In math variation we solved numerous types of problems on variation by using different types of variation like direct variation, inverse variation and joint variation. The problems on variation are mainly related to the questions based on word problems of constant variation, word problems of direct variation, word problems of inverse variation and also word problems of joint variation. Each word problems on variation are explained step by step so that students can understand the question and their solution easily.

**1. The area of an umbrella
varies directly as the square of its radius. If the radius of the umbrella is
doubled, how much will be the area of the umbrella?**

**Solution:**

If the area of the umbrella is C and radius is R then C α R^{2}
or C= KR^{2} where K is the constant of variation.

So the area of the umbrella is KR^{2}.

Now if the radius is doubled the area will be

K(2R)^{2 }= 4KR^{2 }= 4C.

So the area will be by 4 times of normal the area of the umbrella.

Let V be the volume of a globe of radius R.

Then, by problem,

V ∝ R

Therefore V = kR

If v

v

v

v

Let v cubic metre be the volume of the new solid globe. Then,

v = v

or, v = 216k/64

or, v = 27k/8

If the radius of the new solid globe be r metre, then using (1) we get,

v = kr

or, kr

or, r

or, r = 3/2.

**3. If 5 men take 8 days
to type 10 books, apply the principle of variation to find out how many days 8
men will take to type 2 books?**

**Solution:**

If N represents numbers of men, D is number of days and B is number of books the from the principle of variation

N α \(\frac{1}{D}\) or, N is in inverse variation with D as when numbers of men increase it will take less time, so the numbers of days will decrease.

N α B or N is in direct variation with B as when numbers of men increase they can type more, so the numbers of books can be typed will increase.

From the theorem of joint theorem

N α \(\frac{B}{D}\)

or, N = K\(\frac{B}{D}\) where K is constant of variation.

For the given data

5 = K × \(\frac{10}{8}\)

or, K = \(\frac{40}{10}\) = 4.

Substituting the value of K in the variation equation

N = 4\(\frac{B}{D}\)

For 8 men to type 2 books number

8 = 4 x \(\frac{2}{D}\)

or, D = 1.

So it will 1 day.

Let us assume that M men take D days to plough A acres.

Clearly, M ∝ A when D is constant and, M ∝ 1/D when A is constant.

Hence, using the theorem of joint variation, we get, M ∝ A ∙ 1/D when both A and D vary.

Therefore, M = k ∙ A/D ……... (1) [k = constant of variation]

Given M = 5 when A = 10 and D = 9

Putting these values in (1), we get,

5 = k ∙ 10/9 or, l0k = 45

or, k= 9/2

Therefore, the law of variation is [putting the value o k in (1)],

M = 9/2 ∙ A/D …………….… (2)

when M = 25 and A = 30, we get from (2),

25 = 9/2 ∙ 30/D

or, 50 D = 270

or, D = 27/5 = 5

Therefore the required number of days = 5

**5. In X is in indirect variation
with square of Y and when X is 3, Y is 4. What is the value of X when Y is 4?**

From the given problem indirect variation equation can be expressed as

X = \(\frac{K}{Y^{2}}\)

or, K = XY^{2}

For the given case

K
= 3 x 4^{2 }=48.

So when Y is 4,

XY^{2}
= 48

or, X = \(\frac{48}{Y^{2}}\)

= \(\frac{K}{4^{2}}\)

= 3

So the value of X is 3.

The profits are $ 54 per head per month when there are 50 boarders in the boarding house and each boarder pays $ 390 a month.

Hence, the real expense per boarder = $ (390 - 54) = $ 336 a month.

Therefore, the total expenses of the boarding house when there are 50 boarders = $ 50 × 336 = $ 16,800.

Similarly, the total expenses when there are 60 boarders = $ 60 × (390 - 64) = $ 60 × 326 = $ 19,560.

Now, let C

Then, C

or, C

Let C denote the total expenses when there are n boarders.

Then, C = C

Now, C = 16,800 when n = 50;

putting these values in (1) we get,

C

Again, C = 19,560 when n = 60;

hence, from (1) we get,

C

Subtracting (2) from (3) we get,

l0k = 2760

or, k = 276

Putting the value of k in (2) we get,

C

or, C

Putting the values of C

C = 3,000 + 276n ...... (4)

Now let x denote the total expenses of the boarding house when there are 80 boarders.

Then, C = x when n = 80.

Therefore, from (4) we get, x = 3,000 + 276 × 80 = 25,080.

Hence, the actual expenses are $ 25,080 when there are 80 boarders ; but 80 boarders pay $ 390 × 80 = $ 31,200.

Therefore, the total profit of the boarding house when there are 80 boarders

= $ (31,200 - 25,080) = $ 6,120.

**Note: ***Variation is a very important part of algebra in higher grade and college grade. By practicing the problems of variation student get very clear concept on different types of variation.*

**7. If a car runs at a average speed
of 40kmph with some regular intervals and takes 3 hrs to run a distance of 90
km, what time it will take to run at a average speed of 60 kmph with same
intervals to run 120 km?**

If T is the time taken to cover the distance and S is the distance and V is the average speed of the car,

The from the theory of variation

V α S or V varies directly with S when T is constant as when average speed will increase for a fixed time, distance covered by the car will increase.

V α \(\frac{1}{T}\) or, V varies inversely with T when S is constant as when average speed will increase to cover a fixed distance, time taken by the car to will decrease.

So V α \(\frac{S}{T}\)

or, V = K\(\frac{S}{T}\) where K is the constant of variation.

For the case given in the problem

V = K\(\frac{S}{T}\)

40 = K x \(\frac{90}{3}\)

or, K = \(\frac{4}{3}\)

Substituting the value of K in the variation equation

V = \(\frac{4S}{3T}\)

So at a average speed of the car is 60kmph to run 120 km it will take

V = \(\frac{4S}{3T}\)

*
*

or, 60 = \(\frac{4 × 120}{3T}\)

or, T =\(\frac{60}{160}\)

= \(\frac{3}{8}\) hrs

= \(\frac{3}{8}\) × 60 mins

= 22.5 mins.

Calculate the percentage loss thus incurred by the breakage.

Let V be the value of a precious stone of weight W.

Then, from the condition of the problem, we have,

V ∝ W

V ∝ kW

Since the weights of the three broken pieces are proportional to 2 : 3: 5, we assume their weights as 2w, 3w and 5w respectively.

Hence, the weight of the unbroken piece of stone = 2w + 3w + 5w = 10w.

Then, V = 15,600 when W = l0w.

Hence, from (1) we get,

15,600 = k ∙ (10w)

or, kw

Let, v

Then, from (1) we get,

v

v

and v

Therefore the total value of the 3 pieces

= $ (v

= $ (8kw

= $ 160kw

= $ 160 × 15.6

= $ 2,496.

Therefore, the total loss incurred by the breakage

= (the value of the original stone) - (the total value of the 3 pieces)

= $ 15,600 - $ 2,496 = $ 13,104.

{w

Let x and y be the weights of an empty ship and its cargo respectively whose length is 1.

Then, by problem, x ∝ l

Therefore, x = kl

Therefore, x + y = kl

Given, x + y = w

hence, from (1) we get,

w

or, w

Similarly, w

Therefore w

= (k + ml

= k(l

= k × 0 + m × 0

= 0.

Let W be the weight of a sphere of radius R and D be the density of the material of which it is made.

From the condition of the problem, we have,

W ∝ R

or, W = kR

By the conditions of the problem, if 17r be the radius of the first sphere then that for the second is 8r; if 3d is the density of the material of the first sphere, then that for the second is 4d. Let w kg. be the weight of the first sphere. Then, for the first sphere,

W = w when R = 17r and D = 3d.

Therefore from (1) we get, w = k ∙ (17r)

And, for the second sphere we have,

W= 40 kg. when R = 8r and D = 4d.

Therefore, from (1) we get, 40 = k ∙ (8r)

Now, dividing (2) by (3) we get,

w/40 = {k ∙ (17r)

or , w = {(17)

Let l be the illumination at a distance D from the source of light. From the condition of the problem, we have,

I ∝ 1/D

or, I = k/D

Let i and be the illuminations at a distance of 9 cm. and x cm. respectively from the source of light.

Then, I = i when D = 9;

hence, from (1) we get,

i = k/9

Again, I = i/3 when D = x;

hence, from (1) we get,

i /3 = k/x

Dividing (2) by (3) we get,

i ÷ i/3 = k/9

or , i × 3/i = k/9

or, 3 = x

or , x

or, x = 9√3 (since x > 0)

Therefore, the book must be removed further through a distance of (9√3 - 9) cm. = 6.6 cm. (approx.).

Let us assume that the speed of the engine diminishes by v when n wagons are attached with it.

Again, if V be the actual speed of the engine when n wagons are attached, then we must have,

V = 40 - v ……..... (1)

where, v ∝ √n or, v = k√n , k = constant of variation.

Therefore, from (1) we get, V = 40 - k√n ……..... (2)

Given, V = 28 when n = 16;

hence, from (2) we get,

28 = 40 - k√16

or, 4k = l2

or, k = 3

Putting the value of k in (2), we get,

V = 40 - 3√n ………….... (3)

Now, V = 10 when,

10 = 40 - 3√n

or, 3√n = 30

or, √n = 10

or, n = 100.

Hence, the actual speed of the engine is 10km. per hour when 100 wagons are attached with it.

Therefore, the greatest number of wagons with which the speed of the engine do not fall below 10 km. per hour is 100.

Let, w = weight of each ring (in carat)

x = value of one carat of gold (in $)

y = cost of workmanship for each ring (in $)

and $ V be the value of a diamond of weight W carat.

Then, by the condition of the problem,

V ∝ W

V = kW

If V

V

V

V

Since the weight of each ring is w carat and diamonds in them weigh 3, 4, 5 carats respectively, hence golds in them weigh (w - 3), (w - 4) and (w - 5) carats respectively.

Since the price of a ring = price of diamond + price of gold + cost of workmanship, hence we must have,

9k + (w – 3)x + y = a ………… (2) [for the first ring]

16k + (w - 4)x + y = b ….……... (3) [for the second ring]

and 25k + (w - 5)x + y = c …………. (4) [for the third ring]

Now, from (2), (3) and (4) we get,

(a + c)/2 – b = {9k + (w - 3)x + y + 25k + (w - 5)x + y}/2 - [16k + (w - 4)x + y]

or, (a + c)/2 - b = {34k + (2w - 8)x + 2y}/2 - [16k + (w - 4)x + y]

or, (a + c)/2 - b = 17k + (w - 4)x + y - 16k - (w - 4)x - y

or, (a + c)/2 – b = k.

Again, when W = 1, then from (1) we get,

W = k ∙ 1

i.e., the value of one carat of diamond .

= $ k = $ {(a + c)/2 – b) .

Let $ P be the amount of pension of a man whose length of service is n years. Then, by the condition of the problem, we have,

P ∝ √n

or, P = k√n ……….... (1) (k = constant of variation]

Now, assume that the second man has serverd for x years and receives a pension of $ y. Then, the first man has served for (x + 9) years and receives a pension of $ (y + 500).

Hence, using (1) we get,

y = k√x …….. (2) [for the second man]

and y + 500 = k√(x + 9) ……….. (3) [for the first man]

Subtracting (2) from (3) we get,

k{√(x + 9) - √x} = 500 ………..... (4)

Again, if the length of service of the first had exceeded that of the second by 4

Therefore, using (1) we get,

{k ∙ (√(x + 4

or, (x + 17/4)/x = 81/64

or, 81x = 64x + 17 ∙ 16

or, 17x = 17 ∙ 16

or, x = 16

Now, putting the value of x in (4) we get,

k ∙ {(√(l6 + 9) - √16 } = 500

or, k ∙ (5 - 4) = 500

or, k = 500.

Therefore, from (2) we get,

y = 500 ∙ √16 = 2000.

Let $ C be the cost of a metallic ball of radius R mm. Then, by the condition of the problem, we have,

C = x + y …………… (1)

where, x ∝ R

Therefore, x = k

Putting the values of x and y in (1) we get,

C = k

Given, C = 1 when R= 2 ;

hence, from (2) we get,

1 = k

or, 8k1+ 4k2= 1 …….. (3)

Again, C = 3.15 when R = 3 ; hence, from (2) we get,

3.15 = k

or, 27k

From (4) and (3) we get,

k

and k

_________________________________

(Subtracting) k

Therefore, k

or, k

Putting the values of k

C= 0.10 × R

Therefore, when R = 4, then C = 0.10 × 4

Therefore, the required cost of a ball of radius 4 mm. is $ 7.20.

**●** **Variation**

**What is Variation?****Direct Variation****Inverse Variation****Joint Variation****Theorem of Joint Variation****Worked out Examples on Variation****Problems on Variation**

**11 and 12 Grade Math**** ****From Problems on Variation to HOME PAGE**

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