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Question 1 What is the sum of the irrational roots of the equation (x-1)(x-3)(x-5)(x-7)=9 ? a)10 b)8 c)6 d)4 Answer : b)8 Solution: Given that (x - 1) (x - 3) (x - 5) (x - 7) = 9 Let x - 4 = p Then the given eqn becomes (p + 3) (p + 1) (p - 1) (p - 3) = 9 (p 2 - 1) (p 2 - 9) = 9 p 4 - 10p 2 + 9 = 9 p 2 (p 2 - 10) = 0 p 2 =0 or p 2 -10 =0 p = 0 or p = sqrt(10) or p = - sqrt(10) then x - 4 = 0, x - 4 = sqrt(10) or x - 4 = - sqrt(10) Now the roots of the given eqn are 4,4 + sqrt(10) and 4 - sqrt(10) The irrational roots are 4+sqrt(10) and 4 - sqrt(10) The sum of the irrational roots = 4 + sqrt(10) + 4 - sqrt(10) = 8. Hence the answer is 8. Question 2 The product of the distinct roots of the equation (3x)(3x+2)(3x-4)(3x-6)= 64 is: a)-32/27 b)-61/5 c)63/16 d)69/12 Answer : a)-32/27 Solution: Given that (3x)(3x+2)(3x-4)(3x-6)= 64 Let 3x - 2 = p Then the given eqn becomes (p + 2) (p + 4) (p - 2) (p - 4) = 64 (p 2 - 4) (p 2 - 16...

Question 1 If 11 + 21 = 3, 34 + 45 = 32 and 42 + 53 = 23 then 64 + 75 = ? a)59 b)43 c)69 d)53 Answer : a)59 Solution : Let AB be the first number of the addition and CD be the second one. By closely observing, we can conclude that questions are in the general form AB + CD = (A x B) + (C x D). i.e.,11 + 21 = 1 x 1 + 2 x 1 = 1 + 2 = 3 34 + 45 = 3 x 4 + 4 x 5 = 12 + 20 = 32 42 + 53 = 4 x 2 + 5 x 3 = 8 + 15 = 23. Therefore, the result for 64 + 75 = 6 x 4 + 7 x 5 =24 + 35 = 59. Question 2 If 15 + 22 = 32, 19 + 35 = 62 and 23 + 49 = 101 then 34 + 52 = ? a)91 b)101 c)172 d)72 Answer : c)172 Solution: By observing the question we can generalize the equations to, AB + CD = (AB x C) + D Then, 15 + 22 = (15 x 2)+ 2 = 30 + 2 = 32 19 + 35 = (19 x 3)+ 5 = 57 + 5 = 62 23 + 49 = (23 x 4)+ 9 = 92 + 9 = 101 Similarly,34 + 52 = (34 x 5)+ 2 = 170 + 2 = 172 Hence 172 is the required answer. Question 3 If 3 + 5 = 19, 5 + 9 = 61 and 9 + 12 = 117 then 13 + 14 = ? a)1...

Question 1 A manufacturer reduces the size of his machine of Material1 from 30" to 28" and that of Material2 from 24" to 23". He normally spends Rs.1440 for Material1 and Material2. 3/4th of the amount spent would be for Material1. How much will he save under the size of new size (Assuming that the costs incurred by the machines are directly proportional to their sizes)? a)Rs.85 b)Rs.87 c)Rs.86 d)Rs.88 Answer : b)Rs.87 Solution: The manufacturer totally spends Rs.1440 for both material 1 and 2. Out of this 3/4th is entirely for Material1. Amount spent on Material1 = 1440 x 3/4 = Rs.1080 i.e, he spent Rs.1080 for Material1 of 30" and he reduces 2". Size Amount 30" Rs.1080 2" ? Amount saved on Material1 = 1080 x 2/30 = Rs.72 Amount spent on Material2 = total amount - amount for Material1 = 1440 - 1080 = Rs.360. i.e., he spent Rs.360 for Material2 of 24" and he reduces 1". Size Amount 24" Rs.360 ...

1) a) NMI b) Software interrupts c) Hardware interrupts d) Software and hardware interrupt Ans: b 2) Which of the following is error correction and deduction? a) Hamming code b) CRC c) VRC d) None Ans: b 3) When you switch on your computer, which of the following component affect first? a) Mother board b) SMPS c) Peripherals d) None Ans : a 4) Which of the following function transports the data? a) TCP/IP b) Transport layer c) TCP d) None Ans: c 5) Which of the following does not consists address? a) IP b) TCP/IP c) Network d) Transport Ans:a 6) They given like this……. And some conditions? a) pre order b) post order c) In order d) None Ans:c 7) Authentication means…. a) Verifying authority b) Source c) Destination d) Request Ans: a 8) Symorphous is used for a) Analysis b) Synchournization c) Asynchrouns d) None Ans: b 9) There are five nodes. With that how many trees we can make? a) 27 b) 28 c) 30 d) 29 Ans: c (Check the ans not sure) 10) Traverse the given tree using in order, Preord...

1. How many segment registers are there in the 8086 processor? a) 4 b) 6 c) 8 d) none Ans: a 2. What is the addressing mode of this instruction MOV AL, [BX]; a) direct addressing mode b) register indirect addressing mode ANS: b I am not sure of it. 3. What is the type of file system used in the CD ROM? a) VFAT B) 4) About CPU I think but answer is DMA. 5) If we double the clock speed, then? a) It increases the speed of the processor b) It increases the speed of the system bus c) both of the above D) None 6) Data recovery is done in which layer? a) physical b) datalink c) network d) transport 7) What is thrashing? ANS: swapping in and out the pages frequently from memory. 8) By using 3 nodes how many trees can be formed? ANS:5 9) They give one tree, and ask the post order traversal for that tree? ANS:C 10) Page cannibalation is? ANS:C Aptitude section: 1) They give one scenario and ask questions on that. that is easy. Those are 5 questions. 2) They ask 2 questi...

Three sample questions based on bank related transactions. Question 1 Ritika is a student of a famous Engineering College in Ongole. She wanted to buy a mobile phone but she is not inclined to disturb her father who is a farmer. A local Bank was ready to extend a loan for purchase of mobiles by students on simple interest. She chose a phone that costs 1500. The interest rate on the loan is 12%. If the loan is to be paid back in weekly installments over 2 years, calculate: 1. The amount of interest paid over two years 2. The total amount to be paid back 3. The weekly payment amount (a) 300, 1800, 18 (b) 320, 1820, 22 (c) 350, 1850, 25 (d) 360,1860, 17.88 Answer : d) 360,1860, 17.88 Solution: Principal, P=1500 Number of Years, N= 2 years Rate of interest per annum, R=12% Simple Interest,SI= P X N X R / 100 =1500X2X12 / 100=360. Amount of interest to be paid back in 2 years is 360. Total amount to be paid back P + SI= 1500 + 360=1860. This is to be pai...

Below are three easy aptitude questions. They are not from any specific mathematics section, but are simple questions requiring logical thinking. Solving these questions is easy if you understand the questions well. Question 1 Hema malini was enjoying her vacation at her Cousin's home. In the mornings, she used to go for a jogging with her cousin. The evenings were spent on tennis court in nearly sports club. Since the activities of jogging and playing tennis were tiring, they could manage only one per day, i.e., either they went for a jog or played tennis each day. There were days when they felt lazy and stayed home all day long. For every four days they when for jog on first day, went for walk on the second day, did nothing for the third day and went for walk on the fourth day. This cycle continued. In total if they had played tennis for 30 days, how many total number of days Hema stayed in her Cousin's home. (a)15 (b) 14 (c) 18 (d) 60 (e) 15 Answer : d) 60 Solution: This is...