4 Mark Question

Question 14 Marks

Principal $=₹ 1000$, Rate $=8 \%$ per annum. Fill in the following table and find which type of interest (simple or compound) changes in direct proportion with time period.

Time period	1 yr	2 yr	3 yr
Simple interest (in ₹)
Compound interest (in ₹)

Answer

Given, $P=$ ₹ $ 1000, R=8 \%$
Simple interest for different time periods
(i) For $T=1 yr$,
$
\text { Simple interest }=\frac{P R T}{100}=\frac{1000 \times 8 \times 1}{100}=$ ₹ $ 80
$
(ii) For $T=2 yr$,
$
\text { Simple interest }=\frac{P R T}{100}=\frac{1000 \times 8 \times 2}{100}=$ ₹ $ 160
$
(iii) For $T=3 yr$,
$
\text { Simple interest }=\frac{P R T}{100}=\frac{1000 \times 8 \times 3}{100}=$ ₹ $ 240
$
Compound interest for different time periods
(i) For $T=1 yr$
Compound interest $=P\left(1+\frac{R}{100}\right)^T-P$
$=1000\left(1+\frac{8}{100}\right)^1-1000$
$=1000 \times \frac{108}{100}-1000$
$=1080-1000=$ ₹ $80$
(ii) Ans. ₹ 166.40
(iii) Ans. ₹ 259.712
Hence, the complete table is as follows

Time period	1 yr	2 yr	3 yr
Simple interest (in ₹)	80	160	240
Compound interest (in ₹ )	80	166.40	259.712

Now, ratio of simple interest with time period are as follows :
For $1 yr , \frac{80}{1}=80$
For $2 yr , \frac{160}{2}=80$
and for $3 yr , \frac{240}{3}=80$
Here, ratio of simple interest with time period is same for every year. Hence, the simple interest changes in direct proportion with time period.
Ratio of compound interest with time period are as follow
For $1 yr , \frac{80}{1}=80$
For $2 yt , \frac{166.40}{2}=8.2 .20$
and for $3 yt , \frac{259.712}{3}=86.5706$
Here, ratio of compound interest with time period is not same for every year. Hence, the compound interest does not change in direct proportion with time period.

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Question 24 Marks

Two persons could fit new windows in a house in 3 days.
(i) One of the persons fell ill before the work started. How long would the job take now ?
(ii) How many persons would be needed to fit the windows in one day?

Answer

(i) Let the job would take $x$ days.
Then, we have the following table

Number of persons	2	2 – 1 = 1
Number of days	3	x

Here, lesser the number of persons, more would be the number of days required to complete the job. Therefore, it is a case of inverse proportion.
$
\begin{array}{lc}
\text { So, } & 2 \times 3=1 \times x \\
\therefore & x=6
\end{array}
$
Hence, the job would take 6 days when one of the persons fell ill before the work started.
(ii) Let the number of persons needed to fit the windows in one day be $x$.
Then, we have the following table

Number of days	3	1
Number of persons	2	x

Here, lesser the number of days, more will be the number of persons needed to fit the windows. Therefore, it is a case of inverse proportion.
So, $\quad 3 \times 2=1 \times x \Rightarrow x=6$
Here, 6 persons would be needed to fit the windorns In one day.

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Question 34 Marks

A mixture of paint is prepared by mixing 1 part of red pigment with 8 parts of base. In the following table, find the parts of base that need to be added.

Parts of red pigment	1	4	7	12	20
Parts of base	8	…	…	…	…

Answer

Let the number of parts of red pigment be $x$ and the number of parts of the base be $y$.
So, the relation between them is of the type
$
\frac{x_1}{y_1}=\frac{x_2}{y_2}
$
Given, $\quad x_1=1$ and $y_1=8$
(i) For $x_2=4$,
$
\frac{x_1}{y_1}=\frac{x_2}{y_2} \Rightarrow \frac{1}{8}=\frac{4}{y_2} \Rightarrow y_2=\frac{4 \times 8}{1}=32
$
Hence, 32 parts of base should be added to 4 parts of red pigment.
(ii) For $x_3=7$,
$
\frac{x_1}{y_1}=\frac{x_3}{y_3} \Rightarrow \frac{1}{8}=\frac{7}{y_3} \Rightarrow y_3=\frac{7 \times 8}{1}=56
$
Hence, 56 parts of the base should be added to 7 parts of red pigment.
(iii) For $x_4=12$,
$
\frac{x_1}{y_1}=\frac{x_4}{y_4} \Rightarrow \frac{1}{8}=\frac{12}{y_4} \Rightarrow y_4=\frac{12 \times 8}{1} \Rightarrow
$ $y_4=96$
Hence, 96 parts of the base should be added to 12 parts of red pigment.
(iv) For $x_5=20$,
$
\frac{x_1}{y_1}=\frac{x_5}{y_5} \Rightarrow \frac{1}{8}=\frac{20}{y_5} \Rightarrow y_5=\frac{20 \times 8}{1} \Rightarrow y_5=160
$
Hence, 160 parts of the base should be added to 20 parts of red pigment.
Hence, the complete table is

Parts of red pigment (x)	1	4	7	12	20
Parts of base (y)	8	32	56	96	160

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Question 44 Marks

The following table shows the distance travelled by one of the new eco-friendly energy efficient cars travelled on gas.

Litres of gas	1	0.5	2	2.5	3	5
Distance (km)	15	7.5	30	37.5	45	75

Which type of properties are indicated by the table? How much distance will be covered by the car in 8 L of gas?

Answer

The above table is given as

Litres of gas	1	0.5	2	2.5	3	5
Distance (km)	15	7.5	30	37.5	45	75

All the above $x$ and $y$ are in direct proportion.
(ii) Let distance covered by car in 8 litres of gas be x km.

Litres of gas	1	8
Distance (km)	15	x

As if is direct proportion
$\begin{array}{l} \therefore \frac{x_1}{y_1}=\frac{x_2}{y_2} \Rightarrow \frac{1}{15}=\frac{8}{x} \\ x=15 \times 8=120 km\end{array}$

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Question 54 Marks

A volleyball court is in a rectangular shape and its dimensions are directly proportional to the dimensions of the swimming pool given below. Find the width of the pool.

Answer

It is given that a volleyball court and a swimming pool both are rectangular in shape and their dimensions are directly proportional.
Let length be $l$ and breadth be $b$.
$
\begin{array}{ll}
\text { If } & l \propto b \\
\Rightarrow & \frac{l_1}{b_1}=\frac{l_2}{b_2}
\end{array}
$
Here, $l_1=18 m, b_1=9 m, l_2=75 m, b_2=$ ?
Then, $\quad \frac{9}{18}=\frac{b_2}{75}$
$
\therefore \quad b_2=\frac{9 \times 75}{18}=\frac{75}{2}=37.5 m
$
So, the width of the pool is 37.5 m .

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Question 64 Marks

A water tank casts a shadow 21 m long. A tree of height 9.5 m casts a shadow 8 m long at the same time. The lengths of the shadows are directly proportional to their heights. Find the height of the tank.

Answer

It is given that a water tank casts a shadow 21 m long and a tree of height 9.5 m casts a shadow 8 m long at the same time. Here the lengths of the shadows are directly proportional to their heights.
Let height $=h$ and the length of shadow $=l$.
Here, $h_1=9.5 m, l_1=8 m, h_2=x, l_2=21 m$
For direct proportion,
$
\begin{aligned}
\frac{h_1}{l_1} =\frac{h_2}{l_2} \Rightarrow \frac{95}{8}=\frac{x}{21} \\
\therefore =\frac{95 \times 21}{8}=24.9 m
\end{aligned}
$
$\therefore$ The height of the tank is 24.9 m .

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Question 74 Marks

Ashwani has a road map with a scale of $1 cm=24 km$.
(i) Find the distance covered in the map, if he drives on a road for 168 km .
(ii) Find the distance on the road, if corresponding distance covered on map is 11 cm .
(iii) Which mathematical concept is used here?

Answer

(i) Let the distance covered on the map be x cm.
Then, we have the following table :

Actual distance covered on the road (in km )	Distance covered (represented) on the map (in cm )
24	1
168	x

It is the case of direct proportion.
$
\therefore \quad \frac{24}{168}=\frac{1}{x} \Rightarrow x=\frac{168}{24}=7
$
So, distance covered on the map is 7 cm .
(ii) Let the distance covered on the road be x km.
Then, we have the following table :

Actual distance covered on the road (in km )	Distance covered (represented) on the map (in cm )
24	1
x	11

It is case of direct proportion.
$
\begin{array}{l}
\text { So, } \frac{24}{x}=\frac{1}{11} \\
\Rightarrow \quad x=24 \times 11 \Rightarrow x=264
\end{array}
$
Hence, the distance covered on the road is 264 km .
(iii) Here, concept of direct proportion is used.

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Question 84 Marks

The students of Anju's class sold posters to raise money. Anju wanted to apply the concept of variation for finding the amount of money her class would make for different numbers of posters sold. She know that they could raise ₹ 250 for every 60 posters sold.
(i) How much money would Anju's class make for selling 102 posters?
(ii) If Anju's class raise exactly ₹ 2000, then how many posters would they need to sell?

Answer

(i) Let money raised by Anju's class be ₹ x.
The above information can be exhibited in the following table form

Money	250	x
Poster	60	102

Since, more posters sold, more money will be raised. So, it is a case of direct variation.
$\therefore \quad \frac{250}{60}=\frac{x}{102}$
$\begin{array}{l}\Rightarrow 60 \times x=250 \times 102 \\ \Rightarrow \quad x=\frac{250 \times 102}{60} \Rightarrow x=425\end{array}$
Hence, ₹ 425 would make by Anju's class for selling 102 posters.
(ii) Let posters need to sell be $x$.
Then, we have the following table

Money	250	2000
Poster	60	x

It is a case of direct variation.
$
\therefore \quad \frac{250}{60}=\frac{2000}{x} \Rightarrow 250 \times x=60 \times 2000
$
$
\Rightarrow \quad x=\frac{60 \times 2000}{250}=480
$
So, 480 posters need to be sell.

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Question 94 Marks

A bicycle dealer has ₹ 60000 to invest, when a bicycle is available for ₹ 1000. If the price of a bicycle increases by $25 \%$, then find the number of bicycle he can purchase with the same sum?

Answer

If the cost of bicycle is more, dealer can buy less number of bicycles in the given amount.
Given, total money $=$ ₹ $ 60000$
Cost of 1 bicycle $=$ ₹ $ 1000$
$\therefore$ Number of bicycles $=\frac{60000}{1000}=60$
If the cost of bicycle increases by $25 \%$, then
new price of a bicycle $=1000+25 \%$ of 1000
$
=1000+\frac{25 \times 1000}{100}=1000+250=$ ₹ $ 1250
$
Here, $x_1=60, y_1=$ ₹ $ 1000, x_2=?, y_2=$ ₹ $ 1250$
For inverse proportion, we know that
$
\begin{array}{l}
x_1 y_1=x_2 y_2 \\
\therefore \quad 60 \times 1000=x_2 \times 1250 \\
\Rightarrow \quad x_2=\frac{60 \times 1000}{1250}=48
\end{array}
$
So, he can purchase 48 bicycles with the same sum.

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Question 104 Marks

It is given that I varies directly as $m$.
(i) Write an equation which relates I and $m$.
(ii) Find the constant of proportion (k) when $l$ is 6 and $m$ is 18 .
(iii) Find $l$ when $m$ is 33 .
(iv) Find $m$ when $l$ is 8 .

Answer

Given, $l \propto m$
$
\Rightarrow \quad l=k m
$
where, $k$ is any constant.
$
\Rightarrow \quad k=\frac{l}{m}
$
(i) $k=\frac{l}{m}$
(ii) Given, $l=6$ and $m=18$
$
\begin{array}{ll}
\because & k=\frac{l}{m} \Rightarrow k=\frac{6}{18} \\
\therefore & k=\frac{1}{3}
\end{array}
$
(iii) Given, $m=33$ and $l=$ ?
$\because k=\frac{1}{m} \Rightarrow \frac{1}{3}=\frac{1}{33}\qquad$ $[$$\because k=\frac{1}{3}$ from part (ii)$]$
$\therefore l=11$
(iv) Given, $I=8$ and $m=$ ?
$\because \quad k=\frac{1}{m} \Rightarrow \frac{1}{3}=\frac{8}{m}\qquad$ $[$$\because k=\frac{1}{3}$ from part (ii)$]$
$\Rightarrow \quad m=24$

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Question 114 Marks

Here is a keyboard of a harmonium.

(i) Find the ratio of white keys to black keys of the keyboard.
(ii) What is the ratio of black keys to all keys of the given keyboard?
(iii) This pattern of keys is repeated on larger keyboard. How many black keys would you expect to find on a keyboard with 14 such patterns?

Answer

In the given figure, number of white keys $=10$
Number of black keys $=7$
Number of total keys $=10+7=17$
(i) Required ratio $=\frac{\text { Number of white keys }}{\text { Number of black keys }}=\frac{10}{7}=10: 7$
(ii) Required ratio $=\frac{\text { Number of black keys }}{\text { Number of total keys }}=\frac{7}{17}=7: 17$
(iii) In 1 pattern, number of black keys = 7
In 14 such patterns, number of black keys
$=7 \times 14=98$

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Question 124 Marks

Fill in the blanks in the following table by a suitable number, if $x$ and $y$ vary directly.

x	2	-	-	12	4	-
y	3	15	12	-	-	21

Answer

x	2	-	-	12	4	-
y	3	15	12	-	-	21

Since, $x$ and $y$ vary directly.
Hence, $\frac{x}{y}=$ constant
$\Rightarrow \quad \frac{x}{y}=\frac{2}{3}$
We can take $x$ as $x_1=2$ as $y_1=3$ for each case.
Case I
$\begin{array}{l}\text { Here, } x_1=2, y_1=3, x_2=?, y_2=15 \\ \because \quad \frac{x_1}{y_1}=\frac{x_2}{y_2} \Rightarrow \frac{2}{3}=\frac{x_2}{15} \\ \Rightarrow \quad x_2=\frac{2 \times 15}{3}=10\end{array}$
Case II
$
\begin{aligned}
& x_1=2, y_1=3, x_2=?, y_2=12 \\
\because & \frac{x_1}{y_1}=\frac{x_2}{y_2} \Rightarrow \frac{2}{3}=\frac{x_2}{12} \\
\Rightarrow \quad x_2 & =\frac{2 \times 12}{3}=8
\end{aligned}
$
Case III
$
\begin{aligned}
& x_1=2, y_1=3, x_2=12, y_2=? \\
\because \frac{x_1}{y_1} & =\frac{x_2}{y_2} \Rightarrow \frac{2}{3}=\frac{12}{y_2} \\
\Rightarrow & y_2=\frac{3 \times 12}{2}=18
\end{aligned}
$
Case IV
$
\begin{array}{l}
x_1=2, y_1=3, x_2=4, y_2=? \\
\because \frac{x_1}{y_1}=\frac{x_2}{y_2} \Rightarrow \frac{2}{3}=\frac{4}{y_2} \Rightarrow y_2=\frac{4 \times 3}{2}=6
\end{array}
$
Case V
$
\begin{aligned}
& x_1=2, y_1=3, x_2=?, y_2=21 \\
\because & \frac{x_1}{y_1}=\frac{x_2}{y_2} \Rightarrow \frac{2}{3}=\frac{x_2}{21} \\
\Rightarrow & x_2=\frac{2 \times 21}{3}=14
\end{aligned}
$
So, the complete table is shown below :

x	2	10	8	12	4	14
y	3	15	12	18	6	21

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MATHS 4 Mark Question

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