CA Foundation Dec 2022 Past Paper PDF

CA Foundation Dec 2022 1. If log10 2 = y and log10 3 = x ,then the value of log10 15 is (a) x  y 1 (b) x  1 1 (c) x  y 1 (d) y  x 1 Sol. (a) Here, log10 2 = y and log10 3 = x then log10 15 = log10  3  5 ...

CA Foundation Dec 2022 1. If log10 2 = y and log10 3 = x ,then the value of log10 15 is (a) x  y 1 (b) x  1 1 (c) x  y 1 (d) y  x 1 Sol. (a) Here, log10 2 = y and log10 3 = x then log10 15 = log10  3  5  = log10 3  log10 5 = log10 3  log10 10 / 2  = log10 3  log10 10  log10 2 = x  1 y = x  y 1 2. log3 4. log4 5. log5 6. log6 7. log7 8. log8 9 equal to: (a) 3 (b) 2 (c) 1 (d) 0 Sol. (b) Here, log3 4. log4 5. log5 6. log6 7. log7 8. log8 log4 log5 log6 log7 log8 log9 =      log3 log4 log5 log6 log7 log8 log9 log32 2log3 =   2 log3 log3 log3 3. A sum of money is to be distributed among A, B, C, D in the proportion of 5:2:4:3. If C gets ₹ 1,000 more than D, what is B’s share? (a) ₹ 2,000 (b) ₹ 1,500 (c) ₹ 2,500 (d) ₹ 1,000 Sol. (a) Given, A: B: C: D=5:2:4:3 Let, A=5 x , C=4 x B  2x D  3x Cgets ` 1,000more thanD C  D  1,000 4 x  3 x + 1,000 4 x - 3 x + 1,000 x  1,000 share'sof B  2 x 1 CA Foundation = 2  1,000 = ` 2,000   /  4a b  a b , the answer will be: 6 2 4. By simplifying 2a3b4 3 2 2 (a) 4a2b3 (b) 4a6b4 (c) 4a10b10 (d) 4a10b20 Sol. Here,  2a b   2 a b 3 4 6 18 24  4a b    a b   4 a b  a b  2 2 6 2 2 2 3 2 2 64a18b24 = 16a8b4 = 4a10.b20 5. A group of 400 soldiers posted at border area had a provision for 31 days. After 28 days 280 soldiers from this group were called back. Find the number of days for which the remaining ration will be sufficient? (a) 3 (b) 6 (c) 8 (d) 10 Sol. (d) Here, Total men = 400, No. of days = 31 Total No. of unit of food for 400 men in 31 days = 400 x 31 = 12,400 unit Total No. of unit of food for 400 men in 28 days = 400 x 28 = 11,200 unit Rest food = 12,400-11,200 = 1,200 unit Remain men after 28 days =400-280=120 No. of days for which the remaining food will be sufficient Total Re st food = No. of remaining men 1,200 = 120 =10 days 6. If the roots of the equation x2  px  q  0 are in the ration 2:3, then: (a) p2  25q (b) p2  6q (c) 6p3  5q (d) 6p2  25q Sol. (d) If    are the roots of Q.E x2  px  q  0 Given,  :  =2:3 2 CA Foundation Let  = 2k,  = 3k and a=1, b= -p, c= q We know that: b   a   p  2k  3k  1 5k  p p K __________ i  5 c and .   a q 2k. 3k  1 6k2  q 2 p 6    q Putting k  p / 5 from eq. 1 5 p2 6. q 25 6p2  25q 7. What will be the value of k , if the roots of the equation  k  4  x 2  2kx   k  5   0 are equal ? (a) 18 (b) 20 (c) 19 (d) 21 Sol. (b) Given Q.E k  4  x 2  2kx  k  5   0 Computing from ax2  bx  c  0 We get a= (k-4), b=-2k, c= (k+5) If roots of Q.E are equal Then   0 b2  4ac  0  2k   4  k  4 k  5   0 2 3 CA Foundation  4k 2  4 k 2  5k  4k  20  0  4k 2  4k 2  4k  80  0 4k  80 80 k  20 4 8. If the cost of 3 bags and 4 pens is₹ 257 whereas the cost of 4 bags and 3 pens is ₹ 324, then the cost of one bag is: (a) 8 (b) 24 (c) 32 (d) 75 Sol. (d) Let the cost of 1 bag = ₹ x The cost of 1 pen =₹ y According to 1st condition and 2nd condition are 3 x + 4y = 257 1 4 x + 3y = 324 2 Multiply by 3 in equation (1) 9 x +12y = 771_______________ 3  Multiply by (4) in equation (2) 16 x +12y =1296 (4) eq. (4) –e.q (3) 16 x + 12y = 1296 9 x +12y = 771 - - - 7 x = 525 525 x= = 75 7 The cost of one bag =75 9. The solution of the following system of linear equations 2x  5y  4  0 and 2x  y  8  0 will be: (a)  2, 3  (b) 1, 3  (c)  3,2  (d)  2,2  4 CA Foundation Sol. (c) Given Equation 2 x  5y  4  0  2 x  5y  4 ________ 1 and 2 x  y  8  0  2 x  y  8 ________ 2  e.q. (2) –eq.(1) 2x  y  8 2x  5y  4 - + + 6y  12 12 y 2 6 Putting y  2in eq  2  2x  2  8 2x  8  2 2x  6 6 x 3 2 x  3,y  2 10. If 2 x +5 > 3x + 2 and 2 x – 3< =4 x – 5, then ‘ x ’can take which of the following value? (a) 4 (b) -4 (c) 2 (d) -2 Sol. (c) If 2x  5  3 x  2 and if 2x  3  4 x  5 5  2  3 x  2x  3  5  4 x  2x 3  x2  x 2 x3 x 2 x  ,.....1,21  x x 1 x  1,2,3,...... Here,’2’ is satisfied the both equation so solution is 2 5 CA Foundation 11. A machine worth ₹ 4,90,740 is depreciated at 15% on its opening value each year. When its value would reduce to ₹ 2,00,750? (a) 5 years 5 months (b) 5 years 6 months (c) 5 years 7 months (d) 5 years 8 months Sol. (b) Here, original value (P) =₹ 4,90,740 Rate of depreciation (R) = 15% Scrap value (A) = ₹2,00,750 Time (T) =? Scrap value after 'T' years T  R  A=P  1    100  T  15  2,00,750 =4,90,740  1    100  2,00,750   0.85  T 4,90,740   0.85  T 0.4090  0.85    0.85  5.5 T We get T  5.5 years T  5 years & 6months 12. If ₹ 64 Amount to ₹ 83.20 in 2 years, what will ₹ 86 Amount to in 4 years at the same Rate percent per annum? (a) ₹ 127.60 (b) ₹147.60 (c) ₹ 145.34 (d) ₹117.60 Sol. (c) CASE-1 Here, P = 64 A=83.20, R = ? T=2 2  R  A  P 1   100  2  R  83.20  64  1    100  6 CA Foundation 2  R  83.20  1  100   64   2  R   R   1  100   1.3   1  100   1.3 _______ i      and CASE  2 p  86, T  4yrs 2  R  1  100   1.3   We know that T  R  A  P 1   100    4 A  86 1.3 A  86  1.3  1/24 A  86  1.3  2 A  86  1.69 A  145.34 13. A farmer borrowed ₹ 3,600 at the rate of 15% simple interest per Annum. At the end of 4 years, he cleared this account by paying ₹ 4,000 and a cow. The cost of the cow is: (a) ₹ 1,000 (b) ₹ 1,200 (c) ₹ 1,550 (d) ₹ 1,760 Sol. (d) Here, Principal (P) ₹ 3,600 R=15%, T=4 Years PRT 3,600  15  4 S.I =   2,160 100 100 Amount (A) =P+S.I. =3,600+2,160 =₹ 5,760 but he paid ₹ 4,000 and a cow to clear his debt ,then 4,000+the cost of cow =5,760 The cost of cow=5,760-4,000 =₹ 1,760 7 CA Foundation 14. How much amount is required to be invested every year so as accumulate ₹ 5,00,000 at the end of 12 years if interest compounded annually at 10% (Where A (12,0.1) = 21.384284) (a) ₹ 23,381.65 (b) ₹ 24,385.85 (c) ₹ 26,381.65 (d) ₹ 28,362.75 Sol. (a) Given, Future value A = ₹ 5,00,000 n.i   Rate (R) = 10% n = 12 years R 10 i =   0.1 100 100 Annuity (A) = ? Future value A 1  i  1 n A    n.i i  A  1  0.1  1 12 5,00,000  0.1  A  1.1  1 12 5,00,000  0.1   A 5,00,000   2.1384284 0.1 5,00,000=A × 21.384284 5,00,000 A 21.384284 A  23,381.65 15. The effective annual rate of interest corresponding to a normal rate of 6% per annum payable half yearly is: (a) 6.06% (b) 6.07% (c) 6.08% (d) 6.09% 6 Sol. (d) Given ,R = % = 3% 2 R=3%, T=1 × 2 half yearly T=2 Effective rate of interest 8 CA Foundation  R  T  E   1    1  100%  100    3  2  =  1    1  100%  100   = 1.0609  1  100% = 0.0609  100% = 6.09% 16. 10 years ago the earning per share (EPS) of ABC Ltd. was ₹ 5 share. Its EPS for this year is ₹ 22. Compute at what rate, EPS of the company grow annually? (a) 15.97% (b) 16.77% (c) 18.64% (d) 14.79% Sol. (a) 10 years ago EPS of ABC Ltd.  v 0   ` 5 Now EPS OF ABC Ltd.  v1  ` 22 t 2  t1  10 Rate of EPS of the company grow annually  1     t 2  t1  1  100% V (R)=   1  V0      1   22  =    100% 10   5  1    =  4.4   1  100% 0.1   = 1.1597  1  100% = 0.1597  100% = 15.97% 17. Raju invests ₹ 20,000 every year in a deposit scheme starting from today for next 12 years. Assuming that interest rate on this deposit is 7% per annum compounded annually. What will be the future value of this annuity? Given that (1+0.07)12 = 2.25219159. (a) ₹ 5, 40,526 (b) ₹ 3, 82,813 (c) ₹ 6, 43,483 (d) ₹ 3, 57,769 Sol. (b) Here, annual instalment (A) = 20,000 n = 12 R = 7% p.a. 9 CA Foundation 7 i  0.07 100 Future value A ni =? We know that: A Future Value A ni = 1  i   1 1  i  n i   20,000  1  0.07   1 1  0.07  12 = 0.07  20,000  1.07   1 1.07  12 = 0.07   20,000 = 2.25219159  1 1.07  0.07 20,000 =  1.25219159  1.07 0.07 = 3,82,813  approx  18. Mr. A invested ₹ 10,000 every year for next 3 years at the interest rate of 8 percent per annum compounded annually. What is future value of the annuity? (a) ₹ 32,644 (b) ₹ 32,464 (c) ₹ 34,264 (d) ₹ 36,442 Sol. (b) Annual Instalment (A) = ₹ 10,000 T = 3 years R = 8% p.a. 8 i  0.08 100 Future Value: A 1  i  1 n A (n, i) = i   10,000  1  0.08   1 3 = 0.08   10,000  1.08   1 3 = 0.08  =₹ 32,464 19. Mr. Prakash invested money in two schemes 'A' and 'B' offering compound interest at the rate of 8% and 9% per annum respectively. If the total amount of interest accrued through 10 CA Foundation these two schemes together in two years was ₹ 4,818.30 and total amount invested was ₹ 27,000. What was the amount invested in schemes 'A'? (a) ₹ 12,000 (b) ₹ 12,500 (c) ₹ 13,000 (d) ₹ 13,500 Sol. (a) Case-1 Principle (P₁) = ₹ x (R) = 8% p.a. T = 2 years  R  T   C.I1  P1 1    1  100    8  2  = P1  1    1  100   = x 1.08   1 2   = 0.1664 x Case  2 Priciple P2  ` y R  9% T  2years  9  2   C.I2  P2  1    1  100    P2 1.09   1 2    C.I2  y [0.1881]  C.I2  0.1881y Given total compound interest =4,818.30  C.I1   C.I2 =4,818.30 0.1664 x  0.1881y  4,818........... 1 Total investment of both part =27,000 x  y  27,000 y  27,000  x....................  2  Putting value of y in eq. (1) 11 CA Foundation 0.1664 x  0.1881 27,000  x    4,818.30 0.1664 x  5,078.70  0.1881x  4,818.30 5,078.70  4,818.30  0.1881x  0.1664 x 260.40  0.0217 x 260.40 x  0.0217 x  12,000 The amount invested in scheme ‘A’ =₹ 12,000 20. A sum of money invested of compound interest doubles itself in four years. In how many years it become 32 times of itself at the same rate of compound interest? (a) 12 Years (b) 16 Years (c) 20 Years (d) 24 Years Sol. (c) Case-1 Case -2 Let P= ₹ 100 Let p =₹ 100 A= ₹ 200 A= ₹ 3,200 T=4 years T=?  R   1  100    2  1 R=? 4   T T  R   R  A  P 1  Now A =P  1    100   100  4 T  R   R  200  100  1   3,200  100  1    100   100    4 200  R  3,200 1 T  1  24 100  100  100 4  R  T 2  1  32  2 4  100   R   1  100    2  ________ 1 1 T 4 25  2 4   On computing we get T 5  T  20yrs 4 12 CA Foundation 21. The difference between compound interest and simple interest on an amount of ₹ 15,000 for 2 years is ₹ 96. What is the rate of interest per Annum? (a) 9% (b) 8% (c) 11% (d) 10% Sol. (b) Given C.I - S.I = ₹ 96 P = ₹ 15,000 T = 2 years We know that 2  R  C.I – S.I =p   [for two years]  100  2  R  96  15,000    100  R2 96  15, 000  10, 000 96  10  R2 15 64  R2 R  64  8% OR Difference=C.I - S.I.  96  P 1  i  1  2i 2   96  P 1  2i  i2  1  2i  96  Pi2 96  i2  P 96  i2   0.0064 15000  i  0.08  R  8% 13 CA Foundation 22. ₹ 5,000 is invested every month end in an account paying interest @12% per annum compounded monthly. What is the future value of this annuity just after making 11th payment? (Given that (1.01)¹¹ = 1.1156) (a) ₹ 57,800 (b) ₹ 56,100 (c) ₹ 56,800 (d) ₹ 57,100 Sol. (a) Here, annual investment (A) = ₹ 5,000 12 R   %  1% 12 R 1 i    0.01% 100 100 n  11 Future value A  A n,i    1  i   1 n i   5,000  1  0.01  1 11  0.01  5,000   1.01  1 11  0.01   5,000   1.1156  1 0.01 5,000   0.1156 0.01  ` 57,800 23. A sum of money doubles itself in 4 years at certain compound interest rate. In how many years this sum will become 8 times at the same compound interest rate? (a) 12 Years (b) 14 Years (c) 16 Years (d) 18 Years Sol. (a) (a) Case – 1 Case – 2 Let P=100 Let P = 100 A = 200 A = 800  R   1  100    2  1 4 T= 4 years   We know that 14 CA Foundation T  R  A  P 1  T?  100  4 T  R   R  200  100  1   A  P 1   100   100    4 200  R  1 T   1 800  100 2 4 100  100  4  R  800 T  2  1  2 4  100  100  R    2 4 1 T  1  82 4  100  T 23  2 4 3T 4  T  12years 24. Sinking fund factor is the reciprocal of: (a) Present value interest factor of a single cash flow (b) Present value interest factor of an annuity (c) Future value interest factor of an annuity (d) Future value interest factor of a single cash flow. Sol. (b) Sinking fund factor is the reciprocal of present value inter factor of an annuity. 25. There are 20 points in a plane area. How many triangles can be formed by these points if 5 points are collinear? (a) 550 (b) 560 (c) 1,130 (d) 1,140 Sol. (c) Here, Total No. of points in a plane (n) = 20 No. of collinear points (k) = 5 No. of triangle are formed from 'n' points In which ‘k’ points are collinear = nc 3  k c 3 = 20c 3  5c 3 = 1,140  10 = 1,130 15 CA Foundation 26. The number of ways 4 boys and 3 girls can be seated in a row so that they are alternate is: (a) 12 (b) 288 (c) 144 (d) 256 Sol. (c) Total Boys = 4, Total girls = 3 BGBGBGB No. of ways = 4! x 3! = 24 x 6 = 144 27. If n Pr =3,024 and n Cr =126, then find n and r? (a) 9, 4 (b) 10, 3 (c) 12, 4 (d) 11, 4 Sol. (a) If n Pr =3,024 and n Cr =126 We know that n Cr 1 n  Pr r 126 1  3,024 r 3,024 r 126 r  24 r  4r 4 Here nPr  3,024 n  3,024 nr hits and trails  n  9 and r  4 9 9 9876 5    3,024 94 5 5 It is satisfied so n =9, r=4 28. How many 3 digit odd numbers can be formed using the digits 5,6,7,8,9, if the digits can be repeated? (a) 55 (b) 75 (c) 65 (d) 85 Sol. 16 CA Foundation 29. If P th term of an AP is q and its qth term is p, then what will be the value of (p+q)th the term? (a) 0 (b) 1 (c) P+q-1 (d) 2 (p+q-1) Sol. (a) Given Tp  q and Tq  p a   p  1 d  q a   q  1 d  p a  pd  d  q _____ i  a  qd  d  p ____  ii  eq. (1) –eq (i) a  pd d  q a  qd d  p    pd  qd  q  p d (p-q) =-(p-q)   p  q d  p  q d  1 Putting d=-1 in eq. (i) a + p + (-1) - (-1) = q a–p+1=q a  p  q 1 Now Tp q = a + (p + q - 1) d = p +q – 1 + (p + q - 1) (-1) =p+q–1–p–q+1 =0 30. In a GP 5th term is 27 and 8th term is 729 find its 11th term 17 CA Foundation (a) 729 (b) 6,561 (c) 2,187 (d) 19,683 Sol. (d) In G.P Given T5  27 and T8  729 ar 4  27 _____ 1 ar 7  729 _____  2  ar 7 729  ar 4 27 r 3  27  r 3  33  r  3 Putting r=3 in equation (1) , we get a.34  27 a  81  27 27 a 81 1 a 3 Now, T11  ar111  Tn  ar n1  = ar10 1  3 10 = 3 1 =  59049 3 = 19683 31. If A = 1,2,3,4,5,6,7,8,9 and B= 2,4,6,7,9 then how many proper subset of A  B can be created? (a) 16 (b) 15 (c) 32 (d) 31 Sol. (b) A= 1,2,3,4,5,6,7,8,9 B= 2,4,6,7,9 A  B =4 No. of proper subset of  A  B  = 2n  1 = 24  1 = 16  1 18 CA Foundation = 15 32. Let A = 1,2,3 and consider the relation R = 1,1, 2,2,3,3,1,2, 2,3 , 1,3  Then R is: (a) Symmetric and transitive (b) Reflexive but not transitive (c) Reflexive but not symmetric (d) Neither symmetric, nor transitive Sol. (c) Let A = {1, 2, 3 R= 1,1, 2,2,3,3,1,2, 2,3 , 1,3  Then R is reflexive but not symmetric. 33. The number of subsets of the set (0, 1, 2, 3) is: (a) 2 (b) 4 (c) 8 (d) 16 Sol. (d) Here, A= 0,1,2,3 n(A)=4 No.of subset= 2n = 24 =16 34. Find the area under the curve f  x   x 2  5 x  2 with the limits 0 to 1? (a) 3.833 (b) 4.388 (c) 4.833 (d) 3.338 Sol. (c) Here f  x   x 2  5 x  2 Area (A) =  f  x  dx b a  x  1 2  5 x  2 dx 0 1  x3 5 x 2  =   2x 3 2 0   13 5 12    0 3 5  0 2  =    2 1      2  0    3 2   3 2   1 5 2   0 0  =         0    3 2 1   3 2   2  15  12   =    0  6   19 CA Foundation  29  =    4.833  6  35. The maxima and minima of the function y  2x3  15 x2  36 x  10 Occurs respectively at: (a) x  2and x  3 (b) x  1 and x  3 (c) x  3and x  2 (d) x  3and x  1 Sol. (a) Given function y  2 x 3  15 x 2  36 x  10 ______ 1 Diff.w.r.t.’x’ dy  6 x 2  30 x  36 _______  2  dx Again diff.w.r.t.’ x ’ d2 y  12 x  30 ___________  3  dx 2 For maxima and minima dy 0 dx 6 x2  30 x  36  0  6 x2  5 x  6  0  x2  5 x  6  0 x2  3 x  2 x  6  0 x  x  3  2  x  3  0  x  3  x  2   0 If x  3  0 or if x  2  0 x3 x2 Putting x  2 in eq. (3) d2 y  12  2  30  24  30  6 dx 2 d2 y  ve So function is dx 2 Maximum at x  2 20 CA Foundation Putting x =3 in eq. (3) d2 y  12  3  30  36  30  6 dx 2 d2 y   ve So function is minimum at x =3 dx 2 36. If = x x then dy/dx at x =1 is equal to (a) 0 (b) 1 (c) -1 (d) 2 Sol. (b) Given by = x x Taking log on both side Log y=log x x Log y= x log x Diff w.r.t, ‘n’ d d log y   x log x  dx dx 1 dy d d  x. log x   log x  x  y dx dx dx 1 dy 1  x.  log x.1 y dx x 1 dy  1  log x  y dx dy  y 1  log x  dx dy  x x 1  log x  dx  dy   dx   1 1  log1 1   x 1 = 11  0  = 1 1  1   2x  3 5 37. dx is : 2x  3  2x  3 2x  3  2x  3 6 6 6 6 (a) (b) (c) (d) 6 2 12 3 21 CA Foundation Sol. (c)  2x  3 6 1  2x  3 5 dx   6 2  2x  3 6  12 dy 38. If x5  y5  5 xy  0 then is : dx y  x4 y  x4 x  y4 x  y4 (a) (b) (c) (d) x  y4 y4  x x1  y x4  y Sol. (b) Given If x5  y5  5 xy  0 x 5  y 5  5 xy Diff w.r.t ‘ x ’ d 5 d 5 d x  xy   5 xy  dx dx dx dy  dy  5 x 4  5y 4  5 x  y.1 dx  dx  dy dy 5 x 4  5y 4  5x  5y dx dx dy dy 5y 4  5 x  5y  5 x 4 dx dx 5  dy 4 dx y  x   5  y  x4  dy 5 y  x  4   dx 5 y 4  x   dy  y  x 4    dx  y 4  x  4 xdx 39. 2 x2  1 is : 1  17   17  (a) log   (b) 2log   2  5   5  22 CA Foundation 1  5   5  (c) log   (d) 2log   2  17   17  4 xdx 1 4 2x Sol. (a) 2  x 1 2 2  x 2 2 1 dx  1   4  log x 2  1  2 2 log  42  1  log  22  1 1   2 1  log17  log5 2 1  17   log   2  5  40. If 'FROZEN' is decoded as 'OFAPSG'. Tick the right option that depicts 'MOLTEN' written in this way? (a) OFPOMN (b) OFSMPN (c) OFUMPN (d) OFUNPN Sol. (c) 41. Find the odd man out: 34,105,424,2123,12756. (a) 12756 (b) 2123 (c) 424 (d) 34 Sol. (b) Given series 34,105,424,2123,12756 Here: 34  3  3  105 105  4  4  424 424  5  5  2125 2125  6  6  12756 Here, 2123 is odd man out 23 CA Foundation 42. Find the missing number in the following series? 3, 5, 5, 19, 7, 41, 9,?, 11, 109 (a) 71 (b) 61 (c) 69 (d) 79 Sol. (a) Here, 5 + 14 = 19 19 + 22 = 41 41 + 30 = 71 71 + 38 = 109 43. In certain code language, if TOUR, is written as 1234, CLEAR is write 56784 and SPARE is written as 90847, find the code for CARE? (a) 1247 (b) 4847 (c) 5247 (d) 5847 Sol. (d) Given T O U R C L E A R S P A R E 1 2 3 4 5 6 7 8 4 9 0 8 4 7 Then C A R E 5 8 4 7 44. Find the next number in the given sequence? 11, 17, 39, 85,?, 281, 447 (a) 133 (b) 143 (c) 153 (d) 163 Sol. (d) Series 45. If ROSE 'is coded as 6821, CHAIR is coded as 73456 and PREACH is Coded as 961473, what will be the code for SEARCH? (a) 246173 (b) 214673 (c) 216473 (d) 214763 Sol. (b) 24 CA Foundation Here, R O S E C H A I R P R E A C H 6 8 2 1 7 3 4 5 6 9 6 1 4 7 3 Then, S E A R C H 2 1 4 6 7 3 46. Radha moves towards South-East a distance of 7 km, then she moves towards West and travels a distance of 14 km. From here she moves towards North-West a distance of 7 km. and finally she moves a distance of 4 km. towards East. How far is she now from the starting point? (a) 3 km. (b) 4 km. (c) 10 km. (d) 11 km. Sol. (c) 10 km. (c) Here, BC = DA BC = OE + EA 14km. = 4 km. + EA EA = 14 km. - 4 km. EA = 10 km. 47. One morning a boy starts walking in a particular direction for 5 Km. and then takes a left turn and walks another 5 Km. thereafter he again takes left turn and walks another 5 Km. and at last he takes right turn and walks 5 Km. Now he sees his shadow in front of him. What direction he did start initially? (a) South (b) North (c) West (d) east Sol. (b) North 25 CA Foundation North direction he did start initially. 48. It is 3'o clock in a watch. If the minute hand points towards t North-East then the hour hand will point towards the: (a) South (b) South-West (c) North-West (d) South-East Sol. (d) South-East If the minute hand points towards north-east then hour hand will point towards the south east 49. A man is facing west. He turns 45 degree in the clockwise direction and then another 180 degree in the same direction and then 270 degree in the anticlockwise direction. Find which direction he is facing now? (a) South-East (b) West (c) South (d) South-West Sol. (a) south – west directions 50. P,Q.R and S are playing a game of carom. P, R and S, Q are partners, 'S' is to the right of 'R'. If 'R' is facing west, then 'Q' is facing which direction? (a) South (b) North (c) East (d) West 26 CA Foundation Sol.(b) S N P R W E Q S Q is facing North direction. 51. Six persons A, B, C, D, E and F are sitting in two rows with three persons in each row. Both rows are in front of each other. E is not at the end of the any row and D is second left to the F, C is neighbour of E and diagonally opposite to D. If B is neighbour of F, who is in front of C then who is sitting diagonally to F? (a) C (b) E (c) A (d) D Sol. (c) A E C D B F ‘A’ is sitting diagonally ‘F’ 52. P, Q, R, S and T are sitting in a line facing west. P and Q are sitting together. R is sitting at South end and S is sitting at North end. Tis neighbour of Q and R. Who is sitting the middle? (a) P (b) Q (c) R (d) S Sol. (b) In this arrangement Q is sitting in the middle 53. Suresh's sister is the wife of Ram. Ram is Rani's brother. Ram's father is Madhur. Sheetal is Ram's grandmother. Rema is Sheetal's daughter -in-law. Rohit is Rani's brother's son. Who is Rohit to Suresh? 27 CA Foundation (a) Brother-in-law (b) Son (c) Brother (d) Nephew Sol. (d) Here sibiling, Couple Rohit is nephew of suresh 54. There are six children playing football namely A, B, C, D, E and F, A & E are brothers. F is sister of E, C is the only son of A's uncle. B & D are daughters of the brother of C's father. How D is related to A? (a) Uncle (b) Cousin (c) Niece (d) Sister Sol. (d) 55. In a joint family, there are father, mother, 3 married sons and one unmarried daughter. Out of the sons, two have 2 daughters each and one has a son only. How many female members are there in the family? (a) 3 (b) 6 (c) 9 (d) 5 Sol. (c) Therefore, total female members in the family=9 56. When Rani saw Vinit, she recollected that "He is the brother of my grandfather's son". How is Rani related to Vinit? (a) Aunt (b) Daughter (c) Sister (d) Niece Sol. (d) 28 CA Foundation Rani is niece of vinit 57. Annanya is mother of Satya and Shyam is the son of Bhima. Shiva is brother of Annanya. If Satya is sister of Shyam. How Bhima is related to Shiva? (a Son (b) Cousin (c) Brother-in-law (d) Son-in-law Sol. (c) So bhima is brother in law to shiva 58. Suman is daughter-in-law of Rakesh and sister-in-law of Rajesh. Ramesh is the son of Rakesh and only brother of Rajesh. Find the relation of Suman with Ramesh? (a) Sister-in-law (b) Cousin (c) Aunt (d) Wife Sol. (d) So suman is the wife of Ramesh. 59. Pointing to a man in the photograph. Khushi says, "This man's son's sister is my mother- in-law." How is the Khushi's husband related to the man in the photograph? (a) Grandson (b) Son (c) Son in law (d) Cousin Sol. (a) "Khushi husband is the grandson of the man." 29 CA Foundation 60. Which one of the following is a source of primary data? (a) Government Records (b) Research Articles (c) Journals (d) Questionnaire filled by Enumerators Sol. (d) Questionnaire filled by Enumerators is a source of primary data. 61. Which is the left part of the table providing the description of the rows? (a) Caption (b) Box head (c) Stub (d) Body Sol. (c) Stub is the left part of the table providing the description of the rows. 62. The suitable formula for computing the number of class intervals is: (a) 3.322 logN (b) 0.322 logN (c) 1+3.322 logN (d) 1-3.322 logN Sol. (c) Number of class interval (k) = 1 + 3.322 log N When N = Total frequency. 63. Ogive for more than type and less than type distributions interest at: (a) Mean (b) Median (c) Mode (d) Origin Sol (c) The point of intersection of less than ogive and greater than ogive curve gives us median 64.  If mean x is = 10 and mode (Z) is =7, then find out the value of median (M)? (a) 9 (b) 17 (c) 3 (d) 4.33  Sol. (a) Mean x =10, mode (z) =7, Median (Me) =? We know that: Mode =3 median -2mean 7=3 × Me – 2 × 10 7=3× Me -20 7+ 20 = 3Me 27 Me = =9 3 30 CA Foundation Me = 9 65. If the coefficient of variation and standard deviation are 30 and 12 respectively, then the arithmetic mean of the distribution is: (a) 40 (b) 36 (c) 25 (d) 19 Sol. (a) C.V. = 30%, S.D. = 12 find x =? S.D C.V =  100 Mean 12 30 =  100 Mean Mean = 12  100  40 30 66. _______________________________is based on all the observations and _______________________is based on the central fifty percent of the observations. (a) Mean deviation, Range (b) Mean deviation, quartile deviation (c) Range, standard deviation (d) Quartile deviation, standard deviation Sol. (b) Mean Deviation is based on all observations and Quartile Deviation is based on the central fifty percent of the observations. 67. The relationship between two variables x and y is given by 4 x -10y = 20. If the median value of the variable x is 10 then what is median value of variable y? (a) 1.0 (b) 2.0 (c) 3.0 (d) 4.0 Sol. (b) Given Equation: 4 × -10y = 20 Median of x = 10 4 × 10-10y=- 20 40-10y = 20 -10y = 20-40 -10y = -20 y=2 Median of y = 2 68. Which one of the following is not a measures of dispersion? (a) Standard deviation (b) Mean deviation (c) Range (d) Concurrent deviation method Sol. (d) Concurrent deviation is not a method of Measures of dispersion. 69. Mean deviation is minimum when deviations are taken from: 31 CA Foundation (a) Mean (b) Median (c) Mode (d) Range Sol. (b) Mean deviation (M.D) is minimum when deviations are taken from median. 70. If the first quartile in 56.50 and the third quartile is 77.50, then the co-efficient of quartile deviation is: (a) 638.09 (b) 15.67 (c) 63.80 (d) 156.71 Sol. (b) Here, First quartile Q1 = 56.50 Third quartile Q3 = 77.50 Q3  Q1 Coefficient of Q.D. =  100 Q3  Q1  77.50  56.50  =   100  77.50  56.50  21 =  100 134 = 15.67 71. The median of the observations 42,75,35,92,67,85,72,81,51,56 Is: (a) 69.5 (b) 72 (c) 64 (d) 61.5 Sol. (a) Write all observations in ascending order. 35,42,51,56,67,72,72,81,85,92 Here, No of observation (N) =10 Median (Me) =Average of two middle term =  67  72   2   = 139 2 = 69.5 72. If the sum of square of the values equals to 3390, number of observation are 30 and standard deviations. Is 7, what is the mean value of the above observations? (a) 14 (b) 11 (c) 8 (d) 5 Sol. (c) Here: x 2  3390 andS.D  7 N  30 x ? 32 CA Foundation We know that S.D =  x   x 2 2 N 3390  2 7=  x 30 On squaring both side 7 3390  2 2   x 30  2 49 = 113  x  x  =113-49 2  x 2 = 64 x  64  8 Mean  x  8 73. The mean of 50 observations is 36. If two observations 30 and 42 are to be excluded, then the mean of the remaining observations will be (a) 36 (b) 38 (c) 48 (d) 50 Sol. (a) The mean of 50 observations = 36 The sum of all observations = 50  36 = 1800 If two observations 30 and 42 are excluded then the sum of Remaining (50-2 = 48) observations = 1800-30-42 = 1728 1728 The mean of 48 observations = 48 = 36 74. If Arithmetic Mean and Geometric Mean between two numbers are 5 and 4 respectively, then these numbers are: (a) 2&3 (b) 2&8 (c) 4&6 (d) 1 & 16 33 CA Foundation Sol. (b) Here, A.M. 5 and G.M. = 4 HINTS/TRIALS (B) Two observations are 2 and 8 ab 28 A.M. =  5 2 2 G.M = ab  2  8  16  4 So, these nos.are 2 and 8 75. If the variance of random variable 'x' is 17, then what is variance of y = 2x + 5 ? (a) 34 (b) 39 (c) 68 (d) 78 Sol. (c) Given: v  x  = 17 S.D of x = 17 Given Equation y = 2 x + 5 2x  y  5  0 Coeff of x 2 b  2 Coeff of y 1 S.D of y = b S.D of x = 2  17 = 2 17   2 V(y) = 2 17 = 4  17 V(y) = 68 76. If the variance of given data is 12 and their mean value is 40 what is coefficient of variation (C.V)? (a) 5.66% (b) 6.66% (c) 7.50% (d) 8.65% Sol. (d) Variance =12 S.D  12  2 3  Mean x  40 S.D C.V   100 Mean = 2 3  100 = 8.65% 40 34 CA Foundation 77. In a given set if all data are of same value then variance would be: (a) 0 (b) 1 (c) -1 (d) 0.5 Sol. (a) If all data are of same value then S.D is zero and also variance = 0. 78. If Arithmetic mean between two numbers is 5 and Geometric mean is 4 then what is the value of Harmonic mean? (a) 3.2 (b) 3.4 (c) 3.5 (d) 3.6 Sol. (a) Given: A.M = 5, G.M = 4, H. M =?  G.M  4 2 2 16 H. M. =    3.2  A.M 5 5 79. The average age of 15 students in a class is 9 years. Out of them, the average age of 5 students is 13 years and that 8 students is 5 years. What is the average of remaining 2 students? (a) 5 years (b) 9 years (c) 10 years (d) 15 years  Sol. (d) Total student = 15 x  9years  n1  5 n2  8 n3  15  5  8   2 x 1  13 years x 2  5years x3  x Combined mean x =  n1 x 1  n2 x 2  n3 x 3 n1  n2  n3 5  13  8  5  2  x 9 582 9 65  40  2 x  1 15 135=105+2 x 2 x  30 x =15 years The average of remaining 2 students =15 years 80. A machine is made of two parts A and B. The manufacturing process of each part is such that probability of defective in part A is 0.08 and that B is 0.05. What is the probability that the assembled part will not have any defect? (a) 0.934 (b) 0.864 (c) 0.85 (d) 0.874 Sol. (d) P (defective part of A) = 0.08 35 CA Foundation   P A = 0.08 P (defective part of B) = 0.05  P B = 0.05   P  A   1  P A  1  0.08  0.92  P B   1  P B  1  0.05  0.95 P (the Assembled part will not have any defect) = P (A∩B) = P (A). P(B) =0.92 × 0.95 =0.874 1 3 11 B 81. If P  A   ,P B   andP  A  B   thenp   is : 3 4 12 A 1 4 1 1 (a) (b) (c) (d) 6 9 2 8 Sol. (c) We know that: P  A  B   P  A   P B   P  A  B  11 1 3    P  A  B 12 3 4 1 3 11 P  A  B    3 4 12 49 =  11 12 2 1 = 12 6 1 P  A  B  3 P  A  B P B / A   PA 36 CA Foundation 1 = 6 1 3 1 1 =  6 3 1 = 2 82. The probability that a leap year has 53 Monday is: 1 2 2 3 (a) (b) (c) (d) 7 3 7 5 Sol. (c) There are 366 days in a leap year. 7 366 52 35 16 14 odd days 2 2 odd days may be (a) Sunday & Monday (b) Monday & Tuesday (c) Tuesday & Wednesday n (S)=7 (d) Wednesday & Thursday n (A)=2 (e) Thursday & Friday n (A)=2/7 (f) Friday & Saturday (g) Saturday & Sunday 83. Suppose A and B are two independent events with probabilities P(A)  0 and P(B)  0. Let A' and B' be their complements. Which one of the following statements is FALSE? (a) P (A  B) = P (A) P (B) (b) P (A/B) = P (A) (c) P (A  B) = P (A) + P (B) (d) P (A'  B') = P (A') P (B') Sol. (c) If A and B are two independent events. Where P (A)  0 and P (B)  0 and Let 'A' and 'B' be Their complements. Then, P (A  B) = P (A) + P (B) is false and rest of all is true.. 37 CA Foundation 84. The Theorem of Compound Probability states that for any two events A and B. (a) P(A  B)=P(A)  P(B/A) (b) P(A  B) = P(A)  P(B/A) (c) P(A  B) = P(A)  P(B) (d) P(A  B) = P(A) +P(B) - P(A  B) Sol. (a) the theorem of compound probability states that for only two events A an B Given by P  A  B   P  A   P B / A  85. If a number is selected at random from the first 50 natural numbers, what will be the probability that the selected number is a multiple of 3 and 4? (a) 5/50 (b) 2/25 (c) 3/30 (d) 4/25 Sol. (b) There are first 50 natural numbers if one number is selected. Then No. of sample space n(s) = 50 Event (A) = getting no. is multiple of 3 and 4 (i.e. = 12) 50 nA  12 nA  4 nA 4 2 PA    n S 50 25 86. If three coins are tossed simultaneously, what is the probability of getting two heads together? (a) 1/4 (b) 1/8 (c) 5/8 (d) 3/8 Sol. (d) If three coins are tossed simultaneously Then sample space (s) = {HHH, HHT, HTH, HTT, TTT, TTH, THT, TH H N (s) = 8 Event (A) = 'getting exactly two head' = {HHT, HTH, THH} n (A) = 3 nA 3 Then P (A) =  n S 8 87. Skewness of Normal Distribution is: (a) Negative (b) Positive (c) Zero (d) Undefined 38 CA Foundation Sol. (c) Skewness of Normal Distribution is zero because Normal Curve is Symmetrical. 88. If a Poisson distribution is such that P( x = 2) = P( x = 3) then the variance of the distribution is: (a) 3 (b) 3 (c) 6 (d) 9 Sol. (b) In Poisson distribution P  x  2  P  x  3  em ,m2 em ,m3  2! 3! m2 m3   2 6  2m  6 m3 So variance =m=3 89. The Standard Deviation of Binomial distribution is: (a) npq (b) npq (c) np (d) np Sol. (b) S.D. of Binomial Distribution = npq 90. The speeds of a number of bikes follow a normal distribution model with a mean of 83 km/hr and a standard deviation of 9.4 km./hr. Find the probability that a bike picked at random is travelling at more than 95 km/hr.? Given [P (Z > 1.28) = 0.1003] (a) 0.1003 (b) 0.38 (c) 0.49 (d) 0.278 Sol. (a) Mean (M) =83 , S.D.     9.4  x  m 95  83  P  x  95  = p    9.4  = P  Z  1.28  = 0.1003 91. The equations of the two lines of regression are 4 x + 3y + 7 = 0 and 3 x + 4y + 8 = 0. Find the correlation coefficient between x and y? (a) -0.75 (b) 0.25 (c) -0.92 (d) 1.25 Sol. (a) Given two Equations of Regression lines are: 4 x  3y  7  0 And 3 x  4y  8  0 39 CA Foundation coeff.of y coeff.of x bx y  And byx  coeff. coeff.of y 3 3 bx y  byx  4 4 Coeff.of correlation is given by: r   byx  bxy =  3 / 4   3 / 4 3 = 16 3 = 4 r=-0.75 92. The regression equations are 2x  3y  1  0 and 5 x  6y  1  0 then mean of x and y respectively are: (a) -1,-1 (b) -1, 1 (c) 1,-1 (d) 2, 3 Sol. (c) Given regression Equations are: 2 x  3y  1  0  2 x  3y  1________ 1 and5 x  6y  1  0  5 x  6y  1________  2 Multiply by (2) in eq (1) we get 4 x  6 y  2 _____  3  Eq.(2) –eq.(3) 5 x  6y  1 4 x  6y  2    x 1 Putting x =1 in equation (1) 2  1 3y  1 2  3y  1 3y  1 2 40 CA Foundation 3y  3 y  1 Ans. x  1,y  1 93. If b yx  0.5, b yx  0.46 then the value of correlation coefficient r is: (a) 0.23 (b) 0.25 (c) 0.39 (d) 0.48 Sol. (d) Given byx =0.5, bxy =0.46 find r=? Coeff. Of correlation r   byx  bxy =  0.5  0.46 =  0.23 = 0.48 94. The coefficient of rank correlation between the ranking of following 6 students in two subjects Mathematics and Statistics is: Mathematics 3 5 8 4 7 10 Statistics 6 4 9 8 1 2 (a) -0.25 (b) 0.35 (c) 0.38 (d) 0.20 Sol. (a) MATHEMATICS x , STATISTICS Y Table Marks of Rank of ‘ x ’ Marks of Rank of y D=R x - Ry d2 Maths  x  Rx Stats (y) Ry   3 6 6 3 3 9 5 4 4 4 0 0 8 2 9 1 1 1 4 5 8 2 3 9 7 3 1 6 -3 9 10 1 2 5 -4 16 n=6 2 d = 44 Coeff. Of rank correlation 6 d 2 rR  1   n n2  1  6  44  1  6 62  1  41 CA Foundation 6  44  1 6  35 44  1 35 9  35 rR  0.257 rR  0.25 95. Karl Pearson's Correlation coefficient between x and y is:  S xSy  2 cov  x.y  cov 2  x.y  S xSy (a) (b) (c) (d) S xSy S xS y cov  x,y  cov  x.y  Sol. (a) Kari person’s correlation coefficient cov  x.y  r S xSy Where cov ( x , y) Covariance of ( x ,y) Sx S.Dof x Sy S.Dof y 96. From the following data construct the index number by Laspeyre’s 1 1  99, P0Q1  76, P0Q0  73, PQ Method PQ 1 1  96 (a) 130.36 (b) 131.51 (c) 130.59 (d) 76.01 Sol. (b) Here P Q 1 1  99,  P0 Q1  76 P Q 0 0  73 P Q 1 0  96 Laspeyre Index no = P Q 1 0  100 P Q 0 0 96 =  100 73 =131.51 97. Which of the following index measures the change from month to month in the cost of a representative basket of goods and services of the type which are bought by a typical household? 42 CA Foundation (a) Retail Price Index (b) Laspeyre's Index (c) Fisher's Index (d) Paasche's Index Sol. (a) Retail Price Index measures the change from month to month in the cost of a representative basket of goods and services of the type bought by a typical household. 98. Fisher's Index number is called as ideal index number because it is satisfying. (a) Factor reversal test (b) Time reversal test (c) Both factor and time reversal test (d) Circular test Sol. (c) Fisher's Index No. is called as Ideal Index number because it is satisfying Both Factor and Time reversal test. 99. If Laspeyre's Index is 119 and Paasche's Index is 112. Then Fisher's index number will be: (a) 113.99 (b) 115.45 (c) 115.89 (d) 151.98 Sol. (b) Laspeyre's Index No. (L) = 119 Paasche's Index No. (P) = 112 Fisher Index No. (F) =? We know that F = L P = 119  112 = 13328 = 115.45 100. In price index, when a new commodity is required to be added, which of the following index is used? (a) Shifted price index (b) Splicing price index (c) Deflating price index (d) Value price index Sol. (a) Shifted price Index is used. Answers 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 a b a d d d b d c c b c d a d a b b a c 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 b a a b c c a b a d b c d c a b c b a c 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 b a d d b c b d a b c b d d c d c d a d 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 c c c a a b b d b b a c a b c d a a d d 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 c c c a b d c b b a a c d a a b a c b a 43