[vpitc]

1) Of the three-digit positive integers that have no digits equal
to zero, how many have two digits that are equal to each
other and the remaining digit different from the other two?
A. 24
B. 36
C. 72
D. 144

E. 216

I think the answer option is not avl.......

[Nipunbans]

If M = 3 digit, M^3 can be 7 or 8 digit
If M^2 = 5 digit, M lies mainly b/w 100 to 300. Same condition as above.

Let M be smallest 3 digit no. = 100,
M^2=10000 (5 digit) and M^3=1000000 (7 digit)
If M is 300, M^2=90000 (5 digit) and M^3=27000000 (8 digit)
So, we can say if it is 7 or 8 digit. hence E..

Originally Posted by vpitc

If M is a positive integer, then M^3 has how many digits?
(1) M has 3 digits.

(2) M^2 has 5 digits

is there any shortcut.....

[Nipunbans]

lets say ones and tens digit is same:-
Ones place can have any no. from 1-9 = 9 ways
Tens digit is same as ones = 1 way
hundreds can have anything b/w 1-9 but not the no. chosen by ones = 8 ways.
So, total possiblilties = 8*1*9 = 72
Now there are 3 possiblilities ones and tens can be same, tens and 100ths and ones and 100ths.
So, 72*3 = 216, E is the ans..

Originally Posted by vpitc

1) Of the three-digit positive integers that have no digits equal
to zero, how many have two digits that are equal to each
other and the remaining digit different from the other two?
A. 24
B. 36
C. 72
D. 144

E. 216

I think the answer option is not avl.......

[vpitc]

Originally Posted by Nipunbans

lets say ones and tens digit is same:-
Ones place can have any no. from 1-9 = 9 ways
Tens digit is same as ones = 1 way
hundreds can have anything b/w 1-9 but not the no. chosen by ones = 8 ways.
So, total possiblilties = 8*1*9 = 72
Now there are 3 possiblilities ones and tens can be same, tens and 100ths and ones and 100ths.
So, 72*3 = 216, E is the ans..

hi,
im convinced now
i tried to solve like 9C2*3!/2 and got 108 as solution
tell me where i went wrong

[bhavin422]

Originally Posted by vpitc

hi,
im convinced now
i tried to solve like 9C2*3!/2 and got 108 as solution
tell me where i went wrong

Hey Vijay,
Bold part is incorrect ..
If u have used that if p things out of n are alike then, total ways = n! / p!

Then, in that case, ur numerator is incorrect ..
Extending your logic, 2 distinct digits can be chosen in 9C2 ways, 3rd digit is not repetitve, hence it has 2 options ..And these 3 digits can be arranged in 3 ! ways ...

Hence, numerator = 9C2 * 2 * 3 !

Of these, 2 digits are alike, hence total ways = 9C2 * 2 * 3 ! / 2 ! = 216 ...Ans ...

Hope that helps !

[Nipunbans]

If x, y, and z are integers, and x < y < z, is z – y = y – x?

(1) The mean of the set {x, y, z, 4} is greater than the mean of the set {x, y, z}.
(2) The median of the set {x, y, z, 4} is less than the median of the set {x, y, z}.

[Nipunbans]

A certain investment grows at an annual interest rate of 8%, compounded quarterly. Which of the following equations can be solved to find the number of years, x, that it would take for the investment to increase by a factor of 16?

16 = (1.02)^x/4
2 = (1.02)^x
16 = (1.08 )^4x
2 = (1.02)^x/4
1/16 = (1.02)^4x

[first_strike]

Hi All,

Need some help with this question...

A container has 4 different varieties of fruits. In how many ways can you choose 6 fruits from the container ? ( Information on number of fruits in the container isn't available)

Thanks!

[Nipunbans]

Is the Ans 4^6 ??

Originally Posted by first_strike

Hi All,

Need some help with this question...

A container has 4 different varieties of fruits. In how many ways can you choose 6 fruits from the container ? ( Information on number of fruits in the container isn't available)

Thanks!

[first_strike]

Originally Posted by Nipunbans

Is the Ans 4^6 ??

I saw this question in some material, it says the answer is 33 but doesn't explain how, am clueless about the way to arrive at this number...

Any pointers would help...