A sheet of rectangular paper has length $l$ cm and breadth $b$ cm. If identical squares of side 3 cm are to be cut out from this rectangular paper, there will not be any leftovers. If we rearrange some of these identical rectangular papers to form a square, the smallest possible square formed will have side 30 cm. Given that both $l$ and $b$ are greater than 3 cm and $l > b$, find the values of $l$ and $b$. Answer $l =$ ... $b =$ ... The time taken by Shayne to swim a lap is 40% more than the time taken by Kai.
Is it true that the time taken by Kai was 60% of the time by Shayne?
Explain your answers with appropriate workings. A sweet drink is made by mixing concentrated syrup and water together.
The ratio of syrup to water in a 200 ml sweet drink is 9 : 11.
Find the volume of water that must be added to the 200 ml sweet drink so that the percentage concentration of syrup will be reduced by 10%. Answer ... ml

1. Use the 'no leftover' condition to fix divisibility of $l$ and $b$

Since identical $3\text{ cm}\times 3\text{ cm}$ squares can be cut from the paper with no leftovers, both dimensions must be exact multiples of $3$:
$l = 3m,\qquad b = 3n,\quad m,n \text{ positive integers.}$
With $l,b > 3$, both $m$ and $n$ are at least $2$.

2. Use the 'smallest square = 30 cm' condition

The smallest square that can be formed by tiling identical $l\times b$ rectangles has side equal to the lowest common multiple of $l$ and $b$. Hence
$\operatorname{lcm}(l,b) = 30.$
This forces both $l$ and $b$ to be factors of $30$, and (so that the rectangles actually rearrange into a $30\times30$ square in two directions) both must be smaller than $30$.

3. Find the unique pair satisfying all conditions

Factors of $30$ that are multiples of $3$, greater than $3$ and less than $30$:
$6 \;(=3\times2),\qquad 15 \;(=3\times5).$
Check: $\operatorname{lcm}(15,6)=30.$ ✓ (A $30\times30$ square is tiled by a $2\times5$ array of $15\times6$ rectangles.)
Applying $l>b$:
$l = 15,\qquad b = 6.$

1. Express Shayne's time in terms of Kai's time

Let the time taken by Kai be $K$ and the time taken by Shayne be $S$.
Since Shayne takes $40\%$ more time than Kai:
$S = K + 0.40K = 1.4K$

2. Calculate Kai's time as a percentage of Shayne's time to verify the claim

We need to find $\frac{K}{S} \times 100\%$.
$\frac{K}{1.4K} \times 100\% = \frac{1}{1.4} \times 100\% = \frac{10}{14} \times 100\% \approx 71.4\%$
Since $71.4\% \neq 60\%$, the statement is false.

Alternatively, $60\%$ of Shayne's time is $0.6S = 0.6(1.4K) = 0.84K$, which is not equal to $K$.

1. Find the initial volume of syrup and its percentage concentration.

Total parts = $9 + 11 = 20$
Volume of syrup = $\frac{9}{20} \times 200 = 90$ ml
Initial percentage concentration = $\frac{90}{200} \times 100\% = 45\%$

2. Determine the new percentage concentration.

New percentage concentration = $45\% - 10\% = 35\%$

3. Form an equation for the new concentration with $x$ ml of water added.

Let $x$ be the volume of water added.
New total volume = $200 + x$
$\frac{90}{200 + x} = \frac{35}{100}$

4. Solve for $x$.

$\frac{90}{200 + x} = \frac{7}{20}$
$1800 = 7(200 + x)$
$1800 = 1400 + 7x$
$7x = 400$
$x = \frac{400}{7} = 57\frac{1}{7}$