Set-3 Sec-d

CBSE X Mathematics 2009 Solution (SET 3)

CBSE X 2009 Mathematics Section D Question Number 26 to 30 carry 6 marks each.

26.

A juice seller serves his customers using a glass as shown in Figure 6. The inner diameter of the cylindrical glass is 5 cm, but the bottom of the glass has a hemispherical portion raised which reduces the capacity of the glass. If the height of the glass is 10 cm, find the apparent capacity of the glass and its actual capacity. (Use = 3.14)

OR A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is full of water. A solid cone of base diameter 7 cm and height 6 cm is completely immersed in water. Find the volume of (i) (ii)

water displaced out of the cylindrical vessel. water left in the cylindrical vessel. 22 [Take ] 7

Solution:

Apparent capacity of the glass will be the same as the capacity of the cylindrical portion having its base diameter as 5 cm and height as 10 cm. 5 2.5 cm Base radius = 2 Apparent capacity = πr2h = 3.14 × (2.5)2 × 10 = 196.25 cm3

CBSE X Mathematics 2009 Solution (SET 3) Actual capacity of the glass will be the difference between the cylindrical portion and the hemispherical portion. From the figure, it is clear that the base radius of the hemispherical portion is 5 also cm or 2.5 cm. 2 2 3 πr Actual capacity of the glass = πr 2 h 3 2 196.25 cm 3 3.14 (2.5)3 cm 3 3 = 196.25 cm3 – 32.71 cm3 = 163.54 cm3 Thus, the apparent capacity of the glass is 196.25 cm3 and the actual capacity of the glass is 163.54 cm3. OR

(I)

When the solid cone is completely immersed in water, some water will be displaced out of the cylindrical vessel.

The volume of water thus displaced out will be the same as the volume of the solid cone having its height as 6 cm and base diameter as 7 cm. 7 3.5 cm Base radius 2 1 2 πr h Volume of the water displaced out 3

CBSE X Mathematics 2009 Solution (SET 3)

1 22 (3.5) 2 6 3 7 22 3.5 3.5 2 7 3 77 cm Thus, the water displaced out of the cylindrical vessel is 77 cm3 . (II)

In the beginning, the volume of water in the completely-filled cylindrical vessel will be the same as the capacity of this cylindrical vessel. 10 5 cm Base radius 2 Volume of water in the cylindrical vessel = πr2h 22 (5) 2 10.5 7 33 25

825 cm3 Water left in the cylindrical vessel = Total water – Water displaced out = 825 cm3 – 77 cm3 = 748 cm3 Thus, the water left in the cylindrical vessel is 748 cm3 .

27.

Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Using the above, do the following: In Figure 7, O is the centre of the two concentric circles. AB is a chord of the larger circle touching the smaller circle at C. Prove that AC = BC.

Solution: Given: A circle with centre O and AB is a tangent to the circle at a point P. To prove: OP AB Construction: Take a point C on AB, other than point P, and then join OC.

CBSE X Mathematics 2009 Solution (SET 3)

Proof: The point C must lie outside the circle, otherwise AB will intersect the circle at two points and thus, it will become a secant of the circle, which is not tangent to the circle. Now, it is clear that OC > radius OP of the circle This relation is true for every point on AB, except the point P. So, OP is the shortest of all the distances of point O to the points of AB. Hence, OP AB Given: Two concentric circles with centre O. AB is the chord of the larger circle, touching the smaller circle at point C. We now prove that AC = BC: Construction: Join OC, OA, and OB.

Proof: AB is the tangent to the smaller circle at point C. OC AB [By the above proved theorem] In AOC and BOC, we have: OCA = OCB = 90 [OC AB] OA = OB [Being radii] OC = OC [Common] AOC BOC [By RHS congruency axiom] AC = BC [CPCT] Hence, proved.

28.

A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at that instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.

CBSE X Mathematics 2009 Solution (SET 3) Solution:

Let the initial position A of the balloon change to B during the given interval. Let CD be the girl. Then, the given situation can be represented as:

Here, ACG = 60 and BCG = 30 . AF = BH = 88.2 m AE = BG = 88.2 m – 1.2 m = 87 m In ACE, we have: AE tan 60 CE 87 m 3 CE

CE

87 m 3

87 3 m 3

29 3 m

In BCG, we have: BG tan 30 CG 87 m 1 CG 3

CG 87 3 m Thus, distance travelled by the balloon = AB = EG = CG – CE (87 3 29 3) m

58 3 m

CBSE X Mathematics 2009 Solution (SET 3) 29.

Solve the following equation for x: 9x2 – 9(a + b)x + (2a2 + 5ab + 2b2) = 0 OR If (–5) is a root of the quadratic equation 2x2 + px – 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots, then find the values of p and k. Solution: 9x2 – 9(a + b)x + (2a2 + 5ab + 2b2) = 0 2a 2 5ab 2b 2 x 2 ( a b) x 0 9

x2

2a b 3

a 2b 3

x

x2

2a b 3

a 2b 3

x

x2

2a b 3

a 2b 3

x

x x x

2a b 3

x

a 2b 3

2a b 0 or x 3 2a b a 2b or 3 3

a 2b 3

2a 2 4ab ab 2b 2 9 2a a 2b

b a 2b 9

a 2b 2a b 9

0

0

[Using x 2

0

0

a b x ab

0

OR It is given that 2x2 + px – 15 = 0 If (–5) is a root of this quadratic equation, then 2(–5)2 + p(–5) – 15 = 0 2(25) + p(–5) – 15 = 0 50 – 5p – 15 = 0 5p = 35 p=7 … (1) Again, it is given that p(x2 + x) + k = 0 px2 + px + k = 0 On putting p = 7, we get: 7x2 + 7x + k = 0 If this equation has real roots, then its discriminant must be 0. (7)2 – 4 (7) (k) = 0 49 – 28k = 0

x a x b]

CBSE X Mathematics 2009 Solution (SET 3) k=

49 28

7 4

Thus, the values of p and k are 7 and

30.

7 respectively. 4

During the medical check-up of 35 students of a class their weights were recorded as follows: Weight (in kg) 38 – 40

Number of students 3

40 – 42

2

42 – 44

4

44 – 46

5

46 – 48

14

48 – 50

4

50 – 52

3

Draw a less than type and a more than type ogive from the given data. Hence obtain the median weight from the graph. Solution: For the given data, the cumulative frequency distribution of the less than type can be computed as follows. Weight (in kg) Less than 40 Less than 42 Less than 44 Less than 46 Less than 48 Less than 50 Less than 52

Number of students (Cumulative frequency) 3 3+2=5 5+4=9 9 + 5 = 14 14 + 14 = 28 28 + 4 = 32 32 + 3 = 35

To draw a less than ogive, we mark the upper class limits of the class intervals on the x-axis and their corresponding cumulative frequencies on the y-axis by taking a convenient scale. Now, plot the points corresponding to the ordered pairs [(upper class limit, cumulative frequency) – i.e., (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35)] on the graph paper as follows:

CBSE X Mathematics 2009 Solution (SET 3)

Similarly, we can compute the cumulative frequency distribution of the more than type as follows: Weight (in kg) More than 38 More than 40 More than 42 More than 44 More than 46 More than 48 More than 50

Number of students (Cumulative frequency) 35 35 – 3 = 32 32 – 2 = 30 30 – 4 = 26 26 – 5 = 21 21 – 14 = 7 7–4=3

Now, to draw a more than ogive, we mark the lower class limits of the class intervals on the x-axis and their corresponding cumulative frequencies on the y-axis by taking a convenient scale. Now, plot the points corresponding to the ordered pairs [(lower class limit, cumulative frequency) – i.e., (38, 35), (40, 32), (42, 30), (44, 26), (46, 21), (48, 7), (50, 3)] on the graph paper as follows:

CBSE X Mathematics 2009 Solution (SET 3) Now, to obtain the median weight from the graph, we draw both ogives on the same graph paper. They intersect at (46.5, 17.5). 46.5 kg is the median weight of the given data.