English

(English Medium) ICSE Class 10 - CISCE Question Bank Solutions

Advertisements
Subjects
Topics
Subjects
Popular subjects
Topics

Please select a subject first

Advertisements
Advertisements
< prev  14161 to 14180 of 19085  next > 

Given four quantities a, b, c and d are in proportion. Show that: (a – c)b2 : (b – d)cd = (a2 – b2 – ab) : (c2 – d2 – cd)

[6] Ratio and Proportion
Chapter: [6] Ratio and Proportion
Concept: undefined >> undefined

Find two numbers such that the mean proportional between them is 12 and the third proportional to them is 96.

[6] Ratio and Proportion
Chapter: [6] Ratio and Proportion
Concept: undefined >> undefined

Advertisements

Find the third proportional to `x/y + y/x` and `sqrt(x^2 + y^2)`

[6] Ratio and Proportion
Chapter: [6] Ratio and Proportion
Concept: undefined >> undefined

If p : q = r : s; then show that: mp + nq : q = mr + ns : s.

[6] Ratio and Proportion
Chapter: [6] Ratio and Proportion
Concept: undefined >> undefined

If p + r = mq and `1/q + 1/s = m/r`; then prove that p : q = r : s.

[6] Ratio and Proportion
Chapter: [6] Ratio and Proportion
Concept: undefined >> undefined

If `a/b = c/d` prove that each of the given ratios is equal to

`(5a + 4c)/(5b + 4d)`

[6] Ratio and Proportion
Chapter: [6] Ratio and Proportion
Concept: undefined >> undefined

If a/b = c/d prove that each of the given ratio is equal to `(13a - 8c)/(13b - 8d)`

[6] Ratio and Proportion
Chapter: [6] Ratio and Proportion
Concept: undefined >> undefined

If a/b = c/d prove that each of the given ratio is equal to `sqrt((3a^2 - 10c^2)/(3b^2 - 10d^2))`

[6] Ratio and Proportion
Chapter: [6] Ratio and Proportion
Concept: undefined >> undefined

if `a/b = c/d` prove that each of the given ratio is equal to: `((8a^3 + 15c^3)/(8b^3 + 15d^3))^(1/3)`

[6] Ratio and Proportion
Chapter: [6] Ratio and Proportion
Concept: undefined >> undefined

If a, b, c and d are in proportion prove that `(13a + 17b)/(13c + 17d) = sqrt((2ma^2 - 3nb^2)/(2mc^2 - 3nd^2)`

[6] Ratio and Proportion
Chapter: [6] Ratio and Proportion
Concept: undefined >> undefined

If a, b, c and d are in proportion prove that `sqrt((4a^2 + 9b^2)/(4c^2 + 9d^2)) = ((xa^3 - 4yb^3)/(xc^3 - 5yd^3))^(1/3)`

[6] Ratio and Proportion
Chapter: [6] Ratio and Proportion
Concept: undefined >> undefined

If `x/a = y/b = z/c` prove that `(2x^3 - 3y^3 + 4z^3)/(2a^3 - 3b^3 + 4c^3) = ((2x - 3y + 4z)/(2a - 3b + 4c))^3`

[6] Ratio and Proportion
Chapter: [6] Ratio and Proportion
Concept: undefined >> undefined

If (a2 + b2)(x2 + y2) = (ax + by)2; prove that: `a/x = b/y`.

[6] Ratio and Proportion
Chapter: [6] Ratio and Proportion
Concept: undefined >> undefined

If a, b and c are in continued proportion, prove that `(a^2 + ab + b^2)/(b^2 + bc + c^2) = a/c`

[6] Ratio and Proportion
Chapter: [6] Ratio and Proportion
Concept: undefined >> undefined

If a, b and c are in continued proportion, prove that `(a^2 + b^2 + c^2)/(a + b + c)^2  = (a - b + c)/(a + b + c)`

[6] Ratio and Proportion
Chapter: [6] Ratio and Proportion
Concept: undefined >> undefined

Using properties of proportion, solve for x: 

`(sqrt(x + 5) + sqrt(x - 16))/(sqrt(x + 5) - sqrt(x - 16)) = 7/3`

[6] Ratio and Proportion
Chapter: [6] Ratio and Proportion
Concept: undefined >> undefined

Using properties of proportion, solve for x:

`(sqrt(x + 1) + sqrt(x - 1))/(sqrt(x + 1) - sqrt(x - 1)) = (4x - 1)/2`

[6] Ratio and Proportion
Chapter: [6] Ratio and Proportion
Concept: undefined >> undefined

Using properties of proportion, solve for x:

`(3x + sqrt(9x^2 - 5))/(3x - sqrt(9x^2 - 5)) = 5`

[6] Ratio and Proportion
Chapter: [6] Ratio and Proportion
Concept: undefined >> undefined

If `x = (sqrt(a + 3b) + sqrt(a - 3b))/(sqrt(a + 3b) - sqrt(a - 3b))`, prove that: 3bx2 – 2ax + 3b = 0.

[6] Ratio and Proportion
Chapter: [6] Ratio and Proportion
Concept: undefined >> undefined

Find the fourth proportional to 2xy, x2 and y2.

[6] Ratio and Proportion
Chapter: [6] Ratio and Proportion
Concept: undefined >> undefined
< prev  14161 to 14180 of 19085  next > 
Advertisements
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×