Rates and Concentrations - Questions - Class Mathematics School Portal

Rates and Concentrations - Questions

Question 1

25966

A black rhinoceros can cover 100m in 6.5 seconds. What is its speed in km/hr?

Answer

$\text{55.4 km/hr}$

Worked Solution

Question 2

11377

Usain Bolt can sprint the 100$m$ in 9.58 seconds. His speed in $\text{km/h}$ is?

Answer

$\text{37.6 km/h}$

Worked Solution

Question 3

11378

The Earth's radius is approximately 6400 km. What speed does a point on the equator travel?

Answer

$1676\,\text{km/h}$

Worked Solution

Question 4

11357

A hippopotamus can run at $\text{30 km/h}$. Its speed in metres/second is?

Answer

$\text{8.3 m/sec}$

Worked Solution

Question 5

11358

The cost of a chemical is $\$22/100$$ \text{ ml}$. What is the cost of 11.5 litres?

Answer

$\$2530$

Worked Solution

Question 6

10237

Tony earns $24.50 per hour working in a store. How many hours does Tony work to earn $392?

Answer

$\text{16 hours}$

Worked Solution

Question 7

10234

The earth's radius is approximatly 6400$~\text{km}$.

What distance does a point on the equator travel in 1 hour.

Answer

$1676~\text{km}$

Worked Solution

Question 8

10236

The cost of a chemical is $64/100$~\text{ml}$. What is the cost of 20$~\text{ml}$?

Answer

$\$ 12.80$

Worked Solution

Question 9

10188

A car travels 400$~\text{km}$ on 25$~\text{L}$ of petrol. How far does it travel on 42$~\text{L}$

Answer

$672~\text{km}$

Worked Solution

Question 10

10080

$\text{A cheetah can cover 100 m in 4.5 seconds. Its speed in km/hour is?}$

Answer

$\text{80 km/h}$

Worked Solution

Question 11

119245

Potatoes cost $\$ 2.60 / k g$. How many kilograms of potatoes can be bought for $\$ 19.50$ ?

Answer

$7. 5 \text{ kg}$

Worked Solution

Question 12

25965

Anika is buying a used car for $12,000. Finance is available at a deposit of 10% and monthly repayments of $280 for 4 years.

(i) How much in total did she pay for the car? (2 marks)

(ii) What was the flat interest rate per annum? (2 marks)

Answer

(i) $14,640

(ii) 6.1% p.a

Worked Solution

\begin{align}
\text{(i)}\quad \text{Deposit}&= 12000\times 10\%\\
&=\$1200\\
\text{Repayments} & = 280 \times 48 \\
&=\$ 13440\\
\text{Total}&=1200+13440\\
&=\$14540
\end{align}

\begin{align}
\text{(ii)}\quad \text{Amount borrowed} & = 12000 - 1200 \\
& = \$10800\\
\text{Interest} & = 14640 - 12000\\
&=\$2640\\
\%\text{pa}&=\frac{2640}{10800}\times 100\% \div 4\\
&=6.1\% \text{p.a}
\end{align}