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a swimmer plans to swim at least 100 laps during a 6 day period. During this period, the swimmer will increase the number of laps completed each day by one. Whats the least number of laps the swimmer must complete on the first day ?


Im not asking for answer, if someone could just show me how to set up the equation for this question

2007-12-30 03:22:31 · 8 answers · asked by helppmee =] 2 in Science & Mathematics Mathematics

8 answers

The number of laps on day one is x

x+(x+1)+(X+2)+(X+3)+(X+4)+(X+5)
must be equal to or greater than 100

2007-12-30 03:33:10 · answer #1 · answered by frozen 5 · 0 0

100=x+x+1+x+2+x+3+x+4+x+5=6x+15
6x=85 but 85 is not divisible by 6. 84 is but that would give 99 laps so 6x must be at least 90 and x=15. Check
15+16+17+18+19+20=105. If you start with 14 the total will be 99 so the least amount of laps on the first day is 15

2007-12-30 11:31:34 · answer #2 · answered by oldschool 7 · 0 0

x=least number of laps the swimmer must complete on the first day
x+1= laps on the second day
x+2= laps on the third day
x+3= laps on the fourth day
x+4= laps on the fifth day
x+5= laps on the sixth day

therefore,
x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=100
6x+15=100
6x=85
x=14

since the answer is approximately 14.1, we round up because the swimmer plans to swim at least 100 laps,
so we can't have 99, 98, or anything like that.

so the answer is 15 laps.

2007-12-30 11:31:50 · answer #3 · answered by nightowl 2 · 0 0

If you assume that least number of laps on first day as x, then the number of laps on 6th day = x +5

This series can be represented as arithmetic progression with
fitst term as x and last term as x + 5 with a common difference of 1

The sum of n terms in Arithmetic progression is

S = n/2[a1 + an], where a1 = first term, an = last term, n = number of terms and S = sum

here n = 6, a1 = x and an = x+5 and S = 100

so 6/2[x + x + 5] = 100

3(2x+5) = 100

6x + 15 = 100

6x = 85

x = 85/6 = 14 1/6

so minimum laps on first day should be 15

2007-12-30 11:42:15 · answer #4 · answered by mohanrao d 7 · 0 0

If you are learning Arithmetic Series, and know the summation formula, you can set up the equation:
S = N/2(A + (N-1)D)
wherer A-starting number
N= no of terms
D = increment
100 = 6/2(A + 5*1)
= 3A + 15
Therefore A = 14.17
So start with 15 laps on first day and go on. Total steps will be 105..
-TJ

2007-12-30 12:05:46 · answer #5 · answered by T.J. 3 · 0 0

Day 1: x
Day 2: x+1
Day 3: x+2
Day 4: x+3
Day 5: x+4
Day 6: x+5

(x)+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)>/= 100

i think this is right. When you're done, round x up to the nearest whole number since you can't do like a third of a lap.

2007-12-30 11:30:16 · answer #6 · answered by Kavin 2 · 0 0

you should assign a variable for the number of laps.. since the swimmer will be doing a 100 laps in 6 days, and it's incremental by one everyday..
the first day should be X
second day should be x+1
3rd day x+2
until the 6th day -- x+5
sum them up together,and they should be not more than 100
if you will able to get the value for x.. then you can try to test and verify your equation..

2007-12-30 11:51:15 · answer #7 · answered by ces 3 · 0 0

x = laps 1st day
x+1 = laps 2nd day
.........................
...........................
x+5 = laps 6th day
Add them up getting:
6x+15 => 100

There you go

2007-12-30 11:37:03 · answer #8 · answered by ironduke8159 7 · 0 0

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