Today our topic of discussion is Transformation of ratio.
Transformation of ratio
Transformation of ratio
Here, the quantities of ratios are positive numbers.
- 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
- 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
- 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
- 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]
- 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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