HCL
Maths Puzzle
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The number 642 is added to a 3-digit number 7x2 to give a 4 digit 13y4 which is exactly divisible by 11.find xy
Read Solution (Total 22)
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- if a number is divisible by 11,means diff b/w it's digits at odd and even place should be 0 or multiples of 11.
so 13y4 1+y=3+4 y=6 therefore no is 1364
now subtract 642 from 1364=722
therefore x=2
xy=12 - 9 years agoHelpfull: Yes(27) No(0)
- xy=26......a no is divisible by 11 only when the diff btwn sum of digits at even place nd sum of digits at odd place is equal to zero or div by 11.......so..(1+y)-7=0.y=6.......thereaftr x will be 2....
- 9 years agoHelpfull: Yes(12) No(5)
- 642
7x2
--------
13y4
13y4 is divisible by 11
any number divisile by 11 if difference between sum of its digits at odd places and even places is either 0 or a number divisible by 11
so 3+4=7,1+y=7
y=6 x=2 - 8 years agoHelpfull: Yes(5) No(0)
- if a number is divisible by 11. when the diff b/w unit place and ten's place is equal to 0.so 13y4 (4+3)-(1+y)=0 then y=6 therefor no is 1364. now 1364-642=722.
therefor x=2 and y=6. xy=2*6=12. - 9 years agoHelpfull: Yes(3) No(1)
- 642+7x2=13y4
y=6=>1364
x=2
x*y=12 - 9 years agoHelpfull: Yes(2) No(1)
- x=2 y=6
xy=12
- 8 years agoHelpfull: Yes(1) No(0)
- 1+y=3+4, than y=6 and x=2, so xy=12
- 8 years agoHelpfull: Yes(1) No(0)
- As 13y4 is divisible by 11,
so,
(4+3)-(y+1)=0
y=8
and now,
1384-642=742
Therefore,X=4,Y=8 - 9 years agoHelpfull: Yes(0) No(8)
- 1364,x=6,y=6,try to 11 divisible number by looking at last digit and 1st 2digits
- 9 years agoHelpfull: Yes(0) No(0)
- 13y4 must be divisible by 11 so (4+3)-(y+1)=?.and ? must be multiple of 11 or 0.in given number never possible to be 11 multiple so it should be 0. so y=6 then easily calculate the x=2.
SO THE ANSWER WILL BE 26 - 9 years agoHelpfull: Yes(0) No(0)
- x=2, y=6; 13y4 is a no divisible by 11 so y=6((3+4+(1+6))=0 642+7x2=1364 so x=2;
- 9 years agoHelpfull: Yes(0) No(0)
- Ans:12min
54*t=45*(t+s)
s/t=1/5
s=t/5
per hr :s=60/5
s=12min
- 8 years agoHelpfull: Yes(0) No(0)
- 13y4 is divisible by 11 so y = 6 and 642+7x2 = 1364
X=2 - 8 years agoHelpfull: Yes(0) No(0)
- , if a number is divisible by 11,means diff b/w it's digits at odd and even place should be 0 or multiples of 11.
(1+y)-(3+4)=0
so, y=6
subtract 642 from 1364=722
xy=12 - 8 years agoHelpfull: Yes(0) No(0)
- 13y4/11=124 i.e. y=6 and when putting dis value in 642+7x2=1364 we will get x=2
therefore xy=12 - 8 years agoHelpfull: Yes(0) No(0)
- 642+7x2=13y4
take x=2, y=6
then answer is 1364/11 is exactly divisible - 8 years agoHelpfull: Yes(0) No(0)
- x=2 y=6 because 13y4 is exactly divisible by 11 so the difference between sum of odd places and sum of even place must be 0 so (3+4)-(1+y) so y is 6 and x is 2
- 8 years agoHelpfull: Yes(0) No(0)
- 642+7x2=13y4
13y4 by trying to divide this with 11 it will be 124
from this we came to know that y=6
and for x since it is in the tenth place digit in the tenth place in the 642 is 4
then from 642+7x2=13y4
4+x=y;
4+x=6;
x=2;
then xy=12 - 8 years agoHelpfull: Yes(0) No(1)
- x=2,y=6 xy=12
11 is exactly divisible by 1364 - 8 years agoHelpfull: Yes(0) No(0)
- If a number is divisible by 11, means difference between its digits at odd and even place should be 0 or multiples of 11.
So, for 13y4 to be divisible by 11, 1+y should be equal to 3 + 4 i.e.
=> 1+y=3+4
=> y=6
Therefore no is 1364
Now subtract 642 from 1364=722
Therefore x=2
Hence, xy=12.
- 8 years agoHelpfull: Yes(0) No(0)
- For divisible of 11 (sum of odd places)-(sum of even places) 13y4 (1+y)-(3+4)=0 y=6 So x=2 from above given form so answer will be xy=6×2=12
- 7 years agoHelpfull: Yes(0) No(0)
- answer is x=2 and y=6.
solution is:-
642+7x2=13y4
starting from x=1, we get y=5 and number 1354 is not divisible by 11.
then checked for x=2 and 1264 number is divisible by 11, so answer is xy=2*6=12 - 7 years agoHelpfull: Yes(0) No(0)
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