Question

What no. must be added to each of the numbers 6 , 15 , 20 and 43 to make four no.s proportional ?

Answer 1

Answer:

Ans Is 3

Step-by-step explanation:

Let no. be ‘x’

Let no. be ‘x’∴ 6 + x/15 + x = 20 + x/43 + x

Let no. be ‘x’∴ 6 + x/15 + x = 20 + x/43 + x⇒ (6 + x)(43 + x) = (20 + x)(15 + x)

Let no. be ‘x’∴ 6 + x/15 + x = 20 + x/43 + x⇒ (6 + x)(43 + x) = (20 + x)(15 + x)⇒ 258 + 49x + x2 = 300 + 35x + x2

Let no. be ‘x’∴ 6 + x/15 + x = 20 + x/43 + x⇒ (6 + x)(43 + x) = (20 + x)(15 + x)⇒ 258 + 49x + x2 = 300 + 35x + x2⇒ 14x = 42

Let no. be ‘x’∴ 6 + x/15 + x = 20 + x/43 + x⇒ (6 + x)(43 + x) = (20 + x)(15 + x)⇒ 258 + 49x + x2 = 300 + 35x + x2⇒ 14x = 42x = 3

Answer 2

$\bold{ \underline{ \underline \blue{Given:}}}$

↦  The numbers = 6 , 15 , 20 and 43

$\bold{ \underline{ \underline \blue{To \; Find:}}}$

↦  No. which must be added to make them proportional

$\bold{ \underline{ \underline \blue{Solution:}}}$

↦ Let the required no. be x

ATQ :

$\sf \dfrac{6+x}{15+x} = \dfrac{20+x}{43+x}$

$\sf \{ cross \; multiply \}$

$\sf (6+x)(43+x)=(20+x)(15+x)$

$\sf 49x - 35 x = 300 - 258$

$\sf \purple{ \fbox{ \underline{x=3}}}$

∴ The required no. is 3

____________

All Done ! :D