FOUNDATON EXAMINATION

Transcription

1 FOUNDATON EXAMINATION (REVISED SYLLABUS - 008) Paper - 4 : BUSINESS MATHEMATICS & STATISTICS FUNDAMENTALS Section - I [ Arithmetic ] y z Q.. (a) If then show that b c c a ab (b c)( a) + (c a)(y b) + (a b)(z c) = 0 (b) Two numbers are in the ratio of :4. If 0 is subtracted from both of them then the ratio becomes :. The numbers are : (i) 9 and (ii) and 6 (iii) 5 and 0 (iv) none of these Answer. (a) Let y z b c c a ab = k (constant). say Then = k(h + c), y= k(c + a), z = k(a + b) So, (b c) ( a) + (c a) (y b) + (a b) (z c) = [ (b c)+y(c a)+z(a b)] [a(b c)+b(c a)+c(a b)] = [k(b + c) (b c) + k(c + a) (c - a) + k(a + b) (a b)] [ab ac+bc ab+ac bc] = [k(b c ) + k(c a ) + k(a b )] 0 = [k(b c + c a + a b )] 0 = k 0 0 = 0 0 = 0 Proved Answer. (b) Let the numbers be k and 4k k 0 Now 9k 0 4k 0 4k 0 5k 0 k 4 So the numbers are 4 =, and 4 4 = 6

2 [ December 0 ] Revisionary Test Paper (Revised Syllabus-008) Q.. (a) The average marks in Elements of Mathematics of Preliminary students of centres in India is 50. The number of candidates in centres are respectively 00, 0, 50. If the averages of the first two centres are 70 and 40, find the average marks of the third centre. (b) A class has divisions. Average marks of the students of the class, first division, second division and third division are 47, 44, 50 and 45 respectively in Mathematics. If first two division have 0 and 40 students, find the number of students in third division when all the students of the class have Mathematics as a subject. (c) Mean monthly income of 0 workers in factory A is ` 4,000 and that of workers in factory B is `,700. If the mean income of all workers in A and B is,800 per month, find the number of workers in B. Answer. (a) Let the average marks of the third centre be. Then using the formula : n n n n n n, we get 50 = ,800, or, 50 =, or,, = 8,500, or, 50 = 8,500,800 = 6,700 = 6, = Hence the required average marks of the third centre = Answer. (b) Let the no. of students be in the third division. Students have total marks in Mathematics in st division = 0 44 = 0 Students have total marks in Mathematics in nd division = = 000 Students have total marks in Mathematics in rd division = 45 = 45 Total marks in Mathematics in the whole class = Total number of students in the class = = 70 + Then average marks of the students in the class = 0 45 Thus, = So = or, 0 = or, = 5

3 Foundation : Paper-4 : Business Mathematics & Statistics Fundamentals [ December 0 ] Answer. (c) n n Let n be the no. of workers in B. Then. n n Here n = 0, = 4000, =,700 and =,800, n =? n =, or, n 0 n = n, or 00n = 000, or, n = 0. Hence the required no. of workers in B = 0. Q.. (a) A vessel contains a miture of Wine and Water. Had there been a litre more of Wine and a litre less of Water, the ratio of Wine to Water would have been 7 : 8; but had there been a litre more of Water and a litre less of Wine, the ratio would have been :. How many litres does the miture consist of? (b) The proportion of liquid I and II in four samples are :, :, 5: and 7:5. A miture is prepared by taking equal quantities of the samples. Find the ratio of liquid I to liquid II in the final miture. Answer. (a) Let the vessel contain litres of milk and y litres of water. Then the vessel contains ( + y) litres of miture. By the given conditions, 7 y 8 (i) From (i), From (ii), and y. (ii) = 7y 7, or, 8 7y = 5. (iii) = y +, or, y = 5. (iv) Solving (iii) and (iv), we get = and y = 7. Hence the vessel contains + 7, i.e., 0 litres of miture. Answer. (b) Liquid I Liquid II = = Ratio of Liquid I and Liquid II in final miture is 99:6.

4 4 [ December 0 ] Revisionary Test Paper (Revised Syllabus-008) Q. 4. (a) If then prove that ++ = 0 = p + q + r. (b) A sum deposited at a bank fetches `,440 after 5 years at the rate of % simple interest. Find the principal amount. (c) If I ask you for a loan and agree to repay you ` 00 after nine months from today, how much should you loan me if you are willing to make the loan at the rate of 6% p.a.? Answer 4. (a) Let k q r r p p q, or, = k (q r); = k (r p); = k (p q) k (q r r p p q) 0 ; p q r pk (q r) qk (r p) rk (p q) or, k (pq pr + qr pq + pr qr) = 0 0 p q r Answer 4. (b) Let the principal amount be ` 00. Then Simple interest on ` 00 for 5 years at % p.a. = 5 = ` 60. Amount at the end of 5 years = = ` 60. Amount Principal = 00,440,440 = ` 8, Answer 4. (c) If ` 00 be the amount of loan, then interest 6 9 ` 9 and amount with interest 00 9 ` If repayable amount be `, then amount of loan is ` 00. If repayable amount be ` 00, then amount of loan is ` ` Q. 5. (a) A bill was drawn on 4 June 989 at 8 months after date and was discounted on 4 September 984 at 5% p.a. If the banker s gain on the basis of simple interest is `, for what sum the bill was drawn? (b) If the difference between true discount and banker s discount on a sum due in months 4% per annum is ` 0, find the amount of bill. (c) The Bill Value (B.V.) of a bill is `,0,000. Find the Banker s Gain ( B.G.) after 7 days at 5% p.a.

5 Foundation : Paper-4 : Business Mathematics & Statistics Fundamentals [ December 0 ] 5 Answer 5. (a) Date of drawing Period 8 months Nominal due date Days of grace days Legally due date Unepired period = to = = 46 days. Given B. G. = `, or, B.D. T.D. = `... (i) 46 If P. V. = ` 00, then T.D. = 5 = `, B. V. = P. V. + T. D. = 00 + = ` B. D. = Interest on B. V. = `.04. B. G. = B. D. T. D. = `.04 ` = `.04 = `. 5 If B. G. = `, then B. V. = ` If B. G. = `, then B. V. = = ` 7,650. /5 Hence the required Bill Value is ` 7,650. Answer 5. (b) A = Amount due at the end of n years = P ( + ni) where P = Present value, i = rate of interest, n year 4 BD = Ani = P ( + ni) ni, TD = Pni BD TD = P(ni) 0 P = (in `) A = P ( + ni) = = = Amount of Bill : ` Answer 5. (c) BV Pv = 0000 = ni c BD =BV ni = = ` 00 TD= PV ni = = ` 000 BG = BD-TD = ` 0

6 6 [ December 0 ] Revisionary Test Paper (Revised Syllabus-008) Section - II [ Algebra ] Q. 6. Choose the correct option showing necessary reasons/calculations. (a) Solution of 7 4 ) 7 / ( ) ( is (i) =, (ii) =, (iii) = 4, (iv) none of these n n (b) Cr Cr is equal to (i) n n Cr, (ii) Cr n C, (iii) r, (iv) none of these. log log y log z (c) If then the value of yz is y z z y (i), (ii) 0, (iii), (iv) none of these. Answer 6. (a) (iv) ( / ) +7 = ( /4 ) 7+/ Answer 6. (b) (ii) or, = +. or, 8 = 8. or, = 6 =. n r n C C = r n r n-r n r - n-r n nr r = r n- r r(n r ) ( n) n n n = Cr r n r r n r. Answer 6. (c) (iv) log log y log z K y z z y (say) Then log = K(y z) log y = K(z ) log z = K( y)

7 Foundation : Paper-4 : Business Mathematics & Statistics Fundamentals [ December 0 ] 7 Adding log + log y + log z = K (y z + z + y) = K 0 = 0 or, log yz = log or, yz =. Q. 7. (a) If w be an imaginary cube root of unity find out the value of (-w) (-w ) (-w 4 ) (-w 8 ). (b) Simple interest and compound interest in years for same principal are Rs. 00 and Rs. 0 at the same rate of interest per annum. Find the principal amount. (c) The volume of a gas varies directly as the absolute temperature and inversely as pressure. When the pressure is 5 units and the temperature is 60 units the volume is 00 units. What will be the volume when the pressure is 8 units and the temperature is 95 units? Answer 7. (a) (-w) (-w ) (-w 4 ) (-w 8 ) = (-w) (-w ) (-w) (-w ) = (-w) (-w ) = (+w -w) (+w 4 -w ) = (-w) (-w ) = 9w = 9 Answer 7. (b) Let = Principal amount and r % = rate of interest per annum r r The simple interest = Rs. 00 = r r The compound interest = Rs. 0 = 00 = r 00 r 0000r r So, 000 Rs. r 0 r r r Answer 7. (c) Volume = V, Pressure = P, Absolute Temp = T T T V T & V V V K K = constant P P P Then P = 5, T = 60 then V = = K K When P = 5, T = 60 then V 5 units 8

8 8 [ December 0 ] Revisionary Test Paper (Revised Syllabus-008) Q. 8. (a) If p = log 0 0 and q = log 0 5, find and such that log 0 ( + ) = p q (b) If = log a a, y = log a a, z = log 4a a, Show that : yz + = yz. (c) Show that log.... Answer 8. (a) p q = log 0 0 log 0 5 = log 0 (0) log = log log 0 5 = log 0 = log0 6 5 Now, log 0 (+)=log 0 6 or, log 0 (+) = log 0 6 or, ( + ) = 6 = (± 4) or, + = ± 4 =, 5. Answer 8. (b) L. H. S. = log a a. log a a. log 4a a+ = (log 0 a log a 0) (log 0 a log a 0) (log 0 a log 4a 0)+ log0 a log0 a log0 a = log a log a log 4a log0 a = log 4a = log 4a a + log 4a 4a = log 4a (a 4a) = log 4a 4a. 0 R.H.S. = log a a log 4a a = log 4a (a) = log 4a 4a Hence the result. Answer 8. (c) Let,... or... (squaring both sides) or, = or, = 0 or, ( ) = 0 or, = 0 (as 0), = given epression = log =. Q. 9. (a) In a class of 0 students, 5 students have taken Hindi, 0 students have taken Hindi but not English. All the students in the class have taken at least one of the subjects of English and Hindi. Find the number of students who have taken English but not Hindi.

9 Foundation : Paper-4 : Business Mathematics & Statistics Fundamentals [ December 0 ] 9 (b) A student is to answer 8 out of 0 questions on an eamination : (i) How many choice has he? (ii) How many if he must answer the first three questions? (iii) How many if he must answer at least four of the first five questions? Answer 9. (a) n(e H) 0, n(e H) 50 5 n(h) 5 n(e H) 0 n(e H) n(e) n(h) n(e H) n(e) 0 (5 5) 0 n(e H) n(e) n(e H) So, the no. of students who have taken English but not Hindi is 5. Answer 9. (b) (i) The 8 questions out of 0 questions may be answered in 0 C 8 Now 0 0! 09(8)! C ways 8!! 8!! (ii) The first questions are to be answered. So there are remaining 5 (= 8 ) questions to be answered out of remaining 7 (= 0 ) questions which may be selected in 7 C 5 ways. Now, 7 C 5 = 7.6 = 4 ways. (iii) Here we have the following possible cases : (a) 4 questions from first 5 questions (say, group A), then remaining 4 questions from the balance of 5 questions (say, group B). (b) Again 5 questions from group A, and questions from group B. For (a), number of choice is 5 C 4 5 C 4 = 5 5 = 5 For (b), number of ways is 5 C 5 5 C = 0 = 0. Hence, Required no. of ways = = 5. Q. 0. (a) If y, prove that p + qy a + by, where p, q, a, b are fied constants. (b) If + y y, show that a + by p + qy, a, b, p, q being all constants. (c) Find, if ( ). Answer 0. (a) Since y, we have = ky, where k is a constant. p qy pky qy y(pk q) pk q Now constant = k (say), a by aky by y(ak b) ak b or, p + qy = k (a + by), where k is a constant. Hence p + qy a + by.

10 0 [ December 0 ] Revisionary Test Paper (Revised Syllabus-008) Answer 0. (b) If + y y; + y = m( y), where m is a constant, or, + y = m my, or, y + my = m, or, y ( + m) = (m ), m or, y k, where k is a constant. Now see 0(a). m Answer 0. (c) ( ), or,. / ( ), or, or, / / ;, or, 9, 4 or, / / ( ), 9 [ 4 0 ] Section - III [ Mensuration ] Q.. (a) The length, breadth and height of a bo are m, 4 m and m respectively. The length of the largest rod that can be placed in the bo is (i) 5m (ii) m (iii) m (iv) none of these (b) If the hypotenuse of a right angled isosceles triangle is 4 cm then the area of the triangle is (i) sq. cm (ii) 8 sq. cm (iii) 4 sq. cm (iv) none of these Answer. (a) (ii) Length of the largest rod = 4 = m Answer. (b) (iii) Let the lendth of equal sides be d Cm. Length of hypotenuse = d d d cm

11 Foundation : Paper-4 : Business Mathematics & Statistics Fundamentals [ December 0 ] Now d 4 d cm Area of the triangle d ( ) = 4 sq. cm. Q.. (a) The height of the right circular cone is 4 cm and its slant height is 45.5 cm. Find the cost of painting of its total surface at the rate of ` per sq cm. (Take = /7). (b) A right pyramid with height 8 cm stands on a base which is a triangle with sides of lengths cm, 4 cm and 5 cm. Find the volume of the pyramid. Answer. (a) Here r l h = 7.5 cm Total surface area rl r 7.5( ) = 465 sq. cm. 7 So total cost comes to ` 465 Answer. (b) Semi perimeter of base = S = 4 5 = 6 cm Area of base 6(6 )(6 4)(6 5) 6 = 6 sq. cm Volume of the pyramid = X Area of base height = 68 = 6 cu. cm Q.. (a) The perimeter of a rectangle, having area 8 sq. cm and its length being twice its breadth, is (i) 9 cm (ii) 8 cm, (iii) 4 cm, (iv) none of these (b) Find the quantity of water in litre flowing out of a pipe of cross-section area 5 cm in minute if the speed of the water in the pipe is 0 cm/sec. (c) The volumes of two spheres are in the ratio 8:7 and the difference of their radii is cm. Find the radii of both the spheres. Answer. (a) (ii) Given l = b and lb = 8 b = 8 b = cm So l = 6 cm. Perimeter = (+b) = 8 cm. Answer. (b) Volume of water flowing in sec = 5 0 = 50 c.c. Volume of water flowing in min = = 9000 c.c. = 9 litre

12 [ December 0 ] Revisionary Test Paper (Revised Syllabus-008) Answer. (c) 4 r 4 r 8 7 r r k, r k r Now k k = k = So, the radii (r ) of st sphere = 6 cm and the radii (r ) of nd sphere = 9 cm. Q. 4. (a) The circumference of the base of a cylinder is 44 cms. and its height is 0 cms. Find the volume of the cylinder. (b) A solid cylindrical rod of length 80 cms. radius, 5 cms. is melted and made into a cube. Find the side of the cube. Answer 4. (a) r = 44 or, r 44 or, r = 7 cm. 7 Volume = r h 490 = 080 cu. cm. 7 Answer 4. (b) Volume of cube = a cu. cm. [Where a is the side of cube] Volume of cylindrical rod r h 5 80 a, by question. 7 Or, a = or, a cm. Q. 5. (a) The circumference of the base of a cylinder is 44 cm and its height is 0 cm. Find the volume of the cylinder. (b) The curved surface of a cylinder is 000 sq cm and the diameter of the base is 0 cm. Find the volume of the cylinder and its height to the nearest millimeter. Answer 5. (a) If r cm be the radius of the base of the cylinder, then r = 44, or, r 7. Volume of the cylinder r h cu cms. 7

13 Foundation : Paper-4 : Business Mathematics & Statistics Fundamentals [ December 0 ] Answer 5. (b) 0 If r cm be the radius of the base of the cylinder, then r 0 ; curved surface = rh. rh = 000, or, h = 50, where h = height. Volume of the cylinder = r h = 0 h = = 5000 cu cm. Height of the cylinder h cm. Q. 6. (a) The diameter of the base of a conical water tank is 8 m and its height is 8 m. How much water does the tank hold? (b) A conical tent is required to accommodate 5 people, each person must have 6 sq ft of space on the ground and 00 cu ft of air to breathe. Give the vertical height, slant height and width of the tent. Answer 6. (a) 8 Radius of the circular base of the tank 4 m and its height = 8 m. Volume of the water tank = r h (4) = 696 cu m. = decimetres = litres = 696 kilolitres. Answer 6. (b) Space required to accommodate 5 people = 6 5 = 80 sq ft; r = 80, where r is the radius of the base. Again volume of the conical tent r h, where h = height of the tent 80h h = 00 5, or, h 875 ft. 80 From r = 80, we have r 80, 7 or, 560 r 5.45 ; r = 5.04 ft. Width = r = 5.04 = 0.08 ft. l = slant height h r (8.75) ft.

14 4 [ December 0 ] Revisionary Test Paper (Revised Syllabus-008) Section - IV [ Co-ordinate Geometry ] Q. 7. (a) If the point (, 0) is equidistant from the points (-, ) and (6, 4) then value of is (i) (ii) (iii) 4 (iv) none of these (b) In what ratio is the joint of the points (4, ) and (5, ) divided by the line + y 8 = 0? Answer 7. (a) (ii) Distance between (, 0) and (-, ) = Distance between (, 0) and (6, 4) ( ) (0 ) ( 6) (0 4) Answer 7. (b) Let the line + y 8 = 0 divide the join of A (4, ) and B (5, ) at the point C in the ratio m : n. Then the 5m 4n m n co-ordinates of C are C,. mn mn B (5, ) n Since C lies on the line + y 8 = 0; 5m 4n m n 5m 4n 9m n 8m 8n 8 0, or, 0, m n m n m n m 7 or, 6m 7n = 0, or, 6m = 7n, or,, i.e., m : n = 7 : 6. n 6 Hence the required ratio is 7 : 6. m A (4, -) C Q. 8. (a) Prove that the two circles + y + 6y + 5 = 0 and + y + 0 y + = 0 touch each other eternally. (b) Show that the point (, 7) lies inside the circle + y 6 8y = 0. Answer 8. (a) + y + 6y + 5 = 0. > [ + ] + (y ) = 5 Centre is (-, ), Radius = r = 5 + y + 0 y + = 0 = > (+5) + (y ) = 5. Centre is (-5, ), Radius = r = 5 Distance between the two centres ( 5) ( ) 5 Again r r The circle touch each other eternally.

15 Foundation : Paper-4 : Business Mathematics & Statistics Fundamentals [ December 0 ] 5 Answer 8. (b) Here g = 6, f = 8 and c =. g =, f = 4. The centre C of the circle is the pont ( g, f) = (, 4), and the radius of the circle g f The distance of the point P (, 7) from the centre c C (, 4) ( ) (7 4), which is less than 6. Hence the point (, 7) lies inside the circle. Q. 9. (a) Find the co-ordinates of the verte and the focus and the equation of the directri of the parabola y = 6. Find also the length of the latus rectum. (b) The major and minor aes of an ellipse are the and y aes respectively. Its eccenricity is / and the length of the latus rectum is units. Find the equation of the ellipse. Answer 9. (a) We have which is of the form y = 4a. Here 6 y = 6, or, y = 6 4 4a =, or, a. The co-ordinates of the verte are (0, 0) and the co-ordinates of the focus are (a, 0), i.e., (4/, 0). The equation of the directri is 4 a 0, or, 0, or, + 4 = 0. The length of the latus rectum 4a = 6/ units. Answer 9. (b) y Equation of the ellipse is a b Here, b... (i) a e b b a a... (ii) b From (i) and (ii), a a a and b 9 y Equation of the ellipse y /

16 6 [ December 0 ] Revisionary Test Paper (Revised Syllabus-008) Q. 0. (a) Find the co-ordinates of the foci, the eccentricity and the equations of the directrices of the hyperbola 6 9y = 44. y (b) Show that the line y = touches the ellipse. What are the co-ordinates of the 5 6 point of contact? Answer 0. (a) We have 6 9y y 6 9y =44, or,, or, Here a = 9 and b = 6. a =, taking positive sign. Now a b e ; a 9 9 e [ e >.] The co-ordinates of the foci are 5 (± ae, 0) = (±., 0) = (± 5, 0). 5 The eccentricity is e =. The equations of the directrices are 5 e ± a = 0, or, 0 or, 5 ± 9 = 0. Answer 0. (b) y = (i) From (), From (), ( y) 5 y and. 5 6 = + y. y 69 9y 78y y y 48y 5y, or, or,, or, 69y + 48y + 04 = 0, or, (y + 48) = which gives two real and equal values of y, i.e., y,. 8 8 Since the roots are equal, the line (i) intersects the ellipse (ii) in two coincident points. Hence the line (i) touches the ellipse (ii). 48 Substituting y in = + y, we get The point of contact is,.

17 Foundation : Paper-4 : Business Mathematics & Statistics Fundamentals [ December 0 ] 7 Section - V [ Calculus ] Q.. (a) If y = log a then prove that ( a ) y y 0. (b) If y = Ae m + Be m show that y m y = 0. (c) Find the area of the region bounded by curves y = and y =. Answer. (a) y = log a y = ( a ). a a a a y = ( a ). ( a ) a a y y 0 Answer. (b) y = Ae m + Be m y = Ame m Bme m = m (Ae m Be m ) y = m (Ame m + Bme m ) =m (Ae m + Be m ) = m y y m y = 0 Answer. (c) y = and y = cut at (0, 0) and (, ) Required area = 0 d 0 d / / 0 0 ( 0) 0 ( 0) 6 sq. unit

18 8 [ December 0 ] Revisionary Test Paper (Revised Syllabus-008) dy y Q.. (a) If a y b = ( + y) a+b show that where a and b are independent of and y. d m (b) If y show that ( + )y + y = m y. d (c) The value of is 0 (i) log e, (ii) log e, (iii) log e, (iv) none of these Answer. (a) a log + b log y = (a+b) log (+y) Differentiating w.r. to. a b dy ab dy y d y d or, b a b dy ab a y y d y b( y) (a b)y dy (a b) a( y) or, y( y) d ( y) or, dy d b ay y( y) ( y) (b ay) y Answer. (b) y m y m m y m m y my y( ) m y y ( ) y m yy y y( ) y m y ( ) y y m y Proved.

19 Foundation : Paper-4 : Business Mathematics & Statistics Fundamentals [ December 0 ] 9 Answer. (c) d (ii) I = d 0 0 So I = log y loge e dy y where y i.e. d dy Q.. (a) If y log a y + y = 0. (b) If sum of two values is 8 find the maimum value of their product. (c) Find the area of the region bounded by curves y = and y =. (d) If u= +y +z, the value of u + yu y + zu z is (i) u (ii) (iii) -u (iv) none of these Answer. (a) y = log a ( a ). y a a a y. ( a ) ( a ) a ( a )y y 0 a Answer. (b) Let two values be and y Then + y = 8. Let A = y = (8 ) = 8 da = 8. da = 0 give = 4 d d A d d = < 0. So A is maimum at = 4 = 4, then y = 8 4 = 4 y = 4.4 = 4 = 6. So, maimum A = 84 4 = 6. Answer. (c) y = and y = cut at (0, 0) and (, ) Required area d d / / ( 0 0) ( 0) 6 sq. unit

20 0 [ December 0 ] Revisionary Test Paper (Revised Syllabus-008) Answer. (d) (i) u = + y + z u = u y = y u z = z. u + yu y + zu z = () + y (y) + z(z) = ( +y +z ) = u Q. 4. (a) The total cost (C) for output is as follows : 5 C. 5 4 Find (i) Cost when output is 5 units (ii) Average cost of output of 0 units (iii) Marginal cost (C) (b) The cost function (C) for commodity (q) is given by C = q 4q + 6q. Find the AVC and also find the value of q for which AVC is minimum. (c) If y = log e, show that d y 4y d Answer 4. (a) 5 For 5 units units C 5 AC d(c) AC for 0 units ; MC d 5 Answer 4. (b) dy d C AVC q 4q 6, q (in cost function (C), fied cost is absent) d MC (q 4q 6q) q 8q 6. dq For AVC minimum, slope of AC is zero i.e., d (q dq 4q 6) 0 or, q 4 = 0 or, q = units. Answer 4. (c) y = log e dy log d e = + log e d y log d = + log e d y d e dy + 4y = + log e + 4 log e = ( + log e ) =. d

21 Foundation : Paper-4 : Business Mathematics & Statistics Fundamentals [ December 0 ] Section - VI [ Statistics ] Q. 5. (a) Draw a Pie Chart to represent the following data relating to the production cost of manufacture : Cost of Materiala ` 8,400, Cost of Labour ` 0,70; Direct Epenses of Manufacture `,50; Factory Overhead epenses ` 5,60 (b) From the following distribution of wages of workers in a factory, draw a histogram and a frequency polygon Wages (in `) : No. of Workers : Answer 5. (a) We first epress each item as a percentage of the total cost, viz., Rs. 96,000. Percentage Central angles. Cost of Materials 8,400 96,000 00% = 40%.6 40 = 44. Cost of Labour. Direct Epenses 0,70 96,000,50 96,000 00% = %.6 = 5. 00% = %.6 = Factory Overhead 5,60 96,000 00% = 6%.6 6 = 57.6 Total = 00% = 60 A circle of conveient redius is now drawn and the above angles are marked out at the centre of the circle. 4 radii will then divide the whole circle into four required sectors. The different sectors are generally differently shaded. fig. gives the Pie Chart required. The calculations of angles at the centre can be avoided if the circumstance of the circle be divided into 00 equal parts.

22 [ December 0 ] Revisionary Test Paper (Revised Syllabus-008) Answer 5. (b) As the class internal are unequal, we have to find frequency density so that height of the rectangle could be calculated. () () () (4) = () () (5) = (4) 0 Class interval Class width frequency frequency density height of the rectangle Histogram and Frequency Polygon Wages (in `) Q. 6. (a) The mean annual salary of all employees of a company is Rs. 8,500. The mean salaries of male and female employees are Rs. 0,000 and Rs. 5,000, respectively. Find the percentage of males and females employed by the company. (b) Median marks of 50 candidates in mathematics in a test are 6. Frequencies in the ranges 0-0 and 0-40 are missing the following table : Marks obtained : No. of candidates : Determine the missing frequencies. Answer 6. (a) Let n be the number of males and n the number of females n n Then n n n 0,000 n 5,000 8,500 n n

23 Foundation : Paper-4 : Business Mathematics & Statistics Fundamentals [ December 0 ],500n 500 n i n 70 n :n 7: n 0 7 Percentage of males and females = 0. 0 Answer 6. (b) Let the missing frequencies be f and f 5 + f f + 7 = 50 f + f = 8 Class No. of Cumulative (Marks) Candidates frequencies f 5 + f f * median class 0-40 f 5+f +f f +f 5 (5 f ) Median 6 0 0, f f f + f = 8 f = 8 8 = 0 The missing frequencies are 8 and 0. 8 Q. 7. (a) Find the mean deviation about arithmetic mean of the first 0 natural numbers. (b) Following are the marks obtained by 50 students in mercantile law paper : Marks Students Find median graphically. Answer 7. (a) n (n ) 0 Sum of first 0 natural numbers A.M. of first 0 natural numbers

24 4 [ December 0 ] Revisionary Test Paper (Revised Syllabus-008) Answer 7. (b) First 0 natural 5.5 numbers () = Mean deviation Marks Freq Marks more than Marks less than = = = = = = = = = = 50 Median could be found graphically by making more than ogive and less than ogive. Where ever these two ogives meet, draw perpenditular from that point to X-ais to get the value of median. Computation of Median graphically

25 Foundation : Paper-4 : Business Mathematics & Statistics Fundamentals [ December 0 ] 5 Q. 8. (a) Find the mean and standard deviation of the values, (a + b) and (a b). (b) Prove that for any two positive real quantities AM GM HM. (c) Calculate coefficient of variation for the following distribution of marks obtained by 60 students in a test : Marks : Students : Answer 8. (a) a b a b Mean a. ( ) (a b a) (a b a) b b S.D b. n Answer 8. (b) Let and be any two positive real quantities. Now AM GM... (I) Net Combining (I) & (II) AM GM HM Answer 8. (c) HM GM... (II) d Class Mid No. of 5 d 0 Interval Value Students = d fd fd f f 60 fd fd = 0 = 96

26 6 [ December 0 ] Revisionary Test Paper (Revised Syllabus-008) Mean fd A i appro. f S.D. 0 fd f fd f S.D.. C.V Mean.7 Q. 9. (a) Eplain the concept is negatively skewed. For a frequency distribution, the quartiles are ` 0 and ` 50, and the median is 0. Calculate bowley s coefficient of skewness. (b) The means of two samples of sozes 50 and 00 respectively are 54. and 50.. Obtain the mean of the sample of size 50 obtained by combining the two samples. (c) Calculate the arithmetic mean and the median of the frequency distribution given below. Hence claculate the mode using the empirical relation between the three. Class-limits Frequency Answer 9. (a) Firs Quartile = 0 = Q Second Quartile = Meadian = 0 = Q Third Quartile = 50 = Q Bowley s Coefficient of Skewness Q Q Q Q Q Answer 9. (b) Here n = 50, n = 00, 54., 50.. n Mean n n n = 5.57 (appro.) 50

27 Foundation : Paper-4 : Business Mathematics & Statistics Fundamentals [ December 0 ] 7 Answer 9. (c) Calculation of A.M. Class-intervals Mid-values A d= (i = 5) i Frequency f fd = A Total = N = Efd Arithmetic Mean = A fd i N 00 CALCULATION OF CUMULATIVE FREQUENCY Class-boundary Cumulative Frequency (less than) Median = N N Here Median (M) = the value corresponding to cumulative frequency 50. Median class is and Median l N C i f Here l = 44.5, N = 50, C = 48, f = 4, i = 5. Median The empirical relation between Mean, Median and Mode is Mean Mode = (Mean Median), or, 45.5 Mode = ( ) = 0.4, or, = Mode, i.e. Mode =

28 8 [ December 0 ] Revisionary Test Paper (Revised Syllabus-008) Q. 0. (a) (i) Find mean and standard deviation of following frequency distribution of ages : Class of age (yrs) : Total No. of persons : (ii) Find the median and mode of the following grouped frequency distribution : Salaries (in `) per hour : Total No. of persons : (iii) For a group containing 90 observations the mean and standard deviation are 59 and 9 respectively. For 40 observations of them mean and standard deviation are 54 and 6 respectively. Find the mean and standard deviation of the remaining 50 observations. (b) Short notes on : (i) Central Tendency of Data; (ii) Ogive less than type. Answer 0. (a) (i) Mid value () : Total u = ( 5)/0 : 0 freq (f) : fu : fu : fu 0( ) mean yrs. f 0 fu fu s.d yrs. f f 0 0 (ii) Class (`) Frequency Cumulative frequency (< type)

29 Foundation : Paper-4 : Business Mathematics & Statistics Fundamentals [ December 0 ] So median class is since value corresponding to 50 (C.F.) lies in that class Median = = 7.8 `. Modal class is since maimum frequency 0 lies in that class Mode (0 0) (0 5) 0 5 = = 7.8 `. (iii) n n n n i.e., i.e., 6 90 n n n n 6 (59 54) 50 (6 59) i.e., 9 = s.d of remaining 50 observations Answer 0. (b) (i) CENTRAL TENDENCY OF DATA : A given raw statistical data can be condensed to a large etent by the methods of Classification and tabulation. But this is not enough for interpreting a given data we are to depend on some mathematical measures. Such a type of measure is the measure of Central Tendency. By the term of Central Tendency of Data we mean that Central Value of the data about which the observations are concentrated. Since the single value has a tendency to be somewhere at the Centre and within the range of all values, it is also known as the measure of Central Tendency. There are three measures of Central Tendency : (i) Mean (ii) Median (iii) Mode Mean is the most important measure which is of three types : (i) Arithmetic mean (ii) Geometric Mean (iii) Harmonic Mean Mean of a series (usually denoted by X ) is the value obtained by dividing the sum of the values of various items, in a series ( X ) divided by the number of items (N) constituting the series.

30 0 [ December 0 ] Revisionary Test Paper (Revised Syllabus-008) Median : If a set of observations is arranged in order of magnitude, then the middle most or central value gives the median. Median divides the observations into two equal parts, in such a way that the number of observations smaller than median is equal to the number greater than it. Mode : Mode is the value of the variate which occurs with maimum frequency. It represents the most frequent value of a series. In most frequency distributions Mean, Median and Mode obey the approimate relation known as Empirical relation epressed as Mean Mode = (Mean Median). (ii) Ogive less than type : Cumulative frequency corresponding to a given variate value of a distribution is defined to be the sum total of frequencies up to and including that variate value. This is known as cumulative frequency of less than type. In case of a grouped frequency distribution, cumulative frequency (less than) of a class corresponds to the upper class-boundary of that class and it is the sum total of frequencies of classes up to and including that class. For grouped frequency distribution of a continuous variable, cumulative, cumulative frequency distribution of less than type can be represented graphically by means of a cumulative frequency polygon also known as ogive less than type. To draw a cumulative frequency polygon, boundary values of each class are located in the X ais. The cumulative frequency table provides the cumulative frequency (less than type) corresponding to upper class boundary of a class along Y ais. For each pair of values (U i, CF i ), a point is plotted in the graph paper. Joining all these points by straight lines, we get a cumulative frequency polygon of less than type. Cumulative frequency (less than) (y) Class Boundaries ()