Math Ncert Exemplar 2019 Solutions for Class 8 Maths Chapter 10 - Direct & Inverse Proportions

Answer:

Given that u and v vary directly with each other.
$\therefore \frac{u}{v}=k \left(\mathrm{constant}\right)$
If u = 10, v = 15, then
$\begin{array}{rcl}k& =& \frac{u}{v}\\ & =& \frac{10}{15}\\ & =& \frac{2}{3}\end{array}$

Option (a), $\frac{2}{3}=k$
Option (b), $\frac{8}{12}=\frac{2}{3}=k$
Option (c), $\frac{15}{20}=\frac{3}{4}\ne k$
Option (d), $\frac{25}{37.5}=\frac{2}{3}=k$

Hence, the correct answer is option C.

Answer:

Given that x and y vary inversely with each other.
$\therefore x\times y=k \left(\mathrm{constant}\right)$
If x = 10, y = 6, then
$\begin{array}{rcl}k& =& x\times y\\ & =& 10\times 6\\ & =& 60\end{array}$

Option (a), $12\times 5=60=k$
Option (b), $15\times 4=60=k$
Option (c), $25\times 2.4=60=k$
Option (d), $45\times 1.3=58.3\ne k$

Hence, the correct answer is option D.

Answer:

If the land to be uniformly fertile, then the area of land and the yield on it vary directly with each other.

Hence, the correct answer is option A.

Answer:

We cannot predict about the number of teeth with exactly the age of a person.
The number of teeth and the age of a person vary sometimes directly and sometimes inversely with each other.
It changes with person-to-person.

Hence, the correct answer is option D.

Answer:

A truck requires 54 L of diesel to cover a distance = 297 km
In 1 L, a truck can cover = $\frac{297}{54}=5.5 \mathrm{km}$
Diesel required for 550 km = $\frac{550}{5.5}=100 \mathrm{L}$
Hence, the correct answer is option A.

Answer:

Given that, speed of car = 48 km/h
Time taken by car = 10 h
$\begin{array}{rcl}\therefore \mathrm{Distance}& =& \mathrm{Speed}\times \mathrm{Time}\\ & =& 48\times 10\\ & =& 480 \mathrm{km}\end{array}$

Speed of the car, if 480 km is to cover in 8 h = $\frac{480}{8}=60 \mathrm{km}/\mathrm{h}$
Hence, the correct answer is option A.

Answer:

In option (a),
x = 0.5,2, 8, 32 and y = 2, 8, 32,128
$$\begin{gathered} \therefore \frac{x}{y}=\frac{0.5}{2}=\frac{1}{4} \\ \mathrm{and}, \frac{\mathit{x}}{\mathit{y}}=\frac{2}{8}=\frac{1}{4} \\ \mathrm{and}, \frac{\mathit{x}}{\mathit{y}}=\frac{8}{32}=\frac{1}{4} \\ \mathrm{and}, \frac{\mathit{x}}{\mathit{y}}=\frac{32}{128}=\frac{1}{4} \end{gathered}$$

Thus, x and y vary directly from each other.
In all other options, x and y do not vary directly with each other.
Hence, the correct answer is option A.

Answer:

In option (d),
when u increases, v decreases.

Answer:

We know that, when we increase the speed, then the time taken by the vehicle decreases.
$$\begin{gathered} \therefore \mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}} \\ \Rightarrow \mathrm{Speed}\propto \frac{1}{\mathrm{Time}} \end{gathered}$$
Thus, the speed and time taken vary inversely with each other.
Hence, the correct answer is option D.

Answer:

Both x and y are in directly proportion.
$$\begin{gathered} \therefore x\propto y \\ \Rightarrow x=ky \\ \Rightarrow \frac{1}{x}=\frac{1}{ky} \\ \Rightarrow \frac{1}{x}=\frac{l}{y} \left[\because l=\frac{1}{k}\right] \\ \Rightarrow \frac{1}{x}\propto \frac{1}{y} \end{gathered}$$
then $\frac{1}{x} \mathrm{and} \frac{1}{y}$ are in direct proportion.

Hence, the correct answer is option A.

Answer:

Given that the speed of the cycle is 12 km/h.
Time taken to reach school = 20 min = $\frac{1}{3} \mathrm{h}$
Total distance cover = $12\times \frac{1}{3}=4 \mathrm{km}$
Speed of the cycle to reach in 12 min = $\frac{4}{{\displaystyle \frac{12}{60}}}=4\times 5=20 \mathrm{km}/\mathrm{h}$

Hence, the correct answer is option C.

Answer:

100 persons had food provision for 24 days.
1 person had food provision for 24 × 100 = 2400 days

As 20 persons left the place then 80 persons remained.

∴ Food provision for remaining 80 persons = $\frac{2400}{80}=30 \mathrm{days}$

Hence, the correct answer is option A.

Answer:

If two quantities x and y vary directly with each other, then
$$\begin{gathered} x\propto y \\ \Rightarrow x=ky \\ \Rightarrow \frac{x}{y}=k \end{gathered}$$
Since, in direct proportion, both x and y increases or decreases together such a manner that the ratio of their corresponding value remains constant.

Hence, the correct answer is option A.

Answer:

If two quantities p and q vary inversely with each other, then
$$\begin{gathered} p\propto \frac{1}{q} \\ \Rightarrow p=\frac{k}{q} \\ \Rightarrow p\times q=k \end{gathered}$$
Since, in inverse proportion, an increase in p cause a proportional decrease in q and vice-versa.

Hence, the correct answer is option C.

Answer:

Distance traveled in one hour = 10 km
Distance traveled in 60 min = 10 km
Distance traveled in 1 min = $\frac{10}{60}=\frac{1}{6}$ km
Distance traveled in 1 min = $\frac{1000}{6}=\frac{500}{3} \mathrm{m}$
Hence, the correct answer is option D.

Answer:

Given that x and y vary directly with each other.
$\therefore \frac{x}{y}=k \left(\mathrm{constant}\right)$
If x = 10, y = 14, then
$\begin{array}{rcl}k& =& \frac{x}{y}\\ & =& \frac{10}{14}\\ & =& \frac{5}{7}\end{array}$

Option (a), $\frac{25}{35}=\frac{5}{7}=k$
Option (b), $\frac{35}{25}=\frac{7}{5}\ne k$
Option (c), $\frac{35}{49}=\frac{5}{7}=k$
Option (d), $\frac{15}{21}=\frac{5}{7}=k$

Hence, the correct answer is option B.

Answer:

Given: x = 5y
$$\begin{gathered} \frac{x}{y}=5=\mathrm{constant} \\ \Rightarrow x\propto y \end{gathered}$$

Hence, x and y vary directly with each other.

Answer:

Given: xy = 10
$$\begin{gathered} xy=10=\mathrm{constant} \\ \Rightarrow x=\frac{10}{y} \\ \Rightarrow x\propto \frac{1}{y} \end{gathered}$$

Hence, x and y vary inversely with each other.

Answer:

When two quantities x and y are in direct proportion or vary directly, they are written as x ∝ y.x∝y

Answer:

When two quantities x and y are in inverse proportion or vary inversely, they are written as x ∝ $\frac{1}{y}$.

Answer:

$$\begin{gathered} xy=k \\ \Rightarrow x=\frac{k}{y} \\ \Rightarrow x\propto \frac{1}{y} \end{gathered}$$
Both x and y are said to vary inversely with each other, if for some positive number k, xy = k.

Answer:

Given, $x\propto y$
$$\begin{gathered} \Rightarrow x=ky \\ \Rightarrow \frac{x}{y}=k \end{gathered}$$
Hence, x and y are said to vary directly with each other, if for some positive number k, $\underline{\frac{\mathit{x}}{\mathit{y}}}$= k.

Answer:

Two quantities are said to vary directly with each other, if they increase (decrease) together in such a manner that the ratio of their corresponding values remains constant.

Answer:

Two quantities are said to vary inversely with each other, if increase in one cause a decrease in the other in such a manner that the product of their corresponding values remains constant.

Answer:

Time taken by 12 pumps can empty a reservoir is 20 h.
1 pump can empty a reservoir is $20\times 12=240 \mathrm{h}$.
Then, 45 pumps can empty a reservoir is
$\begin{array}{rcl}\frac{240}{45}\mathrm{h}& =& \frac{240}{45}\times 60 \min \\ & =& \frac{240}{3}\times 4\\ & =& 320 \min \\ & =& 5 \mathrm{h} 20 \min \end{array}$

The time required by 45 such pumps to empty the same reservoir is 5 hours 20 min.

Answer:

If x varies inversely as y, then
$$\begin{gathered} x\propto \frac{1}{y} \\ \Rightarrow x=\frac{k}{y} \\ \Rightarrow xy=k \end{gathered}$$

If x = 60 and y = 10
$\begin{array}{rcl}\therefore k& =& xy\\ & =& 60\times 10\\ & =& 600\end{array}$

When y = 2, then
$$\begin{gathered} x\times 2=600 \\ \Rightarrow x=300 \end{gathered}$$

Answer:

If x varies directly as y, then
$$\begin{gathered} x\propto y \\ \Rightarrow x=ky \\ \Rightarrow k=\frac{x}{y} \end{gathered}$$

If x = 12 and y = 48
$\begin{array}{rcl}\therefore k& =& \frac{x}{y}\\ & =& \frac{12}{48}\\ & =& \frac{1}{4}\end{array}$

When x = 6, then
$$\begin{gathered} x=ky \\ \Rightarrow 6=\frac{1}{4}y \\ \Rightarrow y=24 \end{gathered}$$

Answer:

We know that
$$\begin{gathered} \mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}} \\ \Rightarrow \mathrm{Distance}=\mathrm{Speed}\times \mathrm{Time} \\ \mathrm{If} \mathrm{speed} \mathrm{is} \mathrm{constant}, \mathrm{then} \\ \mathrm{Distance}\propto \mathrm{Time} \end{gathered}$$
When the speed remains constant, the distance travelled is directly proportional to the time.

Answer:

We have,
$$\begin{gathered} a\propto b \\ \Rightarrow a=kb \\ \Rightarrow \frac{a}{b}=k \left(k \mathrm{is} \mathrm{constant}\right) \end{gathered}$$
Hence, on increasing a, b increases in such a manner that $\frac{a}{b}$ remains constant and positive, then a and b are said to vary directly with each other.

Answer:

We have,
$$\begin{gathered} a\propto \frac{1}{b} \\ \Rightarrow a=\frac{k}{b} \\ \Rightarrow ab=k \left(k \mathrm{is} \mathrm{constant}\right) \end{gathered}$$
Hence, if on increasing a, b decreases in such a manner that ab remains constant and positive, then a and b are said to vary inversely with each other.

Answer:

We have,
$$\begin{gathered} x\propto y \\ \Rightarrow x=ky \\ \Rightarrow \frac{x}{y}=k \left(k \mathrm{is} \mathrm{constant}\right) \end{gathered}$$

If two quantities x and y vary directly with each other, then ratio of their corresponding values remains constant.

Answer:

We have,
$$\begin{gathered} p\propto \frac{1}{q} \\ \Rightarrow p=\frac{k}{q} \\ \Rightarrow pq=k \left(k \mathrm{is} \mathrm{constant}\right) \end{gathered}$$

If two quantities p and q vary inversely with each other then product of their corresponding values remains constant.

Answer:

Let the radius of the circle be r.
Diameter of a circle = 2r
Perimeter of a circle = $2\mathrm{\pi }r$
$$\begin{gathered} \Rightarrow \mathrm{Perimeter}=\mathrm{\pi }\times \mathrm{Diameter} \\ \Rightarrow \mathrm{Perimeter}\propto \mathrm{Diameter} \left(\because \mathrm{\pi }=\mathrm{constant}\right) \end{gathered}$$

Hence, the perimeter of a circle and its diameter vary directly with each other.

Answer:

Distance travelled by a car in 1 h = 48 km
Distance travelled by the car in 60 min = 48 km
Distance travelled by the car in 1 min = $\frac{48}{60}$ km
Distance travelled by the car in 12 min = $\frac{48}{60}\times 12=\frac{48}{5}=9.6$ km

Hence, the distance travelled by the car in 12 minutes is 9.6 km.

Answer:

An auto rickshaw takes 3 hours to cover a distance of 36 km.
Speed of the auto rikshaw is $\frac{36}{3}=12 \mathrm{km}/\mathrm{h}$
If speed is increased by 4 km/h, then the new speed of auto rikshaw is 12 + 4 = 16 km/h.
Time taken to cover distance of 36 km is
$\begin{array}{rcl}\frac{36}{16}\mathrm{h}& =& \frac{36}{16}\times 60 \min \\ & =& 135 \min \\ & =& 2 \mathrm{h} 15 \min \end{array}$

Hence, the time taken by it to cover the same distance is 2 h 15 min.

Answer:

The thickness of a pile of 12 cardboard sheets = 45 mm
The thickness of a pile of 1 cardboard sheet = $\frac{45}{12}$ mm
The thickness of a pile of 240 cardboard sheets = $\frac{45}{12}\times 240=45\times 20=900$ mm = 90 cm

Hence, the thickness of a pile of 240 sheets is 90 cm.

Answer:

If x varies inversely as y, then
$$\begin{gathered} x\propto \frac{1}{y} \\ \Rightarrow x=\frac{k}{y} \\ \Rightarrow xy=k \end{gathered}$$

If x = 4 and y = 6
$\begin{array}{rcl}\therefore k& =& xy\\ & =& 4\times 6\\ & =& 24\end{array}$

When x = 3, then
$$\begin{gathered} xy=k \\ \Rightarrow 3\times y=24 \\ \Rightarrow y=8 \end{gathered}$$

Hence, when x = 3 the value of y is 8.

Answer:

We have, a is directly proportional to b.
$$\begin{gathered} \therefore a\propto b \\ \Rightarrow a=kb \\ \Rightarrow \frac{a}{b}=k \\ \Rightarrow \frac{a_1}{b_1}=\frac{a_2}{b_2} \end{gathered}$$

Answer:

We have, a is inversely proportional to b.
$$\begin{gathered} a\propto \frac{1}{b} \\ \Rightarrow a=\frac{k}{b} \\ \Rightarrow ab=k \\ \Rightarrow a_1{b}_1=a_2{b}_2 \\ \Rightarrow \frac{a_2}{{\mathit{a}}_{\mathbf{1}}}=\frac{b_1}{{\mathit{b}}_{\mathbf{2}}} \end{gathered}$$

Thus, the statement is False.

Answer:

The area occupied by 15 postal stamps = 60 cm²
The area occupied by 1 postal stamp = $\frac{60}{15}=4$ cm²
The area occupied by 120 postal stamps = $4\times 120=480$ cm²

Hence, the area occupied by 120 such postal stamps will be 480 cm².

Answer:

We have,
45 persons can complete a work = 20 days
1 person can complete a work = $45\times 20=900$ days
75 persons can complete a work = $\frac{900}{75}=12$ days

Hence, the time taken by 75 persons will be 12 hours.

Answer:

Distance travel by Devangi in 75 steps = 50 m
Distance travel by Devangi in 1 step = $\frac{50}{75}$ m
Distance travel by Devangi in 375 steps = $\frac{50}{75}\times 375=250$ m = 0.25 km

Hence, the distance travelled in 375 steps is 0.25 km.

Answer:

False,
If $x\propto y$, then
$$\begin{gathered} \Rightarrow x=ky \\ \Rightarrow \frac{x}{y}=k \end{gathered}$$

Answer:

False,
As,
$$\begin{gathered} \mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}} \\ \Rightarrow \mathrm{Distance}=\mathrm{Speed}\times \mathrm{Time} \\ \Rightarrow \mathrm{Distance}\propto \mathrm{Time} \left(\because \mathrm{Speed} \mathrm{is} \mathrm{fixed}\right) \end{gathered}$$
When the speed is kept fixed, time and distance vary directly with each other.

Answer:

False
As,
$$\begin{gathered} \mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}} \\ \Rightarrow \mathrm{Speed}\propto \frac{1}{\mathrm{Time}} \left(\because \mathrm{Distance} \mathrm{is} \mathrm{fixed}\right) \end{gathered}$$
When the distance is kept fixed, speed and time vary indirectly/inversely with each other.

Answer:

False
Length of a side of a square and its area does not vary directly with each other.

Ex. Let a be length of each side of a square.
So, area of the square = side² = a²

So, if we increase the length of the side of a square, then their area increases but not directly.

Answer:

False
Length of a side of an equilateral triangle and its perimeter vary directly with each other.

Ex. Let a be the side of an equilateral triangle.
So, perimeter = 3 × (Side) = 3 × a = 3a .
$\Rightarrow \mathrm{Perimeter}\propto \mathrm{side}$
So, if we increase the length of the side of the equilateral triangle, then their perimeter will also increase.

Answer:

False

If d varies inversely as t², then we can write dt² = k, where k is some constant.
$$\begin{gathered} d\propto \frac{1}{t^2} \\ \Rightarrow d=\frac{k}{t^2} \\ \Rightarrow d{t}^2=k \end{gathered}$$

Answer:

True

Height of a tree = 24 m
Length of its shadow = 15 m
If the length of the shadow of a pole is 6 m then let the height of the pole be x m.
Since, length and shadow vary directly to each other.
$$\begin{gathered} \therefore \frac{24}{15}=\frac{x}{6} \\ \Rightarrow 15\times x=24\times 6 \\ \Rightarrow x=\frac{24\times 6}{15} \\ \Rightarrow x=9.6 \mathrm{m} \end{gathered}$$

Thus, the height of the pole is 9.6 m.

Answer:

False
Consider x and y are in direct proportion.
$$\begin{gathered} \therefore x\propto y \\ \Rightarrow x=ky \\ \Rightarrow \frac{x}{y}=k \end{gathered}$$
Let x = 12 and y = 4
$\begin{array}{rcl}\therefore k& =& \frac{x}{y}\\ & =& \frac{12}{4}\\ & =& 3\end{array}$

Then, x – 1 = 11 and y – 1 = 3
$\begin{array}{rcl}\therefore \frac{x-1}{y-1}& =& \frac{11}{3}\\ & \ne & k\end{array}$
Hence, if x and y are in direct proportion, then (x – 1) and (y – 1) cannot be in direct proportion.

Answer:

False
If x and y are in inverse proportion, then xy = k (constant)

Let x = 4 and y = 2
∴ xy = 4 × 2 = 8 = k.
Now, x + 1 = 4 + 1 = 5 and y + 1 = 2 + 1 = 3
Then, (x + 1)(y +1) = 5 × 3 = 15 ≠ k [not in inverse proportion]

Hence, if x and y are in direct proportion, then (x + 1)and (y + 1) cannot be in inverse proportion.

Answer:

False
If p and q are in inverse proportion, then pq = k (constant)

Let p = 5 and q = 6
Then, pq = 5 × 6 = 30 = k
Now, p + 2 = 5 + 2 = 7 and q – 2 = 6 – 2 = 4
Thus, (p + 2) (q – 2) = 7 × 4 = 28 ≠ k [not in inverse proportion]

Hence, if p and q are in inverse variation then (p + 2) and (q – 2) cannot be in inverse proportion.

Answer:

False
If one angle of a triangle is kept fixed, then the measure of the remaining two angles cannot vary inversely with each other.

Ex. In ∆BC, we have
∠A + ∠B + ∠C = 180°
If ∠A = 60°, then ∠B + ∠C = 180° − 60° = 120°
So, the remaining two angles cannot depend on any proportion.

Answer:

True

When two quantities are related in such a manner that if one increases, the other also increases, then they always vary directly.
Then both quantities are in direct proportion.

Answer:

True
When, two quantities are related in such a manner that if one increases and the other decreases, then they always vary inversely.
Then both quantities are in inverse proportion.

Answer:

False
If x varies inversely as y, then
$$\begin{gathered} x\propto \frac{1}{y} \\ \Rightarrow x=\frac{k}{y} \\ \Rightarrow xy=k \end{gathered}$$

If x = 6 and y = 8
$\begin{array}{rcl}\therefore k& =& xy\\ & =& 6\times 8\\ & =& 48\end{array}$

When x = 8, then
$$\begin{gathered} xy=k \\ \Rightarrow 8\times y=48 \\ \Rightarrow y=6 \end{gathered}$$

Hence, when x = 8 the value of y is 6.

Answer:

False
The number of workers and the time to complete a job is a case of indirect proportion.

Ex. If 50 workers can complete a work in 8 days.
Then, 100 workers can complete the same work in 4 days.

Answer:

True
$$\begin{gathered} \mathrm{SI}=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100} \\ \Rightarrow \mathrm{SI}=\frac{\mathrm{RT}}{100}\times \mathrm{P} \\ \Rightarrow \mathrm{SI}\propto \mathrm{P} \left[\because \frac{\mathrm{RT}}{100} \mathrm{is} \mathrm{constant}\right] \end{gathered}$$

Answer:

True
The area of cultivated land and the crop harvested is a case of direct proportion.
Since the quantities of crop harvested is depended upon the area of cultivated land.

Answer:

(i) The time taken by a train to cover a fixed distance and the speed of the train are inversely proportional.
e.g. Let a train cover 60 km in 1 h with a speed of 60 km/h.
Then, the same train cover 60 km in 30 min with a speed of 120 km/h.

(ii) The distance travelled by CNG bus and the amount of CNG used are directly proportional.
e.g. Let a CNG bus can travelled 15 km in 1 kg of CNG.
Then, the same CNG bus travelled 30 km in 2 × 1 = 2 kg of CNG.

(iii) The number of people working and the time to complete a given work are inversely proportional to each other.
e.g. Let 10 workers can complete a work in 1 day.
Then, 5 workers can complete the same work in 2 days.

(iv) Income tax and the income are directly proportional to each other.
e.g. Let Mr. X have 5 lakh annual income.
Then, he pays 5% income tax on his income.
But if Mr. X have 6 lakh annual income, then he has to pay 10% income tax on his salary/income.

(v) Distance travelled by an auto-rickshaw and time taken are directly proportional to each other.
e.g. Let an auto-rickshaw takes 1 h to travel 10 km.
Then, it will take 3 h to travel 30 km.

Answer:

(i) Number of students in a hostel and consumption of food are directly proportional to each other.

(ii) Area of the walls of a room and the cost of white washing the walls are directly proportional to each other.

(iii) The number of people working and the quantity of work are directly proportional to each other.

(iv) Simple interest on a given sum and the rate of interest are directly proportional to each other.
$$\begin{gathered} \mathrm{SI}=\frac{\mathrm{PRT}}{100} \\ \Rightarrow \mathrm{SI}\propto \mathrm{R} \end{gathered}$$

(v) Compound interest on a given sum and the sum invested are directly proportional to each other.
$$\begin{gathered} \mathrm{CI}=\mathrm{P}{\left(1+\frac{r}{100}\right)}^n \\ \Rightarrow \mathrm{CI}\propto \mathrm{P} \end{gathered}$$

Answer:

(i) The quantity of rice and its cost are directly proportional to each other.
e.g. Let 1 kg of rice price = ₹60
Then, 5 kg of rice price = ₹5 × 60 = ₹300

(ii) The height of a tree and the number of years are neither directly nor inversely proportional to each other.

(iii) Increase in cost and number of shirts that can be purchased, if the budget remains the same are inversely proportional to each other.
e.g. Let the price of 2 shirts be ₹1500.
After increasing the price of each shirt, the price became ₹1000.

(iv) Area of land and its cost are directly proportional to each other.
e.g. Let the cost of 100 m² land be ₹2500.
Then the cost of 200 m² land be ₹5000.

(v) Sales Tax and the amount of the bill are directly proportional to each other.
e.g. Let amount of bill be ₹800. and sales tax be 10%.
$\therefore \mathrm{Sales} \mathrm{tax}=\frac{10}{100}\times 800=$₹80
If bill amount be ₹2400.
$\therefore \mathrm{Sales} \mathrm{tax}=\frac{10}{100}\times 2400=$₹240

Answer:

If x varies inversely as y, then
$$\begin{gathered} x\propto \frac{1}{y} \\ \Rightarrow x=\frac{k}{y} \\ \Rightarrow xy=k \end{gathered}$$

If x = 20 and y = 600
$\begin{array}{rcl}\therefore k& =& xy\\ & =& 20\times 600\\ & =& 12000\end{array}$

When x = 400, then
$$\begin{gathered} xy=k \\ \Rightarrow 400\times y=12000 \\ \Rightarrow y=\frac{12000}{400}=30 \end{gathered}$$

Hence, when x = 400 the value of y = 30.

Answer:

If x varies directly as y, then
$$\begin{gathered} x\propto y \\ \Rightarrow x=ky \\ \Rightarrow \frac{x}{y}=k \end{gathered}$$

If x = 80 and y = 160
$\begin{array}{rcl}\therefore k& =& \frac{x}{y}\\ & =& \frac{80}{160}\\ & =& \frac{1}{2}\end{array}$

When x = 64, then
$$\begin{gathered} \frac{x}{y}=k \\ \Rightarrow \frac{64}{y}=\frac{1}{2} \\ \Rightarrow y=128 \end{gathered}$$

Hence, when x is 64 the value of y is 128.

Answer:

If l varies directly as m, then
$$\begin{gathered} l\propto m \\ \Rightarrow l=km \\ \Rightarrow \frac{l}{m}=k \end{gathered}$$

If l = 5 and m = $\frac{2}{3}$
$\begin{array}{rcl}\therefore k& =& \frac{l}{m}\\ & =& \frac{5}{{\displaystyle \frac{2}{3}}}\\ & =& \frac{15}{2}\end{array}$

When m = $\frac{16}{3}$, then
$$\begin{gathered} \frac{l}{m}=k \\ \Rightarrow \frac{l}{{\displaystyle \frac{16}{3}}}=\frac{15}{2} \\ \Rightarrow l=\frac{16}{3}\times \frac{15}{2} \\ \Rightarrow l=8\times 5 \\ \Rightarrow l=40 \end{gathered}$$

Hence, when m = $\frac{16}{3}$, the value of l is 40.

Answer:

If x varies inversely as y, then
$$\begin{gathered} x\propto \frac{1}{y} \\ \Rightarrow x=\frac{k}{y} \\ \Rightarrow xy=k \end{gathered}$$

If x = 1.5 and y = 60
$\begin{array}{rcl}\therefore k& =& xy\\ & =& 1.5\times 60\\ & =& 90\end{array}$

When y = 4.5, then
$$\begin{gathered} xy=k \\ \Rightarrow x\times 4.5=90 \\ \Rightarrow x=20 \end{gathered}$$

Hence, when y = 4.5, the value of x = 20.

Answer:

Flour enough for 300 persons for 42 days.

Therefore, flour is enough for 1 person for 42 × 300 days i.e., 12600 days.

If 20 more persons join the camp, then the total number of persons in camp become 320.

Thus, for 320 persons flour is enough for $\frac{12600}{320} \mathrm{i}.\mathrm{e}., 39\frac{3}{8} \mathrm{days}$.

Hence, for $39\frac{3}{8}$ days the flour will last after 20 more persons join the camp.

Answer:

A work of stadium can complete in 9 months = 560 persons
A work of stadium can complete in 1 month = 560 × 9 = 5040 persons
A work of stadium can complete in 5 months = $\frac{5040}{5}=1008$ persons

Thus, the number of extra persons required are 1008 − 560 = 448.

Answer:

In 6 mins, Sobi can type 108 words.
In 1 min, she can type $\frac{108}{6} \mathrm{words} \mathrm{i}.\mathrm{e}., 18 \mathrm{words}$.
Thus, in 30 mins, she can type (18 × 30) words i.e., 540 words.

Answer:

Speed of car to cover a distance in 40 mins = 60 km/h = $60\times \frac{1000}{60}=1000$ m/min
Speed of car to cover a distance in 1 min = $1000\times 40=40000$ m/min
Speed of car to cover a distance in 25 mins = $\frac{40000}{25}=1600$ m/min = $1600\times \frac{60}{1000}=96$ km/h

Hence, the average speed to cover the same distance in 25 minutes is 96 km/h.

Answer:

Given that l varies directly as m.
(a) $l\propto m$
$$\begin{gathered} \Rightarrow l=km, \mathrm{where} k \mathrm{is} \mathrm{constant} \\ \Rightarrow \frac{l}{m}=k \end{gathered}$$

(b) If l = 6, then m = 18
$\begin{array}{rcl}\therefore k& =& \frac{l}{m}\\ & =& \frac{6}{18}\\ & =& \frac{1}{3}\end{array}$

(c) when m = 33,
$$\begin{gathered} \therefore \frac{l}{m}=k \\ \Rightarrow \frac{l}{33}=\frac{1}{3} \\ \Rightarrow l=11 \end{gathered}$$

(d) when l = 8,
$$\begin{gathered} \therefore \frac{l}{m}=k \\ \Rightarrow \frac{8}{m}=\frac{1}{3} \\ \Rightarrow m=24 \end{gathered}$$

Answer:

Interest earned in 3 years at Rs 2000 = Rs 500
Interest earned in 3 years at Rs 1000 = Rs $\frac{500}{2}=\mathrm{Rs} 250$
Interest earned in 3 years at Rs 36,000 = Rs $250\times 36=\mathrm{Rs} 9000$

Answer:

Given that the mass of an aluminum(m) rod varies directly with its length(l).
$$\begin{gathered} \therefore m\propto l \\ \Rightarrow m=kl \\ \Rightarrow \frac{m}{l}=k \end{gathered}$$
If l = 16 cm and m = 192 g
$\begin{array}{rcl}\therefore k& =& \frac{m}{l}\\ & =& \frac{192}{16}\\ & =& 12\end{array}$
When, m = 105 g, then
$$\begin{gathered} k=\frac{m}{l} \\ \Rightarrow 12=\frac{105}{l} \\ \Rightarrow l=8.75 \mathrm{cm} \end{gathered}$$

Hence, the length of the rod whose mass is 105 g is 8.75 cm.

Answer:

Given that a and b are in inverse proportion.

∴ ab = k(constant)

Answer:

Distance travelled to 250 steps = 200 m
Distance travelled to 1 step = $\frac{200}{250}$ m
Distance travelled to 350 steps = $\frac{200}{250}\times 350=280 \mathrm{m}$

Answer:

Petrol needed to travels 225 km by the car = 25 L
Petrol needed to travels 1 km by the car = $\frac{25}{225}=\frac{1}{9}$ L
Petrol needed to travels 540 km by the car = $\frac{1}{9}\times 540=60$ L

Answer:

For direct proportion, $\frac{x}{y}=k\left(\mathrm{constant}\right)$
(i)

Answer:

Given that a and b vary inversely to each other.
$\Rightarrow ab=k\left(\mathrm{constant}\right)$
(i)

Answer:

Cost of 25 m cloth = ₹337.50
Cost of 1 m cloth = $₹\frac{337.50}{25}=₹13.5$
(i) Cost of 40 m cloth = $₹13.5\times 40=₹540$

(ii) Length of cloth bought for ₹1 = $\frac{25}{337.50}\mathrm{m}$
Length of cloth bought for ₹810 = $\frac{25}{337.50}\times 810=60 \mathrm{m}$

Answer:

A swimming pool can be filled in 4 hours by 8 pumps.
Pumps required to fill the swimming pool in 1 h = 4 × 8 = 32 pumps
Pumps required to fill the swimming pool in $2\frac{2}{3}$ h i.e. $\frac{8}{3}$ h = $32÷\frac{8}{3}=12$ pumps

Answer:

Cost of 27 kg of iron = ₹1080
Cost of 1 kg of iron =$₹\frac{1080}{27}=₹40$
Cost of 120 kg of iron = $₹40\times 120=₹4800$

Hence, the cost of 120 kg of iron is ₹4800.

Answer:

Given that, the length of Qutub Minar = 72 m
Length of its shadow = 80 m
Length of shadow of electric pole = 1000 cm = 10 m
Thus, the length of the electric pole = $\frac{72}{80}\times 10=9 \mathrm{m}$

Answer:

For 50 girls food is available for 40 days.

For 1 girl food is available for (50 × 40) days i.e., 2000 days.

If 30 more girls join the hostel, then the total number of girls in the became 80.

For 80 girls food is available for $\frac{2000}{80} \mathrm{days}$ i.e., 25 days.

Answer:

(i) In mixture A,
Number of blue containers = 3
Number of white containers = 4
Thus, ratio of blue to white containers = $\frac{3}{4}=0.75$
In mixture B,
Number of blue containers = 3
Number of white containers = 3
Thus, ratio of blue to white containers = $\frac{3}{3}=1$
Clearly, mixture A is a lighter shade of blue.

(ii) In mixture C,
Number of blue containers = 3
Number of white containers = 3
Thus, ratio of blue to white containers = $\frac{3}{3}=1$
In mixture D,
Number of blue containers = 2
Number of white containers = 5
Thus, ratio of blue to white containers = $\frac{2}{5}=0.4$
Clearly, mixture D is a lighter shade of blue.

(iii) In mixture E,
Number of blue containers = 6
Number of white containers = 1
Thus, ratio of blue to white containers = $\frac{6}{1}=6$
In mixture F,
Number of blue containers = 4
Number of white containers = 2
Thus, ratio of blue to white containers = $\frac{4}{2}=2$
Clearly, mixture F is a lighter shade of blue.

(iv) In mixture G,
Number of blue containers = 3
Number of white containers = 3
Thus, ratio of blue to white containers = $\frac{3}{3}=1$
In mixture H,
Number of blue containers = 4
Number of white containers = 3
Thus, ratio of blue to white containers = $\frac{4}{3}=1.33$
Clearly, mixture G is a lighter shade of blue.

From the above mixtures, mixture D is the lightest among all mixtures.
Total number of containers required for painting = 105
Number of blue containers required for painting = $\frac{2}{7}\times 105=30$
Number of white containers required for painting = $\frac{5}{7}\times 105=75$

Answer:

From the figure,
Purple = 12, Blue = 20 and White = 16
Total squares = 12 + 20 + 16 = 48

Five ratio statement:
I: Purple : Total = 12 : 48 = 1 : 4
II: Blue : Total = 20 : 48 = 5 : 12
III: White : Total = 16 : 48 = 1 : 3
IV: Purple : Blue = 12 : 20 = 3 : 5
V: Purple : White = 12 : 16 = 3 : 4

Answer:

Total number of children = 10
If each child received 4 sweets, then the total number of sweets = 10 × 4 = 40 sweets
If 40 sweets are distributed between 8 children, then each get $\frac{40}{8}$ i.e. 5 sweets.

Answer:

Given that, cows that can graze a field in 9 days = 44
Number of cows that can graze the same field in 1 day = 44 × 9 = 396 cows
Number of cows that can graze the same field in 12 days = $\frac{396}{12}=33$ cows
Hence, (44 − 33) i.e., 11 more cows will graze the same field in 12 days.

Answer:

In 17 days, persons required to reap a field = 30
In 1 day, persons required to reap a field = 30 × 17
In 10 days, persons required to reap a field = $\frac{30\times 17}{10}=51$ persons

Answer:

Distance covered by Subham in 20 mins = 6 km/h = $6\times \frac{1000}{60}\times 20=2000 \mathrm{m}$ 

Shabnam's speed to cover 2000 m distance in 24 mins = $\frac{2000}{24}=\frac{250}{3}$ m/min

= $\frac{250}{3}\times \frac{60}{1000}=\frac{20}{4}=5$ km/h

Hence, the speed of Shabnam to cover the same distance in 24 minutes is 5 km/h.

Answer:

Let the total distance be x km.
Let the time taken by Ravi to reach school be t min.
If the speed of the bicycle is 10 km/h, then he reaches his school late by 8 minutes.
$$\begin{gathered} \therefore \frac{x}{10}=t+\frac{8}{60} \\ \Rightarrow \frac{x}{10}=t+\frac{2}{15} .....\left(1\right) \end{gathered}$$
If the speed of the bicycle is 16 km/h, then he reaches his school early by 10 minutes.
$$\begin{gathered} \therefore \frac{x}{16}=t-\frac{10}{60} \\ \Rightarrow \frac{x}{16}=t-\frac{1}{6} .....\left(2\right) \end{gathered}$$
From (1) and (2), we get
$$\begin{gathered} \frac{x}{10}-\frac{x}{16}=\frac{2}{15}-\left(-\frac{1}{6}\right) \\ \Rightarrow \frac{8x-5x}{80}=\frac{12+15}{90} \\ \Rightarrow \frac{3x}{80}=\frac{27}{90} \\ \Rightarrow x=\frac{27}{90}\times \frac{80}{3} \\ \Rightarrow x=9\times \frac{8}{9}=8 \end{gathered}$$

Substitute x = 8 in (2), we get
$$\begin{gathered} \therefore \frac{8}{16}=t-\frac{1}{6} \\ \Rightarrow t=\frac{1}{6}+\frac{1}{2} \\ \Rightarrow t=\frac{1+3}{6} \\ \Rightarrow t=\frac{2}{3} \mathrm{h} \\ \Rightarrow t=\frac{2}{3}\times 60 \min \\ \Rightarrow t=40 \min \end{gathered}$$

Since starting time of school is (8 : 20 a.m. + 40 min) i.e. 9:00 a.m.

Answer:

Answer:

In 75 g of sauted fish, protein = 20 g
In 1 g of sauted fish, protein = $\frac{20}{75}$ g
In 225 g of sauted fish, protein = $\frac{20}{75}\times 225=60$ g

Answer:

The distance between Jhareda to Ganwari in the map = 3.5 cm
Given scale, 1 cm = 10 km
So, actual distance between the villages = 3.5 × 10 = 35 km

Answer:

Given that, the height of the tree = 9.5 m
Shadow of the tree = 8 m
Since the lengths of the shadows are directly proprotional to their heights.
Shadow of the water tank = 21 m
$$\begin{gathered} \therefore \frac{8}{9.5}=\frac{21}{X} \\ \Rightarrow X=\frac{21\times 9.5}{8} \\ \Rightarrow X\approx 24.9 \mathrm{m} \end{gathered}$$

Hence, the height of the water tank is 24.9 m.

Answer:

Elevator A takes 29 s to cover a distance of 435 m.
Thus, the distance covered by elevator A in 1 s = $\frac{435}{29}=15 \mathrm{m}$

Elevator B takes 28 s to cover a distance of 448 m.
Thus, the distance covered by elevator B in 1 s = $\frac{448}{28}=16 \mathrm{m}$

Elevator C takes 10 s to cover a distance of 130 m.
Thus, the distance covered by elevator C in 1 s = $\frac{130}{10}=13 \mathrm{m}$

Elevator D takes 5 s to cover a distance of 85 m.
Thus, the distance covered by elevator D in 1 s = $\frac{85}{5}=17 \mathrm{m}$

Hence, in 1 s, elevator D covers more distance than any other elevator.
So, elevator D is fastest, while elevator C is slowest.

Now, elevator B cover distance in 140 s = 140 × 16 = 2240 m
Elevator C cover distance in 140 s = 140 × 13 = 1820 m
Thus, elevator B covers more distance than C = 2240 − 1820 = 420 m

Answer:

Length of the volleyball court = 18 m
Breadth of the volleyball court = 9 m
Length of the swimming pool = 75 m
Let, breadth of the swimming pool be x m.
Since the dimensions of the volleyball court and swimming pool are directly proportional.
$$\begin{gathered} \therefore \frac{9}{18}=\frac{x}{75} \\ \Rightarrow x=\frac{75}{2}=37.5 \mathrm{m} \end{gathered}$$
Hence, the width of the pool is 37.5 m.

Answer:

A muffins requires 1 cup of milk and 1.5 cups of chocolates.
Riya has 7.5 cups of chocolates.
Number of cups of milk required = $\frac{7.5}{1.5}=5$ cups

Answer:

In pattern A,
Number of red dots = 4
Number of blue dots = 2
Proportion involving red dots and blue dots in pattern A = $\frac{4}{2}=2$
Since pattern B consists of four tiles like pattern A.

In pattern B,
Number of red dots = 4 × 4 = 16
Number of blue dots = 2 × 4 = 8
Proportion involving red dots and blue dots in pattern B = $\frac{16}{8}=2$

Yes, they are in direct proportion and the constant of proportion is 2.

Answer:

We have,
$\begin{array}{rcl}\therefore \mathrm{Speed} \mathrm{of} \mathrm{the} \mathrm{cricket} \mathrm{ball}& =& 120 \mathrm{km}/\mathrm{h}\\ & =& 120\times \frac{1000}{60\times 60} \mathrm{m}/\mathrm{s}\\ & =& \frac{200}{6} \mathrm{m}/\mathrm{s}\\ & =& \frac{100}{3} \mathrm{m}/\mathrm{s}\end{array}$
Time taken by the ball to cover 20 m = $\frac{20}{{\displaystyle \frac{100}{3}}}=\frac{3}{5}=0.6 \sec$

Answer:

The variable x is inversely proportional to y.
$$\begin{gathered} \therefore x\propto \frac{1}{y} \\ \Rightarrow x=\frac{k}{y} \\ \Rightarrow xy=k\left(\mathrm{constant}\right) \end{gathered}$$

Since, we know that two quantities x and y are said to be in inverse proportion.
If an increase in x cause a proportional decrease in y and vice-versa.
So, we can say y decrease by p%.

Answer:

(a) Total number of black keys = 7
Total number of white keys = 10
Thus, the ratio of white keys to black keys = $\frac{10}{7}$ = 10 : 7

(b) Total number of keys = 10 + 7 = 17
Thus, the ratio of black keys to all keys = $\frac{7}{17}$ = 7 : 17

(c) Number of black keys in 1 keyboard = 7
Number of black keys in 14 keyboard = 14 × 7 = 98 keys

Answer:

From the given table, the distance travelled by one of the new eco-friendly energy-efficient earns travelled on gas.
The car travelled 15 km In 1 L of gas.
The car travelled 7.5 km in 0.5 L of gas.
The car travelled 30 km in 2 L of gas.
This rate shows a direct proportion between litres of gas and the distance covered.
The car can cover the distance in 8 L of gas = 8 × 15 = 120 km

Answer:

Kratika's sister used $1-\frac{1}{6}=\frac{5}{6}$ cup of sugar syrup.
Kratika left with $\frac{1}{6}$ cup of syrup.
She need $\frac{1}{3}$ cup of syrup for one piece of bread.
Thus, ingredient will be in proportion of $\frac{1}{2}$.

Now, the bread recipe will be
$\frac{1}{2}$ cup quick cooking oats
1 cup bread flour
$\frac{1}{6}$ cup sugar syrup
$\frac{1}{2}$ tablespoon cooking oil
$\frac{2}{3}$cups water
$1\frac{1}{2}$ tablespoons yeast
$\frac{1}{2}$ teaspoon salt.

Answer:

Students-teacher ratio = 35 : 1
In school, students increased = 280
Thus, the teacher required for 280 students = $\frac{280}{35}=8$ teachers

Answer:

50 miles per hour is the same as travelling 80 kilometres per hour.

Thus, 1 km = $\frac{50}{80}$ mile

⇒ 200 km = $\frac{50}{80}\times 200=125$ mile

Hence, Kusum has to go 125 miles for 200 km.

Answer:

(a) For 60 posters, amount raised = ₹250
For 1 poster, amount raised = $₹\frac{250}{60}=₹\frac{25}{6}$
For 102 posters, amount raised = $₹\frac{25}{6}\times 102=₹425$
Hence, Anju’s class make ₹425 for selling 102 posters.

(b) For 1 poster, amount raised = $₹\frac{250}{60}=₹\frac{25}{6}$
For ₹2000, posters need to sell = $2000÷\frac{25}{6}=480$ posters