X - y and x - 1-z- if y - 4- z - 5 - then x - 2- Now- if y - 20 and z - 25 -then x - 2-3-4-5-

The correct option is A 2 x ∝ y and x ∝ 1/z ⇒ x ∝ y/z ⇒ x = k.(y/z) Putting the given values of x, y and z we get : 2 = k.(4/5) ⇒ k = 5/2 Now , putting the new ...

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Byju's Answer

Standard X

Mathematics

Joint Variation

x ∝ y and x ∝...

Question

x ∝ y and x ∝ $\frac{1}{z}$ . if y = 4, z = 5 , then x = 2. Now, if y = 20 and z = 25 ,then x =

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Solution

The correct option is A
2

x ∝ y and x ∝ $\frac{1}{z}$

⇒ x ∝ $\frac{y}{z}$

$\Rightarrow x = k . (\frac{y}{z})$

Putting the given values of x, y and z we get :

$2 = k . (\frac{4}{5})$

$\Rightarrow k = \frac{5}{2}$

Now , putting the new value of k along with y and z we get :

x = $(\frac{5}{2}) \times (\frac{20}{25}) = 2$

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Similar questions

Q. If x + y + z = 5 and xy + yz + zx = 7, then x³ + y³ + z³ – 3xyz = _________.

(i) If a(y+z) = b(z+x) = c(x+y) and out of a, b, c no two of them are equal then show that,

\frac{y - z}{a (b - c)} = \frac{z - x}{b (c - a)} = \frac{x - y}{c (a - b)}

(ii) If

\frac{x}{3 x - y - z} = \frac{y}{3 y - z - x} = \frac{z}{3 z - x - y}

and

x + y + z \neq 0

then show that the value of each ratio is equal to 1.

(iii) If

\frac{a x + b y}{x + y} = \frac{b x + a z}{x + z} = \frac{a y + b z}{y + z}

and

x + y + z \neq 0

then show that

\frac{a + b}{2}

.

(iv) If

\frac{y + z}{a} = \frac{z + x}{b} = \frac{x + y}{c}

then show that

\frac{x}{b + c - a} = \frac{y}{c + a - b} = \frac{z}{a + b - c}

.

(v) If

\frac{3 x - 5 y}{5 z + 3 y} = \frac{x + 5 z}{y - 5 x} = \frac{y - z}{x - z}

then show that every ratio =

\frac{x}{y}

Q. Solve the following system of equations by matrix method:
(i) x + y − z = 3
2x + 3y + z = 10
3x − y − 7z = 1

(ii) x + y + z = 3
2x − y + z = − 1
2x + y − 3z = − 9

(iii) 6x − 12y + 25z = 4
4x + 15y − 20z = 3
2x + 18y + 15z = 10

(iv) 3x + 4y + 7z = 14
2x − y + 3z = 4
x + 2y − 3z = 0

(v)

\frac{2}{x} - \frac{3}{y} + \frac{3}{z} = 10 \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 10 \frac{3}{x} - \frac{1}{y} + \frac{2}{z} = 13

(vi) 5x + 3y + z = 16
2x + y + 3z = 19
x + 2y + 4z = 25

(vii) 3x + 4y + 2z = 8
2y − 3z = 3
x − 2y + 6z = −2

(viii) 2x + y + z = 2
x + 3y − z = 5
3x + y − 2z = 6

(ix) 2x + 6y = 2
3x − z = −8
2x − y + z = −3

(x) x − y + z = 2
2x − y = 0
2y − z = 1

(xi) 8x + 4y + 3z = 18
2x + y +z = 5
x + 2y + z = 5

(xii) x + y + z = 6
x + 2z = 7
3x + y + z = 12

(xiii)

\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4, \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1, \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2; x, y, z \neq 0

(xiv) x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12

Find the value of $4 x^{2} + y^{2} + 25 z^{2} + 4 x y - 10 y z - 20 z x$ when x = 4, y = 3 and z = 2.

If x, y and z are variables, verify the cyclic symmetry of the following expressions.

(1) x(y + z) + y(z + x) + z(x + y)

(2) xy(x − y) + yz(y − z) + zx(z − x)

(3) x²y(x + y) + y²z(y + z) + z²x(z + x)

(4) x³(x + y) + y³(y + z) + z³(z + x)

(5) xy²(x − y) + yz²(y − z) + zx²(z − x)

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x ∝ y and x ∝ 1z. if y = 4, z = 5 , then x = 2. Now, if y = 20 and z = 25 ,then x =

The correct option is A 2 x ∝ y and x ∝ 1z ⇒ x ∝ yz ⇒x=k.(yz) Putting the given values of x, y and z we get : 2=k.(45) ⇒k=52 Now , putting the new value of k along with y and z we get : x = (52)×(2025)=2

x ∝ y and x ∝ $\frac{1}{z}$ . if y = 4, z = 5 , then x = 2. Now, if y = 20 and z = 25 ,then x =