PROBLEM & SOLUTION ON PROPORTION -
Example.1) What least number must be added to each of the numbers 6, 15, 20, & 43, so that the resulting numbers are proportional ?
Ans.) Let the required number to be added be x. Then,
(6 + x) : (15 + x) : : (20 + x) : (43 + x)
=> (6 + x) (43 + x) = (20 + x) (15 + x) [so, Product of extremes = Product of means]
=> x² + 49x + 258 = x² + 35x + 300
=> 49x – 35x = 300 – 258
=> 14x = 42
=> x = 3
Hence, the required number is 3 (Ans.)
Example.2) Find the fourth proportional to 7, 13, and 35
Ans.) Let the fourth proportional to 7, 13, and 35 be ‘x’
Then, 7 : 13 : : 35 : x
7 35
=> ------- = -------- [so, Product of extremes = Product of means]
13 x
35 X 13
=> x = ------------ = 65
7
Hence the fourth proportional to 7, 13, & 35 is 65. (Ans.)
Example.3) Find the third proportional to 9, and 15
Ans.) Let the third proportional to 9 & 15 be ‘x’
Then, 9 : 15 : : 15 : x
9 15
=> -------- = -------- [so, Product of extremes = Product of means]
15 x
15 X 15
=> x = ----------- = 25
9
Hence, the third proportional to 9 and 15 is 25. (Ans.)
Example.4) If b is the mean proportion between a & c, then prove that
a² - b² + c²
---------------- = b⁴
a¯² - b¯² + c¯²
Ans.) Since b is the mean proportion between a & c, so as per rules we have b = √ac
=> b² = ac
a² - b² + c²
------------------
a¯² - b¯² + c¯²
a² - b² + c²
=> ----------------------------
1 1 1
------ - ------ + ------
a² b² c²
(a²- b² + c²)
=> ------------------------
b²c² - a²c² + a²b²
--------------------
a²b²c²
a²b²c² (a² - b² + c²)
=> ------------------------
b²c² - a²c² + a²b²
a²(ac)c² (a² - ac + c²)
=> ------------------------ [where ac = b²]
(ac) c² - a²c² + a² (ac)
a³c³ (a² - ac + c²)
=> ----------------------
ac³- a²c² + a³c
a³c³ (a² - ac + c²)
=> ----------------------
ac (a² - ac + c²)
=> a²c² = (b²)² = b⁴ [where ac = b²] (Proven)
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