# Trying to figure out what is wrong with my logic

I have been given this question: Sam’s last two times for the $10,000$ metres added up to $71.4$ minutes. Her second time was $10\%$ slower than her first. Next season Sam wants to improve her first time by $5\%$

I evaluated in this way:

1st try = $x$

2nd try = $y$

$y = x+10\%$

THEREFORE - I think my formula to evaluate should be: $x+(x+10\%)=71.4$

I think this evaluates to $x = 35.65$ minutes SO - when I then take $x+10\%$ I get $39.215$ and this should be $Y$.

BUT $35.65 + 39.215 = 74.865$

What am I doing wrong?

• Hint: $x+10\%$ means $1.1x$. – TonyK Mar 8 '17 at 1:36
• Intervals of time are not slower or faster but they can be larger or smaller, Velocities can be slower. – DanielWainfleet Mar 8 '17 at 3:02
• $y= x + 10%$ is not just wrong but is meaningless. What you want is $y =.9*x$ – fleablood Mar 8 '17 at 3:41

As @TonyK said in the comments $x+10\%$ is equivalent to saying $1.1x$

$$x = \text{ first try}$$ $$y = \text{ second try}$$ $$x + (x\cdot 1.1) = 71.4 \,\,\Rightarrow x=34$$ $$y = 71.4 - x = 37.4$$

How the formula works:

As you said, if $x$ is the first run, then $x + 10\%$ (x plus ten percent of x) is the time of her second run. Well, $10\%$ of $x$ is the same as saying $\frac{10}{100} \cdot x = 0.1x$. Making $x+0.1x=1.1x$ be the time of the second run. So the sum of the first two attempts ($x + 1.1x$) is the total distance, $x + 1.1x = 71.4$ and solve this for $x$ and you get $x=34$

• sooo - x = 37.578947 – kiltannen Mar 8 '17 at 1:42
• If that is correct then y must equal 33.821053 – kiltannen Mar 8 '17 at 1:42
• BUT in the original question y being the 2nd try is actually 10% longer in time than x? – kiltannen Mar 8 '17 at 1:43
• my fault. I labeled it incorrectly – Dando18 Mar 8 '17 at 1:48
• @kiltannen I fixed it. – Dando18 Mar 8 '17 at 1:52