Variation
Class 10th Mathematics West Bengal Board Solution
- Corresponding values of two variables A and B are : |c|c|c|c|c| a&25&30&45&250…
- Corresponding values of two variables x and y |c|c|c|c|c| x&18&8&12&6 y&3& 27/4 & 9/2 &…
- A taxi of Bipin uncle travels 14 km path in 25 minute. Let us calculate by applying…
- A box of sweets is divided among 24 children of class one of our school, they will get…
- 50 villagers had taken 18 days to dig a pond. Let us calculate by using theory of…
- y varies directly with square root of x and y = 9 when x = 9. Let us find the value of…
- x varies directly with y and inversely with z. When y=4, z=5, then x=3. Again, if…
- x varies directly with y and inversely with z. When y=5, z=9 then x = 1/6 Let us find…
- If x ∝ y, let us show that x + y ∝ x - y.
- a proportional 1/c , c proportional 1/b let us show that A ∝ B.
- If a proportional b , b proportional 1/c and c ∝ d, let us write the relation of…
- If x ∝ y, y ∝ z and z ∝ x. Let us find the relation among three constants of…
- x^2 + y^2 ∝ xy If x + y ∝ x - y, let us show that
- x^3 + y^3 ∝ x^3 - y^3 If x + y ∝ x - y, let us show that
- ax + by ∝ px + qy [where a, b, p, q are non zero constant] If x + y ∝ x - y, let us…
- If a^2 + b^2 ∝ ab, let us prove that a + b ∝ a - b.
- If x^3 + y^3 ∝ x^3 - y^3 , let us prove that x + y ∝ x - y.
- If 15 farmers can cultivate 18 bighas of land in 5 days, let us determine by using…
- Volume of a sphere varies directly with the cube of its radius. Three solid spheres…
- y is a sum of two variables, one of which varies directly with x and another varies…
- If a ∝ b, b ∝ c let us show that a^3 b^3 + b^3 c^3 + c^3 a^3 ∝ abc (a^3 + b^3 + c^3).…
- To dig a well of x dcm deep. One part of the total expenses varies directly with x and…
- Volume of a cylinder is in joint variation with square of the length of radius of base…
- An agricultural Co-operative Society of village of Pachla has purchased a tractor.…
- Volume of a sphere varies directly with cube of length of its radius and surface area…
- Q16A1 x proportional 1/y thenA. x = 1/y B. y = 1/x C. xy = 1 D. xy = non zero constant.…
- Q16A2 If x ∝ y thenA. x^2 ∝ y^3 B. x^3 ∝ y^2 C. x ∝ y^3 D. x^2 ∝ y^2…
- Q16A3 If x ∝ y and y = 8 when x = 2; if y = 16, then the value of x ifA. 2 B. 4 C. 6 D. 8…
- Q16A4 If x ∝ y^2 and y = 4 when x = 8; if x = 32, then the value of y isA. 4 B. 8 C. 16 D.…
- Q16A5 If y-z proportional 1/x , z-x proportional 1/y and x-y proportional 1/z sum of three…
- Let us write whether the following statements are true or false : i. If y…
- Let us fill in the blanks : i. If x proportional 1/y and y proportional 1/z then x∝…
- If x ∝ y^2 and y = 2a when x = a; x and y let us find the relation between x and y.…
- If x ∝ y, y ∝ z and z ∝ x, let us find the product of three non zero constants.…
- If x proportional 1/y and y proportional 1/z let us find if there be any relation of…
- If x ∝ yz and y ∝ zx let us show that z is a non zero constant.
- If b ∝ a^3 and a increases in the ratio of 2 : 3, let us find in what ratio b will be…
Let Us Work Out 13
Question 1.Corresponding values of two variables A and B are :
If there is any relation of variation between A and B, let us determine it and write the value of variation constant.
Answer:
Given:
We see in the given table, as the value of A increases or decreases the corresponding value of B is also increasing or decreasing.
Hence, A ∝ B and here value of variation constant is 
.
Here, the value of variation constant is positive. i.e. for value of A increases or decreases the corresponding value of B will also increases or decreases.
Question 2.
Corresponding values of two variables x and y
If there is any relation of variation between x and y, let us write it by understanding.
Answer:
Given:
We see in the given table, as the value of x increases or decreases the corresponding value of y is also decreasing or increasing respectively.
i.e. xy = 54 = constant
Hence, 
and here value of variation constant is 54.
Here, the value of variation constant is positive. i.e. for value of A increases or decreases the corresponding value of B will also decreases or increases respectively.
Question 3.
A taxi of Bipin uncle travels 14 km path in 25 minute. Let us calculate by applying theory of variation how much path he will go in 5 hours by driving taxi with same speed.
Answer:
Given:x1 = 14km and t1 = 25min.
Let the distance cover = x
Time taken = t
2nd time t2 = 5hrs = 5× 60 = 300min.
2nd distance = x2
As speed is constant,
So, x ∝ t
⇒ 25 × x2 = 14 × 300
⇒ x2 = 14× 12
⇒ x2 = 168 km
Hence it will cover 168km in 300min or 5hrs.
Question 4.
A box of sweets is divided among 24 children of class one of our school, they will get 5 sweets each. Let us calculate by applying theory of variation how many sweets would each get, if the number of the children is reduced by 4.
Answer:
Given: number of children (x) = 24
Number of sweets each get (y) = 5
New number of children (x1) = 24 – 4 = 20
Let new Number of sweets each get = y1
As per given condition if the number of students reduced then each will get more sweets than previously given. Hence, it is the question of indirect proportion.
i.e. xy = x1y1
or 24×5 = 20 × y1
⇒ y1 = 6
Hence, each children will get 6 sweets.
Question 5.
50 villagers had taken 18 days to dig a pond. Let us calculate by using theory of variation how many extra persons will be required to dig the pond is 15 days.
Answer:
Given: number of villagers (x) = 50
Number of days to dig a pond (y) = 18
Let New number of villagers (x1) = (x1)
New Number of days to dig = 15
As per given condition if the number of days reduced then number of villagers will increases than the previously given. Hence, it is the question of indirect proportion.
i.e. xy = x1y1
or 50 ×18 = 15 × x1
⇒ x1 = 60 days
Hence, 15 villagers will need 60 days.
Question 6.
y varies directly with square root of x and y = 9 when x = 9. Let us find the value of x when y = 6.
Answer:
Given: y ∝ √x
y = k√x
Acc. To given values y = 9 and x = 9,
9 = k√9
⇒ 9 = k (3)
⇒ k = 3
So, y = 3√x …(1)
Put y = 6 in (1)
⇒ 6 = 3√x
⇒ √x = 2
Square on both sides,
x = 4.
Question 7.
x varies directly with y and inversely with z. When y=4, z=5, then x=3. Again, if y=16, z = 30, let us write by calculating the value of x.
Answer:
Given: 
Acc. To given values y = 4, z = 5 and x = 3,
So, 
…(1)
Put y = 16 and z = 30 in (1)
⇒ x = 2
Question 8.
x varies directly with y and inversely with z. When y=5, z=9 then 
Let us find the relation among three variables x, y and z and if y = 6 and 
let us write by calculating the value of x.
Answer:
Given: 
Acc. To given values y = 5, z = 9 and x = 1/6,
So, 
…(1)
Put y = 6 and z = 1/5 in (1)
⇒ x = 9
Question 9.
If x ∝ y, let us show that x + y ∝ x – y.
Answer:
Given:
x ∝ y
x = ky
Do componendo and dividend,
i.e. x + y ∝ x – y
hence proved.
Question 10.
let us show that A ∝ B.
Answer:
Given:
It can be written as,
Or
i.e. A = constant B
Or A ∝ B
Hence proved.
Question 11.
If 
and c ∝ d, let us write the relation of variation between a and d.
Answer:
Given:
It can be written as,
Put the value of b in a,
Now, put the value of c in a,
We get,
This is the required relation between a and d.
Question 12.
If x ∝ y, y ∝ z and z ∝ x. Let us find the relation among three constants of variation.
Answer:
Given: x ∝ y , y ∝ z and z ∝ x
It can be written as,
x = k y, y = m z and z = n x
where k, m and n are constant of variations.
Put the value of y in x,
We get,
x = k m z
Now put the value of z in x,
⇒ x = k m n x
⇒ k m n = 1
This is the required relation among three constants of variation i.e. k, m and n.
Question 13.
If x + y ∝ x – y, let us show that
x2 + y2∝ xy
Answer:
Given: x + y ∝ x – y
It can be written as,
x + y = k(x – y)
Square on both sides,
⇒ x2 + y2 + 2xy = k(x2 + y2 + 2xy)
⇒ x2 + y2 + 2xy = kx2 + ky2 + k2xy
⇒ x2 - kx2 + y2- ky2 = k2xy - 2xy
⇒ x2 (1-k) + y2 (1-k) = 2xy (k-1)
⇒ (1-k) {x2 + y2} = -2xy(1-k)
⇒ x2 + y2 = -2xy
As -2 is also a constant term.
Hence,
⇒ x2 + y2∝ xy
Hence proved.
Question 14.
If x + y ∝ x – y, let us show that
x3 + y3∝ x3 – y3
Answer:
Given: x + y ∝ x – y
It can be written as,
x + y = k(x – y)
Apply componendo dividend,
Cube on both sides,
Again apply componendo dividend,
i.e. x3 + y3∝ x3 - y3
Hence,
x3 + y3∝ x3 - y3
Hence proved.
Question 15.
If x + y ∝ x – y, let us show that
ax + by ∝ px + qy
[where a, b, p, q are non zero constant]
Answer:
Given: x + y ∝ x – y
ax + by ∝ px + qy
Question 16.
If a2 + b2∝ ab, let us prove that a + b ∝ a – b.
Answer:
Given: a2 + b2∝ ab to prove a + b ∝ a-b
a2 + b2= 2kab
as 2k is a constant.
It can be written as:
Apply componendo dividend,
Square root on both sides,
i.e a + b ∝ a-b
hence proved.
Question 17.
If x3 + y3∝ x3 – y3, let us prove that x + y ∝ x – y.
Answer:
Given:
x3 + y3∝ x3 - y3
Apply componendo dividend,
Cube root on both sides,
Again apply componendo dividend,
i.e. x + y ∝ x-y
Question 18.
If 15 farmers can cultivate 18 bighas of land in 5 days, let us determine by using theory of variation the number of days required by 10 farmers to cultivate 12 bighas of land.
Answer:
Given:
More will be the number of farmers, more bighas of land will be cultivated i.e. direct proportion.
More will be farmers, less number of days will be needed i.e. indirect proportion.
Acc. To given values,
For, F = 10, Bighas of land = 12
⇒ D = 5days
Question 19.
Volume of a sphere varies directly with the cube of its radius. Three solid spheres having length of 
and 
metre diameter are melted and a new solid sphere is formed. Let us find the length of diameter of the new sphere. [let us consider that the volume of sphere remains same before and after melting]
Answer:
Given:
Radius of first sphere = 
Radius of second sphere = 
Radius of third sphere = 
Let volume is v and radius is r
Acc . to given condition:
V ∝ r3
V = kr3
Volume of sphere of radius 3/4 m is = 
Volume of sphere of radius 1 m is = k (1)3 = 1k
Volume of sphere of radius 5/4m is = 
Total volume will be :
If the radius of new sphere is R
Diameter of new sphere will be = 2R = 3 m
Question 20.
y is a sum of two variables, one of which varies directly with x and another varies inversely with x. With x = - 1, then y = 1 and when x = 3, then y = 5. Let us find the relation between x and y.
Answer:
Given:
y = a + b
Acc. To given condition:
a ∝ x
a = kx
Put the values of a and b in y.
x = -1 then y = 1
⇒ 1 = -k – m ….(1)
x = 3 then y = 5
⇒ 15 = 9k + 3m …(2)
Multiply (1) by 3
We get,
3 = -3k – 3m …(3)
Add (2) and (3)
18 = 6k
⇒ k = 3
Put it in (1)
1 = -3 –m
⇒ m = -4
This is the required relation between x and y.
Question 21.
If a ∝ b, b ∝ c let us show that a3b3 + b3c3 + c3a3∝ abc (a3 + b3 + c3).
Answer:
Given: a ∝ b, b ∝ c
a = kb and b = mc
it can be written as:
a = kb = kmc and b = mc
To prove: a3b3 + b3c3 + c3a3∝ abc(a3 + b3 + c3)
Hence, a3b3 + b3c3 + c3a3∝ abc(a3 + b3 + c3)
Question 22.
To dig a well of x dcm deep. One part of the total expenses varies directly with x and other part varies directly with x2. If the expenses of digging wells of 100 dcm and 200 dcm depths are ₹ 5000 and ₹ 12000 respectively, let us write by calculating the expenses of digging a well of 250 dcm depth.
Answer:
Given:
Acc. To given condition:
First part of expense (t1) ∝ x
Where x is the depth of well.
t1 = kx
second part of expense (t2) ∝ x2
t2 = mx2
total expense (t) = t1 + t2
It can be written as:
t = k x + m x2
First depth = 100 dcm and first expense = Rs.5000
5000 = k 100 + m(100)2)
5000 = 100k + 10000m …(1)
First depth = 200 dcm and first expense = Rs.12000
12000 = k 200 + m(200)2)
12000 = 200k + 40000m …(2)
Multiply (1) by 2 :
10000 = 200k + 20000m …..(3)
Subtract (3) from (2)
2000 = 20000m
Put m in (1)
5000 = 100k + 1000
100k = 5000 – 1000
100k = 4000
⇒ k = 40
For x = 250 dcm
⇒ t = 10000 + 6250
t = Rs16250
So total expene will be Rs. 16250
Question 23.
Volume of a cylinder is in joint variation with square of the length of radius of base and its height, Ratio of radii of bases of two cylinders is 2 : 3 and ratio of their heights is 5 : 4, let us find the ratio of their volumes.
Answer:
Given:
Acc. To given condition:
Volume of cylinder ∝ (radius)2(height)
V ∝ (r)2(h)
Radius of two cylinders = r1 : r2 = 2 : 3
Heights of two cylinders = h1 : h2 = 5 : 4
Volume of two cylinders = v1 : v2
Then, the ratio of their volumes = 5 : 9.
Question 24.
An agricultural Co-operative Society of village of Pachla has purchased a tractor. Previously 2400 bighas of land were cultivated by 25 ploughs in 36 days. Now half of the land can be cultivated only by that-tractor in 30 days. Let us calculate by using the theory of variation, the number of ploughs work equally with one tractor.
Answer:
Given:
More will be the number of ploughs, more bighas of land will be cultivated i.e. direct proportion.
More will be ploughs, less number of days will be needed i.e. indirect proportion.
Acc. To given values,
For, D = 30, Bighas of land = 1200
So, 15 ploughs are needed for 1200 bighas of land in 30 days.
Question 25.
Volume of a sphere varies directly with cube of length of its radius and surface area of sphere varies directly with the square of the length of radius. Let us prove that the square of volume of sphere varies directly with cube of its surface area.
Answer:
Given:
Acc. To given conditions:
Volume of sphere ∝ (radius)3
V ∝ (r)3
V = k (r)3 ….(1)
Surface area of sphere ∝ (radius)2
S ∝ (r)2
S = m (r)2
Put the value of r in (1)
Square on both sides,
Where k and m are non zero constants.
So,
v2∝ s3
hence proved.
Question 26.
then
A. 
B. 
C. xy = 1
D. xy = non zero constant.
Answer:
Given:
⇒ xy = k = non zero constant
Option (d) is correct.
Question 27.
If x ∝ y then
A. x2∝ y3
B. x3∝ y2
C. x ∝ y3
D. x2∝ y2
Answer:
Given:
x ∝ y
it can written as,
x = ky
square on both sides,
x2 = k2 y2
as k2 is also a constant.
x2 = k y2
or
x2∝ y2
option (D) is correct.
Question 28.
If x ∝ y and y = 8 when x = 2; if y = 16, then the value of x if
A. 2
B. 4
C. 6
D. 8
Answer:
Given:
x ∝ y
it can written as,
x = ky
y = 8 and x = 2
2 = k (8)
When y = 16
⇒ x = 4
Option (b) is correct.
Question 29.
If x ∝ y2 and y = 4 when x = 8; if x = 32, then the value of y is
A. 4
B. 8
C. 16
D. 32
Answer:
Given:
x ∝ y2
it can written as,
x = ky2
y = 4 and x = 8 (given)
8 = k (4)2
When x = 32,
⇒ y2 = 64
⇒ y = 8
Option (b) is correct.
Question 30.
If 
and 
sum of three variation constants is
A. 0
B. 1
C. -1
D. 2
Answer:
Given:
Where k, m and n are constants,
Adding all the three equations we get,
y-z + z – x + x – y = 
⇒ 
Now, k, m, and n are also constants.
Then, the sum of these variation constants will also be 0
Question 31.
Let us write whether the following statements are true or false :
i. If 
non-zero constant.
ii. If x ∝ z and y ∝ z then y ∝ z
Answer:
(i) Given:
⇒ xy = k = non zero constant
Hence, given statement is not true.
(ii) Given: x ∝ z , y ∝ z
It can be written as,
x = k y, y = m z
where k, m are constant of variations.
Put the value of y in x,
We get,
x = k m z
⇒ x ∝ z
Multiply both sides by y.
We get,
xy ∝ yz
Hence, given statement is not true.
Question 32.
Let us fill in the blanks :
i. If 
and 
then x∝ _______
ii. If x ∝ y, xn∝ ________
iii. If x ∝ y and x ∝ z, then (y + z)∝ ________
Answer:
(i) Given:
Put the value of y in x.
As k and m are constant.
⇒ x ∝ z
Hence, If and 
then x ∝ _z.
(ii) Given: x ∝ y
x = ky
xn = kn.yn
where kn is a constant.
Hence xn ∝ yn
Hence, If 
__yn_
(iii) Given: x ∝ y and x ∝ z
x = ky and x = mz
Hence, y + z = ?
So, y + z ∝ x
If 
and 
then 
__x__
Question 33.
If x ∝ y2 and y = 2a when x = a; x and y let us find the relation between x and y.
Answer:
Given:
x ∝ y2
x = ky2 ……(1)
when x = a then y = 2a
put the value of x and y in equ.(1)
we get,
a = k (2a)2
a = 4a2k
Put the value of k in (1)
This is the required relation between x and y.
Question 34.
If x ∝ y, y ∝ z and z ∝ x, let us find the product of three non zero constants.
Answer:
Given: x ∝ y , y ∝ z and z ∝ x
It can be written as,
x = k y, y = m z and z = n x
where k, m and n are constant of variations.
Put the value of y in x,
We get,
x = k m z
Now put the value of z in x,
⇒ x = k m n x
⇒ k m n = 1
Thus the product of three non zero constants is 1.
Question 35.
If 
and 
let us find if there be any relation of direct or inverse variation between x and z.
Answer:
Given:
Put the value of y in x.
As k and m are constant.
⇒ x ∝ z
So, x and z are in direct variation.
Question 36.
If x ∝ yz and y ∝ zx let us show that z is a non zero constant.
Answer:
Given: x ∝ yz and y ∝ zx
x = kyz
y = mzx
Put the value of y in x.
x = k(mzx)z
x = kmxz2
As k and m are constant.
⇒ z ∝ non zero constant
Hence proved.
Question 37.
If b ∝ a3 and a increases in the ratio of 2 : 3, let us find in what ratio b will be increased.
Answer:
Given: b ∝ a3
b = ka3
Hence, b must increase in ratio 8 : 27.