A few days ago
can you help please with these two math problems?
the first one is use 3 consecutive numbers that add to 201
the second is use 2 consecutive numbers and a number which is double 1 of them
again they all have to add to 201
thanks
Top 8 Answers
A few days ago
Favorite Answer
Can I assume you’re talking about consecutive whole positive numbers?
let Y = the first number
let y + 1 = the next number
let (y + 1) + 1 = the last number
Add them all together, set them equal to 201 and solve for y.
1
A few days ago
the 3 consecutive numbers are 66+67+68!
0
A few days ago
first problem:
x + (x + 1) + (x +2) = 201
3x + 3 = 201
3x = 198
x = 66
therefore numbers are: 66, 67 and 68
second problem:
x + (x + 1) + (2x) = 201
4x + 1 = 201
4x = 200
x = 50
therefore the numbers are 50 and 51
1
A few days ago
u should try and do it yourself, but i’ll give you some clues…
for the first one, the first number is in the 60’s…
for the second one, the number that doesn’t need a double in the sum is in the fifties…
1
A few days ago
1st question: 66+67+68=201.
For the second question, it’s 50+51+100=201.
0
A few days ago
first one is: 66 + 67 + 68
second on is: 50 + 51 + (50*2)
easy!!
1
A few days ago
66,67,68 is the first one
50, 51, 100, is the second
1
A few days ago
the answer to first is 66,67,68 the second i dont get the second.LATER!!!!!!!!!!
0
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