5 Mark Question

Question 15 Marks

A car with speed $60 \mathrm{~km} / \mathrm{hr}$ takes 8 hours to travel some distance. What should be the increase in the speed if the same distance is to be covered in $7 \frac{1}{2}$ hours?

Answer

Let v represent the speed of car in km/hr and t represent the time required.
Since, speed of a car varies inversely as the time required to cover a distance.
$\begin{aligned} & \therefore v \propto \frac{1}{t} \\ & \therefore \mathbf{v}=\mathbf{k} \times \frac{1}{\mathbf{t}}\end{aligned}$
where, k is the constant of variation.
∴ v × t = k …(i)
Since, a car with speed 60 km/hr takes 8 hours to travel some distance.
i.e., when v = 60, t = 8
∴ Substituting v = 60 and t = 8 in (i), we get
v × t = k
∴ 60 × 8 = t
∴ k = 480
Substituting k = 480 in (i), we get
v × t = k
∴ v × t = 480 …(ii)
This is the equation of variation.
Now, we have to find speed of car if the same distance is to be covered in $7 \frac{1}{2}$ hours.
i.e., when $\mathrm{t}=7 \frac{1}{2}=7.5, \mathrm{v}=$ ?
$\therefore$ Substituting, $t=7.5$ in (ii), we get
$
\begin{aligned}
& \mathrm{v} \times \mathrm{t}=480 \\
& \therefore \mathrm{v} \times 7.5=480 \\
& v=\frac{480}{7.5}=\frac{4800}{75} \\
& \therefore \mathrm{v}=64
\end{aligned}
$
The speed of vehicle should be 64 km/hr to cover the same distance in 7.5 hours.
∴ The increase in speed = 64 – 60
= 4km/hr
∴ The increase in speed of the car is 4 km/hr.

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Question 25 Marks

120 bags of half litre milk can be filled by a machine within 3 minutes find the time to fill such 1800 bags ?

Answer

Let $b$ represent the number of bags of half litre milk and $t$ represent the time required to fill the bags.
Since, the number of bags and time required to fill the bags varies directly.
$
\begin{aligned}
& \therefore \mathrm{b} \propto \mathrm{t} \\
& \therefore \mathrm{b}=\mathrm{kt} \ldots \text { (i) }
\end{aligned}
$
where $k$ is the constant of variation.
Since, 120 bags can be filled in 3 minutes
i.e., when $b=120, t=3$
$\therefore$ Substituting $b=120$ and $t=3$ in (i), we get
$\mathrm{b}=\mathrm{kt}$
$\therefore 120=\mathrm{k} \times 3$
$\therefore \mathrm{k}=\frac{120}{3}$
$\therefore \mathrm{k}=40$
Substituting $k=40$ in (i), we get
$\mathrm{b}=\mathrm{kt}$
$\therefore \mathrm{b}=40 \mathrm{t}$...(ii)
This is the equation of variation.
Now, we have to find time required to fill 1800 bags
$\therefore$ Substituting $b=1800$ in (ii), we get
$\mathrm{b}=40 \mathrm{t}$
$\therefore 1800=40 \mathrm{t}$
$\therefore \mathrm{t}=\frac{1800}{40}$
$\therefore \mathrm{t}=45$
$\therefore 1800$ bags of half litre milk can be filled by the machine in 45 minutes.

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Question 35 Marks

If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours ?

Answer

Let, $n$ represent the number of workers building the wall and t represent the time required.
Since, the number of workers varies inversely with the time required to build the wall.
$
\begin{aligned}
& \therefore n \propto \frac{1}{\mathrm{t}} \\
& \therefore n=\mathrm{k} \times \frac{1}{\mathrm{t}}
\end{aligned}
$
where $\mathrm{k}$ is the constant of variation
$
\therefore \mathrm{n} \times \mathrm{t}=\mathrm{k........(i)}
$
15 workers can build a wall in 48 hours,
i.e., when $n=15, t=48$
$\therefore$ Substituting $n=15$ and $t=48$ in (i), we get
$
\begin{aligned}
& \mathrm{n} \times \mathrm{t}=\mathrm{k} \\
& \therefore 15 \times 48=\mathrm{k} \\
& \therefore \mathrm{k}=720
\end{aligned}
$
Substituting $k=720$ in (i), we get
$
\begin{aligned}
& n \times t=k \\
& \therefore n \times t=720 \text {...(ii) }
\end{aligned}
$
This is the equation of variation.
Now, we have to find number of workers required to do the same work in 30 hours.
i.e., when $t=30, \mathrm{n}=$ ?
$\therefore$ Substituting $\mathrm{t}=30$ in (ii), we get
$\mathrm{n} \times \mathrm{t}=720$
$\therefore \mathrm{n} \times 30=720$
$\therefore n=\frac{720}{30}$
$\therefore \mathrm{n}=24$
$\therefore 24$ workers will be required to build the wall in 30 hours.

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Question 45 Marks

Which of the following statements are of inverse variation?
i. Number of workers on a job and time taken by them to complete the job.
ii. Number of pipes of same size to fill a tank and the time taken by them to fill the tank.
iii. Petrol filled in the tank of a vehicle and its cost.
iv. Area of circle and its radius.

Answer

i. Let, $x$ represent number of workers on a job, and $y$ represent time taken by workers to complete the job. As the number of workers increases, the time required to complete the job decreases. $ \therefore x \propto \frac{1}{y} $ ii. Let, $n$ represent number of pipes of same size to fill a tank and t represent time taken by the pipes to fill the tank. As the number of pipes increases, the time required to fill the tank decreases. $ \therefore \mathrm{n} \propto \frac{1}{\mathrm{t}} $ iii. Let, $p$ represent the quantity of petrol filled in a tank and $\mathrm{c}$ represent the cost of the petrol. As the quantity of petrol in the tank increases, its cost increases. $ \therefore \mathrm{p} \propto \mathrm{c} $ iv. Let, A represent the area of the circle and $r$ represent its radius. As the area of circle increases, its radius increases. $\therefore \mathrm{A} \propto \mathrm{r}$ $\therefore$ Statements (i) and (ii) are examples of inverse variation.

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Question 55 Marks

x varies inversely as y, when x = 15 then y = 10, if x = 20, then y = ?

Answer

Given that,
$
\begin{aligned}
& x \propto \frac{1}{\sqrt{y}} \\
& \therefore x=\mathrm{k} \times \frac{1}{\sqrt{y}}
\end{aligned}
$
where, $\mathrm{k}$ is the constant of variation.
$
\begin{aligned}
& \therefore \mathrm{x} \times \mathrm{y}=\mathrm{k} \ldots \text { (i) } \\
& \text { When } \mathrm{x}=15, \mathrm{y}=10 \\
& \therefore \text { Substituting, } \mathrm{x}=1 \\
& \mathrm{x} \times \mathrm{y}=\mathrm{k} \\
& \therefore 15 \times 10=\mathrm{k} \\
& \therefore \mathrm{k}=150
\end{aligned}
$
$\therefore$ Substituting, $x=15$ and $y=10$ in (i), we get
Substituting, $k=150$ in (i), we get
$
\begin{aligned}
& x \times y=k \\
& \therefore x \times y=150 \text {...(ii) }
\end{aligned}
$
This is the equation of variation.
When $x=20, y=$ ?
$\therefore$ substituting $x=20$ in (ii), we get
$
\begin{aligned}
& x \times y=150 \\
& \therefore 20 \times y=150 \\
& \therefore y=\frac{150}{20} \\
& \therefore y=7.5
\end{aligned}
$

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Question 65 Marks

$x \propto \frac{1}{\sqrt{y}}$ and when $\mathrm{x}=40$ then $\mathrm{y}=16$. If $\mathrm{x}=10$, find $\mathrm{y}$.

Answer

$
\begin{aligned}
& x \propto \frac{1}{\sqrt{y}} \\
& \therefore x=\mathrm{k} \times \frac{1}{\sqrt{y}}
\end{aligned}
$
where, $\mathrm{k}$ is the constant of variation.
$
\begin{aligned}
& \therefore \mathrm{x} \times \sqrt{ } \mathrm{y}=\mathrm{k} \ldots \text { (i) } \\
& \text { When } \mathrm{x}=40, \mathrm{y}=16 \\
& \therefore \text { Substituting } \mathrm{x}=40 \text { and }=16 \text { in (i), we get } \\
& \mathrm{x} \times \sqrt{ } \mathrm{y}=\mathrm{k} \\
& \therefore 40 \times \sqrt{ } 16=\mathrm{k} \\
& \therefore \mathrm{k}=40 \times 4 \\
& \therefore \mathrm{k}=160
\end{aligned}
$
Substituting $k=160$ in (i), we get
$
\begin{aligned}
& x \times \sqrt{ } y=k \\
& \therefore x \times \sqrt{ } y=160 \text {...(ii) }
\end{aligned}
$
This is the equation of variation.
When $\mathrm{x}=10, \mathrm{y}=$ ?
$\therefore$ Substituting, $x=10$ in (ii), we get
$
\begin{aligned}
& x \times \sqrt{ } y=160 \\
& \therefore 10 \times \sqrt{y}=160 \\
& \therefore \sqrt{y}=\frac{160}{10} \\
& \therefore \sqrt{y}=16
\end{aligned}
$
$\therefore \mathrm{y}=256 \ldots$ [Squaring both sides]

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Question 75 Marks

The boxes are to be filled with apples in a heap. If 24 apples are put in a box then 27 boxes are needed. If 36 apples are filled in a box how many boxes will be needed?

Answer

Let x represent the number of apples in each box and y represent the total number of boxes required.
The number of apples in each box are varying inversely with the total number of boxes.
$\begin{aligned} & \therefore x \infty \frac{1}{y} \\ & \therefore x=k \times \frac{1}{y}\end{aligned}$
where, k is the constant of variation,
∴ x × y = k …(i)
If 24 apples are put in a box then 27 boxes are needed.
i.e., when x = 24, y = 27
∴ Substituting x = 24 and y = 27 in (i), we get
x × y = k
∴ 24 × 27 = k
∴ k = 648
Substituting k = 648 in (i), we get
x × y = k
∴ x × y = 648 …(ii)
This is the equation of variation.
Now, we have to find number of boxes needed
when, 36 apples are filled in each box.
i.e., when x = 36,y = ?
∴ Substituting x = 36 in (ii), we get
x × y = 648
∴ 36 × y = 648
$
\begin{aligned}
& \therefore y=\frac{648}{36} \\
& \therefore y=18
\end{aligned}
$
$\therefore$ If 36 apples are filled in a box then 18 boxes are required.

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Question 85 Marks

Find constant of variation and write equation of variation for every example given below :
i. $p \propto \frac{1}{q}$; if $\mathrm{p}=15$ then $\mathrm{q}=4$.
ii. $z \propto \frac{1}{w}$; when $z=2.5$ then $w=24$.
iii. $s \propto \frac{1}{t^2}$; if $\mathrm{s}=4$ then $\mathrm{t}=5$.
iv. $x \propto \frac{1}{\sqrt{y}} ;$ if $\mathrm{x}=15$ then $\mathrm{y}=9$.

Answer

$
\begin{aligned}
& \text { i. } p \propto \frac{1}{q} \ldots \text {..[Given] } \\
& \therefore \mathrm{p}=\mathrm{k} \times \frac{1}{q}
\end{aligned}
$
where, $\mathrm{k}$ is the constant of variation.
$
\therefore \mathrm{p} \times \mathrm{q}=\mathrm{k}
$
When $p=15, q=4$
$\therefore$ Substituting $p=15$ and $q=4$ in (i), we get
$
\begin{aligned}
& \mathrm{p} \times \mathrm{q}=\mathrm{k} \\
& \therefore 15 \times 4=\mathrm{k} \\
& \therefore \mathrm{k}=60
\end{aligned}
$
Substituting $k=60$ in (i), we get
$
\begin{aligned}
& \mathrm{p} \times \mathrm{q}=\mathrm{k} \\
& \therefore \mathrm{p} \times \mathrm{q}=60
\end{aligned}
$
This is the equation of variation.
$\therefore$ The constant of variation is 60 and the equation of variation is $p q=60$.

ii. $z \propto \frac{1}{w} \ldots$ Given]
$
\therefore \mathrm{z}=\mathrm{k} \times \frac{1}{w}
$
where, $\mathrm{k}$ is the constant of variation,
$
\therefore \mathrm{z} \times \mathrm{w}=\mathrm{k}
$
When $z=2.5, w=24$
$\therefore$ Substituting $z=2.5$ and $w=24$ in (i), we get
$
\begin{aligned}
& \mathrm{z} \times \mathrm{w}=\mathrm{k} \\
& \therefore 2.5 \times 24=\mathrm{k} \\
& \therefore \mathrm{k}=60
\end{aligned}
$
Substituting $k=60$ in (i), we get
$
\begin{aligned}
& \mathrm{z} \times \mathrm{w}=\mathrm{k} \\
& \therefore \mathrm{z} \times \mathrm{w}=60
\end{aligned}
$
This is the equation of variation.
$\therefore$ The constant of variation is 60 and the equation of variation is $z W=60$.

iii. $s \propto \frac{1}{t^2} \ldots$ [Given]
$
\therefore s=k \times \frac{1}{t^2}
$
where, $k$ is the constant of variation,
$
\therefore \mathrm{s} \times \mathrm{t}^2=\mathrm{k}
$
When $\mathrm{s}=4, \mathrm{t}=5$
$\therefore$ Substituting, $\mathrm{s}=4$ and $t=5$ in (i), we get
$
\begin{aligned}
& \mathrm{s} \times \mathrm{t}^2=\mathrm{k} \\
& \therefore 4 \times(5)^2=\mathrm{k} \\
& \therefore \mathrm{k}=4 \times 25 \\
& \therefore \mathrm{k}=100
\end{aligned}
$
Substituting $k=100$ in (i), we get
$
\begin{aligned}
& \mathrm{s} \times \mathrm{t}^2=\mathrm{k} \\
& \therefore \mathrm{s} \times \mathrm{t}^2=100
\end{aligned}
$
This is the equation of variation.
$\therefore$ The constant of variation is 100 and the equation of variation is st ${ }^2=100$.

iv. $x \propto \frac{1}{\sqrt{y}} \ldots$. [Given]
$
\therefore x=\mathrm{k} \times \frac{1}{\sqrt{y}}
$
where, $\mathrm{k}$ is the constant of variation,
$
\therefore \mathrm{x} \times \sqrt{ } \mathrm{y}=\mathrm{k}
$
When $x=15, y=9$
$\therefore$ Substituting $x=15$ and $y=9$ in (i), we get
$
\begin{aligned}
& \mathrm{x} \times \sqrt{ } \mathrm{y}=\mathrm{k} \\
& \therefore 15 \times \sqrt{ } 9=\mathrm{k} \\
& \therefore \mathrm{k}=15 \times 3 \\
& \therefore \mathrm{k}=45
\end{aligned}
$
Substituting $k=45$ in (i), we get
$
\begin{aligned}
& x \times \sqrt{y}=\mathrm{k} \\
& \therefore x \times \sqrt{y}=45 .
\end{aligned}
$
This is the equation of variation.
$\therefore$ The constant of variation is $k=45$ and the equation of variation is $x \sqrt{ } y=45$

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Question 95 Marks

The information about number of workers and number of days to complete a work is given in the following table. Complete the table.

Number of workers	30	20	__	10	__
Days	6	9	12	__	36

Answer

Let, n represent the number of workers and d represent the number of days required to complete a work.
Since, number of workers and number of days to complete a work are in inverse poportion.
$\begin{aligned} & \therefore \mathbf{n} \propto \frac{1}{\mathrm{~d}} \\ & \therefore \mathbf{n}=\mathrm{k} \times \frac{1}{\mathrm{~d}}\end{aligned}$
where k is the constant of variation.
∴ n × d = k …(i)

i. When n = 30, d = 6
∴ Substituting n = 30 and d = 6 in (i), we get
n × d = k
∴ 30 × 6 = k
∴ k = 180
Substituting k = 180 in (i), we get
∴ n × d = k
∴ n × d = 180 …(ii)
This is the equation of variation

ii. When d = 12, n = 7
∴ Substituting d = 12 in (ii), we get
n × d = 180
∴ n × 12 = 180
$\therefore n=\frac{180}{12}$
∴ n = 15

iii. When n = 10, d = ?
∴ Substituting n = 10 in (ii), we get
n × d = 180
10 × d = 180
$\therefore d=\frac{180}{10}$
∴ d = 18

iv. When d = 36, n = ?
∴ Substituting d = 36 in (ii), we get
n × d = 180
∴ n × 36 = 180
$\therefore n=\frac{180}{36}$
∴ n = 5

Number of workers	30	20	15	10	5
Days	6	9	12	18	36

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Question 105 Marks

If the rate of notebooks is Rs 240 per dozen, what is the cost of 3 notebooks ? Also find the cost of notebooks, 24 notebooks and 50 notebooks and complete the following table.

Number of notebooks (x)	12	3	9	24	50	1
Cost (In Rupees) (y)	240	__	__	__	__	2

Answer

As the number of notebooks increases their cost also increases.
∴ Number of notebooks and cost of notebooks are in direct proportion.

i.
$
\begin{aligned}
& \frac{12}{240}=\frac{3}{y} \\
& \therefore \quad \frac{12}{12 \times 20}=\frac{3}{y} \\
& \therefore \quad \frac{1}{20}=\frac{3}{y} \\
& \therefore y=3 \times 20 \\
& \therefore y=60
\end{aligned}
$
ii.
$
\begin{aligned}
& \frac{12}{240}=\frac{9}{y} \\
& \therefore \quad \frac{1}{20}=\frac{9}{y} \\
& \therefore \mathrm{y}=9 \times 20 \\
& \therefore y= 180
\end{aligned}
$
iii.
$
\begin{aligned}
& \frac{12}{240}=\frac{24}{y} \\
& \therefore \quad \frac{1}{20}=\frac{24}{y} \\
& \therefore \mathrm{y}=24 \times 20 \\
& \therefore \mathrm{y}=480
\end{aligned}
$

iv.
$
\begin{aligned}
& \frac{12}{240}=\frac{50}{y} \\
\therefore \quad & \frac{1}{20}=\frac{50}{y} \\
\therefore \mathrm{y}= & 50 \times 20 \\
\therefore \mathrm{y}= & 1000
\end{aligned}
$

Number of notebooks (x)	12	3	9	24	50	1
Cost (In Rupees) (y)	240	60	180	480	1000	20

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Question 115 Marks

The total remuneration paid to laborers, employed to harvest soybean is in direct variation with the number of laborers. If remuneration of 4 laborers is Rs 1000, find the remuneration of 17 laborers.

Answer

Let, m represent total remuneration paid to laborers and n represent number of laborers employed to harvest soybean.
Since, the total remuneration paid to laborers, is in direct variation with the number of laborers.
∴ m ∝ n
∴ m = kn …(i)
where, k = constant of variation
Remuneration of 4 laborers is Rs 1000.
i. e., when n = 4, m = Rs 1000
∴ Substituting, n = 4 and m = 1000 in (i), we get m = kn
∴ 1000 = k × 4
$\therefore k=\frac{1000}{4}$
∴ k = 250
Substituting, k = 250 in (i), we get
m = kn
∴ m = 250 n …(ii)
This is the equation of variation
Now, we have to find remuneration of 17 laborers.
i. e., when n = 17, m = ?
∴ Substituting n = 17 in (ii), we get
m = 250 n
∴ m = 250 × 17
∴ m = 4250
∴ The remuneration of 17 laborers is Rs 4250.

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Question 125 Marks

Complete the following table considering that the cost of apples and their number are in direct variation.

Number of apples (x)	1	4	__	12	__
Cost of apples (y)	8	32	56	__	160

Answer

The cost of apples (y) and their number (x) are in direct variation.
∴y ∝ x
∴y = kx …(i)
where k is the constant of variation

i. When, x = 1, y = 8
∴ Substituting, x = 1 and y = 8 in (i), we get y = kx
∴ 8 = k × 1
∴ k = 8
Substituting k = 8 in (i), we get
y = kx
∴ y = 8x …(ii)
This the equation of variation

ii. When, $y=56, x=$ ?
$\therefore$ Substituting $y=56$ in (ii), we get $y=8 \mathrm{x}$
$\therefore 56=8 \mathrm{x}$
$\therefore x=\frac{56}{8}$
$\therefore \mathrm{x}=7$

iii. When, x = 12, y = ?
∴ Substituting x = 12 in (ii), we get
y = 8x
∴ y = 8 × 12
∴ y = 96

iv. When, $y=160, x=$ ?
$\therefore$ Substituting $y=160$ in (ii), we get $y=8 x$
$\therefore 160=8 x$
$\therefore x=\frac{160}{8}$
$\therefore x=20$

Number of apples (x)	1	4	7	12	20
Cost of apples (y)	8	32	56	96	160

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