Mathathon 2020 Problem 3 $-$ Medium

• The three weights are $(H, K, T)$ . Let them correspond to $(a,b,c)$ for clarity in the solution.

$\begin{cases} a+b &= 85 \quad &\ldots (1) \\ a+c &= 75 \quad &\ldots (2) \\ b+c &= 60 \quad &\ldots (3) \end{cases}$

• Adding $(1), (2), (3)$

$\begin{aligned}2(a+b+c) &= 220 \\ a+b+c &= 110 &\ldots(4) \end{aligned}$

• Subtract $(1), (2), (3)$ one at a time from equation $(4)$ to get the values

$\begin{cases} (4) - (1) \implies c = 25 \\ (4) - (2) \implies b = 35 \\ (4) - (3) \implies a = 50 \end{cases}$

• $(50, 35, 25) \implies (H,K,T)$ . Now we evaluate the expression

$\dfrac{K}{\sqrt{T}} - H \quad \implies \quad \dfrac{35}{\sqrt{25}} - 50 \quad \implies \quad \boxed{-43}$

I used substituting:

From the sums (H + K = 85, H + T = 75 and K + T = 60), we can conclude that T + 10 = K and K + 15 = H

Therefore, K + T = T + 10 + T = 2T + 10 = 60

Subtracting 10 from both sides: 2T = 60 - 10 = 50

Dividing 2 from both sides: T = 50 / 2 = 25

Square-rooting 25 makes 5

From T + 10 = K, we get K = 35

From K + 15 = H, we get H = 50

35 / 5 - 50 = 7 - 50 = -43

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Author's Notes

I don't know how to use LaTeX

H means HH's weight and same with the others

H + K = 85 (1)

H + T = 75 (2)

T + K = 60 (3)

Adding all equations...

2H + 2T + 2K = 220

H + T + K = 110 (4)

According to (4) - (1):

T = 25

According to (2) & (3):

H = 50, K = 35

K/\sqrt{1} - H

= 35 / \sqrt{25}} - 50

= 35 / 5 - 50

= -43

$\text{ As a first step we can write the given combinations of their weights as a system of linear equations.}$

$\textit{I}\quad H + K = 85kg$

$\textit{II}\quad H + T = 75kg$

$\textit{III}\quad K + T = 60kg$

$T = K - 10kg$

$K = H - 15kg$

$T = H -25kg$

$\text{Now we can take our 2nd equation and substitute T with H - 25. After that we can solve for H.}$

$\textit{II}\quad H + (H - 25kg) = 75kg$

$2 \cdot H - 25k = 75kg$

$2 \cdot H = 100kg$

$H = 50kg$

$K = H - 15kg = 35kg$

$T = H - 25kg = 25kg$

$\frac{K}{ (\sqrt{T}) } - H = \frac{35kg}{\sqrt{25kg}} - 50 kg$

$\text{For the solution to work out we now need to neglect units.}$

$\frac{35} {\sqrt{25} } - 50 = (\frac{35}{5}) - 50$

$7 - 50 = -43$

$\frac{K}{\sqrt{T}} - H = -43$

$\text{We know that that when Tim got on and Kate got off, the reading went down 10 kilograms. So } T = K - 10. \\ \text{We also know that when Kate got on and Horace got off, the reading went down 15 kilograms. So } K = H - 15. \\ \text{Finally, we know that } H + K = 85.$

$\begin{aligned} \color{#20A900}T \color{#20A900}&= \color{#20A900}K \color{#20A900}- \color{#20A900}10\\ \color{#D61F06}K \color{#D61F06}&= \color{#D61F06}H \color{#D61F06}- \color{#D61F06}15\\ \color{#3D99F6}H \color{#3D99F6}+ \color{#3D99F6}K \color{#3D99F6}&= \color{#3D99F6}85\\ \text{Substituting the red equation into the } \color{#3D99F6}\text{blue} \text{ equation:}\\ 2H - 15 &= 85\\ H &= 50\\ \text{Substituting this result into the } \color{#D61F06}\text{red} \text{ equation:}\\ K &= 50 - 15\\ K &= 35\\ \text{Substituting this result into the } \color{#20A900}\text{green} \text{ equation:}\\ T &= 35 - 10\\ T &= 25\\ \text{Substituting these answers into the required answer format:}\\ \displaystyle \frac{K}{\sqrt{T}} -H &= \displaystyle \frac{35}{\sqrt{25}} - 50\\ &= 7 - 50\\ &= \boxed{-43} \end{aligned}$

$h + k = 85$ $h + t = 75$ $t + k = 60$

$\text{When Tiny Tim got on instead of Curly Kate the weight dropped by 10, which means:}$

$t + 10 = k$

$\text{Substitute k for t + 10 in the 3rd equation}$

$t + t + 10 = 60$ $2t = 50$ $\boxed{t = 25}$

$\text{Now that we know Tiny Tim is 25 kg:}$

$h + t = 75$ $h +25 = 75$ $\boxed{h = 50}$

$t + k = 60$ $25 + k = 60$ $\boxed{k = 35}$

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$\Large{\frac{k}{\sqrt{t}} - h = \frac{35}{\sqrt{25}} - 50 = 7 - 50}$

$\huge{= -43}$

This is equal to $\begin{cases} H+K=85,(1)\\ H+T=75,(2)\\ T+K=60.(3) \end{cases}$ Now calculate $(2)-(3)$ . You get $H-K=15\Rightarrow H=K+15.$
Plug this in $(1)$ to get $2K+15=85$ , i.e. $K=35$ . Plug this in $(1)$ again to get $H=50$ . The value of T comes immediately: $T=25$ .
So the answer is $\dfrac{35}{\sqrt{25}}-50=\boxed{\colorbox{#CEBB00}{-43}}.$