2Common Proficiency Test Model Paper 4

5 x2 ∫ 171. I = dx ... (i) x2 +(5−x)2 0 5 (5−x)2 ∫ I = dx ... (ii) [f(x) = f(a–x)] (5−x)2 +(x)2 0 5 2I = ∫ dx = [x]5 0 0 2I = 5 I = 5/2 Ans. (d) None of these (2x −1) 1+x +1 172. Lt . x→0 1+x −1 1+x +1 (2x −1) ( ) lt . lt 1+x+1 x→0 x x→0 App lt. ⇒ log 2 × 2 ⇒ 2 log 2 Ans. (a) 2log 2 |x−1| 173. Lt x→0 x−1 (x−1) RHL … Lt = 1 x→0+ (x−1) −(x−1) LHL Lt = –1 x→0− (x−1) LHL ≠ RHL ∴ f(x) does not exist (c) Does not exist x−|x| 174. f(x) = x x−x RHL Lt = 0 x→0+ x x−(−x) 2x LHL Lt = = 2 x→0− x x f (0) = 2 Common Proficiency Test (CPT) Volume - II 331 © The Institute of Chartered Accountants of India ANSWERS ⇒ RHL ≠ LHL ≠ f(0) ∴ f(x) is not continuous at x = 0 Ans. (b) No. 175. y = log (log x) 3 3 dy d d = log t. (log x) 3 3 dx dt dx 1 1 1 = . ⇒ log 3 x.log 3 x.log3 x. logx .(log3)2 log3 1 = x.log3.logx 1 Ans. (a) = x.log3.logx 176. CI if calculated annually. ⎡ ⎛ 20 ⎞2 ⎤ 11P CI = P ⎢⎜1+ ⎟ −1⎥ = 1 ⎢⎝ 100 ⎠ ⎥ 25 ⎣ ⎦ CI if calculated semi annually. ⎡ ⎛ 10 ⎞4 ⎤ 4641 CI = P ⎢⎜1+ ⎟ −1⎥= P 2 ⎢⎝ 100 ⎠ ⎥ 10000 ⎣ ⎦ 4641P 11P 241P ∴ − =482⇒ =482 10000 25 10000 ∴ P = 20,000 Ans. (a) Rs. 20,000 ⎡( 1+i ) n −1 ⎤ 177. Value of annuity (A) = P ⎢ ⎥ ⎢⎣ i ⎥⎦ ⎡( 1+0.09 ) 3 −1 ⎤ = 3000 ⎢ ( ) ⎥ ⎢⎣ 0.09 ⎥⎦ = 9833.33 Ans. (c) Rs. 9833.33 332 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India 10000 178. Present Value of Annuity = ( ) 1+0.05 10 = Rs. 7724 Ans. (a) Rs. 7724 [ ] [ ] P ( ) 45000 ( ) 179. A = 1+in −1 ⇒ 1+0.06 10 −1 i 0.06 [ ] ( ) 750000 1+0.06 10 −1 {Solve by taking log} A = 517500 ∴ Surplus = 517500 – 5,00,000 = 17,500 Ans. (c) Rs. 17,500 180. Present value of P (rest) of the annuity. ⎡ 1−(1+i) −n⎤ ⎡ 1−(1+0.1) −5⎤ P = A ⎢ ⎥⇒ 2000 ⎢ ⎥ ⎣ i ⎦ ⎣ (0.10) ⎦ [ ] P = 20000 1−(1.1) −5 P = 7294 which is less than the Purchase Price. ∴ leasing is preferable. Ans. (a) leasing is preferable. 181. The sum of deviations of the given values from their Arithmetic Mean is 0. Ans. (a) Arithmetic Mean 182. The sum of squares of the deviations of the given values from their Arithmetic Mean is minimum. Ans. (a) Arithmetic Mean 183. Which is greatly affected by the extreme values – Arithmatic mean Ans. (a) Arithmetic Mean 184. Which is not amenable to further algebric treatment – Mode and Median Ans. (d) Both (b) and (c) a+b 185. = 15 ⇒ a + b = 30 … (i) 2 b – a = 4 – (ii) [By eq (i) & (ii)] 2b = 34 ⇒ b = 17 Common Proficiency Test (CPT) Volume - II 333 © The Institute of Chartered Accountants of India ANSWERS ∴ a = 13 lower limit = 13 Ans. (c) 13 186. Ans. (b) Refer Properties 187. Ans. (a) Refer Properties 188. Ans. (c) Refer Properties 189. Given, consumer price index in, (say), period I = 120 and consumer price index in (say), period II = 215. The wages of the worker in period I and II are given to be Rs. 1,680 and Rs. 3000 respectively. The real wages of the worker in the current period II with respect to the period I as base, are given by: 120 Rs. ×3000 =Rs.1674.42 215 Since this wage (Rs. 1674.42) is less than the wages of the worker in the period I (Viz. Rs. 1680) the workers is not better off but worse off R. 5.58 as compared to the period I. Ans. (a) 190. Purchasing power (P.P) of a rupee in 1994 with respect to the base period 1980 is given by 100 P.P. of a rupee = Consumer Price Index for 1994 w.r.t.base 1980 100 = Rs. = Re. 0.40 250 Ans. (a) 191. Let B , B and B be the events of drawing a boy from the 1st, 2nd and 3 rd group 1 2 3 respectively and G , G and G be the events of drawing a girl from the 1st, 2nd, and 3rd 1 2 3 group respectively then P (B ) = 1/4, P (B ) = 2/4, P(B ) = 3/4 and P(G )=3/4, P(G ) = 2/4, 1 2 3 1 2 P(G ) = 1/4. 3 The required event of getting 1 girl and 2 boys in a random selection of 3 children can materialize in the following mutually exclusive cases. (i) Girl from the first group and boys from the 2nd and 3rd group i.e. the event G ∩B 1 2 ∩B 3 (ii) Girl from the 2nd group and boys from 1st and 3rd groups, i.e. the event B ∩G ∩ B 1 2 3 happens. (iii) Girl from the 3rd group and boys from the 1st and 2 nd groups. i.e., the event B ∩ B 1 2 ∩ G happens. 3 Hence by the addition theorem of probability, required probability ρ is given by: P = P(i) + P(ii) + P(iii) 334 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India = P(G ∩B ∩B ) + P (B ∩G ∩B ) + P(B ∩B ∩G ) 1 2 3 1 2 3 1 2 3 = P(G ) P(B ) P(B ) + P(B ) P(G ) P(B ) + P(B ) P(B ) P(G ) 1 2 3 1 2 3 1 2 3 3 2 3 1 2 3 1 2 1 = × × + × × + × × 4 4 4 4 4 4 4 4 4 18+6+2 26 13 = = = 64 64 32 192. Let P(A) = x 3 3 ∴ P(B) = P(A) = x 2 2 1 1 ⎛3 ⎞ 3 and P(C) = P(B) = ⎜ x⎟ = x 2 2 ⎝2 ⎠ 4 The events A, B and C are exhaustive ∴ P (A or B or C) = 1 ⇒ P(A) + P(B) + P(C) = 1 ( ∵ A, B, C are mutually exclusive) 3 3 x+ x+ x=1 2 4 ⎡4+6+3⎤ x⎢ ⎥ =1 ⎣ 4 ⎦ 4 ∴ x = 13 4 ∴ P(A) = 13 Ans. (b) 193. There are 3+4+2+1 = 10 members in all and a Committee of 4 out of them can be formed in 10C ways. Hence exhaustive number of Cases is: 4 10×9×8×7 10C = =210 4 4! The probability 'p' that the Committee Consists of the doctor and at least one economist is given by p = P [One doctor, One economist, 2 others] + P [One doctor, Two economist, 1 others] + P[One doctor, Three economist] Common Proficiency Test (CPT) Volume - II 335 © The Institute of Chartered Accountants of India ANSWERS 1C ×3C ×6C 1C ×3C ×6C 1C ×3C p= 1 1 2 + 1 2 1 + 1 3 10C 10C 10C 4 4 4 1 ⎡⎛ 6×5⎞ ( ) ( ) ⎤ = ⎢⎜1×3× ⎟+ 1×3×6 + 1×1⎥ 210 ⎣⎝ 2 ⎠ ⎦ 1 [ ] = 45+18+1 210 64 32 = = =0.3048 210 105 Ans. (a) 194. Let A = event that the company executive travel by plane. ∴ P(A) = 2/3 Let B = event that the Company executive travel by train. ∴ P(B) = 1/5 Now the events A and B are mutually exclusive, because he cannot travel by plane and train at the same time. ∴ The prob. of his traveling by plane or train Ans. (a) = P (A or B) = P(A) + P(B) = 2/3 + 1/5 13 = 15 Ans. (b) 195. Let A and B denote the events that the contractor will get a 'plumbing' Contract and 'Electric' Contract respectively. Then we are given: P(A) = 2/3, P( B) = 5/9 ∴ P(B) = 1 P(B) = 4/9 and P(A∪B) = Probability that Contractor gets at least one contract. = 4/5 ⇒ P(A) + P(B) P (A∩B) = 4/5 2/3 + 4/9 P(A∩B) = 4/5 ⇒ P(A∩B) = 2/3 + 4/9 4/5 336 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India 30+20−36 = 45 14 = 45 Hence, the probability that the Contractor will get both the contracts in 14/45 Ans. (a) 196. Standard deviation = σ P(x>60)=0.05 ⇒ 1 − P (x < 60) = 0.05 ∴ P (x < 60) = 0.95 ⎡x−60 60−50⎤ ∴ P ⎢ ≤ ⎥= 0.95 ⎣ σ σ ⎦ ⎛ 10⎞ ⎛10⎞ ∴ P ⎜≤ ⎟= 0.95 ⇒ ϕ ⎜ ⎟ = ϕ (1.64) ⎝ σ ⎠ ⎝ σ ⎠ ⎛ 10 ⎞ ⇒ σ = ⎜ ⎟ = 6.7 ⇒ S.D. = 6.7 ⎝1.64⎠ Ans. (a) 6.7 ⎛ x−10⎞ 197. P ⎜Z≥ ⎟ = 0.10 ⎝ 20 ⎠ 100−x ∴ = 1.28 20 100 − x = 25.6 x = 74.40 Ans. (c) 74.40 ⎛ x−100 ⎞ 198. P ⎜x≤ ⎟ = 0.10 ⎝ 20 ⎠ x−100 ∴ = 1.28 20 x = 25.6 + 100 = 125.6 Ans. (b) 125.6 200. P = P(x>70) = 1 − P (x < 70) Common Proficiency Test (CPT) Volume - II 337 © The Institute of Chartered Accountants of India ANSWERS ⎡x−65 70−65⎤ = 1 − P ⎢ ≤ ⎥ ⎣ 25 5 ⎦ = 1 − P (z < 1) = 10− .0 41 P = 0.06 Ans. (c) 0.06 Model Test Paper – BOS/CPT – 4 100A 100×21315 151. P = ⇒ 100+RT 100+0.045× 4 12 P = 21000 Ans. (a) Rs. 21000 8 152. I = 500 × ×1=40 1 100 8 3 I =1000 × × =60 2 100 4 8 1 I =1000 × × =40 3 100 2 ∴ Total Amount = 500 + 1000 + 1000+40 + 60 + 60 = 2640 Ans. Rs. 2640 ⎛ 5 ⎞4n 153. 2000 = 1200 ⎜1+ ⎟ ⎝ 4×100 ⎠ ⎛81⎞4n 5 = 3 ⎜ ⎟ After taking log both side and solved. ⎝80⎠ log 5 = log 3 + 4n [log 81 log 80] n = 10 years 3 months Ans. (a) 10 years 3 months 338 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India ⎛ r ⎞4 154. 26500 = 20000 ⎜1+ ⎟ ⎝ 100 ⎠ After taking log and solving it r = 7.5% Ans. (c) 7.5% ⎡⎛ 7 ⎞⎛ 8 ⎞⎛ 85 ⎞ ⎤ 155. CI = 7000 ⎢⎜1+ ⎟⎜1+ ⎟⎜1+ ⎟−1⎥ ⎣⎝ 100 ⎠⎝ 100 ⎠⎝ 100 ⎠ ⎦ CI = 1776 3 6P 157. I = P × ×2 = 1 100 100 8 24P I = P × × 3 = 2 100 100 10 10P I = P × × 1 = 3 100 100 Total interest = 1520 40P 100×1520 ∴ = 1520 ⇒ P = 100 40 P = 3800 Ans. Rs. 3800 (a) [ ] 158. A = 7500 (1+i)n [I = 0.01 n = 2] = 7500 [1+0.01)2 ⇒ 7500 × (1.01)2 A = 7650.75 Ans. (a) Rs. 7650.75 [ ] 159. 512.50 = P (1+0.05)2 −1 512.50 = P × 0.1025 ∴ P = 5000 Ans. (b) Rs. 5000 ⎡ r ⎤3 160. 1331 = 1000 ⎢1+ ⎥ ⎣ 100⎦ ⎛11⎞3 ⎛ r ⎞3 r ⎜ ⎟ = ⎜1+ ⎟ ⇒ 1.1 = 1+ ⎝10⎠ ⎝ 100 ⎠ 100 Common Proficiency Test (CPT) Volume - II 339 © The Institute of Chartered Accountants of India ANSWERS ∴ 0.1 = r / 100 ⇒ r = 10% Ans. (a) 10% 161. Range = L − S L … S = 20 → (i) If each item is increased by 15 Range = (L + 15) − (S + 15) Range = L − S = 20 [from eg (i)] Ans. (a) 20 162. Range = L − S ∴ L − S 20 If each item is divided by− 2 L S −1 Range = − = (L−S) −2 −2 2 −1 Range = ×20 = − 10 [(− ) sign ignored] 2 Range = 10 (because it is difference between largest and smallest data) Ans. (b) 10 164. In grouped frequency distribution, if the class interval is unequal then quartile deviation is more appropriate. Ans. (a) Q.D. ∑ ∑ d2 d2 ∑ 165. SD = ⇒ (4)2 = ⇒ d2 = 160 n 10 If each item divided by − 2 ∑ 160 Corrected (d')2 = =40 (−2)2 ∑ (d')2 40 ∴ Corrected S.D. = = = 2 n 10 S.D. = 2 Ans. (a) 2 340 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India (5+10+15.......... ....+125 166. x= 25 25[ ] 2×5+(25−1)5 2 = 130 =65 25 2 Average = 65 Ans. (a) 65 167. a, b, c, d, e are five add integers. a + b + c + d + e a+(a+2)+(a+4)+(a+6)+(a+8) average = = 5 5 ⇒ (a + 4) Ans. (d) a+4 180+258+x 168. Av = 3 438+x 230 = ⇒690−438 =x 3 ∴ x = 252 he should score 252 runs. Ans. (d) None of these (x+2)60+x.120 +(x−2)180 169. 100 = (x+2)+x+x−2 300x = 360x − 240 ∴ 60x = 240 x = 4 Ans. (a) 4 ∑ x ∑ 170. 16 = ⇒ x = 400 25 ∑ x1 ∑ 15 = ⇒ x1 = 360 ∴Age of Teacher = 400 − 360 = 40 24 Ans. 40 Years 171. Ans. (b) Refer Properties 172. Ans. (b) Refer Properties 173. Given two regression lines are Common Proficiency Test (CPT) Volume - II 341 © The Institute of Chartered Accountants of India ANSWERS 3x+2y=26 → (1) 6x+y=31→ (2) ( ) Since the two lines of regression intersect at the point x,y , replacing x and y by and respectively in the given regression equation, we get. (1) ⇒ 2 y = 26 − 3 x 3 y= 13 − x → (3) 2 3 (2) ⇒ 6 x + 13 − x = 31 2 12x+26−3x =31 2 9x+26 =62 9x=62−26 = 36 x = 4 3 ∴ (3) ⇒ y = 13 − (4) 2 = 13 − 6 = 7 ∴ x = 4, y = 7 Ans. (a) 174. Let us assume that 3x + 24 = 26 → (1) represent the regression line of y on x and 6x + y = 31 →(2) represent the regression line of x on y. (1) ⇒ 2y = 26 − 3x 3 y = 13 − − x 2 3 ∴ byx = − 2 (2) ⇒ 6x = 31 − y 31 1 x = − y 6 6 1 ∴ bxy = − 6 342 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India ⎛ 3⎞⎛ 1⎞ 1 ∴ r2 =byx×bxy =⎜− ⎟⎜− ⎟= ⎝ 2⎠⎝ 6⎠ 4 1 1 r = = ± = ± 0.5 4 2 Ans. (b) We take the sign of n as negative since both the regression coefficients are negative). 175. Ans. (a) Refer Properties 176. Let E , E , E denote the events that the probability is solved by X,Y and Z respectively. 1 2 3 Then we have P(E ) = 1/3 ⇒ P(E ) = 1 − P(E ) = 2/3 − 1 1 1 P(E ) = 1/4 ⇒ P(E ) = 1 − P(E ) = 3/4 2 2 2 P(E ) = 1/5 ⇒ P(E ) = 1 − P(E ) = 4/5 3 3 3 Problem will be solved if at least one of the three is able to solve it. Hence, the required probability that the problem will be solved is given by P(E ∪E ∪E ) 1 2 3 = 1−P(E ∩E ∩E ) 1 2 3 [ ( ) ( ) ( )] = 1− P E .P E .P E ) 1 2 3 = 1− 2/3 x 3/4 x 4/5 [Since E , E , E are independent] 1 2 3 = 1 − 2/5 = 3/5 Ans. (c) 1 177. Given P(A) = 2 1 P(B) = 3 ( ) 1 P A∩B = 4 1 ( ) ∴ P(A|B) = P A∩B = 4 = 1 × 3 = 3 P(B) 1 4 1 4 3 Ans. (a) Common Proficiency Test (CPT) Volume - II 343 © The Institute of Chartered Accountants of India ANSWERS 1 1 ( ) 1 178. Given P(A) = , P(B) = and P A∩B = 2 3 4 ( ) ( ) ∴ P A∩B = P(B) − P A∩B 1 1 4−3 1 = − = = 3 4 12 12 Ans. (c) 1 1 ( ) 1 179. Given P(A) , P(B) = , P A∩B = 2 3 4 ( ) ( ) P A∩B = 1 − P A∪B { ( )} = 1 − P(A)+P(B)−P A∩B ( ) = 1 − P(A) − P(B) + P A∩B 1 1 1 = 1− − + 2 3 4 12−6−4+3 5 = = 12 12 Ans. (a) 1 ( ) 1 180. P (A) = ½, P(B) = , P A∩B = 3 4 ( ) ( ) = P A∪B =P A∩B ( ) = 1−P A∩B − = 1 ¼ 4−1 3 = = 4 4 Ans. (b) 181. Given x = 1, x =2, x = 3 1 2 3 1 ( ) 1 ( ) P(x )= ,P x = ,P x =1/6 1 2 3 2 3 ∴ E(x) = x P(x ) + x P(x ) + x P(x ) 1 1 2 2 3 3 ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ 1 2 1 = ⎜1× ⎟+⎜2× ⎟+⎜3× ⎟= + + ⎝ 2⎠ ⎝ 3⎠ ⎝ 6⎠ 2 3 2 344 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India 3+4+3 10 5 = = = =1.666.... 6 6 3 = 1.67 Ans. (c) 182. Given x =1, x =2, x =3 1 2 3 1 1 ( ) 1 P(x ) = ,P(x ) = ,P x = 1 2 2 3 3 6 ∴ V(x) = E(x2) … [E(x) ] 2 E(x) = x P(x )+x P(x )+ x P(x ) 1 1 2 2 3 3 ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ = ⎜1× ⎟+⎜2× ⎟+⎜3× ⎟ ⎝ 2⎠ ⎝ 3⎠ ⎝ 6⎠ 1 2 1 E(x) = + + 2 3 2 3+4+3 10 5 = = = 6 6 3 ( ) ( ) ( ) ( ) E x2 =x 2 P x + x 2P x +x 2 P x 1 1 2 2 3 3 ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ = ⎜1× ⎟+⎜4× ⎟+⎜9× ⎟ ⎝ 2⎠ ⎝ 3⎠ ⎝ 6⎠ 1 4 3 = + + 2 3 2 3+8+9 20 10 = = = 6 6 3 10 ⎛5⎞2 ∴ V(x) = −⎜ ⎟ 3 ⎝3⎠ 10 25 = − 3 9 30−25 5 V(x) = = =.5556 9 9 Ans. (a) 183. Let x denote the number of defective lamps. X can assume the values 0, 1, 2, 3 Common Proficiency Test (CPT) Volume - II 345 © The Institute of Chartered Accountants of India ANSWERS p(X=0) p:probably having 0 bad orange out of 4 bad orange and 3 good orange out of 8 good orange. 4C ×8C 56 P(x=0) = 0 3 = 12C 55 3 4C ×8C 28 P(x=1)= 1 2 = 12C 55 3 4C ×8C 12 P(x=2)= 2 2 = 12C 55 3 P(x=3) = Probability having 3 bad orange out of 4 bad orange and 0 good orange out of 8 good orange. 4C ×8C 1 = 3 0 = 12C 55 3 Probability that at least one orange out of three oranges is good = 1 − P (x=3) − = 1 1/55 55−1 54 = = 55 55 Ans. (a) 184. Given P(A) = 0.5, P (AB) < 0.3 By Addition thereon, P(A or B) = P(A) + P(B) P (AB) ∴ P(A) + P(B) P(AB) < 1 [∴ P(A or B) < 1] ∴ P(B) < 1 P(A) + P (AB) < 1 0.5 + 0.3 P(B) < 0.8 Ans. (a) 185. Let the given events be A, B and P(A) = 2/3 P(B) Let P(B) = x ∴ P(A) = 2/3 x The events A and B are exhaustive ∴ P(A or B) = 1 P(A) + P(B) = 1 ⇒ 2/3 x + x = 1 346 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India 5/3 x = 1 x = 3/5 ∴ P(B) = 3/5 P(A) = 2/3 x 3/5 = 2/5 P(B) = 3/5 ⇒ odds in favourof B are − 3 : 5 3 = 3: 2 Ans. (b) 186. Given person variates with parameter = 1 i.e. λ = 1 By the poison distribution e −λ .λx p(x) = , x>0 x! ∴ The required probability P(3 < × <5) = P(x = 4) e −λ .λ4 = 4! e −1(1)4 = 4! 0.36783 ×1 = 24 P(3 < x <5) = 0.015326 Ans. (a) 187. Given p = 2% = 2/100 = .02 n = 200 ∴ λ = np = 200 × .02 = 4 The probability of at least 5 defective means. − P(x > 5) = 1 P (x < 5) = 1 − {P(x=0) + P(x=1) + P(x=2) + p(x=3) + P(x=4)} ⎧ e −4(4)0 e −4(4)1 e −4(4)2 e −4(4)3 e −4(4)4⎫ = 1−⎨ + + + + ⎬ ⎩ 0! 1! 2! 3! 4! ⎭ Common Proficiency Test (CPT) Volume - II 347 © The Institute of Chartered Accountants of India ANSWERS = 1−e −4 ⎪ ⎨ ⎧ 1+4+ 42 + 43 + 44⎪ ⎬ ⎫ ⎪ ⎩ 2! 3! 4!⎪ ⎭ = 1−e −4⎨ ⎧ 1+4+ 16 + 64 + 256 ⎬ ⎫ ⎩ 2 6 24 ⎭ = 1−(0.183) {5+8+10.6667 +10.667} = 1−(0.83)(34.3334) = 1 0.6 283 P (x > 5) = 0.3717 Ans. (a) − 188. After a man is dealt 4 spade cards from an ordinary pack of 52 cards, there are 52 4 = 48 cards left in the pack, out of which 9 are spade cards and 39 are no spade cards. Now, 3 more cards can be dealt to the same man out of the 48 cards in 48C ways, which 3 determines the exhaustive number of ways. If none of these 3 additional cards is a spade cards, then the 3 additional cards must be drawn out of the 39 non-spade cards, which can be done in 39C ways. 3 The probability that none of the three additional cards dealt to the man is a spade card = 39C 3 48C 3 Hence, the required probability, 'P' that at least one of the additional cards is a spade cards is given by: 39C p= 1− 3 48C 3 39×38×37 3! = 1− × 3! 48×47×46 13×19×37 = 1− 16×47×23 9136 = 1− 17296 = 10.5282 p = 0.4718 Ans. (c) 190. Given P (x = 1) = P (x = 2) Given x is a poison variable. 348 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India e −λ (λ)(1) e −λ (λ)(2) ∴ = 1! 2! λ2 λ= 2 λ=2 = variance Ans. (b) 65 65 191. P = Q = 1 − P = 1 − 500 500 435 Q = 500 n = 500 P Q SE of Proportion of defectives = n 65 435 1 = × × 500 500 500 SE = 0.015 Ans. (a) 0.015 σ 0.75 192. Standard Error of Mean (SE) = = n 100 SE = 0.075 95% Confidence Limit for population mean are given by : x ± 1.96 SE = 5.6 ± 1.96 × 0.75 = 5.6 ± 0.147 The Confidence level are 5.453 and 5.747 Ans. (a) 5.453 and 5.747 σ 4 193. Standard Error (SE) = = =0.353 n 128 96% confidence limit for population mean are ⇒ x + 2.05 × SE = 28 ± 2.05 × 0.353 ⇒ 28 ± 0.72 The confidence level are 27.272 and 28.728 Ans. (b) 27.272 and 28.728 Common Proficiency Test (CPT) Volume - II 349 © The Institute of Chartered Accountants of India ANSWERS 65 65 435 194. P = , Q = 1 − P = 1 − ⇒ Q = 500 500 500 PQ SE of proportion of defectives = n 65 435 1 = × × 500 500 500 SE = 0.015 Confidence limits for the population are = P ± 3 × SE 65 = ±3×0.015 ⇒ 0.13 ± 0.045 500 Levels are 0.085 and 0.175 or Levels are 8.5% and 17.5% Ans. (a) 8.5% and 17.5 196. Variance = 4 σ = 4 = ± 2 Statement is true Ans. (a) True 197. n = 10 P = 0.3 ∴ Q = (1 − P) = 1 − 0.3 = 0.7 ∴ σ = npq ∴ Variance = npq = 10×0.3×0.7 Variance = 2.1 Ans. (a) 2.1 198. When the cost of living increases, the standard of living improves. Ans. (b) false ⎛ σ ⎞ ( ) 199. The 95% confidence limit for the sample mean x is x± 1.96 ⎜ ⎜ ⎟ ⎟ which is not given ⎝ n ⎠ Ans. (b) False 350 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India 200. Mean and variance never be equal ∴ Statement is false Ans. (b) False Model Test Paper – BOS/CPT – 5 151. Let the fraction be x/y. Then according to the given condition of the problem, 3x 18 = y−3 11 33x = 1 8y−54 33 x −18y+54 = 0 11x − 6y+18 = 0 .................... (i) x+8 2 and = 2y 5 ⇒ 5x + 40 = 4y 5x − 4y+40 = 0 .....................(ii) (i) × 2 ⇒ 22x − 12y+36 = 0 .....................(iii) (ii) × 3 ⇒ 15x − 12y+120 = 0 ....................(iv) (i ii) −(iv),w e get 7x − 84 = 0 7x= 84 x = 84/7 = 12 (i) ⇒ (11) (12) − 6y + 18 = 0 132 − 6y + 18 = 0 6y = 150 y =150/6 = 25 Hence, the required fraction is 12/25 ∴Ans. (c) 152. Let the two numbers are x and y Given x + y = 150 % of y Common Proficiency Test (CPT) Volume - II 351 © The Institute of Chartered Accountants of India ANSWERS 150 = ×y 100 x + y = 1.5y x = 0.5 y 1 x = y 2 Ans. (a) 153. Let three consecutive even numbers are x, x + 2, x + 4. 3 Given condition is x + (x+2) + (x+4) = 60× −15 4 3x + 6 = 30 3x = 24 24 x = = 8 3 ∴ The middle number = x + 2 = 8 + 2 = 10 Ans. (b) 154. Suppose my present age is x years and my sons present age is y years Five years ago my age = (x − 5) years my son's age = (y − 5) years According to the first condition of the problem, x−5= 3 (y −5) x −5 = 3y −15 ⇒ x 3y = 15 + 5 ⇒ x− 3y = − 10 .................(i) Ten years later my age = (x+10) years my son's age = (y+10) years According to the second condition of the problem, x+10 = 2(y+10) x+10 = 2y+20 − x-2y = 20 1 0 352 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India x − 2y = 10 ..................(i i) (i) − (ii)⇒ y = 20 (i) ⇒ x – 60 = –10 x = 60 –10 = 50 Hence, my presence age = 50 years and my son•s present age = 20 years ∴Ans.(a) 155. The compound ratio of 4:3, 9:13, 26:5 and 2:15 is 4×9×26×2 = 3×13×5×15 16 = 25 Ans. (b) n! 156. We know nPr = (n−4)! 56! ∴ 56 = P r+6 {56−(r+6)}! 56! = (50−r)! 54! 54P = r+3 {54−(r+3)}! 54! = (51−r)! 56P 56! (51−r)! Thus, r+3 = × 54P (50−r)! 54! r+3 56×55×54! (51−r)(50−r)! = × (50−r)! 54! 56×55×(51−r) = 1 But we are given the ratio as 30800 : 1 56×55×(51−r) 30800 ∴ = 1 1 Common Proficiency Test (CPT) Volume - II 353 © The Institute of Chartered Accountants of India ANSWERS 30800 (or) (51 – r)! = = 10 ∴ r = 41 56×55 Ans. (b) 157. He can arrange his schedule in 8P6 = 8 × 7 × 6 × 5 × 4 × 3 = 20160 ways. Ans. (b) 158. The two Indians can stand together in 2P = 2! = 2 ways. 2 So is the case with the two Americans and the two Russians. Now these 3 groups of 2 each can stand in a row in 3P = 3×2 = 6 ways. Hence by the 3 generalized fundamental principle, the total number of ways in which they can stand for a photograph under given conditions is 6 × 2× 2× 2 = 48 Ans. (c) 159. This is the number of combination of 52 cards taken five at a time. Now applying the formula. 52! 52C = 5 5!(52−5)! 52! = 5!47! 52×51×50×49×48×47! = 5×4×3×2×1×47! = 2598960 Ans. (a) 160. Let the unit's digit of the number be x and the ten's digit by y. Then x + y = 9 → (1) and the number = 10y + x Reversing the order of digits of the given number, Unit's digits becomes y and ten's digits becomes x ∴ Now number = 10x + y According to the given condition of the problem, 354 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India (10x + y) – (x+10y) = 27 10x + y – x – 10y = 27 9x – 9y = 27 x – y = 3 → (2) Adding (1) and (2), 23 get 2 × = 12 12 x = = 6 2 (1) ⇒ 6 + y = 9 y = 9 – 6 = 3 ∴ The given number is 36 Ans. (b) ⎡ (9x −1)−(3x −1) ⎤ ⎢ ⎥ 161. Lt 9x −3x ⇒ Lt ⎢ x ⎥ x→0 4x −2x x→0 ⎢(4x −1)−(2x −1)⎥ ⎢ ⎥ ⎣ x ⎦ ⎢ ⎡ ⎜ ⎛ 9x −1⎟ ⎞ −⎜ ⎛ 3x −1⎟ ⎞ ⎥ ⎤ ⎜ ⎟ ⎜ ⎟ ⎢⎝ x ⎠ ⎝ x ⎠⎥ log 9 −log 3 Lt ⎢ ⎥ ⇒ x→0 ⎢⎜ ⎛ 4x −1⎟ ⎞ −⎜ ⎛ 2x −1⎟ ⎞ ⎥ log 4 −log 2 ⎢⎜ ⎟ ⎜ ⎟⎥ ⎣⎝ x ⎠ ⎝ x ⎠⎦ 2log 3 − log 3 log3 ⇒ = 2log 2 − log 2 log2 log3 Ans. (a) log2 (5x −1)2 52x −2.5x +1 162. Lt = x→0 log(1+x) log(1+x) ⎛ 25x −1 ⎞ ⎛ 5x −1 ⎞ ⎜ ⎜ ⎟ ⎟−2⎜ ⎜ ⎟ ⎟ (25x −2.5x +1) ⎝ x ⎠ ⎝ x ⎠ ⇒ Lt ⇒ Lt x→0 log(1+x) x→0 log(1+x) x App Lt Common Proficiency Test (CPT) Volume - II 355 © The Institute of Chartered Accountants of India ANSWERS log25−2.log5 log25−log25 ⇒ = = 0 1 1 Ans. (d) None of these 163. Lt f(x) ⇒ Lt (x+1) x→1 x→1 App Lt ⇒ 1 + 1 = 2 Ans. (a) 2 164. LHL Lt (x – 1) x→2 App Lt ⇒ 2 – 1 = 1 ∴ LHL = 1 RHL Lt (2x – 3) x→2 App. Lt LHL 2.2 – 3 = 1 f(2) = 2.2 – 3 = 1 ∴ LHL = RHL = f(2) ∴ f(x) Continuous at x = 2 Ans. (a) Continuous x = 2 3x2 +2x+7 3x2 +2x+7 165. f(x) = = x2 −3x+2 (x−2)(x−1) To be continuous (x – 2) ≠ 0 & (x – 1) ≠ 0 ∴ x ≠ 2 & x ≠ 1 ∴ Points of discontinuity = 1, 2 Ans. (a) 1, 2 166. Let z = log x 1 dz = dx x dx = x dz ∴ I = ∫ 1 dx = ∫ 1 xdz x log x x.z 1 ∫ = dz z 356 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India = log z = log (log x) + c ans. (b) 167. Let I = ∫ log x dx 10 = ∫ logx .loge dx e 10 = loge ∫ log x.1dx 10 = loge ⎢ ⎡ logx.x − ∫1 × dx⎥ ⎤ 10 ⎣ x ⎦ [ ] I = loge x log x − x +c 10 ∫4ex +6e −x 168. Let I = dx 9ex −4e −x ∴ I = ∫4e2x +6 dx 9e2x −4 Let t = e2x ∴ dt = 2 e2x dx = 2t dx ∴ I = ∫ 4t+6 dt =∫ 2t+3 dt 9t−4 2t t(9t−4) ( ) = ∫ ⎢ ⎡ 0+3 + 2 4 ( /9 +3 )⎥ ⎤ dt ⎣t(0−4) 4/9 9t−4 ⎦ = ∫ ⎢ ⎡ − 3 + ( 35 )⎥ ⎤ dt ⎣ 4t 4 9t−4 ⎦ ( ) 3 35 log 9t−4 = − log t + . +c 4 4 9 ( ) 3 35 = − log e2x + log 9 e2x − 4 + c 4 36 Ans. (a) 169. See formula from the text book Ans. (c) Common Proficiency Test (CPT) Volume - II 357 © The Institute of Chartered Accountants of India ANSWERS 170. Put x2−6x+100 = t ∴ x2 −6x+100 =t2 (2x – 6) dx = 2t dt (x – 3) dx = t dt ∫ (x−3) x2 −6x+100 dx=∫ x2 −6x+100 (x−3)dx ∫ = t.tdt = ∫ t2 dt = ∫t3 +c 3 1 = (x2 −6x +100)3/2 +c 3 Ans. (c) 171. No. of ways in which one or more friends may invited = 6 +6 +6 +6 +6 +6 C C C C C C 1 2 3 4 5 6 =26 −1 = 63 ways. Ans. (a) 63 ways. 172. No. of ways of failure of candidate. = 4 +4 +4 +4 C C C C 1 2 3 4 = 24 −1 = 15 ways. Ans. (c) 15 358 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India 173. A voter can vote inthe following ways 254 = n +n +n +n +n +n +n ....n C 1 C 2 C 3 C 4 C 5 C 6 C 7 C n−1 ∴ 254 = 2n −(n +1)=2n −(1+1) C n ∴ 256 = 2 n ∴ 28 = 2n ⇒ ∴ n = 8 Total candidates = 8 Ans. (a) 8 174. No. of words of 3 consonants and 2 vowels among 17 consonants and 5 vowels are = 17 ×5 ×5! C C 3 2 = 816000 Ans. (b) 81,6000 176. The present value of annual profit V = A.P. (ni) = 34000 × 3.7079 V = 128886 which is less than initial cost of machine.M achine must not be purchased Ans. (a) Machine should not be purchased. r 177. 40 = 2000 × ×4 100 ∴ r = -0.5% Ans. (b) 0.5% Common Proficiency Test (CPT) Volume - II 359 © The Institute of Chartered Accountants of India ANSWERS 4 178. I = 2000 × ×1 = 80 1 100 14 I = 3000 × × 1 = 420 2 100 Total Interest = 500 500×100 ∴ rate of interest = = 10% 5000×1 r = 10% Ans. (a) r = 10% R R 179. I – I = 30 ⇒ 1200 × ×3 – 1 000 ×3=30 1 2 100 100 ⇒ 36R – 30R = 30 6R = 30 ∴ R = 5% Ans. (c) 5% 2 180. 40 = 2000 × × n 100 n = 1 yr. Ans. (a) = 1 yr. a+b 181. =20 ⇒ a + b = 40 → (1) 2 a−b SD = 5 → =5 2 ∴ a … b = 10 → (2) ⇒ 2a = 50 ⇒ a = 25 ∴ b = 15 Ans. (a) 25, 15 360 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India ∑ x 182. Mean = n 1+2+6+a+b 4.4 = 5 ∴ a + b = 13 → (1) ∑ x2 σ2 = −x2 N ∑ x2 8.24 = −(4.4)2 5 ∑ x2 = 138 1+4+9+a2 +b2 =138 ∴ a2 + b2 = 138 ⇒ a2 + (13 – a) 2= 138 ⇒ a2 – 13a – 36 = 0 a = 9, 4 ∴ Nos → 9, 4 ⎛N+1⎞th 183. For individual series, the rank of the median is = ⎜ ⎟ term ⎝ 2 ⎠ ⎛N+1⎞th Ans. (b) ⎜ ⎟ term ⎝ 2 ⎠ 184. Rank of the median of the series 2, 3, 4, 5, 6, 7 ⎛N+1⎞th ⎛6+1⎞th = ⎜ ⎟ term = ⎜ ⎟ term ⎝ 2 ⎠ ⎝ 2 ⎠ = 3.5 th term Ans. (a) 3.5 185. Regression Eq. 2x + 3y – 10 = 0 If y = 50 ∴ 2x = +10 – 3 × 50 = – 140 x = – 70 None of these Ans. (d) None of these. Common Proficiency Test (CPT) Volume - II 361 © The Institute of Chartered Accountants of India ANSWERS 186. Ans. (c) Refer Properties 187. Ans. (c) Refer Properties 188. Given r (x, y) = 0.4 → (1) a×c We know that r (a X, cY) = .r(x,y) → (2) |a|×|c| Using (2) in (1), we get r (2x, – y) = r (2x, – 1y) 2×(−1) = .r(x,y) |2|×|−1| −2×0.4 = 2×1 r (2x, – y) = – 0.4 Ans. (b) 189. Computation of Correlation Coefficient x y xy x2 y2 69 70 4830 4761 4900 85 87 7395 7225 7569 Total 154 157 12225 11986 12469 154 157 x = =77, y = =78.5 2 2 ∑ xy 12225 i i Cov (x,y) = – x y = – (77) (78.5) n 2 = 68 ∑ x2 11986 i −x2 = −(77)2 Sx = n 2 = 8 ∑ y2 12469 Sy = i −y2 = −(78.5)2 = 8.5 n 2 ( ) Cov x,y 68 68 ∴ n = = = =1 Sx Sy 8×8.5 68 Ans. (a) 362 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India 190. Computation of correlation Co – efficient. x y xy x2 y2 102 50 5100 10404 2500 109 48 5232 11881 2304 Total 211 98 10332 22285 4804 211 98 x = =105.5, y = =49 2 2 ∑ xiyi 10332 ( )( ) Cov (x, y) = − x y = − 105.5 49 n 2 = 5166 – 5169.5 = – 3.5 ∑ xi2 22285 ( ) Sx = −x2 = − 105.5 2 = 3.5 n 2 ∑ yi2 4804 ( ) Sy = Sx = −y2 = − 49 2 =1 n 2 ( ) Cov x,y −3.5 ∴ n = = ( ) ( ) = −1 Sx Sy 3.5 × 1 Ans. (b) 191. Ans. (a) … Refer Properties 192. Ans. (a) … Refer Properties 193. Ans. (b) … Refer Properties 194. Given X ~ N (μ, σ2), where μ = 2 and σ2 = 9 σ = 3 We want x so that P(2 < × < x) = 0.4115 → (1) x−μ 2−2 When X = 2, Z = = =0 σ 3 x−2 When x = x, Z = = Z1 (Say) → (2) 3 From (1), we get P (0 < Z < Z ) = 0.4115 1 ⇒ Z = 1.35 (from Normal Table) 1 Common Proficiency Test (CPT) Volume - II 363 © The Institute of Chartered Accountants of India ANSWERS Substituting in (2), we get x−2 = 1.35 3 ∴ x = 2+3 (1.35) x = 6.05 Ans. (b) 195. Mean = First moment about origin = 35 (given) → (1) Second moment about 35 = 10 (given) ⇒ Second moment about mean = 10 μ2 = 0 → (2) Since the given distribution is normal, β1 = 0 and β2 = 3 μ32 ∴β1 = = 0 ⇒ μ3 = 0 μ23 μ4 β2 = = 3 ⇒ μ4 = 3 μ2 = 3×102 = 300 μ22 2 ∴ μ1=0 (always), μ2 = 10, μ3 = 0, μ4 = 300 Ans. (c) 196. The most commonly used confidence limit is → 95% Ans. (c) 95% 197. Sample mean is statistic Ans. (b) Statistic 198. Deliberate sampling is – Non \random sampling Ans. (b) Non random sampling 199. Stratified random sampling issued for Non – Homogeneous population. Ans. (b) Non–homogeneous 200. Random Sampling is also called lottery sampling Ans. True 364 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India Model Test Paper – BOS/CPT – 6 151. Let given number is x 1⎛1⎛1 ⎞⎞ Then the condition ⎜ ⎜ ⎜ x⎟ ⎟ ⎟ =15 5⎝3⎝2 ⎠⎠ x =15 30 x = 450 152. Let the number be = x. 3⎛1 ⎞ Then given the condition = ⎜ x⎟ = 60 4⎝5 ⎠ 3x = 60 20 60×20 x= =400 3 Ans. (b) 153. Let the number be x. 4 ⎡3( )⎤ Given ⎢ x ⎥ = 24 5 ⎣8 ⎦ 3 x = 24 10 10 x=24× = 80 3 250 ∴ 250% of x = 250% of 80 = ×80 =200 100 Ans. (d) 154. Let the number be x Given condition x+x2 =182 x2 +x −182 = 0 ( )( ) x+14 x−13 = 0 x=−14, x=13 ∴ x = 13 (negative reflected) Ans. (a) Common Proficiency Test (CPT) Volume - II 365 © The Institute of Chartered Accountants of India ANSWERS 155. Let the unit's digit of the number be x and ten's digit by y Then x + y = 12 → (1) and the number = 10y + x Reversing the order of digits of the given number, Unit's digit becomes y and ten's digits becomes x ∴ New number = 10 x + y According to the given condition of the problem (10x + y) – (x + 10y) = 18 9x – 9y = –18 x–y= –2→ (2) Adding (1) and (2) ⇒ 2x = 10 x = 5 ∴ ⇒y = 7 ∴ The number is 75 Ans. (a) 156. Let the number of coins is x 14 18 Given 10x+ x+ x = 430 2 4 40x+28x+18x = 430 4 86x = 430 4 430×4 x = = 20 86 ∴ The one Rupee coins = 10x = 10×20 = 200 The 50 paise coins = 14x = 14×20 = 280 The 25 paise coins = 18x = 18×20 = 360 Ans. (a) 157. First Vessels Contain Milk Ratio 5 First Vessels Contain Water Ratio 2 366 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India Second Vessels Contain Milk Ratio 6 Second Vessels Contain Water Ratio 1 Both the Vessels Milk = 5 + 6 = 11 Both the Vessels Water = 2+ 1 = 3 ∴ The new Ratio = 11:3 Ans. (b) 158. Let the two numbers are x and y Given 8x = 5y → (1) and x + 27 = y → (2) (2) ⇒ x = y – 27 (1) ⇒ 8 (y – 27) = 5y 8y – 216 = 5y 3y = 216 216 y = = 72 3 ∴ x = 72 – 27 = 45 ∴ Sum of two number = x + y = 72 + 45 = 117 Ans. (c) 159. Let their monthly incomes be Rs. 9x and Rs. 7x respectively. Let their monthly expenditures be Rs. 4 y and Rs. 3 y respectively. According ot the given condition of the problem, 9x – 4y = 200 → (1) 7x – 3y = 200 → (2) Multiply (1) by 3, we get 27x – 12y = 600 Multiply (2) by 4, we get 28x – 12y = 800 Subtracting (3) from (4), we get x = 200 Hence their monthly income are Rs. (9×200=1800) and Rs. (7 × 200 = 1400). Ans. (a) Common Proficiency Test (CPT) Volume - II 367 © The Institute of Chartered Accountants of India ANSWERS 160. Let x be the distributed amount of A, B and C Given ∴ 5x + 11x + 3x = 950 19x = 950 950 x = = 50 19 ∴ The amount of A = 5x = 5 × 50 = 250 The amount of B = 11x = 11 × 50 = 550 ∴ The difference of A and B = 300 Ans. (a) e −x −e −1 e1−x −1 161. Lt ⇒ Lt x→1 x−1 x→1 e(x−1) let 1 + h → x where h → 0 eh −1 −1 ∴ Lt = x→0 e(−h) e Ans. (b) –1/e (1+x)n −1 162. Lt x→0 x n(n−1)x2 (1+nx+ +......) −1 2! = Lt x→0 x ⎡ n(n−1)x n(n−1)(n−2)x2 ⎤ x⎢n+ + ......⎥ ⎢⎣ 2! 3! ⎥⎦ = Lt x→0 x App Lt n + 0 = n Ans. (c) n (x+2)5/3 −(a+2)5/3 163. Lt x→0 x−a (x+2)5/3 −(a+2)5/3 Lt x→0 [(x+2)−(a+2)] App Lt 368 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India 5 5 ⇒ .(a+2)5/3 – 1 = (a+2)2/3 3 3 5 Ans. (a) (a+2)2/3 3 2x −3x (2x −1)−(3x −1) 165. Lt ⇒ Lt x→0 x x→0 x 2x −1 3x −1 ⎡ ax −1 ⎤ Lt – Lt ⎢ Lt =logea⎥ x→0 x x→0 x ⎢⎣x→0 x ⎥⎦ ⇒ log 2 – log 3 ⎛2⎞ ⇒ log ⎜ ⎟ ⎝3⎠ Ans. (b) ( ) 166. f ′ x = 3x2 + 2 ( ) ∫ f ′ ( x ) dx =∫ 3x2 + 2 dx ( ) 3x3 f x +c = +2x+c 3 When f (0) = 0 ⇒ c = 0 ∴ f(x) = x3 +2x ∴ f(2) = 23+ 2 (2) = 8 + 4 = 12 Ans. (c) x+3 ∫ 167. Let I = x2 +6x+4 Put x2+6x+4 = t ∴(2x+6)dx = dt dt (x+3)dx = 2 x+3 dt ∴∫ dx=∫ x2 +6x+4 2t 1 1 ∫ = dt 2 t Common Proficiency Test (CPT) Volume - II 369 © The Institute of Chartered Accountants of India ANSWERS 1 log(t) 2 x+3 1 ∴∫ dx= log(x2+6x+4) + c x2 +6x+4 2 x−1 x+1−2 168. ∫ ex dx = ∫ exdx (x+1)3 (x+1)3 ⎡ ⎤ 1 2 = ∫ ex⎢ − ⎥dx ⎣(x+1)2 (x+1)3⎦ = ∫ ex { f(x)+f'(x) } dx 1 = ex f(x) where f(x) = (x+1)2 (x−1) ex ∫ ex dx= +c (x+1)3 (x+1)2 Ans.(a) 169. ∫( 3x+5 ) 4 dx = ( ( 3x+5 ) ) ( 4+ ) 1 +c 4+1 3 ( ) 3x+5 5 = + c 15 Ans. (b) 170. ∫ 7x+5 dx = ∫( 7x+5 )1 dx 2 ( 7x+5 )1 +1 = 2 + c ⎛1 ⎞( ) ⎜ +1⎟ 7 ⎝2 ⎠ ( )3 7x+5 = 2 + c ⎛3⎞( ) ⎜ ⎟ 7 ⎝2⎠ 370 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India ( )3 2 7x+5 2 + c 21 171. Voter has option (i) Two candidates from gentlemen = 3 =3 c 2 (ii) Two candidates from ladies = 3 =3 c 2 (iii) One from each ladies & gentlemen = 3 ×3 = 9 c1 C1 Total options = 3 + 3 + 9 = 15 Ans. (c) 15 172. Total hand shakes in the party = 40 = 780 c 2 Ans. (a) 780 173. Total triangle formed by m sides = m C 3 m(m−1)(m−2)(m−3)! ⇒ (m−3)!3! m(m−1)(m−2) ⇒ 6 m(m−1)(m−2) Ans. (a) 6 174. Cricket team of 11 among 14 players out of which one wicket keeper = 12 × 2 = 66 ×2 C C 10 1 ⇒ 132 Ans. (b) 132 175. No. of ways in which a particular child goes to circus = 7 ×1 = 21 C 2 Ans. (c) 21 176. ax = by = cz = k (let) ∴ log ak = x, log k = y, log k = z b c ∴ log a = 1/x, log b = 1/y, log c = 1/z k k k x, y, z in GP ∴ y2 = xz (log k)2 = (log k).(log k) b a c Common Proficiency Test (CPT) Volume - II 371 © The Institute of Chartered Accountants of India ANSWERS log k log k log a log b ∴ b = c ⇒ k = k log k log k log k log c a b k k ∴ log , log and log in GP a b c Ans. (b) G.P. 1 ⎛1⎞n−1 ⎛1⎞13 ⎛1⎞n−1 177. =8.⎜ ⎟ ⇒ ⎜ ⎟ = ⎜ ⎟ 1024 ⎝2⎠ ⎝2⎠ ⎝2⎠ ∴ n = 14 6th team from end = (14 – 6+1) from beginning = 9th term ⎛1⎞9−1 ⎛1⎞8 1 T =8.⎜ ⎟ =8.⎜ ⎟ = 9 ⎝2⎠ ⎝2⎠ 32 Ans. (c) 1/32 178. Product of 2nd term from start & last 2nd term from end = (ar) × a(r)n…2= a2 r(n…1) Product of first & last term = a×arn−1 =a2rn−1 Hence proved the statement. It is true statement Ans. (a) True 179. a, b, c in GP ∴ b2 = ac a+b a, x, b in AP ⇒ x = 2 b+c b, y, c in AP ⇒ y = 2 a c 2a 2c 2[ab +ac +ac +bc] ∴ + = + ⇒ x y a+b b+c ab +b2 +ac +bc 2[ab +ac +ac +bc] ⇒ {b2 =ac} ab +ac +ac +bc = 2 Ans. (c) 2 a+b b+c + 180. 1 + 1 = x+y ⇒ 2 2 x y xy (a+b)(b+c) 4 2(a+2b+c) 2(a+2b+c) ⇒ = (ab +b2 +ac +bc) ab +b2 +b2 +bc 372 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India 2(a+2b+c) 2 = = b(a+2b+c) b 2 Ans. (b) b 181. The number of times a particular item occurs in a given data is called its frequency. Ans. (b) Frequency 182. Lower class (s) = 10.6 Width = 2.5 (Class interval) Upper Class (L) of Lightest Class = S = 10 × C.I. = 10.6 + 10 x 2.5 = 35.6 Ans. (a) 35.6 LowerClass +Upper Class 183. = Mid Value 2 L+U.Class =m 2 ∴ Upper class = (2m – L) Ans. (c) (2m–L ) 1.x+2.2x+3.3x+....n.nx 184. Mean = n(n+1).x/2 (12 +22 +32 +......n2)x = n(n+1).x/2 (12 +23 +32 +......n2)x = n(n+1).x/2 n(n+1)(2n+1)x×2 (2n+1) = = 6n(n+1)−x 3 (2n+1) Mean = 3 2n+1 Ans. (c) 3 Common Proficiency Test (CPT) Volume - II 373 © The Institute of Chartered Accountants of India ANSWERS 185. x = 10, n = 4 Σx = n. x = 4 × 10 = 40 Corrected Σx = (40 + 4a) 40+4a Corrected x = 4 40+4a 13 = 4 4a = 12 a = 3 Ans. (c) 3 186. From the given data, we observe that 20 + 5 = 25 21 + 4 = 25 and 22 + 3 = 25 Thus, x and y are connected by the linear relation: x + y = 25 → (1) ⇒ There is perfect correlation between x and y ⇒ r = ±1 → (2) From (1) / We get y = 25 – x ∴ As x increases, y decreases (by the same amount) ⇒ x and y are negatively correlated → (3) From (2) and (3), we conclude that r = r (x, y) = –1 Ans. (c) 187. Ans. (b) Refer Properties 188. Ans. (b) Refer Properties 189. Ans. (b) Refer Properties 190. Ans. (b) Refer Properties 191. Lt X … B (n=6, p). When X denotes the number of successes. Then, by binomial probability law, the probability of r successes is givenby p(r) = P(x=r) = 6C Prq6 – r → (1) r r = 0, 1, 2, ........... 6 Put r = 3 and 4 in (1) 374 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India (1) ⇒ p(3) = 6C p3q3 = 20p3q3 = 0.2457 (given) 3 p(4) = 6C p4 q2 = 15 p4q2 = 0.0819 (given) 4 p(4) 15p4q2 0.0819 1 = = = p(3) 20p3q3 0.2457 3 3 p 1 ⇒ . = 4 q 3 ∴ 9p = 4q = 4 (1 – p) ∴ 13p = 4 p = 4/13 ∴ q = 1 – p = 1 – 4/13 = 9/13 192. Ans. (a) – Refer Properties 193. Ans. (b) – Refer Properties 194. Ans. (a) – Refer Properties 195. Ans. (b) – Refer Properties 196. Which measure of dispersion has some desirable mathematical properties → Standard Deviation. Ans. (a) Standard Deviation. 197. x 1+y + y 1+x =0 ⇒ x 1+ y =− y 1+ x Eg. Both side x2(1+y)=y2(1+x) (x2 −y2)=y2x−x2y (x+y)(x−y) = −xy(x−y) −x ∴ x + y +xy = 0 ⇒ y = 1+x dy (1+x)(−1) −(−x)(1) −1−x+x ∴ = = dx (1+x)2 (1+x)2 dy −1 dy ∴ = ⇒(1+x2) =−1 dx (1+x)2 dx Ans. (c) (–1) Common Proficiency Test (CPT) Volume - II 375 © The Institute of Chartered Accountants of India ANSWERS ⎛ ⎞ 198. y = x x2 +1 + log ⎜x+ x2 +1⎟ ⎝ ⎠ ⎛ ⎞ dy = x. 1 .2x + x2 +1.1 + 1 ⎜ 1+ 1 .2x ⎟ ⎜ ⎟ dx 2 x2 +1 x+ x2 +1 ⎝ 2 x2 +1 ⎠ x2 x2 +1+x 1 = + x2 +1+ x2 +1 ⎜ ⎛ x+ x2 +1⎟ ⎞ x2 +1 ⎝ ⎠ x2 +x2 +1+1 2(x2 +1) = = =2 1+x2 x2 +1 x2 +1 Ans. (c) 2 1+x2 199. y = aemx +be −mx dy ∴ =amemx −bme −mx dx d2y d ( ) = = amemx −bme −mx dx2 dx = am2emx +bm2e −mx ( ) = m2 aemx +be −mx d2y = m2y dx2 Ans. (c) m2y 200. 12 +2.12 +12 =14 C5 C4 C3 Cx ( ) ( ) 12 +12 + 12 +12 =14 C C C C C 5 4 4 3 x 13 +13 = 14 C C C 5 4 x [ ] nC +nC =n+1C r r−1 r 14 = 14 C C 5 x x = 5 but value 14 = 14 C C 9 5 ∴ 14 = 14 C C x 9 ∴ x = 9 376 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India x = 5 or 9 Ans. (c) 5 or 9. Model Test Paper – BOS/CPT – 7 151. Let the number be x. Then according to the given condition of the problem, x x+1 = +1 3 4 x x+1 ⇒ − =1 3 4 4x−3(x+1) ⇒ =1 12 x−3 ⇒ =1 12 ⇒ x – 3 = 12 ⇒ x = 3 + 12 = 15 Hence the required number is 15 152. Let the fraction = x And Correct answer = y 16 ∴ Given x = y → (1) 17 x 33 and = y + 16 340 17 17 33 i.e. x = y+ → (2) 16 340 17 (1) ⇒ x = y 16 Substitute x the value of x in equation (2) 17 17 33 × y = y+ 16 16 340 289 33 y = y + 256 340 Common Proficiency Test (CPT) Volume - II 377 © The Institute of Chartered Accountants of India ANSWERS 289 33 y − y = 256 340 ⎛289 −256 ⎞ 33 ⎜ ⎟y = ⎝ 256 ⎠ 340 33 33 y = 256 340 33 256 y = × 340 33 64 y = 85 Ans. (a) 153. Let the number be x 5 ⎡ 4 ( )⎤ 2⎡4( )⎤ Given ⎢ x ⎥ = 8+ ⎢ x ⎥ 7 ⎣15 ⎦ 5⎣9 ⎦ 4 8 x = 8+ x 21 45 4 8 x − x = 8 21 45 ⎛180 −168 ⎞ ⎜ ⎟ x = 8 ⎝ 945 ⎠ 12 x = 8 945 945 x = 8 × 12 x = 708.75 Ans. (d) 154. Let two numbers are x and y Given condition x + y = 14 → (1) y –x = 1 0 → (2) Adding (1) and (2) ⇒ 2x = 24 x = 12 (1) ⇒ 12 + y = 14 378 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India y = 2 ∴ Product of two numbers = x × y = 12 × 2 = 24 Ans. (a) 155. Let two numbers are x and y. Given x – y = 11 → (1) x+y And = 9 5 i.e. x + y = 45 → (2) Adding (1) and (2) ⇒ 2x = 56 X = 28 (1) ⇒ 28 – y = 11 – y = 11 – 28 = –17 y = 17 ∴ The two numbers are 28, 17. Ans. (d) 156. Sub duplicate Ratio of 16:49 = 16 : 49 = 4 : 7 Ans. (a) 157. Duplicate Ratio of 4 : 5 = 42 :52 = 16 : 25 Ans. (a) 158. Triplicate Ratio of 3 : 5 = 33 :53 = 27 : 125 Ans. (a) 159. The sub – triplicate Ratio of 8 : 125 = 3 8 :3 125 = 2 : 5 Ans. (b) 6 15 160. 4th Proportion of 6, 8 and 15 is = 8 x 6x = 15 × 8 Common Proficiency Test (CPT) Volume - II 379 © The Institute of Chartered Accountants of India ANSWERS 15×8 x = 6 = 20 Ans. (c) 161. Let the two numbers be x and y. According to the First condition of the problem, x 4 = y 1 ⇒ x = 4y → (1) According to the second condition of the problem, x+5 3 = y+5 1 x + 5 = 3 (y + 5) x + 5 = 3y + 15 x – 3y = 15 – 5 = 10 → (2) Put x = 4y from (1) in (2), we get 4y – 3y = 10 y = 10 (1) ⇒ x = 4 (10) = 40 Hence the required numbers are 40 and 10. Ans. (b) 162. Let A having money = 3x B having money = 4x C having money = 5x Given 3x = 300 x = 100 ∴ C = 5x = 5 × 100 = 500 Ans. (c) 163. Let the two numbers be x and y. According to the first condition of the problem. x 5 = y 6 6x = 5y 6x – 5y = 0 → (1) 380 Common Proficiency Test (CPT) Volume - II © The Institute of Chartered Accountants of India