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A has twice as much money as B. They play together, and at the end of the first game, B wins one third of A's money from A ; what fraction of the sum that B now has, must A win back in the second game so that they may have exactly equal money?​

Answer» <html><body><h3><u>☆</u><u> </u><u><a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a> <a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> : 1/10</u></h3><h3><u>☆</u><u> </u><u>Explanation</u><u> </u><u>:</u></h3><p><a href="https://interviewquestions.tuteehub.com/tag/let-11597" style="font-weight:bold;" target="_blank" title="Click to know more about LET">LET</a> A has ₹ 200 at the starting of game, then B has ₹ 100 at the starting of game.</p><p>After the game, money left with A = 200 - 200/3 = <strong>₹ 400/3</strong></p><p>∴ Total money with B = 100 + 200/3 =<strong> ₹ 500/3</strong></p><p>Let the <a href="https://interviewquestions.tuteehub.com/tag/fraction-458259" style="font-weight:bold;" target="_blank" title="Click to know more about FRACTION">FRACTION</a> of money lost by B = N</p><p>Then, 500/3 - N x 500/3 = 400/3 + N x 500/3</p><p>⇒ 500/3 - 400/3 = 500N/3 + 500N/3</p><p>⇒ 100/3 = <strong>1000N/3</strong></p><p>∴ N = (100/3) x (3/1000) = 1/10</p><p>so, <strong><u>the fraction of money lost is 1/10.</u></strong></p></body></html>


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