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Finish Line & Beyond

CBSE MODEL QUESTION PAPERS WITH ANSWERS SET 2 MATHEMATICS Time Allowed: 3 Hrs

CLASS X 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 0 10 carry 1 mark each. 1. Write a rational number between

2 and

3.

Solution: A rational number between

2 and

3 is

2.25 = 1.5 =

3 2

2. Write the number of zeroes of the polynomial y=f(x) whose graph is in the given figure.

Y

A X’

O

B

Y’

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X

Finish Line & Beyond Solution: The number of zeroes is 3 as the graph intersects the x-axis at three points, viz., A, O, and B in the given figure. 3. Is x = -2 a solution of the equation x² - 2x + 8 = 0? Solution: Substituting x = -2 on the LHS of the equation we get x² - 2x + 8 = 4 +4+8 = 16

 0 =RHS

Thus, x = -2 is not a solution of the given equation. 4. Write the next term of the AP.

8,

18 ,

32

Solution: The given AP can be be written as

2  2² , 2  3² , 2  4² Or, 2 2 , 3 2 , 4 2 So, the next term will be 5

2 =

2  25 =

50

5. D, E and F are the mid points of the sides AB, BC and CA respectively of ∆ ABC. Find

ar (DEF ) ar (ABC )

Solution: Since D, E and F are the mid points of the sides AB, BC and CA, therefore, DE ll CA, FE ll AB and DF ll BC.

C

F

A

E

D

B

Thus, DECF and ADEF are parallelograms. Now, in ∆ DEF and ∆ ABC, we have


Finish Line & Beyond FDE = ACB (Opposite angles of llgm DECF) DEF  BAC (Opposite angles of llgm ADEF) So, by AA-similarity criterion ∆ DEF ~ ∆ ABC Now,

ar (DEF ) DE ² = ar (ABC ) AC²

1   AC ² 2  = 1 = AC² 4 6. In figure, if ATO  40 , find AOB

A

O

T B

Solution: In ∆ ATO TAO = 90°  AOT = 180° -(90°+40°) = 50° Now, In ∆ ATO and ∆ BTO AO = BO (Radii of the circle) OT = OT (common side for both triangles) TA = TB (Tangent from one point to a circle are always equal)

 ∆ BTO  AOT  BOT  AOB  100 So, ∆ ATO

7.If sin  = cos  , find the value of



Solution: sin  = cos 

sin  =1 cos   tan  = 1   = 45°




Finish Line & Beyond 8.Find the perimeter of the figure, where rectangle.

A

AED is a semi-circle and ABCD is a

D

20 cm E B

C 14 cm

Solution: Radius of semi-circle = 7 cm

22  7 = 22 cm 7 Perimeter of the given figure = 2  20  14  22 Perimeter of semi-circle = r =

= 76 cms 9. A bag contains 4 red and 6 black balls. A ball is taken out of the bag at random. Find the probability of getting a black ball. Solution: Total number of outcomes = 4+6 = 10 Number of favourable outcomes = 6 P (of getting black balls) =

6 3 = 10 5

10.Find the median class of the following data: Marks Obtained Frequency

0-10 8

10-20 10

20-30 12

30-40 22

40-50 30

50-60 18

Solution: Marks Obtained Frequency Cumulative Frequency

Total frequency n =

0-10 8 8

10-20 10 18

20-30 12 30

30-40 22 52

40-50 30 82

 fi = 100


50-60 18 100

Finish Line & Beyond n = 50 2



Now, 30-40 is the class where frequency 52 is greater than

n = 50 2

Therefore, Median Class = 30-40 SECTION B Question numbers 11 to 15 carry 2 marks each. 11.Find the quadratic polynomial sum of whose zeroes is 8 and their product is 12. Hence, find the zeroes of the polynomial. Solution: As you know, f(x)= k  x² - (Sum of the zeroes)x + Product of the zeroes

 , where k is constant.

F(x) = k(x² - 8x + 12) = 0 Or, x² - 8x + 12 = 0 Or, x² - 6x – 2x +12 = 0 Or, x(x – 6) – 2(x – 6) = 0 Or, (x – 6)(x – 2) = 0 Now, x – 6 = 0  x = 6 Or, x – 2 = 0  x = 2 Thus, the zeroes of the equation are: 6 and 2 12. In figure, OP is equal to diameter of the circle. Prove that ABP is equilateral triangle.

A

O

Q

P

B

Solution: Join A to B. We have OP= Diameter Or, OQ + OP= diameter Or, Radius + PQ = diameter ( OQ=radius)


Finish Line & Beyond Or, PQ = diameter – radius Or, PQ = radius Or, OQ = PQ = radius Thus Op is the hypotenuse of the right angle triangle AOP. So, in ∆AOP sin  P =

AO 1 = OP 2

P = 30° Hence, APB = 60° Now, As in ∆AOP, AP=AB So, PAB  PBA =60°  ∆ABP is equilateral triangle. 

13. Wihtout using trigonometric tables, evaluate the following: (sin²25° + sin²65°) +

3 (tan5°tan15°tan30°tan75°tan85°)

Solution:

3 (tan5°tan15°tan30°tan75°tan85°) = (sin²25°+sin²(90°-25°) + 3  tan5°tan15°tan30°tan(90°-15°)tan(90°-5°)

(sin²25° + sin²65°) +

= (sin²25°+cos²25°) +



3 (tan5°tan15°tan30°cot15°cot5°)

3 (tan5°cot5°)(tan15°cot15°)tan30° = 1 + 3 1  1  tan30° 1 = 1+ 3  3 =1+

=1+1=2 14. For what value of k are the points (1,1), (3, k) and (-1, 4) collinear ? Solution: Given three points will be collinear if the area of the triangle formed by them is zero. Area of triangle = 0

1 1k  4  34  1   11  k   = 0 2 1 Or, k  4  12  3  1  k  = 0 2 1 (2k+4) = 0 Or, 2

Or,

Or, k+2 = 0 Or, k = -2


Finish Line & Beyond Alternate Question: Find the area of the ∆ABC with vertices A(-5, 7), B(-4, -5), and C(4, 5) Solution: The area of the triangle formed by the vertices A(-5, 7), B(-4, -5) and C(4, 5) is given by

1 x1 y 2  y3  x2( y3  y1)  x3( y1  y 2) 2 Here, x1 = -5, y1 = 7, x2 = -4, y2 = -5, x3 = 4 and y3 = 5. Area = =

1  5(5  5)  4(5  7)  4(7  (5)) 2

1 (50+8+48) = 53 square units 2

15.Cards, marked with number 5 to 50 are placed in a box and mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the taken out card is (a) a prime number less than 10 (b) a number which is a perfect square. Solution: Total number of cards in the box is 50-4 = 46 So, total number of outcomes = 46 (a)Number of prime numbers less than 10 in the card is 2 (5 & 7) So, number of favourable outcomes = 2 P(number of prime numbers less than 10) =

2 1 = 46 23

(b) Number of perfect squares = 5, which are as follows: 9=(3²), 16= (4²), 25=(5²), 36=(6²) and 49=(7²) P (Number of favourable outcomes =

5 46

SECTION C Question numbers 16 to 25 carry 3 marks each. 16. Prove that

3 is an irrational number.

Solution: Let us assume to the contrary, that is

3 =

3 is rational. Then,

p , where p and q are integers and q  0. q

Suppose p and q have a common factor other than 1, then we can divide by the common factor, to get

3 =

a where a and b are co-prime. b


Finish Line & Beyond So, 3b = a Squaring on both sides we get 3b²=b²  a² is divisible by 3  a is also divisible by 3 Let a= 3m, where m is an integer. Substituting a = 3m we get 3b²= 9m²  b²=3m² This means that b² is divisible by, and so b is also divisible by 3. Therefore, a and b have at least 3 as a common factor. But this contradicts the fact that a and b are coprime. This contradiction has arisen because of our incorrect assumption that rational.

3 is

17.Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m. Solution: Let x be any positive integer then it is of the form 3q, 3q+1 or 3q+2 When x= 3q, then by squaring, we have x² = (3q) ² = 9q² = 3(3q²) = 3m, where m = 3q² When x = 3q+1, then by squaring we get x² = (3q+1) ² = 9q²+6q+1 = 3q(3q+2)+1 = 3m+1, where m = q(3q+2) When x = 3q+2, then by squaring we get x² = (3q+2) ² = 9q²+12q+4 = (9q²+12q+3)+1 = 3(3q²+4q+1)+1= 3m+1, where m=3q²+4q+1 Hence, it can be said that the square of any positive integer, say x, is either of the form 3m or 3m+1 for some integer m. 18.The sum of the 4th and 8th term of an AP is 24 and the sum of the 6th and 10th term is 44. Find the first three terms of the AP. Solution: Let a be the first term and let d be the common difference of the AP. Then as per question, t4+t8 =24 Or, (a+3d)+(a+7d) = 24  2a+10d=24  a+5d = 12-------------------------- (1) Similarly, (a+5d)+(a+9d) = 44  2a+14d = 44  a+7d = 22 -------------------------- (2) from equations (1) and (2) (a+7d)-(a+5d) = 22 – 12 Or, 2d = 10 Or, d = 5


Finish Line & Beyond Putting the value of d in equation 2 we get a +35=22 Or, a = -13 So, first three terms of the AP are as follows: -13, -8, -3 19.Solve for x and y: (a-b)x + (a+b)y = a² - 2ab - b² (a+b)(x+y) = a²+b² Solution: The given linear equations in x and y are (a-b)x + (a+b)y = a² - 2ab - b² ---------------------- (1) (a+b)(x+y) = a² + b² --------------------------------- (2) Re-writing (2) we get (a+b)x + (a+b)y = a² + b² --------------------------- (3) Subtracting (1) from (3) we get (a+b)x + (a+b)y – (a-b)x – (a+b)y = a² + b² - a² +2ab + b² Or, (a+b)x – (a-b)x = 2b²+ 2ab = 2b(b+a) Or, ax + bx – ax +bx = 2b(b+a) Or, 2bx = 2b(b+a) Or, x = b+a Substituting x = b+a in equation (1) we get (a-b)(a+b) + (a+b)y = a² - 2ab - b² Or, a² - b² + (a+b)y = a² - 2ab - b² Or, (a+b)y = a² - 2ab - b² -a² +b² Or, (a+b)y = -2ab Or, y =

 2ab ab

Hence, the solution is x = a+b y=

 2ab ab

Alternate Question: Solve for x and y: 37x+43y = 123 43x+37y = 117 Solution: We have, 37x+43y = 123 -------------------- (1) 43x+37y = 117 -------------------- (2) Adding (1) and (2) we get 80x+80y = 240 Or, x+y = 3 Or, x = 3-y Substituting x= 3-y in equation (1) we get 37(3-y)+43y =123 Or, 111-37y+43y = 123


Finish Line & Beyond Or, 111+6y = 123 Or, 6y = 12 Or, y = 2  x = 3-y  x= 3-2 = 1 Hence solution is X=1 Y =2 20. Prove that (sin  +cosec  )²+(cos  +sec  )² = 7+tan²  +cot²  Solution: LHS (sin  +cosec  )²+(cos  +sec  )² = (sin²  +cosec²  +2.sin  .cosec  )+(cos²  +sec²  +2.cos  .sec  ) = sin²  +cos²  +cosec²  +sec²  +2+2 =5+(1+cot²  )+(1+tan²  ) =7+tan²  +cot²  =RHS proved Alternate Question: Prove that sin  (1+tan  )+cos  (1+cot  ) = sec  +cosec  Solution: LHS = sin  (1+tan  )+cos  (1+cot  ) = (sin  +sin  .tan  )+(cos  +cos  ,cot 

sin   +  cos   cos  cos  cos  sin     sin ² cos ²  = (sin  +cos  )+    sin    cos  sin ³  cos ³ = (sin  +cos  ) + sin  . cos  (sin   cos  )(sin ²  cos² - sin .cos ) = (sin  +cos  )+ sin .cos ( a³+b³=(a+b)(a²+b²-ab)) =  sin  +sin 

 sin ²  cos² - sin .cos   sin .cos    1  sin  . cos   = (sin  +cos  ) 1  sin  . cos     sin  . cos    sin  . cos   = (sin  +cos  )   sin  . cos    = (sin  +cos  ) 1 


Finish Line & Beyond sin   sin  sin  . cos  sin  cos  = + sin  . cos  sin  . cos  1 1 = + cos  sin  = sec  +cosec  =

=RHS Proved 21. If the point P(x, y) is equidistant from the points A(3, 6) and B(-3, 4), prove that 3x+y-5=0. SOLUTION: P(x, y), A(3, 6) and B(-3, 4) are given points such that AP = BP Or, AP² = BP² Or, (x-3) ² + (y-6) ² = (x+3) ² + (y-4) ² Or, (x²+9-6x) + (y²+36-12y) = (x²+9+6x) + (y²+16-8y) Or, (x²+9-6x)-(x²+9+6x) = (y²+16-8y)-(y²+36-12y) Or, -12x = 4y-20 Or, -12x-4y+20 = 0 Or, -3x-y+5 = 0 22. The point R divides the line segment AB, where A9-4, 0) and B(0,6) are such that AR =

3 AB. Find the coordinates of R. 4

SOLUTION: We have AR =

3 AB 4

So, 4AR = 3AB Or, 4AR = 3(AR+RB) Or, 4AR-3AR = 3RB Or, AR = 3RB Or,

AR =3 RB

Or, AR:RB = 3:1 So, it is clear that R divides AB in the ratio of 3:1

A(-4,0)

3:1 * R

B(0, 6)



The coordinates of R are given by

 3  0  1  (4)   3  6  1  0   ,   3 1    3 1   0  4   18  0  Or,  ,    4   4  9  Or,   1,  2  23.In figure, ABC is a right angled triangle, right angled at A. Semicircles are drawn on AB, AC and BC as diameters. Find the area of the shaded region.

A 8 6 B

C

SOLUTION: Area of Shaded Region = (Area of mid-sized semicircle+ Area of smallest semicircle) –(Area of Biggest semicircle-Area of Triangle)

 r² = 8  (r = 4) 2 9 Area of smallest semicircle = (r=6) 2 25 Area of biggest semicircle = 2

Area of Mid-sized semicircle =


Finish Line & Beyond 8²  6² =10; r=5) 1 Area of triangle ABC =  8  6 = 24 2 9 25 Required Area = 8  + +24 2 2 16  9  25 = +24 2 0 = +24 = 24 sq units 2

(diameter =

 ABC = 60°. Construct a 3 ∆AB’C’ similar to ∆ABC, such that sides of ∆AB’C’ are of the corresponding 4 24.Draw a ∆ABC with sides BC=6cm, AB=5cm and

sides of ∆ABC.

X C C’

6 cm

60° A

A1

5 cm

B

A2 A3 A4 Y


Finish Line & Beyond SOLUTION: 1. Draw a line segment AB=5 cm. 2. At B make  ABX = 60° 3. With B as centre and radius equal to 6 cm draw an arc intersecting BX at C 4. Join AC. Then ABC is the required triangle. 5. Draw any ray making an acute angle with AB on the opposite side of the vertex C. 6. Locate 4 points (the greater of ¾ is 4) A1, A2, A3 and A4 on AY so that AA1=A1A2=A2A3=A3A4. 7. Join A4B and draw a line through A3 (the third point being smaller in ¾) parallel to A4B intersecting AB. 8. Draw a line through B parallel to AB intersecting AC at C’. 9. The AB’C’ is the required triangle. 25. D and E are points on the sides CA and CB respectively of ∆ABC right angled at C. Prove that AE²+BD²=AB²+DE² SOLUTION:

A

D

C

E

B

Given: A triangle ABC right angled at C. D and E are points on AC and BC respectively. To prove : AE²+BD² = AB²+DE² Proof: In ∆ ABC AB² = AC²+BC² ----------- (1) In ∆ ACE AE²= AC²+CE² ----------- (2) In ∆BCD BD²= CD²+BC² ----------- (3) In ∆ECD DE²=CD²+CE² ----------- (4) Adding (2) and (3), we get AE²+BD² = AC²+CE²+CD²+BC² Or, AE²+BD² = AC²+BC²+CD²+CE² Proved Or, AE²+BD² = AB²+DE²



SECTION D Question numbers 26 to 30 carry 6 marks each. 26. A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. SOLUTION: Let us assume the speed of stream = x km/h So, Speed of boat upstream = 18-x km/h Speed of boat downstream = 18+x km/h

24 hrs 18  x 24 = hrs 18  x

Hence time taken in going upstream = Time taken in going downstream

So, As per question, Or, Or,

24 24 = 1 hr  18  x 18  x 24(18  x)  24(18  x) =1 (18  x)(18  x) 18  x  18  x 24  =1 18²  x ²

Or, 48x = 324 -x² Or, x² + 48x-324 = 0 Or, x²+54x-6x-324 = 0 Or, x(x+54)-6(x+54)=0 Or, (x+54)(x-6)=0 So, x = -54 and x= 6 Discarding the negative value we have x= 6 So, speed of stream = 6 km/h 27. Two water taps together can fill a tank in

9

3 hours. The tap of larger 8

diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. SOLUTION: Let us assume that the time taken by the tap of smaller diameter = x hrs So, the time taken by the tap of larger diameter = x-10 hrs Now, work done by thinner pipe in one hour = Work done by thicker pipe in one hour =

1 x 1 x  10


Finish Line & Beyond Hence, work done by both pipes in 1 hour

2 x  10 1 1 =  x x  10 x( x  10) x ²  10x 75 Or, = 2x  10 8 Or, Or, Or, Or, Or, Or, Or,

8x²-80x = 150x-750 8x²-80x-150x+750 = 0 8x²-230x-750 = 0 4x²-115x-375 = 0 4x²-100x-15x-375 = 0 4x(x-25)-15(x-25) = 0 (4x-15)(x-25) = 0

So, x=

15 and x=25 4

When x =15/4 , then time taken by thicker pipe to fill the tank is 3.75-10 = -6.25 hrs, as the value is negative so let us discard this value of x When x =25, then time taken by the thicker pipe is 25-10 =15 hrs 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

BX  AB

2 1 2

D

A

B

G

C

Solution: Given : ∆ ABC

E

H

F

 ∆ DEF.

AreaABC BC ² AB ² AC ²   = AreaDEF EF ² DE ² AF ² Construction: Draw AG  BC and DH  EF To Prove:


Finish Line & Beyond 1  BC  AG ar (ABC ) 2 Now, ar (DEF ) = 1  EF  DH 2 ar (ABC ) BC AG  Or, = ar (DEF ) EF DH Now, In ∆ ABG & ∆ DEH

B  E AGB  DHE Hence, ∆ ABG  ∆ DEH AB AG  So, DE DH AB BC  But, DE EF AG BC  So, DH EF ar (ABC ) BC BC BC ²   So, ar (DEF ) EF EF = EF ² Similarly it can be proved that

ar (ABC ) AB ² AC ²   ar (DEF ) DE² DF²

Second Part of the Question:

AXY and ABC , we have AXY  B (Corresponding Angles on Parallel Lines) A  A  So AXY  ABC ar (AXY ) AX ² So, = ar (ABC ) AB² ar (AXY )  AX ²    Or, 2ar ( AXY ) =  AB ²  AX ² 1  Or, AB² 2 AX 1  Or, AB 2 AB  BX 1  Or, AB 2 BX 1  Or, 1AB 2 In


Finish Line & Beyond BX 1  1 AB 2 BX 2 1  Or, Proved AB 2 Or,

29. Prove that the lengths of tangents drawn from an external point to a circle are equal. In the given figure also prove that TA+AR = TB+BR

P A R

O

Q

T

B

SOLUTION: Given a circle with centre O and two tangents are drawn from an external point T to the circle. Prove = TP =TQ Proof: In ∆TPO and ∆TQO TPO  TQO (radius makes right angle with tangent) PO=QO (radius) TO=TO (common side) Hence, ∆TPO ≈ ∆TQO SO, TP = TQ Proved Now, draw a line AB which touches both tangents and the circle as shown in figure. Prove that, TA+AR = TB+BR Using above proof AP=AR (tangents from same external points) BQ=BR (tangents form same external points) TP = TA+AP = TA+AR TQ = TB+BQ =TB+BR So, TA+AR = TB+BR Proved


Finish Line & Beyond 30. A tent consists of a frustum of a cone, surrounded by a cone. If the diameters of the upper and lower circular ends of the frustum be 14 m and 26 m respectively, the height of the frustum be 8 m and the slant height of the surrounded conical portion be 12 m, find the area of the canvas required to make the tent. Assume that the radii of the upper end of the frustum and the base of surmounted conical portion are equal. SOLUTION: A tent consists of a frustum of a cone of height(h) 8 m with diameter 2r and 2R of its upper and lower circular ends respectively are 14 m and 26 m. h= 8 m, r=7m, R=13m. Slant height of the frustum of cone is given by

=

h²  (R - r)²

=

8²  (13 - 7)²

=

64  36)

=

100 =10

12m

7m 8m

13 m



Let the slant height of the cone be  1=12 m Area of the canvas required =Curved surface of the tent =Curved surface of the frustum + Curved surface of the cone =  (R+r)    r 1

22  22   (13  7)  10   7  12 7 7  22 = (200  84) 7 =

=892.57 m² 31. The angle of elevation of a jet fighter from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet is flying at a speed of 720 km/h, find the constant height at which the plane is flying. SOLUTION: Let B and C be two positions of a jet fighter as observed from a point A on the ground. Let APQ be the horizontal line through A. It is given that the angles of elevation of a jet fighter in two positions B and C as observed from a point A are 60° and 30°, respectively.

C

B

h km

60° Q

P

30° A

  BAP = 60° AND  CAQ = 30°

Let the constant height of a jet fighter be h km, i.e., BP = CQ = h km. It is given that a jet is flying at a speed of 720 km/hour.


Finish Line & Beyond In right-angled Δ APB, we have

BP AP h 3 = AP h

tan 60° =

 

AP =

[ QC = PB = h km] ……………..(1)

3

In right-angled Δ AQC, We have

CQ AQ 1 h = 3 AQ

tan 30° =

  Now,

AQ = h 3 BC = PQ = AQ-AP



BC =

Or,

BC =

Or,

BC =

3h 

h 3

3h  h 3 2h 3

km

In 15 seconds the plane has travalled 720 

15 = 3 Kms = BC 60  60 2h

Or, BC = 3 = Or,

h=

3

3 3 =2.598 kms 2

32. Find the mean, mode and median of the following data: Classes 0-10 10-20 20-30 30-40 40-50 50-60 60-70

Frequency 5 10 18 30 20 12 5

SOLUTION: The cumulative frequency distribution table with the given frequency becomes:


Finish Line & Beyond Classes

Frequency fi

0-10 10-20 20-30 30-40 40-50 50-60 60-70 TOTAL

5 10 18 30 20 12 5 n=

 fi  100

From the table, n =

Cumulative Frequency Cf 5 15 33 63 83 95 100

Class Marks xi

di=xi35

ui=

5 15 25 35=a 45 55 65

-30 -20 -10 0 10 20 30

-3 -2 -1 0 1 2 3

xi - 35 10

fiui -15 -20 -18 0 20 24 15

 fiui  6

n

 fi  100  2 = 50, a=35, h=10

Using the formula for calculating the mean

Mean = a+

 fiui  h  fi

= 35+

6  10 = 35.6 100

Since the maximum frequency is 30 , therefore, the modal class is 30-40. Thus, the lower limit(l) of the modal class = 30. Using the formula for calculating the mode:

  fi  fo   h l    2 fi  fo  f 2  30  18   = 30+    10  2  30  18  20  12 = 30+  10 = 35.45 22

Mode =

Now, 30-40 is the class whose cumulative frequency 63 is greater than

n  50 2

Therefore, 30-40 is the median class. Thus, the lower limit (l) of the median class is 30.

n  cf 2 Median = l  h f


Finish Line & Beyond 50  33  10 30 17 = 30+ 3 = 30+

=35.67
