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Text Solution

Solution :

Let x litres of a 3% acid solution be added to 460 litres of 9% acid solution. Then, <br> total quantity of mixture`=(460+x)` litres. <br> Total acid content in (460 + x) litres of mixture <br> `={(460xx(9)/(100))+(x xx (3)/(100))}"litres"=((207)/(5)+(3x)/(100))`litres. <br> Now, the acid content in the resulting mixture must be more than 5% and less than 7%. <br> `therefore5%of(460+x)lt((207)/(5)+(3x)/(100))lt7% "of"(460+x)` <br> `rArr(5)/(100)xx(460+x)lt(4140+x)/(100)lt(7)/(100)xx(460+x)` <br> `rArr5(460+x)lt4140+3xlt7(460+x)` <br> `rArr 2300+5xlt4140+3xand 4140+3xlt3220+7x` <br> `rArr 5x=3xlt4140-2300 and 4140+3x lt3220 +7x` <br> `rArr2xlt1840and 920 lt4x` <br> `rArr xlt920and 230ltx` <br> `rArr 230ltxlt920` Hence, the required quantity of 3% acid solution to be added must be more than 230 litres and less than 920 litres.