West Bengal Board Class 10 Math Book Solution In English Chapter 13 Variation Exercise 13.2
Question 1. Corresponding values of two variables A & B are :
If there is any relation of variation between A and B, let us determine it and write the value of the variation constant.
Solution: Here when values of A increase/decrease, the values of B increase/decrease.
A/B = 25/10 = 30/12 = 45/10 = 250/100 = 5/2
∴ A B. & Variable Constant =
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Question 2. The corresponding values of two variables x & y are :
Solution: Here when the value of A increases/decreases the value of B decreases/ increases.
x.y = 18x³=8x 24/7
= 12 x 9/2
= 6×9
= 54
χ α 1/y
& Variation Constant = 54.
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Question 3.
1. A taxi from Bipin’s uncle travels a 14 km path in 25 minutes. Let us calculate by applying the theory of variation how many paths he will go in 5 hours by driving a taxi at the same speed.
Solution: Let required time = T & required distance = S.
When time increases, distance increases when speed is constant.
∴ T & S are in direct variation.
∴ T α S
or, T = KS where K is constant.
Here T = 25 & = 14
∴ 25 K x 14 or, 25/14
∴ T = 25/14 x s
Now when T = 300 [Show 300 minutes]
∴ 300 = 25/14
∴ S=300×14 / 25
= 168.
∴ He will go 168 km in 5 hours.
2. 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 the of variation how many sweets would each get if the number of children is reduced by 4.
Solution: Let no. of children = C & no. of sweet each will get = S
As total no. of sweets are fixed.
When C increases/decreases, S decreases/increases.
∴ C & S are in increased variation
∴ C α 1/S or, K/S [Where K is a Constant]
Here C = 24; & S = 5
24 = K/5
∴ 24 x 5
= 120
2nd case, C = 24-420, K = 120, S = ?
20 = 120/S or, S = 120/20= 6
∴ Each student will get 6 sweets.
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3. 50 villagers had taken 18 days to dig a pond. Let us calculate by using the theory of variation how many extra persons will be required to dig the pond in 15 days.
Solution: No. of villagers = V & no. of days = D
When V increases/decreases, D decreases/increases.
∴ V & D are in inverse variation.
∴ V α 1/D
or, V = K/D [Where K is a constant]
Here V = 50, D = 18
∴ 50 K/12
∴ K = 50 x 18 = 900
2nd case V = ? D= 15, K = 900
∴ V = 900 / 15
= 60
∴ Extra villagers required = 60-50
= 10.
Question 4.
1. y varies directly with the square root of x and y = 9 when x = 9. Let us find the value of x when y = 6.
Solution: y varies directly with the square root of x
ie, y α √ x
∴y = K√x
When y = 9, x=9 9 = K√9 .. K = 9/√9
=9/3
=3
∴y = 3√x
When y = 6,
∴6 = 3√x
∴√x = 6/3 =2
∴ x = 2² = 4
2. 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.
Solution: Here x x y when z is constant
xα1/z when y is constant.
xαy/z when y & z are both variables.
x = K. y/z (Where K is a Constant)
Given x = 3, y=4 & z=5
3= k 4/3 or, K = 3x 5 / 4
=15/4
When y = 16, z = 30, x = ?
X = K.y/z
X = 15/4 x 16/30
= 2
∴ x = 2
3. x varies directly with y and inversely with z. When y = 5, z = 9 then x == 1/6 Let us find the relation among three variables x, y and z and if y = 6 and z = 1/5, let us write by calculating the value of x.
Solution: x α y when z is constant
χ α 1/z when y is constant.
.. x α y/z, when both y & z are variables.
or, x = K, y x 1/z =Ky/z
Given x = ⅙, y=5 & z=9
..⅙ = K.5/9
..k = 9/5 x 6
=3/9
Now, x = 3/10 . y/z
X = 3y/10z
Question 5.
1. If x α y, let us show that x + y α x-y.
2. If aα b, b α 1/c, and c c d, let us write the relation of variation between a and b.
3. If x α y, y a z, and z α x, let us find the relation among three constants of variation.
6. If x + y α x – y, let us show that
1. x²+ y² α xy
Solution: x+yαx-y.
or, x + y = K(xy) [Where K is a constant]
Or,x+y = K (x-y)
∴ x+y/x-y = k/1
or, (x+y)² / (x-y)² = k²/1
or, (x+y)²+(x-y)² / (x+y)-(x-y) = K²+1 / K²-1
Or, 2x/2y = K+1/K-1
Or, 2x² + 2y² / 4xy = K²+1/K²-1
or, 2(x²+y²)/24xy = K²+1/K²-1
∴x²+y² = 2(K²+1)/k²-1
∴ x²+y² α xy [Where 2(K²+1)/ K2-1is Constant.]
2. x³+ y³ α x³-y³
Solution: x + y x-y
∴ x+y = K(x-y) (Where K is constant)
x+y/x-y = K/1
or, x+y+x-y/x+y-x+y = K+1/K-1
or, 2x/2y = K+1/K-1 = p(let)
∴x/y = p
∴x = py (Where P is Constant)
Now, x³+y³/x³-y³ = (py)³ + y³ / (py)³ – y³
=y³+(p³+1) / y³-(p³-1)=Constant.
3. ax + by α px + ay [where a, b, p, q are non-zero constants]
Solution: x + y α x-y or, x + y = K(x – y)
or, x+y/x-y = K/1
or, x+y+x-y / x+y-x+y = K+1 / K-1
or, 2x/2y = K+1 / k-1 = R(let) (Where K & R are Constants)
∴ x = Ry
ax + by / Px+qy = aRy+by / PRy+ay
= y(aR+b) / y(PR+q)
= aR+b / PR+q = Constant
∴ ax + by α px + qy.
Question 7.
1. If a² + b² α ab, let us prove that a + bx a-b.
Solution: a² + b² α ab
or, a²+ b² Kab
or, a² + b² = K/2
a²+b² / 2ab = K/2
or, a² + b² +2ab / a²+b²-2ab = K+2 /K-2
= (a+b)²/ (a – b)² = K+2/K-2
or, a+b / a-b = √K+2 / √K-2=Constant
∴a+b α a-b Proved.
2. If x³+ y³ α x³-y³, let us prove that x + y α x-y.
Solution: x³+ y³ α x³- y³
∴ x³+ y³ = K(x³-ya) (where K is a Constant)
X³+y³ / x³-y³ = K/1
x³ +y³ +x³-y³ / x³+y³-x³+y³ = K+1 / K-1
or, 2x³/ 2y³ = K+1/K-1
or, x³+y³ = K+1 / K-1
x+y = 3√K+1 / 3√k-1 R (let) (Where Ris Constant)
∴ (x+y) α (x-y) Proved.
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Question 8. If 15 farmers can cultivate 18 bighas of land in 5 days, let us determine by using the theory of variation the number of days required by 10 farmers to cultivate 12 bighas of land.
Solution: No. of farmers A, no. of days = B & area of land = C
No. of days is in inverse variation with no. of farmers when area of land remains constant
i.e., B α 1/A when C is constant
Again, No. of days is an indirect variation with an area of land, when No. of farmers remains constant
∴ B α C when A is Constant.
According to the theorem on joint variation,
Β α C/A when C & A both vary
∴ B = K C/A where K is a constant of variation.
Given A = 15, B = 5, & C = 18.
5 = K 18/15
or, K = 15 x 5 / 18 = 25/6
B = K C/A
=25/6 x 12/10
=5
∴ No. of days 5. Ans.
WBBSE Solutions Guide Class 10
Question 9. The volume of a sphere varies directly with the cube of its radius. Three solid spheres having lengths of 1 1/2, 2 and 2 1/2 metre diameter are melted and a new solid sphere is formed. Let us find the length of the diameter of the new sphere. [let us consider that the volume of the sphere remains the same before and after melting]
Solution: Let the volume of the sphere of radius r be v.
Given v α r³or v = Kr³ where K is the constant of variation.
The radii of 3 spheres are 3 / 4, 1 & 5/4 m.
∴ The volume of 1st sphere = K X (3/4)³ cu.m.
The volume of the 2nd sphere Kx (1)³ cu.m.
The volume of the 3rd sphere = K x (5/4)³ cu.m.
∴ If the radius of the new sphere = R
∴ Its volume = KR³
According to the problem,
KR³= K(3/4)³ + K(1)³ + Kx(5/4)³
KR³ = [27/64+1+125/64]
∴ R³ = 216+64
R = 6/4
R= 3/2
∴ Diameter of the new sphere = 2R = 2x 3/2 =3m.
Question 10. y is a sum of two variables, one of which varies directly with x and another varies inversely with x. When x=-1, then y = 1 and when x = 3, then y = 5. Let us find the relation between x and y.
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Question 11. If a x b, b x c, let us show that a³b³ + b³c³ + c³a³ α abc(a³ + b³ + c³)
Question 12. To dig a well of x dcm deep one part of the total expenses varies directly with x and the other part varies directly with x2. If the expenses of digging wells of 100 dcm and 200 dcm depths are Rs. 5,000 and Rs. 12,000 respectively, let us write by calculating the expenses of digging a well of 250 dcm depth.
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Question 13. The volume of a cylinder is in joint variation with the square of the length of the radius of the base and its height. The ratio of radii of bases of two cylinders is 2: 3 and the ratio of their heights is 54, let us find the ratio of their volumes.
Question 14. An agricultural Co-operative Society of the 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.
Solution: Let no. of ploughs = N, area of Land A, no. of days = D.
∴ N α A & Nα1/D
According to the theorem of joint variation, Nα1/D
or, N = K 1/D (Where K is a constant of variation)
Now, A = 2400, N = 25; D = 36
∴ 25 = K 2400/36
∴ K = 25x 36 / 2400 = 3/8
Now, N = K A/D
= 3/8 x 1200/30
As, A = 2400/2 = 1200 & D = 300
N = 15
∴ No. of ploughs = 15
∴ 15 plough work equally with one tractor.
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Question 15. Volume of a sphere varies directly with the 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 the volume of a sphere varies directly with cube of its surface area.
Solution: Let volume of sphere V, radius = R & surface area of sphere = S
∴V α R³ & S = R
or, V = mR³ or, S = nR² (Where m & n are constants of variation)
or, R³ = V/m or, R² = S/n
R = (V/m)⅓
∴R² = V/m)⅔
S/n = (V/m)⅔
or, S = n.(V/m)⅔
or, S³ = n³/m³.V²
∴V² = m²/n³ S³
∴V² α S³(as m²/n³ = Constant)
Chapter 13 Variation Exercise 13.2 Multiple Choice Question
Question 1. χ α 1/y, then
1. X = 1/y
2. y = 1/x
3. xy = 1
4. xy = non-zero constant.
Answer. 4. xy = non-zero constant.
Question 2. If x α y, then
(a) x²α ya
(b) x α y
(c) x α y³
(d) x² α y²
Answer. (d) x2 α y2
Question 3. If x α y and y = 8 when x = 2; if y = 16, then the value of x is
1. 2
2. 4
3. 6
4. 8
Answer. (b) 4
Question 4. If x α y2 and y = 4 when x = 8; if x = 32, then the value of y is
1. 4
2. 8
3. 16
4. 32
Answer. 2. 8
Question 5. If y-z α 1/x, x-x α 1/y and x – y α 1/z, the sum of three variation constants is
1. 0
2. 1
3. -1
4. 2
Answer. 1. 0
Chapter 13 Variation Exercise 13.2 True Or False
1. If y α 1/x, y/x = non-zero constant.
True
2. If x α z and y α z then xy α z.
False
Chapter 13 Variation Exercise 13.2 Fill In The BLanks
1. If xα1/y and y α 1/z, then xα z.
2. If.x α y, x\(x^n\) α y\(y^n\).
3. If x α y and x α z, then (y + z) α x.
Chapter 13 Variation Exercise 13.2 Short Answer
Question 1. If x a y² and y = 2a when x = a; x and y let us find the relation between x and y. Ky² [K = Constant of variation]
Question 2. If x α y, y α z and z α x, let us find the product of three non-zero constants.
Question 3. If x α 1/y and y α 1/z, let us find if there be any relation of direct or inverse variation between x and z.
Question 4. If x α yz and y α zx, let us show that z is a non-zero constant.
Question 5. If b c z³ and an increase in the ratio of 2: 3, let’s find in what ratio b will be increased.
