Problems on Variation
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 volume of a globe varies as the cube of its radius. Three solid globes of diameters 1
^{1}/
_{2}, 2 and 2
^{1}/
_{2} metres are melted and formed into a new solid globe. Find the diameter of the new globe.
Solution:
Let V be the volume of a globe of radius R.
Then, by problem,
V ∝ R
^{3}
Therefore V = kR
^{3} ..... (1) [k = constant of variation ]
If v
_{1}, v
_{2} and v
_{3} cubic metres be the respective volumes of globes having radii ¾, 1 and 5/4 metres then using (1) we get,
v
_{1} = k ∙ (3/4)
^{3} = 27k/64 ;
v
_{2} = k ∙ l
^{3} = k;
v
_{3} = k ∙ (5/4)
^{3} = 125k/64
Let v cubic metre be the volume of the new solid globe. Then,
v = v
_{1} + v
_{2} + v
_{3}
or, v = 27k/64 + k + 125k/64
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
_{3}
or, kr
^{3} = 27k/8
or, r
^{3} = (3/2)
^{3}
or, r = 3/2.
Therefore, the diameter of the new globe = 2r = 2 ∙ 3/2 = 3 metres.
2. Apply the principle of variation to find how long 25 men will take to plough 30 acres, if 5 men take 9 days to plough 10 acres.
Solution:
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
^{2}/
_{5}
Therefore the required number of days = 5
^{2}/
_{5}
3. The expense of a boarding house is partly fixed and partly vary with the number of boarders. Each boarder pays $ 390 a month. The profits are $ 54 per head per month when there are 50 boarders and $ 64 when there are 60 boarders. What is the profit per head per month when there are 80 boarders?
Solution:
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
_{1} and C
_{2} denote the fixed part and variable part of expenses respectively when there are n boarders in the boarding house.
Then, C
_{2} ∝ n
or, C
_{2} = kn where k = constant of variation.
Let C denote the total expenses when there are n boarders.
Then, C = C
_{1} + C
_{2} = C
_{1} + kn ..... (1)
Now, C = 16,800 when n = 50;
putting these values in (1) we get,
C
_{1} + 50k = 16,800 …... (2)
Again, C = 19,560 when n = 60;
hence, from (1) we get,
C
_{1} + 60k = 19,560 ……... (3)
Subtracting (2) from (3) we get,
l0k = 2760
or, k = 276
Putting the value of k in (2) we get,
C
_{1} + 50 × 276 = 16,800
or, C
_{1} = 16,800 - 13,800 = 3,000
Putting the values of C
_{1} and k in (1), we get,
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.
Hence the required profit per head per month = $ 6120/80 = $ 76.50.
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.
4. A precious stone worth $ 15,600 is accidentally dropped and broken into three pieces the weights of which are respectively proportional to 2 : 3 : 5. The value of a stone of this variety varies as the cube of its weight.
Calculate the percentage loss thus incurred by the breakage.
Solution:
Let V be the value of a precious stone of weight W.
Then, from the condition of the problem, we have,
V ∝ W
^{3}
V ∝ kW
^{3} ……..... (1) [k = constant of variation]
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)
^{3} = 1000kw
^{3}
or, kw
^{3} = 15,600/1000 = 15.6.
Let, v
_{1}, v
_{2} and v
_{3} be the values of the 3 pieces of weights 2w, 3w and 5w respectively.
Then, from (1) we get,
v
_{1} = k(2w)
^{3} = 8kw
^{3};
v
_{2} = k(3w)
^{3} = 27kw
^{3}
and v
_{3} = k(5w)
^{3} = 125kw
^{3}.
Therefore the total value of the 3 pieces
= $ (v
_{1} + v
_{2} + v
_{3})
= $ (8kw
^{3} + 27kw
^{3} + 125kw
^{3})
= $ 160kw
^{3}
= $ 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.
Hence, the required percentage loss (13,104/l5,600) × 100 % = 84 %.
5. The weight of an empty ship and its cargo vary as the square and the cube of the length respectively. If w
_{1} be the weight with cargo of a ship of length l
_{1}, and w
_{2}, w
_{3} the similar weights corresponding to lengths 1
_{2} and l
_{3} respectively, prove that,
{w
_{1}/l
_{1}^{2}} (l
_{2} - l
_{3}) + {w
_{2}/l
_{2}^{2}}(l
_{3} - l
_{1})+ w
_{3}/l
_{3}^{2} (l
_{1} - l
_{2}) = 0.
Solution:
Let x and y be the weights of an empty ship and its cargo respectively whose length is 1.
Then, by problem, x ∝ l
^{2} and y ∝ l
^{3}
Therefore, x = kl
^{2} and y = ml
^{3} where k and m are constants of variation.
Therefore, x + y = kl
^{2} + ml
^{3} …….... (1)
Given, x + y = w
_{1} when l = l
_{1};
hence, from (1) we get,
w
_{1} = kl
_{1}^{2} + ml
_{1}^{3}
or, w
_{1}/l
_{1}^{2} = k + m1
_{1}
Similarly, w
_{2} /l
_{2}^{2} = k + ml
_{2}
and w
_{3}/l
_{3}^{2} = k + ml
_{3}
Therefore w
_{1}/l
_{1}^{2} (l
_{2} - l
_{3}) + w
_{2}/l
_{2}^{2} (l
_{3} - l
_{1}) + w
_{3}/l
_{3}^{2} (l
_{1} – l
_{2})
= (k + ml
_{1}) (l
_{2} - l
_{3}) + (k + ml
_{2}) (l
_{3} - l
_{1}) + (k + ml
_{3}) (l
_{1} - l
_{2})
= k(l
_{2} - l
_{3} + l
_{3} - l
_{1} + l
_{1} - l
_{2}) + m[l
_{1} (l
_{2} - l
_{3})
+ l
_{2} (l
_{3} - l
_{1}) + l
_{3} (l
_{1} - l
_{2})]
= k × 0 + m × 0
= 0.
Proved.
Solved examples on application problems on joint variation are explained here step-by-step with detailed explanation:
6. The weight of a sphere varies jointly as the cube of its radius and the density of the material of which it is made. The radii of two spheres are as 17 : 8 and the densities of materials as 3 : 4. If the weight of the second sphere be 40 kg. find the weight of the first.
Solution:
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
^{3}D
or, W = kR
^{3} D ….... (1) [k = constant of variation ]
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)
^{3} ∙ 3d ………. (2)
And, for the second sphere we have,
W= 40 kg. when R = 8r and D = 4d.
Therefore, from (1) we get, 40 = k ∙ (8r)
^{3} ∙ 4d ……... (3)
Now, dividing (2) by (3) we get,
w/40 = {k ∙ (17r)
^{3} ∙ 3d}/{k ∙ (8r)
^{3} ∙ 4d} = {(17)
^{3} ∙ 3}/{8
^{3} ∙ 4}
or , w = {(17)
^{3} ∙ 3 ∙ 40}/{8
^{3} ∙ 4} = 287.87 (approx.)
Therefore the required weight of the first sphere = 287.87 kg.
7. The illumination from a source of light varies inversely as the square of the distance. A book is at a distance of 9 cm. from a lamp. Find how much further the book is to be removed so that it receives one-third as much light.
Solution:
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
^{2}
or, I = k/D
^{2} …………... (1) [k = constant of variation]
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
^{2} ……….. (2)
Again, I = i/3 when D = x;
hence, from (1) we get,
i /3 = k/x
^{2} .………. (3)
Dividing (2) by (3) we get,
i ÷ i/3 = k/9
^{2} ÷ k/x
^{2}
or , i × 3/i = k/9
^{2} × x
^{2}/k
or, 3 = x
^{2}/9
^{2}
or , x
^{2} = 9
^{2} ∙ 3
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.).
More solved problems on variation are explained here step-by-step with detailed explanation:
8. A locomotive engine without any wagon can go at a rate of 40 km. per hour and its speed diminishes by a quantity which varies as the square root of the number of wagons attached. If with 16 wagons its speed is 28 km. per hour, what is the greatest number of wagons that can be attached if the speed is not to fall below 10 km. per hour.
Solution:
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.
9. The price of a diamond varies as the square of its weight. Three rings of equal weight, each composed of a diamond set in gold, have values $ a, $ b, $ c, the diamonds in them weighing 3, 4, 5 carats respectively. Show that the value of a diamond of one carat is $ {(a + c)/2 – b}, the cost of workmanship being the same for each ring.
Solution:
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
^{2} or,
V = kW
^{2} ……... (1) [k = constant of variation.]
If V
_{1}, V
_{2} and V
_{3} be the values of 3, 4 and 5 carats of diamonds respectively; then from (1) we get,
V
_{1} = k ∙ 3
^{2} = 9k;
V
_{2} = k ∙ 4
^{2} = 16k;
V
_{3} = k ∙ 5
^{2} = 25k.
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
^{2} = k.
i.e., the value of one carat of diamond .
= $ k = $ {(a + c)/2 – b) .
Proved.
10. Two persons are awarded pensions in proportion to the square root of the number of years they have served. One has served 9 years longer than the other and receives a pension greater by $ 500. If the length of service of the first had exceeded that of the second by 4
/_{4} years their pensions would have been in the proportion of 9 : 8. How long had they served and what were their respective pensions?
Solution:
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/_{4} years, their pension would have been in the proportion 9 : 8.
Therefore, using (1) we get,
{k ∙ (√(x + 4/_{4})}/(k ∙ √x) = 9/8
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.
Therefore, the two men served for 16 years and (16 + 9) = 25 years and their respective pensions were $ 2000 and $ (2000 + 500) = $ 2500.
11. The cost of a kind of metallic balls is partly proportional to the cube of its radius and the rest is proportional to the square of its radius. If the cost of such balls of radii 2 mm. and 3 mm. are respectively $ 1 and $ 3.15 each, find the cost of a ball of the same kind of radius 4mm.
Solution:
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^{3} and y ∝ R^{2}
Therefore, x = k_{1} R^{3} and y = k_{2}R^{2} where k_{1} and k_{2} are constants of variation.
Putting the values of x and y in (1) we get,
C = k_{1}R^{3} + k_{2}R^{2} ………… (2)
Given, C = 1 when R= 2 ;
hence, from (2) we get,
1 = k_{1} ∙ 2^{3} + k22^{2}
or, 8k1+ 4k2= 1 …….. (3)
Again, C = 3.15 when R = 3 ; hence, from (2) we get,
3.15 = k_{1} ∙ 3^{3} + k_{2} ∙ 3^{2}
or, 27k_{1} + 9k_{2} = 3.15 …….(4)
From (4) and (3) we get,
k_{2} + 3k_{1} = 0.35 [dividing (4) by 9]
and k_{2} + 2k_{1} = 0.25 [dividing (3) by 4]
_________________________________
(Subtracting) k_{1} = 0.10
Therefore, k_{2} + 2 × 0.10 = 0.25
or, k_{2} = 0.05.
Putting the values of k_{1} and k_{2} in (2) we get,
C= 0.10 × R^{3} + 0.05 × R^{2}
Therefore, when R = 4, then C = 0.10 × 4^{3} + 0.05 × 4^{2} = 6.40 + 0.80 =7.20
Therefore, the required cost of a ball of radius 4 mm. is $ 7.20.
● Variation
11 and 12 Grade Math
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