MATHEMATICS 2009

                                                                    MATHEMATICS
                                                                 Set I - Outside Delhi 

Q. 9. Write the lower limit of the median class in the following frequency distribution .

   

Classes.	0–10	10–20	20–30	30–40	40–50
Frequency.	9	12	5	16	8

Ans. 20

Sol .

Class	F	c.f
0 - 10	8	9
10 - 20	12	12
20 - 30	5	26
30- 40	16	42
40- 50	8	50

Median Class is 20-30
Lower limit of the median class is 20

Q. 10. Cards each marked with one of the numbers 6, 7, 8, …, 15 are placed in a box and mixed thoroughly. One card is drawn at random from the box. What is the probability of getting a card with number less than 10?

Ans . 

Sol. Total no. of cards = 10
Number of cards less than 10 = 6, 7, 8, 9 i.e. 4
Reqd. probability =

SECTION B

Questions number 11 to 15 carry 2 marks each.

Q. 11. What real number should be subtracted from the polynomial 3x³ + 10x² – 14x + 9 so that the polynomial 3x – 2 divides it exactly?

Sol.

5 should be subtracted from 3x³ + 10x²– 14x + 9 So that the polynomial (3x – 2) divides it exactly

Q. 12. Evaluate .

Sol.

Q. 13. Find a relation between x and y if the points (2, 1), (x, y) and (7, 5) are collinear.

Sol . Let A(2, 1), B(x, y) and C(7, 5) Pts. A, B, C are collinear … (Given)
2(y – 5) + x(5 – 1) + 7(1 – y) = 0
2y – 10 + 5x – 5 + 7 – 7y = 0
5x – 5y – 8 = 0
or
5x – 5y = 8

Q. 14. ABC is a right triangle, right angled at A, and D is the mid-point of AB. Prove that
BC² = CD² + 3BD²

Sol.

Inrt

∆ BAC,

BC² = CA² + AB².
…(Pythagoras’ theorem)
BC² = (CD² – AD²) + (2AD)²

BC² = CD² – AD² + 4AD²
= CD² + 3AD²
BC² = CD² + 3BD²

…….

Q. 15. Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is more than 9.

Sol. Two dice can be thrown as 6 X 6 = 36 ways
“a total more than 9” can be obtained as
(4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6) i.e. 6 ways

Required Probability =  

Or

A game consists of tossing a one-rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e. three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Sol. S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
P (Hanif wins) = P (three heads or three tails)

=
P (Hanif loses) = 1 – P(Hanif wins)

SECTION C

Q. 18.  Check graphically whether the pair of equations 3x – 2y + 2 = 0 and x – y + 3 = 0, is consistent. Also find the coordinates of the points where the graphs of the equations meet the y-axis.
3x – 2y + 2 = 0       x – y + 3 = 0
3x + 2 = 2y                x + 3 = y

Sol. 3x – 2y + 2 = 0       x – y + 3 = 0

3x + 2 = 2y                x + 3 = y

x	0	2	-2
y	1	4	-2

x	0	2	-2
y	3	6	0

 By plotting the points and joining them, the lines donot intersect anywhere, i.e., they are parallel.
Given pair of equations is not consistent, i.e., inconsistent.
The equation 3x – 2y + 2 = 0 meets the y-axis at A(0, 1)
The equation x – y + 3 = 0 meets the y-axis at B(0, 3).

Or

A fraction becomes , if 2 is added to both numerator and denominator. If 3 is added to both numerator and denominator, it becomes. Find the fraction.
Sol. Let the fraction by
According to the question

3x + 6 = y + 2
3x – y = 2 – 6
3x – y = –4
3x + 4 = y …(i)

5x + 15 = 2y + 6
5x – 2y = 6 – 15
5x – 2y = – 9
5x – 2(3x + 4) = – 9
…[From (i)
5x – 6x – 8 = – 9
– x = – 9 + 8
– x = – 1
x = 1
From (i), y = 3(1) + 4 = 7
The fraction x/y =4/7

 Q. 19. Find the middle term of the A.P. 10, 7, 4, …, (– 62) .
Sol. 1 ^st term, a = 10
Common difference, d = 7 – 10 = – 3
a_n = – 62
a + (n – 1)d = – 62
10 + (n – 1) (– 3) = – 62
(n – 1) (– 3) = – 62 – 10 = – 72
n – 1 == 24
n = 24 + 1 = 25

Middle term = term term=13^th term

a_n = a + (n – 1)d a₁₃ = a + 12d
a₁₃ = 10 + 12(– 3) = 10 – 36 = – 26

Q. 20. For an acute angle show that

(sin – cosec ) (cos – sec ) =

Sol. LHS = (sin – cosec) (cos – sec )
=
=
=
=

RHS=

=

=…..[ cos2 +sin2 = 1]

= cos + sin …(ii)

From (i) and (ii) LHS = RHS

Q. 21. Point P, Q, R and S in that order are dividing a line segment joining A(2, 6) and B(7, – 4) in five equal parts. Find the coordinates of P and R.

AP : PB = 1 : 4

P

=
AR:RB=3:2

R
=

Q. 22. A (– 4, – 2), B(– 3, – 5), C(3, – 2) and D(2, k) are the vertices of a quadrilateral ABCD. Find the value of k, if the area of the quadrilateral is 28 sq. units.

Sol.

ar(ABC)

= + (-3)[ -2 – (-2 )] + 3[ -2-(-5)] (-4 , -2)

= -3 (-2+2 + 3 (-2 + 5) ]

=-3(0) + 3 (3 ) ]

=

ar ( ABC)ar( ABD) =28

+ ar (ACD)= 28

ar ( ACD) =28 - = =

[-4(-2-k) + 3 (k- (-2)) + 2(-2-(-2))] =

8 + 4k + 3k + 6 = 35

7k + 14 = 35

k + 2 = 5 …(Dividing both sides by 7)

k = 5 – 2

k = 5 – 2 or k = – 5 – 2

k = 3 or k = – 7

Q. 23. In Fig. 4, AB is a chord of length 9.6 cm, of a circle with centre O and radius 6 cm. The tangents at A and B intersect at P. Find the length of PA.

Sol. Join OP. Let it intersect AB at point M.

Then PAB is an isosceles and PO is the angle bisector of APB
So, OP  AB and therefore, OP bisects AB which gives

AM=MB=
Also OAP=900………….
In rt. AMO,OM=
=
=
=

LetPA =xcm and PM = Y cm
In rt AMP, x2 = y2 + (4.8)² (Pythagoras’ theorem)
X²= y² + 23.04 …(i)
In rt. ΔPAO, OP² = PA² + AO²
…(Pythagoras’ theorem)
(y + 3.6) 2= x² + (6) ²

y² + (3.6) ² + 2(y) (3.6) = y² + 23.04 + 36 …[From (i)]
12.96 + 7.2y = 59.04

7.2y = 59.04 – 12.96 = 46.08

From(i),x² = y ²+23.04
=(6.4)²+23.04
=40.96+23 .04 = 64
Tangent, PA ,x ==+8 cm

Q. 24. Construct an isosceles triangle whose base is 9 cm and altitude is 5 cm. Then construct another triangle whose sides are 3/4of the corresponding sides of the first isosceles triangle.

Sol.

Given ABC is an isosceles , in which base
BC = 9 cm and altitude AD = 5 cm .
A′BC′is the required .

Q. 25. In a circular table cover of radius 70 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. 5. Find the total area of the design.

[ Use = 1.73, = ]

Sol. Join OA, OB and OC. O is the centre.
ABC is an eq. .

AOC = BOC = BOA = = =1200
Total area of the design = 3(area of segment)
= 3[area of sector – area of ]

=3

=3
……..( sin² =2 sin cos )

=3

=3

=3x(70)2
=3x 4900
= 14700

=175 x 51.67 = 9042.25cm²

2nd Method : Let O be the centre.
From O draw OD BC

  AOC =   BOC = COA =  =  = 1200
BOD = COD =600 ………[From(i)]
Now,inrt . ODA,cos60 =

1/2 =

2OD=70 OD=35 cm

In rt. ODB,sin 600 = BD/OB

√3/2  = BD/70

2BD= 70     BD=35cm

BC=2BD=2(35)=70

Total area of the design= a rod circle–ar of ABC

= 22 x/7 (70)² -

=22x / 7 (70)^{2 -} √3/4(70 √3 )²

………..

=(70)2

=4900

= 175(88–36.33)

=175 X 51.67=9042.25 cm² )

Or

Calculate the area other than the area common between two quadrants of circles of radius 16 cm each, which is shown as the shaded region in Fig. 6.

Soln . Area of quad =

=

Area of the shaded region
= 2[ ar of square-ar of quadrant]

=2[(16)² -22/7 x 64]

=2 [256 – 1408/7] = 2 (256 – 201.4)

=2(54.86)=109.72cm²

SECTION D

Questions number 26 to 30 carry 6 marks each.

Q. 26 A motorboat whose speed in still water is 5 km/h, takes 1 hour more to go 12 km upstream
than to return downstream to the same spot. Find the speed of the stream.

Sol. Let the speed of the stream = x km/hr
Then the speed of motorboat upstream = (5 – x) km/hr
and the speed of motorboat downstream = (5 + x) km/hr
Distance = 12 km
According to the question,

=

 

24x = 25 – x²             x² + 24x – 25 = 0
(x² + 25x – 1x – 25 = 0
x (x + 25) – 1 (x + 25) = 0
(x + 25) (x – 1) = 0
x + 25 = 0 or x – 1 = 0
x = – 25 or x = 1
But speed can not be –ve
Speed of the stream = 1 km/hr.

Q. 27. A man on the deck of a ship 14 m above water level, observes that the angle of elevation ofthe top of a cliff is 60° and the angle of depression of the base of the cliff is 30°. Calculate the distance of the cliff from the ship and the height of the cliff. [Take 3 = 1.732]

Sol. Let AB be the deck of a ship and CE be the cliff.

Or

The angles of depression of the top and the bottom of a 9 m high buildling from the top of a tower are 30° and 60° respectively. Find the height of the tower and the distance between the building and the tower.

Sol. Let AC be the tower
and DE be the building.
Let AC = y m
and DC = EB = x m

Q. 28. Prove that, in a triangle, if the square on one side is equal to the sum of squares of the other two sides, then the angle opposite to the first side is a right angle. Using the above, show that in an isosceles triangle ABC with AC = BC, if AB² = 2AC² then ABC is a right triangle.

 Sol.

Part I : See Theorem 4,
Page (xxxii).
Part II : AB² = 2AC²
…(Given)
AB² = AC² + AC²
AB² = AC² + BC²
…[AC = BC (given)]

 Or

Prove that the lengths of tangents drawn from an external point to a circle are equal. Use the above result in the following :

A circle is inscribed in a ΔABC, touching AB, BC and AC at P, Q and R
respectively, as shown in

Fig. 7. If AB = 10 cm,
AR = 7 cm and RC = 5
cm, then find the length of BC.

Sol.

Part I : See Theorem 6, Page (xxxii).

Part II : AP = AR = 7 cm …(Given theorem)
QC = RC = 5 cm …(Given theorem)
BP = AB – AP
= 10 – 7 = 3 cm
BQ = BP = 3 cm …(Given theorem)
 BC = BQ + QC = 3 + 5 = 8 cm

Q. 29. A bucket made up of a metal sheet is in the form of a frustum of a cone. The depth of the bucket is 24 cm and the diameters of the top and the bottom are 30 cm and 10 cm respectively. Find the cost of milk which can completely fill the bucket at the rate of Rs. 25 per litre.

Sol. Lower radius of the bucket,

Q. 30. The following table gives production yield per hectare of wheat of 100 farms of a village :

Production yield (kg/hectare)	Number of farms
40 -45	4
45 -45	6
50 -55	16
55-60	20
60 -65	30
65 -70	24

Change the distribution to a ‘more than type’distribution, and draw its ogive.

Sol.

Production yield (kg/hectare)	Number of farms	Production more than	c.f.
40 -45	4	40	100
45 -50	6	45	96
50 -55	16	50	90
55 -60	20	55	74
60 -65	30	60	54
65 -70	24	65	24