L.C. MATHS

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## Question 1

Mary threw a ball onto level ground from a height of 2 m. Each time the ball hit the ground it bounced back up to $$\dfrac{3}{4}$$ of the height of the previous bounce, as shown.

(a) Complete the table below to show the maximum height, in fraction form, reached by the ball on each of the first four bounces.

Bounce $$1$$ $$2$$ $$3$$ $$4$$

Height (m)

$$\dfrac{2}{1}$$

Bounce $$1$$ $$2$$ $$3$$ $$4$$

Height (m)

$$\dfrac{2}{1}$$

$$\dfrac{3}{2}$$

$$\dfrac{9}{8}$$

$$\dfrac{27}{32}$$

Solution
Bounce $$1$$ $$2$$ $$3$$ $$4$$

Height (m)

$$\dfrac{2}{1}$$

$$\dfrac{3}{2}$$

$$\dfrac{9}{8}$$

$$\dfrac{27}{32}$$

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(b) Find, in metres, the total vertical distance (up and down) the ball had travelled when it hit the ground for the $$5$$th time. Give your answer in fraction form.

$$\dfrac{653}{53}\mbox{ m}$$

Solution

\begin{align}d&=2+2\left(\frac{3}{2}+\frac{9}{8}+\frac{27}{32}+\frac{81}{128}\right)\\&=\frac{653}{53}\mbox{ m}\end{align}

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(c) If the ball were to continue to bounce indefinitely, find, in metres, the total vertical distance it would travel.

$$14\mbox{ m}$$

Solution

\begin{align}d&=2+2S_{\infty}\\&=2+2\left(\frac{a}{1-r}\right)\\&=2+2\left(\frac{\frac{3}{2}}{1-\frac{3}{4}}\right)\\&=14\mbox{ m}\end{align}

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## Question 2

Solve the equation $$x^3-3x^2-9x+11=0$$.
Write any irrational solution in the form $$a+b\sqrt{c}$$, where $$a,b,c\in\mathbb{Z}$$.

$$x=1$$, $$x=1-2\sqrt{3}$$ and $$1+2\sqrt{3}$$.

Solution

\begin{align}f(1)&=1^3-3(1^2)-9(1)+11\\&=1-3-9+11\\&=0\end{align}

\begin{align}\downarrow\end{align}

$$(x-1)$$ is a factor

\begin{align}\downarrow\end{align}

$\require{enclose} \begin{array}{rll} x^2-2x-11\phantom{000000}\, \\[-3pt] x-1 \enclose{longdiv}{\,x^3-3x^2-9x+11} \\[-3pt] \underline{x^3-x^2\phantom{0000000000}\,\,} \\[-3pt] -2x^2-9x\phantom{000}\,\\[-3pt]\phantom{0000}\underline{-2x^2+2x\phantom{00}\,}\\[-3pt]\phantom{00}-11x+11\\[-3pt]\phantom{00}\underline{-11x+11}\\[-3pt]0 \end{array}$

\begin{align}\downarrow\end{align}

\begin{align}(x^2-2x-11)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-11)}}{2(1)}\\&=\frac{2\pm\sqrt{48}}{2}\\&=\frac{2\pm4\sqrt{3}}{2}\\&=1\pm2\sqrt{3}\end{align}

Therefore, the solutions are $$x=1$$, $$x=1-2\sqrt{3}$$ and $$1+2\sqrt{3}$$.

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## Question 3

Let $$f(x)=-x^2+12x-27$$, $$x\in\mathbb{R}$$.

(a)

(i) Complete Table $$1$$ below.

$$x$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$

$$f(x)$$

$$0$$

$$5$$

$$8$$

(ii) Use Table $$1$$ and the trapezoidal rule to find the approximate area of the region bounded by the graph of $$f$$ and the $$x$$-axis.

(i)

$$x$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$

$$f(x)$$

$$0$$

$$5$$

$$8$$

$$9$$

$$8$$

$$5$$

$$0$$

(ii) $$35\mbox{ square units}$$

Solution

(i)

$$x$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$

$$f(x)$$

$$0$$

$$5$$

$$8$$

$$9$$

$$8$$

$$5$$

$$0$$

(ii)

\begin{align}A&=\frac{h}{2}[y_1+y_n+2(y_2+y_3+…+y_{n-1})]\\&=\frac{1}{2}[0+0+2(5+8+9+8+5)]\\&=35\mbox{ square units}\end{align}

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(b)

(i) Find $$\int_3^9f(x)\,dx$$.

(ii) Use your answers above to find the percentage error in your approximation of the area, correct to one decimal place.

(i) $$36\mbox{ square units}$$

(ii) $$2.8\%$$

Solution

(i)

\begin{align}\int_3^9f(x)\,dx&=\int_3^9(-x^2+12x-27)\,dx\\&=\left.-\frac{x^3}{3}+6x^2-27x\right|_3^9\\&=\left(-\frac{9^3}{3}+6(9^2)-27(9)\right)-\left(-\frac{3^3}{3}+6(3^2)-27(3)\right)\\&=(-243+486-243)-(-9+54-81)\\&=36\mbox{ square units}\end{align}

(ii)

\begin{align}\mbox{Error}&=\frac{36-35}{36}\times100\\&=2.8\%\end{align}

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## Question 4

(a) The complex numbers $$z_1$$, $$z_2$$ and $$z_3$$ are such that $$\dfrac{2}{z_1}=\dfrac{1}{z_2}+\dfrac{1}{z_3}$$, $$z_2=2+3i$$ and $$z_3=3-2i$$, where $$i^2=-1$$. Write $$z_1$$ in the form $$a+bi$$, where $$a,b\in\mathbb{Z}$$.

$$5+i$$

Solution

\begin{align}\frac{2}{z_1}&=\frac{1}{2+3i}+\frac{1}{3-2i}\\&=\frac{3-2i+2+3i}{(2+3i)(3-2i)}\\&=\frac{5+i}{6-4i+9i-6i^2}\\&=\frac{5+i}{12+5i}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}z_1&=\frac{24+10i}{5+i}\\&=\frac{24+10i}{5+i}\times\frac{5-i}{5-i}\\&=\frac{120-24i+50i-10i^2}{25-5i+5i-i^2}\\&=\frac{130+26i}{26}\\&=5+i\end{align}

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(b) Let $$\omega$$ be a complex number such that $$w^n=1$$, $$\omega\neq 1$$ and $$S=1+\omega+\omega^2+…+\omega^{n-1}$$. Use the formula for the sum of a finite geometric series to write the value of $$S$$ in its simplest form.

$$S=0$$

Solution

\begin{align}a=1&&r=\omega\end{align}

\begin{align}\downarrow\end{align}

\begin{align}S&=\frac{a(1-r^n)}{1-r}\\&=\frac{1(1-\omega^n)}{1-\omega}\\&=\frac{1(1-1)}{1-\omega}\\&=0\end{align}

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## Question 5

(a) Solve the equation $$x=\sqrt{x+6}$$, $$x\in\mathbb{R}$$.

$$x=3$$

Solution

\begin{align}x=\sqrt{x+6}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x^2=x+6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x^2-x-6=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(x+2)(x-3)=0\end{align}

\begin{align}\downarrow\end{align}

$$x=-2$$ or $$x=3$$

$\,$

$$\mathbf{x=-2}$$

\begin{align}-2=\sqrt{-2+6}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-2=\sqrt{4}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-2=2\end{align}

\begin{align}\downarrow\end{align}

Ignored

$\,$

$$\mathbf{x=3}$$

\begin{align}3=\sqrt{3+6}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3=\sqrt{9}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3=3\end{align}

Therefore, the only solution is $$x=3$$.

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(b) Differentiate $$x-\sqrt{x+6}$$ with respect to $$x$$.

$$1-\dfrac{1}{2\sqrt{x+6}}$$

Solution

\begin{align}y(x)=x-(x+6)^{1/2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y'(x)&=1-\frac{1}{2}(x+6)^{-1/2}(1)\\&=1-\frac{1}{2\sqrt{x+6}}\end{align}

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(c) Find the co-ordinates of the turning point of the function $$y=x-\sqrt{x+6}$$, $$x\geq-6$$.

$$\left(-\dfrac{23}{4},-\dfrac{25}{4}\right)$$

Solution

\begin{align}y'(x)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}1-\frac{1}{2\sqrt{x+6}}=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2\sqrt{x+6}=1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\sqrt{x+6}=\frac{1}{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x+6=\frac{1}{4}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=\frac{1}{4}-6\\&=-\frac{23}{4}\end{align}

and

\begin{align}y\left(-\frac{23}{4}\right)&=-\frac{23}{4}-\sqrt{-\frac{23}{4}+6}\\&=-\frac{23}{4}-\sqrt{\frac{1}{4}}\\&=-\frac{23}{4}-\frac{1}{2}=-\frac{25}{4}\end{align}

\begin{align}\downarrow\end{align}

Turning point: $$\left(-\dfrac{23}{4},-\dfrac{25}{4}\right)$$

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## Question 6

(a) Donagh is arranging a loan and is examining two different repayment options.

(i) Bank A will charge him a monthly interest rate of $$0.35\%$$. Find, correct to three
significant figures, the annual percentage rate (APR) that is equivalent to a monthly
interest rate of $$0.35\%$$.

(ii) Bank B will charge him a rate that is equivalent to an APR of $$4.5\%$$. Find, correct to three significant figures, the monthly interest rate that is equivalent to an APR of $$4.5\%$$.

(i) $$4.28\%$$

(ii) $$0.367\%$$

Solution

(i)

\begin{align}(1+i)^t&=(1+0.0035)^{12}\\&=1.042818…\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\mbox{APR}\approx4.28\%\end{align}

(ii)

\begin{align}(1+i)^{12}=1.045\end{align}

\begin{align}\downarrow\end{align}

\begin{align}i&=\sqrt[12]{1.045}-1\\&=0.0036748…\\&\approx0.367\%\end{align}

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(b) Donagh borrowed €$$80{,}000$$ at a monthly interest rate of $$0.35\%$$, fixed for the term of the loan, from Bank $$A$$. The loan is to be repaid in equal monthly repayments over ten years. The first repayment is due one month after the loan is issued. Calculate, correct to the nearest euro, the amount of each monthly repayment.

$$818\mbox{ euro}$$

Solution

\begin{align}A&=P\left[\frac{i(1+i)^t}{(1+i)^t-1}\right]\\&=80{,}000\left[\frac{(0.0035)(1.0035)^{120}}{(1.0035)^{120}-1}\right]\\&\approx818\mbox{ euro}\end{align}

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## Question 7

A plane is flying horizontally at $$P$$ at a height of $$150\mbox{ m}$$ above level ground when it begins its descent.
$$P$$ is $$5\mbox{ km}$$, horizontally, from the point of touchdown $$O$$. The plane lands horizontally at $$O$$.

Taking $$O$$ as the origin, $$(x,f(x))$$ approximately describes the path of the plane’s descent where $$f(x)=0.0024x^3+0.018x^2+cx+d$$, $$-5\leq x\leq 0$$, and both $$x$$ and $$f(x)$$ are measured in $$\mbox{km}$$.

(a)

(i) Show that $$d=0$$.

(ii) Using the fact that $$P$$ is the point $$(-5,0.15)$$, or otherwise, show that $$c=0$$.

Solution

(i)

\begin{align}f(0)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0.0024(0^3)+0.018(0^2)+c(0)+d=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}d=0\end{align}

as required.

(ii)

\begin{align}f(-5)=0.15\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0.0024(-5)^3+0.018(-5)^2+c(-5)=0.15\end{align}

\begin{align}\downarrow\end{align}

\begin{align}c&=\frac{0.15-0.0024(-5)^3-0.018(-5)^2}{-5}\\&=0\end{align}

as required.

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(b)

(i) Find the value of $$f'(x)$$, the derivative of $$f(x)$$, when $$x=-4$$.

(ii) Use your answer to part (b) (i) above to find the angle at which the plane is descending when it is $$4\mbox{ km}$$ from touchdown. Give your answer correct to the nearest degree.

(i) $$-0.0288$$

(ii) $$2^{\circ}$$

Solution

(i)

\begin{align}f'(x)=0.0072x^2+0.036x\end{align}

\begin{align}\downarrow\end{align}

\begin{align}f'(-4)&=0.0072-4^2)+0.036(-4)\\&=-0.0288\end{align}

(ii)

\begin{align}\tan\theta=-0.0288\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\theta=178.35…^{\circ}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\alpha&=180^{\circ}-178.35…^{\circ}\\&\approx2^{\circ}\end{align}

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(c) Show that $$(-2.5,0.075)$$ is the point of inflection of the curve $$y=f(x)$$.

Solution

\begin{align}f^{\prime\prime}(x)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0.0144x+0.036=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=-\frac{0.036}{0.0144}\\&=-2.5\end{align}

and

\begin{align}f(-2.5)&=0.0024(-2.5)^3+0.018(-2.5)^2\\&=0.075\end{align}

as required.

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(d)

(i) If $$(x,y)$$ is a point on the curve $$y=f(x)$$, verify that $$(-x-5,-y+0.15)$$ is also a point
on $$y=f(x)$$.

(ii) Find the image of $$(-x-5,-y+0.15)$$ under symmetry in the point of inflection.

(ii) $$(x,y)$$

Solution

(i)

\begin{align}f(-x-5)&=0.0024(-x-5)^3+0.018(-x-5)^2\\&=0.0024(-x^3-15x^2-75x-125)+0.018(x^2+10x+25)\\&=-(0.0024x^3+0.018x^2)+0.15\\&=-y+0.15\end{align}

as required.

(ii)

\begin{align}(-x-5,-y+0.15)&\rightarrow(-2.5,0.075)\\&\rightarrow(-x-5+(x+2.5),0.075+(y-0.075))\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(-2.5,0.075)&\rightarrow(-2.5+(x+2.5),0.075+(y-0.075))\\&\rightarrow(x,y)\end{align}

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## Question 8

An oil-spill occurs off-shore in an area of calm water with no currents. The oil is spilling at a rate of $$4\times10^6\mbox{ cm}^3$$ per minute. The oil floats on top of the water.

(a)

(i) Complete the table below to show the total volume of oil on the water after each of the first $$6$$ minutes of the oil-spill.

Time (minutes) $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$

Volume ($$10^6 \mbox{ cm}^3$$)

$$8$$

(ii) Draw a graph to show the total volume of oil on the water over the first 6 minutes.

(iii) Write an equation for $$V(t)$$, the volume of oil on the water, in $$\mbox{cm}^3$$, after $$t$$ minutes.

(i)

Time (minutes) $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$

Volume ($$10^6 \mbox{ cm}^3$$)

$$4$$

$$8$$

$$12$$

$$16$$

$$20$$

$$24$$

(ii)

(iii) $$V(t)=(4\times10^6)t$$

Solution

(i)

Time (minutes) $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$

Volume ($$10^6 \mbox{ cm}^3$$)

$$4$$

$$8$$

$$12$$

$$16$$

$$20$$

$$24$$

(ii)

(iii)

\begin{align}V(t)=(4\times10^6)t\end{align}

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(b) The spilled oil forms a circular oil slick 1 millimetre thick.

(i) Write an equation for the volume of oil in the slick, in $$\mbox{cm}^3$$, when the radius is $$r\mbox{ cm}$$.

(ii) Find the rate, in cm per minute, at which the radius of the oil slick is increasing when the radius is $$50\mbox{ m}$$.

(i) $$0.1\pi r^2$$

(ii) $$1{,}273.3\mbox{ cm/min}$$

Solution

(i)

\begin{align}V&=\pi r^2h\\&=\pi r^2(0.1)\\&=0.1\pi r^2\end{align}

(ii)

\begin{align}V=0.1\pi r^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{dV}{dr}=0.2\pi r\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{dr}{dt}&=\frac{dr}{dV}\times\frac{dV}{dt}\\&=\frac{1}{0.2\pi r}\times(4\times10^6)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{dr}{dt}(r=5{,000})&=\frac{1}{0.2\pi (5{,}000)}\times(4\times10^6)\\&\approx1{,}273.3\mbox{ cm/min}\end{align}

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(c) Show that the area of water covered by the oil slick is increasing at a constant rate of $$4\times10^7\mbox{ cm}$$ per minute.

Solution

\begin{align}A=\pi r^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{dA}{dr}=2\pi r\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{dA}{dt}&=\frac{dA}{dr}\times\frac{dr}{dt}\\&=(2\pi r)\left(\frac{4\times10^6}{0.2\pi r}\right)\\&=4\times10^7\mbox{ cm}^2\mbox{/min}\end{align}

as required.

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(d) The nearest land is $$1\mbox{ km}$$ from the point at which the oil-spill began. Find how long it will take for the oil slick to reach land. Give your answer correct to the nearest hour.

$$13\mbox{ hr}$$

Solution

\begin{align}A&=\pi r^2\\&=\pi(10^5)^2\\&=\pi10^{10}\mbox{ cm}^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t&=\frac{\pi10^{10}}{4\times10^7}\\&=785.398..\mbox{ min}\\&\approx13\mbox{ hr}\end{align}

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## Question 9

The approximate length of the day in Galway, measured in hours from sunrise to sunset, may be calculated using the function

\begin{align}f(t)=12.25+4.75\sin\left(\frac{2\pi}{365}t\right)\end{align}

where $$t$$ is the number of days after March 21st and $$\left(\frac{2\pi}{365}t\right)$$ is expressed in radians.

(a) Find the length of the day in Galway on June 5th ($$76$$ days after March 21st). Give your answer in hours and minutes, correct to the nearest minute.

$$16\mbox{ hr }50\mbox{ min}$$

Solution

\begin{align}f(76)&=12.25+4.75\sin\left(\frac{2\pi}{365}(76)\right)\\&=16.837…\mbox{ hr}\\&\approx16\mbox{ hr }50\mbox{ min}\end{align}

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(b) Find a date on which the length of the day in Galway is approximately $$15$$ hours.

$$26$$th of April

Solution

\begin{align}f(t)=15\end{align}

\begin{align}\downarrow\end{align}

\begin{align}12.25+4.75\sin\left(\frac{2\pi}{365}t\right)=15\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4.75\sin\left(\frac{2\pi}{365}t\right)=2.75\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\sin\left(\frac{2\pi}{365}t\right)=\frac{2.75}{4.75}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{2\pi}{365}t=\sin^{-1}\left(\frac{2.75}{4.75}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t&=\frac{365}{2\pi}\sin^{-1}\left(\frac{2.75}{4.75}\right)\\&\approx36\mbox{ days}\end{align}

Therefore, the date is $$36$$ days after the $$21$$st of March, i.e. the $$26$$th of April.

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(c) Find $$f'(t)$$, the derivative of $$f(t)$$.

$$f'(t)=\dfrac{9.5\pi}{365}\cos\left(\dfrac{2\pi}{365}t\right)$$

Solution

\begin{align}f'(t)&=4.75\left(\frac{2\pi}{365}\right)\cos\left(\frac{2\pi}{365}t\right)\\&=\frac{9.5\pi}{365}\cos\left(\frac{2\pi}{365}t\right)\end{align}

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(d) Hence, or otherwise, find the length of the longest day in Galway.

$$17\mbox{ hr}$$

Solution

\begin{align}\mbox{Maximum}&=12.25+4.75(1)\\&=17\mbox{ hr}\end{align}

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(e) Use integration to find the average length of the day in Galway over the six months from March 21st to September 21st ($$184$$ days). Give your answer in hours and minutes, correct to the nearest minute.

$$15\mbox{ hr }15\mbox{ min}$$