The document contains solutions to several math problems from a CBSE X Mathematics exam in India from 2012. 1) One problem involves finding the fraction where the numerator is 3 less than the denominator, and adding 1 to the denominator decreases the fraction by 1/15. The solution finds the fraction to be 2/5 or 6/9. 2) Another involves calculating the original duration of a flight given the distance, original and reduced speeds due to weather, and increase in time. The original duration is found to be 3 hours 30 minutes. 3) A third determines the common difference of an arithmetic progression where the first term is 5 and the sum of the first four terms is half the sum of the next

1. CBSE X Mathematics All India 2012 Solution (SET 1)
Section D
Q29. The numerator of a fraction is 3 less than its denominator. If 1 is added to the
denominator, the fraction is decreased by
1
15
. Find the fraction.
Solution:
Let the denominator of the fraction be x then the numerator will be (x – 3).
 Original fraction =
3x
x

It is given that the new fraction is obtained after adding 1 to the denominator.
 New fraction =
3
1
x
x


According to the given condition,
     
 
 2 2 2
3 3 1
1 15
3 3 1
1 15
3 1 3 1
1 15
15 2 3 3
x x
x x
x x
x x
x x x x
x x
x x x x x x
 
  

 
  

   
 

      
 15(x – 3) = x2
+ x
 15x – 45 = x2
+ x
 x2
+ x – 15x + 45 = 0
 x2
– 14x + 45 = 0
 x2
– 9x – 5x + 45 = 0
 x(x – 9) – 5(x – 9) = 0
 (x – 5) (x – 9) = 0
 x = 5 or x = 9
 Denominator = 5 or 9
 Numerator = 5 – 3 = 2 or 9 – 3 = 6
2 6
Hence, the fraction is or .
5 9
OR
In a flight of 2800 km, an aircraft was slowed down due to bad weather. Its average speed
is reduced by 100 km/h and time increased by 30 minutes. Find the original duration of
the flight.
Solution:

2. CBSE X Mathematics All India 2012 Solution (SET 1)
Let the original speed of the plane be x km/hr.
If the speed of the plane is reduced by 100 km/hr, then
Reduced speed of the plane = (x – 100) km/hr
Time taken by the plane to reach its destination at original speed, 1
2800
hrt
x

Time taken by the plane to reach its destination at reduced speed, 2
2800
hr
100
t
x


Given,
Time taken by the plane to reach its destination at reduced speed – Time taken by the
plane to reach its destination at original speed = 30 minutes
 
 
   
   
2 1
2
2
2
1
hr
2
2800 2800 1
100 2
2800 2800 100 1
100 2
2800 100 2 100
100 560000 0
800 700 560000 0
800 700 800 0
800 700 0
800 0 or 700 0
800 or 700
800 Speed cannot b
t t
x x
x x
x x
x x
x x
x x x
x x x
x x
x x
x x
x
  
  

 
 

    
   
    
    
   
    
   
   e negative
Original speed of flight = 800 km/hr
Original duration of flight, 1
2800
800
t 
7 1
3 Hrs
2 2
 
Thus, the original duration of the flight is 3 hours 30 minutes.
Q30. Find the common difference of an A. P. whose first term is 5 and the sum of its first four
terms is half the sum of the next four terms.
Solution:
Let the common difference of the given A. P. be d.
First term (a) = 5 (Given)
We know that sum of the first n terms of an A.P. is given as:

3. CBSE X Mathematics All India 2012 Solution (SET 1)
 
   
 
4
S 2 1
2
4
Sum of first four terms S 2 5 4 1
2
2 10 3
20 6
n
n
a n d
d
d
d
    
      
 
 
And, sum of next four terms = S8 – S4
 
8
2 5 8 1 (20 6 )
2
40 28 20 6
20 22
d d
d d
d
       
   
 
According to the given condition,
 
 
4 8 4
1
S S S
2
1
20 6 20 22
2
20 6 10 11
11 6 20 10
5 10
10
5
2
d d
d d
d d
d
d
d
 
   
   
   
 
 
 
Hence, the common difference of the given A.P. is 2.
Q31. Prove that the length of tangents drawn from an external point to a circle are equal.
Solution:
Here is the link for the solution.
http://www.meritnation.com/discuss/question/1465554/how-to-prove-that-the-
length-of-tangents-drawn-from-an-external-point-to-a-circle-are-equal
Q32. A hemispherical tank, full of water, is emptied by a pipe at the rate of
25
7
litres per sec.
How much time will it take to empty half the tank if the diameter of the base of the tank
is 3 m?
Solution:
It is given that, diameter of base of tank = 3 m

4. CBSE X Mathematics All India 2012 Solution (SET 1)
3
Radius, m
2
r 
Volume of water in the hemispherical tank
3
3
3
2
π
3
2 22 3
m
3 7 2
99
m
14
r
 
   
 

Rate of flow of water out of the pipe
25
litres / sec
7

Let the time taken to empty half the tank be t sec.
 Rate of flow of water × t sec
1
Volumeof waterin thehemisphericaltank
2
 
3
3 3 3
25 1 99
litre m
7 2 14
25 1 1 99 1
m m 1 litre m
7 1000 2 14 1000
990
t
t
t
   
 
      
 
 

 Time taken to empty half the tank is (960 + 30) sec = 16 min 30 sec
OR
A drinking glass is in the shape of the frustum of a cone of height 14 cm. The diameters
of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
22
Use
7

 
  
Solution:
Here is the link for the solution.
http://cbse.meritnation.com/study-
online/solution/math/avhMQQvRVi@qLo29qijJoA!!
Q33. A military tent of height 8.25 m is in the form of a right circular cylinder of base diameter
30 m and height 5.5 m surmounted by a right circular cone of same base radius. Find the
length of the canvas use in making the tent, if the breadth of the canvas is 1.5 m.
Solution:

5. CBSE X Mathematics All India 2012 Solution (SET 1)
It is given that, radius of cylinder (r1)
30
m
2
 = 15 m
Height of cylinder (h1) = 5.5 m
And, height of the tent (H) = 8.25 m
So, height of cone (h2) = 8.25 m – 5.50 m = 2.75 m
And, radius of cone (r2) = 15 m
Let the slant height of the cone be l m.
   
2 22
2 2
15m 2.75m
(225 7.5625)m
232.5625 m
15.25m
l
l
l
l
  
  
 
 
Curved surface area of the tent = Curved surface area of cylinder + Curved surface area
of cone
1 1 2
2 2
2
2π π
22 22
2 15 5.5 cm 15 15.25 cm
7 7
(518.57 718.93)cm
rh r l 
 
       
 
 
= 1237.5m2
Now, curved surface area of the tent is equal to the area of rectangular piece of canvas.
It is given that, breadth of canvas = 1.5 m
Let l be the length of canvas.
 l × 1.5 = 1237.5

6. CBSE X Mathematics All India 2012 Solution (SET 1)
1237.5
825m
1.5
l  
Hence, the length of canvas used in making the tent is 825 m.
Q34. The angles of elevation and depression of the top and bottom of a light-house from the
top of a 60 m high building are 30° and 60° respectively. Find
(i) the difference between the heights of the light-house and the building.
(ii) the distance between the light-house and the building.
Solution:
Here is the link for the solution.
http://www.meritnation.com/discuss/question/1865658/the-angles-of-elevation-and-
the-depression-of-the-top-and-bottom-of-a-light-house-from-the-top-of-a-60-m-
high-building-are-30-and-60-respectively