Solution cbse-board-class-10-previous-paper-2007

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CBSE CLASS 10th
2007: MATHEMATICS SOLVED PAPER
Time Allowed: 3 Hours Max Marks: 80
General Instruction:
(i) All the questions are compulsory
(ii) The question paper consists of 25 questions divided into three sections- A, B, and C.
Section A contains 7 questions of 2 marks each. Section B contains 12 questions of 3
marks each. Section C contains 6 questions of 5 marks each.
(iii) There is no overall choice. However, an internal choice has been provided in two
questions of two marks each, two questions of three marks each and two questions of
five mark each.
(iv) In questions on construction, the drawing should be neat and exactly as per the given
measurement.
(v) Use of the calculator is not permitted. However, you may ask for Mathematical tables.
Section A: Question No 1 to 7 carry 2 marks each.
Ques [1] : Find the GCD of the following polynomials
12x4
+ 324x and 36x3
+ 90x2
– 54x
Solution:
12x4
+ 324x = 12x(x3
+ 27)
= 6 × 2x(x3
+ (3)3
)
= 6 × 2x (x + 3)(x2
+ 9 –3x)
36x3
+ 90x2
– 54x = 18x(2x2
+ 5x – 3)
= 6 × 3x (2x2
+ 6x – x – 3)
= 6 × 3x (2x (x+ 3) – 1(x + 3))
= 6 × 3x (x+ 3)(2x – 1)
Thus, the GCD of the given polynomials is 6x (x + 3).
Ques [2] : Solve for x and y:
2
2 and 4
x y x y
a b a b
   
Solution:
We are given the equations as
2
2
x y
a b
  …(1)
4
x y
a b
  …(2)

From (1) + (2), we get
2
2 4
3
6
2
x y x y
a b a b
x
a
x a
     
 
 
Now, putting the value of x in the equation (1), we get
2 2
2
4 2
2
2
a y
a b
y
b
y
b
y b

 
  
  
  
Thus, the solution of the given equations are x = 2a and y = –2b.
OR
Solve for x and y:
31x + 29y = 33 and 29x + 31y = 27
Solution:
31x + 29y = 33 …(1)
29x + 31y = 27 …(2)
31x + 29y + 29x + 31y = 33 + 27
60x + 60y = 60
60(x + y) = 60
x + y = 1 …(3)
From (1) – (2), we get
31x + 29y – 29x – 31y = 33 – 27
2x – 2y = 6
2(x – y) = 6
x – y = 3 …(4)
2x = 4 x = 2
From (3) – (4), we get
2y = –2 y = –1
Thus, we get x = 2 and y = –1
Ques [3]: Find the sum of all the three digits whole numbers which are multiple of 7
Solution:

All the three-digits whole numbers are 100, 101, 102,…, 999.
Out of which, the multiples of 7 are 105, 112, 119, …, 994
Now,
994 = 105 + (n – 1)(7)
994 –105 = 7(n –1)
889 = 7(n –1)
889
1 127
7
127 1 128
n
n
   
   
Thus, the required sum of the whole numbers which are multiples of 7 is
 
 128
2
128
105 994 64 1099 70336
2
n
n
S a l
S
 
     
Ques [4]: In the figure (1), PQ || AB and PR || AC. Prove that QR || BC.
Solution:
Since PQ || AB in ∆OAB, we get
OP OQ
AP BQ
 …(1) [ By Basic Proportionally Theorem ]
Since PR || AC in ∆OAC, we get
OP OR
AP RC
 …(2) [ By Basic Proportionally Theorem ]
From the equations (1) and (2), we get
||
OQ OR
QR BC
BQ RC
  [ By Converse of Basic Proportionally Theorem ]
OR
In the figure (2) incircle of ∆ABC touches its sides AB, BC and CA at D, E
and F respectively. If AB = AC, prove that BE = EC

Solution:
We know that the tangents drawn from an external point to the circle are equal in
length. Thus, we get
Now, we have
AB = AC x + y = x + z y = z BE = EC.
Ques [5]: If the mean of the following frequency distribution is 49, find the missing
frequency p:
Class Frequency
0-20 2
20-40 6
40-60 p
60-80 5
80-100 2
Solution:

Class Class-Mark
(x)
Frequency
(f)
x×f
0-20 10 2 20
20-40 30 6 180
40-60 50 p 50p
60-80 70 5 350
80-100 90 2 180
Total f = 15 + p xf =730 + 50p
Now, we know that the mean of the grouped frequency is
 
730 50
49
15
49 15 730 50
735 49 730 50
5
xf
M
f
p
p
p p
p p
p


 

   
   
 


Ques [6]: A wrist watch is available for Rs. 1000 cash or Rs. 500 as cash down payment
followed by three equal monthly instalments of Rs. 180. Calculate the rate of
interest charged under the installment plan.
Solution:
Here, we have
P = Rs. 1000,
A = Rs. (500 + 180 × 3) = Rs. (500 + 540) = Rs 1040
T = 3 Months =
3 1
12 4
 years
SI = A – P = Rs. (1040 – 1000) = Rs. 40.
Thus, we need to find the rate of interest R = ?
We know that
100
1000 1
40
100 4
16%
PRT
SI
R
R

 
 

 
Therefore, the rate of the interest is R = 16% per year.

Ques [7]: An unbiased die is tossed once. Find the probability of getting
(i) A multiple of 2 or 3
(ii) A prime number greater than 2.
Solution:
In an unbiased die, we have {1,2,3,4,5,6}
Thus, we have n(S) = 6
(i) A = multiple of 2 = {2,4,6} n(A) = 3
B = multiple of 3 = {3,6} n(B) = 2
A B = multiple of both 2 and 3 = {6}   1n A B  
Thus, the required probability of getting a multiple of 2 or 3 is
       
 
 
 
 
 
 
3 2 1
6 6 6
4
6
2
3
P A B P A P B P A B
n A n B n A B
n S n S n S
    

  
  


(ii) A = A prime number greater than 2 = {3,5} n(A) = 2
Thus, the required probability of getting a prime number greater than 2 is
 
 
 
2
6
1
3
2
3
n A
P A
n S




Section B: Question No 8 to 19 carry 3 marks each.
Ques [8] : Solve the following system of equations graphically
2x + y = 8 and x + 1 = 2y
Solution:
2x + y = 8 …(1)
x + 1 = 2y …(2)
From the equation (1), we get

y = 8 – 2x
On taking values of x, we get the corresponding values of y as given in the table.
From the equation (2), we get
y =
1
2
x 
On taking values of x, we get the corresponding values of y as given in the table.
On plotting these points of the two tables, we get graph of the given equations as
below.
Now, we see that both the line intersect at the point (3, 2). Thus, the solution of the
given equation is x = 3 and y = 2
Ques [9] : Simplify the following rational expression in the lowest terms
2 3 2 2 2 2
3 3 2 2 4 4 4
2ax x a ax x a ax x
a x a x x a x
     
  
   
Solution:
 
  
 
 
2 3 2 2 2 2
3 3 2 2 4 4 4
2 22 4 4
2 22 2 2 2 2
2
2
ax x a ax x a ax x
a x a x x a x
a ax xx a x a x
a ax xa x a ax x x a x
     
  
   
   
   
       
x 1 2 3 4
y 6 4 2 0
x 1 2 3 4
y 1 1.5 2 2.5

 
   
  
 
   
   
 
2 2 2 2
22 2
2 2
2 2
1
1 1
1
a x a xa x
a x a x a x
a x a x a x
a x a xa x
 
  
  
  
  
 

Ques [10] : If the sum to first n terms of an AP is given by Sn = n(n + 1), find the 20th
term
of the AP.
Solution:
We know that if the sum of n terms of an AP is Sn, then the nth
term of the AP is
Tn = Sn – S(n – 1)
Thus,
T20 = 20(20 + 1) – 19(19 + 1) = 20 × 21 – 19 × 20 = 20(21 – 19) = 20 × 2 = 40.
Ques [11] : In a cyclic quadrilateral ABCD, diagonal AC bisects angle C. Prove that the
tangent to circle at A is parallel to the diagonal BD.
Solution:
First, we draw the cyclic quadrilateral ABCD with
tangent AM at the point A.
In the given cyclic quadrilateral ABCD, the
diagonal AC bisects  BCD.
Thus,  BCA =  DCA = θ …(1)
We know that the angles in the same segment of
the circle are equal.
Thus,  BDA =  BCA = θ …(2)
 DBA =  DCA = θ …(3)
Now, according to angle segment theorem, the
angle made by the tangent with the chord is equal
to the angle made in the opposite segment of the circle. Thus,
 BAM =  BDA = θ =  DBA [From equation (3)]
Since  BAM =  DBA, we get the tangent to circle at A is parallel to the
diagonal BD. Proved.
OR
In the figure 3, O is any point in the interior of ∆ABC. OD, OE and OF are
drawn perpendiculars to the sides BC, CA and AB respectively. Prove that

AF2
+ BD2
+ CE2
= OA2
+ OB2
+ OC2
– OD2
– OE2
– OF2
Solution:
First, we draw OA, OB and OC.
In right triangle OAF, by Pythagorus Theorem, we have
AF2
= OA2
– OF2
…(1)
In right triangle OBD, by Pythagorus Theorem, we have
BD2
= OB2
– OD2
…(2)
In right triangle OCE, by Pythagorus Theorem, we have
CE2
= OC2
– OE2
…(1)
On adding the equations (1), (2) and (3), we get
AF2
+ BD2
+ CE2
= OA2
+ OB2
+ OC2
– OD2
– OE2
– OF2
Ques [12] : Construct a ∆ABC in which base BC = 6 cm,  B = 45 degree and  C = 60
degree. Draw a circumcircle of ∆ABC.
Solution:

Ques [13] : The diameter of solid copper sphere is 18 cm. It is melted and drawn into a
wire of uniform cross section. If the length of the wire is 108 m, find its diameter.
Solution:
We know that if a solid is melted and recast into another solid then its volume is
constant. Thus, we have
Volume of Solid Copper Sphere = Volume of Wire
 
 
3
2
2
4 18
108 100
3 2
4
9 9 9 108 100
3
r
r
 
 
  
 
     
2
2 4 9 9 9 3
3 108 100 10
3
0.3
10
r
r
    
    
   
  
Thus, the diameter of the wire is d = 2r = 0.6.
Ques [14] : The expenditure (in rupees) of a family for a month is as follows:
Items Rent Food Education Electricity &
Water
Others
Expenditure 800 3000 1200 400 1800
Solution:
Items Expenditure Central Angle in Pie
Chart
Rent 800 800
360 40
7200
  
Food 3000 3000
360 150
7200
  
Education 1200 1200
360 60
7200
  
Elec & Water 400 400
360 20
7200
  
Others 1800 1800
360 90
7200
  
Total 7200

Ques [15] : From a pack of 52 cards, red face cards are removed. After that a card is
drawn at random from the pack. Find the probability that the card drawn is
(i) A queen
(ii) A red card
(iii) A spade card
Solution:
In a pack of 52 cards, there are 6 red face cards, which are removed. Thus, there are
46 cards are left in the pack of the cards.
Thus, we have n(S) = 46
(i) Now, in the pack of 46 cards, we have 6 queen cards. Thus, we have n(Q) = 6
Thus, the required probability that the card drawn is a queen is
 
 
 
6 3
46 23
n Q
P Q
n S
  
(ii) In the pack of 46 cards, we have 20 red cards. Thus, we have n(R) = 20
Thus, the required probability that the card drawn is red card is
 
 
 
20 10
46 23
n R
P R
n S
  
(iii) In the pack of 46 cards, we have 13 spade cards. Thus, we have n(Sp) = 13
Thus, the required probability that the card drawn is spade card is
 
 
 
13
46
n Sp
P Sp
n S
 
Ques [16] : Prove that 2 2 2 4
1 1 1
1 1
tan cot sin sin   
  
    
  
Solution:
  
2 2
2 2
2 2
1 1
1 1
tan cot
1 cot 1 tan
cos sec
LHS
ec
 
 
 
  
    
  
  
 

 
2 2
2 2
2 4
1
sin cos
1
sin 1 sin
1
sin sin
RHS
 
 
 



 
 

OR
If A, B and C are the interior angles of a triangle ABC, show that
cos sin
2 2
B C A 
 
 
Solution:
In ∆ABC, we have
A + B + C = 1800
B + C = 180 – A
180
2 2
90
2 2
B C A
B C A
 
 

  
Taking cos of both angles, we get
cos cos 90
2 2
cos sin
2 2
B C A
B C A
   
     
   
 
  
 
Ques [17]: The coordinates of the mid-points of the sides of a triangle are (4, 3), (6, 0) and
(7, –2). Find the coordinates of the centroid of the triangle.
Solution:
Let the coordinate of the vertices of the ∆ABC are A(x1, y1), B(x2, y2) and
C(x3, y3), whose the mid-points of the sides are D(4, 3), E(6, 0) and F(7, –2).
Thus, we have
2 3
2 34 8
2
x x
x x

    …(1)
3 1
3 16 12
2
x x
x x

    …(2)
1 2
1 27 14
2
x x
x x

    …(3)
 1 2 3
1 2 3
3 8 12 14
34
3
x x x
x x x
    
  

Now, we have
2 3
2 33 6
2
y y
y y

    …(4)
3 1
3 10 0
2
y y
y y

    …(5)
1 2
1 22 4
2
y y
y y

      …(6)
 1 2 3
1 2 3
3 6 0 4
2
3
y y y
y y y
    
  
We know that the coordinate of the centroid of ∆ABC is 1 2 3 1 2 3
,
3 3
x x x y y y
G
    
 
 
.
Thus, the coordinate of the centroid of the given ∆ABC is
34 2
34 23 3, ,
3 3 9 9
G G
 
   
   
  
 
Ques [18]: If the distance of P(x, y) from two points with coordinates (5, 1) and (–1, 5) is
equal, prove that 3x = 2y.
Solution:
We are given that the distance of the point P(x, y) from two points A(5, 1) and
B(–1, 5) are equal. Thus, we have
       
       
     
     
2 2
2 2 2 2
2 2 2 2
5 1 1 5
5 1 5 1
5 1 5 1 5 1 5 1
2 4 6 2 6 4
3 2
AP BP
AP BP
x y x y
x x y y
x x x x y y y y
x y
x y

 
       
       
             
     
 
Ques [19]: A loan of amount of Rs. 24600 is to be paid in two equal semi-annual
installment. If the interest is charged at 10% per annum, compounded semi-
annual, find the installment.
Solution:
If P is the amount that is to be paid in n equal installment x at R% per annum
compounded annually, then we have

1 2 3
...
1 1 1 1
100 100 100 100
n
x x x x
P
R R R R
    
       
          
       
Here we have, P = Rs. 24600, R = 5% as the amount is paid in two equal
installment semi-annually. Thus, we get
1 2
24600
5 5
1 1
100 100
13230
x x
x
 
   
    
   
 
Thus, the installment is Rs. 13230.
Section C: Question No 20 to 25 carry 5 marks each.
Ques [20] : Prove that the angle subtended by an arc at the center is double the angle
subtended by it at any point on the remaining part of the circle.
Using the above, prove the following:
In figure 4, O is the center of the circle. If  BAO = 30 degree and
 BCO = 40 degree, find the value of  AOC.
Solution:
Let ACB be the arc. Draw the radii
OA, OB and OC. Draw AC and BC.
In ∆OAC, OA = OC,
  OAC =  OCA = θ (let)
Thus, by exterior angle of ∆OAC,
 AOD = θ + θ = 2θ
In ∆OBC, OB = OC,
  OBC =  OCB = α (let)
Thus, by exterior angle of ∆OBC,
 BOD = α + α = 2α

Now, the angle made by the arc ACB at the center O is
 AOB =  AOD +  BOD = 2θ + 2α = 2(θ + α) = 2( ACB) Proved.
In the given figure, first we draw OB.
In ∆OAB, OA = OB, we have OAB =  OBA = 30
In ∆OBC, OB = OC, we have OBC =  OCB = 40
Thus,  OBC = 30 + 40 = 70
From the above proof, we get  AOC = 2 × 70 = 140.
Ques [21] : State and prove Pythagoras Theorem.
Use the above to prove to prove the following:
ABC is an isosceles right triangle, right angled at C. Prove that AB2
= 2AC2
.
Solution:
Pythagoras Theorem:
In a right angled triangle ABC right angled at C, AB2
= AC2
+ BC2
.
Proof:
Draw CD AB.
In ∆ACD and ∆ABC, we have
∠CAD = ∠BAC = 90 – θ
∠ACD = ∠ABC = θ
Thus, we get ∆ACD and ∆ABC are similar.
We get
AC CD AD
AB BC AC
 
From first and last, we get AC2
= AB × AD …(1)
In ∆CBD and ∆ABC, we have
∠BCD = ∠BAC = 90 – θ
∠CBD = ∠ABC = θ
Thus, we get ∆CBD and ∆ABC are similar.
We get
BC CD BD
AB AC BC
 
From first and last, we get BC2
= AB × BD …(2)

AC2
+ BC2
= AB × AD + AB × BD = AB × (AD + BD) = AB × AB = AB2
Thus, we get
In a right angled triangle ABC right angled at C, AB2
= AC2
+ BC2
.
Now, since ABC is an isosceles triangle right angled at C, we have
AC = BC.
Thus, by Pythagoras theorem, we get
AB2
= AC2
+ BC2
= AC2
+ AC2
= 2AC2
Ques [22] : The side of the square exceeds the side of the another square by 4 cm and the
sum of areas of two squares is 400 sq cm. Find the dimension of the square.
Solution:
Let the two squares are ABCD and PQRS, whose sides are x cm and (x + 4) cm
respectively. Thus, we are given
x2
+ (x + 4)2
= 400
 x2
+ x2
+ 8x + 16 = 400
 2x2
+ 8x + 16 = 400
 x2
+ 4x – 192 = 0
 x2
+ 12x – 16x – 192 = 0
 x(x + 12) – 16(x + 12) = 0
 (x + 12)(x – 16) = 0
 x = –12 and 16
The length of the side of the square cannot be negative. Thus, x = 16 cm
Therefore, the dimensions of the squares are 16 cm and 20 cm.
OR
A fast train takes 3 hours less than a slow train for a journey of 600 km. If the
speed of the slow train is 10km/h less than that of the fast train, find the
speeds of the two trains.
Solution:
Let the speed of the fast train is x km/h and the speed of the slow train is (x – 10)
km/h. Thus, we are given
 
600 600
3
10x x
 

x2
– 10x – 2000 = 0
x2
– 50x + 40x – 2000 = 0
x(x – 50) + 40(x – 50) = 0
(x + 40)(x – 50) = 0
x = –40 or 50.
Since x is the speed of the train it cannot be negative, x = 50

Thus, the speed of the fast train is 50 km/h and the speed of the slow train is 40
km/h.
Ques [23] : A hollow copper sphere of external and internal diameter 8 cm and 4 cm
respectively is melted into a solid cone of base diameter 8 cm. Find the height
of the cone.
Solution:
When a solid is melted to form another solid, then their volume remains constant.
Thus,
Vcone = Vsphere
2 3 3
1 8 4 8 4
3 2 3 2 2
1
16 4 56
3
4 56 3
42
16
h
h
h
 
      
        
       
    
 
  
Thus, the height of the cone is 42 cm.
OR
If the radii of the circular end of a bucket 45 cm high are 28 cm and 7 cm,
find the capacity and surface area of the bucket.
Solution:
The bucket is a form of frustum of the cone.
We are given R = 28 cm, r = 7 cm and h = 45 cm.
Thus, the volume of the bucket is
 
   
 
2 2
2 2
3
1
3
1 22
45 28 28 7 7
3 7
1 22
45 784 196 49
3 7
1 22
45 1029
3 7
48510cm
V h R Rr r
V
V
V
V
  
        
 
      
    
 
The surface area of the bucket is
   
2 2
A R r R r h   

     
2 2
2
28 7 28 7 45
22
35 49.65
7
5461.5 cm
A
A
A
    
   
 
Ques [24] : An observer in a lighthouse observes two ships on the same side of the
lighthouse and in the same straight line with the base of the lighthouse. The
angles of depression of the ships approaching it are 30 and 60 degrees. If the
height of the lighthouse is 150 m, find the distance between the ships.
Solution:
Let AB be the lighthouse and C and D be
the two ships in the same side of the
lighthouse.
Thus, in ∆ABC,
150
tan 60
150
3
150
50 3
3
x
x
x

 
  
Now, in ∆ABD,
150
tan30
1 150
3 50 3
50 3 150 3
100 3
x d
d
d
d


 

  
 
Thus, the distance between the two ships is 100 3 m.

Solution cbse-board-class-10-previous-paper-2007

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Solution cbse-board-class-10-previous-paper-2007