Proportion

Given:

Amount invested: ₹10000 and ₹15000

Interest after 1 year: ₹ 900 for first amount and ₹1200 for second amount.

i). To check whether interest proportional to the investments.

Interest ratio

Investment ratio

If Interest ratio is equal to investment ratio, then only both are in proportion.

As we can observe from above solution, interest ratio is not equal to investment ratio.

i.e.

ii). Ratio of interest to investment

In first case, amount invested is ₹10000 and interest earned is ₹900

∴ Ratio

Now,

In second case, amount invested is ₹15000 and interest earned is ₹1200

∴ Ratio

iii). Annual rate of the scheme

Rate

In first scheme, rate

⇒

∴ Annual rate in this scheme is 9% per annum.

In second scheme, rate

⇒

⇒

⇒ 2 × 4 = 8%

∴ Annual rate in this scheme is 8% per annum.

Given:

Area of A0 paper = 1m²

Area of A1 paper = half of 1

Area of A2 paper = half of

Area of A3 paper = half of

Area of A2 paper = half of

Now we calculate the sides of A4 sizes:

Let the length be l and width be w

Area = Length × width

⇒

We know that, l = √2 × w

⇒

⇒

⇒

⇒ w² = 0.04419

⇒ w = √0.04419

⇒ w = 0.2102 m

⇒ w = 210.2 mm

So, length = √2 × 210.2 mm

= 1.414 × 210.2

= 297.3 mm

Given:

Ratio of masses of calcium, carbon and oxygen in calcium carbonate is 10: 3: 12

Now,

In 150 grams, 60 gram of calcium, 20gram of carbon and 70gram of oxygen

Ratio of masses in 150gram of a compound

⇒ 60: 20: 70

⇒ 6: 2: 7

As we observe that ratio given for calcium carbonate is 10: 3: 12 but in 150 grams of compound the ratio is 6: 2: 7.

∴ 150gram compound is not the calcium carbonate.

Perimeter and radius of circle.

Let the perimeter be y and radius of circle be r.

Perimeter = 2πr

⇒ y = 2 × π × r

⇒ y = 2πr

We can whatever value we like for x on right side of the above equation. So, x is a variable.

Whatever value we give for r, that the value will be multiplied by a constant 2π. So, y will always be proportional to nr.

Area and radius of circle

Let the area be y and radius be r

Area = πr²

⇒ y = π × r²

⇒ y = πx²

We can whatever value we like for r on right side of the above equation. So, r is a variable.

Whatever value we give for r, that the value will be multiplied by a constant π and the variable r. That means r is not multiplied by the same constant every time.

So, y is not proportional to r.

The distance travelled and the number of rotations of a circular ring moving along a line.

Let the distance travelled be d and the number of rotation be n.

We know,

The distance travelled in one rotation = 2πr

where r is the radius of the circle

so, distance travelled in n rotation =d = 2nπr

We can whatever value we like for n on right side of the above equation. So, n is a variable. But r is a constant, we are considering the movement of a single ring.

Whatever value we give for n, that the value will be multiplied by a constant 2πr. So, d will always be proportional to n.

The interest got in any year and the amount deposited in a scheme in which interest in compounded annually.

Let the interest got in year be ‘i' and the amount deposited be p.

We know that,

i = p × r (where r is the rate of interest)

We can whatever value we like for p on right side of the above equation. So, p is a variable. But r is a constant for a scheme.

Whatever value we give for p, that the value will be multiplied by a constant r. So, i will always be proportional to p.

The volume of water poured into a hollow prism and the height of the water level.

Let the volume of hollow prism be v and height of water level be h.

Let the base area of the prism be ‘a’

The poured volume = volume of the water column formed in the prism

But volume for water column in the prism = base × height

⇒ v = a × h

⇒ v = ah

⇒

We can whatever value we like for v on right side of the above equation. So, v is a variable. But a is a constant for a prism.

Whatever value we give for n, that the value will be multiplied by a constant . So, d will always be proportional to n.

Let us consider a surface with three squares of height h and side 1m on a region.

On this surface, we can mark a random number of squares, each side of side 1m. So, each of such squares will have an area of 1m².

It is stated in the question that, “the volume of water falling in each square metre may be considered equal”.

So, each of the three squares will receive equal volume of water.

If this water is not allowed to sun off and if there is no sewage into the ground, the water received will become a water column. There will be three water columns.

These water columns are prism with square base. And they have the same volume (V), because, according to volume of water in each square should be considered equal.

We know that,

Volume = base area × height

But base area of all prism = 1m²

⇒ V = 1 × h

⇒ V = h

But V is same for all three prisms.

Let us consider two hollow prisms kept near to one another with different base.

We can think of random number of water columns inside these prisms. Let the base of these columns be squares of area 1cm².

Then each water column will be a square of prism of base area 1cm².

We have seen in above part, that the heights of all water prisms be equal. Regardless of whether they are in the first hollow prism or second hollow prism.

Let the height be h.

Let the base area of first prism = a₁

Let the base area of second prism = a₂

Then total number of water prisms (each with base area 1cm²) in first prism

Then total number of water prisms (each with base area 1cm²) in second prism

All the a₁ and a₂ in the first prism and second prism will have the same heights h.

That means, the water level in both prism will be h.

So, whatever number of hollow prism (of different base sizes) we place near one another, after the rainfall, the height of water in all will be the same.

Consider two paddy fields in a locality. Let their areas be a₁ and a_2. If the water is allowed to run off and if there is no sewage into the ground, there will be two water prisms. Each will cover the entire area of the respective field.

The volume of first field = v₁ = a₁h

The volume of second field = v₂ = a₂h

h is a constant. So, if area increases, volume increases and if area decreases, volume decreases.

That means volume is proportional to the area.

Let the different position of spring 0, 1 and 2.

Position 0 shows the situation when no load is applied on the spring

Position 1 shows the situation when a load of w₁ is applied on the spring.

We can see that there is an extension of x cm from the initial position.

Position 2 shows the situation when a load of w₂ is applied on the spring.

We can see that there is extension of x₂ cm from the initial position.

It is given in the question that; the extension is proportional to the applied load.

So, any extension x will be proportional to the weight w that produces that extension.

We can write: x = a constant k × w

⇒ x = kw

⇒

The constant ‘k’ is called the spring constant. Every spring will have a particular value of spring constant. We want to find this constant for our spring.

For that, we adopt the following procedure:

Put a known weight w₁. Measure the extension x₁.

Then

Put another known weight w₂. Measure the extension x₂.

Then

Since k is a constant, we will get the same value for k in both the weight.

Repeat the trial with several known weights. In all cases we will get the same k.

Once k is obtained, we can make the markings on the spring balance.

The spring is fixed inside an outer casting. Markings are made on this casing.

When the spring is at Zero load position, mark that position on the casing as 0kg.

Now we want to mark the 1kg position

Let the extension for a load of 1kg be x₁ kg

x₁ = k × 1 kg = k

we have already calculated k. so we get x₁ kg.

Measure this x₁ kg from the 0kg mark on the casing and mark it as 1kg.

Now we want to mark the 2kg position

Let the extension for a load of 2kg be x₂ kg

X₂ = k × 2 kg = 2k

we have already calculated k. so we get x₂ kg.

Measure this x₂ kg from the 0kg mark on the casing and mark it as 2kg.

i). The given angle in question figure is reproduced in below figure.

Let the blue object which moves along the slanting line is marked as Q and perpendicular is dropped from Q to the horizontal leg of the given angle. The foot of perpendicular is marked as P.

So, we get a triangle Δ APQ

There is another right-angle ΔABC. This is our base triangle. That is, we are going to calculations based on Δ ABC:

We assume that BC is fixed at its position and also that all sides of Δ ABC are known.

But PQ is not fixed. Because, Q can be at any point along the slanting line.

Now, we find relation between Δ ABC and Δ APQ:

∠ A is denoted as x° (It is same for both the Δ)

∠ P and ∠ B are 90°

∠ c and ∠ Q will both be equal to (90 – x) °

So, both the triangles have the same angles. Therefore, they are similar.

i.e. Δ ABC ≅ Δ APQ

⇒

From the above,

∵ BC and PQ are heights and AC and AQ are distance.

⇒

⇒

⇒

is constant bcause Δ ABC is fixed.

⇒ AQ = k × PQ

That means, the distance of Q from the vertex A is proportional to the height from Q from the horizontal line.

ii). In this part we explore the cases when the angle x at vertex A is 30°, 60° and 45°.

First, we will take 30°.

A base triangle Δ ABC is drawn such that AC = 2cm, BC = 1cm and AB = √3

As we see in figure, here also Δ ABC and Δ APQ are similar

So, we get:

∵ BC and PQ are heights and AC and AQ are distance.

⇒

⇒

⇒

⇒ AQ = 2 PQ

In the above result, ‘2’ is constant. So AQ is proportional to PQ.

That means the distance of Q from the vertex A is proportional to the height Q from the horizontal line.

Now take x°= 60°,

A base triangle Δ ABC is drawn such that AC = 2cm, AB = 1cm and BC = √3

As we see in figure, here also Δ ABC and Δ APQ are similar

So, we get:

∵ BC and PQ are heights and AC and AQ are distance.

⇒

⇒

⇒

In the above result, is constant. So AQ is proportional to PQ.

That means the distance of Q from the vertex A is proportional to the height Q from the horizontal line.

Now take x°= 35°,

A base triangle Δ ABC is drawn such that AC = √2cm, AB = 1cm and BC = 1cm

As we see in figure, here also Δ ABC and Δ APQ are similar

So, we get:

∵ BC and PQ are heights and AC and AQ are distance.

⇒

⇒

⇒

⇒ AQ = √2 PQ

In the above result, is constant. So AQ is proportional to PQ.

That means the distance of Q from the vertex A is proportional to the height Q from the horizontal line.

We know that area of equilateral triangle is given by a

Where s is the length of side.

Put s² = q

Then

Here, is constant

When ‘q’ increases ‘a’ increases

When ‘q’ decreases ‘a’ decreases

So, ‘a’ is proportional to q. That means, ‘a’ is proportional to the square of the side

The constant of proportionality is .

Let us consider a square of side ‘s’ and its area ‘a’

Let s = 1cm

Area = (side)²

∴ a = 1 × 1 = 1cm²

Now, let us change the side and see how it affects the area:

Let s = 2 cm, now a = (2)² = 4 cm²

Let s = 2.25 cm, now a = (2.25)² = 5.0625 cm²

Let s = 3 cm, now a = (3)² = 9 cm²

Let s = 0.4 cm, now a = (0.4)² = 0.16 cm²

Now we will write the above result in a tabular form, and calculate a/s ratio in each case:

From the table, we can see that a/s ratio is not a constant. We will not get ‘a’ by multiplying ‘s’ by a fixed number. So, a is not proportional to s.

Let the area of rectangle be a, length of the rectangle be x and breadth be y.

Then the area of rectangle = Length × breadth

⇒ a = x × y

⇒ a = xy

Area is given as 1m², which is constant.

⇒ xy = 1

This is the algebraic equation which gives the relation between length and breadth of a rectangle whose area is 1m².

Now we see if there is any proportionality between length and breadth

We have xy = 1. So, length or breadth increases, the other decreases.

Also, if length or breadth decreases, the other increases

Their product will remain constant only if this simultaneous increase and decrease take place.

We can write:

⇒

So, length is proportional to the reciprocal of the breadth.

That means, length and width are inversely proportional.

The constant of proportionality is 1.

Let us consider triangles of same area.

That means area is constant. Let it be ‘a’.

Let the length of the longest side be x and length of the perpendicular from the opposite vertex to this longest side be y.

Then we have,

⇒

⇒ xy = 2a

Here 2a is constant. So, if x or y increases, the other decreases.

Also, if x or y decreases, the other increases.

Their product will remain constant only if this simultaneous increase and decrease take place.

We can write:

⇒

So, x is inversely proportional to y

If we change the shape of the triangle while keeping the area the same, the longest side that we considered may become the shortest side. Then y should increase proportionately so that area will remain the same.

Let us consider a regular polygon.

We know how to calculate the sum of all its interior angles.

s = 180 (n – 2)

Where, s = sum

N is the number of sides of the regular polygon

Let us use this formula for a triangle:

For a triangle, n = 3

So, s = 180 × (3 – 2)

= 180 × 1

= 180

Let us use this formula for a square:

For a square, n = 4

So, s = 180 × (4 – 2)

= 180 × 2

= 360

Let us use this formula for a pentagon:

For a square, n = 5

So, s = 180 × (5 – 2)

= 180 × 3

= 540

Let us tabulate the results:

From the above table we can see that s/n is not a constant. So, s is not proportional to n.

Let us modify the formula

Let s = 180 × m

Where s = sum

M = (n-2)

N is the number of side of the regular polygon

We can see that s is proportional to m. The constant of proportionality is 180

In ordinary language, we can say this:

The sum of interior angle of a regular polygon is proportional to ‘2 less than the number of sides’.

i). Let the rate of flow be ‘r’ m³/s.

That means, in 1 second, ‘r’ m³ of water will enter the tank.

Let the base area of the tank be ‘a’m².

Then in the 1^st second, that is, when t = 1, the height of water level will be:

[∵ after 1 second, the volume in the tank will be r m³]

In the 2^nd second, that is, when t = 2, the height of water level will be:

[∵ after 2 seconds, the volume in the tank will be 2r m³]

In the 3^rd second, that is, when t = 3, the height of water level will be:

So, we can write:

The height of water after the n^th second = h

⇒

n is a constant because we will put a particular value of n. We want the height of water at that n

a is also constant

So, is a constant.

Thus, we have a relation between two quantities:

Height ‘h’ at the n^th second

Rate of flow ‘r’.

We can write: h = kr.

This is the algebraic equation.

Where k

From the equation, we can see that h is directly proportional to r.

The constant of proportionality is .

ii). We can use the same equation used above, i.e.

The height of water after the n^th second = h

⇒

In this case, h is a constant because of the following two reasons:

1. A fixed volume of water is flowing into the tank.

2. When the tank is filled, it will have a particular value of ‘h’.

n is the number of seconds required to fill the tank. It will change if the rate ‘r’ is increased or decreased.

So, n and r are the variables. Let us bring them to opposite side of the ‘=’ sign:

⇒

⇒

This is the algebraic equation.

‘ah’ is a constant. Let it be ‘k’.

We can write:

So, r is proportional to the reciprocal of n. That means n is inversely proportional to r.

If r increases and ‘n’ decreases, indicating a lesser time sufficient to fill up the tank.

If r decreases and ‘n’ increases, indicating a greater time required to fill up the tank.