2016 10 mathematics_sample_paper_sa2_04_ans_w9tafd6

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CBSE Sample Paper-04
SUMMATIVE ASSESSMENT –II
MATHEMATICS
Class – X
(Solutions)
SECTION-A
1. (d)
2. (d)
3. (d)
4. (c)
5. According to the question,
Area of new circle = Sum of areas of two circles
⇒ 2 2 2
1 2r r rπ π π= +
⇒ ( )2 2 2
1 2r r rπ π= +
⇒ 2 2 2
1 2r r r= +
⇒ 2 2 2
3 4 9 16 25r = + = + = ⇒ r =5 cm
6. Let n cones be formed. Then,
n× Volume of one cone = Volume of sphere
⇒ 2 3
1
1 4
.
3 3
n r h rπ π= ⇒ 2 3
1. 4n r h r=
⇒ ( ) ( ) ( )
2 3
4 3 4 12n× = × ⇒
4 12 12 12
4 4 3
n
× × ×
=
× ×
⇒ 144n =
7. Volume of water that flows out through the pipe in 1 hour.
=
2
22 14
15 1000 100
7 2
 
× × × × 
 
cm3
Volume of tank = 50 x 44 x 21 x 100 x 100 cm3
∴ Required time = 2
50 44 21 100 100
22 14
15 1000 100
7 2
× × × ×
 
× × × × 
 
= 2 hours
8. 2
6 9 0x x+ + =
⇒ ( ) ( )
2 2
2.3. 3x x+ +
⇒ ( )
2
3x + ⇒ 3, 3x = − −
Hence 3x = − is a solution.

9. 21, 42 21 21a d= = − =
( )1na a n d= + −
⇒ 420 = ( )21 1 21n+ − ⇒ ( )1 21 399n − =
⇒ 1 19n − = ⇒ 20n =
∴ 20th term is 420.
10. ∠ BPA = 90 and ∠ PBA = 35°
∴ ∠ PAB = ( )180 90 35 55°− °+ ° = °
∵ OA = OP
∴ ∠ OAP = ∠ OPA = 55°
∠ OPT = 90
⇒ ∠ OPA = ∠ APT = 90
⇒ 55° + ∠ APT = 90
⇒ ∠ APT = 35°
∴ ∠ PTA = 55 35 20°− ° = °
11. For collinear,
( ) ( ) ( )1 2 3 2 3 1 3 1 2x y y x y y x y y− + − + −
⇒ ( ) ( ) ( ) ( )1 1 2 1 3 5 3 0p p− − − + − − + − =  
⇒ ( )1 2 6 15 5 0p p− − − − + − =
⇒ 6 6 0p− + =
⇒ 1p =
12. Let A ( )3,0 ,→ B ( )6,4→ and C ( )1,3 .→ −
AB = ( ) ( )
2 2
6 3 4 0− + − = 2 2
3 4 25 5+ = = units
BC = ( ) ( )
2 2
1 6 3 4− − + − = ( ) ( )
2 2
7 1 50 5 2− + − = = units
CA = ( ) ( )
2 2
3 1 0 3+ + − = ( ) ( )
2 2
4 3 25 5+ − = =
∴ ∆ ABC is an isosceles triangle.
AB2 + CA2 = BC2
⇒ ( )
2
2 2
5 5 5 2+ =
⇒ 25 + 25 = 25 x 2
⇒ 50 = 50
∴ ∆ ABC is a right angled triangle.
13. Side of square = 121 = 11 cm
∴ Perimeter of square = 4 x 11 = 44 cm
∴ Circumference of circle = 44 cm
⇒ 2 rπ = 44

⇒
44 7
2 22
r
×
=
×
= 7 cm
∴ Area of the circle = 2
rπ
=
22
7 7
7
× × = 154 cm2
14. ∵ 2 rπ = 220
⇒ r = 35 m
Width of track = 15 m
∴ External radius = 35 + 15 = 50 m
∴ Area of circular track = 2 2
R rπ π−
= ( ) ( )
2 2
50 35π π−
= ( )2500 1225π −
=
22
1275
7
×
=
28050
7
m2
15. Volume of hemispherical bowl = 32
3
rπ
= ( )
32
9
3
π cm3
Volume of one bottle = 2
r hπ
= ( )
2
3
4
2
π
 
 
 
∴ Number of bottles required =
Volume of hemisphere
Volume of 1 bottle
=
( )
( )
3
2
2
9
3
3
4
2
π
π
 
 
 
= 54
16. Given,
1
, 11, 1
2
a b c= = − =
∵
2
4
2
b b ac
x
a
− ± −
=
∴
( ) ( )
2 1
11 11 4. .1
2
1
2
2
x
− − ± − −
=
 
 
 
⇒
11 11 2
1
x
± −
= ⇒ 11 3x = ±

17. According to questions, the number are 10, 13, 16, 19, ……..97
Here, 10, 3, 97a d l= = =
∵ ( )1l a n d= + −
∴ ( )97 10 1 3n= + − 97 ⇒ ( )3 1 87n − =
⇒ 1 29n − = ⇒ 30n =
∵ ( )
2
n
n
S a l= +
∴ ( )
30
10 97
2
nS = + = 15 x 107 = 1605
18. ∵ Area of triangle =
1
2
Base x Height
∴ Area of ∆ ABC = Area of ∆ AOB + Area of ∆ BOC + Area of ∆ COA
⇒
2 2
6 8 6 6 8 8
2 2 2 2
r r r× × + × ×
= + +
⇒ 24 3 5 4r r r= + +
⇒ 24 12r= ⇒ 2r = cm
19. In right triangle ABD,
AB
tan 60
BD
=
⇒
200
3
BD
=
⇒ BD =
200
3
= 115.33 m
In right triangle AEC,

AE
tan30
EC
=
⇒ 1 AE
2003
3
=
 
 
 
[∵ EC = BD]
⇒ AE =
200
3
= 66.67 m
∴ CD = EB = AB – AE
= 200 – 66.67 = 133.33 m
20. There are 52 cards in a pack, out of which six cards are removed.
∴ Total number of possible outcome = 52 – (2 + 2 + 2) = 46
(i) Number of kings = 2
Hence, Required probability =
2 1
46 23
=
(ii) Number of red cards = 26 – 6 = 20
Hence required probability =
20 10
46 23
=
(iii) Number of spades = 3
Hence required probability =
13
46
21. Steps of construction:
(a) Draw a right angled triangle ABC with given measurements.
(b) Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
(c) Locate 5 points B1, B2, B3, B4, B5 on BX so that BB1 = B1B2 = B2B3 = B3B4 = B4B5.
(d) Join B5C and draw a line through B4 parallel to B5C, intersecting the extended line segment
BC at C’.
(e) Draw a line through C’ parallel to CA intersecting the extended line segment BA at A’.
The A’BC’ is the required triangle.
22. In right triangle PBA,

4000
tan 60
AB
=
⇒ AB =
4000
3
……….(i)
In right triangle ABQ,
BQ
tan 45
AB
=
⇒ AB = BQ [∵ tan 45 1= ]
∴ BQ =
4000
3
……….(ii)
∴ PQ = PB – BQ
⇒ PQ =
4000
4000
3
− =
4000 3 4000
3
−
= 1690.4 m
23. Total number of balls in the bag = 4 + 6 + 7 + 3 = 20
∴ Number of all possible outcomes = 20
(i) Number of white balls = 4
∴ Required probability =
4 1
20 5
=
(ii) Number of balls ‘not black’ = 4 + 6 + 3 = 13
13
20
(iii) Number of balls ‘neither white nor black = 6 + 3= 9
9
20
(iv) Number of balls ‘red or white’ = 6 + 4 = 10
10 1
20 2
=
24. B ( )0,0→
∵ BC lies on x − axis.
∴ C ( ),0a→

Draw AD = BC, then BD = DC =
2
a
In right ∆ ADB, AB2 = AD2 + BD2 [Pythagoras theorem]
⇒
2
2 2
AD
2
a
a
 
= +  
 
⇒ AD =
3
2
a
∴ A
3
,
2 2
a a 
→   
 
25. Total surface area of the solid
= Curved surface area of cylinder + 2 x Curved surface area of the hemisphere
= 2
2 2 2rh rπ π+ ×
= ( )2 2r h rπ +
= ( )
22
2 14 58 14 14 2 14
7
× × − + + ×  
= [ ]88 58 28 28− +
= 5104 cm2
26. Volume = 12308.8 cm3
1r = 20 cm
2r = 12 cm
Volume (V) = ( )2 2
1 2 1 2
1
3
h r r rrπ + +
⇒ ( ) ( )
2 21
12308.8 3.14 20 12 20 12
3
h  = × × + + ×
 
⇒ [ ]
3.14
12308.8 400 144 240
3
h= + +
⇒ 12308.8 1.05 784h= × ×
⇒ h = 15 cm
And ( )
22
1 2l h r r= + − = ( ) ( )
2 2
15 20 12+ − = 225 64− = 17 cm
∴ Area of metal sheet used = ( ) 2
1 2 2l r r rπ π+ +
= ( ) ( )
2
3.14 17 20 12 3.14 12× + + ×
= 3.14 x 17 x 32 + 3.14 x 144
= 2160.32 cm2

27. 1 1 1 1
p q x p q x
= + +
+ +
⇒ 1 1 1 1
p q x x p q
− = +
+ +
⇒ ( )
( )
x p q x p q
p q x x pq
− + + +
=
+ +
⇒
( )
1 1
p q x pq
−
=
+ +
⇒ ( ) 0p q x x pq+ + + =
⇒ ( )2
0x p q x pq+ + + = ⇒ ( )( ) 0x p x q+ + =
⇒ ,x p q= − −
28. Let he buys x articles.
C.P. of x articles = Rs. 900
∴ C.P. of 1 article = Rs.
900
x
No. of articles damaged = 5
∴ No. of articles not damaged = ( )5x −
S.P. of 1 article = Rs. 900
2
x
 
+ 
 
∴ S.P. of ( )5x − articles = Rs. 900
2
x
 
+ 
 
( )5x −
According to the question,
900
2
x
 
+ 
 
( )5x − − 900 = 80
⇒ ( )( )900 2 5 980x x x+ − = ⇒ 2
900 4500 2 10 980x x x x− + − =
⇒ 2
2 90 4500 0x x− − = ⇒ 2
45 2250 0x x− − =
⇒ 2
75 30 2250 0x x x− + − = ⇒ ( ) ( )2
75 30 75 0x x x− + − =
⇒ ( )( )75 30 0x x− + = ⇒ 75, 30x = −
30x = − is inadmissible as x is the number of articles.
Hence, No. of articles = 75
29. (i) Nidhi saves on first day = Rs.2, on second day = Rs.4, on third day = Rs.4 and so on.
Thus, savings form an AP, whose first term 2a =
Common difference ( )d = 4 – 2 = 2
We know that year 2012 is a leap year. So there are 29 days in month of February.
So, we have to find her savings in 29 days, i.e. 29n =
Now, ( )2 1
2
n
n
S a n d= + −  
⇒ ( )29
29
2 2 29 1 2
2
S = × + −   = [ ]
29
4 28 2
2
+ ×
=
29
60
2
× = 870
(ii) Economy, Saving

30. ∵ Tangent segments from an external point to a circle are equal in length.
∴ TP = TQ
And ∠ TQP = ∠ TPQ ……….(i) [Angles opposite to equal sides of a ∆ are equal]
∵ Tangent is perpendicular to the radius through the point of contact.
∴ ∠ OPT = 90 ………(ii)
⇒ ∠ OPQ + ∠ TPQ = 90
⇒ ∠ TPQ = 90 −∠OPQ ……….(iii)
In ∆ TPQ,
∠ PTQ + ∠ TPQ + ∠ TQP = 180° [Angle sum property of a ∆ ]
⇒ ∠ PTQ + 2∠ TPQ = 180° ……….(iv) [From eq. (ii)]
⇒ ∠ PTQ + 2(90 − ∠ OPQ) = 180° [From eq. (iii)]
⇒ ∠ PTQ = 2∠ OPQ
31. First part:Given : A circle with centre O and radius r and a tangent AB at a point P.
To Prove : OP⊥ AB
Construction: Take any point Q, other than P on the tangent AB. Join OQ. Suppose OQ
meets the circle at R.
Proof : Clearly OP = OQ [Radii]
Now, OQ = OR + RQ
⇒ OQ> OR
⇒ OQ> OP [OP = OQ]
⇒ OP <OQ
Thus, OP is shorter than any segment joining O to any point of AB.
So, OP is perpendicular to AB.
Hence,
OT = OT’ ……. (Radii of the same circle)
and OP = OP ……..(Common)
∴ ∆ OTP ≅ ∆ OT’P ……..(RHS congruency)
Hence, OP⊥ AB
Second part: Using the above, we get,
∠ OPQ = 90
OP2 + PQ2 = OQ2 [By Pythagoras theorem]
⇒ 122 = 52 + PQ2
⇒ PQ = 119 cm

2016 10 mathematics_sample_paper_sa2_04_ans_w9tafd6

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2016 10 mathematics_sample_paper_sa2_04_ans_w9tafd6