| 20th November 2015 02:32 PM | |
| Saksham | Re: Oracle Solved Placement Papers The Oracle Corporation is an American global computer technology corporation, headquartered in Redwood City, California. Every year oracle will conduct on campus and off campus placement exam for hiring candidate for his company As you want to known about previous year placement paper of oracle here is the detail of it Question 1 Let X be an integer such that i. X is a sum of four consecutive integers ii. X is divisible by 5 iii. X is less than 1000. Find the number of possible X. a) 25 b) 17 c) 50 d) 152 Answer : c) 50 Solution : From statement (i), X is a sum of four consecutive integers. i.e., X = n + n+1 + n+2 + n+3 i.e., X = 4n+6...(1) From statement (ii), X is divisible by 5. X = 4n+6 must be divisible by 5. 4n+6 must end in a 5 or a 0. If 4n+6 end with 5, then 4n must end in 9. But this is not possible, since, multiples of 4 cannot end with 9. So, we have only one case that 4n+6 ending with 0. Then 4n will end in 4. And the unit’s digit of n must be 1 or 6. The possibilities of n are 1, 6, 11, 16, 21, 26, and so on. If, n = 1 then the sum of the consecutive numbers = X = 1 + 2 + 3 + 4 = 10 (divisible by 5) If, n = 6 then the sum of the consecutive numbers = 6 + 7 + 8 + 9 = 30 (divisible by 5) If, n = 11 then the sum of the consecutive numbers = 11 + 12 + 13 + 14 = 50 (divisible by 5) If, n = 16 then the sum of the consecutive numbers = 16 + 17 + 18 + 19 = 70 (divisible by 5) ..... Proceeding like this, the possibilities of the sum are 10, 30, 50, 90,..., 950, 970, 990, 1010, 1030,...and so on. From statement (iii),X is less than 1000 Required X values are 10, 30, 50, 70,..., 950, 970, 990. We have to find the number of numbers in the above series. This is an arithmetic progression with first term = a = 10, common difference = d = 20 and last term n = 990. We know that the nth term of A.P. = a+(n1)d Here, a+(n1)d = 990 10+(n1)20 = 990 (n1)20 = 980 (n1) = 49 n = 50. Hence, 50 numbers under 1000 that are divisible by 5 and can be made from the sum of four consecutive integers. Alternately, Required X values; 10, 30, 50, 70, 90,..., 950, 970, 990 Note that, there are 5 numbers (10, 30, 50, 70, 90) under X are divisible by 5 and can be made from the sum of four consecutive integers. Therefore, under 1000, we have 5 x 10 = 50 numbers. Question 2 How many numbers less than 150 that are divisible by 15 can be made from the sum of six consecutive integers? a) 4 b) 6 c) 7 d) 8 Answer : a) 4 Solution : Let the six consecutive integers be n, n+1, n+2, n+3, n+4 and n+5. Given that, n + n+1 + n+2 + n+3 + n+4 + n+5 = 6n+15 We have to find number of possibilities of 6n+15 are less than 150 and divisible by 15. i.e., 6n+15 divisible by 15. so, that, 6n must be divisible by 15. The multiples of 6 which are divisible by 15 are, 6x5 = 30, 6x10 = 60, 6x15 = 90,... Therefore, the possibilities of n are 5, 10, 15, 20,... If, n = 5, then, 6n+15 = sum of 6 consecutive integers = 5+6+7+8+9+10 = 45 (divisible by 15 and less than 150) If, n = 10, then, 6n+15 = sum of 6 consecutive integers = 10+11+12+13+14+15 = 75 (divisible by 15 and less than 150) If, n = 15, then, 6n+15 = sum of 6 consecutive integers = 15+16+17+18+19+20 = 105 (divisible by 15 and less than 150) If, n = 20, then, 6n+15 = sum of 6 consecutive integers = 20+21+22+23+24+25 = 135 (divisible by 15 and less than 150) If, n = 25, then, 6n+15 = sum of 6 consecutive integers = 25+26+27+28+29+30 = 165 (divisible by 15 and greater than 150). Hence, 4 numbers under 150 are divisible by 15 and can be made from the sum of six consecutive integers. Question 3 How many numbers less than 350 and greater than 100 that are divisible by 10, can be made from the sum of four consecutive even integers? a) 4 b) 6 c) 8 d) 10 Answer : b) 6 Solution : We know that, any even number can be in the form 2k. Let the four consecutive even integers be 2k, 2k+2, 2k+4 and 2k+6. Given that, 2k + 2k+2 + 2k+4 + 2k+6 = 8k + 12 We have to find number of possibilities of 8k+12 which is less than 350 and greater than 100 and divisible by 10. i.e., 8k+12 is divisible by 10. So, 8k+12 must end in 0 and 8k should end with 8. The multiples of 8 which are ending with 8 are, 8x1 = 8, 8x6 = 48, 8x11= 88, 8x16=128,.... Therefore, the unit’s digit of k must be 1 or 6. If, k = 1 then 8k+12 = sum of four consecutive even integers = 2+4+6+8 = 20 If, k = 6 then 8k+12 = sum of four consecutive even integers = 12+14+16+18 = 60 If, k = 11 then 8k+12 = sum of four consecutive even integers = 22+24+26+28 = 100 If, k = 16 then 8k+12 = sum of four consecutive even integers = 32+34+36+38 = 140 which is greater than 100. Note that, the sum are increased by 40. Therefore, the required sum of 8k+12 is 140, 180, 220, 260, 300 and 340. (100 < sum < 350) Hence, 6 numbers are less than 350 and greater than 100 are divisible by 10 and can be made from the sum of 4 consecutive even integers. |
| 20th November 2015 01:42 PM | |
| Unregistered | Oracle Solved Placement Papers Hi sir I had completed my B Tech degree from computer science and appearing for the off campus placement exam of oracle please provide me some previous year placement paper of oracle which help me in preparation? |
