M4maths Previous Puzzles

14 Aug 2015 Number

A number more than 1200 and less than 1300 is such that all its digits except one digit are factor of 9. What is the probability that number is more than 1250.

OPtion

1) 1/2
2) 1/3
3) 1/4
4) 4/9
5) 25/99
6) 8/25
7) 1
8) 0
9) 1/9
10)2/5

Solution

Let x be the number
If 1200 < x < 1300
Then thousands digit will be 1
Hundreds digit will be 2
Tens digit has 3 choices 1, 3, 9
Units digit has 3 choice 1, 3, 9
Total four digit numbers = 1*1*3*3 = 9
If 1250 < x then
Thousands digit will be 1
Hundreds digit will be 2
Tens digit will be 9
Units digit has 3 choice = 1, 3, 9
Total four digit numbers = 1*1*1*3 = 3
Probability = 3/9 = 1/3

Correct Option: 2 Best Solution (22)

Devendra Marghade   9 years AGO

The number can be 1211, 1233, 1266
So 1266 can be the number between 1250 & 1300.
Probability=1/3
Option 2)

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anand vijay   9 years AGO

numbers greater than 1200 but less than 1300 whose digits except one digit are factor of 9 :-1211,1213,1219,1231,1233,1239,1291,1293,1299
numbers greater than 1250 but less than 1300 whose digits except one digit are factor of 9 :-
1291,1293,1299
probability=3/9=1/3
option 2)

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rahul shaw   9 years AGO

since the no is after 1200 so...1st two digit value is fix...
and the factors of 9 are 1,3,9.
2 is non factor
the no are 1211,1213,1219,1231,1233,1239, after 1250....the no are 1291,1293,1299
so probability is=3/9=1/3
option 2

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PAWAN KUMAR   9 years AGO

Total no of such nos b/w 1200 and 1300 = 9.
Nos above 1250 = 3
So, probability = 3/9 = 1/3.

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Vishal   9 years AGO

factors of 9: 1,3,3,9
for no. to be between 1200-1300, 12_ _ is necessary, now 1 is factor of 9 while 2 is not, hence the only digit which is not factor of 9 is 2.
last 2 digit can be 33,39,93,99, each 3 and 9 being factor of 9.
1200-1250: 1211,1213,1219,1231,1233, 1239
1250-1300: 1291,1293,1299
probability of no more than 1250: 3/9 = 1/3.

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Santosh Kumar Kasaudhan   9 years AGO

all digits are factor of 9 except one digit
here all no are starts from 12_ _
so 2 is not factor of 9
now, we can have last two digit 1,3,9
so total choice= 3!=9
no greater then 50
129_= have 3 choice
so prob=3/9
ans=1/3

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Ajay Kumar Gupta   9 years AGO

possible no. are-1211,1213,1219,1231,1233,1239,1291,12931299,
total nos. are=9
nos. more than 1250 =3
so p(more than 1250)=3/9=1/3

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nidhi tiwari   9 years AGO

numbers are:
1211
1213
1219
1231
1233
1239
1291
1293
1299
total no=9
but greater than 1250 are only 3
so probability
3/9=1/3

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apoorv gupta   9 years AGO

the number have to be 12_ _. Andfactors of 9 are 1,3,9.
since 2nd digit (i.e. 2) is not a factor of 9. therefore last two digits have to be factor of 9.
Therefore possible numbers are 1211, 1213, 1231, 1219, 1233, 1239, 1291, 1293, 1299.
probability that number is more than 1250 is 1/3.

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Deepak kumar kumawat   9 years AGO

factors of 9=1,3,9
total numbers=[1][2]�3�3=9
1211
1213
1219...........3
1231
1133
1239.............3
1291
1293
1299..............3
so required probability = 3/9=1/3
hence option 2

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Narendra Gopi   9 years AGO

Total numbers are 1211,1213,1219
1231,1233,1239
1291,1293,1299
so th probability is more than 1250 so 3/9=1/3

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vini   9 years AGO

Factors of 9 are 1,3,9
One digit is not factor of 9 that must be hundred position number ie., 12_ _
The possible numbers are
1211
1213
1219
1231
1233
1239
1291
1293
1299
Probability=Favorable event/Sample space
Probability of the number are greater than 1250 = 3/9=1/3

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Deepanshu Chaturvedi   9 years AGO

total no. with factors of 9 between 1200 and 1300 are 9(1211,1213,1219,1231,1233,1239,1291,1293,1299) selecting no. greater than 1250 is 3.therefore probability is 3/9=1/3

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Santosh Kumar   9 years AGO

1st find the total number of cases=3*3=9
(because 1st fix 1 and 2 and lst two digit will be 1,3,9)
no of favourable cases=fix pstn of 1,2 and 9 so lst digit will be filled by any three of them1 or 3 0r 9=3
hence probablity=3/9=1/3

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ritu yadav   9 years AGO

1211
1213
1219..3
1231
1233
1239..3
1291
1293
1299..3
only 1291 ,1293 and 1299 are greater than 1250
so probability is 3/9=1/3

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NITESH KUMAR SINGH   9 years AGO

total no possible is 9 and no greater than 1250 is 3 so probability is 1/3

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siddhi datri   9 years AGO

total nos.- 1211,1213,1219,1231,1233,1239,1291,1293,1299
p=3/9=1/3

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kapil gupta   9 years AGO

here we know that 1st and 2nd digits will be 1 and 2.. and for next three possibilites are 1,3 and 9.so for 2 places 3 numbers are there.so it will be 3/9=1/3

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Azhar Syed   9 years AGO

The possible numbers are 1211, 1213, 1219, 1231, 1233, 1239, 1291, 1293, 1299
so out of 9 only 3 are above 1250 hence probability is 3/9 i.e 1/3

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Aurobinda Mahapatra   9 years AGO

1211
1213
1219
1231
1233
1239
1291
1293
1299
so 3/9=1/3

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Math Lovers   9 years AGO

Number is greater than 1200 and less than 1300.
It means number lies between 1201 to 1299

Now factors of 9 = 1, 3, 9

As in number 12XY, 2 is already not a factor of 9 so x and y must be either of 1, 3 or 9

Hence total such numbers are = 1*1*3*3 = 9

now number greater than 1250 = 1291, 1293, 1299

prob = 3/9 = 1/3

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shivjee yadav   9 years AGO

its clear from question 1st and 2nd position are fixed that 1 and 2nd,at third and 4th position among (1,3,9) that means 3*3=9 sample space if i calculate more than 1250 that means third position are also fixed with 9 and 4th among(1,3,9) =3 that means probality=3/9=1/3

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