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102 is 3 parts, 34 is 1 part

                                    X     :       Y   (drains all)     ,  X left

      Scenario 1,          3u      :        1u                          , 420l

      Scenario 2,          1p       :        3p                         , 1176l

also Scenario 1,        9p      :         3p                        , 420l   —> compare to (Scenario 2), difference of 8p is 1176-420 = 756. So 1p is 94.5l

X has 6p+420 more water = 987l more# than Y.

 

Checking for X : Y

S2)   94.5*1  +1176  :    94.5 *3  

S1)  94.5 * 9 +420 :   94.5 *3

0 Replies 1 Like ✔Accepted Answer

Scenario 1 : If Tank X drains water at a rate of 102 litres per hour and Tank Y drains water at a rate of 34 litres per hour, Tank X would be left with 420 litres of water by the time Tank Y is fully drained.
Tank X : 102, 204, 306, … , 420 left
Tank Y : 34, 68, 102, … 0

Scenario 2 : If Tank X drains water at a rate of 34 litres per hour and Tank Y drains water at a rate of 102 litres per hour, Tank X would be left with 1176 litres of water by the time Tank Y is fully drained.
Tank X : 34, 68, 102,, … , 1176 left
Tank Y : 102, 204, 306, … 0

Method 1 :
1176 – 420 = 756
306 – 34 = 272
(756/272) x 34 + 1176 = 1270.5
(756/272) x 102 = 283.5
1270.5 – 283.5 = 987

Method 2 :
34 x number of hours in Scenario 1 = 102 x number of hours in Scenario 2 (Tank Y)
number of hours in Scenario 1 = 3 x number of hours in Scenario 2 (Tank Y)
102 x number of hours in Scenario 1 + 420 = 34 x number of hours in Scenario 2 + 1176
306 x number of hours in Scenario 2 + 420 = 34 x number of hours in Scenario 2 + 1176

number of hours in Scenario 2 = (1176 – 420)/(306 – 34) = 756/272
(756/272) x 34 + 1176 = 1270.5 or (756/272) x 3 x 102 + 420 = 1270.5
(756/272) x 102 = 283.5 or (756/272) x 3 x 34 = 283.5
1270.5 – 283.5 = 987

Ans : 987 litres more.

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