1. No category

1
i
ANSWER OF SAMPLE PAPER
ANSWER OF SAMPLE PAPER – 01
SECTION – A
1.
2.
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]
3.
Perimeter of the given figure
Now
⇒
= 28 + π . 14 + 28 + π . 14
5.
6.
13 12
13
5
=
⇒
=
3 EA
3 ED
⇒ EA = 36/13 cm.⇒ ED = 15/13 cm.
In ∆ ABC, AC = 5 cm, BC = 12 cm ⇒ AB = 13 cm.
SECTION – B
+ CD + BEC
= AB + AED
4.
AB BC
AB AC
=
and
=
AD EA
AD ED
[1]
11. Given the zeroes are 4 and -3
2
So the quadratic equation is x – x – 12
22
22
= 56 + .14 + .14
7
7
= 56 + 44 + 44 = 144 cm.
Median
[1]
[1]
9
1
or
18
2
[1]
Given an = 3 – 4n
a1 = 3 – 4 = -1
a2 = 3 – 8 = -5
[1]
So sum of zeroes = 4+(-3) = 1 =
−1 −coefficient of x
=
1 coefficient of x 2
[½]
Product of zeroes = 4(-3) = -12 =
−12
cons tan t term
=
1
coefficient of x 2
[½]
12.
cos 70°
+ cos 57° · cosec 33° - 2cos60°
sin 20°
13.
cos(90° − 20°)
+ cos 57°icos ec(90° − 57°) − 2 cos 60°
sin 20°
[1]
sin 20°
=
+ cos 57°isec 57° − 2cos 60°
sin 20°
1
= 1 + 1 – 2 · = 2- 1 = 1
[1]
2
12
3
(i)
P(face card) =
=
[1]
52 13
24
6
(ii)
P(neither king nor a red card) =
=
[1]
52 13
OR
1
(i)
P(a queen) =
[1]
5
1
(ii)
(a) P(ace) = (b) P(queen) = 0
[1]
4
∴ d = -5 – (-1) = -4
∴ S 25 =
=
7.
25
[2.(−1) + 24.(−4)]
2
=
25
25
[−2 − 96] =
( −98) = 25 (-49) = - 1225.
2
2
A
Here OA = OP2 − AP2
15 cm
= 172 − 152
= 289 − 225
P
17 cm
= 64 = 8 cm
8.
O
[1]
The system of equations is inconsistent if it has no
solution.
For no solution; if
a1 b1 c 1
=
≠
a2 b2 c 2
[½]
2 k 11
2 k
=
≠
⇒ =
⇒ 5k = -14
5 −7 5
5 −7
⇒ k = -14/5.
∴ for k = -14/5, the system of equations is inconsistent.
[½]
16 cot A = 12. ⇒ cot A = 12/16.
sin A cos A
sin A + cos A sin A + sin A
=
(divide by sin A)
sin A − cos A sin A cos A
−
sin A sin A
12
1 + cot A 1 + 16 28
=
=
=
= 7.
[1]
1 − cot A
12 4
1−
16
∴
9.
[1]
10. In ∆ ABC and ∆ ADE,
∠ACB = ∠AED (= 90°)
∠BAC = ∠EAD (Common)
∴ ∆ ABC ~ ∆ ADE (By AA criterion)
14. Suppose the line 2x + y – 4 = 0 divides the line segment
joining A(2, -2) and B(3, 7) in the ratio k : 1 at point C.
 3k + 2 7k − 2 
,
[½]
Then the coordinates of C are 
.
 k +1 k +1 
But C lies on the line 2x + ky – 4 = 0
 3k + 2  7k − 2
⇒ 2
−4 =0
+
 k +1  k +1
⇒ 6k+4+7k – 2–4k – 4 = 0
⇒ 9k – 2 = 0 ⇒ k = 2/9
[1]
So the required ratio is 2 : 9
 24 4 
Now the coordinate of C are  , − 
 11 11 
[½]
15. Given ABC is an isosceles triangle in which AB = AC
206, Aggarwal Mall, 2nd Floor, Plot No. 3, Sector – 5, Dwarka, New Delhi - 110075
2
i
ANSWER OF SAMPLE PAPER
A
We know that AD = AF
[1]
BD = BE; CE = CF
We have AB = AC
D
F
B
C
E
[1]
[1]
-2
0
0
1
2
y
-1
0
1
The quadratic equation has equal roots when D = 0 [1]
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 [1]
⇒ k . 2 = ± 2√2⇒ k = 2 ±2√2
So k = 2 + 2√2 or 2 – 2√2
19. Let ‘a’ be the 1
st
[1]
term of the given A.P.
Given a=4, d=7
Now S12 =
=
[1]
[1]
n st
(1 term + last term)
2
12
( 4+81)=6(85)=510
2
[1]
20. L.H.S = sinθ(1+tanθ) + cosθ(1+cotθ)
1 

= sinθ(1+tanθ) + cosθ  1 +

tan
θ

[1]
 tan θ + 1 
= sinθ(1+tanθ) + cosθ 

 tan θ 
1 

= (1+tanθ)  sin θ + cos θ

tan
θ

6
5
4
3
2
1
-3 -2 -1
2
2x – (k – 2) x + 1 = 0
So there are 12 term in the given AP
Y
-5 -4
[1]
OR
⇒ 7n = 84⇒ n = 12
Here x = 4, y = 4
X'
⇒ (x – 2) (x – 1) = 0⇒ x –2 = 0 x – 1 = 0
⇒ a +(n-1)d = 81⇒ 4 + (n-1)7 = 81
x – y = 1⇒ y = x – 1
x–y=1⇒y=x-1
x = 0 ⇒ y = -1
x=1⇒y=0
x=2⇒y=1
x
[1]
Let an be the last term
17. 2x – y = 4 ⇒ y = 2x – 4
x = 0 ⇒ y = -4
x = 1 ⇒ y = -2
x=2⇒y=0
-4
2
⇒ x=2 or x = 1
⇒ 3 + 5√2 is an irrational number
y
[1]
2
16. Let 3 + 5√2 is a rational number.
a
⇒ 3 + 5√2 = , [a, b are co-primes and b≠0]
[1]
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
[1]
5b
a − 3b
But √2 is an irrational number ⇒
is irrational.
5b
Which is a contradiction. This contradiction has arisen
due to our wrong assumption.
2
11
30
2
SECTION – C
1
x − 3x − 28
=
⇒ x – 2x – x + 2 = 0⇒ x(x – 2) – 1 (x – 2) = 0
⇒ BC is bisected at the point of contact.
0
11
2
⇒ x –3x – 28=-30 x – 3x + 2 = 0
⇒ BE = CE ( ∵ BD = BE and CE = CF)
x
1
1
11
x − 7 − x − 4 11
−
=
⇒
=
x + 4 x − 7 30
(x + 4)(x − 7) 30
⇒
⇒ AB – AD = AC – AF
⇒ BD = CF
18.

cos2 θ 
= (1+tanθ)  sin θ +

sin θ 

[1]
sin θ 1 
 1   1
.
= (1+tanθ) 
+
 =

 sin θ   sin θ cos θ sin θ 
O
1
2
3
4
5
6 X
-1
-2
-3
-4
-5
Y'
[3]
nd
206, Aggarwal Mall, 2
=cosecθ + secθ = R.H.S
OR
L.H.S. = (1 + cot θ - cosec θ) (1 + tan θ + sec θ)
1 
sin θ
1 
 cos θ
= 1 +
−
+

 1+
 sin θ sin θ   cos θ cos θ 
 sin θ + cos θ − 1  cos θ + sin θ + 1
=


sin θ
cos θ



Floor, Plot No. 3, Sector – 5, Dwarka, New Delhi - 110075
[1]
[1]
3
i
=
ANSWER OF SAMPLE PAPER
(sin θ + cos θ − 1)(sin θ + cos θ − 1)
sin θ.cos θ
[1]
(sin θ + cos θ) 2 − 1 1 + 2 sin θ.cos θ − 1
=
=
= 2. R.H.S.
sin θ.cos θ
sin θ.cos θ
22.
A
[1]
A'
21. Given : In ∆ ABC, ∠C = 90° and D is the mid point of
BC.
A
D
B
2
B
C
2
[1]
1.
Construction ∆ ABC such that BC = 6 cm, AB =
5 cm and ABC = 60°
2.
Draw any ray BX making an acute angle with
BC
3.
Take points B1,B2, B3,B4 on BX such that
BB1=B1B2=B2B3=B3B4
4.
Jain B4C
5.
Through B3, draw B3,C’||B4C and let it intersect
BC at C’
6.
Through C’, draw C’A’||CA and let it intersect
BA at A’.
Proof : In ∆ ACB, ∠C = 90°.
2
2
∴ AB = AC + BC {By Pythagoras theorem}
2
[1]
2
= AC + (2 CD) [ ∵ BC = 2CD]
2
2
= AC + 4 CD
2
2
2
= AC + 4 (AD – AC )
2
2
[1]
= 4 AD – 3 AC .
OR
In the given figure, DB ⊥ BC, DE ⊥ AB
D
and AC ⊥ BC
To prove :
A
Now ∆ A’BC’ is the required triangle whose sides are
BE AC
=
DE BC
of the corresponding sides of ∆ABC.
E
Proof : Since DB ⊥ BC
⇒ ∠DBC = 90°
C
Steps of construction :
2
To Prove : AB = 4AD – 3 AC .
2
C'
6 cm
B
⇒∠DBE + ∠CBE = 90°….(1)
C
[1]
Again in ∆ BDE, DE ⊥ AB
⇒ ∠DEB = 90°
⇒∠BDE + ∠DBE = 90° ………(2)
[1]
From (1) and (2), ∠DBE + ∠CBE = ∠BDE + ∠DBE
⇒ ∠CBE = ∠BDE
23. 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
Now in ∆ BDE and ∆ABC
∠DEB = ∠ACB (=90°)
∠BDE = ∠CBE
2
P
⇒ ∆BDE ~ ∆ ABC (By A – A criterion)
BE DE
BE AC
⇒
=
⇒
=
AC BC
DE AC
1
[1]
3
4
[3]
B (3,6)
D
(5,4) C
206, Aggarwal Mall, 2nd Floor, Plot No. 3, Sector – 5, Dwarka, New Delhi - 110075
[1]
[1]
[2]
4
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ANSWER OF SAMPLE PAPER
24. Since the point (x, y) is equidistant from the points A(3,
6) and B(-3, 4)
⇒ PA = PB
[1]
2
2
2
2
2
⇒ PA = PB ⇒ (x – 3) + (y – 6) = (x + 3) + (y – 4)
2
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
Since ∆ ABC ~ ∆ DEF
∴
AB BC AC
=
=
……..(3)
DE EF DF
∴
AB BC
=
………(4)
DE EF
[1]
2
ar ABC BC × AD  AB AB   AB 
=
=
×
=
ar DEF EF × DM  DE DE   DF 
using (2) and (4)
[½]
25. Here radius = 3.5 cm ⇒ OA = OB = 3.5 cm
From (1)
OD = 2cm
Area of the quadrant OACB = ¼ πr
∴ (sides are proportional)
2
2
Now
A
2
2
ar ABC  AB 
 BC 
 AC 
=
=
=


 using (3)
ar DEF  DE 
 EP 
 DF 
C
[½]
D
Hence, proved
2nd Part
A
4
O
B
1
7 1 22 35 35 77
2
2
(3.5 × 2) = =
cm =9.625 cm [1]
× ×
=
2
2 4 7 10 10 8
Area of ∆ BOD = ½ Base × Height
1
7
2
= ½ OB × OD = (3.5 × 2) = = 3.5 cm
2
2
[1]
O
3
1
D
Since AB || CD
C
∴ ∠1 = ∠2
∠3 = ∠4 (all interior angles are equal) ……….(1)
Area of the shaded region = 9.625 – 3.5
2
= 6.125 cm
SECTION – D
[1]
In ∆ ABO and ∆ COD
∠1 = ∠2
26. Given - ∆ ABC ~ ∆ DEF
A
B
2
∠3 = ∠4 Using (1)
∴ By AA condition ∆ ABO ~ ∆COD
D
[1]
As we know that ratio of the area of 2 similar triangles
are equal to ratio of the square of their corresponding
sides
B
C
D
To prove :
E
ar ( ABC )
M
2
F
2
 AB 
 BC 
 AC 
=
=
 =  DF 
ar (DEF )  DE 
EP




Construction – Draw AD ⊥ BC and DM ⊥ EF
∴
2
∴
2
[1]
In ∆ ABD and ∆ DEM
∠B = ∠E ∵ (∆ ABC ~ ∆ DEF)
∠ADB = ∠DME (Each 90°)
∴ By AA condition ∆ ABD ~ ∆ DEM
∴
AB AD
=
DE DM
………(2)
ar AOB  2 CD 
4
=
 =
ar COD  CD 
1
[1]
Hence 4 : 1
27. Let the speed of the stream be x km/hr
Given speed of the boat in still water is 18 km/hr
1
× EF × DM
2
1
ar ABC 2 × BC × AD BC × AD
=
=
……..(1)
1 × EF × DM
ar DEF
EF × DM
2
2
Since AB = 2CD ∴
1
Proof – Here area of ∆ ABC = × BC × AD
2
area of ∆ DEF =
ar AOB  AB 
=
ar COD  CD 
∴ speed in up stream = (18-x) km/hr
∴ speed in down stream = (18+x) km/hr
[1]
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
206, Aggarwal Mall, 2nd Floor, Plot No. 3, Sector – 5, Dwarka, New Delhi - 110075
[1]
[1]
5
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ANSWER OF SAMPLE PAPER
24
24
−
=1
18 − x 18 + x
⇒
432 + 24x − 432 + 24x
324 − x 2
2
=1
=
[1]
60° 30°
E
2
⇒ 48x = 324 – x ⇒ x + 48x – 324 = 0
⇒x=
D
A
[1]
[1]
∠AEB = 60° and ∠CED = 30°.
Given speed = 648 km/hr = 180 m/s
−48 ± 2304 + 1296
−48 ± 60
=
2
2
After a flight 10 seconds, the distance traveled BC =
180 × 10 = 1800 m.
[1]
12
108
or
(rejected) = 6 Km/hr
2
2
Hence speed of the stream be 6 Km/hr
C
B
Now in ∆ AEB, tan 60° =
[1]
AB
BE
⇒ AB = BE √3 ….. (i)
OR
Suppose the faster pipe takes x minutes to fill the
cistern. Therefore, the slower pipe will take
(x + 3) minutes to fill the cistern.
Again in ∆ CDE, tan 30° =
1
⇒
3
Since the faster pipe takes x minutes to fill the cistern.
∴ Portion of the cistern filled by the faster pipe in one
minute = 1/x.
⇒ Portion of the cistern filled by the faster pipe in 40/13
1 40
40
minutes = ×
=
[2]
x 13 13x
Similarly, portion of the cistern filled by the slower pipe
in 40/13 minutes.
=
[1]
=
CD
CE
[1]
BR 3
⇒ CE = 3BE⇒ BE + BC = 3BE
CE
⇒ BC = 2BE ⇒ BE = 900 m.
Hence AB = 900 √3 m. (the height of the flight from
ground).
[2]
OR
Let CE (= 8 m) = BD be the tall building and AD = x m
be the multi – storeyed building.
Let BC = DE = y m, be the distance between the two
buildings.
[1]
Then, AB = AD – BD = AD – CE = (x – 8) m be the
difference of the height between the two buildings.
1
40
40
×
=
x + 3 13 13(x + 3)
It is given that the cistern is filled in 40/13 minutes.
∴
40
40
+
=1
13x 13(x + 3)
⇒
1
1
13
+
=
x x + 3 40
⇒
x + 3 + x 13
=
x(x + 3) 40
[1]
[1]
[1]
2
⇒ 13x – 41x – 120 = 0
⇒ (x – 5) (13x + 24) = 0
In right triangle ABC, we have tan 30° = AB/BC
[1]
⇒ x = 5 or, x = -24/13
⇒ x = 5 [ ∵ x > 0]
Hence, the faster pipe fills the cistern in 5 minutes and
the slower pipe takes 8 minutes to fill the cistern.
[1]
28. Here height of the flight from the ground AB = CD
E be the point of observation.
⇒
1
3
=
x−8
⇒ y = √3 (x – 8) m ………… (1)
y
[1]
In right triangle ADE, we have tan 45° = AD/DE
⇒ 1=
x
⇒ y = x ……………. (2)
y
[1]
From (1) and (2), we get √3 (x – 8) = x
⇒ √3 x – 8√3 = x, ⇒ √3x – x = 8√3, ⇒ (√3 – 1) x = 8√3
206, Aggarwal Mall, 2nd Floor, Plot No. 3, Sector – 5, Dwarka, New Delhi - 110075
6
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ANSWER OF SAMPLE PAPER
⇒x=
8 3
3 −1
⇒x=
8 3( 3 + 1)
( 3 − 1)( 3 + 1)
= 576 π + 168 π = 744 π
[1]
= 744 ×
8(3 + 3)
⇒x=
⇒ x = 4 (3 + √3) m
3 −1
22
2
= 2338.28 m
7
[2]
30.
From (2) and (3), we get y = 4 (3 + √3) m
C.I
fi
xi
di
ui
uifi
cf
So, the height of the multi-storeyed building is 4(3 + √3)
0-10
3
5
-30
-3
-9
3
m and distance between the two buildings is also
10-20
4
15
-20
-2
-8
7
4 (3 + √3) m.
20-30
7
25
-10
-1
-7
14
30-40
15
35
0
0
0
29
40-50
10
45
10
1
10
39
50-60
7
55
20
2
14
46
60-70
4
65
30
3
12
50
[1]
29. Given height of the frustrum = 8m.
Upper radius of the frustrum = r1 = 14 m.
Lower radius of the frustrum = r2 = 26m
∑ uifi = 12
A
12m
Here, mean = A +
X
D
∑ uifi × h
∑ fi
E
14 m
B
8m
C
Y
26 m
64 + 144
=
208 = 14.4 m.
[1]
Slant height of the conical portion=12m. and its
[1]
radius=14 m.
So total canvas required = C.S.A of frustrum + C.S.A
of cone.
= π(r1 + r2)l + πr1l = π(40)
12
× 10 = 35+ 2.4 = 37.4
50
Again
N
= 25
2
[2]
The median class is 30-40
∴ l = h2 + (r2 − r1)2
=
= 35 +
144
+ π × 14 × 12
10
So l = 30, c.f = 14, f = 15, h = 10
N
− cf
25 − 14
Median = l + 2
× h = 30 +
× 10
f
15
11
22
= 30 +
× 10 = 30 +
= 30 + 7.3 = 37.3
15
3
[2]
Modal class = 30 -40
So l = 30, f1 = 15, f0 = 7, f2= 10 and h = 10
[1]
So mode = l +
[1]
= 30 +
f1 − f0
×h
2f1 − f0 − f1
8
80
× 10 = 30 +
= 30 + 6.15 = 36.15
13
13
206, Aggarwal Mall, 2nd Floor, Plot No. 3, Sector – 5, Dwarka, New Delhi - 110075
[2]
7
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ANSWER OF SAMPLE PAPER
ANSWER OF SAMPLE PAPER – 02
1.
SECTION – A
2
3 tan θ + 3 = 3 (tan θ + 1)
⇒ x = 3.
[1]
2
SECTION – B
2
= 3 sec θ
2
27 
1
3
3
= 3  =
sec θ =
=
4 
cos θ 2 
2
2.
We have
This
[1]
11. Since √3 and - √3 are zeroes of f(x), therefore (x - √3)
and (x + √3) are two factors of f(x).
2
⇒ (x - √3) (x + √3) = x – 3 is also a factor of f(x).
13
13
=
3125 2° × 5 5
2
Now, divide f(x) by x – 3 to find other zeros.
2
3
2
[1]
2
x – 3 2x – 3x – 5x + 9x – 3 2x – 3x + 1
that the prime factorization of the
13
m
n
denominator of
is of the form 2 × 5 . Hence it
3125
has terminating decimal expansion.
[1]
3.
4
shows
4
2
2x
–
– 6x
+
3
2
– 3x + x + 9x – 3
3
– 3x
+
The system of equations will have no solution, if
a1 b1 c 1
=
≠
a2 b2 c 2
+ 9x
–
2
x –3
2
x –3
– +
1 2 5
⇒ = ≠
3 k 15
⇒
4.
0
1 2
2 5
= and ≠
3 k
k 15
2
2
⇒ k = 6 and k ≠ 6 which is not possible.
= [(x + √3) (x - √3)] [2x – 2x – x + 1]
Hence, there is no value of k, for which the given system
[1]
of equations has no solution.
= [(x + √3) (x - √3)] (2x – 1) (x – 1)
12.
CE = CD = AC – AE = 7 cm
∴ BC = BD + CD = 9 + 7 = 16 cm.
[1]
sin θ − 2 sin 3 θ
− tan θ
2 cos 3 θ − cos θ
=
sin θ − 2 sin 3 θ
− tan θ
2 cos 3 θ − cos θ
=
sin θ(1 − 2 sin 2 θ)
− tan θ
cos θ(2 cos 2 θ − 1)
=
sin θ[1 − 2(1 − cos 2 θ)]
− tan θ
cos θ(2 cos 2 θ − 1)
=
sin θ(2cos 2 θ − 1)
− tan θ
cos θ(2 cos 2 θ − 1)
In ∆ ADE and ∆ ABC,
∠ADE = ∠B (given)
∠A = ∠A (common)
So ∆ ADE ~ ∆ ABC. (by A – A criterion)
AD DE
3.8 DE
⇒
=
⇒
=
AB BC
5.7 4.2
⇒ DE = 2.8 cm.
[1]
6.
-3, 0, 2.
[1]
7.
Median = 35.
[1]
8.
h 2
=
r 1
[1]
= tan θ - tan θ = 0.
[1]
No. of non-face cards = 40.
(i) Probability of getting a non-face card =
Let h be the height of the cone.
Given that volume of cone = Volume of hemisphere.
40 10
=
.
52 13
[1]
(ii) Since there are 2 black kings and 2 red queens in 52
playing cards ⇒ P (getting a black king or a red
4
1
queen) =
=
.
[1]
52 13
1
2
⇒ πr 2h = πr 3
3
3
9.
[1]
13. No. of face cards in a deck of playing cards = 12.
Let the radius of both cone and hemisphere be r cm.
⇒ h = 2r, ⇒
[1]
So the other zeros are 1 and 1/2.
Here AF = AE = 8 cm
BF = BD = 9 cm.
5.
2
∴ f(x) = (x – 3) (2x – 3x + 1)
h 2
= .
r 1
OR
[1]
2/49.
10. Since x + 1, 3x and 4x + 2 are in A.P.
⇒ 3x – (x + 1) = 4x + 2 – 3x
Since the king, queen and jack are removed, the cards
remains are 49.
So the probability of getting a card of heart = 12/49
[1]
Probability of getting a card of queen = 3/49
Probability of getting a card of clubs = 9/49.
⇒ 2x – 1 = x + 2
nd
206, Aggarwal Mall, 2
Floor, Plot No. 3, Sector – 5, Dwarka, New Delhi - 110075
[1]
8
i
ANSWER OF SAMPLE PAPER
14.
A
(3,-4)
P
(p,-2)
B
(1,2)
Q
5 
 ,q 
3 
⇒ 3=
Let A (3, -4) and B (1,2)
17. 3x + y – 5 = 0
 1× 1 + 2 × 3 1× 2 + 2 × ( −4) 
∴P 
,

1+ 2
 1+ 2

2x – y – 5 = 0
⇒ y = 5 – 3x
[½]
x
… (Using section formulae)
7 6
⇒ P  , −  = P (p, -2)
3 3
0
y
5
⇒ y = 2x - 5
[1]
1
x
0
1
2
y
-5
-3
[½]
… (Given)
5
7
3
[½]
(0,5)
4
3
Again AQ : QB = 2 : 1.
(1,2)
2
 2 × 1 + 1× 3 2 × 2 + 1× ( −4) 
∴Q 
,

2 +1
 2 +1

[½]
1
… (Using section formulae)
5 
5 
⇒ Q  ,0  = Q  ,q 
3


3 
a2 + 1
is rational.
2a
⇒ √3 is rational. Which is a contradiction as √3 is
irrational. Hence √2 + √3 is irrational.
[1]
∴ AQ : QB = 2:1
Now AP: PB = 1:2
∴p=
[1]
Since a & 1 are rational ⇒
The line segment AB is trisected at P and Q
∴ AP: PB = 1: 2
a2 +1
2a
-4
-3
-2
-1
1
2
-1
… (Given)
-3
15. Let the incircle touch the sides AB, BC and CA at D, E,
F respectively. Let O be the centre of the circle.
-4
-5
4
3x+y-5 = 0
-2
[½]
∴q=0
3
2x-y-5 = 0
(0,-5)
A
[2]
∴ Hence the two lines meet at (0, 5) and (0, -5) with y[½]
axis.
18. Clearly it is an AP .with first term a = 54, and common
difference d = -3.
F
8 cm
x
D
x
Let the sum of n terms be 513.
O
Then Sn = 513
x
B
E
⇒
C
6 cm
Then, OD = OE = OF = x cm.
Since the tangents to a circle from an external point are
equal, we have
[1]
2
n – 37n + 342 = 0
(n- 18) (n – 19) = 0
AF = AD = (8 – x) cm, and CF = CE = (6 – x) cm.
th
Here the common difference is negative. So 19 term is
given by a19 = 54 + (19 - 1) x 3 = 0.
2
Now, AC = AB + BC .
2
2
2
Thus, the sum of 18 terms as well as that of 9 terms is
513.
[1]
2
⇒ (14 – 2x) = 8 + 6 = 100 = (10) .
⇒ 14 – 2x = ± 10 ⇒ x = 2 or x = 12.
⇒ x = 2 [neglecting x = 12]
Hence, the radius of the incircle is 2 cm.
19.
[1]
SECTION – C
16. Let us suppose that √2 + √3 is rational.
Let √2 + √3 = a, where, a is rational.
Therefore √2 = a - √3
2
⇒ 2 = a + 3 – 2√3 a (squaring both sides)
[1]
n = 18 or 19
∴ AC = AF + CF = (8 – x) cm + (6 – x) cm = (14 – 2x)
cm.
2
[1]
n
[108 + (n-1) x 3] = 513
2
Also, AB = 8 cm and BC = 6 cm.
2
n
{2a + (n-1) d} = 513
2
[1]
10
2
+
= 4 …….… (1)
x+y x−y
15
5
−
= −2
…..… (2)
x+y x−y
1
1
Let
= p and
=q
x+y
x−y
then we get, 10p + 2q = 4
….… (3)
206, Aggarwal Mall, 2nd Floor, Plot No. 3, Sector – 5, Dwarka, New Delhi - 110075
[1]
9
i
ANSWER OF SAMPLE PAPER
21.
15p – 5q = –2
…..… (4)
n
eq 3 × 5, 50p + 10q = 20
n
eq 4 × 2, 30p – 10q = –4
____+__________
80p = 16
⇒p=
1
5
[½]
n
Putting the value of p in eq 3, we get
10.
1
+ 2q = 4
5
⇒ 2q = 2 ⇒ q = 1
[½]
1
1
Now we have,
=
x+y 5
⇒x+y=5
… (5)
Steps of Construction :
1
=1
x−y
⇒x–y=1
(i)
Take BC = 8 cm
(ii)
At B construct a right angle and cut BA = 6 cm.
(iii)
Join BC . ABC is the given triangle.
(iv)
Draw BD perpendicular to AC
[1]
(v)
Bisect BC at M . With M as centre and radius
BM = MC = DM, draw a circle.
[1]
(vi)
This circle is passing through B, C, D.
(vii)
AB is already one tangent. Cut AE = AB and
join AE which is second tangent.
[3]
… (6)
n
Adding eq 5 and 6, 2x = 6
⇒x=3
n
putting the value of x in eq 5, we get y = 2
20. L.H.S. =
=
cos A
sin A
+
1 − tan A 1 − cot A
cos A
sin A
cos2 A
sin2 A
+
=
+
sin A
cos A
cos A − sin A sin A − cos A
1−
1−
cos A
sin A
=
cos2 A
sin2 A
−
sec A − sin A cos A − sin A
=
cos2 A − sin2 A (cos A − sin A)(cos A + sin A)
=
cos A − sin A
(cos A − sin A)
[1]
2
[1]
= cosA + sinA = R.H.S.
OR
L.H.S. =
cot A − cos A
22. Let ABCD be a square and radius of each circle at
corners be a cm such that the side of the square be 2a
cm
2
2
∴ Area of square = (2a) = 4a sq cm
90
Area of circular corners = 4 ×
× πa2
360
=
D
cos A − cos A.sin A
cos A + cos A.sin A
cos A(1 − sin A )
=
2
[1]
[1]
cot A + cos A
cos A
− cos A
sin A
= cos A
+ cos A
sin A
=
2
∴ Shaded portion = 4a - πa = (4 - π)a
3 24
But given area = 3 =
sq cm
7
7
C
B
A
[1]
1 − sin A
[1]
1 + sin A
cos A(1 + sin A )
cos ec A − 1
R.H.S. =
cos ec A + 1
According to question,
2
∴ (4 - π)a =
1
−1
1 − sin A
sin A
=
=
1
1 + sin A
+1
sin A
⇒
24
7
22  2 24

⇒ 4 −
a =
7 
7

6 2 24
a =
⇒a=2
7
7
∴ Radius of each circle = 2 cm
∴ L. H. S. = R. H. S
[1]
nd
206, Aggarwal Mall, 2
23. To prove:
area ( ∆ABC )
area ( ∆DEF )
=
AP2
DQ2
Floor, Plot No. 3, Sector – 5, Dwarka, New Delhi - 110075
[1]
10
i
ANSWER OF SAMPLE PAPER
∴ AF = BE
D
A
[1]
∴ 2BC.BE = 2AD.AF
B
C
P
F
Q
E
BD2 = AB2 + AD2 – 2BC.BE
Adding, we get
Proof: ‹ ∆ABC ~ ∆DEF
AB BC AC
AB BC
∴
=
=
or
=
DE EF DF
DE EF
⇒
AB BP + PC
=
DE EQ + QF
⇒
AB 2BP
=
DE 2EQ
AC2 + BD2 = AB2 + BC2 + AB2 + AD2
= AB2 + BC2 + DC2 + AD2 [ ( AB = DC]
Hence the result.
24. Here the points are A (-3, 0), B (1, -3) and C (4, 1).
[‹ BP = PC, EQ = QF]
Now, in ∆ABP and ∆DEQ
AB BP
=
DE EQ
area ( ∆DEF )
=
BC2
EF
2
=
AP2
DQ
2
(from iii)
[1]
OR
D
F
[½]
AC = (4 + 3)2 + 12 = 49 + 1 = 50 = 5 2 unit
[½]
Hence the triangle is an isosceles triangles.
2
2
2
Again AC = 50 = AB + BC
⇒ The triangle is a right angle triangle.
B
E
But BC = AD [ ‹ opposite sides of a ||gm are equal]
Also from triangles ∆ADF and ∆BCE, AD = BC
∠DAF = ∠CBE [Each = 90°]
[corresponding ∠s since AD || BC and ABE is a
transversal]
[1]
[½]
OR
Let the point X, divide the line segment joining the points
P (-3, 10) and Q (6, - 8) in the ratio k : 1.
 6k − 3 −8k + 10 
Then, the co-ordinates of X are 
,

k +1 
 k +1
k
[1]
1
P (-3,10) X
Q (6, -8)
But the co-ordinates of X are given (-1, 6)
∴
6k − 3
−8k + 10
= −1 and
=6
k +1
k +1
[1]
⇒ 6k – 3 = -k – 1 and – 8k + 10 = 6k + 6
⇒ 7k = 2
⇒ - 14k = -4
⇒ k = 2/7
⇒ k = 2/7
Hence, the point X divides PQ in the ratio 2 : 7.
[1]
25. Let the co-ordinates of B and C be (a, b) and (x, y)
respectively.
A (1, -4)
C
Let ABCD be a parallelogram with ∠B as obtuse and
∠A as acute.
2
2
2
Then (AC) = (AB) + (BC) + 2BC.BE
……(1)
2
2
2
and (BD) = (AB) + (AD) – 2AD.AF ………….. (3) [1]
∴ ∆ADF ≅ ∆BCE
[½]
BC = (4 − 1)2 + (1 + 3)2 = 25 = 5 unit
(2, -1) P
A
[1]
Here AB = BC = 5 unit
∠ABP = ∠DEQ
( ‹ ∆ABC ~ ∆DEF)
∴ ∆ABP ~ ∆DEQ
(SAS criterion)
Their corresponding sides must be proportional
AB BP AP
∴
=
=
DE EQ DQ
AB BC
⇒
=
…(i)
[1]
DE EF
since, ∆ABC ~ ∆DEF
AB BC
∴
=
…. (ii)
DE EF
From (i) and (ii),
AP BC
=
..(iii)
DQ EF
We know that the areas of two similar triangles are in
the ratio of the square of their corresponding sides.
area ( ∆ABC )
EMBED Equation.DSMT4
[1]
AB BP
⇒
=
DE EQ
i.e.
[ ( BC = AD, BE = AF]
From (1) and (2), we get
AC2 = AB2 + BC2 + 2BC.BE
B
(a, b)
Q (0, -1)
C
(x, y)
 1 + a −4 + b 
Then (2, -1) = 
,
2 
 2
1+ a
−4 + b
Therefore
= 2 and
= −1
2
2
⇒ a = 3.
⇒ b = 2.
1
+
x
−
4
+
y


Also (0, −1) = 
,
2 
 2
1+ x
−4 + y
Therefore
= 0 and
= −1
2
2
⇒ x = -1
⇒ y = 2.
206, Aggarwal Mall, 2nd Floor, Plot No. 3, Sector – 5, Dwarka, New Delhi - 110075
[1]
[1]
11
i
ANSWER OF SAMPLE PAPER
The co-ordinates of ∆ ABC are A (1, -4), B (3, 2) and
C (-1, 2).
1
Area of ∆ ABC = [1 (2 – 2) + 3 (2 + 4) – 1 (-4 – 2)]
2
1
=
(18 + 6) = 12 sq. units.
[1]
2
SECTION – D
26. Given : A right triangle ABC, right angled at B.
2
2
[½]
To prove : (Hypotenuse) = (Base) + (Perpendicular)
2
2
2
2
i.e., AC = AB + BC
[½]
Construction : Draw BD ⊥ AC
[½]
B
⇒
63x + 378 + 72x
=3
x(x + 6)
⇒
135x + 378
=3
x 2 + 6x
2
⇒ 3x + 18 x = 135 x + 378.
[1]
2
⇒ 3x – 117 x – 378 = 0
2
⇒ x – 39 x – 126 = 0
⇒ (x + 3) (x – 42) = 0
⇒ x = -3 or x = 42.
[2]
Since x is the average speed of the train, x cannot be
negative.
[1]
Therefore x = 42.
So the original average speed of the train is 42 km/h. [1]
OR
C
D
A
[½]
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].
AD AB
=
[Sides are proportional]
So,
AB AC
2
⇒ AD. AC = AB . ……………. (1)
Also, ∆ BDC ~ ∆ ABC
CD BC
So,
=
[Sides are proportional]
BC AC
2
⇒ CD. AC = BC ………….. (2)
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
For the Second Part :
D
[2]
B
Here ∆ ACD ∼ ∆ ABC
⇒
AC AD
=
AB AC
2
⇒ AC = AB . AD ………….. (1)
Again ∆ BCD ~ ∆ ABC. (By A – A criterion)
BC BD
⇒
=
AB BC
2
⇒ BC = AB . BD …………. (2)
BC 2 AB.AD AD
From (1) & (2),
=
=
.
AC 2 AB.BD BD
27. Let the original speed of the train be x km/h.
63 72
Therefore according to the question
+
=3.
x x+6
Similarly, the time taken to go downstream =
24
hrs.
18 + x
24
24
−
= 1.
18 − x 18 + x
i.e. 24(18 + x) – 24 (18 – x) = (18 – x) (18 + x).
2
i.e. x + 48 x – 324 = 0
using the quadratic formula, we get
According to the question,
[1]
[1]
−48 ± 48 2 + 1296 −48 ± 3600
=
2
2
−48 ± 60
=
= 6 or -54.
[1]
2
Since x is the speed of the stream, if cannot be
negative.
So we ignore the root x = -54.
Therefore x = 6 gives the speed of the stream 6 km/h.[1]
28. Here r1 = 10 cm, r2 = 20 cm, h = 30 cm.
x=
C
A
Let the speed of the stream be x km/hr
Therefore, the speed of the boat upstream = (18 – x)
km/h and the speed of the boat downstream = (18 + x)
km/h.
[1]
dis tance
24
=
hrs .
The time taken to go upstream =
speed
18 − x
[½]
[½]
[1]
[1]
πh 2 2
Capacity of the bucket =
(r + r2 + r1r2 )
[½]
3 1
(3.14)30
=
(100 + 400 + 200)
3
3
= (3.14) 10 × 700 cm
3
= 21980 cm .
= 21.98 litres.
[1]
Cost of 1 litre of milk = Rs. 25.
Cost of 21.98 litres of milk Rs. 25 × 21.98 = Rs. 549.50.
[1]
Slant height l = h 2 + (r2 − r1) 2 = 900 + 100 = 31.62 cm.
[1]
[½]
2
Therefore surface area of the bucket = πl (r1 + r2) + πr1 .
206, Aggarwal Mall, 2nd Floor, Plot No. 3, Sector – 5, Dwarka, New Delhi - 110075
12
i
ANSWER OF SAMPLE PAPER
2
= 3.14 × 31.62 (20 + 10) + 3.14 (10)
[1]
= 3.14 (948.6 + 100)
= 3.14 (1048.6)
3
= 3292.6 cm (approx).
[1]
29. Here BC be the tower on which CD pole stands.
Let height of the tower be h metres and AB = x metres.
[½]
PQ
.
AQ
⇒ PQ = AQ = h m ……… (1)
PQ
Again in ∆ PBQ, tan 60° =
.
BQ
h
⇒ 3=
h − 100
⇒ h = √3 (h – 100)
100 3
⇒h=
= 50(3 + 3)
3 −1
Hence, the height of the balloon is 50 (3 + √3) m.
In ∆ PAQ, tan 45° =
D
5m
C
[½]
[1]
[½]
[1]
[½]
30.
45°
CI
fi
xi
ui
fiui
C.f.
0 – 10
4
5
-3
-12
4
10 – 20
5
15
-2
-10
9
20 – 30
7
25
-1
-7
16
[1]
30 – 40
10
35
0
0
26
[1]
40 – 50
12
45
1
12
38
50 – 60
8
55
2
16
46
60 – 70
4
65
3
12
50
hm
45° 60°
B
A
x
BC
In ∆ ABC, tan 45° =
AB
h
⇒1=
x
⇒ x = h ………….. (1)
[½]
Again in ∆ ABD, tan 60° =
BD
.
AB
Total = 50
[1]
Here, a = 35, Σfi = 50, Σfiui = 11, h = 10.
Now Mean = a +
[½]
5( 3 + 1) 5(2.73) 13.65
=
=
= 6.825 m.
2
2
2
Hence height of the tower is 6.825 m.
OR
Let the height of the balloon at P be 'h' metres.
Let A and B be the two cars. Thus AB = 100 m.
⇒h=
[1]
[½]
[½]
P
= 35 +
= 30 +
hm
25 − 16
× 10 = 30 + 9 = 39 .
10
= 40 +
Q
[½ ]
[1]
[½ ]
 12 − 10 
= 40 + 
 × 10
 24 − 10 − 8 
60°
B
[½ ]
11
× 10 = 35 + 2.2 = 37.2
50
 f1 − f 0 
Mode = l + 
×h
 2f1 − f 0 − f 2 
60°
100 m
∑ fiu i
×h
∑ fi
n

 2 − Cf 
Median = l + 
×h
f 




45°
45°
Σfiui = 11
[1½ ]
h+5
⇒ 3=
x
⇒ h + 5 = √3 x = √3 h (As x = h)
⇒ (√3 – 1) h = 5
5
⇒h=
3 −1
A
[1]
20
= 43.33 .
6
[1]
206, Aggarwal Mall, 2nd Floor, Plot No. 3, Sector – 5, Dwarka, New Delhi - 110075
[1]
13
i
ANSWER OF SAMPLE PAPER
ANSWER OF SAMPLE PAPER – 03
SECTION – A
⇒ 5k = 15
31. Given A + B = 90°
⇒ k = 3.
[1]
⇒ A = 90° - B.
SECTION – B
⇒ sin A = sin (90° - B)
4
32.
2
4
3
2
4
3
2
x + 2x – 3 4x + 2x – 2x + x – 1 4x2– 6x + 22
8
.
5
3
2 k 5
= ≠
6 3 15
⇒
1 k
k 1
= and ≠
3 3
3 3
2
22x + 44 x – 66
–
–
+
– 61 x + 65
Here the remainder is – 61 x + 65.
According to the question ax + b = - 61x + 65.
∴ a = - 61 and b = 65.
⇒ k = 1 and k ≠ 1.
Thus there is no value of k, for which the given system
of equations has no solution.
[1]
42.
cos ec θ
cos ec θ
+
−2
cos ec θ − 1 cos ec θ + 1
=
34. Here CP = CQ = 12 cm.
BC = 9 cm ⇒ BQ = 12 – 9 = 3 cm.
2 cos ec 2θ
2 cos ec 2θ
−2 =
−2
(cos ec 2θ − 1)
cot 2 θ
2
[1]
But BQ = BR = 3 cm.
2
= 2 cosec θ . tan θ - 2
2
= 2 sec θ - 2
35. Given DE || AB.
2
2
= 2 (sec θ - 1) = 2 tan θ.
AD BE
⇒
=
DC EC
43. If the ten, jack, queen, king and ace of diamonds are
removed, then the total of cards left is 47.
3x + 19 3x + 4
=
x+3
x
2
2
– 6x + 10x + x – 1
3
2
– 6x – 12x + 18 x
+
+
–
2
22x – 17 x – 1
[1]
33. The system of equations will have no solution, if
a1 b1 c 1
=
≠
.
a2 b2 c 2
⇒
2
4x + 8x – 12x
– –
+
[1]
7
7
7 × 53
7 × 125 875
= 4
= 4
=
= 4 = 0.0875
80 2 × 5 2 × 5 4
10 4
10
⇒
2
41. Divide 4x + 2x – 2x + x – 1 by x + 2x – 3.
= cos B.
∴ cos B =
3
Now the queens left in the deck is 3.
2
[1]
⇒ 3x + 19 x = 3x + 13 x + 12
So P (getting a queen) = 3/47.
⇒ 6x = 12
If the jack, queen and king of diamonds are removed,
then the face cards left = 9.
⇒ x = 2.
[1]
36. 4.
[1]
37. 6.
[1]
38. Let the radius of the sphere which fits exactly into a
cube be r units. Then length of the edge of the cube = 2r
units.
Let V1 and V2 be the volumes of the cube and sphere
respectively.
3
⇒ V1 : V2 = 6 : π.
40. Since 2k – 1, 7 and 3k are in A.P.
⇒ 7 – (2k – 1) = 3k – 7
⇒ 8 – 2k = 3k – 7
OR
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.
[1]
44. Suppose the point P(-3, k) divides the line segment
joining points A(-5, -4) and B(-2, 3) in the ratio x : 1.
V1
8r 3
6
=
=
V2 4 3 π
πr
3
39. 3/6 or ½.
[1]
When a coin is tossed 3 times, the possible outcomes
are : HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. [1]
So P (lose the game) = 6/8 = 3/4.
3
The V1 = (2r) , V2 = 4/3 πr .
∴
So P (getting a face card) = 9/47.
[1]
 −2x − 5 3x − 4 
Then, the coordinates of P are 
,

x +1 
 x +1
[1]
But, the coordinates of P are given as (-3, k)
∴
−2x − 5
3x − 4
= −3 and
=k
x +1
x +1
⇒ -2x – 5 = -3x – 3
206, Aggarwal Mall, 2nd Floor, Plot No. 3, Sector – 5, Dwarka, New Delhi - 110075
[1]
14
i
ANSWER OF SAMPLE PAPER
⇒ x = 2 and k = 2/3
[1]
Hence, the ratio is 2 : 1 and k = 2/3.
45. In the given figure ∠B = 90°. AD = 23 cm, AB = 29 cm,
DS = 5 cm.
DS = DR = 5 cm. ∴ AR = AQ = 23 – 5 = 18 cm
[1]
Again BQ = BP = 29 – 18 = 11cm.
Now we have OP ⊥ BC and OQ ⊥ AB.
⇒ OPBQ is a square. Hence radius is 11 cm.
[1]
A
R
Q
O
D
B
S
P
[2]
C
The co-ordinate of the point where the lines meet the xaxis is (4, 0).
[0.5]
SECTION – C
46. Let us assume that √P is rational. Then there exists
a
positive co-primes a and b. Such that p = , b ≠ 0
b
a2
⇒p=
2
⇒ pb = a
b2
⇒ p divides a
48. Let a be the first term and 'd' be the common
difference of the given A.P.
th
⇒ a 4 + a 8 = 24
2
⇒ a + 3d + a + 7d = 24
⇒ 2a + 10d = 24
2
⇒ p divides a …………… (1)
⇒ a + 5d = 12
[1]
terms is 44
⇒ a + 5d + a + 9d = 44
2
⇒ p c = pb
⇒ b = pc
[1]
⇒ a 6 + a10 = 44
2 2
2
th
Again the 6 and 10
Now a = p c
2 2
….....(1)
th
⇒ a = pc for some positive integer c.
2
th
The 4 and 8 terms of an AP is 24
⇒ 2a + 14d = 44
2
⇒ p divides b
⇒ a + 7d = 22
2
⇒ p divides b …………. (2)
….....(2)
Subtracting (1) from (2) we get 2d = 10
[1]
From (1) and (2), p is a common factor of a and b.
⇒d=5
This contradicts that a and b are co-prime.
So first term a = -13
This is due to our wrong assumption.
Second term = -13 + 5 = -8
Hence √p is an irrational number.
[1]
Third term = -8 + 5 = -3
[1]
Hence the first three terms of the AP are -13, -8,
-3.
[1]
47. 2x + y =8
⇒ y = 8 – 2x
x
1
2
3
4
y
6
4
2
0
49.
1
x+ 4
⇒
3x – 2y = 12
⇒ 2y = 3x – 12
−
1
x−7
=
11
30
x −7 −x −4
11
−11
11
=
⇒
=
2
( x + 4)(x − 7) 30
x − 3x − 28 30
[1]
2
⇒ x – 3x – 28 = -30
3x − 12
⇒y=
2
[0.5]
2
⇒ x – 3x + 2 = 0
2
x
4
2
0
6
y
0
-3
-6
3
⇒ x – 2x – x + 2 = 0
[1]
⇒ x (x - 2) - 1 (x - 2) = 0
⇒ (x – 2) (x – 1) = 0
⇒ x = 2, 1.
nd
206, Aggarwal Mall, 2
Floor, Plot No. 3, Sector – 5, Dwarka, New Delhi - 110075
[1]
15
i
50.
ANSWER OF SAMPLE PAPER
cos ecθ
cos ecθ
+
cos ecθ − 1 cos ecθ + 1
=
cos ecθ(cos ecθ + 1) + cos ecθ(cos ecθ − 1)
(cos ecθ − 1)(cos ecθ + 1)
[1]
=
2 cos ec 2θ 2cos ec 2θ
=
cos ec 2θ − 1
cot 2 θ
[1]
2
2
2
4.
Draw any ray AX making an acute angle with AB
on the side opposite to the vertex C.
[1]
5.
Locate 7 points (the greater of 7 and 5 in 7/5) A1,
A2, A3, A4, A5, A6 and A7 on AX so that AA1 = A1A2
= A2A3 = A3A4 = A4A5 = A5A6 = A6A7.
6.
Join A5 (5 being smaller of 5 and 7 in 7/5) to B and
draw a line through A7 parallel to A5B, intersecting
the extended line segment AB at B'.
7.
Draw a line through B' parallel BC intersecting the
extended line segment AC at C' (see figure). Then
AB'C' is the required triangle.
[1]
2
= 2 cosec θ tan θ = 2 (cot θ + 1) tan θ
2
2
2
2
= 2 cot θ . tan θ + 2 tan θ = 2 + 2 tan θ = R.H.S. [1]
OR
L.H.S. =
cot A − cos A
cot A + cos A
cos A
cos A − sin A.cos A
− cos A
sin
A
sin A
=
=
cos A
cos A + cos A.sin A
+ cos A
sin A
sin A
C'
[1]
C
cos A − sin A.cos A cos A(1 − sin A)
=
=
cos A + sin A.cos A cos A(1 + sin A)
[1]
1
sin A
1 − sin A sin A − sin A cos ec A − 1
=
=
=
= R.H.S.
1 + sin A
1
sin A cos ec A + 1
+
sin A sin A
[1]
6 cm
51. Given BL and CM are medians of the ∆ ABC in which
∠A = 90°.
2
2
2
In ∆ ABC, BC = AB + AC ……….. (1)
2
2
2
2
7 cm
B
A
A1
[1]
A2
2
In ∆ ABL, BL = AL + AB = (AC/2) + AB .
2
2
2
⇒ 4BL = AC + 4 AB …………. (2)
B'
5 cm
A3
A4
A5
A6
[1]
A7
B
X
[1]
53. Given P(x, y) is any point on the line joining the points
A(a, 0) and B (0, b).
M
Let P(x, y) divide AB internally at the ratio k : 1.
A (a, 0)
A
2
2
x=
2
In ∆ CMA, CM = AC + AM
2
= AC + (AB/2)
2
2
2
⇒
2
⇒ 4CM = 4AC + AB …………. (3)
Adding (2) and (3), we get
2
2
2
2
4(BL + CM ) = 5 AB + 5 AC
2
2
2
[1]
a
x
1
kb
⇒ =
and y =
k +1 a k +1
k +1
y
k
=
b k +1
Now
= 5(AB + AC ) = 5 BC .
B (0, b)
Using the section formula, we get
L
C
P (x, y)
[1]
52. Steps of Construction :
[1]
x y
1
k
1+ k
+ =
+
=
=1
a b k +1 k +1 k +1
Hence
x y
+ = 1.
a b
[1]
54. Let the required ratio be k : 1.
1.
Draw a line segment AB = 5 cm.
Then by section formula, the co-ordinates of P are
2.
With A as centre and radius = 6 cm, draw and arc.
3.
Again, B as centre and radius = 7 cm, draw
another are cutting the arc in step 2 at C. Join AC
and BC. This ABC is the required triangle.
 4k − 3 −9k + 5 
 k +1 , k +1 


∴
4k − 3
−9k + 5
= 2 and
= −5
k +1
k +1
⇒ 4k – 3 = 2k + 2 and -9k + 5 = - 5k – 5
nd
206, Aggarwal Mall, 2
[1]
Floor, Plot No. 3, Sector – 5, Dwarka, New Delhi - 110075
[1]
16
i
ANSWER OF SAMPLE PAPER
⇒ k = 5/2
A
⇒ 4k = 10
M
N
⇒ k = 5/2
OR
E
D
[1]
Hence the ratio is 5 : 2.
C
B
[1.5]
Let the required ratio be k : 1
Let P(x, y) be the point on the given line dividing AB in
the ratio k : 1, then
Construction : Join BE and CD.
Draw DM ⊥ AC and EN ⊥ AB.
Proof : Consider ∆ ADE.
3k + 2
7k − 2
x=
and y =
k +1
k +1
[1]
The point P(x, y) lies on the line 2x + y – 4 = 0
ar(ADE) =
1
AD × EN.
2
In ∆BAE, DF || AE
 3k + 2  7k − 2
∴2
−4 = 0
+
 k +1  k +1
∴
⇒ 6k + 4 + 7k – 2 = 4k + 4
[1]
BD BF
=
......(1)
DA FE
Again in ∆ BAC, DE || AC
[0.5]
BD BF
∴
=
......(2)
DA CE
⇒ 9k = 2
⇒ k = 2/9.
[1]
Hence the required ratio is 2 : 9.
55. Required Area = area of semicircle with diameter AB +
area of semicircle with diameter AC + area of the
triangle ABC – area of the semicircle with diameter BC.
From (1) and (2) we get
Similarly, ar(BDE) =
ar(ADE) =
[1]
1
DB × EN
2
[1]
1
AE × DM
2
and ar(DEC) =
A
BF BF
=
.
FE CE
1
EC × DM.
2
 1
Therefore,
6 cm
8 cm
 
 2 
B
C
∴ Required area =
[1]
= 24 sq units.
SECTION – D
and,
2
56. Volume of earth dug out = πr h = π (1.5) × 14 =
3
31.5πm .
[2]
2
2
Area of circular ring = π(R – r ) = π[(5.5) – (1.5) ]
2
= 28 π m .

[1]

Also, ∆ BDE and ∆ DEC are on the same base DE, and
between the same parallels BC and DE.
Therefore, from (1), (2) and (3),
AD AE
=
DB EC
[1]
A
[2]
D
⇒ 28 π × h = 31.5 π
[1]
⇒ h = 31.5/28 = 1.125 m.
[1]
B
57. Given: A triangle ABC in which a line parallel to side BC
intersects other two sides AB and AC at D and E
respectively.
AD AE
=
.
DB EC

( ar ( ADE ) ) =   2  AE × DM  = AE . … (2)
( ar (DEC ) )   1  EC × DM  EC
 
 2 
Let the height of the embankment be h metres.
To prove:
…..(1)
Therefore, ar(BDE) = ar(DEC) …..(3)
2
2
 1
[1]
1
1
1
1
2
2
2
π(3) + π(4) +
4 × 8 - π(5) .
2
2
2
2
2

( ar ( ADE ) ) =   2  AD × EN  = AD
( ar (BDE ) )   1  DB × EN  DB
F
E
C
In ∆BAE, DF || AE
BD BF
∴
=
......(1)
DA FE
Again in ∆ BAC, DE || AC
BD BF
∴
=
......(2)
DA CE
206, Aggarwal Mall, 2nd Floor, Plot No. 3, Sector – 5, Dwarka, New Delhi - 110075
17
i
ANSWER OF SAMPLE PAPER
4 9 1
+ = ………… (2)
x y 2
and
[1]
[1]
Let 1/x = u and 1/y = v.
Then u + v = 1/12 ⇒ 12 u + 12 v = 1 ………… (3)
and 4u + 9v = 1/2 ⇒ 8u + 18 v = 1 …………. (4)
-
[1]
1 1
⇒ =
y 30
⇒ x = 20
⇒ y = 30.
D
[1]
1
[1]
= BD ………….. (1)
AE AE
=
CE BD
=
AB − 50
BD
[1]
⇒ BD = √3 (AB – 50) …………. (2)
[1.5]
From equation (1)
AB
3
= 3 (AB – 50)
⇒ AB = 3 AB – 150
132
x
[1]
⇒ AB = 75 m.
[1]
and BD = 75/√3 = 25√3 m.
Hence height of the tower is 75 m and the horizontal
distance between the building and the tower is 25√3 m.
OR
B
 132 
Time taken by express train to cover 132 km = 

 x + 11
hours.
[1]
132 132
−
=1
x
x + 11
[1]
1452
=1
x 2 + 11x
[1]
⇒
[1]
⇒ 2 AB = 150
Let the average speed of the passenger train be x
km/hr. Then the average speed of the express train = (x
+ 11) km/hr.
∴
[1]
AB
BD
Again in ∆ ACE, tan 30° =
3
OR
hrs.
AB
⇒
So the pipe of larger diameter alone can fill the pool in
20 hours and the pipe of smaller diameter alone can fill
the pool in 30 hours.
[0.5]
Time taken by passenger train to cover 132 km =
60°
3
[1]
C
AB
Here in ∆ ABD, tan 60° =
.
BD
⇒
Putting value of v in equation (3) we get u = 1/20
So u = 1/20 and v = 1/30
1 1
⇒ =
x 20
30°
E
⇒ 3=
-
- 30 v = - 1
⇒ v = 1/30.
60°
B
equation (3) × 2 : 24 u + 24 v = 2
equation (4) × 3 : 24 u + 54 v = 3
-
30°
Building
50 m
1 1 1
According to the question : + =
………… (1)
x y 12
A
Tower
BF BF
=
.
[1]
FE CE
58. Let the time taken by the pipe of larger diameter to fill
the pool be x hours and that taken by the pipe of smaller
diameter pipe alone be y hours.
From (1) and (2) we get
28.5 m
30 m
30°
D
F
60°
G
2
⇒ x + 11x – 1452 = 0
C
2
⇒ x + 44x – 33x – 1452 = 0
⇒ (x + 44) (x – 33) = 0
⇒ x = - 44 or x = 33
[1]
∴ x = 33 ( ∵ speed cannot be negative)
Hence, the average speed of passenger train is 33
km/hr and the average speed of the express train is 44
km.
[1]
59. In the given figure AB is the tower and CD is the building
of 50 m.
xm
E
ym
A
[1]
Let AB = 30 m be the height of the building CD = 1.5 m
be the height of the boy.
At D, the angle of elevation is 30° and at F, the angle of
elevation is 60°.
[1]
We have AB = 30 m, AG = CD = EF = 1.5 m.
∴ BG = 30 – 1.5 = 28.5 m.
Let CE = DF = x m and EA = FG = ym.
In ∆ BFG, tan 60° = BG/FG.
206, Aggarwal Mall, 2nd Floor, Plot No. 3, Sector – 5, Dwarka, New Delhi - 110075
[1]
18
i
ANSWER OF SAMPLE PAPER
⇒ 3=
28.5
28.5
⇒y=
= 9.5 3
y
3
[1]
In ∆ BDG, tan 30° = BG/DG.
Mean x =
∑ fix i 2830
=
= 35.375 years
∑ fi
80
[1.5]
Here N = 80 ⇒ N/2 = 40. So Median class is 35 – 45.
1
28.5
⇒
=
x
+y
3
[1]
⇒ x + y = 28.5 √3
⇒ x = 28.5 √3 – 9.5 √3 = 19√3 m.
[1]
60.
l = 35, Cf = 38, f = 23, h = 10.
[0.5]
N

 − Cf 
 40 − 36 
× h = 35 + 
Now Median = l +  2
 × 10
f 
 23 




CI
fi
xi
fixi
C.f.
5 – 15
6
10
60
6
15 – 25
11
20
220
17
25 – 35
21
30
630
38
35 – 45
23
40
920
61
45 – 55
14
50
700
75
 f1 − f 0 
Therefore Mode = l + 
×h
 2f1 − f 0 − f 2 
55 – 65
5
60
300
80
Putting l = 35, f1 = 23, f0 = 21, f2 = 14, h = 10, we get
Σfi = 80
= 35 +
20
= 35.87 years
23
[1]
Again the modal class is 35 – 45 as the highest
frequency is 23.
[0.5]
Σfixi = 2830
[1.5]
Mode = 35 +
2
× 10 = 36.81 years .
11
206, Aggarwal Mall, 2nd Floor, Plot No. 3, Sector – 5, Dwarka, New Delhi - 110075
[1]