3 Marks Question

Question 13 Marks

If Naresh walks $250$ steps to cover a distance of $200$ metres, find the distance travelled in $350$ steps.

Answer

Naresh walks $250$ steps to cover distance $= 200m$
In $1$ step he covers the distance$=\frac{200}{520}\text{m}$
In $350$ steps, he covers$=\frac{200}{250}\times350=\frac{20\times350}{25}=\frac{7000}{25}=280\text{m}$

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Question 23 Marks

Here is a key board of a harmonium:

$a.$ Find the ratio of white keys to black keys on the keyboard.
$b.$ What is the ratio of black keys to all keys on the given keyboard.
$c.$ 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

$a.$ The total number of black keys $= 7$
The total number of white keys $= 10$
Hence, ratio of white keys to black eyes on the keyboard $=\frac{10}{7}$
$b.$ The total number of all keys $= 10 + 7 = 17$
The ratio of black keys to all keys on the given keyboard $= 7/17$
$c.$ Black keys in $1$ keyboard $= 7$
Black keys in $14$ such keyboards $= 14 \times 7 = 98$ keys

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Question 33 Marks

A contractor undertook a contract to complete a part of a stadium in $9$ months with a team of $560$ persons. Later on, it was required to complete the job in $5$ months. How many extra persons should he employ to complete the work?

Answer

In $9$ months, a part of stadium can complete by $560$ persons. In $1$
month, the work can be complete by $9 \times 560 = 5040$ persons In $5$ months,
the work can be complete by $\frac{5040}{5}=1008$ persons
Now, number of extra persons required to complete the work in $5$ months $= 1008 - 560 = 448$

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Question 43 Marks

Find the values of $x$ and $y$ if $a$ and $b$ are in inverse proportion:
$a. 12 x 8$
$b. 30 5y$

Answer

Given,

$a$	$12$	$x$	$8$
$b$	$30$	$5$	$y$

Here, we see that in part, $(b)$ when we divide $30$ by $6,$ we get $5$
So, in part $(a)$
we will get the value of $x \text{x}=12\times6=72$
I.e., similarly in part $(a)$ when we divide $x$
i.e., $72$ by $9,$ we will get
so, in part $(b)$ we will get the value of $y$.
I.e., $y = 5 \times 9 = 45$

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Question 53 Marks

Many schools have a recommended students-teacher ratio as $35:1.$ Next year, school expects an increase in enrolment by $280$ students. How many new teachers will they have to appoint to maintain the students-teacher ratio?

Answer

Students teachers ratio $= 35:1$ It show every $35$ students one teacher should available in the school. In the school, number of students increases $= 280$ students The number of teachers required for $280$ students $\frac{280}{35}=8\text{ teachers}$

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Question 63 Marks

The variable $x$ varies directly as $y$ and $x = 80$ when $y$ is $160$. What is $y$ when $x$ is $64?$

Answer

If x varies directly y. $\frac{\text{x}}{\text{y}}=\text{k}$ (constant) $x = 80$ and $y = 160 \frac{\text{x}}{\text{y}}=\frac{80}{160}=\frac{1}{2}$
$\text{k}=\frac{1}{2}$ When $x = 64$,
then from eq. $(i) \frac{64}{\text{y}}=\frac{1}{2} y = 64 \times 2 = 128$

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Question 73 Marks

A bowler throws a cricket ball at a speed of $120km/h$. How long does this ball take to travel a distance of $20$ metres to reach the batsman?

Answer

The speed of the cricket ball $= 120km/h =\frac{120\times1000}{60}\text{m/min.}$
$=2\times1000=2000\text{m/min}$
Now, speed in m/s $=\frac{2000}{60}=\frac{200}{6}=\frac{1000}{33}\text{m/s}$
So, $20$ m can be cover in $=\frac{\frac{20}{100}}{3}=\frac{60}{1000}=0.6\text{s}$

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Question 83 Marks

Shabnam takes $20$ minutes to reach her school if she goes at a speed of $6 \ km/h$. If she wants to reach school in $24$ minutes, whatshould be her speed?

Answer

Shabnam’s speed = 6km/h$=\frac{6\times100}{60}\times20$
Total distance covered by Shabnam in $20$ min$=\frac{6\times1000}{60}\times20$
$=\frac{1000}{10}\times20 = 100 \times 20 = 2000m$
If she want to reach the school in $24$ min,
then she should maintain the speed $=\frac{2000}{24}$
$=\frac{1000}{12}=\frac{500}{6}\text{m/min}$
$=\frac{500\times60}{1000\times6}=5\text{km/h}$

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Question 93 Marks

Kusum always forgets how to convert miles to kilometres and back again. However she remembers that her car’s speedometer shows both miles and kilometres. She knows that travelling $50$miles per hour is same as travelling $80$ kilometres per hour. To cover a distance of $200\ km$, how many miles Kusum would have to go?

Answer

$50$ miles per hour is same as travelling $80\ km$ per hour.
So, $1 \ km$ covers distance in $1$ h to $\frac{5}{80}\text{miles.}$
To cover a distance of $200km=\frac{50}{80}\times200=\frac{5\times200}{8}=125\text{miles}$
Hence, $125$ miles Kusum has to go for $200\ km$

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Question 103 Marks

If $x$ varies inversely as $y$ and $x = 20$ when $y = 600$, find y when $x = 400.$

Answer

Ifx varies inversely as $y. $
$xy = k$ (costant) If $x = 20$ and $x = 600 xy = 20 \times 600 = 12000 k = 12000$
When$ x = 400$, then from eq. $(i)$

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Question 113 Marks

Kritika is following this recipe for bread. She realises her sister used most of sugar syrup for her breakfast. Kritika has only $\frac{1}{6}$ cup of syrup, so she decides to make a small size of bread. How much of each ingredient shall she use? Bread recipe $1$ cup quick cooking oats $2$ cups bread flou $\frac{1}{3}$cup sugar syrup $1$ tablespoon cooking oi $1\frac{1}{3}$cups water $3$ tablespoons yeast $1$ teaspoon salt.

Answer

The remaining sugar is after used $=\frac{1}{6}\text{cup}$ Thus, its means $1-\frac{1}{6}=\frac{5}{6}$ has been used. She need $1/3$ cup of sugar syrup for $1$ piece of bread. So, new quantity of ingredient will be in proportion of $1/2$. Now, the bread recipe will be look like $1/2$ cup quick cooking oats $1$ cup bread flour $\frac{1}{6}$cup sugar syrup $1/2$ tablespoon cooking oil $\frac{2}{3}$cups water $3/2$ tablespoons yeast $1/2$ teaspoon salt.

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Question 123 Marks

Pattern $B$ consists of four tiles like pattern A. Write a proportion involving red dots and blue dots in pattern $A$ and $B$. Are they in

direct proportion? If yes, write the constant of proportion.

Answer

No. of red dots in pattern $(A) = 4$ No. of blue dots in pattern $(A’) = 2$ Pattern $B$ consists of four tiles like pattern $A$
i.e, Pattern $A × 4$ =Pattern $B$ Proportion in pattern $=\frac{2}{6}=\frac{1}{3}$
Now, proportion of blue dots & red dots in pattern $B =\frac{8}{32}=\frac{1}{4}$

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Question 133 Marks

The mass of an aluminium rod varies directly with its length. If a $16\ cm$ long rod has a mass of $192 \ g$, find the length of the rod whose mass is $105g.$

Answer

According to the question, The mass $(m)$ of an aluminium rod varies directly with its length $(i).$
Here, we use the direct proportion.
In direct proportion,$\frac{\text{m}}{\text{l}}=\text{k}$(constant) $=\frac{\text{m}}{\text{l}}=\frac{192}{16}=12 = k = 12$
If mass of the rod $= 105g$
Then, $\frac{\text{m}}{\text{l}}=\text{k}$
$\frac{105}{\text{l}}=12$
$\text{l}=\frac{105}{12}=8.75\text{cm}$

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Question 143 Marks

In a camp, there is enough flour for $300$ persons for $42$ days. How long will the flour last if $20$ more persons join the camp?

Answer

For $300$ persons flour is enough for $42$ days. For $1$ person flour enough $= 300 \times 42 = 12600$ days
Now, $20$ more persons join the camp. So, total persons$=300 + 20 = 320$ For $320$ persons flour enough $=\frac{12600}{320}=\frac{3}{8}\text{days}$

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Question 153 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

Length of volleyball court $= 18M$
Breadth of volleyball court$= 9M$
Length of pool $= 75M$
Let width of swimming pool $= XM$
The size of volleyball court & swimming pool are in direct proportion to each other. $\frac{9}{18}=\frac{\text{x}}{75}$ $\text{x}=\frac{75\times9}{8}=\frac{75}{2}=37.5\text{m}$
Hence, the width of the swimming pool is $37.5m$

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Question 163 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 $8$ litres of gas?

Answer

On the basis of given table, the distance travelled by one of the new eco-friendly energy efficient earns travelled on gas.
The car travelled $15km$ In L of gas. The car travelled $7.5\ km$ in $0.5L$ of gas.
The car travelled $30km$ in $2L$ of gas.
This rate shows direct proportion between litres of gas and the distance cover.
The car can cover the distance in $8$ L of gas $= 8 \times 15 = 120$

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Question 173 Marks

A car covers a distance in $40$ minutes with an average speed of $60\ km/ h$. What should be the average speed to cover the same distance in $25$ minutes?

Answer

A car covers a distance in $40$ min with an average speed $=60\text{km/ h}=\frac{60\times1000}{60}\text{m/min}$ In $1$ min.,
the same distance can be cover with speed $=\frac{60\times1000\times40}{60}=40000\text{m/min}$ In $25$ min.
the same distance can be cover with speed $=\frac{4000}{25}=1600\text{m/min}$
$=\frac{1600}{1000}\times60=16\times6=96\text{km/ h}$

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Question 183 Marks

Campus and Welfare Committee of school is planning to develop a blue shade for painting the entire school building. For this purpose various shades are tried by mixing containers of blue paint and white paint. In each of the following mixtures, decide which is a lighter shade of blue and also find the lightest blue shade among all of them.

If one container has one litre paint and the building requires $105$ litres for painting, how many container of each type is required to paint the building by darkest blue shade?

Answer

$1.$ In mixture $E:$
$2.$ The mixture of blue container $= 6$
The number of white container $= 1$
Ratio of blue $\&$ white containers $=\frac{6}{1}=6$
$3.$ In mixture $F:$
$4.$ The number of blue container $= 4$
The number of white containers $= 2$
Ratio of blue $\&$ white $=\frac{4}{2}=2$
Clearly, mixture $F$ would be lighter shade,
since for lighter shade white container should be equal or more than or nearest number of blue container.

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Question 193 Marks

If one container has one litre paint and the building requires $105$ litres for painting, how many container of each type is required to paint the building by darkest blue shade?

Answer

$1.$ Inmixture $C:$
The number of blue containers $= 3$
The number of white containers $= 3$
Ratio of blue $\&$ white $=\frac{3}{3}=1$
$2.$ Inmixture $D:$
The number of blue container $= 2$
The number of white container $= 5$
Ratio of blue $\&$ white $=\frac{2}{5}=0.4$
Clearly, mixture $D$ would be lighter shade,
since for lighter shade white container should be more than blue container.

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Question 203 Marks

If one container has one litre paint and the building requires $105$ litres for painting, how many container of each type is required to paint the building by darkest blue shade?

Answer

$1.$ In mixture $G:$
The number of blue container $= 3$
The number of white containers $= 3$
Ratio of blue $\&$ white $=\frac{3}{3}=1$
$2.$ In mixture $H:$
The number of blue container $= 4$
The number of white container $= 3$
Ratio of blue $\&$ white $=\frac{4}{3}=1.33$
Clearly, mixture $G$ would be lighter shade,
since for lighter shade white container should be more than blue container.
Above all mixtures, mixture $D$ is lightest among them.
The total number of containers required for painting $= 105$
Number of blue containers required for painting $=\frac{2}{7}\times105=2\times15=30$
Number of white containers required for painting $=\frac{5}{7}\times105=5\times15=75$

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Question 213 Marks

If one container has one litre paint and the building requires $105$ litres for painting, how many container of each type is required to paint the building by darkest blue shade?

Answer

$1.$ In mixture $A:$
The number of blue containers $= 3$
The number of white containers $= 4$
Ratio of blue $\&$ white $=\frac{3}{4}=0.75$
$2.$ In mixture $B:$
The number of blue containers $= 3$
The number of white containers $= 3$
Ratio of blue $\&$ white $=\frac{3}{3}=1$
Clearly, mixture $A$ would be lighter shade.

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Question 223 Marks

The students of Anju’s class sold posters to raise money. Anju wanted to create a ratio for finding the amount of money her class would make for different numbers of posters sold. She knew they could raise $Rs. 250$ for every $60$ posters sold.
$a.$ How much money would Anju’s class make for selling $102$ posters?
$b.$ Could Anju’s class raise exactly $ Rs.2,000?$ If so, how many posters would they need to sell? If not, why?

Answer

$a.$ Every $60$ posters , Anju’s class students raise $= Rs. 250$
so, from $1$ poster, Anju’s class students raise $=\frac{25}{60}=\text{Rs.}\frac{25}{6}$
If Anju’s sell $102$ posters, then they raise$=\frac{25}{6}=102$
$= 17 \times 25 = Rs. 425$
Hence, Anju’s class make $Rs.425$ for selling of $102$ posters
$b.$ Since, by selling $1$ poster, Anju’s class raise $=\text{Rs}\frac{25}{6}$
For raise exactly $Rs.2000$ they needs to sell $=2000\div\frac{25}{6}$
$=2000\times\frac{6}{25}$
$=480\text{posters}$

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Question 233 Marks

Work with a partner to write at least five ratio statements about this quilt, which has white, blue, and purple squares.

How many squares of each colour will be there in $12$ such quilts?

Answer

Basis of given figure, white, blue & purple squares are given: Purple $= 12$, Blue $= 20$ & white $= 16$ Total squares $= 12 + 20 +16 = 48$
Statement I Purple: total $= 12 : 48 = 1 : 4$
Statement II Blue : total $= 20 : 48=5 : 12$
Statement III White : total $=16 : 48 = 1 : 3$
Statement IV Purple : Blue $= 12 : 20 = 3 : 5$
Statement V Purple : White $= 12 : 16 = 3 : 4$

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Maths 3 Marks Question

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