DAY 4 - RIDDLES (Parents and Cheerleaders)
Another joint challenge. Easy as pie today. Just solve the riddles and submit your answers using the form at the bottom of the page. There are 9 to guess in total. Even though this is a joint challenge, we will still need separate enteries from Parents and Cheerleaders.
RIDDLES 4 & 9 REQUIRE TO ADVISE WHY YOU HAVE PUT THE ANSWER YOU HAVE GIVEN
Please make sure your answers are correct before submitting as your first submission will be the one that is scored. If you don't know the answer....have a guess 😁
Which is the odd shape out?
Which is the odd number
out and why?
Solve this mathematical equation
Four cheerleaders are lined up (as per the diagram) each of which are standing above a trap door. A wall separates the 3rd and 4th cheerleader (which you are unable to see through) and each cheerleader is only able to see the person(s) in front of them. Coach Ewan has advised the cheerleaders that they each have a hat on which is either black or white and that there is a total of 2 black hats and 2 white hats being worn. The cheerleaders are told that one of them must shout out the colour of their hat correctly (not by guessing) otherwise all 4 cheerleaders will have their trap door opened beneath them. They are not allowed to communicate with each other, they are not allowed to turn around, they are not allowed to look at their own hat.
Which cheerleader calls out the correct colour of their hat and why?
594 AS THIS IS THE ONLY NUMBER WHERE THE FIRST NUMBER MULTIPLIED BY THE THIRD NUMBER DOES NOT EQUAL THE 2ND NUMBER
14. MONKEY = 5, BEAR = 4 AND SINGLE BANANA = 1. WHEN BODMAS IS APPLIED, THE FINAL LINE IS CALCULATED AS 4 + (2 X 5)
120. WE ARE LOOKING FOR THE TOTAL NUMBER OF NEW CUPS WE CAN POSSIBLY CREATE. TO WORK THIS OUT WE DO:
961 DIVIDED BY 9 = 106 WITH 7 REMAINDER
(106+7) DIVIDED BY 9 = 12 WITH 5 REMAINDER
(12+5) DIVIDED BY 9 = 1 WITH 8 REMAINDER
9 DIVIDED BY 9 = 1
THEN YOU NEED TO ADD ALL THE TOTAL USED CUPS TO GET THE FINAL ANSWER SO 106+12+1+1=120
IF COACH EWAN OVERTAKES THE PERSON IN SECOND PLACE HE WILL NOW BE IN SECOND PLACE
S. ALTERNATE LETTERS ARE THE ALPHABET INCRESING/DECREASING BY ONE LETTER
CHEERLEADER 2 IS CERTAIN OF HIS HAT COLOUR.
CHEERLEADER 4 IS UNABLE TO SEE ANYONE SO IS NOT ABLE TO KNOW THE COLOUR OF HIS HAT
CHEERLEADER 3 IS UNABLE TO SEE ANYONE SO IS NOT ABLE TO KNOW THE COLOUR OF HIS HAT
CHEERLEADER 1 CAN SEE THAT 2 AND 3 HAS A WHITE AND BLACK HAT BUT CANNOT SEE 4 SO IS NOT CERTAIN IF HIS HAT IS BLACK OR WHITE
CHEERLEADER 2 KNOWS THAT 1 CAN SEE 2 AND 3'S HAT. IF 2 AND 3 HAD THE SAME COLOUR HAT 1 WOULD KNOW THAT HIS HAT IS THE OPPOSITE COLOUR. AS THIS IS NOT THE CASE, 2 KNOWS THAT THE COLOUR OF HIS HAT MUST BE THE OPPOSITE COLOUR OF 3