MODEL QUESTION PAPERS - study materials notes … · · 2009-02-11Finish Line & Beyond 5. In the given figure, AD= 4 cm, BD = 3cm and CB = 12 cm, find the value of cotθ. Solution:

Finish Line & Beyond

MODEL QUESTION PAPERSWITH ANSWERS

SET 1

MATHEMATICS CLASS XTime Allowed: 3 Hrs Max. Marks : 80

General Instructions:(1) All questions are compulsory.(2) The question paper consists of 30 questions divided into 4 sections: A, B, C

and D. Section A comprises of 10 questions of 1 marks each. Section B comprises of 5 questions of 2 marks each. Section C comprises of 10 questions of 3 marks each and section D comprises of 5 questions of 6 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 2 marks each, three questions of 3 marks each and two questions of 6 marks each. You have to attempt only one of the alternatives in all scuh questions.

(5) In questions on constructions, the drawing should be neat and clean and exactly as per the given measurements.

(6) Use of calculator is not permitted. Section ‘A’Question numbers 1 to 10 carry 1 mark each.

1. If qp

is a rational number ( 0≠q ), what is the condition on q so that

the decimal representation of qp

is terminating.

Solution: If the prime factorization of q is of the form 2n.5m, where, n, and m are non-

negative integers, then the decimal representation of qp

is terminating.

2. Write the zeroes of the polynomial x²+ 2x + 1.

Solution:x² + 2x + 1 = (x+1) (x+1)So, the value of x² + 2x + 1 will be zero When, x+1=0 i.e., x= -1Or, x+1=0 i.e., x= -1So, the zeroes of the given equation are -1 and -1

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3. Find the value of k so that the following system of equations has no solution:3x – y – 5 = 0; 6x – 2y – k = 0

Solution:

Here, 21

aa

= 63

= 21

21

bb

= 21

−−

= 21

21

cc

= k−

− 5 =

k5

If, 21

aa

= 21

bb ≠

21

cc

Then the system of equations will have no solution.

If, k= 10 , then k5

= 21

So, if k ≠ 10 , then the equations will have no solution.

4. The nth term of an AP is 7 – 4n. Find its common difference.

Solution:Let tn denote the nth term of the AP.Then tn = 7 – 4n

Common Difference d = tn - tn -1

Or, d = 7 – 4n – [ 7 – 4(n – 1)]

Or, d = 7 – 4n – 7 + 4n – 4Or, d = -4

Alternate Method,t1 = 7-4 = 3t2 = 7-8 = -1t3 =7-12= -5It is clear that common difference is -4




5. In the given figure, AD= 4 cm, BD = 3cm and CB = 12 cm, find the value of cotθ .

Solution:In ∆ABD, using Pythagoras theorem value of AB can be calculated as follows:AB² = AD² + BD² = 4² + 3² = 16 + 9 = 25Or, AB = 5

Now, In ∆ ABC, cotθ = ABBC

= 5

12

6. In the given figure, P and Q are points on the sides AB and AC respectively of ∆ABC such that AP= 3.5 cm, PB = 7 cm AQ = 3 cm and QC = 6 cm. If PQ = 4.5 cm, find BC.


A

θ3

4

12

D

CB

A

QP

CB



Solution:

ABAP

= 5.105.3

= 31

ACAQ

= 93

= 31

So, ∆APQ ≈ ∆ABC

So, BCPQ

= 31

∴ BC = PQ× 3 = 4.5× 3 = 13.5

7. In figure, PQ = 24 cm, QR = 26 cm, ∠ PAR = 90°, pa = 6 CM AND ar = 8 CM. Find ∠ QPR.

Solution. In right ∆ PAR, we have PR² = PA² + AR² = (6)² + (8)² = 36 + 64 = 100 = (10)²

⇒ PR = 10 cm∆QPR,

PQ² + PR² = (24)² + (10)² = 576 + 100 = 676⇒ PQ² + PR² = (26)² = QR²⇒ ∠ QPR = 90°

8. In figure, O is the centre of a circle. The area of sector OAPB is 185

of

the area of the circle. Find x.


Q

A

R

P



Solution. Area of a sector OAPB = 185

x Area of a circle

⇒°360

x X Πr² =

185

Πr², where r = OA = OB

⇒ x = 185

x 360° = 100°

Thus, x = 100°.

9. Which measure of central tendency is given by the x-coordinate of the point of intersection of the “more than ogive” and “less than ogive”?

Solution. The median of a grouped data of central tendency is given by the x-coordinate of the point of intersection of the “more than ogive” and “less than ogive”.

10.From a well-shuffled pack of card is drawn at random. Find the probability of getting a black queen.

Solution. Well-shuffling ensures equally likely outcomes. Since there are 52 cards in a pack, therefore, the total numbers of possible outcomes = 52. There are Two black queens in a pack of 52 cards. Let E be the event “getting a black queen”, then the number of outcomes favourable to E = 2.

So, P(E) = 522

= 261

Section ‘B’Question numbers 11 to 15 carry 2 marks each.

11.Find the zeroes of the quadratic polynomial 6x² - 3 – 7x and verify the relationship between the zeroes and the coefficients of the polynomial.


A

O

B

P



Solution. We have 6x² - 3 – 7x = 6x² - 7x – 3⇒ 6x² - 3 – 7x = 6x² - 9x + 2x – 3⇒ 6x² - 3 – 7x = 3x(2x – 3) + (2x – 3)⇒ 6x² - 3 – 7x = (2x – 3)(3x + 1)

So, the value of 6x² - 3 – 7x is zero when 2x – 3 = 0 or 3x + 1 = 0, i.e, when x = 23

or x = 31−

. Therefore, the zeroes of 6x² - 3 – 7x are 23

and 31−

.

Now, Sum of zeroes = 23

+ (31−

) = 6

29 − =

67

= - 6

)7(− =

x²of Coeff) xof Coeff(−

Product of zeroes = 23

x (31−

) = - 21

= 6

)3(− =

x²of Coeff.TermConst

Without using the trigonometric tables, evaluate the following :

711

. °°

20cos70sin

- 74

. °°°°

°°75tan55tan35tan15tan

37cos53cos ec

Solution. We have

711

. °°

20cos70sin

- 74

. °°°°

°°75tan55tan35tan15tan

37cos53cos ec

= 711

. °

°−°20cos

)2090sin( -

74

. )1590tan().3590tan(.35tan15tan

37cos).3790cos(°−°°−°°°

°°−° ec

= 711

. °°

20cos20cos

- 74

. °°°°

°°15cot.35cot.35tan15tan

37cos.37sin ec

[ θθθθθθ cot)90tan(,sin)90cos(,cos)90sin( =−°=−°=−° ]

= 711

. °°

20cos20cos

- 74

. )35cot.35)(tan15cot.15(tan

37cos.37sin°°°°

°° ec

= 711

. (1) - 74

. )1)(1(

1

= 711

- 74




=77

= 1.

12.For what value of p, are the points (2,1), (p, -1) and (-1, 3) collinear ?Solution. Since the given points are collinear, therefore, the area of the triangle formed by them must be zero, i.e.

21

[x1 (y2 – y3) + x2(y3 –y1) + x3(y1 – y2)] = 0, where

x1 = 2, y1 = 1, x2 = p, y2 = -1, x3 = -1, y3 = 3

⇒ 21

[ 2(-1 -3) + p(3 – 1) + (-1)(1 + 1)] = 0

⇒ 21

[- 8 + 2p – 2] = 0

⇒ 21

[- 10 + 2p] = 0

⇒ -5 + p = 0 ⇒ p = 5Verification :

Area of V = 12

[2(-1 -3) + 5(3 – 1) + (-1)(1 +1)]

= 21

[- 8 + 10 – 2] = 0

13.ABC is an isosceles triangle, in which AB = AC, circumscribed about a circle. Show that BC is bisected at the point of contact.

Solution. ABC is an isosceles triangle, in which AB = AC, circumscribed about a circle with centre O. Since tangents drawn from an external point to a circle are equal in length. ∴ AF = AE …..(1) [Tangents from A] BF = BD …..(2) [Tangents from B] CD = CE …..(3) [Tangents from C]

Adding (1), (2) and (3), we get AF + BF + CD = AE + BD + CE⇒ AB + CD = AC + BDBut AB = AC (given)⇒ CD = BD ⇒ BC is bisected at the point of contact D.

Or


D

O

B

EF

A

C



In figure, a circle is inscribed in quadrilateral ABCD in which ∠ B = 90°. If AD = 23 cm, AB = 29 cm and DS = 5 cm, find the radius ® of the circle.

Solution. Since tangents to a circle is perpendicular to the radius through the point. ∠ OPB = ∠ OQB = 90°

It is given that ∠ B = 90°. Also, OP = OQ. Therefore, OPBQ is a square.Since tangents drawn from an external point to a circle are equal in length.

DR = DS [Tangents from D] AR = AQ [Tangents from A] And BP = BQ [Tangents from B] Now, DR = DS

⇒ DR = 5 [Q DS = 5 cm (given)]⇒ AD – AR = 5⇒ 23 – AR = 5⇒ AR = 23 – 5 = 18⇒ AQ = 18 [AR = AQ]⇒ AB – BQ = 18⇒ 29 – BQ =18 [Q AB = 29 cm (given)]⇒ BQ = 29 -18 = 11


D

R

S

A

Q

P

B

C

O

∴



⇒ BQ = 11 cmBut OPBQ is a square, therefore, OP = OQ = BP = BQHence, OP = 11 cm, i.e., r = 11 cm.

14.A die is thrown once. Find the probability of getting(i) a prime number(ii) a number divisible by 2.

Solution. As we know that a die has six faces with 1, 2, 3, 4, 5 and 6 written on them. Thus, the total number of outcomes when a die is thrown once are 6, i.e., 1, 2, 3, 4, 5, 6.

(i) Since there are 6 equally likely outcomes : 1, 2, 3, 4, 5 or 6 in a single throw of a die and there are 3 ways of getting a prime number, namely, 2, 3 or 5.

∴ P(getting a prime number) = 36

= 12

(ii) Since there are 6 equally likely outcomes : 1, 2, 3, 4, 5 or 6 in a single throw of a die and there are 3 ways of getting a number divisible by 2, namely, 2, 4 or 6.

∴ P(getting a number divisible by 2) = 36

= 12

Section ‘C’Question numbers 16 to 25 carry 3 marks each.

15. Show that 5 - 2 3 is an irrational number.

Solution. Let us assume, to contrary, that 5 - 2 3 is rational.

That is, we can find coprime a and b (b ≠ 0) such that

5 - 2 3 = ab

Therefore, 2 3 = 5 - ab

⇒ 2 3 = 5b a

b−

⇒ 3 = 5

2b a

b−

Since a and b are integers, 5

2b a

b−

is rational, and so 3 is rational.

But this contradicts the fact that 3 is irrational.

This contradiction has arisen because of our incorrect assumption that 5 - 2 3 is

rational.

So, we conclude that 5 - 2 3 is irrational.

16. Find the roots of the following equation :




14x +

- 1

7x − =

1130

, x ≠ -4, 7.

Solution. We have1

4x + -

17x −

= 1130

As x ≠ -4, 7, multiplying the equation by (x + 4)(x – 7), we get

(x – 7) – (x + 4) = 1130

(x + 4)(x – 7)

⇒ x – 7 –x -4 = 1130

(x + 4)(x – 7)

⇒ - 11 = 1130

(x + 4)(x – 7)

⇒ - 30 = (x + 4)(x – 7)⇒ - 30 = x² - 7x + 4x -28⇒ x² - 3x + 2 = 0So, the given equation reduces to x² - 3x + 2 = 0, which is a quadratic equation.Here a = 1, b = -3, c = 2.So, b² - 4ac = (-3)² - 4(1)(2) = 9 – 8 = 1 > 0

Therefore, x = 3 1

2±

= 3 1

2±

i.e., x = 2 or x = 1

So, the roots are 1 and 2.

18. Represent the following system of linear equations graphically. From the graph, find the points where the lines intersect y-axis.

3x + y – 5 = 0; 2x – y - 5 = 0Solution. Given equations are :

3x + y – 5 = 0 ⇒ y = 5 – 3x …..(1)2x – y - 5 = 0 ⇒ y = 2x – 5 …..(2)

Let us draw the graphs of the equations (1) and (2) by finding two solutions for each of these equations. They are given in tables :

Y = 5 – 3x y = 2x – 5


X 0 2Y 5 -1 A B

X 0 3y -5 1

C D

Y



In figure, we observe that the two lines representing the two equations are intersecting at the point B (2, -1).Hence, x = 2 and y = -1.The line AB cuts the y-axis at the point A(0,5) and the line CD cuts the y-axis at the point C(0,-5).

19. The sum of n terms of an A.P. is 5n² - 3n. Tind the A.P. Hence, find its 10th term.

Solution. Let Sn denote the sum of first n terms of an A.P., thenSn = 5n² - 3n

And Sn-1 = 5(n – 1)² - 3(n – 1) tn = Sn - Sn-1

= 5n² - 3n – [5(n – 1)² - 3(n – 1)] = 5[n² - (n – 1)²] – 3[n – (n – 1)] = 5[n² - n² + 2n – 1] – 3(1) = 5(2n – 1) -3 = 10n – 8

Putting n = 1, 2, 3, …, we get


OX’

Y’

X1 2 3 4 5 6

123456

-6

-6 -1-1-2

-2

-3

-3

-4

-4

-5

-5

-3x + y – 5 = 0

2x – y – 5 = 0

C(0, -5)

B(2,-1)

D(3,1)

A(0,5)



t1 = 10 x 1 – 8 = 2, t2 = 10 x 2 – 8 = 12, t3 = 10 x 3 – 8 = 22, ……Clearly, d = t2 –t1 = 12 – 2 = 10 and d = t3 – t2 = 22 -12 = 10Thus, the A.P. is 2, 12, 22, 32, ……..Now, 10th term of the A.P.

= t10 = a + (10 – 1 )d = 2 + 9(10) = 2 + 90 = 92

Hence, the 10th term of the A.P. is 92.

20.Prove that :

cot coscot cos

A AA A

−+

= cos 1cos 1

ecAecA

−+

Solution. We have

L.H.S = cot coscot cos

A AA A

−+

- cos A =

cossin

AA

+ cos A

cos A1 1

sin A −

=

cos A1 1

sin A +

1 1

sin A −

=

1 1

sin A +

= cos 1cos 1

ecAecA

−+

= R.H.S

OrProve that: (1 + cot A – cosec A)(1 + tan A + sec A) = 2

Solution. We have




L.H.S = (1 + cot A – cosec A)(1 + tan A + sec A)

= cos 11sin sin

AA A

+ − sin 11cos cos

AA A

+ +

= sin 1

sinA cosA

A+ −

cos sin 1

cosA A

A+ +

= [(sin cos ) 1][(sin cos ) 1]

sin cosA A A A

A A+ − + +

= AA

AAAAcossin

(1)²- ²)cos)(sincos(sin ++

=AA

AAAAcossin

1)cossin2²cos²sin −++

=AAAA

cossin1)cossin21 −+

= AAAA

cossincossin2

= 2 RHS

21.Determine the ratio in which the line 3x+4y-9=0 divides the line segment joining the points (1, 3) and (2, 7).

Solution: Let the required ratio be k:1 in which the line segment joining the points (1, 3) and (2, 7) be divided by the point R. Then the coordinates of R are

++

++

137,

112

kk

kk

k:1• * (1, 3) R (2, 7)

Since the line 3x+4y-9 = 0Divides the line segment joining the points (1,3 ) and (2, 7), therefore R lies on the line 3x+4y-9=0


** *



∴ 091374

1123 =−

+++

++

kk

kk

⇒ (6k+3)+(28K+12)-(9k-9)= 0⇒ (6k+28K-9k)+(3+12-9)= 0⇒ 25k+6 = 0

⇒ k= -256

Hence the required ratio is -256

:1,

i.e., -6:25 internallyOr, 6:25 externally

22. Construct a ∆ABC in which AB=6.5 cm, B∠ =60° and BC = 5.5 cm. Also construct a triangle ∆AB’C’ similar to ABC, whose

each side is 23

times the corresponding side of the ∆ABC.


C’



Solution:Steps of Construction:

1. Draw a line segment AB=6.5 cm2. At B construct ∠ ABX= 60°3. With B as centre and radius BC=5.5 cm draw an arc intersecting BX at C.4. Join AC. Triangle ABC so formed is the required triangle.5. Construct an acute angle BAY at A on opposite side of vertex C of ∆ABC.

6. Locate three points (the greater of 3 and 2 in 23

) A1, A2, A3 on AY such that

AA1=A1A2=A2A3.

7. Join A2 (the 2nd point, 2 being smaller of 2 and 3 in 23

) to B and draw a line

through A3 parallel to A2B, intersecting the extended line segment AB at B’.8. Draw a line through B’ parallel to BC intersecting the extended line segment

AC at C’. Triangle AB’C’ so obtained is the required triangle.


A

X

C

B’B

YA3

A2

A1

5.5 cm

6.5 cm



23. If the diagonals of a quadrilateral divide each other proportionally, prove that it is a trapezium.

Solution:Given a quadrilateral ABCD whose diagonals AC and BD intersect each other at O such that

ODBO

OCAO =

To Prove: Quadrilateral ABCD is a trapezium, i.e., AB||DC

Construction: Draw OE ll BA, meeting AD in E.Proof: In ∆ABD, we have

OE ll BA

So, ODBO

EDAE =

But, ODBO

OCAO =

From above equations it is clear that

OCAO

EDAE =

So, from Parallel line’s theorem it can be said thatEO ll DCEO ll BA

So, DC ll BA or, AB ll CDHence, ABCD is a trapezium.

Alternate Question: Two ∆’s ABC and DBC are on the same base BC and on the same side of BC in which °=∠=∠ 90DA . If CA and BD meet each other at E , show that

AE. EC= BE.ED


A

EO

D C

B



Solution: In ∆ABC and ∆BCD∠ BAC = ∠ BDC (Right Angle)∠ AEB = ∠ DEC (Opposite Cone)

So, By AA theorem of similar triangles, ∆ABC ≈ ∆DBC

So,ECBE

EDAE =

Or, AE.EC = BE.EC proved

24. If the distances of P(x,y) from the points A(3,6) and B(-3, 4) are equal, prove that 3x+y = 5.

Solution:Here P(x, y), A(3. 6) and B(-3, 4) are given points. It is given that distances of P(x, y) from A(3, 6) and B(-3, 4) is equal.

So, AP= BPOr, AP2=BP2

Or, (x-3)²+(y-6)²= (x+3)²+(y-4)²Or, (x²-6x+9)+(y²-12y+36)= (x²+6x+9)+(y²-8y+16)Or, -6x-12y+45 = 6x-8y+25Or, 6x+6x-8y+12y=45-25Or, 12x+4y = 20Or, 3x+y = 5 proved

25. In the given figure, find the perimeter of the shaded region where ADC, AEB and BFC are semi-circles on diameters AC, AB and BC respectively.


A

ED

CB

F

D

E

CBA 2.8 cm 1.4 cm



Solution: The sum total of perimeters of semicircles AEB, ADC and BFC will give the perimeter of the shaded figure.

Perimeter of Semicircle AEB = ∏2d

= 4.1722 × =4.4

Perimeter of Semicircle ADC = 22.4

722 × = 6.6

Perimeter of Semicircle BFC = 7.0722 × = 2.2

So, the required perimeter = 4.4 + 6.6 + 2.2 = 13.2 cms

Alternate Question: Find the area of the shaded region in the figure, where ABCD is a square of side 14 cm.


A

D C

B



Solution: Areas of four circles subtracted from the area of square will give the area of the shaded region. Radius of one circle will be equal to 1/4th of the side of the square.

Area of Square = Side² = 14² = 196

Area of one circle = Πr² = 27

27

722 ××

So, Area of four circles = 27

27

722 ×× 4× = 722 × = 154

So, Area of shaded region = 196 – 154 = 42 cm²

SECTION DQuestion numbers 26 to 30 carry 6 marks each.

26. In a class test, the sum of the marks obtained by P in the Mathematics and Science is 28. Had he got 3 more marks in Mathematics and 4 less marks in Science, the product of marks obtained by him would have been 180. Find the marks obtained in two subjects separately.

Solution: Let us assume that marks obtained by P in Math = M and in Science = SSo, as per question,

M+S = 28 ----------------------------- (1)(M+3)(S-4) = 180 --------------------------- (2)

From equation (1) M = 28-SPutting value of M in equation (2) we get

(28-S+3)(S-4) = 180(31-S)(S-4) = 18035S -S²-124 = 180

Or, -S²+35S = 304Or, S²-35S+304 = 0Or, S²-19S-16S+304 =0Or, S(S-19)-16(S-19)= 0Or, S=16 or S=19If, S= 16 then M=28-16 = 12Then (12+3)(16-4)= 15× 12 = 180If, S=19 then M= 28-19 =9Then (9+3)(19-4) = 12× 15 = 180

Alternate Question:The sum of Areas of two squares is 640 m². If the difference in their perimeters be 64 m, find the sides of the two squares.

Solution:Let us assume that the side of one square is S1 and that of another square is S2




Then, S1²+S2² = 640And, 4S1-4S2 = 64Or, S1-S2 = 16 ----------------------------- (1)Or, S1 = 16-S2

Then, (16-S2)²+( S2)² = 640Or, 256+S2²-32S2+ S2² = 640Or, S2²+16S2+128 = 320Or, S2²+16S2-192 = 0Or, S2²+24 S2-8 S2-192 = 0Or, S2(S2+24)-8(S2+24)= 0Or, S2= -24 or +8As side of a square can’t be negative so lets take 8 as the side of one of the squares.From equation (1) it is clear that the side of another square is 24 m.Sum of Areas = 8²+24²= 64+576= 640Difference in Perimeters = 96-32 = 64Note: Always cross check your answers by testing if they fulfill the conditions given in the question.

27. A statue 1.46 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal. (use 3 =1.73)

Solution:


A

60°

DC

B

45°

14.6 m



In ∆ BCD tan 45° = CDBC

Or, 1= CDBC

Or, BC = CD

In ∆ ACD tan 60° = CDAC

= CD

BC+46.1

Or, 3 = BC

BC+46.1

Or, 1.73 BC = 1.46 + BCOr, 1.73 BC –BC = 1.46Or, 0.73 BC=1.46

Or, BC = 73.046.1

= 2 m

28. Prove that the ratio of areas of two triangles is equal to the ratio of the squares of their corresponding sides.Using the above results, prove the following:




In ∆ ABC, XY is parallel to BC and it divides ∆ ABC into two equal parts areawise.

Prove that 2

12 −=ABBX

Solution: Given : ∆ ABC ≈ ∆ DEF.

To Prove: DEFAreaABCArea

∆∆

= ²²

²²

²²

AFAC

DEAB

EFBC ==

Construction: Draw AG ⊥ BC and DH ⊥ EF

Now, )()(

DEFarABCar

∆∆

= DHEF

AGBC

××

××

2121

Or, )()(

DEFarABCar

∆∆

= DHAG

EFBC ×

Now, In ∆ ABG & ∆ DEHEB ∠=∠

DHEAGB ∠=∠Hence, ∆ ABG ≈ ∆ DEH

So,DHAG

DEAB =

But, EFBC

DEAB =

So,EFBC

DHAG =

So, EFBC

EFBC

DEFarABCar ×=

∆∆

)()(

=²²

EFBC

Similarly it can be proved that


HG FE

D

CB

A



DF²

²DE²

²)()( ACAB

DEFarABCar ==

∆∆

Second Part of the Question:

In AXY∆ and ABC∆ , we have BAXY ∠=∠ (Corresponding Angles on Parallel Lines)

AA ∠=∠So ABCAXY ∆≈∆

So, )()(

ABCarAXYar

∆∆

= AB²

²AX

Or, )(2)(

AXYarAXYar ∆

=

²²

ABAX

Or, 21

AB²² =AX

Or, 2

1=ABAX

Or, 2

1=−AB

BXAB

Or, 1-2

1=ABBX

Or, 2

11 −=ABBX

Or, 2

12 −=ABBX

Proved

29. A gulab jamun, when ready for eating, contains sugar syrup of about 30% of its volume. Find approximately how much syrup would be found in 45 such gulab jamuns, each shaped like a cylinder with two hemispherical ends if the complete length of each of them is 5 cm and its diameter is 2.8 cms.

Solution:The volume of gulab jamun can be calculated by adding volumes of two hemispheres and one cylinder. Here the radius of cylinder and that of hemisphere is 1.4 cms and height of cylinder is 5 cms.Volume of syrup in one gulab jamun will be 30% of its volume.


5 cms

2.8 cms



Volume of 2 hemispheres = ∏×34

r

= 4.14.14.1722

34 ××××

Volume of cylinder= ∏ r²h= 2.2

So, volume of gulabjamun = 4.14.1722 ×× ×

+ 2.2

36.5

= 223

2.124.12.0 ×××

Hence, volume of syrup in 45 gulabjamuns

= 223

2.124.12.0 ××× 45%30 ×× =338.184 cm

30.A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice-cream. This ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with icecream.

Solution:Radius of Cylinder = 6 cmHeight of Cylinder = 15 cmVolume of Cylinder = ∏ r²h

= ∏ 6²× 15 = ∏ 540

Volume of Cones = 31

Π r²h

= 31

Π3²× 12 = 36 Π

Volume of Hemispherical top = 32

Πr³= 32

Π3³= 18Π

So, volume of ice-cream= Π(36+18)= 54Π




Number of cones required = ∏∏

54540

= 10 cones

31. A survey regarding the heights (in cms) of 50 girls of class X of a school was conducted and the following data was obtained.

Height in cms 120-130 130-140 140-150 150-160 160-170 TotalNumber of Girls 2 8 12 20 8 50

Find the mean, median and mode of the above data

Solution: The cumulative frequency distribution with the given frequency becomes:

Height (In cms)

NumberOf Girls(f1)

Cumulative Frequency(cf)

ClassMark(x1)

d1=x1-145U1=

10145-1x f1u1

120-130 2 2 125 -20 -2 -4130-140 8 10 135 -10 -1 -8140-150 12 f0 22 cf 145=a 0 0 0150-160 20 f1 42 155 10 1 20160-170 8 f2 50 165 20 2 16Total N= ∑ fi=50 ∑ fiui

From the table, n= ∑ fi= 50 ⇒2n

= 25

a= 145 h= 10

Using the formula for calculating the mean:

Mean = a+ hfi

f×

∑∑ iui

= 145 + 105024 ×

= 149.8

Now, 150-160 is the class whose frequency 42 is greater than 2n

= 25

Therefore, 150-160 is the median class. Thus, the lower limit (l) of the median class is 150.




Median= hf

cfn

l ×

−+ 2

= 150+ 1020

2225 ×−

=151.5

Since the maximum number of girls is 20, therefore, the modal class is 150-160. Thus, the lower limit (l) of the modal class is 150.Using the formula for calculating the mode:

Mode= hffofi

fofil ×

−−

−+22

= 10812202

1220150 ×−−×

−+

= 150+ 10208 ×

=154



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MODEL QUESTION PAPERS - study materials notes … · · 2009-02-11Finish Line & Beyond 5. In the given figure, AD= 4 cm, BD = 3cm and CB = 12 cm, find the value of cotθ. Solution: