2016 10 Lyp Maths Set 02 Delhi Ans Oieosi1
2016 10 Lyp Maths Set 02 Delhi Ans Oieosi1
2016 10 Lyp Maths Set 02 Delhi Ans Oieosi1
Set-2
Time: 3 hrs M.M: 90
Section A
Ans1: It is given that the box contains cards marked with numbers 3, 4, 5, ..., 50.
Total number of outcomes = 48
Between the numbers 3 and 50, there are six perfect squares, i.e. 4, 9, 16, 25, 36 and 49.
Number of favourable outcomes = 6
Probability that a card drawn at random bears a perfect square
Number of favourable outcomes 6 1
= = =
Total number of outcomes 48 8
Ans2: In the given figure,
AB = AD + DB = 6 m
Given: AD = 2.54 m
2.54 m + DB = 6 m
DB = 3.46 m
BD
= sin 60
Now, in the right triangle BCD, CD
3.46 m 3
=
CD 2
3.46 m 1.73
=
CD 2
2 3.46 m
CD =
1.73
CD = 4 m
Thus, the length of the ladder CD is 4 m.
Ans3: Common difference, d, of the AP = 9 5 = 4
Last term, l, of the AP = 185
We know that the nth term from the end of an AP is given by l (n 1)d.
Thus, the 9th term from the end is
185 (9 1)4
= 185 4 8
= 185 32
= 153
Ans4:
Section B
Ans5: Let the y-coordinate of the point P be a.
Then, its x-coordinate will be 2a.
Thus, the coordinates of the point P are (2a, a).
It is given that the point P (2a, a) is equidistant from Q (2, 5) and R (3, 6).
Thus, we have
( 2a 2 ) + (a (5) 2 ) = (2a (3)2 + (a 6) 2
2
Let the four shaded regions be I, II, III and IV and the centres of the semicircles be P, Q, R
and S, as shown in the figure.
It is given that the side of the square is 14 cm.
Now,
Area of region I + Area of region III = Area of the square Areas of the semicircles with
centres S
1
and Q. = 14 14 2 7 2 ( Radius of the semicircle=7 cm)
2
22
196 49
= 7
=196154
= 42 cm 2
Similarly,
Area of region II + Area of region IV = Area of the square Areas of the semicircles with
1
centres P and R. = 14 14 2 7 2 ( Radius of the semicircle=7 cm)
2
22
196 49
= 7
=196154
= 42 cm 2
= 84 cm 2
Ans14:
Surface area of the block = Total surface area of the cube Base area of the hemisphere +
Curved surface area of the hemisphere
= 6 ( Edge ) r 2 + 2 r 2
2
= ( 63 + r 2 )
22 3.5 3.5
= 216 +
7 2 2
= (216+9.625)
2
= 225.625 cm
Ans15:
360 4
60
(102 )
3
= (10 )
2
360 4
2 2
Area of the shaded region (part III)
25 3
= 100
2 6 4
25 100
= + 25 3
2 6
25
= 25 3
6
= 25 3 sq units
6
Hence proved.
Ans17:
Let the height of the tower AB be h m and the horizontal distance between the tower and
the building BC be x m.
So,
AE=(h50) m
InAED,
AE
tan 45 =
ED
h 50
1=
x
x = h 50 ..... (1)
In ABC ,
AB
tan60 =
BC
H
3=
x
x 3 = h ..... ( 2 )
50( 3 + 1)
x= = 25 2.73 = 68.25 m
2
Substituting the value of x in (1), we get
68.25 = h 50
h = 68.25 + 50
h = 118.25 m
Hence, the height of tower is 118.25 m and the horizontal distance between the tower
and the building is 68.25 m.
x +1 x 2 2x + 3
Ans18: + = 4
x 1 x + 2 x2
( x + 1)( x + 2 ) + ( x 1)( x 2 )
( x 1)( x + 2 )
4 ( x 2 ) ( 2 x + 3)
=
x2
( x 2
+ 2 x + x + 2) + ( x 2 2 x x + 2)
x2 + 2x x 2
4x 8 2x 3
=
x2
x + 3x + 2 + x2 3x + 2
2
x2 + x 2
2 x 11
=
x2
2 x2 + 4 2 x 11
=
x2 + x 2 x2
( 2x + 4) ( x 2)
2
= ( 2 x 11) ( x 2 + x 2 )
2 x3 4 x 2 + 4 x 8
= 2 x 3 + 2 x 2 4 x 11x 2 11x + 22
2 x3 4 x 2 + 4 x 8
= 2 x 3 9 x 2 15 x + 22
2 x 3 2 x 3 4 x 2 + 9 x 2 + 4 x + 15 x 8 22 = 0
5 x 2 + 19 x 30 = 0
5 x 2 + 25 x 6 x 30 = 0
5 x ( x + 5) 6 ( x + 5 ) = 0
Let h and l be the height and slant height of the cone, respectively.
CSA of the cone = 47.1 cm
2
rl = 47.1
3.14 3 l = 47.1
47.1
l =
9.42
l = 5 cm
Section D
Ans21: Let the usual speed of the plane be x km/h.
Let the time taken by the plane to reach the destination be t1.
1500
t1 =
x
To reach the destination on time, the speed of the plane was increased to (x + 250)km/h.
1500
t2 =
x + 250
t t = 30 min
Given: 1 2
Now,
1500 1500 30
=
x x + 250 60
1500 ( x + 250 x ) 1
=
x ( x + 250 ) 2
750000 = x 2 + 250 x
x 2 + 250 x 750000 = 0
x 2 + 250 x 750000 = 0
On solving the equation, we get
x=750
Thus,
Usual speed of the plane = 750 km/h
The value depicted in this question is that of humanity. The pilot has set an example of a
good and responsible citizen of the society.
Ans22: From the given figure, we have
TP = TQ (Two tangents, drawn from an external point to a circle, have equal
length.)
And
TQO = TPO = 90 (Tangent to a circle is perpendicular to the radius through the
point of contact.)
In TOQ,
x 2 + 64 = 144 + x 2 24 x
24 x = 80
80 10
x= =
24 3
20
AB = 2 x = cm
3
20
Thus, the length of AB is cm.
3
Ans23: Given: A circle with centre O, a point P lying outside the circle and PQ and PR as the two
tangents
To prove: PQ = PR
Construction: Join OP, OQ and OR.
Proof:
Given:
OA = 5 cm
OP = 10 cm
We know that the tangent at any point of a circle is perpendicular to the radius through
the point of contact.
Therefore, OAP is a right-angled triangle.
OAP = 90
Now,
OP 2 = OA2 + AP 2
102 = 52 + AP 2
AP 2 = 75
AP = 5 3 cm
1200
Area of sector OACB = 3.14 ( 5 )
2
360
1
= 3.14 25 = 26.17 cm 2 ( Approx.)
3
Area of the shaded region = (Area of OAP + Area of OBP) Area of the sector OACB
= 43.25 cm 2 26.17 cm 2
= 17.08 cm 2 ( Approx.)
Ans27: Consider the following figure:
Given:
3
Volume of the frustum is 12308.8 cm .
Radii of the top and bottom are r1 = 20 cm and r2 = 12 cm, respectively.
l = 152 152 + ( 20 12 )
2
= 225 + 64 = 289 = 17 cm
Curved surface area of the frustum
= ( 20 + 12 )17
= 544 3.14
= 1708.16 cm2
Area of the base = 122 = 144 3.14 = 452.16 cm2
Metal sheet required to make the frustum=1708.16+452.16=2160.32 cm2
Ans28: Let the height of the tower be h m.
In ABP,
h
tan 600 =
4
h
3=
4
h = 1.73 4 = 6.92 cm