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CBSE X
SET - I
MATHEMATICS PAPER 2009.

Time allowed: 3 hours; Maximum Marks: 80

General Instructions:
1) All questions are compulsory.
2) The question paper consists of thirty questions divided into 4 sections A, B, C and D. Section A comprises of ten questions of 01 mark each, Section B comprises of five questions of 02 marks each, Section C comprises ten questions of 03 marks each and Section D comprises of five questions of 06 marks each.
3) All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
4) There is no overall choice. However, internal choice has been provided in one question of 02 marks each, three questions of 03 marks each and two questions of 06 marks each. You have to attempt only one of the alternatives in all such questions.
5) In question on construction, drawing should be near and exactly as per the given measurements.
6) Use of calculators is not permitted.
SECTION A

Question number 1 to 10 carry 1 mark each.

Question1

1.Write whether the rational number will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

Solution:

The rational number given in the question is . It can also be written as

The above number is of type , in which x=2 and y=3.

Since x and y are positive integers, the rational number or will have a terminating decimal expansion.

Question2

2.Write the polynomial, the product and sum of whose zeroes are and respectively.

Solution:

Let us assume that and are the zeroes of the polynomial.

The product of the zeroes = =

The sum of the zeroes = =

The required polynomial equation can be written as

=

=

If we take k=2, we will get the polynomial as .

One polynomial where the product of zeroes and sum of zeroes , is

Question3

3.

Write whether the following pair of linear equations is consistent or not:

x + y = 14

x ? y = 4

Solution:

The equation x + y = 14 can also be written as x + y - 14 = 0

Similarly, x ? y = 4 can be written as x - y - 4 = 0

x + y -14 = 0 can be represented by where, = 1 and = - 14

x + y -14 = 0 can be represented by where , = - 1 and = - 4

=

=

We see from above that .

Therefore the pair of linear equations given in the question is consistent.

Question4

4.Write the nature of roots of quadratic equation

Solution:

The given equation can be represented by where a=4, b= and c=3.

The discriminant or D is given by . Substituting a=4, b= and c=3 in this, we get

= ()2 - 4 = 48 - 48 =0.

From above, we see that there are two equal and real roots for the given quadratic equation.

Question5

5.For what value of k, are the numbers x, 2x + k and 3x + 6 three consecutive terms of an A.P.

Solution:

Let us assume that a, b and c are the three consecutive terms of the A.P. where a=x, b= 2x + k and c = 3x + 6.

For an A.P. , 2b= a + c

Substituting a=x, b= 2x + k and c = 3x + 6 in this, we get

2 = x + 3x + 6

4x + 2k = 4x + 6

2k = 6

k = 3

Thus x, 2x + k and 3x + 6 are three consecutive terms of an A.P. if k=3.

Question6

6.In a ?ABC, DE||BC. IF DE = BC and area of ?ABC = 81 cm2, find the area of ?ADE.

Solution:

?DAE = ?BAC as it is common angle

?ADE = ?ABC since they are corresponding angles

Hence according to the AA similarity condition, ?ADE ?ABC

For similar triangles, the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

= = =

Since DE = BC, =

=

= = .

area of ?ADE = area of ?ABC = 81 cm2 = 36 cm2.

Hence area of ?ADE = 36 cm2.

Question7

7.If sec A = and A + B = 90°, find the value of cosec B.

Solution:

Since A + B = 90°, A = 90° - B

sec A = sec(90° - B)

As cosec = sec(90° - ),

sec A = cosec B

Given: sec A =

cosec B =

Hence value of cosec B is .

Question8

8.If the mid-point of the line segment joining the points P (6, b ? 2) and Q (?2, 4) is (2, ?3), find the value of b.

Solution:

Using mid-point formula, = - 3

= - 3

b + 2 = -6

b = -8

Hence we get the value of b as - 8.

Question9

9.The length of the minute hand of a wall clock is 7 cm. How much area does it sweep in 20 minutes?

Solution:

In 60 minutes, the angle that will be swept by the minute hand =

in 20 minutes, the angle that will be swept by the minute hand = 20 =

Area swept by the minute hand in 20 minutes = (7 cm)2 .

=

=

Hence the area swept by the minute hand in 20 minutes is

Question10

10.

What is the lower limit of the modal class of the following frequency distribution?

Age in (years)

0 -10

10 -20

20 - 30

30 - 40

40 - 50

50 - 60

Number of patients

16

13

6

11

27

18

Solution:

A modal class is the class interval having maximum frequency.

In the above frequency distribution, 27 is the maximum frequency and it comes in class interval 40 - 50.

Hence 40 - 50 is the modal class

40 is the lower limit of the modal class in the given frequency distribution.

Question11

11.

Without drawing the graph, find out whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincident:

Solution:

If and are two linear equations.

1. If , the above equations will represent intersecting lines

2. If , the equations will represent coincident lines

3. If , the equations will represent parallel lines

The equations given are :-

or

or

Here and

Similarly , and


and

Hence , which means that the given pair of equations represent coincident lines

Question12

12.The 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.

Solution:

For an A.P. its nth term or an is an = a + (n-1) d where a is the first term and d is the common difference between the terms.

The 17th term of an A.P. or

The 10th term of the A.P or

As given in the question, the difference between 17th term and 10th term of A.P. = 7.

7d = 7

Hence the common difference or d between the terms of the A.P. = 1

Question13

13.

Without using trigonometric tables, evaluate:

Solution:

and



=

=

= 5

The value of the given expression is 5.

Question14

14.

Show that the points (?2, 5); (3, ?4) and (7, 10) are the vertices of a right angled isosceles triangle.

OR

The centre of a circle is (2? ? 1, 7) and it passes through the point (?3, ?1). If the diameter of the circle is 20 units, then find the values(s) of ?.

Solution:

Using distance formula, we get

AB = = = =

BC = = = =

CA = = = =

As AB = CA, we see that the triangle is isosceles.

AB2 + CA2 = 106 + 106 =212 = BC2 .

using Pythagoras theorem, we find that the triangle is right-angled at A.

Hence we have proved that the given points (?2, 5), (3, ?4) and (7, 10) are the vertices of a right-angled triangle.

OR

The center of the circle is given as (. The point it passes through is given as (-3, -1)

O(2?-1, 7),P(-3,-1)



Using distance formula, the radius of the circle or OP

=

=

=

Since the diameter of circle is given as 20 units and diameter =

=

=

When , then

When , then

Question15

15.If C is a point lying on the line segment AB joining A (1, 1) and B (2, ?3) such that 3AC = CB, then find the coordinates of C.

Solution:

Let us assume that the coordinates of C are m and n. Point C can be represented as C(m, n).

Given: 3AC = CB

Using section formula,

=

The coordinates of point C are ()

Question16

16.

Show that the square of any positive odd integer is of the form 8m + 1, for some integer m.

OR

Prove that is not a rational number.

Solution:

Let us represent a positive integer with 'a'

It can be represented as

where and is an integer

Hence a will be

Positive odd integer will be of the form

Let us consider the above 4 positive odd integer forms seperately

Case 1 :

is an integer

Case 2:

is an integer

Case 3:

is an integer

Case 4:

is an integer

From all these cases we see that the square of a positive odd integer is of the form where is any integer

OR

Let us assume that is a rational number

If so it can be represented using integers a and b ( as

a and b are rational numbers. So should also be rational and should also be a rational number. But we know that is an irrational number and so our assumption is wrong.

Since our assumption is false it is proved that is not a rational number.

Question17

17.If the polynomial 6x4 + 8x3 ? 5x2 + ax + b is exactly divisible by the polynomial 2x2 ? 5, the find the values of a and b.

Solution:

p(x) = 6x4 + 8x3 ? 5x2 + ax + b

q(x) = 2x2 - 5

Dividing p(x) by q(x)

We get the remainder as .

As p(x) is exactly divisible by q(x), remainder should be 0.

= 0

From and

From

Therefore the value of and

Question18

18.If 9th term of an A.P. is zero, prove that its 29th term is double of its 19th term.

Solution:

Let a be the 1st term and d be the common difference of the A.P.

nth term of A.P. =

9th term of A.P =

Given:

19th term of A.P. =

29th term of A.P. =

Hence the 29th term of the A.P. is double the 19th term of the A.P.

Question19

19.Draw a circle of radius 3 cm. From a point P, 6 cm away from its centre, construct a pair of tangents to the circle. Measure the lengths of the tangents.

Solution:

With radius as 3 cm, draw a circle. Mark the center point as O. Take a point P at a distance of 6 cm from O and join O and P.

Draw a perpendicular bisector to the line OP. Mark the point at which it intersects the circle as point Q.

Using Q as center and QP as radius, draw a circle. This will intersect the first circle at 2 points. Mark the points as M and N.

Join M with P to obtain line MP and N with P to obtain the line NP. MP and NP are the tangents.

On measuring, the length of the tangents, MP and NP is 5.20 cm .

Question20

20.In figure 1, two triangles ABC and DBC lie on the same side of base BC. P is a point on BC such that PQ || BA and PR || BD. Prove that QR || AD.

Solution:

Consider and

PQ || BA and hence according to the basic proportionality theorem, we get

--------------------- (1)

Consider and

PR || BD and hence according to the basic proportionality theorem, we get

-----------------------(2)

Taking (1) and (2) together we get

, from above, using converse of proportionality theorem, we get QR|| AD.

Question21

21.

In figure 2, a triangle ABC is right angled at B. Side BC is trisected at points D and E. Prove that 8

http://img1.mnimgs.com/img/paper/1/10/1/239/1C_html_5a2efa0f.jpg

OR

In figure 3, a circle is inscribed in a triangle ABC having side BC = 8 cm, AC = 10 cm and AB = 12 cm. Find AD, BE and CF.

http://img1.mnimgs.com/img/paper/1/10/1/239/1C_html_m5769abc4.jpg

Solution:

http://img1.mnimgs.com/img/paper/1/10/1/239/1C_html_5a2efa0f.jpg

Given: BC is trisected at D and E.

Consider . Using Pythagoras Theorem, we get

---------------------(1)

Consider . Using Pythagoras theorem, we get

-----------------------(2)

Consider . Using Pythagoras theorem, we get

-----------------------(3)

We have to prove that

From (2), we know that . Hence substituting in above, we get

Hence proved.

OR

AD and AF are tangents drawn from point A to the circle. Similarly BD and BE are tangents from point B to the circle and CE and CF are tangents from point C

http://img1.mnimgs.com/img/paper/1/10/1/239/1C_html_m5769abc4.jpg

The tangents drawn to the circle from any point external to the circle are equal.

AD=AF

BD=BE and

CE =CF

Let us take AD = a, BE = b and CF = c

-----(1)

-----(2)

------(3)

------(4)

Substituting (1) in (4), we get

Substituting (2) in 4, we get

Substituting (3) in (4), we get

Thus we get

Question22

22.

Prove that

Solution:


Question23

23.Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

Solution:

Since the points (x, y), (1, 2) and (7, 0) are collinear, the area of the triangle formed with these three points as vertices will be zero.

Area of triangle with points (x1,y1) , (x2,y2) , (x3,y3) =

Thus the area of the triangle is . Since this area is zero, we get

is the equation which shows the relation between x and y if the points have to be collinear.

Question24

24.

In figure 4, the shape of the top of a table in a restaurant is that of a sector of a circle with centre O and ?BOD = 90°. If BO = OD = 60 cm, find

(i) the area of the top of the table.

(ii) the perimeter of the table top.

(Take ? = 3.14)

http://img1.mnimgs.com/img/paper/1/10/1/239/1C_html_a8f6f97.jpg

OR

In figure 5, ABCD is a square of side 14 cm and APD and BPC are semicircles. Find the area of shaded region. (Take )

http://img1.mnimgs.com/img/paper/1/10/1/239/1C_html_m2b0910ee.jpg

Solution:

http://img1.mnimgs.com/img/paper/1/10/1/239/1C_html_a8f6f97.jpg

(i) Area of table top

Area of a sector can be obtained by where is the angle of the sector and r is the radius.

Given : .

Hence area of table top =

Hence area of table top =

(ii) Perimeter of table top

= length of the arc BD + OB + OD

Length of the arc BD =

Perimeter of the table top =

OR

http://img1.mnimgs.com/img/paper/1/10/1/239/1C_html_m2b0910ee.jpg

Area of the shaded region = Area of the square ABCD - (area of semicircle APD + area of semicircle BPC)

Area of square ABCD

(since length of sides of the square = 14 cm)

Area of semicircle APD = where r = 7 cm (AD = 14 cm, )

Area of semicircle BPC = since r= 7 cm

Area of the shaded region =

Hence area of shaded region =

Question25

25.

A box has cards numbered 14 to 99. Cards are mixed thoroughly and a card is drawn from the bag at random. Find the probability that the number on the card, drawn from the box is

(i) an odd number,

(ii) a perfect square number,

(iii) a number divisible by 7.

Solution:

The probability of an event =

Number of cards in box = 99 -14 + 1 = 86

(i)

Number of cards having odd number =

Hence probability of getting odd number =

(ii)

Number of cards having perfect squares

( Perfect squares between 14 and 9 are 16, 25, 36, 49, 64 and 81)

Hence probability of getting perfect square

(iii)

Number of cards with numbers divisible by 7

= 13

(Numbers divisible by 7 between 14 and 99 are 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91 and 98)

Hence probability of getting number divisible by 7 =

Question26

26.

A trader bought a number of articles for Rs 900. Five articles were found damaged. He sold each of the remaining articles at Rs. 2 more than what he paid for it. He got a profit of Rs. 80 on the whole transaction. Find the number of articles he bought.

OR

Two years ago the man's age was three times the square of his son's age. Three years hence his age will be four times his son's age. Find their present ages.

Solution:

Let the total number of articles bought be 'x'

Cost of x articles = Rs 900

Cost price of 1 article =Rs

Number of articles sold = x-5

Selling price of 1 article = Rs

Profit after selling x-5 articles = Rs 80

Since number of articles will not be negative, x= 75

Hence the number of articles bought by the trader = 75

OR

Let the son's age be 'x'

Son's present age =

Son's age 2 years ago =

Man's age 2 years ago =

Man's present age =

Son's age 3 years later from present =

Man's age 3 years later from present =

Man's present age =

Since age cannot be fraction, x = 5

Son's age = 5 years

Man's age = years

Question27

27.

Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Using the above theorem prove the following:

The area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal.

Solution:

Let the 2 similar triangles be .

To prove:

Let us draw AP ? BC and XQ ? YZ.

From above, we get

Using (4) and (5), we get

------------------------------------(6)

Substituting (6) in (3), we get

From (6), we get

To prove:

Let the sides of the square be of length 'a'

(since BC is diagonal)

All the angles of equilateral triangle are and all the sides of an equilateral triangle are equal.

Hence all equilateral triangles are similar.

Hence =

Therefore the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal.

Q28 The angle of elevation of the top of a building from the foot of a tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Question28

28.The angle of elevation of the top of a building from the foot of a tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Solution:

http://img1.mnimgs.com/img/paper/1/10/1/239/1D_html_311c7947.jpg

Let the building be AB and the tower CD.

Length of CD = 50 m

Consider

Consider

Thus the height of the building is

Question29

29.

A spherical copper shell, of external diameter 18 cm, is melted and recast into a solid cone of base radius 14 cm and height cm. Find the inner diameter of the shell.

OR

A bucket is in the form of a frustum of a cone with a capacity of 12308.8 . The radii of the top and bottom circular ends of the bucket are 20 cm and 12 cm respectively. Find the height of the bucket and also the area of metal sheet used in making it.

Solution:

Let 'r' be the inner radius and 'R' be the external radius of the copper shell.

Given : External diameter of copper shell = 2R = 18 cm

Volume of the spherical copper shell =

The spherical copper shell is made into a solid cone

Hence volume of solid cone = volume of spherical shell

Volume of cone =

Hence

Therefore inner radius of the spherical copper shell = 8 cm and hence inner diameter = 16 cm

OR

Let the top radius of the bucket be 'r' and bottom radius of bucket be 'R'

r = 20 cm and R = 12 cm

Let the height of the bucket be 'h'

Volume of the bucket =

Let 's' be the slant height of the metal sheet

= = = 17 cm

The area of the metal sheet used in making the bucket

= CSA of the bucket + area of the bottom circular end of the bucket

=

Question30

30.

Find the mode, median and mean for the following data:

Marks obtained

25 ? 35

35 ? 45

45 ? 55

55 ? 65

65 ? 75

75 ? 85

Number of students

7

31

33

17

11

1

Solution:

(a). Mode

Marks obtained

25?35

35?45

45?55

55?65

65?75

75?85

Number of students

7

31

33

17

11

1

The maximum frequency is 33 and the class having the maximum frequency is 45?55.

modal class = 45?55.

Lower class limit (l) of modal class = 45

Frequency (f1) of modal class = 33

Frequency (f0) of class preceding the modal class = 31

Frequency (f2) of class succeeding the modal class = 17

Class size (h) = 10

= 46.11(approximate)

(B). Median

Let us calculate the cumulate frequency of the data in another column:-

Marks obtained

Number of students

Cumulative frequency

25 ? 35

7

7

35 ? 45

31

38

45 ? 55

33

71

55 ? 65

17

88

65 ? 75

11

99

75 ? 85

1

100

We obtain n as 100

Cumulative frequency (cf) just greater than 50 = 71

71 comes under in the class interval 45-55. Hence median class = 45-55

Lower limit (l) of median class = 45

Class size (h) = 10

Frequency (f) of median class = 33

Cumulative frequency (cf) of class preceding median class = 38

Median =

(C). Mean

The class marks for each interval can be calculated by using the formula

Taking 60 as assured mean (a), di, ui, and fiui can be calculated as follows:

Marks obtained

Number of students (fi)

xi

di = xi ? 60

http://img1.mnimgs.com/img/paper/1/10/1/239/1D_html_33d3dafc.gif, where h = 10

fiui

25­?35

7

30

? 30

? 3

? 21

35?45

31

40

? 20

? 2

? 62

45?55

33

50

? 10

? 1

? 33

55?65

17

60

0

0

0

65?75

11

70

10

1

11

75?85

1

80

20

2

2

Total

100

? 103

Mean =

mode = 46.11, median = 48.63 and mean = 49.7

Section A Section B Section C Section D