WBBSE Solutions For Class 10 Maths Chapter 13 Variation Exercise 13.1

Maths WBBSE Class 10 Solutions Chapter 13 Variation Exercise 13.1

Question 1. If the square art paper having sides of 60 cm length is covered with coloured paper on its four sides, then let us write by calculating how much length of coloured paper is required.

Solution:

Given

If the square art paper having sides of 60 cm length is covered with coloured paper on its four sides,

Coloured paper is required = 4 x 60 cm of length = 240 cm in length.

If the length of one side of the square art paper is 50 cm, then the coloured paper is required to cover the art paper of 4 x 50 = 200 cm length.

Question 2. Similarly, let us write by understanding that the number of pens (x) and the total cost price of the pen (y) are in direct variation. 

Solution: No. of pens (x)

The total cost of the pen (y)

Here No. of the pen (x) & Total cost price (y) are in direct variation such that,

= Constant, i.e., x $ y.

Read and Learn More WBBSE Solutions For Class 10 Maths

Question 3. Let us write four examples related to two variables, where the variables are in direct variations.

We have got values of interrelated variables A and B :

Solution: Four examples related to two variables, where the variables are in direct variation are:

(a) Distance (x) & Time (t) is in direct variation when speed in constant.

(b) Base (b) & Area (A) of the triangle are in direct variation when height is constant.

(c) Area of base (b) & Volume (V) of the cylinder is in direct variation when height is constant.

(d) Temperature (T) in kelvin scale & volume of gas (V) are in direct variation.

Question 7. Let us find if there is any variation relation between P and Q.

Solution: Here the value of P increases/decreases, and the value of Q increases/decreases.

Here, P/Q =35/15 = 49/21 = 56/24 = 14/6 = 7/3

PaQ & the constant of variation = 7/3

“WBBSE Class 10 Maths Variation Exercise 13.1 solutions”

Application 1. y varies directly with the square root of x and y = 9 when x = 4, let us write the value of variational constant; and let us express y as a function of x and if y = 8, let us write the value of x.

Solution:

Given

y varies directly with the square root of x and y = 9 when x = 4,

Here, y varies directly with the square root of x

i.e., y $ √x

or, y = K√x

y = 9, when x = 4

∴ 9 K√4   k = 9/2

..y = 9/2 √x

or, 8 = 9/2 √x

or, √x = 8x 2 /9 

..X= 256/81

.. When y = 8, x = 256 / 81

The value of x= 256 / 81

Application 2. By counting, we see that there are 24 coconut laddus in the tiffin box. We shall divide them equally without breaking any laddu, let us find how many laddus we shall each get.

Solution:

Given

By counting, we see that there are 24 coconut laddus in the tiffin box. We shall divide them equally without breaking any laddu

We shall each get 24 ÷ 2 = 12 laddus.

But Shibani too has joined us in this activity.

So, we shall divide 24 laddus among 3 persons. Now each will get 8 laddus.

But it was expected that three friends would come to the club room. If they will come we shall be 6 people in total.

So, if 24 laddus are divided among 6 persons then each will get 4 laddus. 

“West Bengal Board Class 10 Maths Chapter 13 Variation Exercise 13.1 solutions”

Application 3. I arranged 36 buttons in a rectangular size and let us write by understanding in each case if there is any inverse variation between buttons lying along the length and those of breadth. 

Solution:

Here no. of buttons lying in length increases but no. of buttons lying in breadth decreases. 

But x x y = (4 x 9); (2 x 18); (3 x 12); (9 x 4)

.. x α 1/y & variation constant = 36

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Application 4. Let us write two examples with two interrelated variables where the variables are in inverse variation. 

Solution: 1. Pressure (P) & Volume (V) gas are in inverse proportion.

2. Resistance (R) & Cross Section (A) of wire are in inverse proportion.

Application 5. The values of interrelated variables x and y are

“WBBSE Class 10 Variation Exercise 13.1 solutions explained”

Let us determine if there is any variation relation between x and y. 

Solution: Here x.y = Constant (54)

Application 6.

Solution: Let speed = 4 & Time = T

∴ V α 1/T

or, V = K. 

1/T [Where K is a constant]

or, 60= K.1/2

∴ K = 120.

When V = 80, V = K/T

or, 80 = 120 / T

∴ T = 120/80

T = 3/2

∴ Samir babu takes 1 1/2 hours to reach the station from his house.

Application 7. If A²+ B² x A² – B², let us prove that A & B 

Solution:

Given

A²+B²/ A² -B² = k/1  (Constant)

A²+B²+A² -B² / A²+B²-A²2+B²

or, 2A² / 2B² = K+1 / K-1

A/B = √K+1 / √K-1 = P (Constant)

Application 8. Here, is V in joint variation with T and 1/P? V = R. [R = constant] in this relation we can say V is in joint variation with P. number of persons engaged in that work and number of days they worked. 

Solution: Let total earning (A); No. of men B & No. of days = C.

.. Aα B when C is constant

& A αC when B is constant.

.. Aα BC when B & C both vary.

We can say, A is in joint variation with B & C.

“WBBSE Class 10 Maths Exercise 13.1 Variation problem solutions”

Application 9. If 5 men can cultivate 10 bighas of land in 9 days, let us calculate by the theory of variation how long will be taken 25 men for cultivating 10 bighas of land. 

Solution:

Given

5 men can cultivate 10 bighas of land in 9 days

Let No. of men = A, 

No. of days = B, & Area of land = C

No. of men is in direct variation with an area of land when no. of days remains constant, 

i.e., A αC when B is constant.

Again, No. of men is in inverse variation with no. of days when the area of land remains constant,

i.e., A α 1/B when C is constant.

.. According to the theorem on joint variation,

Αα C/B when B & C both vary

.. A K. C/B where K is a constant.

Given A=5, B= 9, C = 10

5=K 10/9 or, 

10K = 9×5 = 9/2

A = K C/B

25= 9/2 x 10/B

or, B = 9×10 / 2×25

=9/5

=1 4/5

“Class 10 WBBSE Maths Exercise 13.1 Variation step-by-step solutions”

Application 10. If x x y and y o z, let us prove that x2 + y2+ z2 α xy + yz + zx

Solution: x α y ∴ x = K₁y

& y α z ∴ y = K₁z

∴ x = K₁y = K₁K₂z

Now, \(\frac{x^2+y^2+z^2}{x y+y z+z x}=\frac{\left(K_1 K_2 z\right)^2+\left(K_2 z\right)^2+z^2}{K_1 K_2 z \cdot K_2 z+K_2 z \cdot z+z \cdot K_1 K_2 z}\)

= \(\frac{K_1^2 K_2^2 z^2+K_2^2 z^2+z^2}{K_1 K_2^2 z^2+K_2 z^2+K_1 K_2 z^2}=\frac{z^2\left(K_1^2 K_2^2+K_2^2\right)}{z^2\left(K_1 K_2^2+K_2+K_1 K_2\right.}\) = P (Constant)

∴ \(x^2+y^2+z^2 \propto x y+y z+z x\) Proved.

“WBBSE Class 10 Chapter 13 Variation Exercise 13.1 solution guide”

Application 11. the volume of a sphere varies directly with the cube of the length of its radius. The length of the diameter of a solid sphere of lead is 14 cm. If the sphere is melted, let us find by applying the theory of variation how many spheres having a length of 3.5 cm radius can be made. (let the volume remain unchanged before or after melting.)

Solution:

Let the volume of a sphere of radius r is v.

Given

the volume of a sphere varies directly with the cube of the length of its radius. The length of the diameter of a solid sphere of lead is 14 cm. If the sphere is melted,

v α r³

or v = Kr³. Where K is a constant?

Diameter of lead sphere = 14 cm

∴ its radius (r) = 7  cm

∴ Volume = K(7³) cucm

Volume of small sphere of radius 3.5 cm = K x (3.5)³ cucm.

As the volume remains same before & after melting,

no. of small spheres = \(\frac{K \times 7^3}{K \times(3.5)^3}=\frac{K \times 7 \times 7 \times 7}{K \times 3.5 \times 3.5 \times 3.5}=8 .\)