ANSWERS TO EASY PUZZLES
PUZZLE- E1
A REAL BOILER
It
can be done in exactly 9 minutes - as much time as it will take if you have a
real clock. Let us see how.
First,
let us start the timing by flipping both the hourglasses together, starting our
unique timer. Simultaneously drop the egg into the water for boiling.
When
the four-minute hourglass runs out of sand flip it over and similarly when the
seven-minute hourglass also has run out of sand. Now the egg has been cooking
for 7 minutes and 2 more minutes are left.
Now when the four-minute hourglass
runs out of sand, i.e., after 8 minutes from the time we started boiling, flip
the seven-minute hourglass, which has now 1 minute of sand in the bottom. When
this sand runs out after a minute, your nine minutes are over and the egg is
ready to be taken out of the boiling pot.
PUZZLE-E2
AN AGELESS QUESTION
Under normal circumstances, it will look
as if we need to treat this problem as if it is in a integer number line – with
zero in the center and B.C in the left hand side (- ve side) and A. D in the
right hand side (+ ve side). Then the number of years the man lived would have
been a ripe 90 years. But the Christian calendar has one major difference-
there is no year 0. Hence he would have lived would have been 89 years.
PUZZLE:
E3
A WEIGHTY QUESTION
The important
condition is that we are allowed to weigh only once. Since we need to check all
the ten lots, we need to draw samples from all the heaps and weigh the whole
sample lot once; but the samples should be drawn in such a way so that we can
trace back the rogue lot from the weight reading.
This can be
achieved by drawing 1 coin from heap 1, 2 coins from heap2, 3 coins from heap3
etc till 10th heap. We would have drawn a sample size of 55 coins.
Supposing there had been no rogue lot the weight would have been 550 grams. Now
by seeing the weight we can locate the rogue heap. Supposing the weight of the
samples chosen happens to be 549 grams then the rogue lot obviously is the
heap1. {As only one coin is from heap1}. If the weight turns out to be 548,
then the rogue lot is heap2 and so on.
PUZZLE:
E4
STRANGE MULTIPLIER
Y R S Y G Y B MULTIPLIED BY 5
= B N R L E G R
The answer can be
got by looking at the two sides of the equation. Since both sides have 7 digits
each, Y has to be 1. If y is more than 1, then the number of digits in the
right side would be 8.
Now R can be
either 0 or 5, as any digit multiplied by 5 will have only one of these two
digits. We need to take R as 0 and work through and then take R as 5 and work
through. By eliminating the various possibilities one by one we will get the
answer as under:
1531817
X 5 = 7659085
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