Three gold coins of weight 780gm, 840gm and 960gm are cut into small pieces, all of which have the equal weight.Each piece must be heavy as possible. If one such piece is shared by two persons, then how many persons are needed to give all the pieces of gold coins?
Answer: A HCF(780, 840, 960) = 60 Thus total number of pieces => 780/60 + 840/60 +960/60 = 13+14+16 = 43 Total number of person required = 43*2 = 86.
Q. No. 8:
The three words "GPL Assets Rock" are flashed such that the words individually are switched on at regular intervals of 4,7 and 9 seconds respectively and after they are switched on, the words are switched off after 2 sec, 4 sec and 5 sec respectively. If at time 't' all the words happened to switch off simultaneously, find the least time 't' which all three words will switch on simultaneously.
Answer: C The first word will switch on after 4x+ (4-2) seconds from t The second word will switch on after 7y+ (7-4) seconds from t The third word will switch on after 9z+ (9-5) seconds from t. => If at t', all the switches on together t' - t = 4x+2 = 7y+5 = 9z+4 We want the least value of t'-t for which x, y and z are integers. Check the options we get 94 the least option.
P =1234567891011.............9989991000
Q. No. 1:
Find the remainder when the number P is divided by 16?
Answer: D P is obtained by writing the first 1000 natural numbers from left to right. We need last four digits of the number formed by the first thousand digits to check the divisibility of 16. The first 99 natural number = 9+(90*2) = 189 digits. We need another 1000-189 = 811 digits. => another 811/3 = 270(approx) i.e 99+270 +one digit i.e end of the required number will look like .........................3683693 => remainder will be same as the remainder of 3693/16 = 13.
Answer: C We need to find sum of S of the digits of P. Till 999 each of the digits from 1 to 9 occurs equal number of times. Total digit of all the numbers from 1 to 999 will be multiple of Simga(9) = 45. Hence, S= 45k+(1+0+0+0) => Required remainder = 1.
Q. No. 10:
How many ordered pairs of integer (x,y)are there such that their product is a positive integer less than 100.
Answer: D Given 0 < xy <100 and x,y are integers. x and y are either both positive or both negative. Also given (x,y) is not equal to (y,x). If x =1 y can take values from 1 to 99 =>we have 99*2 = 198 pairs but (1,1) is repeated Thus can take 198-1 = 197 pairs. If x=2 , y can take values from 2 to 49 [(2,1) and (1,2) are also covered in 197 pairs above]. => 48*2 - 1 = 95 pairs [(2,2) is repeated] Similarly, If x=3 or y=3 we have 61 pairs If x=4 or y=4 we have 41 pairs If x=5 or y=5 we have 29 pairs If x=6 or y=6 we have 21 pairs If x=7 or y=7 we have 15 pairs If x=8 or y=8 we have 9 pairs If x=9 or y=9 we have 5 pairs We have total 473 pairs when x and y are positive. We will have 473 pairs when a and b are negative. => We have a total of 946 ordered pairs.