Question

Two numbers are such that the ratio between them is 3:5 but if each is increased by 10, the ratio between them becomes 5 : 7, the numbers are

Answer 1

let the nos. be 3x and 5x respectively.
3x+10/5x+10=5/7
7 (3x+10)=5 (5x+10)
21x+70=25x+50
21x - 25x=50 - 70
-4x= -20
x=20/4=5
first number=15
second number=25

Answer 2

$\circ\circ\circ\circ\underline{\fbox{\fbox{\fbox{\fbox{\fbox{\Large{\underline{\textbf{SOLUTION}}}}}}}}}\circ\circ\circ\circ$

$\\\\Ratio\:of\:two\:number\:is\:3:5.\\\\Let\:us\:take\:the\:common\:multiple\:as\:'x'\\\\Then\:the\:numbers\:are\:3x\:and\:5x\\\\According\:to\:the\:question,\\\\each\:numbers\:are\:increased\:by\:10\\\\Then\:the\:new\:number\:be\:(3x+10)\:and\:(5x+10)\\\\By\:the\:given\:condition,\\\\(3x+10):(5x+10)=5:7\\\\\implies\frac{(3x+10)}{(5x+10)}=\frac{5}{7}\\\\\implies7(3x+10)=5(5x+10)\:[By\:cross\:multiplication]\\\\\implies21x+70=25x+50\\\\\implies25x-21x=70-50\\\\\implies4x=20\\\\\implies x=\frac{20}{4}$

$\\\\\implies x=5\\\\\therefore The\:two\:numbers\:are\\\\(3\times5)\:and\:(5\times5)\\\\=\boxed{\boxed{15\:and\:25\:is\:answer}}$