We will discuss here how to solve the problems on Ratio.
1. What should be added to each term of the ratio a : b to make x : y?
Solution:
Let p is to be added to each term of a : b to get the ratio x : y
Therefore,
\(\frac{a + p}{b + p}\) = \(\frac{x}{y}\)
⟹ x(a + p) = y(b + p)
⟹ ax + px = by + py
⟹ py – px = ax – by
⟹ p(y  x) = ax  by
⟹ p = \(\frac{ax  by}{y  x}\)
Therefore, the required number is \(\frac{ax  by}{y  x}\).
2. If a : b = 2 : 3, then find (4a  b) : (2a + 3b)?
Solution:
Given a : b = 2 : 3, then a = 2k, y = 3k (k ≠ 0 is a common multiplier)
Therefore, (4a  b) : (2a + 3b) = \(\frac{4a  b}{2a + 3b}\) = \(\frac{4 ∙ 2k  3k}{2 ∙ 2k + 3 ∙ 3k}\)
= \(\frac{8k  3k}{4k + 9k}\)
= \(\frac{5k}{13k}\)
= \(\frac{5}{13}\)
= 5 : 13
3. If x : y = 2 : 5, y : z = 4 : 3 then find x : z.
Solution:
x : y = 2 : 5 ⟹ \(\frac{x}{y}\) = \(\frac{2}{5}\) .......................... (i)
y : z = 4 : 3 ⟹ \(\frac{y}{z}\) = \(\frac{4}{3}\) .......................... (ii)
Multiplying (i) and (ii), we get
\(\frac{x}{y}\) × \(\frac{y}{z}\) = \(\frac{2}{5}\) × \(\frac{4}{3}\)
Therefore, \(\frac{x}{z}\) = \(\frac{8}{15}\)
Therefore, x : z = 8 : 15.
4. If (3x + 5y) : (7x  4y) = 7 : 4 then find the ratio x : y
Solution:
Given, (3x + 5y) : (7x  4y) = 7 : 4
⟹ \(\frac{3x + 5y}{7x  4y}\) = \(\frac{7}{4}\)
⟹ 4(3x + 5y) = 7(7x – 4y)
⟹ 12x + 20y = 49x – 28y
⟹ 12x  49x = 28y  20y
⟹  37x =  48y
⟹ 37x = 48y
⟹ \(\frac{x}{y}\) = \(\frac{48}{37}\)
⟹ x : y = 48 : 37
5. If a : b = 5 : 12, b : c = 8 : 3 and c : d = 9 : 16 , what is a : d?
Solution:
a : b = 5 : 12 ⟹ \(\frac{a}{b}\) = \(\frac{5}{12}\) .......................... (i)
b : c = 8 : 3 ⟹ \(\frac{b}{c}\) = \(\frac{8}{3}\) .......................... (ii)
c : d = 9 : 16 ⟹ \(\frac{c}{d}\) = \(\frac{9}{16}\) .......................... (iii)
Multiplying (i), (ii) and (iii), we get
\(\frac{a}{b}\) × \(\frac{b}{c}\) × \(\frac{c}{d}\) = \(\frac{5}{12}\) × \(\frac{8}{3}\) × \(\frac{9}{16}\) = \(\frac{5}{8}\)
Therefore, \(\frac{a}{d}\) = \(\frac{5}{8}\)
Therefore, a : d = 5 : 8
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