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Maths 5 Mark Question

Direct and Inverse Variations · Maharashtra Board STD 8 (MSBSHSE - English Medium). 3 questions.

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5 Mark Question

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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 × 4 = k
⇒ 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 × 6 = k
⇒ 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:
$v\left(\right.$ in $\left.cm ^3\right)$
...
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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