Latest Elitmus Aptitude Question SOLUTION: How many 4 digit can be formed by using 2,3,4,5,6,7 only once divisible by 25 remember only once* divisible by a 40 b 12 c 20 d 24

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(#M40037926) ELITMUS Numerical Ability question Keep an EYE

How many 4 digit can be formed by using 2,3,4,5,6,7 only once divisible by 25
remember only once* divisible by
a 40
b 12
c 20
d 24

Asked In Elitmus Gaurav Malhotra (9 years ago)

Unsolved Read Solution (9)

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Read Solution (9) :

Mani kandan  (9 years ago)

Ans : 24

For any number to be divided by 25, it should be ended with {25,50,75,100)...
Here 0 is not in the list... so d possibilities are the numbers ended with (25 nd 75)

if v take the no. to end with 25, the remaining 2 digits can be any 2 among {3,4,6,7} . so the possibities are 4!/2= 24/2= 12

if v take the no. to end with 75, the remaining 2 digits can be any 2 among {2,3,4,6} . so the possibities are 4!/2= 24/2= 12

so 12+12=24

Is this solution Helpfull? Yes (6) | No (1)  

LOKESH SAMANT  (9 years ago)

units place= 5
tens place= 2 or 7(as tens place must have 2 or 7 to be exactly divisible by 25)
remaining places= 3 or 4 or 6 or one from 2/5
so total numbers formed= 3*2*2*2*1=24

Is this solution Helpfull? Yes (4) | No   

Yati Sawhney  (9 years ago)

Correct ans is 20

Soln


- - - -

unit place can be filled in 1 way using 5

tens place can be filled in 2 ways uing 2 or 7 in order to make it divisible by 25 exactly once

hundred and thousand place can be filled in 4 and 3 ways using remainin digits

Total no = 24

Now , here we have include no of the form X625 where X= 3,4,7
Thus these 3 numbers must be subtracted

Also 625 * 7 = 4375 is include above and hence must be subtracted

Hence the total number of digits = 24-4 = 20 ANS

Is this solution Helpfull? Yes (4) | No (7)  

Prakash Singh Latwal  (9 years ago)

24, unit place must be 5, and 10's place must be 2 or 7, then remaining digits are either (2,3,4,6) or (3,4,6,7).

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Ritesh Sharma  (9 years ago)

yeah 24 it is... but in the remaining places it should be 3 or 4 or 6 or one from 2/7 not 5... as 5 must present at units place

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Gourab Dutta  (9 years ago)

Unit place should be 5 here & to make it divisible by 25 tens place should be either 2 or 7.
As multiple of 25 are 25,50,75,100,..,175,..225 like this
Now if last two digit 2 & 5 , rest of two digits are selected from 4 digit(3,4,6,7)
So the total number of digits which contain 25 as their last two digits are:- 4P2*1!*1!=12
In the same way when the last two digit is 75, the total number of digit should be 12 ( according to the previous logic)
Hence total no of digits= 12+12=24..

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Liti  (9 years ago)

ans 24.
when 2 is in tens place we got total= 4*3*1*1=12 ways

when 7 is in tens place we got total= 4*3*1*1=12 ways

12+12=24

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Ritesh Sharma  (9 years ago)

units place= 5
tens place= 2 or 7(as tens place must have 2 or 7 to be exactly divisible by 25)
remaining places= 3 or 4 or 6
so total numbers formed= 3*2*1*2*1=12

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tirupathi reddy  (9 years ago)

What about 275 and 725 in the last three digits
@ritesh sharma
In my view the possibilities in ends
25, 75, 275, 325, 375, 425, 475, 625, 675, 725.
For 25 and 75----------4p2*2=12
For remain 8-----------3*8=24
Total 36

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