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ONE TRAIN WAS RUNNING TWICE AS FAST AS THE OTHER. LET: SPEED OF THE FAST TRAIN = F SPEED OF THE SLOW TRAIN = S TIME IT TAKES FOR THE TRAINS TO MEET (PASS EACH OTHER) = T SINCE BOTH TRAINS TRAVEL THE SAME TOTAL DISTANCE AND DISTANCE = TIME X SPEED: F(T+1) = S(T+4) WE'RE TRYING TO FIGURE OUT F/S WHICH IS EQUAL TO (T+4) / (T+1) FROM THE EQUATION ABOVE. SO WE NEED TO FIGURE OUT THE VALUE OF T. AFTER THEY MEET, THE FAST TRAIN TRAVELS ONE MORE HOUR AT SPEED F AND COVERS THE SAME DISTANCE THE SLOW TRAIN COVERED IN T HOURS: F1 = ST OR F = ST AFTER THEY MEET, THE SLOW TRAIN TRAVELS FOR 4 MORE HOURS AND COVERS THE SAME DISTANCE THE FAST TRAIN COVERED IN T HOURS: S4 = FT SUBSTITUTING ST FROM THE FIRST EQUATION IN FOR F IN THE 2ND EQUATION: 4S = STT 4 = TT 2 = T SUBSTITUTE 2 IN FOR T IN THE (T+4) / (T+1) EQUATION TO GET 6/3 OR 2. THE FAST TRAIN IS GOING TWICE AS FAST AS THE SLOW TRAIN THE ELDEST IS 8 YEARS OLD AND THE 2 YOUNGER ONES ARE 3 YEARS OLD. LET'S BREAK IT DOWN. THE PRODUCT OF THEIR AGES IS 72. SO THE POSSIBLE CHOICES ARE: 2, 2, 18 - SUM(2, 2, 18) = 22 2, 4, 9 - SUM(2, 4, 9) = 15 2, 6, 6 - SUM(2, 6, 6) = 14 2, 3, 12 - SUM(2, 3, 12) = 17 3, 4, 6 - SUM(3, 4, 6) = 13 3, 3, 8 - SUM(3, 3, 8 ) = 14 1, 8, 9 - SUM(1,8,9) = 18 1, 3, 24 - SUM(1, 3, 24) = 28 1, 4, 18 - SUM(1, 4, 18) = 23 1, 2, 36 - SUM(1, 2, 36) = 39 1, 6, 12 - SUM(1, 6, 12) = 19 THE SUM OF THEIR AGES IS THE SAME AS YOUR BIRTH DATE. THAT COULD BE ANYTHING FROM 1 TO 31 BUT THE FACT THAT JACK WAS UNABLE TO FIND OUT THE AGES, IT MEANS THERE ARE TWO OR MORE COMBINATIONS WITH THE SAME SUM. FROM THE CHOICES ABOVE, ONLY TWO OF THEM ARE POSSIBLE NOW. 2, 6, 6 - SUM(2, 6, 6) = 14 3, 3, 8 - SUM(3, 3, 8 ) = 14 SINCE THE ELDEST KID IS TAKING PIANO LESSONS, WE CAN ELIMINATE COMBINATION 1 SINCE THERE ARE TWO ELDEST ONES. THE ANSWER IS 3, 3 AND 8 AS THE PROBLEM SAYS THE APPRENTICE MIXED UP THE HANDS SO THAT THE MINUTE HAND WAS SHORT AND THE HOUR HAND WAS LONG. THE FIRST TIME THE APPRENTICE RETURNED TO THE CLIENT WAS ABOUT 2 HOURS AND 10 MINUTES AFTER HE HAD SET THE CLOCK AT SIX.THE LONG HAD MOVED OLNY FROM TWELVE TO A LITTLE PAST TWO. THE LITTLE MADE TWO WHOLE CIRCLES AND AN ADDITIONAL 10 MINUTES. THUS THE CLOCK SHOWED THE CORRECT TIME. THE NEXT DAY AROUND 7:O5 A.M.HE CAME A SECOND TIME,13 HOURS AND 15 MINUTES AFTER HE HAD SET THE CLOCK FOR SIX. THE LONG HAD, ACTING AS THE HOUR HAND,COVERED 13 HOURS TO REACH 1. THE SHORT HAND MADE 13 FULL CIRCLES AND 5 MINUTES, REACHING 7, SO THE CLOCK SHOWED THE CORRECT TIME AGAIN AT FIRST IT MIGHT SEEM LIKE NO MATTER WHAT YOU DO, YOU'RE JUST A MINUTE OR TWO SHORT OF TIME, BUT THERE IS A WAY. THE KEY IS TO MINIMIZE THE TIME WASTED BY THE TWO SLOWEST PEOPLE BY HAVING THEM CROSS TOGETHER. AND BECAUSE YOU'LL NEED TO MAKE A COUPLE OF RETURN TRIPS WITH THE LANTERN, YOU'LL WANT TO HAVE THE FASTEST PEOPLE AVAILABLE TO DO SO. SO, YOU AND THE LAB ASSISTANT QUICKLY RUN ACROSS WITH THE LANTERN, THOUGH YOU HAVE TO SLOW DOWN A BIT TO MATCH HER PACE. AFTER TWO MINUTES, BOTH OF YOU ARE ACROSS, AND YOU, AS THE QUICKEST, RUN BACK WITH THE LANTERN. ONLY THREE MINUTES HAVE PASSED. SO FAR, SO GOOD. NOW COMES THE HARD PART. THE PROFESSOR AND THE JANITOR TAKE THE LANTERN AND CROSS TOGETHER. THIS TAKES THEM TEN MINUTES SINCE THE JANITOR HAS TO SLOW DOWN FOR THE OLD PROFESSOR WHO KEEPS MUTTERING THAT HE PROBABLY SHOULDN'T HAVE GIVEN THE ZOMBIES NIGHT VISION. BY THE TIME THEY'RE ACROSS, THERE ARE ONLY FOUR MINUTES LEFT, AND YOU'RE STILL STUCK ON THE WRONG SIDE OF THE BRIDGE. BUT REMEMBER, THE LAB ASSISTANT HAS BEEN WAITING ON THE OTHER SIDE, AND SHE'S THE SECOND FASTEST OF THE GROUP. SO SHE GRABS THE LANTERN FROM THE PROFESSOR AND RUNS BACK ACROSS TO YOU. NOW WITH ONLY TWO MINUTES LEFT, THE TWO OF YOU MAKE THE FINAL CROSSING. AS YOU STEP ON THE FAR SIDE OF THE GORGE, YOU CUT THE ROPES AND COLLAPSE THE BRIDGE BEHIND YOU, JUST IN THE NICK OF TIME