Latest Elitmus Aptitude Question SOLUTION: HAT *CUP ------ EIUI EART- EUPI-- ---------- HIEEEI ---------

345
876
----
2070
2415
2760
-----
302220
-----

T*P= number trailing with I
T*U= number trailing with T
T*C= number trailing with I

So T=5, I=0;

So P and C are EVEN i.e. any of 2, 4, 6, 8.
and U is ODD i.e any of 3, 7, 9.

Now see column IRI->E. So R+1= number in format as 1E or a single digit E.
Now RE, RI, E+U+1(carry generated from column EAP->E)=10. So E+U=9 and from column E->H we get E+1=H. We have derived that U can be any of 7, 9. So for E+U to be 9 U must be

HAT

* CUP

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

EIUI

EART

EUPI

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

HIEEEI

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

Explanation:

Looking at the multiplicant HAT the partial multipication T*U gives T which says T=5 and U is odd. Also T with other two values i.e c & p yields is i. Which means i=0 because when odd number multiplied with 5 gives 5 itself which is not the case. Hence c & p are even which gives i=0. Hence the problem looks like this,

HA5

* CUP

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

E0U0

EAR5

EUP0

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

H0EEE0

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

now looking at the third column from right R+1=E ( Here there is no carry bcz for this column to carry the maximum value should be 9 which makes E=0 which cant be bcz I is already 0) and the second column from right is U+5=E+10. The maximum value of U is >5 and it must be odd for this condition. Which makes values for U as either 7 or 9. If we take U=9 E will be 4. Then the second column from the left which is 4+9+carry=0+1 or 2 for which the carry has to be 7 which is not possible by adding four numbers.Hence U=7 and E=2. Which reduces our problem to,

HA5

* C7P

-----------

2070

2AR5

27P0

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

H02220

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

Now considering R+1=E makes R=1 since E=1. considering second column from left the equation 7+2+carry=0+1or 2. Here the possible carry form prior column could be one so that 7+2+1=10. So H becomes (E+1=H) H=3. Therefore the problem becomes,

3A5

*C7P

-----------

2070

2A15

27P0

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

302220

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

now considering partial multiplication of 7*3A5=2A15 makes the value of A=4. And considering third column from left i.e 2+A+P=2+10 makes P=6. Hence the problem becomes

345

*C76

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

2070

2415

2760

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

302220

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

Finally to find C we can us partial multiplication of C*345=2760, which makes C=8.

Hence the solution is

345

*876

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

2070

2415

2760

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

302220

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



Thank you..

can anybdy explain me how to solve such kind of ques.?
expln wth d hlp of propr basics

Plz explain the method

Please someone explain such type of questions ? i mean the method of solving?

743
469
----
6687
4458
2972
-----
348467

Could anybody please explain in proper manner...?How re you estimating those values?

rajesh paul ur explanation was gud can u explain still more detailed

345
*876
----------
2070
2415
2760
-----------------
302220
hint is :- HIEEEI take any value of e from 2-9 (hit and trial) and HAT*U=EART ie T*U=T (T is either 5 and u is odd no(3,7,9) or U is 6 and T is (2,4,8,0)any even no)