Question

• A person loses 75% of his money in the first
bet, 75% of the remaining in the second and
75% of the remaining in the third bet and re-
turns home with * 2 only. His initial money
was​

Answer 1

Answer:

let the total money be x then X x 75 ÷ 100

Answer 2

Question :

A person loses 75% of his money in the first bet, 75% of the remaining in the second and 75% of the remaining in the third bet and returns home with 2 only. His initial money was ?

Answer :

$\sf\pink{Given:}$

• in first bet the person loses 75% of his money.

• in second loses 75% of the remaining.

• in third 75% of the remaining after second bet .

• returns home with 2 .

$\sf\pink{To\:find:}$

• his initial money ?

Explanation :

• let his initial money be "x"

therefore his remaining amount after first bet =

➡️ $\sf{x - \dfrac{75}{100}\times x}$

$\\ \longrightarrow\sf{ x - \dfrac{75x}{100}}$

$\\ \longrightarrow\sf{ \dfrac{100x - 75x}{100}}$

$\\ \longrightarrow\sf{ \dfrac{25x}{100}}$

• Therefore remaining amount after first bet is

➡️$\sf{ \dfrac{25x}{100}}$

his remaining amount after second bet =

➡️$\sf{\dfrac{25x}{100} - \dfrac{75}{100}\times \dfrac{25x}{100}}$

$\\ \longrightarrow\sf{ \dfrac{25x}{100} - \dfrac{75}{100}\times \dfrac{25x}{100}}$

$\\ \longrightarrow\sf{ \dfrac{25x}{100} - \dfrac{75}{100}\times \dfrac{5x}{20}}$

$\\ \longrightarrow\sf{ \dfrac{25x}{100} - \dfrac{15}{100}\times \dfrac{5x}{4}}$

$\\ \longrightarrow\sf{ \dfrac{25x}{100} - \dfrac{15}{20}\times \dfrac{x}{4}}$

$\\ \longrightarrow\sf{ \dfrac{25x}{100} - \dfrac{15x}{80}}$

$\\ \longrightarrow\sf{ \dfrac{25x\times 80}{100\times 80} - \dfrac{15x\times 100}{80\times 100}}$

$\\ \longrightarrow\sf{ \dfrac{2000x}{8000} - \dfrac{1500x}{8000}}$

$\\ \longrightarrow\sf{ \dfrac{2000x - 1500x}{8000}}$

$\\ \longrightarrow\sf{ \dfrac{500x}{8000}}$

$\\ \longrightarrow\sf{ \dfrac{5x}{80}}$

• Therefore remaining amount after second bet is

➡️ $\sf{\dfrac{5x}{80}}$

his remaining amount after second bet =

➡️$\sf{\dfrac{5x}{80} - \dfrac{75}{100}\times \dfrac{5x}{80}}$

$\\ \longrightarrow\sf{ \dfrac{5x}{80} - \dfrac{75}{100}\times \dfrac{5x}{80}}$

$\\ \longrightarrow\sf{ \dfrac{5x}{80} - \dfrac{75}{20}\times \dfrac{x}{80}}$

$\\ \longrightarrow\sf{ \dfrac{5x}{80} - \dfrac{15}{20}\times \dfrac{x}{16}}$

$\\ \longrightarrow\sf{ \dfrac{5x}{80} - \dfrac{15x}{320}}$

$\\ \longrightarrow\sf{ \dfrac{5x\times 4}{80\times 4} - \dfrac{15x}{320}}$

$\\ \longrightarrow\sf{ \dfrac{20x}{320} - \dfrac{15x}{320}}$

$\\ \longrightarrow\sf{ \dfrac{20x - 15x}{320}}$

$\\ \longrightarrow\sf{ \dfrac{5x}{320}}$

• Therefore his remaining amount after third bet is

➡️$\sf{\dfrac{5x}{320}}$

Here he returned home with 2 this means.

$\\ \longrightarrow\sf{\dfrac{5x}{320} = 2}$

$\\ \longrightarrow\sf{5x = 320\times 2}$

$\\ \longrightarrow\sf{5x = 640}$

$\\ \longrightarrow\sf{x = \dfrac{640}{5}}$

$\\ \longrightarrow\sf{x = 128}$

• Therefore the initial money was 128 .