Latest GMAT Aptitude Question SOLUTION: Q. How many odd numbers less than 1000 can be formed using the digits 0, 2, 5, 7? (repetition of digits is allowed ?

• Placement Papers
• /
• GMAT
• /
• Numerical Ability
• /
• Number System

GMAT Question Papers:

(#M40031263) GMAT Numerical Ability question Keep an EYE

Q. How many odd numbers less than 1000 can be formed using the digits 0, 2, 5, 7? (repetition of digits is allowed ?

Asked In GMAT pramila (11 years ago)

Unsolved Read Solution (11)

Is this Puzzle helpful?   (26)   (12) Submit Your Solution Number System

Read Solution (11) :

RAKESH  (11 years ago)

odd numbers less than 1000
3 digit no.=3*4*2=24
2 digit no.=2*3=6
1 digit no.=2
so total odd no. less than 1000 =24+6+2=32
ans 32

Is this solution Helpfull? Yes (10) | No (2)  

madhuri gupta  (11 years ago)

i think 14

Is this solution Helpfull? Yes (2) | No (6)  

Vivek  (11 years ago)

3digit no.- 3*4*2=24
2digit no.- 3*2=6
1digit no.- 2(odd, 5 nd 7)

Total= 24+6+2=32

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

mahima srivastava  (11 years ago)

3*4*2=24
any digit except 0 can come at first place so 3 ways
then any digit at tens place so 4 ways
either 0 or 2 at units place so 2 ways
3*2*4=24

Is this solution Helpfull? Yes | No (3)  

elavendan  (10 years ago)

32
For three digit odd numbers _ _ _ the first place cannot be filled by 0. other three can fill. Second place can be filled by all 4. Third place cannot be filled by either 2 or 0. So, 3*4*2=24.
For two digit odd numbers _ _. First place cannot be filled by 0. Other three can fill. Second place can be filled either by 5 or 7. So 3*2=6.
For single digit odd numbers it can either be 5 or 7. so 2.
24+6+2=32

Is this solution Helpfull? Yes | No (1)  

RAHUL KUMAR  (10 years ago)

ans=48
3 digit=3
2nd digit=4
1st digit=4
so total odd no=4*4*3=48

Is this solution Helpfull? Yes | No (2)  

don mahmud  (10 years ago)

very easy, 1c1*4c1*4c1*2c1 =32.
1c1 bwcause anything without 0 will make it more than 1000 & 2c1 because with only 5 or 7 the number will be odd.

Is this solution Helpfull? Yes | No   

sonu  (10 years ago)

32 :)
confirm !

Is this solution Helpfull? Yes | No   

santhi   (10 years ago)

three digit number:
last no must be 5,7 only 2 option is possible.
way=4*4*2=32.

Is this solution Helpfull? Yes | No   

Aniket  (10 years ago)

Places: 4th place 3rd place 2nd place unit pace
nos can be placed: 0 0,2,5,7 0,2,5,7 5,7
possibilities: 1 4P1=4 4P1=4 2P1=2
Total no of odd nos = 1+4+4+2 = 11

Is this solution Helpfull? Yes | No   

Ikbal Ahmed Laskar  (7 years ago)

Its a based on P & C, 4 boxes are there.
1.first box can be filled in 1 ways ( 0 )
2. Second box can be in 4 ways
3. Third box can be in 4 ways.
4.Fourth box can be in 2 ways (5 & 7)

Thus total permutation = 1*4*4*2
= 32

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.