5 Marks Questions

Question 15 Marks

It $x$ and $y$ vary inversely, fill in the following blanks:

$x$	$16$	$32$	$8$	$128$
$y$	$4$	$...$	$...$	$0.25$

Answer

Since x and y vary inversely, we have: $xy = k$ For $x = 16$ and $y = 4$,
we have: $16 \times 4$ $= k \Rightarrow k = 64$ For $x = 32$ and $k = 64$,
we have: $xy = k$ $\Rightarrow32\text{y}=64$
$\Rightarrow\text{y}=\frac{64}{32}$
$=2$ For $x = 8$ and $k = 64\ xy = k$
$\Rightarrow8\text{y}=64$
$\Rightarrow\text{y}=\frac{64}{8}$
$=8$

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Question 25 Marks

It $x$ and $y$ vary inversely, fill in the following blanks:

$x$	$12$	$16$	$...$	$8$	$...$
$y$	$...$	$6$	$4$	$...$	$0.25$

Answer

Since $x$ and $y$ vary inversely, we have: $xy = k$ For $x = 16$ and $y = 6$,
we have: $16 \times 6 = k \Rightarrow k = 96$ For $x = 12$ and $k = 96$,
we have: $xy = k$ $\Rightarrow12\text{y}=96$
$\Rightarrow\text{y}=\frac{96}{12}$
$=8$ For $y = 4$ and $k = 96$,
we have: $xy = k$ $\Rightarrow4\text{x}=96$
$\Rightarrow\text{x}=\frac{96}{4}$
$=24$ For $x = 8$ and $k = 96$,
we have: $xy = k$ $\Rightarrow\text{8y}=96$
$\Rightarrow\text{y}=\frac{96}{8}$
$=12$ For $y = 0.25$ and $k = 96$,
we have: $xy = k$ $\Rightarrow0.25\text{x}=96$
$\Rightarrow\text{x}=\frac{96}{0.25}$
$=384$

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Question 35 Marks

It is known that for a given mass of gas, the volume $v$ varies inversely as the pressure $p$. Fill in the missing entries in the following table:

v (in $cm^3$)	$...$	$48$	$60$	$...$	$100$	$...$	$200$
p (in atmospheres)	$2$	$...$	$\frac{3}{2}$	$1$	$...$	$\frac{1}{2}$	$...$

Answer

Since the volume and pressure for the given mass vary inversely,
we have: $vp = k$
For $v = 60$ and $\text{p}=\frac{3}{2},$
we have: $\text{k}=60\times\frac{3}{2}$
$=90$
For $p = 2$ and $k = 90$,
we have: $2v = 90$
$\Rightarrow\text{v}=\frac{90}{2}$
$=45$
For $v = 48$ and $k = 90$,
we have: $48p = 90$
$\Rightarrow\text{p}=\frac{90}{48}$
$=\frac{15}{8}$
For $p = 1$ and $k = 90$,
we have: $1v = 90$
$\Rightarrow\text{v}=\frac{90}{1}$
$=90$
For $v = 100$ and $k = 90$,
we have: $100p = 90$
$\Rightarrow\text{v}=\frac{90}{100}$
$=\frac{9}{10}$
For $\text{p}=\frac{1}{2}$ and $k = 90$,
we have: $\frac{1}{2}\text{v}=90$
$\Rightarrow\text{v}=90\times2$
$=180$
For $v = 200$ and $k = 90$,
we have: $200p = 90$
$\Rightarrow\text{p}=\frac{90}{200}$
$=\frac{9}{20}$

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Question 45 Marks

Raghu has enough money to buy $75$ machines worth $Rs\ 200$ each. How many machines can he buy if he gets a discount of $Rs\ 50$ on each machine?

Answer

Let x be the number of machines he can buy if a discount of $Rs.\ 50$ is offered on each machine.

Cost of a cycle (in Rs.)	$500$	$625$
Number of cycles	$25$	$x$

Since Raghu is getting a discount of $Rs\ 50$ on each machine, the cost of each machine will get decreased by $Rs.\ 50$ If the price of a machine is less, he can buy more number of machines. It is a case of inverse variation.
Therefore, we have: $75\times200 =\text{x}\times150$
$\Rightarrow\text{x}=\frac{75\times200}{150}$
$=\frac{15000}{150}$
$=100$
$\therefore$ The number of machines he can buy is $100$.

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Question 55 Marks

It $x$ and $y$ vary inversely, fill in the following blanks:

$x$	$9$	$...$	$81$	$243$
$y$	$27$	$9$	$...$	$1$

Answer

Since $x$ and $y$ vary inversely, we have: $xy = k$ For $x = 9$ and $y = 27 9 \times 27 = k$
$\Rightarrow k = 243$ For $y = 9$ and $k = 243,$
we have: $xy = k$
$\Rightarrow9\text{x}=243$
$\Rightarrow\text{y}=\frac{243}{9}$
$=27$ For $x = 81$ and $k = 243$,
we have: $xy = k$
$\Rightarrow81\text{y}=243$
$\Rightarrow\text{y}=\frac{243}{81}$
$=3$

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Question 65 Marks

Complite the following tables given that $x$ varies directly as $y$.

$x$	$5$	$...$	$10$	$35$	$25$	$...$
$y$	$8$	$12$	$...$	$...$	$...$	$32$

Answer

Here, $x$ and $y$ vary directly.
$\therefore\text{x}=\text{ky}$
$x = 5$ and $y = 8$
i.e., $5 = k × 8$
$\Rightarrow\text{k} =\frac{5}{8}=0.625$
For $y = 12$ and $k = 0.625$,
we have: $x = ky$
$\Rightarrow x = 12 \times 0.625 = 7.5$
For $x = 10$ and $k = 0.625$,
we have:$x = ky$
$\Rightarrow 10 = 0.625 \times y$
$\Rightarrow\text{y}=\frac{10}{0.625}=16$
For $x = 35$ and $k = 0.625$,
we have: $x = ky$
$\Rightarrow 35 = 0.625 \times y$
$\Rightarrow\text{y}=\frac{35}{0.625}=56$
For $x = 25$ and $k = 0.625$,
we have: $x = ky$
$\Rightarrow 25 = 0.625 \times y$
$\Rightarrow\text{y}=\frac{25}{0.625}=40$
For $y = 32$ and $k = 0.625$,
we have: $x = ky$
$\Rightarrow x = 0.625 \times 32 = 20$

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Question 75 Marks

A car can finish a certain journey in $10$ hours at the speed of $48 \ km/hr$. By how much should its speed be increased so that it may take only $8$ hours to cover the same distance?

Answer

Let the increased speed be $x$ $km/h$.

Time (in h)	$10$	$8$
Speed (km/h)	$48$	$x + 48$

Since speed and time taken are in inverse variation, we get:
$10\times48 = 8(\text{x+48})$
$\Rightarrow480=8\text{x}+384$
$\Rightarrow8\text{x}=480-384$
$\Rightarrow8\text{x}=96$
$=12$
Thus, the speed should be increased by $12\ km/h$.

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Question 85 Marks

$1200$ soldiers in a fort had enough food for $28$ days. After $4$ days, some soldiers were transferred to another fort and thus the food lasted now for $32$ more days. How many soldiers left the fort?

Answer

It is given that after $4$ days, out of $28$ days, the fort had enough food for $1200$ soldiers for $(28 - 4 = 24)$ days. Let $x$ be the number of soldiers who left the fort .

Number of soldiers	$1200$	$1200-x$
Number of days for which food lasts	$24$	$32$

Since the number of soldiers and the number of days for which the food lasts are in inverse variation,
we have: $1200\times24 =(1200-\text{x})\times32$
$\Rightarrow\frac{1200\times24}{32}=1200-\text{x}$
$\Rightarrow900=1200-\text{x}$
$\Rightarrow\text{x}=1200-900$
$=300$ Thus, $300$ soldiers left the fort.

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Question 95 Marks

Complite the following tables given that $x$ varies directly as $y$.

$x$	$4$	$9$	$...$	$...$	$3$	$...$
$y$	$16$	$...$	$48$	$36$	$...$	$4$

Answer

Here, $x$ and $y$ vary directly.
$\therefore\text{x}=\text{ky}$ $x = 4$ and $y = 16$ i.e., $4 = k \times 16$
$\Rightarrow\text{k} =\frac{4}{16}=\frac{1}{4}$ For $x = 9$ and $\text{k}=\frac{1}{4},$
we have: $9 = ky$
$\Rightarrow y = 4 \times 9 = 36$ For $y = 48$ and $\text{k}= \frac{1}{4},$
we have: $x = ky$ $=\frac{1}{4}\times48=12$ For $y = 36$ and $\text{k}=\frac{1}{4},$
we have: $x = ky$ $=\frac{1}{4}\times36 =9$ For $x = 3$ and $\text{k}=\frac{1}{4},$
we have: $x = ky$
$\Rightarrow3 =\frac{1}{4}\times\text{y}$
$\Rightarrow y = 12$ For $x = 4$ and $\text{k}=\frac{1}{4},$
we have: $x = ky$ $=\frac{1}{4}\times4=1$

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MATHS 5 Marks Questions

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