2007-08-18 20:52:40 · 19 answers · asked by Surender k 1 in Science & Mathematics ➔ Mathematics
let number of bugs for first person be x
therefore number of bugs for second is x + 2
hence total number of bugs = 2x + 2 = 100
hence 2x = 98
hence x = 49
thus first person gets 49
second gets 51
2007-08-18 20:55:14 · answer #1 · answered by trop de choses 1 · 2⤊ 1⤋
If we remove 2 bugs from the larger share both the persons will have equal quantity. From 100 remove 2 and the balance should be equally devided. Hence both will get 49 each. Initially removed 2 should be given to one of them. So he will get 49+2=51, and the other one will have the same 49 only. Hence shares are 49 and 51
2007-08-19 10:40:58 · answer #2 · answered by Joymash 6 · 0⤊ 0⤋
I know the answer easily! 51 and 49.
To explain, let x be the person getting the 2 less bugs.
Therefore x + 2 will be the person getting 2 bugs more.
so,
x + x + 2 = 100
2x + 2 = 100
2x = 98
x = 49.
That will be the person getting lesser bugs.
x + 2 = 49 + 2
= 51
Therefore the other person will get 51 bugs.
Hope this helps!
2007-08-19 06:50:16 · answer #3 · answered by noel 2 · 0⤊ 0⤋
Let x = number of bugs of the person which has 2 more than the other.
x + x -2 = 100
2x = 102
x = 51
Answer: the 1st person must have 51 and the other must have 49 (51 - 2):
Proof:
51 + (51 -2) = 100
51 + 49 = 100
100 = 100
2007-08-19 04:17:29 · answer #4 · answered by Jun Agruda 7 · 3⤊ 0⤋
easy.... just half 100 and add one for the second person and minus one for the first...person 1 gets 49 person 2 gets 51.
LONGER WAY
first person's share = x
second person's share = (x + 2).........because they get 2 more!
Both their shares MUST = 100 right? so that means:
x + (x + 2) = 100
and solve
x + x + 2 = 100
2x + 2 = 100
2x = 98
x = 49
so that means
first person has x which is 49 bugs
second person has (x + 2) which is 51 bugs!
2007-08-19 06:10:40 · answer #5 · answered by Anonymous · 0⤊ 0⤋
51 and 49
2007-08-19 08:02:54 · answer #6 · answered by Anonymous · 0⤊ 0⤋
Obviously, you could simply count out 49 bugs for one person, and give the remaining 51 to the other. A simpler procedure is to start by giving one person two bugs, then give a bug to each until you run out. That way, you don't even have to count them.
2007-08-19 03:57:35 · answer #7 · answered by Anonymous · 0⤊ 0⤋
Let th frist person's bugs be= x
x+x+2=100
x=49
so the solution is 49 & 51
2007-08-19 08:11:19 · answer #8 · answered by GB 2 · 0⤊ 0⤋
the distribution must be 51 and 49
2007-08-19 04:09:36 · answer #9 · answered by Calito Way 2 · 0⤊ 0⤋
the bugs of one person = x
and other = x+2
x+(x+2)=100 => 2x+2=100 => 2x=98 =>x=49
the bugs of first person is 49+2 = 51
2007-08-19 04:02:26 · answer #10 · answered by soha 2 · 0⤊ 0⤋
you have 100 bugs.
consider first person has 'x' bugs with him,
then 2nd one will have x+2 with him.
therefore, x+x+2=100,
2x+2=100
2x=100-2
2x=98,
x=49. so,1st has 49&2nd has 51.
2007-08-20 09:48:42 · answer #11 · answered by kesav 1 · 0⤊ 0⤋