Latest HCL Aptitude Question SOLUTION: 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

• Placement Papers
• /
• HCL
• /
• undefined

HCL Question Papers: HCL Interview Experience(1) HCL Company Profile

(#M40152973) HCL question Number Keep an EYE

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

Asked In HCL ABHIRUCHI SHARMA (9 years ago)

Unsolved Read Solution (22)

Is this Puzzle helpful?   (38)   (9) Submit Your Solution undefined

Read Solution (22) :

selvi  (9 years ago)

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

Is this solution Helpfull? Yes (27) | No   

Rupen Raj  (9 years ago)

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....

Is this solution Helpfull? Yes (12) | No (5)  

priyamvada  (9 years ago)

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

Is this solution Helpfull? Yes (5) | No   

kishor kumar sahu  (9 years ago)

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.

Is this solution Helpfull? Yes (3) | No (1)  

ramakrishna  (9 years ago)

642+7x2=13y4
y=6=>1364
x=2
x*y=12

Is this solution Helpfull? Yes (2) | No (1)  

shekhar  (9 years ago)

x=2 y=6
xy=12

Is this solution Helpfull? Yes (1) | No   

gyan  (9 years ago)

1+y=3+4, than y=6 and x=2, so xy=12

Is this solution Helpfull? Yes (1) | No   

Obaid Alam  (9 years ago)

As 13y4 is divisible by 11,
so,
(4+3)-(y+1)=0
y=8

and now,
1384-642=742
Therefore,X=4,Y=8

Is this solution Helpfull? Yes | No (8)  

sowmyashree.k  (9 years ago)

1364,x=6,y=6,try to 11 divisible number by looking at last digit and 1st 2digits

Is this solution Helpfull? Yes | No   

Ankit kumar tripathi  (9 years ago)

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

Is this solution Helpfull? Yes | No   

Saumya Bhasin  (9 years ago)

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;

Is this solution Helpfull? Yes | No   

Mal reddy  (9 years ago)

Ans:12min
54*t=45*(t+s)
s/t=1/5
s=t/5
per hr :s=60/5
s=12min

Is this solution Helpfull? Yes | No   

aditi ojha  (8 years ago)

13y4 is divisible by 11 so y = 6 and 642+7x2 = 1364
X=2

Is this solution Helpfull? Yes | No   

anjali mehta  (8 years ago)

, 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

Is this solution Helpfull? Yes | No   

Arpita Mehrotra  (8 years ago)

13y4/11=124 i.e. y=6 and when putting dis value in 642+7x2=1364 we will get x=2
therefore xy=12

Is this solution Helpfull? Yes | No   

kakumanu Gowthami  (8 years ago)

642+7x2=13y4
take x=2, y=6
then answer is 1364/11 is exactly divisible

Is this solution Helpfull? Yes | No   

santhoshi  (8 years ago)

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

Is this solution Helpfull? Yes | No   

sailakshmi  (8 years ago)

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

Is this solution Helpfull? Yes | No (1)  

manohar  (8 years ago)

x=2,y=6 xy=12
11 is exactly divisible by 1364

Is this solution Helpfull? Yes | No   

akbarmuddasir  (8 years ago)

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.

Is this solution Helpfull? Yes | No   

Madhullapalli Suhasini  (7 years ago)

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

Is this solution Helpfull? Yes | No   

Divya Batra  (7 years ago)

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

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.