Numerical Reasoning Test 2 Nlng Job

First, determine the profit made from the Treeline model sales.

Profit per Treeline unit is \(22\% \times 39,000 = 8,580\).
For 14 units, total sales profit is \(14 \times 8,580 = 120,120\).

Next, calculate the warranty bonus from Treeline sales.
Warranty bonus per unit is \(9\% \times 39,000 = 3,510\).
For 7 warranties, total warranty bonus is \(7 \times 3,510 = 24,570\).

Now calculate the warranty bonus from the Lumberman model.
Warranty bonus per Lumberman unit is \(8\% \times 48,000 = 3,840\).
For 12 warranties, total warranty bonus is \(12 \times 3,840 = 46,080\).

Adding all known profits gives:
\(120,120 + 24,570 + 46,080 = 190,770\).

Subtract this from the total company profit to get the profit made from Lumberman unit sales only:
\(363,570 - 190,770 = 172,800\).

Profit per Lumberman unit is \(20\% \times 48,000 = 9,600\).
Number of Lumberman units sold is therefore:
\(\dfrac{172,800}{9,600} = 18\).

This shows that 18 Lumberman units were sold.

First, find the selling price of one RV.
A markup of 22% means the selling price is the cost plus 22% of the cost:
\(\text{Markup per RV} = 0.22 \times 25,250 = 5,555\).
So, selling price per RV is:
\(25,250 + 5,555 = 30,805\).

Next, calculate the total revenue from selling 14 RVs:
\(30,805 \times 14 = 431,270\).

Now calculate the total cost of purchasing the 14 RVs:
\(25,250 \times 14 = 353,500\).

The gross profit before business expenses is:
\(431,270 - 353,500 = 77,770\).

The total cost of doing business for the month is \$45,000, which must be subtracted from the gross profit:
\(77,770 - 45,000 = 32,770\).

This is the actual profit made after all costs are accounted for.

Let Dan’s painting rate be \(d\) square feet per hour.
Since Betty paints twice as fast as Dan, her rate is \(2d\).

Working together, Dan and Betty paint at a combined rate of:
\(d + 2d = 3d\).

They paint 2,400 square feet in 4 hours, so their combined rate is:
\(\dfrac{2400}{4} = 600\) square feet per hour.
Thus:
\(3d = 600 \Rightarrow d = 200\).

So, Dan paints 200 square feet per hour, and Betty paints 400 square feet per hour.

When Sue joins them, the three of them paint 3,600 square feet in 3 hours, giving a combined rate of:
\(\dfrac{3600}{3} = 1200\) square feet per hour.

The combined rate of Dan and Betty is already 600 square feet per hour, so Sue’s rate is:
\(1200 - 600 = 600\) square feet per hour.

Now convert 4 hours 27 minutes to hours:
\(27\text{ minutes} = \dfrac{27}{60} = 0.45\) hours.
Total time = \(4 + 0.45 = 4.45\) hours.

The area Sue can paint alone in this time is:
\(600 \times 4.45 = 2670\) square feet.

This shows that Sue can paint 2,670 square feet in 4 hours and 27 minutes.

He started with \(\dfrac{3}{4}\) cup of water.
Fifty per cent more than what he started with means \(1.5 \times \dfrac{3}{4}\).\[\begin{flalign} &1.5 \times \dfrac{3}{4} = \dfrac{3}{2} \times \dfrac{3}{4} &&\\[6pt] &= \dfrac{9}{8} && \end{flalign}\]So, the target amount of water is \(\dfrac{9}{8}\) cup.

He poured half of the water into the batter:\[\begin{flalign} &\dfrac{1}{2} \times \dfrac{3}{4} = \dfrac{3}{8} && \end{flalign}\]Water left in the cup after pouring is:\[\begin{flalign} &\dfrac{3}{4} - \dfrac{3}{8} = \dfrac{3}{8} && \end{flalign}\]He then spilled \(\dfrac{1}{8}\) cup on the floor, leaving:\[\begin{flalign} &\dfrac{3}{8} - \dfrac{1}{8} = \dfrac{1}{4} && \end{flalign}\]To reach \(\dfrac{9}{8}\) cup from \(\dfrac{1}{4}\) cup, the extra water needed is:\[\begin{flalign} &\dfrac{9}{8} - \dfrac{1}{4} = \dfrac{9}{8} - \dfrac{2}{8} = \dfrac{7}{8} && \end{flalign}\]This means he must add \(\dfrac{7}{8}\) cup of water to what is left in the measuring cup.

To find the cheapest option, the cost per pen for each box must be calculated by dividing the total cost by the number of pens.

For the box of 25 pens:\[\begin{flalign} &\dfrac{\$7.19}{25} = \$0.2876 \text{ per pen} && \end{flalign}\]For the box of 50 pens:\[\begin{flalign} &\dfrac{\$10.78}{50} = \$0.2156 \text{ per pen} && \end{flalign}\]For the box of 75 pens:\[\begin{flalign} &\dfrac{\$14.53}{75} \approx \$0.1937 \text{ per pen} && \end{flalign}\]For the box of 100 pens:\[\begin{flalign} &\dfrac{\$22.50}{100} = \$0.225 \text{ per pen} && \end{flalign}\]
Comparing these values, the box of 75 pens has the lowest cost per pen, making it the cheapest option overall.

Having \(\dfrac{1}{4}\) fewer computers than employees means the number of computers is \(\dfrac{3}{4}\) of the number of employees.

Current number of computers is:\[\begin{flalign} &\dfrac{3}{4} \times 80 &&\\[6pt] &= 60 && \end{flalign}\]So, the company currently has 60 computers.

They want to hire 5% more employees. Five per cent of 80 is:\[\begin{flalign} &\dfrac{5}{100} \times 80 &&\\[6pt] &= 4 && \end{flalign}\]This increases the number of employees to \(80 + 4 = 84\).

To ensure that each employee has one computer, they will need 84 computers in total. Since they already have 60 computers, the number of computers they must buy is:\[\begin{flalign} &84 - 60 &&\\[6pt] &= 24 && \end{flalign}\]This means they need to buy 24 additional computers.

A discount of 20% means Joe pays 80% of the marked prices.

The discounted prices are:
Paper: \(0.8 \times 21.99 = 17.592\) dollars per box
Toner: \(0.8 \times 52.90 = 42.32\) dollars each
Folder: \(0.8 \times 0.99 = 0.792\) dollars each

Cost of 3 boxes of paper:\[\begin{flalign} &3 \times 17.592 = 52.776 && \end{flalign}\]Cost of 2 toner cartridges:\[\begin{flalign} &2 \times 42.32 = 84.64 && \end{flalign}\]Total spent on paper and toner:\[\begin{flalign} &52.776 + 84.64 = 137.416 && \end{flalign}\]Money left from the \$175 budget:\[\begin{flalign} &175 - 137.416 = 37.584 && \end{flalign}\]Number of folders he can buy is:\[\begin{flalign} &\dfrac{37.584}{0.792} \approx 47.47 && \end{flalign}\]Since Joe can only buy whole folders, he can buy 47 folders without exceeding his budget.

Jack’s initial typing speed at the beginning of the course was \(80\) words per minute.

At the end of the course, he typed a \(1,680\)-word document in \(20\) minutes, so his final typing speed is calculated as:\[\text{Final speed} = \dfrac{1680}{20} = 84\;\text{words per minute}\]
The increase in typing speed is the difference between the final and initial speeds:\[84 - 80 = 4\;\text{words per minute}\]
To find the percentage increase relative to the initial speed, divide the increase by the original speed and multiply by \(100\%\):\[\text{Percentage increase} = \dfrac{4}{80} \times 100\% = 5\%\]
This shows that from the beginning of the course to the end, Jack’s typing speed increased by \(5\%\).

It takes the old computer:
22 secs to load 1 graphic and, 9 seconds to load 1 paragraph.
Thus, it would talke 22 secs $\times$ 2 = 44 secs  to load 2 graphics.
Load time for 2 graphic and 1 paragraph:
= 44 + 9 = 53 secs.
The old computer takes 53 secs to load 2 graphic and 1 paragraph. 
Since the new computer is 17% faster, it means that it will take (100 - 17 = 83%) less time to load same document. 
Thus new computer load time will be 83% load time of old computer
= 0.83 $\times$ 53 secs = 43.99 secs

First, determine the production rate of the machine.

The machine produces 1,000 pieces in 2 hours, so the number of pieces produced per hour is:\[\frac{1000}{2} = 500\ \text{pieces per hour}\]
Since the machine runs for 24 hours each day, the total number of pieces produced per day is:\[500 \times 24 = 12{,}000\ \text{pieces per day}\]
Each box contains 50 pieces, so the number of boxes produced per day is:\[\frac{12{,}000}{50} = 240\ \text{boxes per day}\]
Each box sells for \$3.00, therefore the daily revenue is:\[240 \times 3 = \$720\ \text{per day}\]
The machine costs \$9,000. To find how many days it will take to recover this cost, divide the cost by the daily revenue:\[\frac{9000}{720} = 12.5\ \text{days}\]
This shows the machine will generate enough income to pay for itself after 12.5 days of continuous operation.