4 Mark Question - Maths STD 7 Questions - Vidyadip

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4 Mark Question

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17 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks

The ratio of the radii of two circles is $4: 5$. Find the ratio of their areas.

Answer

Ratio in the radii of two circles $=4: 5$
Let radius of first circle $\left(r_1\right)=4 x$
and radius of the second circle $\left(r_2\right)=5 x$
$\therefore$ Area of first circle $=\pi r _1^2$
$=\frac{22}{7} \times 4 x \times 4 x$
$=\frac{352}{7} x^2$
and area of second circle $=\pi r _2^2$
$=\frac{22}{7} \times 5 x \times 5 x=\frac{550}{7} x^2$
Now Ratio between their areas
$=\frac{352}{7} x^2: \frac{550}{7} x^2$
$=352: 550 \text { (Dividing by } 22 \text { ) }$
$=16: 25$

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Question 24 Marks

Find the area of a rhombus having each side equal to 13 cm and one of the diagonals equal to 24 cm .

Answer

Let $A B C D$ be the rhombus whose diagonals intersect at $O$.
Then, $A B=13 cm$
$AC=24 cm$
The diagonals of a rhombus bisect each other at right angles.
Therefore, $\triangle AOB$ is a right-angled triangle, right angled at O , such that:
$OA=\frac{1}{2} AC=12 cm$
$AB=13 cm$
By Pythagoras theorem :
$(A B)^2=(O A)^2+(O B)^2$
$\Rightarrow(13)^2=(12)^2+(O B)^2$
$\Rightarrow(O B)^2=(13)^2-(12)^2$
$\Rightarrow(O B)^2=169-144=25$
$\Rightarrow(O B)^2=(5)^2$
$\Rightarrow O B=5 cm$
$B D=2 \times O B=2 \times 5 cm=10 cm$
Area of the rhombus $ABCD =\frac{1}{2} \times AC \times BD cm ^2$
$=\frac{1}{2} \times 24 \times 10$
$=120 cm^2$

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Question 34 Marks

The length and breadth of a park in the ratio $2: 1$ and its perimeter is 240 m . A path 2 m wide runs inside it, along its boundary. Find the cost of paving the path at Rs 80 per $m ^2$.

Answer

Ratio in length and breadth of the park $=2: 1$
Its perimeter $=240 m$

Let length $=2 x$
then breadth $=x$
Perimeter $=2(2 x+x)=2 \times 3 x=6 x$
$6 x=240$
$\Rightarrow x =\frac{240}{6}=40$
Length $=2 x =2 \times 40=80 m$
and breadth $=x=40 m$
Area $= L \times B =80 \times 40 m^2=3200 m^2$
Width of path inside the park $=2 m$
Inner length $(I)=80-2 \times 2=80-4=76 m$
and breadth (b) $=40-2 \times 2=40-4=36 m$
Inner area $=76 \times 36=2736 m^2$
Area of path $=$ Outer area - Inner area $=3200-2736=464 m^2$
Rate of paving the path $=$ Rs. 80 per $m ^2$

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Question 44 Marks

In the given figure, a circle of diameter 21cm is given. Inside this circle, two circles with diameters $\frac{2}{3}$ and $\frac{1}{3}$ of the diameter of the big circle have been drawn, as shown in the given figure. Find the area of the shaded region.

Answer

Diameter of largest circle (outer circle) }=$21 cm$
$\therefore \text { Radius }(R)=\frac{21}{2} cm$
$\text { Area }=\pi R^2=\frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}=\frac{693}{2} cm^2$
$=346.5 cm^2$
Diamerer of bigger circle $=\frac{2}{3}$ or 21
$=14 cm$
$\therefore \operatorname{Radius}\left(r_1\right)=\frac{14}{2}=7 cm$
and area $=\pi r_1^2=\frac{22}{7} \times 7 \times 7=154 cm^2$
Diameter of smaller circle $=\frac{1}{3}$ of 21
$=7 cm$
$\therefore \text { Radius }\left(r_2\right)=\frac{7}{2} cm$
and area $=\pi r_2^2=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$
$=\frac{77}{2} cm=38.5 cm^2$
Area of shaded portion $=346.5-(154+38.5)=(346.5-192.5)=154 cm^2$

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Question 54 Marks

The diameter of the wheel of a car is 77cm. How many revolutions will it make to travel 121km?

Answer

Diameter of the wheel $=77\text{cm}$
⇒ Radius of the wheel $=\Big(\frac{77}{2}\Big)\text{cm}$
Circumference of the wheel $=2\pi\text{r}$
$=\Big(2\times\frac{22}{7}\times\frac{77}2{}\Big)\text{cm}=(22\times11)\text{cm}=242\text{cm}$
$=\Big(\frac{242}{100}\Big)\text{m}=\Big(\frac{121}{50}\Big)\text{m}$
Distance covered by the wheel in 1 revolution $=\Big(\frac{121}{50}\Big)\text{m}$
Now, $\Big(\frac{121}{50}\Big)\text{m}$ is covered by the car in 1 revolution.
(121 × 1000) m will be covered by the car in $\Big(1\times\frac{50}{121}\times121\times1000\Big)$ revolutions, i.e. 50000 revolutions.
$\therefore$ Required number of revolutions = 50000

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Question 64 Marks

A well of diameter 140cm has a stone parapet around it. If the lenght of the outer edge of the parapet is 616cm, find the width of the parapet.

Answer

Given: Diameter of the well (d) = 140cm. Radius of the well (r) $=\Big(\frac{140}{2}\Big)\text{cm}=70\text{cm}$

Let the radius of the outer circle (including the stone parapet) be R cm. Length of the outer edge of the parapet $=616\text{cm}$ $\Rightarrow2\pi\text{R}=616$ $\Rightarrow\Big(2\times\frac{22}{7}\times\text{R}\Big)=616$ $\Rightarrow\text{R}=\Big(\frac{616\times7}{2\times22}\Big)\text{cm}=98\text{cm}$ Now, width of the parapet = {Radius of the outer circle (including the stone parapet) - Radius of the well} = {98 - 70}cm = 28cm Hence, the width of the parapet is 28cm.

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Question 74 Marks

The inner circumference of a circular track is 330m. The track is 10.5m wide everywhere. Calculate the cost of putting up a fence along the outer circle at the rate of Rs 20 per metre.

Answer

Let the inner and outer radii of the track be r metres and (r + 10.5) metres, respectively.

Inner circumference $=330\text{m}$
$\therefore2\pi\text{r}=330=2\times\frac{22}{7}\times\text{r}=330$
$\Rightarrow\text{r}=\Big(330\times\frac{7}{44}\Big)=52.5\text{m}$
Inner radius of the track = 52.5m
$\therefore$ Outer radii of the track = (52.5 + 10.5)m = 63m
$\therefore$ Circumference of the outer circle $=\Big(2\times\frac{22}{7}\times63\Big)\text{m}=396\text{m}$
Rate of fencing = Rs. 20 per metre
$\therefore$ Total cost of fencing the outer circle = Rs. (396 × 20) = Rs. 7920

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Question 84 Marks

The base of an isosceles triangle is 12cm and its perimeter is 32cm. Find its area.

Answer

Perimeter of an isosceles triangle $=32 cm$
Base $=12 cm$
Sum of two equal sides $=32-12=20 cm$
Length of each equal side $=\frac{20}{2}=10 cm$
Let $AD \perp BC$
$BD=DC=\frac{12}{2}=6 cm$

Now, in right $\triangle ABD$
$A B^2=A D^2+B D^2$
$\Rightarrow(10)^2=A D^2+(6)^2$
$\Rightarrow 100=A D^2+36$
$\Rightarrow A D^2=100-36=64=(8)^2$
$\therefore A D=8 cm$
Now, area of $\triangle ABC =\frac{1}{2} BC \times AD$
$=\frac{1}{2} \times 12 \times 8 cm^2$
$=48 cm^2$

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Question 94 Marks

The hour and minute hands of a clock are 4.2cm and 7cm long respectively. Find the sum of the distances covered by their tips in 1 day.

Answer

Length of the hour hand (r) $=4.2\text{cm}$
Distance covered by the hour hand in 12 hours $=2\pi\text{r}=\Big(2\times\frac{22}{7}\times4.2\Big)\text{cm}=26.4\text{cm}$
$\therefore$ Distance covered by the hour hand in 24 hours $=(2\times26.4)=52.8\text{cm}$
Length of the minute hand (R) $=7\text{cm}$
Distance covered by the minute hand in 1 hour $=2\pi\text{R}=\Big(2\times\frac{22}{7}\times7\Big)\text{cm}=44\text{cm}$
$\therefore$ Distance covered by the minute hand in 24 hours $=(44\times24))\text{cm}=1056\text{cm}$
$\therefore$ Sum of the distances covered by the tips of both the hands in 1 day $= (52.8 + 1056)\text{cm}$
$=1108.8\text{cm}$

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Question 104 Marks

A horse is tied to a pole in a park with a string 21m long. Find the area over which the horse can graze.

Answer

Length of rope $(r)=21 m$
Area of the circle $=\pi r ^2=\frac{22}{7} \times 21 \times 21 m^2=1386 m^2$
The horse will graze on $1386 m^2$ area

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Question 114 Marks

A school has a hall which is 22m long and 15.5m broad. A carpet is laid inside the hall leaving all around a margin of 75cm from the walls. Find the area of the carpet and the area of the strip left uncovered. If the width of the carpet is 82cm, find its cost at the rate of Rs 60 per m.

Answer

Length of hall (l) = 22m Breadth (b) = 15.5m

Space left along the walls $=75\text{m}=\frac{3}{4}\text{m}$ Inner length (l) $=22-2\times\frac{3}{4}=20.5\text{m}$ Inner breadth (b) $=15.5-2\times\frac{3}{4}=15.5-1.5=14\text{m}$ Area of carpet = Inner area $=20.5\times14\text{m}^2=287\text{m}^2$ Outer area $=22\times15.5=341\text{m}^2$ Area of strip left out $=341-287=54\text{m}^2$ Width of carpet $=82\text{cm}=\frac{82}{100}\text{m}$ Length of carpet $=287\div\frac{82}{100}$ $=\frac{287\times100}{82}=350\text{m}$ Rate of carpet = Rs. 60 per metre Total cost = Rs. 60 × 350 = Rs. 21000

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Question 124 Marks

A rectangular grassy lawn measuring 38 m by 25 m has been surrounded externally by a 2.5 -m-wide path. Calculate the cost of gravelling the path at the rate of Rs 120 per $m ^2$.

Answer

Inner length of lawn $(1)=38 m$
and breadth $(b)=25 m$
Width of path $=2.5 m$

$\text { Outer length }(L)=38+2 \times 2.5=38+5=43 m$
$\text { and outer breadth }(B)=25+2 \times 2.5=25+5=30 m$
$\text { Area of path }=\text { Outer area - Inner area }$
$=(43 \times 30-38 \times 25) m ^2$
$=(1290-950) m ^2=340 m^2$
$\text { Rate of gravelling the path }=\text { Rs. } 120 \text { per } m ^2$
$\text { Total cost }=\text { Rs. } 120 \times 340=\text { Rs. } 40800$

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Question 134 Marks

A saree is 5 m long and 1.3 m wide. A border of width 25 cm is printed along its sides. Find the cost of printing the border at Rs 1 per $10 cm^2$.

Answer

Let ABCD be the saree and EFGH be the part of saree without border.

Length, $A B=5 m$

Breadth, $BC =1.3 m$

Width of the border of the saree $=25 cm=0.25 m$

$\therefore \text { Area of } ABCD =5 m \times 1.3 m=6.5 m^2$

$\text { Length, } GH =\{5-(0.25+0.25)\} m =4.5 m$

$\text { Breadth, FG }=\{1.3-0.25+0.25\} m =0.8 m$

$\therefore \text { Area of } EFGH =4.5 m \times .8 m=3.6 m^2$

$\text { Area of the border }=\text { Area of } ABCD -\text { Area of } EFGH$

$=6.5 m^2-3.6 m^2$

$=2.9 m^2=29000 cm^2\left[\text { since } 1 m^2=10000 cm^2\right]$

$\text { Rate of printing the border }= Rs 1 \text { per } 10 cm^2$

$\therefore \text { Total cost of printing the border }= Rs \left(\frac{1 \times 29000}{10}\right)$

$=\text { Rs } 2900$

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Question 144 Marks

In the given figure a rectangular plot of land measures 8m by 6m. In each of the corners, there is a flower bed in the form of a quadrant of a circle of radius 2m. Also, there is a flower bed in the area of the remaining plot.

Answer

Length of plot (l) = 8m
and breadth (b) = 6m
Area of plot $=1 \times b =8 \times 6=48 m^2$
Radius of each quadrant at the corner = 2m
$\therefore$ Area of 4 quadrants $=4\times\frac{1}{4}\pi\text{r}^2$
$=\pi\text{r}^2=\frac{22}{7}\times2\times2=\frac{88}{7}\text{m}^2$
Radius of circle at the centre $=2\text{m}$
and area of circle $=\pi\text{r}^2=\frac{22}{7}\times2\times2$
$=\frac{88}{7}\text{m}^2$
$\therefore$ Area of the remaining plot
$=48-\Big(\frac{88}{7}+\frac{88}{7}\Big)=\Big(48-\frac{176}{7}\Big)\text{m}^2$
$=48.00-25.14=22.86\text{m}^2$

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Question 154 Marks

Find the area of a rhombus each side of which measures 20cm and one of whose diagoanls is 24cm.

Answer

Let ABCD be the rhombus, whose diagonals intersect at O
.

$AB=20 cm \text { and } AC=24 cm$

The diagonals of a rhombus bisect each other at right angles.

Therefore, $\triangle AOB$ is a right angled triangle, right angled at 0 .

Here, $OA =\frac{1}{2} AC =12 cm$

$A B=20 cm$

By Pythagoras theorem:

$(A B)^2=(O A)^2+(O B)^2$

$\Rightarrow(20)^2=(12)^2+(O B)^2$

$\Rightarrow(O B)^2=(20)^2-(12)^2$

$\Rightarrow(O B)^2=400-144=256$

$\Rightarrow(O B)^2=(16)^2$

$\Rightarrow O B=16 cm$

$\therefore BD=2 \times OB=2 \times 16 cm=32 cm$

$\therefore$ Area of the rhombus $ABCD =\left(\frac{1}{2} \times AC \times BD \right) cm ^2$

$=\left(\frac{1}{2} \times 24 \times 32\right) cm^2$

$=384 cm^2$

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Question 164 Marks

A rectangular plot of land measures 95 m by 72 m . Inside the plot, a path of uniform width of 3.5 m is to be constructed all around. The rest of the plot is to be laid with grass. Find the total expenses involved in constructing the path at Rs 80 per $m ^2$ and laying the grass at Rs 40 per $m ^2$.

Answer

Outer length of the plot $(L)=95 m$
and breadth $(B)=72 m$
Width of path $=3.5 m$

Inner length (1)=95-2 * 3.5=95-7=88m
and breadth = 72-2\times3.5=72-7=65m
Outer area = L * B = 95 × 72m
and inner area = I * b = 88 * 65m
Area of path = outer area - inner area = 6840 - 5720 = 1120m
Rate of constructing it = Rs. 80 per m
Total cost = Rs. 1120 }\times80=\mathrm{ Rs. 89600
and rate of laying grass = Rs. 40 per m
Total cost = Rs. 40 × 5720 = Rs. 228800
Total cost = Rs. 89600 + Rs. 228800 = Rs. 318400

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Question 174 Marks

A bicycle wheel makes 5000 revolutions in moving 11km. Find the circumference and the diameter of the wheel.

Answer

It may be noted that in one revolution, the bicycle covers a distance equal to the circumference of the wheel.
Total distance covered by the bicycle in 5000 revolutions = 11km
⇒ 5000 × Circumference of the wheel = 11000m [since 1km = 1000m]
Circumference of the wheel $=\Big(\frac{11000}{5000}\Big)\text{m}=2.2\text{m}=220\text{cm}$ [since 1m = 100cm]
Circumference of the wheel = $\pi$ × Diameter of the wheel
$\Rightarrow220\text{cm}=\frac{22}{7}$ × Diameter of the wheel
⇒ Diameter of the wheel $=\Big(\frac{220\times7}{22}\Big)\text{cm}=70\text{cm}$
Hence, the circumference of the wheel is 220cm and its diameter is 70cm.

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