How to solve questions which involves arthematic operations on alphabets like xxxx yyyy +zzzz====== yxxxz======

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buzz

  • May 14th, 2006
 

x=9y=1z=8

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vaibhav chauhan

  • May 24th, 2006
 

   i dont know how u solv that but here is my approach

           x  x  x  x

           y  y  y  y

           +  z  z  z  z

---------------------------------

           y  x  x  x  z

---------------------------------

so we can say that

  z = z from zero's place

     x = x+y+z; for ten's place so x from both LHS and RHS get cancel and we will get y = -z;

same for hunderads and thousand place

for Ten Thousand place we get y = x+y; i.e. x = 0;

so we get x = 0; y = -z;

 

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Samriti Jindal

  • Jun 12th, 2006
 

x=9y=1z=8x=8z=8/19y=x9y=8y=8/9z=9yz=9*8/9z=8/1proved for correctness.

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naveen

  • Jul 13th, 2006
 

Absolutely fantastic

Your analysis is superb.

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SRINIVAS_SRIT

  • Jul 18th, 2007
 

R4 R3 R2 R1
x x x x
y y y y
+ z z z z
---------------------------------
y x x x z
A4 A3 A2 A1
--------------------------------
FIRST EQ: FROM ROW R1
X+Y+Z=10+Z ------>(1)
FROM ROW R2:
X+Y+Z+1= X+10;
THEN Y+Z=9------->(2)
FROM ROW R3:
SAME AS ABOVE R2 Y+Z=-1;
FROM ROW R4:
HERE Y VALUE IS 1.
SUM OF X,Y,Z IS LESS THEN 20 WHY BECAUSE X,Y,Z VALUES ARE BELOW TEN AND NOT REPEATABLE.
*X+Y+Z<20 MEANS 20=10 SO WE CAN GET Y VALUE FROM
ROW R4 : Y IS THE CARRY OF (X+Y+Z+1) THIS IS LESS THEN 20 GREATER
THAN 10 SO WE CAN TAKE Y=1;
SUBSTITUTE Y VALUE IN EQ (1): X+Y+Z=Z+10;
1+X=10;
THEN
X=9;
THEN SUBSTITUTE Y VALUE IN EQ(2):
Y+Z=9:
1+Z=9;
THEN Z=8:
RESULT IS:X=9,Y=1,Z=8.

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