
Time taken to travel 24 Km up steam
 24 
=
 Km / hr
 18 − x 
Time taken to travel 24 Km down
 24 
stream = 
 Km / hr
 18 + x 
According to the questions
24
24
432 + 24x − 432 + 24x
−
=1 ⇒
=1
18 − x 18 + x
324 − x 2
2
2
⇒ 48x = 324 – x ⇒ x + 48x – 324 = 0
−48 ± 2304 + 1296
−48 ± 60
⇒x=
=
2
2
12
108
=
or
(rejected) = 6 Km/hr
2
2
Hence speed of the stream be 6 Km/hr
OR
Let the tap of smaller diameter takes x hours to fill the
tank
⇒ Tap of larger diameter will take (x + 10) hrs to fill the
tank.
In 1 hour the small tap can fill in 1/x of the tank and the
 1 
larger tap can fill 
 of the tank.
 x + 10 
1
1
8
According to the question +
=
x x + 10 75
2x + 10
8
2
⇒
=
⇒ 150x + 750 = 8x + 10x
2
75
x + 10x
28.
Here height of the flight from the ground AB = CD
E be the point of observation.
water takes 1 hour more to go 24 km upstream
than to return downstream to the same spot.
Find the speed of the stream.
OR
Two water taps together can fill a tank in 9 3
8
hours. The tap of larger diameter takes 10
hours less than the smaller one to fill the tank
separately. Find the time in which each tap can
separately fill the tank.
28. The angle of elevation of a jet fighter from a
point A on the ground is 60°. After a flight of 10
seconds, the angle of elevation changes to
30°. If the jet is flying at a speed of 648
km/hour, find the constant height at which the
jet is flying. [Use √3 = 1.732].
A
E
60° 30°
B
D
C
∠AEB = 60° and ∠CED = 30°.
Given speed = 648 km/hr = 180 m/s
After a flight 10 seconds, the distance traveled BC =
180 × 10 = 1800 m.
AB
Now in ∆ AEB, tan 60° =
BE
⇒ AB = BE √3 ….. (i)
CD
Again in ∆ CDE, tan 30° =
CE
BR 3
=
⇒ CE = 3BE⇒ BE + BC = 3BE
CE
3
⇒ BC = 2BE ⇒ BE = 900 m.
Hence AB = 900 √3 m. (the height of the flight from
ground).
⇒
1
QUESTION
ANSWER
29. A tent consists of a frustrum of a cone,
29.
surmounted by a cone. If the diameters of the
upper and lower circular ends of the frustrum
be 14 m and 26 m respectively, the height of
the frustrum be 8 m and the slant height of the
surmounted conical portion be 12 m, find the
area of the canvas required to make the tent.
(Assume that the radii of the upper circular end
of the frustrum and the base of surmounted
conical portion are equal).
Given height of the frustrum = 8m.
Upper radius of the frustrum = r1 = 14 m.
Lower radius of the frustrum = r2 = 26m
A
X
E
D
∴ l = h2 + (r2 − r1)2
8
=
64 + 144
=
208 = 14.4 m.
B
C
Y
Slant height of the conical portion=12m. and its
radius=14 m.
So total canvas required = C.S.A of frustrum + C.S.A of
cone.
144
= π(r1 + r2)l + πr1l = π(40)
+ π × 14 × 12
10
22
2
= 576 π + 168 π= 744 π= 744 ×
= 2338.28 m
7
30. Find the mean, mode and median of the
30.
C.I
0-10
10-20
20-30
30-40
40-50
50-60
60-70
following data:
Classes
Frequency
0 – 10
3
10 – 20
4
20 – 30
7
30 – 40
15
40 – 50
10
50 – 60
7
60 – 70
4
fi
3
4
7
15
10
7
4
xi
5
15
25
35
45
55
65
Here, mean = A +
di
-30
-20
-10
0
10
20
30
ui
-3
-2
-1
0
1
2
3
uifi
-9
-8
-7
0
10
14
12
∑ uifi = 12
∑ uifi × h
∑ fi
12
= 35 +
× 10 = 35+ 2.4 = 37. 4
50
N
Again
= 25
2
The median class is 30-40
So l = 30, c.f = 14, f = 15, h = 10
N
− cf
25 − 14
Median = l + 2
× h = 30 +
× 10 =
f
15
11
30 +
× 10
15
22
= 30 +
= 30 + 7.3= 37. 3
3
Modal class = 30 -40
So l = 30, f1 = 15, f0 = 7, f2= 10 and h = 40
f1 − f0
So mode = l +
×h
2f1 − f0 − f1
= 30 +
8
80
× 10 = 30 +
= 30 + 6.15 = 36.15
13
13
cf
3
7
14
29
39
46
50