For regular updates Like our Page : facebook/synergyias/likes C-SAT – SOLUTIONS 1. C 2. B 3. C 4. D 5. B 6. C 7. C 8. C 9. C 10. C 11. A 12. C 13. B 14. A 15. A Explanation – • Let the profit made during the year be Rs. 3X Therefore, Rs. X would have been shared equally and the remaining Rs. 2X would have been shared in the ratio 7 : 3. i.e., 70% of 2X would go to Maya and 30% of 2X would go to Mayank. Hence, Maya would get (70 - 30)% of 2X more than Mayank Or 40% of 2x = Rs. 800 i.e., (40/100) * 2X = Rs. 800 or 2X = 2000 Hence, the profit made by the company during the year Rs.3x = Rs.3000. 16. B Explanation – • 1, 3 and 4 figures are made with three straight lines whereas 2 is made with four straight lines. 17. B Explanation – • Number of ‘X’ remain same in all the figures i.e. 2 whereas number of ‘o’ are in 3,4,3,4 order. 18. B Explanation – • Bigger circle will have its pointed peak in the north direction as the order of its peak is East, South, West and North. Smaller circle will have the shaded region in South-West direction as the order of the shaded region is South-East, North-East, North-West and South West. Hence, only possible solution is B. 19. A Explanation – • First half of the letters are reversed and then second half of the letters are reversed. SESAME – first 3 letters ‘SES’ are reversed as ‘SES’ only and next 3 letters ‘AME’ are reversed as ‘EMA’ and then combined again i.e. ‘SESEMA’. Same way in ‘NEARER’, first 3 letters are reversed as ‘AEN’ and ‘RER’ is reversed as ‘RER’ only. Combination is ‘AENRER’. Hence, for ‘PROBABLE’, it is going to be combination of ‘BORP’ and ‘ELBA’ i.e. ‘BORPELBA’. 20. C Explanation – • We see that in 13529 and 35293, 3529 is common. Hence, it should be either “RAIN” or “RIVE” which is common in “BRAIN” and “DRAIN” or in “RIVER” and “DRIVE” respectively. Hence, remaining 1 and 3 in 13529 and 35293 would be codes for two of the letters among these three - “B”, “D” or “R”. In the same way, 13754 and 83754 have 1 and 8 as different digits. Hence, we see that 1, 3 and 8 are the codes for “B”, “D” and “R” (not in the same order). But, 1 is repeated twice and “D” is also repeated twice. Hence, the code for “D” is 1. 21. B Explanation – • Let M1s 1 day work be X and M2s 1 day work be Y ...(1) • Given that, the time taken to complete the order by (M1 + M2) = 10 days. • Then (M1 + M2)s 1 day work = 1/10 …(2) • Therefore, from (1) and (2), X+Y = 1/10 …(3) • Suppose, M3 works twice as M1 then M3s 1 day work 2X. • And, M4 works 1/3 as much as M2 then M4s 1 day work be Y/3. • Now, the time taken to complete the order by (M3 + M4) = 6 days • 2X+Y/3 = 1/6 • 12X+2Y = 1 ….(4) • Solving the above two equations(3) and (4), we get X = 4/50. • Thus, M1s 1 day work = 4/50. • Hence M1 alone can complete the entire order of work in 50/4 days = 12.5 days. 22. C Explanation – • Instead of man-hours, here we want to interact plow-minutes. Feet and minutes are already compared, so all we have to is add “plows” to the expression. If we divide 123 ft/min by 3 plows, we get: • 123 ft/minute/3 plows = 41 ft/plow-minute • At this rate, if we want to increase minutes to 5 and plows to 8, we can simply insert these into the existing rate. Note the absolute rate does not change, since we are multiplying top and bottom by 40, so the value is constant. • 41*40 feet / 40 plow-minutes = 1640 feet / 40 plow-minutes. 23. C Explanation – • Probability problems require the knowledge of all possible combinations (mathematicians call this the probability space), the total number of combinations that satisfy the condition you are looking for and the probability that the combination will occur. The die is fair all possibilities are equally likely. Let A, B, C represent all possible values for the 3 die tosses. A = {1,2,3,4,5,6} , B = {1,2,3,4,5,6}, C = {1,2,3,4,5,6} and a,b,c represent possible outcomes of A,B,C respectively. (mathematicians say a,b,c are elements of the set A,B,C). We are to solve for the sum of 3 die tosses to equal 16. In other words, we are looking for all possible combinations such that a + b + c = 16. 1. Each die toss has six possible outcomes. Tossing the die once gives six possible values. Tossing the die twice gives a total of ( 6 possibilities for the first toss) times ( 6 possibilities for the second toss) = 36 possible outcomes . Similarly, for three die tosses one has (6)(6)(6) = 216 possible outcomes. 2. The first die toss is from 1 to 6. Since the maximum sum of the final 2 die tosses is 6 + 6 = 12, the first dice toss must be at least 4 for the sum of all three tosses to be 16. 3. If the first die toss is 4 then the last two must add up to 12. This will occur for the case {4,6,6} which is a total of one possibility. 4. If the first die toss is 5 then the last two must add up to 11. This will occur for the cases {5,5,6}, {5,6,5} which is a total of two possibilities. 5. If the first die toss is 6 then the last two must add up to 10. This will occur for the case {6,4,6}, {6,5,5}, {6,6,4} which is a total of three possibilities. 6. Out of the 216 total possible combinations, 1+2+3=6 add up to 6. 7. The answer is given by (number of outcomes that add to 16 ) divided by (the total number of possible outcomes) which is 6 / 216 or 1 / 36 24. C Explanation – 1. Evaluate each statement one by one, starting with the first. 2. Evaluate Statement I 1. In order to be divisible by 6, a number must have the prime factors of 6, which are 2 and 3. 2. In a set of 3 consecutive integers, at least one of them will be even. Any even number has a 2 as one of its prime factors. Thus, the product of the 3 consecutive positive integers will have a factor of 2. 3. Any consecutive series of 3 integers has a multiple of 3 in it, since every third integer is a multiple of 3. Thus, either x, y, or z is a multiple of 3 and therefore has 3 as one of its prime factors. Thus, the product xyz will have a factor of 3. 4. Since xyz will have a 3 and a 2 in its prime factorization tree, xyz must be divisible by 6. Therefore, I is always true. 3. Evaluate Statement II 1. Since x, y, and z are consecutive integers such that x < y < z, we can rewrite y and z in terms of x: y=x+1, and z=x+2. 2. Substitute these values in the equation: (z-x)(y-x+1)=4 ([x+2]-x)([x+1]-x+1)=4 3. Simplify the equation: (x-x+2)([x-x+1+1)=4 (2)(1+1) = 2(2) = 4 4. It is clear that II will always be true. 4. Evaluate Statement III 1. Since x, y, and z are consecutive numbers such that x < y < z, if y is even, then x and z will both be odd. The product of two odd numbers is odd so xz would be odd in this case. But, xy = (odd)(even) = even. 2. But, if y is odd, then x and z will be even. The product of two even numbers is even, so xz is even in this case and xy is also even. 3. Since there is no way to guarantee that both x and y are odd, we cannot conclude that statement III is always true. 4. Note: Since x and y are consecutive integers, either x or y will always be even. Consequently, xy will always be even: either (even)(odd) = even OR (odd)(even) = even. 5. Since I and II must be true, but III is not always true, the correct answer is C. 25. D Explanation – 1. The number of individuals that had infection A is 1/3 of N, or N/3. Infection A: N/3 2. The number of individuals that also had infection B is 1/5 of the number that had infection A. In other words, it is (1/5)A = (1/5)(N/3) = N/15. Infection B: (1/5)A = (1/5)(N/3) = N/15 3. If N/15 of the N individuals have infection A and B, then N – (N/15) did not have both infection A and B. Consequently, 14N/15 individuals did not have both infection A and infection B. Neither A or B: N - (N/15) = 14N/15 26. D Explanation – 1. One approach to solve the problem is to list the different possibilities for a toss of coin three times. Because there are two outcomes and the coin is tossed three times, the table will have 2*2*2 or 8 rows. 2. Next add the resulting rows together to find the sum (the fourth column in the table below). Toss 1 Toss 2 Toss 3 Sum 1 1 1 3 1 1 2 4 1 2 1 4 1 2 2 5 2 1 1 4 2 1 2 5 2 2 1 5 2 2 2 6 3. From the table we see that there are 4 situations where the sum of the tosses will be greater than 4. And there are 8 possible combinations resulting in a probability of 4. 4/8 or a probability of1/2. So the correct answer is D. 27. D Explanation – • The distance formula is Distance = Rate × Time. To solve this question, set x as the length of time (in hours) for the first trip. The distance to the destination city is 60 × x, where rate = 60 and time = x. The return trip was 4 hours longer, so its distance is 40(x + 4). Since we know the distances are the same, we can set them equal to each other and solve for x: 60x = 40(x + 4) • 60x = 40x + 160 • 20x = 160 • x = 8 • So, the trip took 8 hours, and its distance is 8 × 60 = 480 kilometers. The question asks for the roundtrip distance, so we double the length of the one-way trip: 480 × 2 = 960 kilometers. The correct answer is choice (D). 28. B Explanation – • If the total journey is for D kms. • He covers 2/15 of the journey by train i.e. 2D/15 • He covers 9/20 of the journey by tonga i.e. 9D/20. • Total distance covered by tonga and train together is 2D/15+9D/20 = 35D/60 • Remaining distance is D- 35D/60 = 25D/60 • This distance is given as 10 kms. Hence, • 25D/60 = 10 => D = 60*10/25 => D = 24kms. • Hence, correct answer is B. 29. C Explanation – • The numbers between 100 and 1000 that are multiples of 6 are 102,108,114,……996 This forms an A.P series with First term: a=102, Common Difference:d=6, last term=996. • Using formula for A.P series: • a+(n-1)d = nth term • 102 + (n-1)6 = 996 • n=150 Hence answer is C. 30. A Explanation - • The minute hand moves 360o in an hour. In 40 minutes , it moves by (40/60)*360= 240o. The hour hand moves by 30 degrees in an hour. • So angle between hour and minute hand (at 14:40)will be somewhere around 240-60=180. However an hour hand will be ahead of 2 at 14:40. In 40 minutes it will move by (30/60)*40 = • 20 degrees. • Hence the angle between hour and minute hand will be 180 - 20 = 160o 31. C Explanation – • Any term is sum of all previous terms + previous term • Third term i.e. 5 = sum of (1 and 2) + 2 • Same way 13 = sum of (1, 2 and 5) + 5 • Hence, next term ‘?’ = sum of (1, 2, 5 and 13) + 13 = 34 • To confirm, we can see that 89 = sum of (1, 2, 5, 13 and 34) + 34 • Hence, Correct answer is choice C. 32. Ans. A Explanation – • Middle term in any row is sum of the difference of the digits of right and left numbers. • In 1st row, left number is 19. Difference of the digits are 9 – 1 = 8 • Same way, right number is 37, Difference of the digits are 7 – 3 = 4 • Hence, the middle term is 8 + 4 = 12 • Same way is applicable for row 3 i.e. (7-1) + (8-1) = 13 Ans. 4.5% profit – Not given in options. Sorry for the typo. Explanation – • Let the marked price be S and the cost price of the article be C • When the merchant sells at 80% of marked price he sells at 0.8S • This results in a loss of 12%. • Loss is always computed as a percentage of cost price. • Therefore, the loss incurred by the merchant = 0.12C • Hence, he will be selling the article at C - 0.12C = 0.88C when he sells at 80% of his marked price. • Equating the two sides of the relation, we get 0.8S = 0.88C • Or S = • Or S = 1.1C • Now, if the merchant sells at 95% of the marked price, he will be selling at 95% of 1.1C = 1.045C • Hence, the merchant will make a profit of 4.5%. 33. D Eplanation – • Total revenue generated is Rs. 13500. • 70% of it is generated by regular tickets i.e. 0.7 * 13500 = Rs. 9,450 • Price of each regular ticket is Rs. 30. Hence, total number of regular tickets are Rs.9450/30 = 315 • Remaining 30% revenue is generated by VIP tickets i.e. 0.3*13500 = Rs. 4,050 • Price of each VIP ticket is Rs. 50. Hence, total number of VIP tickets are Rs.4050/Rs.50 = 81 • Hence, total number of tickets are 315 + 81 = 396 34. C Explanation – • Rotate the second figure in such a way that B comes on the top side. You will realize that 2 has gone on the back side and 3 on the right side. Now, if you merge first and second figure then 3 would be opposite to 1. Hence, correct answer is choice C. 35. Ans. D Explanation – litres in solution Proportion of milk total litres milk Milk 10 1 10 cereal 40 0.30 (40)(0.30) = 12 mix 50 ? 10 + 12 = 22 From the last row, you see that there are 22 litres of milk in the 50 litres of solution, or 22/50. Simplify, and then convert to a percentage. 36. A 37. B 38. C 39. D 40. C 41. C 42. A 43. B • Blood feuds and customary fights are types of resolving disputes based on kinship. 44. B • At the highest pastoral and agricultural levels systems of public justice are established (even in cases of stateless society). 45. B Personal response on DI questions – As most of you said last time that the DI was lengthy. I tried to give simpler questions with easier calculations. I hope that all of you did well in DI. Expect little tougher questions in the next paper. 46. D Explanation – Add values for Thermal Power Production from 1990-1995 i.e. 140 + 140 + 120 + 150 + 145 +153 = 848. Divide it by 6 to get an average i.e. 848/6 = 141.3 47. B Explanation – Clearly visible that it was in 1991 i.e. 180+140 = 320 48. A Explanation – In 1991, percentage increase is (180-140/140)*100 = 28.57%. This is highest among all the years. 49. D Explanation – Let the thermal power production be X MW in year 1989. It has reduced by 30% in 1990, i.e. production in 1990 is 0.7 X which is equal to 140 MW. 0.7X = 140 => X = 200 MW 50. C Explanation – Profit in 1990 = Rs. 30 lakkhs Profit in 1989 = Rs. 20 lakhs Increase in profit percentage = (30-20)/20 = 50% which is highest. 51. B Explanation – • Just add all the revenue figures, total revenue comes as Rs. 1165 lakhs. Divide it by 7. Rs. 1165/7 = Rs. 166.42 lakhs 52. A Explanation – • In 1993, expenditure has increased from Rs. 120 lakhs to Rs. 140 lakhs. Percentage increase is (140-120)/120 *100 = 16.67%. This is highest among all the years. 53. A Explanation – • In 1994, profit is Rs. 60 lakhs and in 1995, profit is Rs. 70 lakhs. Hence, the growth rate is (70-60)/60 * 100 = 16.67% or 1/6 • Same way in 1996, it will be 70 + 16.67% (or 1/6) of 70 = 70 + 11.67 = Rs. 81.67 lakhs = Rs. 82 lakhs (approx.) 54. B 55. B 57. B 58. D 59. A 60. A 61. A 62. B 63. A 64. C 65. A 66. B 67. A 68. D • Dilip is taught Indian Painting on Wednesday. 69. A • If david cames on Tuesday, Daniel must come on Monday and learn modern painting. 70. B • Medieval painting is on Tuesday. With the condition given in the question, Pencil sketches should be on Friday and portraits on Tuesday. 71. B • Medieval Painting is taught on Tuesday. 72. C • Clearly, only either I or III follows. 73. D • Clearly, all given conclusions follows 74. A • Clearly, none follows 75. A 76. D 77. A 78. B 79. A 80. C
Posted on: Thu, 17 Jul 2014 08:33:12 +0000
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