MT EDUCARE LTD. EXTRA HOTS PROBLEMS HOTS SUMS CHAPTER : - ARITHMETIC PROGRESSION AND GEOMETRIC PROGRESSION. If 3 d tem of an A.P. is p and the 4 th tem is q. Find its n th tem and hence find its 0 th tem. (5 maks) Given : t 3 = p t 4 = q To find : t n and t 0 Sol. t n = a + (n ) d t 3 = a + (3 ) d p = a + d a + d = p...(i) t 4 = a + (4 ) d q = a + 3d a + 3d = q...(ii) Subtacting (ii) fom (i), a + d = p a + 3d = q ( ) ( ) ( ) d = p q Multiplying both sides by d = q p Substituting d = q p in (i), a + (q p) = p a + q p = p a = p + p q a = 3p q t n = a + (n )d t n = 3p q + (n ) (q p) t 0 = 3p q + (0 ) (q p) = 3p q + 9 (q p) = 3p q + 9q 9p t 0 = 7q 6p. Find thee consecutive tems in a G.P. such that thei sum is 38 and poduct is 78. (5 maks) Sol. Let the thee consecutive tems of G.P. be a, a and a. Fom fist given condition, a + a + a = 38...(i) Fom second given condition, a a a = 78 a 3 = 78 a = [Taking cube oots on both sides] Substituting a = in (i), + + = 38 Multiplying thoughout by, SCHOOL SECTION 339
+ + = 38 + 38 + = 0 6 + = 0 Dividing thoughout by, 6 3 + 6 = 0 6 9 4 + 6 = 0 3 ( 3) ( 3) = 0 ( 3) (3 ) = 0 3 = 0 o 3 = 0 = 3 o 3 = MT EDUCARE LTD. = 3 o = 3 If = 3 then If = 3 then a a = 3 = 3 = 8 a a = 3 = 3 = 8 3 a = a = a = 8 a = 8 3 The thee consecutive tems of a G.P. ae 8,, 8 o 8,, 8. 3. The 5 th, 8 th and th tem of a G.P. ae p, q and s. Show that q = ps. (4 maks) Given : Fo a G.P. t 5 = p t 8 = q t = s To find : q = ps Poof : Fo a G.P. t n = a n t 5 = a 5 p = a 4...(i) t 8 = a 8 q = a 7...(ii) t = a s = a 0...(iii) q = a 7 [Fom (ii) L.H.S. = q = (a 7 ) = a 4...(iv) R.H.S. = ps = a 4. a 0 = a 4+0 = a 4...(v) Fom (iv) and (v), L.H.S. = R.H.S. q = ps 340 SCHOOL SECTION
MT EDUCARE LTD. 4. If each tem of A.P. is doubled, is the esulting sequence also an A.P.? If it is wite its fist tem, common diffeence and n th tem. (4 maks) Sol. Let the A.P. be a, a + d, a + d, a + 3d.... If each tem of the A.P. is doubled the new sequence will be a, a + d, a + 4d, a + 6d [Multiplying each tem by 0] t = a t = a + d t 3 = a + 4d t 4 = a + 6d t t = a + d a = d...(i) t 3 t = a + 4d (a + d) = a + 4d a d = d...(ii) t 4 t 3 = a + 6d (a + 4d) = a + 6d a 4d = d...(iii) The diffeence between two consecutive tem is d emaining constant the new sequence is an A.P. with Fist tem (A) = a Common diffeence (D) = d n th tem (T n ) = A + (n )D = a + (n ) d = [a + (n ) d T n = t n 5. Thee consecutive tems in G.P. ae such that thei sum is 6 and sum of thei squaes is 364. Find the thee consecutive tems in G.P.(5 maks) Sol. Let the thee consecutive tems in G.P. be a, a, a Accoding to the fist condition, a + a + a = 6...(i) Accoding to the second condition, a + a + a = 364...(ii) Fom (i) a + a = 6 a Squaing both sides, a a = (6 a) a a a (a) = 6 6 a + a a a a = 676 5a + a a a a a = 676 5a a a a = 676 5a Substituting (ii) in (iii),...(iii) 364 = 676 5a 5a = 676 364 SCHOOL SECTION 34
5a = 3 a = 3 5 a = 6 Substituting a = 6 in (i) we get, 6 + 6 + 6 = 6 Multiplying thoughout by 6 + 6 + 6 = 6 6 + 6 6 + 6 = 0 6 0 + 6 = 0 Dividing though by, 3 0 + 3 = 0 3 9 + 3 = 0 3 ( 3) ( 3) = 0 ( 3) (3 ) = 0 3 = 0 o 3 = 0 MT EDUCARE LTD. = 3 o = 3 If = 3 If = 3 a = 6 3 = a = 6 3 = 8 a = a = 6 = 6 3 = 8 3 6 = 3 The thee consecutive tems in a G.P. ae, 6, 8 o 8, 6,. 6. If sum of m tems is n and sum of n tems is m then show that sum of (m + n) tems is (m + n). (5 maks) Given : S m = n S n = m To find : S m+n = (m + n) Poof : S m = m [a + (m ) d] n = m [a + (m ) d]...(i) S n = n [a + (n ) d] m = n [a + (n ) d]...(ii) Subtacting (ii) fom (i), n m = m [a + (m ) d] n [a + (n ) d] Multiplying thoughout by, (n m) = m [a + (m ) d] n [a + (n ) d] (n m) = m [a + md d] n [a + nd d] 34 SCHOOL SECTION
MT EDUCARE LTD. (n m) = am an + m d n d md + nd (n m) = a (m n) + d (m n ) d (m n) (m n) = a (m n) + d (m n) (m + n) d (m n) Dividing thoughout by m n = a + d (m + n) d = a + (m + n ) d...(iii) Now, S m+n = m n [a + (m + n ) d] S m+n = m n [ ] S m+n = m + n ( ) S m+n = (m + n) 7. 00 logs of wood ae stacked in the following manne 0 logs in the bottom ow, 9 in the next ow, 8 in the ow next to it and so on. In how many ows 00 logs ae placed and how many logs ae thee in the top ow. (5 maks) Sol. Thee ae 0 logs in the fist ow, 9 in the second ow and 8 in the ow next to it This aangement of logs 0, 9, 8,... foms an A.P. with a = 0, d = Let 00 logs be aanged in n ows S n = 00 We know, S n = n [a + (n ) d] 00 = n [ (0) + (n ) ( )] 400 = n [40 n + ] 400 = n (4 n] 400 = 4n n n 4n + 400 = 0 n 5n 6n + 400 = 0 n (n 5) 6 (n 5) = 0 n 5 = 0 o n 6 = 0 n = 5 o n = 6 If n = 5 t n = a + (n ) d t 5 = 0 + (5 ) ( ) t 5 = 0 + 4 ( ) t 5 = 0 4 t 5 = 4 No. of logs in the 5th ow cannot be negative n = 5 is not acceptable n = 6 t n = a + (n ) d t 6 = 0 + (6 ) ( ) t 6 = 0 + 5 ( ) t 6 = 0 5 t 6 = 5 00 logs ae placed in 6 ows and thee ae 5 logs in the top ow. SCHOOL SECTION 343
MT EDUCARE LTD. CHAPTER : - QUADRATIC EQUATIONS. Two pipes unning togethe can fill a cisten in Sol. 3 3 minutes. If one pipe takes 3 minutes moe than the othe to fill it, find the time in which each pipe would fill the cisten. (5 maks) Let the time taken by othe pipe to fill the cisten be x minutes Time taken by fist pipe to fill the cisten = (x + 3) minutes Potion of cisten filled by othe pipe in one minute = x Potion of cisten filled by fist pipe in one minute = x 3 Potion of cisten filled by both the pipes in one minute = 3 3 = = 40 3 3 40 Fom the given condition, x x 3 = 3 40 x 3 x x (x 3) = 3 40 x 3 = 3 x 3x 40 40 (x + 3) = 3 (x + 3x) 80x + 0 = 3x + 39x 3x 4x 0 = 0 3x 65x + 4x 0 = 0 3x (x 5) + 4 (x 5) = 0 (x 5) (3x + 4) = 0 x 5 = 0 o 3x + 4 = 0 x = 5 o x = 4 3 x = 4 3 x = 5 x + 3 = 5 + 3 = 8 is not acceptable because time cannot be negative. Time taken by two pipes to fill the cisten sepaately ae 5 minutes and 8 minutes.. If the equation ( + m ) x + m cx + (c a ) = 0 has equal oots, pove that c = a ( + m ). (4 maks) Sol. ( + m ) x + mcx + (c a ) = 0 Compaing with Ax + Bx + C = 0 we get, A = + m, B = mc, C = c a The equation has equal oots, B 4AC = 0 (mc) 4( + m ) (c a ) = 0 4m c 4( (c a ) + m (c a )] = 0 344 SCHOOL SECTION
MT EDUCARE LTD. Dividing thoughout by 4, m c [c a + m c m a ] = 0 m c c + a m c + m a = 0 c + a + m a = 0 a + m a = c c = a ( + m ) Hence poved. 3. Out of a goup of swans, 7 times the squae oot of the numbe ae playing on the shoe of a tank. The two emaining one ae playing with amoous fight in the wate. What is the total no. of swans? (5 maks) Sol. Let total no. of swans be x then, 7 No. of swans playing on the shoe of the tank = x No. of swans with amoous fight = Fom the given condition, x = 7 x Multiplying thoughout by we get, x = 7 x x 7 x 4 = 0 x 7 x 4 = 0 Substituting x = m m 7m 4 = 0 m 8m + m 4 = 0 m (m 4) + (m 4) = 0 (m 4) (m + ) = 0 (m 4) (m + ) = 0 m 4 = 0 o m + = 0 m = 4 o m = Resubstituting m = x we get, x = 4 o x = Squaing thoughout, x = 4 o x = x = 6 o x = 4 x = is not acceptable because no. of swans cannot be a faction. 4 x = 6 Total no. of swans is 6. 4. Two tains leave a ailway station at the same time. The fist tain tavels due west and the second tain due noth. The fist tain tavels 5km/h faste than second tain. If afte two hous, they ae 50 km apat. Find the speed of each tain. (5 maks) Sol. Let speed of second tain be x km/h Speed of fist tain = (x + 5) km/h SCHOOL SECTION 345
Distance = Speed time Distance coveed by second tain = x = (x) km Distance coveed by fist tain = (x + 5) = (x + 0) km In ABC, m ABC = 90º By Pythagoas theoem, AB + BC = AC (x) + (x + 0) = (50) 4x + (x) + + x 0 + 0 = 500 4x + 4x + 40x + 00 500 = 0 8x + 40x 400 = 0 Dividing thoughout by 8, x + 5x 300 = 0 x + 0x 5x 300 = 0 x (x + 0) 5 (x + 0) = 0 (x + 0) (x 5) = 0 x + 0 = 0 o x 5 = 0 x = 0 o x = 5 x = 0 is not acceptable because speed cannot be negative x = 5 x + 5 = 5 + 5 = 0 The speed of the tains ae 5 km/h and 0 km/h. MT EDUCARE LTD. 5. Thee is a squae field whose side is 44 m. A squae flowe bed is pepaed in its cente leaving a gavel path all ound the flowe bed. The total cost of laying flowe bed and gavelling the path at Rs..75 and Rs..50 pe squae mete espectively is Rs. 4904. Find the with of the gavel path. (5 maks) Sol. Length of the squae field = 44 m x Aea of the squae field = 44 44 = 936 sq.m. 44 x Let width of the gavel path be x m Length of the squae flowe bed = (44 x)m Aea of the squae flowe bed = (44 x) sq.m. Aea of the gavel path = Aea of the field Aea of the squae flowe bed = 936 (44 x) = 936 [44 44 x + (x) ] = 936 (936 76x + 4x ) = 936 936 + 76x 4x = (76x 4x ) sq.m. Cost of laying the flowe bed = (Aea of flowe bed) (Rate pe sq. m.) = [44 44 x + (x) ] 75 00 = [936 76x + 4x ] 75 00 = [484 44x + x ] 4 75 00 = [484 44x + x ] = Rs. (534 484x + x ) Cost of laying the gavel path = (Aea of gavel path) (Rate pe sq. m.) = (76x 4x ) (.50) C x 50 km x + 0 Flowe Bed x 44 x A x B x 346 SCHOOL SECTION
MT EDUCARE LTD. = (76x 4x ) 50 00 = (88x x ) 3 = Rs. (64x 6x ) As pe the given condition, 534 484x + x + 64x 6x = 4904 5x 0x + 40 = 0 Dividing thoughout by 5, x 44x + 84 = 0 x 4x x + 84 = 0 x (x 4) (x 4) = 0 (x 4) (x ) = 0 x = 4 o x = 0 x = 4 o x = x = 4 is not acceptable because side of field is 44 m. x = The width of the gavel path is m. 6. Solve : x 4 9x 3 + 4x 9x + = 0 (5 maks) Sol. x 4 9x 3 + 4x 9x + = 0 Dividing thoughout by x we get, 4 3 x 9x 4x 9x x x x x x = 0 x 9x + 4 9 x + x = 0 x + x 9x 9 x + 4 = 0 x 9 x 4 = 0 x x Substituting x + = m x Squaing both sides x = m x x + x x + x = m x + + x = m x + x = m (m ) 9m + 4 = 0 m 4 9m + 4 = 0 m 9m + 0 = 0 m 4m 5m + 0 = 0 m (m ) 5(m ) = 0 (m ) (m 5) = 0 m = 0 o m 5 = 0 m = o m = 5 m = o m = 5 Resubstituting m = x x SCHOOL SECTION 347
MT EDUCARE LTD. x = x...(i) x = 5 x....(ii) x x [Fom (i)] Multiplying thoughout by x, x + = x x x + = 0 (x ) = 0 x = 0 x = x = 5 [Fom (ii)] x Multiplying thoughout by x we get, x + = 5x x 5x + = 0 x 4x x + = 0 x (x ) (x ) = 0 (x ) (x ) = 0 x = 0 o x = 0 x = o x = Solution set =,, CHAPTER : 3 - LINEAR EQUATIONS IN TWO VARIABLES. Solve : x + 3 y = 7, x+ 3 y+ = 5. (4 maks) Sol. x + 3 y = 7...(i) x+ 3 y+ = 5 x. 3 y. 3 = 5 4. x 3.3 y = 5...(ii) Substituting x = a and 3 y = b in (i) and (ii), a + b = 7...(iii) 4a 3b = 5...(iv) Multiplying (iii) by 3 3a + 3b = 5...(v) Adding (iv) and (v), 4a 3b = 5 3a + 3b = 5 7a = 56 a = 8 Substituting a = 8 in (iii), 8 + b = 7 b = 7 8 b = 9 Resubstituting the values of a and b a = x 8 = x 3 = x x = 3 [ Bases ae equal, powes ae also equal] 348 SCHOOL SECTION
MT EDUCARE LTD. b = 3 y 9 = 3 y 3 = 3 y [ Bases ae equal, powes ae also equal] y = x = 3 and y = is the solution.. Solve : Sol. (a b)x + (a + b)y = a ab b...(i) (a + b) (x + y) = a + b (a + b)x + (a + b)y = a + b...(ii) Subtacting (ii) fom (i), (a b)x + (a + b)y = a ab b (a + b)x + (a + b)y = a + b ( ) ( ) ( ) ( ) [a b (a + b)]x = ab b (a b a b) x = b (a + b) bx = b (a + b) b (a b) x = b x = (a + b) Substituting x = a + b in (i), (a b) (a + b) + (a + b)y = a ab b a b + (a + b)y = a ab b (a + b) y = ab y = ab a b x = a + b and y = ab a b (4 maks) is the solution of given simultaneous equations. 3. Find the aea of the egion bounded by the following lines and X-axis. 4x 3y + 4 = 0, 4x + 3y 0 = 0. (5 maks) Sol. 4x 3y + 4 = 0 4x + 4 = 3y 4x 4 = y 3 y = 4x 4 3 x 5 4 y 4 8 4 (x, y) (, 4) (5, 8) ( 4, 4) 4x + 3y 0 = 0 3y = 0 4x 0 4x y = 3 x 5 y 4 8 0 (x, y) (, 4) (, 8) (5, 0) SCHOOL SECTION 349
MT EDUCARE LTD. Y Scale : cm = unit on both the axes (, 8) 8 7 (5, 8) 6 5 4 A (, 4) 3 4x + 3y 0 = 0 X -5-4 -3 - - 0 3 4 5 X 4x 3y + 4 = 0 ( 4, 4) - - -3-4 Y A (ABC) = b h = 6 4 A (ABC) = sq. units 4. Places A and B 00 km apat on the highway. One ca stats fom A and anothe fom B at the same time. If the cas tavel in the same diection at diffeent speed, they meet in 5 hous. If they tavel towads each othe, they meet in hou. What ae the speeds of the two cas? (5 maks) Sol. Let the speed of fist ca, stating fom A be x km/h Let the speed of second ca, stating fom B be y km/h Distance = Speed Time 350 SCHOOL SECTION
MT EDUCARE LTD. Distance tavelled by fist ca in 5 hous = 5x km Distance tavelled by second ca in 5 hous = 5y km 5x km A 00 km B C 5y km AC BC = AB Fom fist given condition, 5x 5y = 00 Dividing thoughout by 5, x y = 0...(i) Distance tavelled by fist ca in hou = x = x km Distance tavelled by second ca in hou = y = y km x km y km AC + BC = AB Fom second given condition, x + y = 00...(ii) Adding (i) and (ii), x y = 0 x + y = 00 x = 0 x = 60 Substituting x = 60 in (ii), 60 + y = 00 y = 00 60 y = 40 A B C 00 km The speed of cas ae 60 km/h and 40 km/h. 5. Find the values of p and q fo which the following system of equations has infinite solutions. (4 maks) Sol. x + 3y = 7 (p + q) + (p q) = Compaing with a x + b y = c a =, b = 3, c = 7 Compaing with a x + b y = c a = p + q, b = p q, c = The equation have infinite solutions a a = b b = c c 3 p q = p q = 7 p q = 7 4 = 7 (p + q) p + q = 6...(i) 3 p q = 3 3 = 3 (p q) 3 = p q 3 p q =...(ii) SCHOOL SECTION 35
MT EDUCARE LTD. Adding (i) and (ii), p + q = 6 p q = 3p = 7 p = 9 Substituting p = 9 in (i), 9 + q = 6 q = 6 9 q = 3 p = 9 and q = 3. 6. The atio of incomes of two pesons is 9 : 7 and the atio of thei expenditues is 4 : 3. If each of them saves Rs. 00 pe month. Find thei monthly incomes. (4 maks) Sol. Let monthly incomes of two pesons be Rs. x and Rs. y Fom fist given condition, x y = 9 7 7x = 9y 7x 9y = 0...(i) Each one of them saves Rs. 00 Expenditue of fist peson = Rs. (x 00) Expenditue of second peson = Rs. (y 00) Fom second given condition, x 00 y 00 = 4 3 3 (x 00) = 4 (y 00) 3x 600 = 4y 800 3x 4y = 00...(ii) D = 7 9 3 4 = (7 4) ( 9 3) = 8 + 7 = D x = 0 9 00 4 = (0 4) ( 9 00) = 0 800 = 800. D y = 7 0 3 00 = (7 00) (0 3) = 400 0 = 400 By Came s ule, x = Dx = 800 D = 800 y = Dy = 400 D = 400 The income of two peson ae Rs. 800 and Rs. 400. CHAPTER : 4 - PROBABILITY. A coin is tossed twice. If the second toss esults in a tail a die is thown. Wite the sample space. Sol. S = { HH, TH, HT, HT, HT3, HT4, HT5, HT6, TT, TT, TT3, TT4, TT5, TT6 } n (S) = 4 35 SCHOOL SECTION
MT EDUCARE LTD.. Two dice ae thown togethe. What is the pobability that sum of the numbes on two dice is 5 o numbe on the second die is geate than o equal o equal to the numbe on the fist die. (4 maks) Sol. When two dice ae thown S = { (, ), (, ), (, 3), (, 4), (, 5), (, 6), (, ), (, ), (, 3), (, 4), (, 5), (, 6) (3, ), (3, ), (3, 3), (3, 4), (3, 5), (3, 6) (4, ), (4, ), (4, 3), (4, 4), (4, 5), (4, 6) (5, ), (5, ), (5, 3), (5, 4), (5, 5), (5, 6) (6, ), (6, ), (6, 3), (6, 4), (6, 5), (6, 6) } n (S) = 36 Let A be the event that the sum of the numbes on the two dice is 5 A = { (, 4), (, 3), (3, ), (4, ) } n (A) = 4 P (A) = n (A) n (S) 4 P (A) = 36 Let B be the event that numbe on the second die is geate than o equal to the numbe on the fist die B = { (, ), (, ), (, 3), (, 4), (, 5), (, 6), (, ), (, 3), (, 4), (, 5), (, 6) (3, 3), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6) (5, 5), (5, 6), (6, 6) } n (B) = P (B) = n (B) n (S) P (B) = 36 A B is the event that sum of the numbes on the two dice is 5 and the numbe on the second die is geate than o equal to the numbe on the fist die. A B = { (, 4), (, 3) } n (A B) = P (A B) = (A B) n (S) P (A B) = 36 A B is the event that sum of the numbe on two dice is 5 o numbe on the second die is geate than o equal to the numbe on the fist die. P (A B) = P (A) + P (B) P (A B) = P (A B) = 3 36 4 36 36 36 = 4 36 SCHOOL SECTION 353
MT EDUCARE LTD. 3. On the disc show below, a playe spins the aow twice. The faction a b is fomed whee a is the numbe of the secto whee the aow stops afte the fist spin and b is the numbe of the secto whee the aow stops afte the second spin. On evey spin each of the numbeed secto has ae equal pobability of being the secto on which the aow stops. What Sol. is the pobability that the faction a is geate than? b Since the aow can stop in any one of the six sectos. So a and b both ca assume values fom to 6. Thus, the odeed pain (a, b) can be as follows : 3 S = { (, ), (, ), (, 3), (, 4), (, 5), (, 6), (, ), (, ), (, 3), (, 4), (, 5), (, 6) (3, ), (3, ), (3, 3), (3, 4), (3, 5), (3, 6) 4 (4, ), (4, ), (4, 3), (4, 4), (4, 5), (4, 6) (5, ), (5, ), (5, 3), (5, 4), (5, 5), (5, 6) (6, ), (6, ), (6, 3), (6, 4), (6, 5), (6, 6) } n (S) = 36 Fo the faction a b (4 maks) to be geate than, a should be geate than B. 5 6 Let A be the event that the faction a b A = { (, ), (3, ), (3, ), (4, ), (4, ), (4, 3), (5, ), (5, ), (5, 3), (5, 4), (6, ), (6, ), (6, 3), (6, 4), (6, 5) } n (A) = 5 P (A) = n (A) n (S) P (A) = 5 36 P (A) = 5 4. In the adjoining figue, a dat thown lands in the inteio of the cicle. What is the pobability that the dat will land in the shaded egion? Sol. We have l (AB) = l (CD) = 8 l (BC) = l (AD) = 6 ABCD is a ectangle In ABC, m ABC = 90º By Pythagoas theoem, AC = AB + BC AC = 8 + 6 AC = 64 + 36 AC = 00 AC = 0 Aea of cicle = (OA) = (5) = 5 = 5 3.4 = 78.5 sq. units 354 D A 6 8 O 8 C (4 maks) 6 B SCHOOL SECTION
MT EDUCARE LTD. Aea of ABCD = AB BC = 8 6 = 48 sq. units Aea of shaded eagin = Aea of cicle Aea of ectangle = 78.5 48 = 30.5 sq. units Let A be the event that the dat lands in the shaded egion Aea of shaded egion P (A) = Aea of cicle P (A) = = 30.5 78.5 = 305 785 6 57 5. A lette is chosen at andom, fom the lette in the wod ASSASSINATION. Find the pobability that the lette chosen is a (i) vovel (ii) consonant. (3 maks) Sol. Thee ae 3 lettes in the wod ASSASSINATION out of which one lette can be chosen in n (S) = 3 (i) Let A be the event that the lette chosen is a vowel Thee ae 6 vowels n (A) = 6 P (A) = n (A) n (S) P (A) = 6 3 (ii) A is the event that the lette chosen is a consonant. P (A) + P (A) = P (A) = P (A) 6 3 P (A) = P (A) = 3 6 3 P (A) = 7 3 6. A numbe x is selected fom the numbes,, 3 and then a second numbe y is selected fom the numbes, 4, 9. What is the pobability that the poduct xy of the two numbes will be less than 9. (3 maks) Sol. Numbe x can be selected in thee ways and coesponding to each such way thee ae thee ways of selecting numbe y. Two numbes can be selected in 9 ways as listed below : S = { (, ), (, 4), (, 9), (, ), (, 4), (, 9), (3, ), (3, 4), (3, 9) } n (A) = 5 P (A) = n (A) n (S) P (A) = 5 9 SCHOOL SECTION 355
MT EDUCARE LTD. CHAPTER : 5 - Statistics - I. The mean of the following fequency distibution is 50. Find the value of f : (5 maks) Class inteval 0-0 0-40 40-60 60-80 80-00 Fequency 7 f 3 4 9 Sol. Class Class maks Fequency inteval (xi ) (f i ) 0-0 0 7 70 0-40 30 f 30f 40-60 50 3 600 60-80 70 4 680 80-00 90 9 70 Total 9 + f 560 + 30f Mean x = f ixi f i 50 = 560 30 f 9 f 50 (9 + f) = 560 + 30f 4600 + 50f = 560 + 30f 50f 30f = 560 4600 0f = 560 f = 560 0 f = 8. An incomplete fequency distibution is given as follows : Classes inteval Fequency 0-0 0-30 30 30-40? 40-50 65 50-60? 60-70 5 70-80 8 Total 9 Given that median value is 46, detemine the missing fequencies using the medians fomula. (5 maks) Sol. Median is 46 it lies in the class 40-50 and the coesponding fequency is 65. Classes Fequency Cumulative fequency c.f. 0-0 0-30 30 4 30-40 f 4 + f 40-50 65 f 4 + f + 65 = 07 + f 50-60 f 07 + f + f 60-70 5 07 + f + f + 5 = 3 + f + f 70-80 8 3 + f + f + 8 = 50 + f + f = 9 Total 9 f i x i 356 SCHOOL SECTION
MT EDUCARE LTD. Fom the last c.f. 50 + f + f = 9 f + f = 9 50 f + f = 79 f = 79 f...(i) Median = N h L c. f. f Whee L = 40, N = 9, c.f. = 4 + f, h =0, f = 65 Median = 9 0 40 (4 f ) 65 46 = 9 40 4 f 3 46 40 = 9 4 f 3 3 6 = 9 4 f 39 = 9 4 f 78 = 9 84 f f = 9 84 78 f = 67 f = 67 f = 33.5 34 f = 79 f [Fom (i)] f = 79 34 f = 45 Hence f = 34 and f = 45. 3. The following data gives the infomation on the obseved life time (in hou) of 5 electical components : Sol. Lifetime (in hou) Fequency 0-0 0 0-40 35 40-60 5 60-80 6 80-00 38 00-0 9 Detemine the modal life time of the components. (5 maks) The class having maximum fequency is 60-80 is the modal class. Lifetime (in hou) Fequency 0-0 0 0-40 35 40-60 5 f 60-80 6 f m 80-00 38 f 00-0 9 The modal class is 60-80 and L = 60, f = 5, f m = 6, f = 38, h = 0 SCHOOL SECTION 357
MT EDUCARE LTD. Mode = = = f m f L f f f m h 6 5 60 0 (6) 5 38 9 60 0 90 = 9 60 0 3 Mode = 60 + 5.65 Mode = 65.65 hous. CHAPTER : 6 - Statistics - II. The following pie-diagam shows the pecentage distibution of the expenditue incued in publishing a book study the diagam and answe the questions : Sol. (a) (b) If fo cetain quantity of books, the publishe has to pay Rs. 30,600 as pinting, then what will be amount of oyalty to be paid fo these books. What is the cental angle of the secto coesponding to the expenditue incued on Royalty. (3 maks) Pomotion cost (a) Let the total cost fo cetain quantity of books be Rs. x Pinting cost = 0% of x 30600 = 0 x 00 30600 00 = x 0 x = 53000 Royalty paid fo these book = 5% of total cost 5 = 00 53000 = 5 530 = 950 5 (b) Measue of cental angle fo Royalty = 360 00 = 54º Royalty 5% Pinting cost 0 %. Daw histogam and fequency polygon both in one figue shaing the following infomation in a city. (3 maks) Age (yeas) 5-30 30-35 35-40 40-45 45-50 No. of doctos 6 58 5 36 0 0% Taspoation 0 % Binding 0 % Pape cost 5% 358 SCHOOL SECTION
MT EDUCARE LTD. Y Scale : On X = axis : cm =.5 yeas On Y = axis : cm = 5 doctos 75 70 65 60 55 50 45 40 No. of doctos 35 30 5 0 5 0 5 X 0 5 30 35 40 45 50 X Y 4. The annual pofits eaned by 30 shops if a shopping complex in a locality give ise to the following distibution. Pofit (in lakhs of Rs.) Age in yeas No. of shops (fequency) Moe than o equal to 5 30 Moe than o equal to 0 8 Moe than o equal to 5 6 Moe than o equal to 0 4 Moe than o equal to 5 0 Moe than o equal to 30 7 Moe than o equal to 35 3 (5 maks) SCHOOL SECTION 359
MT EDUCARE LTD. Daw both ogive cuves fo the above data and hence obtain the median. Sol. Pofit in lakh Classes No. of shops fequency c.f. less of Rs. type Moe than o equal to 5 5-0 30 Moe than o equal to 0 0-5 8 4 Moe than o equal to 5 5-0 6 6 Moe than o equal to 0 0-5 4 4 0 Moe than o equal to 5 5-30 0 3 3 Moe than o equal to 30 30-35 7 4 7 Moe than o equal to 35 35-40 3 3 30 Cumulative fequency (No. of shops) Y 30 8 6 4 0 8 6 4 0 8 6 (5, 30) (0, 8) (5, 0) (5, 6) (0, 6) (5, 4) (0, 4) (5, 0) (35, 7) (40, 30) Scale : On X = axis : cm = Rs. 5 lakhs On Y = axis : cm = shops (30, 3) (30, 7) Median = Rs. 7.5 lakhs 4 (0, ) (35, 3) X 0 Y (5, 0) (40, 0) 5 0 5 0 5 30 35 40 Classes (Pofit in lakhs of Rs.) X 360 SCHOOL SECTION
MT EDUCARE LTD. 3. Pove that the total aea of the ectangles in a histogam is equal to the total aea bounded by the coesponding fequency polygon and the X-axis. (5 maks) Poof : Conside the following histogam and the coesponding fequency polygon. Let the total aea of the ectangles of the histogam be A sq. units and the aea enclosed by the fequency polygon and the X-axis be A sq. units. Then, we find the aea of tiangles numbeed,,..., 0 Let us denote these tiangles as,,..., 0 Then,, 4, 6, 8 and 0 ae included in ae A but not in aea A...(i), 3, 5, 7 and 9 ae included in aea A but not in aea A...(ii) The emaining aea is common to both the figues,...(iii) Now, conside, i.e., ABC and, i.e., EDC In these tiangles, B D [Each is a ight angle] seg AB = seg DE [Each of length 5 units] ACB ECD [Vetically opposite angles] ABC EDC [A-S-A test fo conguency] A (ABC) = A (EDC) [Conguent tiangles ae of equal aea] i.e., A ( ) = A ( )...(iv) Y 45 40 35 Scale : cm = 5 units on Y axis 6 0 Fequency 30 5 0 5 0 D E 3 4 5 9 8 X 5 A C B 0 0 0 30 40 50 X Y Class Similaly, we can pove that A ( 3 ) = A ( 4 ), A ( 5 ) = A ( 6 ), A ( 7 ) = A ( 8 ) and A ( 9 ) = A ( 0 )...(v) Fom (iv) and (v), A ( ) + A ( 3 ) + A ( 5 ) + A ( 7 ) + A ( 9 ) = A ( ) + A ( 4 ) + A ( 6 ) + A ( 8 ) + A ( 0 )...(vi) Fom (i), (ii), (iii) and (vi) Aea of the histogam = aea of the fequency polygon with the X-axis. 7 SCHOOL SECTION 36