document.write( "Question 1185110: Jenny had 3/5 as many beads as Kelly. Kelly gave Jenny 24 beads. Then, Jenny had $\"3%2F4\"$ of all the beads. How many beads did they have? \n" ); document.write( "

Algebra.Com's Answer #815848 by ikleyn(52908) $\"\"$ $\"About$

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\n" ); document.write( "Jenny had 3/5 as many beads as Kelly. Kelly gave Jenny 24 beads.
\n" ); document.write( "Then, Jenny had 3/4 of all the beads. How many beads did they have $\"highlight%28at_first%29\"$ ?
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document.write( "        After reading the post, I do not understand the problem's question.\r\n" );
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document.write( "        How many beads did they have TOGETHER ? SEPARATELY ?\r\n" );
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document.write( "        If SEPARATELY, then AT THE BEGINNING or AT THE END ?\r\n" );
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document.write( "        In this sense, the post is DEFECTIVE  (and deserves serious punishment).\r\n" );
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document.write( "        In order for it would make sense, I will consider the question in this form:\r\n" );
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document.write( "              How many beads did EACH of THEM have AT THE BEGINNING ? \r\n" );
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document.write( "Since \"Jenny had 3/5 as many beads as Kelly\", we can think that Kelly had 5x beads, while Jenny had 3x beads.\r\n" );
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document.write( "After Kelly gave Jenny 24 beads, they have:  Kelly  5x-24 beads;  Jenny  3x+24 beads.\r\n" );
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document.write( "The total beads is 5x + 3x = 8x.\r\n" );
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document.write( "The last condition of the problem states that\r\n" );
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document.write( "    3x + 24 = .\r\n" );
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document.write( "Simplify this equation and find x\r\n" );
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document.write( "    3x + 24 = 3*(2x)\r\n" );
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document.write( "    3x + 24 = 6x\r\n" );
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document.write( "         24 = 6x - 3x \r\n" );
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document.write( "         24 = 3x\r\n" );
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document.write( "          x = 24/3 = 8.\r\n" );
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document.write( "ANSWER.  Originally, they had  8*5 = 40 beans (Kelly)  and  8*3 = 24 beans (Jenny). \r\n" );
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document.write( "         If you want to know the total, you may add these numbers.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Solved and thoroughly explained.\r
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