APTITUDE FOR ITITES PART II On a 26

APTITUDE FOR IT/ITES – PART II

On a 26 question test, five points were deducted for each wrong answer and eight points were added for each correct answer. If all the questions were answered, how many were correct, if the score was zero ? a. 10 b. 12 c. 11 d. 13

Answer: (a)Take options and check. If 10 are correct, his score is 10 x 8 = 80. But 16 are wrong. So total negative marking is 16 x 5 = 80. So final score is zero.

Mr. Bean chooses a number and he keeps on doubling the number followed by subtracting one from it, if he chooses 3 as initial number and he repeats the operation for 30 times then what is the final result? a. (2^30) – 1 b. (2^30) – 2 c. (2^31) + 1 d. (2^31) + 2

Answer : (c) Step 1: (3 x 2) - 1 = 5 ( 2^2 + 1) Step 2: (5 x 2) - 1 = 9 (2^3 + 1) Step 3: (9 x 2) - 1 = 17 (2^4 + 1) Step 4: (17 x 2) - 1 = 33 (2^5 + 1). . . So After 30 steps we have 2^31 + 1

If pq=28 where p and q are whole numbers then which of the following will be (p 2 - q 2)? a) 32 b) 83 c) 783 d) cannot be determined.

Answer : (c) Given that, pq=28 and p, q are integers. The factors of 28 are 1, 2, 4, 7, 14 and 28. Then the possibilities of pq = 28 are (p, q) = (1, 28), (2, 14), (4, 7), (7, 4), (14, 2) and (28, 1) We would have p > q since the given options are positive. (That is, if p < q then (p 2 - q 2) will be negative). Therefore, the possibilities of (p, q) reduced to (7, 4), (14, 2) and (28, 1). If (p, q) = (28, 1) then (p 2 - q 2) = 282 - 12 = 784 - 1 = 783. If (p, q) = (14, 2) then (p 2 - q 2) = 142 - 22 = 196 - 4 = 192 If (p, q) = (7, 4) = (p 2 - q 2) = 72 - 42 = 49 - 16 = 33 From the given options, required answer is option c.

A father has divided his properties in such a way that one-half of total properties goes to A, two-third of the remaining shared equally to B, C and D and the rest to E, F, G and H such they equally gets Rs. 30, 000 by their sharing property. Then the amount will D get is: a) Rs. 80, 000 b) Rs. 68, 000 c) Rs. 24, 000 d) Rs. 72, 000

Answer : (a) A's share = 1/2 Then remaining = 1 - 1/2 = 1/2. . . (A) 2/3 of this remaining 1/2 (as in eq A) goes to B, C and D. Total share of B, C and D = 2/3 of 1/2 = 1/3 Therefore, now remaining property = Total share of E, F, G and H = value of (A) - total share of B, C and D = 1/2 - 1/3 = 1/6 Then, E, F, G and H 's individual shares = (1/6) / 4 = 1/6 x 1/4 = 1/24 Given 1/24 = Rs. 30, 000 Then, total property amount = Rs. 24 x 30, 000 = Rs. 7, 20, 000 Total share of B, C and D = 1/3 x 7, 20, 000 Then their individual share amount = (1/3 x 7, 20, 000) / 3 = Rs. 80, 000. Hence D's share is Rs. 80000.

How many three digit numbers exist with the property that first digit of the 3 -digit number is the product of last two digits? a) 19 b) 17 c) 12 d) 23

First digit (100’s place) of a 3 -digit number cannot be zero, let us start with 1. If the first digit is 1, then the possible last two digits to get the product 1 is (1, 1)Therefore, the number is 111. If the first digit is 2, then the possible last two digits to get the product 2 are (1, 2) and (2, 1). Therefore, the numbers are 212 and 221. Proceeding like this, we get, First digit Corresponding 3 -digit numbers 111 212, 221 3 313, 331 4 414, 441, 422 5 515, 551 6 616, 661, 623, 632 7 717, 771 8 818, 881, 842, 824 9 919, 991, 933 Thus, the required number of such 3 -digit numbers is 23

Indices and Surds [(74)/(493/2)](1/6) x [(73 /(3434/3)](1/2) = ? / [(74 / 49)1/6] a) 1 b) 724 c) 76 d) 74

Given expression is, [(74)/(493/2)](1/6) x [(73 /(3434/3)](1/2) = ? / [(74 / 49)1/6] = [(74)/(493/2)](1/6) x [(73 /(3434/3)](1/2) x [(74 / 49)1/6] = ? = [(74)/(72)(3/2)](1/6) x [(73/(73)(4/3)](1/2) x [(74/72)1/6] = (74)/(73)](1/6) x [(73)/(74)](1/2) x [(74/72)1/6] = (74 -3)(1/6) x [(73 -4)](1/2) x [(74 -2)1/6 = (71)(1/6) x [(7 -1)](1/2) x [(72)1/6 = (7)(1/6) x (7)(-1/2) x (7)1/3 = (7)(1/6 -1/2 + 1/3) = (7)(2 -6+4/12) = (70 )(1/12) = 1

Men and Work Jake can dig a well in 16 days. Paul can dig the same well in 24 days. Jake, Paul and Hari together dig the well in 8 days. Hari alone can dig the well in a. 96 days b. 48 days c. 32 days d. 24 days

Answer : (b) Simple one. Let the total work to be done be 48 meters. Now Jake can dig 3 mts, Paul can dig 2 mts a day. Now all of them combined dug in 8 days so per day they dug 48/8 = 6 mts. So Of these 8 mts, Hari capacity is 1 mt. So he takes 48 /1 = 48 days to complete the digging job.

Algebraic Equations Mark told John "If you give me half your money I will have Rs. 75. John said, "if you give me one third of your money, I will have Rs. 75/- How much money did John have ? a. 45 b. 60 c. 48 d. 37. 5

Answer : (b) Let the money with Mark and John are M and J respectively. Now M + J/2 = 75 & M/3 + J = 75 Solving we get M = 45, and J = 60.

The difference between a two digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number? A. 3 B. 4 C. 9 D. Cannot be determined E. None of these

Answer: (B) Let the ten's digit be x and unit's digit be y. Then, (10 x + y) - (10 y + x) = 36 9(x - y) = 36 x - y = 4.

A man spent 5/16 of his age plus two as student, 1/40 plus 1 as a husband, 1/4 as a good politician and 3/40 as a father. And the remaining 6 years as a good grand father. Then the living days of the man is: a) 72 b) 89 c) 80 d) 69

Answer : (c) Lets say his age is X. So years spent as student = 5 X / 16 + 2, as Husband = X/4 + 1, as Politician = X/4, and as Father = 3 X/40. Remaining 6 years (as grandfather) = X - { 5 X/16 + 2 + X/4 + 1 + X/4 + 3 X/40} i. e. , 6 = X - { 5 X / 16 + 2 + X / 4 + 1 + X / 4 + 3 X / 40} 6 = X - [13 X / 16 + 3 X / 40] 6 = X - [71 X / 80 + 3] 9 X/80 = 9 X = 80. Hence the age of the man is 80 years.

Depreciation The value of a scooter depreciates in such a way that its value of the end of each year is 3/4 of its value of the beginning of the same year. If the initial value of the scooter is Rs. 40, 000, what is the value at the end of 3 years ? a. Rs. 13435 b. Rs. 23125 c. Rs. 19000 d. Rs. 16875

Ans: D 40, 000 x (3/4)^3=16875

Logarithms Which of the following statements is not correct? A. log 10 10 = 1 B. log (2 + 3) = log (2 x 3) C. log 10 1 = 0 D. log (1 + 2 + 3) = log 1 + log 2 + log 3

Answer : (b) (a) Since loga a = 1, so log 10 10 = 1. (b) log (2 + 3) = log 5 and log (2 x 3) = log 6 = log 2 + log 3 Hence log (2 + 3)≠ log (2 x 3) (c) Since loga 1 = 0, so log 10 1 = 0. (d) log (1 + 2 + 3) = log 6 = log (1 x 2 x 3) = log 1 + log 2 + log 3. So, (b) is incorrect.

The value of 1 + 1 is: log 360 log 460 log 560 A. 0 B. 1 C. 5 D. 60

Answer: (B) Using logarithm base switch rule logb(c) = 1 / logc(b) Given expression = log 60 3 + log 60 4 + log 60 5 = log 60 (3 x 4 x 5) = log 60 60 = 1

To be specific, the logarithm of a number x to a base b is just the exponent you put onto b to make the result equal x. For instance, since 5² = 25, we know that 2 (the power) is the logarithm of 25 to base 5. Symbolically, log 5(25) = 2. If log 27 = 1. 431, then the value of log 9 is: A. 0. 934 B. 0. 945 C. 0. 954 D. 0. 958

Answer: (C) log 27 = 1. 431 log (3^3 ) = 1. 431 3 log 3 = 1. 431 / 3 = 0. 477 log 9 = log(3^2 ) = 2 log 3 = (2 x 0. 477) = 0. 954.

Averages In the first 10 overs of a cricket game, the run rate was only 3. 2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs? A. 6. 25 B. 6. 5 C. 6. 75 D. 7

Answer: (a) Required run rate = ( 282 - (3. 2 x 10) ) / 40 = 250 / 40 = 6. 25

A family consists of two grandparents, two parents and three grandchildren. The average of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. What is the average of the family? A. 28 4/7 years B. 31 5/7 year C. 32 1/7 years D. None of these

Answer: (b) Required average = (67 x 2 + 35 x 2 + 6 x 3 )/(2 + 3) = (134 + 70 + 187 )/ 7 = 222/7 = 31 5/7

A pupil's marks were wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half (1/2). The number of pupils in the class is: A. 10 B. 20 C. 40 D. 73

Answer : (b) Let there be x pupils in the class. Total increase in marks =X/2 X / 2= (83 - 63) X /2 = 20 X = 40.

If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, then the average marks of all the students is: A. 53. 33 B. 54. 68 C. 55 D. None of these

Answer: (b) Required average = (55 x 50 + 60 x 55 + 45 x 60) / (55 + 60 + 45) = (2750 + 3300 + 2700) / 160 = 8750 / 160 = 54. 68

Simple Interest and Compound Interest The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 1. The sum (in Rs. ) is: A. 625 B. 630 C. 640 D. 650

CI = X (1+ 4/100) ^ 2 – X = (676 X / 675 ) – X = 51 x /625 SI = X * 4 * 2 /100 = 2 X/25 51 X /625 – 2 X/25 = 1 51 X/ 625 – 50 X/625 = 1 X = 625.

Ratio & Proportion Two containers of equal sizes are filled with the mixture of two variety rice. The proportion of two varieties in first container is 2: 3 and the proportion of two varieties in second container is 5: 9. The content of the two containers are emptied into another container and the quantity of mixture in new container is 280 kg. What is the quantity of variety 1 rice in the new container? a) 53 kg b) 106 kg c) 87 kg d) 174 kg

Answer : (b) Given that, the ratio of variety 1 and variety 2 in each container is 2: 3 and 5: 9 Quantity of variety 1 in 1 st container = 2/(2 + 3) = 2/5 (that is, 2 parts of 5 parts is variety 1) Quantity of variety 1 in 2 nd container = 5/(5 + 9) = 5/14. Therefore, total quantity of variety 1 in new container = 2/5 + 5/14 = (28 + 25)/70 = 53/70 Similarly, quantity of variety 2 in two containers are 3/5 and 9/14 respectively. And, the total quantity of variety 2 in new container = 3/5 + 9/14 = (42 + 45)/70 = 87/70 Now, the ratio of variety 1 and variety 2 in new container = (53/70) : (87/70) = 53: 87 i. e. , 53 parts of (53 + 87) parts are variety 1 => 53 parts of 140 parts are variety 1. Since the total quantity in new container is 280 kg then the kg of variety 1 is (53/140) x 280 = 2 x 53 = 106. Hence, the answer is 106 kg

The ratio of numbers of boys and girls of three sections A, B and C are 2: 3, 3: 7 and 4: 11 respectively. The total number of students in A, B, C is 30, 40 and 60 respectively. If students of all the three sections are assembled together then what will be the ratio of boys to girls? a) 20: 45 b) 12: 35 c) 15: 42 d) 21: 38

Answer : (a) Given that, total number of students in A, B, and C is 30, 40 and 60 respectively. And the respective ratio of boys and girls is 2: 3, 3: 7 and 4: 11. Number of boys in A = [[2/(3+2)] x 30 = 2/5 x 30 = 12 So, the remaining = number of girls in A = 30 -12 = 18. Number of boys in B = [[3/(3+7)] x 40 = 3/10 x 40 = 12 So, the remaining = number of girls in B = 40 -12 = 28. Number of boys in C = [[4/(4+11)] x 60 = 4/15 x 60 = 16 So, the remaining = number of girls in C = 60 -16 = 44. Therefore, total number of boys = 12 + 16 = 40 And, total number of girls = 18 + 28 + 44 = 90 Now, the required ratio = number of boys : number of girls = 40: 90 = 20: 45

A motorboat can cover 10 1/3 km in 1 hour in still water. And it takes twice as much as time to cover up than as to cover down the same distance in running water. The speed of the current is: a)3 4/9 km/hr b) 2 1/3 km/hr c) 4 km/hr d) none of these

Answer (a) 3 4/9 km/hr Let the speed of upstream be X km/hr. Then, speed in downstream = 2 X km/hr (since boat takes twice as much as time to cover up than as to cover down the same distance in running water). Speed in still water = (2 X+X)/2 km/hr. (formula 3) = 3 X/2 km/hr. Given that, boat covers 10 1/3 km in 1 hour in still water. Therefore, 3 X/2 = 10 1/3 X = 62/9 So, speed in upstream = 62/9 km/hr. And, speed in downstream = 2 x 62/9 = 124/9 km/hr Hence, speed of the current = [(124/9 - 62/9)]/2 km/hr = 62/9 x 2 = 34/9 = 3 4/9 km/hr x--------------x