Download CBSE X Mathematics 2012 Solution (SET 1)

CBSE X Mathematics 2012 Solution (SET 1)
Section B
Q11. Find the value(s) of k so that the quadratic equation x2 – 4kx + k = 0 has equal roots.
Solution:
Given equation is x2  4kx  k  0
For the given equation to have equal roots, D = 0.
 b 2  4ac  0
  4k   4 1  k   0
2
 16k 2  4k  0
 4k  4k  1  0
 4k  0 or 4k  1  0
1
 k  0 or k 
4
1
Hence, for k = 0 or , the given equation will have equal roots.
4
Q12. Find the sum of all three digit natural numbers, which are multiples of 11.
Solution:
The smallest and the largest three digit natural numbers, which are divisible by 11 are 110 and
990 respectively.
So, the sequence of three digit numbers which are divisible by 11 are 110, 121, 132, …, 990.
Clearly, it is an A.P. with first term, a = 110 and common difference, d = 11.
Let there be n terms in the sequence.
So, an = 990
 a   n  1 d  990
 110   n  1 11  990
 110  11n  11  990
 11n  99  990
 11n  990  99
 11n  891
 n  81
n
Now, required sum   2a   n  1 d 
2
CBSE X Mathematics 2012 Solution (SET 1)
81
2 110   81  1 11
2 
81

220  880
2
81
 1100
2
 44550



Hence, the sum of all three digit numbers which are multiples of 11 is 44550.
Q13. Tangents PA and PB are drawn from an external point P to two concentric circles with
centre O and radii 8 cm and 5 cm respectively, as shown in Fig. 3. If AP = 15 cm, then
find the length of BP.
Solution:
Given that: OA = 8 cm, OB = 5 cm and AP = 15 cm
To find: BP
Construction: Join OP.
 Tangent to a circle is perpendicular to the 
Now, OA  AP and OB  BP 

radius through the point of contact


 OAP  OBP  90
On applying Pythagoras theorem in OAP, we obtain:
(OP)2 = (OA)2 + (AP)2
 (OP)2 = (8)2 + (15)2
 (OP)2 = 64 + 225
CBSE X Mathematics 2012 Solution (SET 1)
 (OP)2 = 289
 OP = 289 
 OP = 17
Thus, the length of OP is 17 cm.
On applying Pythagoras theorem in OBP, we obtain:
(OP)2 = (OB)2 + (BP)2
 (17)2 = (5)2 + (BP)2
289 = 25 + (BP)2
 (BP)2 = 289 – 25
 (BP)2 = 264
 BP = 16.25 cm (approx.)
Hence, the length of BP is 16.25 cm.
Q14. In Fig. 4, an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that the
point of contact P bisects the base BC.
Solution:
Given: An isosceles ABC with AB = AC, circumscribing a circle.
To prove: P bisects BC
Proof: AR and AQ are the tangents drawn from an external point A to the circle.
 AR = AQ (Tangents drawn from an external point to the circle are equal)
Similarly, BR = BP and CP = CQ.
It is given that in  ABC, AB = AC.
 AR + RB = AQ + QC
 BR = QC (As AR = AQ)
 BP = CP (As BR = BP and CP = CQ)
 P bisects BC
Hence, the result is proved.
OR
In Fig. 5, the chord AB of the larger of the two concentric circles, with centre O, touches the
smaller circle at C. Prove that AC = CB.
CBSE X Mathematics 2012 Solution (SET 1)
Solution:
Given: Two concentric circles C1 and C2 with centre O, and AB is the chord of C1 touching C2 at
C.
To prove: AC = CB
Construction: Join OC.
Proof: AB is the chord of C1 touching C2 at C, then AB is the tangent to C2 at C with OC as its
radius.
We know that the tangent at any point of a circle is perpendicular to the radius through the point
of contact.
 OC  AB
Considering, AB as the chord of the circle C1. So, OC  AB.
 OC is the bisector of the chord AB.
Hence, AC = CB (Perpendicular from the centre to the chord bisects the chord).
Q15. The volume of a hemisphere is 2425
1
cm3. Find its curved surface area.
2
Solution:
1
4851
cm3 
cm3
2
2
Let the radius of the hemisphere be ‘r’ cm.
Given, volume of hemisphere  2425
22 

 Use   7 
CBSE X Mathematics 2012 Solution (SET 1)
2
Volume of hemisphere   r 3
3
2 3 4851
 r 
3
2
2 22
4851
  r3 
3 7
2
4851  3  7
 r3 
2  2  22
441  21
 r3 
2 2 2
21  21  21
 r3 
2 2 2
21  21  21
r 3
2 2 2
21
 r  cm
... 1
2
Curved surface area of hemisphere = 2πr2
2
22  21 
 
7  2
22 21 21
 2 
7
2 2
2
= 693 cm
 2
 using 1 
Q16. In Fig. 6, OABC is a square of side 7 cm. If OAPC is a quadrant of a circle with centre O,
22 

then find the area of the shaded region.  Use  
7 

Solution:
It is given that OABC is a square of side 7 cm
 Area of square OABC = (7)2 cm2 = 49 cm2
CBSE X Mathematics 2012 Solution (SET 1)
Also, it is given that OAPC is a quadrant of circle with centre O.
Radius of the quadrant of the circle = OA = 7 cm
1
 Area of the quadrant of circle  πr 2
4
1
  π  7 2  cm 2
4
49π

cm 2
4
49 22
 
cm 2
4 7
77

cm 2
2
 
Area of the shaded region = Area of Square – Area of Quadrant of circle.
77 

  49   cm 2
2 

 98  77  2

 cm
 2 
21

cm 2
2
= 10.5 cm2
Thus, the area of the shaded region is 10.5 cm2.
Q17. If a point A (0, 2) is equidistant from the points B (3, p) and C (p, 5), then find the value of
p.
Solution:
The given points are A (0, 2), B, (3, P) and C (P, 5).
According to the question, A is equidistant from points B and C.
 AB = AC

3  0   p  2

 3   p  2 
2
2
2
2

 9  p2  4  4 p 

p 2  4 p  13 

 p  0  5  2
2
 p    3
2
p2  9
p2  9
2
2
CBSE X Mathematics 2012 Solution (SET 1)
On squaring both sides, we obtain:
 p 2  4 p  13  p 2  9
 4 p  4
 p 1
Q18. A number is selected at random from first 50 natural numbers. Find the probability that it is
a multiple of 3 and 4.
Solution:
Total number of outcomes = 50
Multiples of 3 and 4 which are less than or equal to 50 are:
12, 24, 36, 48
Favorable number of outcomes = 4
Probability of the number being a multiple of 3 and 4
Number of favourable outcomes

Total number of outcomes
4

50
2

25