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# (ii) $$\frac{105}{a} < \frac{100}{b}$$

Jan 10, 2016

#2
+15

If a and b are positive numbers and 100/a =95/b,

prove that:

(i)  a > b

$$\frac{100}{a}=\frac{95}{b}\\ 100b=95a\\ 95a=100b\\ a=\frac{100b}{95}\\ a=\frac{95b}{95}+\frac{5b}{95}\\ a=b+more\\ a>b$$

i)  prove $$\frac{105}{a}<\frac{100}{b}$$

$$\frac{100}{a}=\frac{95}{b}\\ \frac{100}{a}\times 1.05=\frac{95}{b}\times 1.05\\ \frac{105}{a}=\frac{99.75}{b}\\ \frac{105}{a}=\frac{100}{b}-\frac{0.25}{b}\\ \frac{105}{a}=\frac{100}{b}-\;\;a\; bit\\ so\\ \frac{105}{a}<\frac{100}{b}$$

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Jan 10, 2016

#1
0

If a and b are positive numbers and 100/a =95/b,

prove that:

(i)  a > b

(ii) 105/a <100/b

Jan 10, 2016
#2
+15

If a and b are positive numbers and 100/a =95/b,

prove that:

(i)  a > b

$$\frac{100}{a}=\frac{95}{b}\\ 100b=95a\\ 95a=100b\\ a=\frac{100b}{95}\\ a=\frac{95b}{95}+\frac{5b}{95}\\ a=b+more\\ a>b$$

i)  prove $$\frac{105}{a}<\frac{100}{b}$$

$$\frac{100}{a}=\frac{95}{b}\\ \frac{100}{a}\times 1.05=\frac{95}{b}\times 1.05\\ \frac{105}{a}=\frac{99.75}{b}\\ \frac{105}{a}=\frac{100}{b}-\frac{0.25}{b}\\ \frac{105}{a}=\frac{100}{b}-\;\;a\; bit\\ so\\ \frac{105}{a}<\frac{100}{b}$$

Melody Jan 10, 2016
#3
+15

If a and b are positive numbers and 100/a =95/b,

prove that:

(i)  a > b

(ii) 105/a <100/b

(i)  if 100/a = 95/b, this implies that 95a = 100b, which implies that a = (100/95)b......Then a must be greater than b since we have to multiply b by a quantity > 1 to get a

(ii)  105/a < 100/b..... Cross-multiplying........

105b < 100a........but, by definition, a =(100/95)b....so....

105b < 100(100/95)b  .......divide both sibes by 100 →

(105/100)b < (100/95)b ........reduce the fractions →

(21/20)b < (20/19)b ........divide both sides by b  →

(21/20) < (20/19)  →    cross-multiply, again

19*21 < 20*20  →

399 < 400......and since the left side is less than the right side......then the left side of the original inequality is also less than the right   Jan 10, 2016