Transformation of ratio

Today our topic of discussion is Transformation of ratio.

Transformation of ratio

 

 

Transformation of ratio

Here, the quantities of ratios are positive numbers.

1. a: b = c :d , then b: a = d : c [Invertendo]

Proof: Given that,

a:b = c:d

or, ad = bc [multiplying both the sides by bd]

or, (ad)/(ac) = (bc)/(ac) [dividing both the sides by ac where a and c are non-zero]

or, d/c = b/a

i.c z. b : a = d : c

2. If a : b = c : d then, a : c = b : d [Alternendo]

Proof: Given that,

a : b = c : d

or, ad = bc  [multiplying both the sides by bd]

or, (ad)/(cd) = (bc)/(cd)  [dividing both the sides by cd where c, d are non-zero]

or, a/c = b/d

i .e.,a:c=b:d

3. if a : b = c : d then, (a + b)/b = (c + d)/d  [Componendo]

Proof: Given that,

a/b = c/d

or, a/b + 1 = c/d + 1 [adding 1 to both the sides]

i.e, (a + b)/b = (c + d)/d 

4. If a:b = c:d then (a – b)/b = (c – d)/d [dividendo]

Proof: Given that,

a:b = c:d

or, a/b – 1 = c/d – 1 [subtracting 1 from both the sides]

i.e., (a – b)/b = (c – d)/d

5. If a:b = c:d then (a + b)/(a – b) = (c + d)/(c – d) 

Proof: Given that, a: b =c: d [componendo dividendo]

or, a/b = c/d

By componendo, (a + b)/b = (c + d)/d

Again, by dividendo, (1) (a – b)/b = (c – d)/d

or, b/(a – b) = d/(c – d) [[by invertendo]]… (2)

a+b Therefore, X b c+d = a-b d X c-d [multiplying(1) & (2) ]

i.e., (a + b)/(a – b) = (c + d)/(c – d) here a ne b,c ne d]

6. If a:b = c:d = e:f = g:h then each of the ratio = (a + c + e + g)/(b + d + f + h)

Proof: let,

a:b = c:d = e:f = g:h = k *. a = bk , c = dk e = fk q = hk

therefore (a + c + e + g)/(b + d + f + h) = (bk + dk + fk + hk)/(b + d + f + h) = (k(b + d + f + h))/(b + d + f + h) = k

But, k is equal to each of the ratio. therefore a/b = c/d = e/f = g/h = (a + c + e + g)/(b + d + f + h)

Work:

1) The sum of ages of mother and sister is s years. The ratio of their ages, before t years was r: p. After years, what will be the ratio of their ages?

2) Let a man be standing at p metre distance from a light-post, r be the height of the man, s be the shadow length. Determine the height of the light-post in terms of p, r and s.

Example 2. The ratio of present ages of father and son is 7: 2, and after 5 years, the ratio will be 8: 3. What are their present ages?

Solution: Let the present age of father be a and that of the son be b.

So, by the conditions of first and second of the problems, we have,

a/b = 7/2 …(1)

(a + 5)/(b + 5) = 8/3 (2)

From equation(1), we get, a = (7b)/2 (3)

From equation(2), we get,

3(a + 5) = 8(b + 5)

or, 3a + 15 = 8b + 40

or, 3a – 8b = 40 – 15

or, 3 * (7b)/2 – 8b = 25 by using(3)]

or, (21b – 16b)/2 = 25 r, 5b = 50

* b = 10

In equation (3), by putting b = 10 , we get a = (7 * 10)/2 = 35

∴ The present age of father is 35 years, and that of the son is 10 years.

Example 3. If a :b = b:c prove that, ((a + b)/(b + c))² = (a² + b²)/(b ² + c²)

Solution: Given that, a :b = b :c

therefore b² = ac

Now, (a+b/b+c)²= (a+b)² /(b+c)²

= (a² + 2ab + b²)/(b² + 2bc + c²)

= (a² + 2ab + ac)/(ac + 2bc + c²)

= (a(a + 2b + c))/(c(a + 2b + c)) = a/c

And, (a² + b²)/(b² + c ²)

= (a² + ac)/(ac + c²)

= (a(a + c))/(c(a + c))

= a/c

((a + b)/(b + c)) ² = (a ² + b²)/(b² + c²)

Example 4. If  a :b = c:d show that, (a² + b ²)/(a² – b ²) = (ac + bd)/(ac – bd)

Solution: Let, a:b = c:d = k

∴a = bk and c = dk

Now And, , a² +b²a² -b²

= (bk)²+b² (bk)² -b²

= b² (k² +1) b² (k² -1)

= k²+1 k²-1 (ac + bd)/(ac – bd)

= (bkdk + bd)/(bkdk – bd)

= (bd(k ²+ 1))/(bd(k² – 1))

= (k² + 1)/(k ² – 1)

* (a² + b²)/(a ² – b ²)

= (ac + bd)/(ac – bd)

 

 

Example 5. Solve: (1 – ax)/(1 + ax) * √((1 + bx)/(1 – bx)) = 1 where 0 < b < 2a < 2b

Solution: Given that, (1 – ax)/(1 + ax) * √((1 + bx)/(1 – bx)) = 1

or, √((1 + bx)/(1 – bx)) = (1 + ax)/(1 – ax)

or, (1 + bx)/(1 – bx) = ((1 + ax)²)/((1 – ax)²) [squaring both the sides]

or, (1 + bx)/(1 – bx) = (1 + 2ax + a² * x²)/(1 – 2ax + a² * x²)

or, (1 + bx + 1 – bx)/(1 + bx – 1 + bx) = (1 + 2ax + a² * x²+ 1 – 2ax + a ²* x ²)/(1 + 2ax + a² * x² – 1 + 2ax – a² * x²)  [by componendo and dividendo]

2/(2bx) = (2(1 + a² * x²))/(4ax)

or, 2ax = bx(1 + a² * x²)

or, x\{2a – b(1 + a ² * x²)\} = 0

x = 0

Again, 2a – b(1 + a² * x²) = 0

or, b(1 + a² * x ²) = 2a

or, 1 + a² * x² = (2a)/b

or, a² * x² = (2a)/b – 1

or, x² = 1/(a²) * ((2a)/b – 1)

∴x = ± 1/a * √((2a)/b – 1)

Thus, the required solution is

x = 0, ± 1/a * √((2a)/b – 1)

Example 6. If (√(1 + x) + √(1 – x))/(√(1 + x) – √(1 – x)) = p prove that, p² – (2p)/x + 1 = 0

Solution: Given that, (√(1 + x) + √(1 – x))/(√(1 + x) – √(1 – x)) = p

or, √(1 + x) + √(1 – x) + √(1 + x) – √(1 – x))/(√(1 + x) + √(1 – x) – √(1 + x) + √(1 – x)) = (p + 1)/(p – 1) [by componendo  dividendo]

or, (2√(1 + x))/(2√(1 – x)) = (p + 1)/(p – 1)

or, (√(1 + x))/(√(1 – x)) = (p + 1)/(p – 1)

or, (1 + x)/(1 – x) = ((p + 1) ²)/((p – 1)²) = (p² + 2p + 1)/(p² – 2p + 1) squaring both the sides]

or, (1 + x + 1 – x)/(1 + x – 1 + x) = (p² + 2p + 1 + p² – 2p + 1)/(p² + 2p + 1 – p² + 2p – 1) [by componendo  dividendo]

or, 2/(2x) = (2(p² + 1))/(4p)

or, 1/x = (p² + 1)/(2p)

or, p²+ 1 = (2p)/x

p² – (2p)/x + 1 = 0

Example 7. [ if * (a³+ b³)/(a – b + c) = a(a + b) , prove that, a, b, c are continued

proportion.

Solution: Given that, (a² + b²)/(a – b + c) = a(a + b)

or, ((a + b)(a² – ab + b²))/(a – b + c) = a(a + b)

or, (a² – ab + b²)/(a – b + c) = a dividing both the sides by (a + b) ]

or, a²– ab + b² = a ² – ab + ac

or, b² = ac

therefore a ,b , c are continued proportion.

Example 8. If * (a + b)/(b + c) = (c + d)/(d + a) prove that, c = a or a + b + c + d = 0

Solution: Given that, (a + b)/(b + c) = (c + d)/(d + a)

or, (a + b)/(b + c) – 1 = (c + d)/(d + a) – 1 subtracting 1 from both the sides]

or, (a + b – b – c)/(b + c) = (c + d – d – a)/(d + a)

(a – c)/(b + c) = – (a – c)/(d + a)

(a – c)/(b + c) + (a – c)/(d + a) = 0

or, (a – c)(1/(b + c) + 1/(d + a)) = 0

or, (a – c) * (d + a + b + c)/((b + c)(d + a)) = 0

or, (a – c)(d + a + b + c) = 0

.. a – c = 0ord + a + b + c = 0

c = a or a+b+c+d=0

Example 9. If * x/(y + z) = y/(z + x) = z/(x + y) and x y z are not mutually equal, prove that the value of each ratio is either equal to -1 or equal to 1/2

Solution: Let, x/(y + z) = y/(z + x) = z/(x + y) = k

therefore x= k(y + z) …(1)

y= k(z + x) …(2)

z= k(x + y) …(3)

By subtracting equation (2) from (1),

x – y = k(y – x) * or, k(y – x) = – (y – x)

k = – 1

Again, adding equations (1), (2) and (3), we get, x + y + z = k(y + z + z + x + x + y) = 2k(x + y + z)

or, k = (x + y + z)/(2(x + y + z))

k = 1/2

.. The value of each of the ratio is -1 or 1/2

Example 10. If ax = by = cz show that, (x²)/(yz) + (y²)/(zx) + (z²)/(xy) = (bc)/(a²) + (ca)/(b²) + (ab)/(c²)

Solution: Let, ax = by = cz = k x = k/a, y = k/b, z = k/c

Now, (x²)/(yz) + (y²)/(zx) + (z²)/(xy) = (k²)/(a²) * (bc)/(k²) + (k²)/(b²) * (ca)/(k²) + (k²)/(c²) * (ab)/(k²) = (bc)/(a²) + (ca)/(b²) + (ab)/(c²)

That is, (x²)/(yz) + (y²)/(zx) + (z²)/(xy) = (bc)/(a²) + (ca)/(b²) + (ab)/(c²)

 

 

Example 11. If a, b, c and d are continued proportional, and x = (10pq)/(p + q)

1) Show that, a/c = (a² + b²)/(b² + c²)

2) Prove that, (a² + b² + c²)(b² + c ²+ d²) = (ab + bc + cd) ^ 2

3) Determine (x + 5p)/(x – 5p) + (x + 5q)/(x – 5q) ,where p ne q

Solution:

1) Given that, a: b = b : c ,or,a/b=b/c or,ac=b²

RHS= (a² + b²)/(b ² + c²)

=(a² + ac)/ac + c² 

={a(a+c)}/{c(a+c)}

=a/c

=LHS

2) Given that, a, b, c and d are continued proportional.

a/b = b/c = c/d

Let, a/b = b/c = c/d = k  where k is a constant proportional.

c/d = k

or, c = dk

b/c = k or, b = ck = dkk = d * k²

a/b = k or, a = bk = d * k²* k = d * k³

LHS= (a²+b²+c²²)(b² + c² + d²)

= \{(d * k³)² + (d * k²)²+ (dk)²\}\{(d * k²)² + (dk)² + d ²\}

= (d ² * k ^ 6 + d²* k⁴ + d² * k ²)(d² * k⁴ + d ² * k² + d²)

= d² * k ²* (k⁴ + k² + 1) * d ² * (k⁴ + k ²+ 1) .

= d ⁴ * k ² * (k⁴ + k ²+ 1) ²

RHS= (ab+be+cd)²

= (d * k³ * d * k² + d * k ² * dk + dkd) ²

= (d ² * k⁵ + d² * k ³ + d² * k)²

= \{d ² * k(k⁴ + k ² + 1)\} ²

= d⁴* k²* (k⁴ + k² + 1) ²

= LHS

∴ (a² + b² + c ²)(b² + c² + d²) = (ab + bc + cd) ² 

3) Given that, x = (10pq)/(p + q)

or, x/(5p) = (2q)/(p + q)

(x + 5p)/(x – 5p) = (2q + p + q)/(2q – p – q) – (1) [by componendo dividendo]

or, (x + 5p)/(x – 5p) = (p + 3q)/(q – p)

Again, x = (10pq)/(p + q)

x/(5q) = (2p)/(p + q)

or, (x + 5q)/(x – 5q) = (2p + p + q)/(2p – p – q) [by componendo dividendo]

or, (x + 5q)/(x – 5q) = (3p + q)/(p – q) (2)

Here, by adding equations (1) and (2) we get, (x + 5p)/(x – 5p) + (x + 5q)/(x – 5q) = (p + 3q)/(q – p) + (3p + q)/(p – q) = (p + 3q)/(q – p) – (3p + q)/(q – p)

= (p + 3q – 3p – q)/(q – p) = (2q – 2p)/(q – p) = (2(q – p))/(q – p) =2    [ ∴q-p ≠0]

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