There were three measuring cups: A, B and C. Tuan poured water into 3 cups in the ratio of 9 : 7 : 16. Then he transfered 25% of the water from cup C

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1. Solution: `960 (ml)`

   Step by step:

   Let the amount of water in cup `A, B, C` at first `= a, b, c  (a, b, c >0; a>b)`

   Given that,

   `+)` Tuan transferred `25%` of the water from cup `C` to cup `A.`

    `∴` The amount of water in cup `A, B, C` after that was:

       `a+25%c  ; b  ; 75%c`

   `+)` Then, Tuan continued to transfer `25%` of the remaining water from cup `C` to cup `B.`

    `∴` The amount of water in cup `A, B, C` after that was: 

       `a+25%c ;`

       `b+(25%×75%c) = b+3/16c ;`

       `75%c-(25%×75%c)=75%c-3/16c`

   `+)` There was `90 ml` more water in cup `A` than cup `B` at the end.

    `⇒ (a+25%c)-(b+3/16c) = 90 (ml)`

    `⇒ a + 1/4c-b-3/16c = 90 (ml)`

    `⇒ a-b+1/16c = 90`

    `⇒ 16a-16b+c = 1440`

   `+)` Tuan poured water into cup `A, B, C` in the ratio of `9 : 7 : 16.`

    `⇒ a/9 = b/7 = c/16  (1)`

   Using theorem on equal ratios:

   `a/9 = b/7 = c/16 = (16a)/144 = (16b)/112 = {16a – 16b + c}/{144-112+16}`

                                                                `= 1440/48 (∵16a-16b+c=1440)`

                                                                `=30`

   `∴ a = 270; b = 210; c = 480`

   `∴` Total water in `3` cups: `270+210+480=960 (ml)`

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