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QUESTION PAPER CODE 30/1/1
EXPECTED ANSWERS/VALUE POINTS
SECTION - A
Q.No.
1.
–9
4
2.
1:3
3.
21
26
Marks
4.
25
o
1×4 = 4 m
SECTION - B
5.
 ABQ 
1
 AOQ  29o
2
1m
 ATQ  180o –  ABQ   BAT   180 o – 119o  61o
6.
The given quadratic equation can be written as
4x
2

– 4a 2 x  a 2 – b 4  0
½m
 – b 
1m

or 2x – a 2
2

2 2
 2x – a 2  b 2

7.
1m
x
0
 2x – a
2
– b2

2





0
a 2 – b2 a 2  b2
,
2
2
In Δs TPC and TQC
TP  TQ
TC  TC
 1   2 (TP and TQ are equally
inclined to OT)
½m







1m
 Δ TPC  Δ TQC
 PC  QC and  3   4
2
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½m
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 3   4  180 o   3   4  90 o 

 OT is the right bisector of PQ

But
8.
9.
The given A.P. is 6, 13, 20, ---, 216
Let n be the number of terms, d = 7, a = 6
½m
 216 = 6 + (n – 1) .7 
½m
n = 31
 Middle term is 16th
½m
 a16 = 6 + 15 × 7 = 111
½m
ABC is right triangle

2
2
2
AC = BC + AB
2
2
AB 2  5 – 2   2  2   25  AB  5 

2
2
2
BC 2  2  2   t  2   16  t  2 


2
2
2
2
AC  5  2   2 – t   49  2 – t 


49  2 – t   41  t  2 
2
t  22 – 2 – t 2 
2
8
4  2t  8  t  1
10.
½m





1m
Let P divide AB in the ratio of k : 1
1
2  3  8 K  2  3K  3
K 1
4
 K1
5
2K

 Required ratio = 1 : 5
3
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1m
½m






1m
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SECTION - C
11.
P is the mid-point of AB
x+1=4  x=3

similarly y = 2
B (3, 2)

1m
similarly finding C (–1, 2)

12.
Area Δ ABC 
½m
1
1 2 – 2  3 2  4  – 1 – 4 – 2   1  24  12 sq.u.
2
2
1½ m
The given quadratic eqn. can be written as
(k + 1)x2 – 2(k – 1)x + 1 = 0
1m
For qual roots 4(k – 1) 2 – 4(k  1)  0
or k 2 – 3k  0
 k  0, 3
 Non-zero value of k = 3 : Roots are
1 1
,
2 2
13.
14.



1m
½+½ m
Fiqure
½m
(i)
30
 tan 45o  1  y  30
y
1m
(ii)
x
1
y
30
 tan 30 o 
 x

 10 3
y
3
3
3
1m

Height of building is 10 3 m
½m
The possible outcomes are (2, 3), (3, 2), (1, 4), (4, 1) : Number : 4
1m
Total possible out comes = 36
(i)

Required Probabilit y 
4 1

36 9
4
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(ii)
The possible outcomes are
(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)
1m
their number is 9

15.
16.
Required Probabilit y 
9 1

36 4
Let a be the first term and d the common difference
S12 = 6 [2a + 11d] = 12a + 66d
1m
S8 = 4 [2a + 7d] = 8a + 28d
½m
S4 = 2 [2a + 3d] = 4a + 6d
½m
3 (S8 – 34) = 3 (4a + 22d) = 12a + 66d = S12
1m
Let OA = OB = r


40 
22 r 22
 
 r  r  280  40r
7 2 7
r7




 1 22 7 7 1 22

shaded area   
   
 7  7  cm 2
2 7 2 2 2 7

5
385
1

  77   or
cm 2  96 cm 2
4
4
4

17.
½m
ARQ ~ ADC

1m
1m
1m
½m
x 4

 x  2
6 12
½m
82  4 2  4 5
½m
QC 
5
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Total surface area of frustum PQCB
 π

18.
6  2  4
5  6   2
2
2

22
32  2.236  40  22 111.552   22  15.936
7
7
 350.592




1m





1m
Volume of solid wooden toy
5
2 22 7 7 7 1 22 7 7
 
    
  h
6
3 7 2 2 2 3 7 2 2
1001 22 7 7
or

  7  h 
6
7 2 2
166

7 h 
1001 7
 13
22  7

h  6 cm
Area of hemispherical part of toy   2  22  7  7  cm 2
7 2 2

 77 cm 2

19.
1m





Cost of Paenting = Rs. (77 × 10) = Rs. 770
½+½ m
½m
½m
Total surfacearea of solid cuboidal block
= 2 (15 × 10 + 10 × 5 + 15 × 5) cm2 = 550 cm2
Area of two circular bases = 2 
22 7 7
   77 cm 2
7 2 2
Area of curved surface of cylinder = 2rh  2 
22 7
  5  110 cm 2
7 2
Reqd - area = (550 + 110 – 77) cm2 = 583 cm2
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1m
½m
1m
½m
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20.
Area of Sq. ABCD = 142 or 196 cm2
½m
Area of Small Sq. = 42 or 16 cm2
½m
Area of 4 semi circles  4. 1 3.14 (2) 2  cm 2
 2

 25.12 cm 2
 Reqd. area  (196 – 16 – 25.12) cm 2 


 154.88 cm 2





1m
1m
SECTION - D
21.
Let the fraction be
x –3
x
½m
By the given condition, new fraction



x – 3 2 x –1

x2
x2
x – 3 x – 1 29


x
x  2 20
20 x – 3 x  2  x x – 1  29 x 2  2x


2

2

2
20 x – x – 6  x – x  29x  58x







½m
1m
or 11x2 – 98x – 120 = 0
or 11x2 – 110x – 12x – 120 = 0
(11x + 12) (x – 10) = 0
 The Fraction is
22.

1m
x = 10
7
10
1m
1m
Money required for Ramkate for admission of daughter = Rs. 2500
A.P. formed by saving
1m
7
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(i)
= 100, 120, 140, --- upto 12 terms
12
2 100  11 20  6 420
2
 Rs. 2520




Sum of AP (i) 

23.
She can get her doughter admitied
½m
Value : Small saving can fulfill your big desires or any else
1m
2
3
23


x  1 2x – 2
5x
or 5x 4 x – 2  3x  3  46 x  1x – 2 

1½ m

5x 7x – 5  46 x 2 – x – 2  11x 2 – 21x – 92  0

x
25.
1m
21  441  4048
21  67

22
22
1m
– 23
11
½m
 4,
24.
1½ m
Correctly stated
Given, to Prove, Construction and correct figure
2m
correct Proof
2m
PR  PQ   PRQ   PQR 
180 – 30o
2
 75o
SR | | QP and QR is a transversal   SRQ  75o
  ORQ   RQO  90o – 75o  15o
1m



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1m
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  QOR  180 – 2  15  150o   QSR  75o
o
1m
 RQS  180 o –  SRQ   SQR   30o
26.
1m
Correctly drawn  ABC
1½ m
Correctly drawn a triangle similar to  ABC of given scale factor
2½ m
27.
figure
1m
(i)
x
1
 tan 30o 
 y 3x
y
3
1m
(ii)
x 5
x 5
 tan 60 o  3 or
 3
y
3x
Writing the trigonometric equations


28.
(i)
3x  x  5
or x  2.5
Height of Tower  2.5 m






Required Probabilit y 
13
20
Prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19 : 8 in number

29.
½m
Numbers divisible by 2 or 3 from 1 to 20 are
2, 4, 6, 8, 10, 12, 14, 16, 18, 3, 9, 15 Their number is 13
(ii)
1½ m
Required Probabilit y 
8
2
or
20
5
Area ABC

1
– 4 – 4  5 – 3 – 5 – 8  0 8  4 
2
9
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1m
1m
1m
1m
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
1
2
– 4  39 
35
2
1½ m
Area of ACD

1
– 4 – 5 – 6  0 6 – 8  5 8  5
2

109
2
1½ m
35 109

 72 sq.u.
2
2
1 m.
 22

   2  2  14  cu.m  176 cu.m ................. (i)
 7

1m
 Area of Qurd. ABCD =
30.
Volume of earth taken out after digging the well
Let x be the width of embankment formed by using (i)
Volume of embankment 

22
2  x 2 – 22  40  176
7
100

x2 + 4x – 140 = 0


31.


(x + 14) (x – 10) = 0
x = 10
1½ m





1½ m
Width of embankment = 10 m
Let x m be the internal radius of the pipe
Radius of base of tank = 40 cm =
2
m
5
Level of water raised in the tank = 3.15 or
315
100
2.52 km/hour  1.26 km in half hour = 1260 m
10
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1m
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
Getting the equation
2 2 315
π x 2 . 1260  π . . 
5 5 100
1m
4
315
1
1
.


25 100
1260
2500
1
 x
m  2 cm
50
 x2 






Internal diameter of pipe = 4 cm
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1½
½m