Problem Solving

Uri wrote:let, b--> no. of boys
g--> no. of girls and
c--> no. of chairs
we are given, b + g = c + 1............(1)
and (b/2) + g = c - 3..............(2)
Solving we get, b = 8
Ans. (C)



sureshbala, plz provide the answer of the subset (S50) problem also, when you are back from vacation!

Here is the solution for the subset problem......

S1 contains 1 element
S2 contains 2 elements
S3 contains 3 elements
.................................
.................................
.................................
S49 contains 49 elements

Hence total number of elements used in the first 49 subsets =
1+2+3+...........+49 = (49 * 50)/2 = 1225.

So S1 starts with the number 1 and S49 ends with the number 1225.

Hence S50 contains 50 elements starting with 1226.

S50={1226,1127,..................1275}

Hence sum of these numbers = 50/2 (1226+1275) = 25 * 2501 =62525

sureshbala wrote:Folks, sorry for the delay (I am on a vacation).

Anyway, here is the next one.....

There are some boys and some girls. If each child sits on one chair, one boy will have no chair to sit. If two boys sit on one chair and one girl occupies one chair, then three chairs are unoccupied. How many boys are there?

A. 4

B. 6

C. 8

D. 10

E. Cannot be determined

Folks, this can be answered very quickly if you can conclude the following two results.

Result 1: Since the number of chairs occupied by the girls in both the cases is same there is no need to consider the number of girls.

Result 2: In the second case we are able to seat 2 boys per chair and no boy is left unseated, hence the number of boys must be even.

Let the number of boys be 2x.

Seats occupied in the first case = 2x-1

Seats occupied in the second case = x

Given that 2x-1-x = 3

So x = 4

Hence the number of boys 2x = 8.

Here is the next one.....

If x and y are positive integers such that 5 * x^3 = 6 * y^4, find the minimum possible value of x+y.

A. 1240

B. 1260

C. 1060

D. 1080

E. None of these

My answer is 1260, B. Will post the solution if it is correct, Suresh please confirm.

Solid question....I have B as the answer as well.

Can I post my solution....or are we supposed to give everyone a chance to have a crack at it?

gabriel wrote:My answer is 1260, B. Will post the solution if it is correct, Suresh please confirm.

Hi Gabriel,

You are right...Please post your solution.

Musiq wrote:Solid question....I have B as the answer as well.

Can I post my solution....or are we supposed to give everyone a chance to have a crack at it?

Hi,

You can post your solution.

sureshbala wrote:

Musiq wrote:Solid question....I have B as the answer as well.

Can I post my solution....or are we supposed to give everyone a chance to have a crack at it?

Hi,

You can post your solution.

Thanks Suresh.

When Positive integers are the ONLY exponents being considered (as is the case with this problem), then for equality to hold, the PRIME numbers on the left hand side of the equation MUST match the PRIME numbers on the right hand side.

So we rewrite the question as:

5 * X^3 = 2^1 * 3^1 * Y^4

Or, we can think of this as Y^ 4 = (5 * X^3) / ( 2^1 * 3^1 )

Since Y is a positive integer, the Right hand side of the above equation MUST give us a perfect quadrupled power.

Intuitively, this means X has to be such that the 2 and the 3 in the denominator get cancelled (so X should have properties of 2*3). Also, it should have properties of 5 since there is a single 5 in the numerator (which is not a quadrupled power).

Therefore X is equal to 2^3 * 3^3 *5^1 ( or numerically this value is 1080)

If you check back with this value in the right hand side, after simplification we get, 2^8 * 3^ 8 *5^4. This is the perfect quadrupled power of an integer.

Plug this value of X into the original equation to get value of Y as 180.

Since we want X+Y; the answer is B = 1260.

My solution. We have to have a combination of 5 and 6's in both the side. So we have

5 (5^a*6^b)^3 = 6(5^m*6^n)^4, over here x = 5^a*6^b and y = 5^m*6^n.

Now, 5^(3a+1) = 5^4m

and

6^3b = 6^(4n+1)

So,

3a+1 = 4m and 4n+1=3b

Now, choose the smallest value of a, that makes 3a+1 a multiple of 4. And choose the smallest value of n that makes 4n+1 a multiple of 3.

Substitute these values in the initial equation and you have your value for x and y.

Cheers.

sureshbala wrote:Keeping in mind the feedback(I agree to the point that these questions are a bit tough but at the same time much within the basic concepts) that I got, here is the next one...

Edusoft, a software firm has engaged certain number of men to finish a project in the stipulated time. But after working for 14 days, 1/4 of them have left and to complete the remaining project the remaining members have taken as many days as the initial number of men would have taken to complete the entire project. In how many days was the project completed?

A. 48 B. 60 C. 56 D. 40 E. None of these

I am getting answer as E. and here is the solution.

Lets say the later # of days (after 1/4 left) = D
Lets say 1 man in 1 days does X work.
=> In 14 days N man does 14NX work
=> In D days N man does DNX work
=> work left after 14 days DNX-14NX
=> output per day there onwards = 3/4NX
=> since "the remaining members have taken as many days as the initial number of men would have taken to complete the entire project"
=> (DNX-14NX)/3/4NX = D
=> D = 56
Total days = 56+14 = 70
Answer E

(2. 3 Y^4)/X^3=5

This is only true if X has 2 and 3 as prime factors. X must have 5 also to cancel the 5’s in the numerator otherwise we end up with 5^4. Since X is cubed and has prime factors 2 and 3 , Y must have the factor of 2 and 3 to cancel the remaining 2's and 3's in the denominator. And of course y must have 5 as a prime factor or else we wouldn't have 5 on the RHS. So we are able to reduce Y and X to their respective fundamental prime factorizations! The problem is we don’t know what powers each will have so we have to assign variables.

X= 2^a . 3^b . 5^c

Y =2^d . 3^e . 5^f

Putting these in the above we have

2^1 . 3^1 . (2^d . 3^e . 5^f)^4)/( 2^a . 3^b . 5^c)^3 =5

(2^(4d+1) . 3^(4e+1) . 5^4f)/ (2^3a . 3^(3b) . 5^3c) =5

5 on the RHS of equation must mean that 4d+1= 3a, 4e+1=3b, since 2 and 3 are not on the RHS. And 4f -3c=1, since there is a singe 5 on RHS.

The smallest positive integer values in which these hold is d=2, a=3, e=2, b=3 , f=1 c=1

Plugging these in our prime factorization equations gives

X= 2^3 . 3^3 . 5^1=1080
Y =2^2 . 3^2 . 5^1=180

X+Y=1260

5 * x^3 = 6 * y^4 RHS is 2*3*y^4 since LHS is a multiple of 5 so should the RHS be , so y is a multiple of 5 and we have one 5 left out on the LHS after getting the perfect cube , so y = 2^2*3^2*5 and x = 2^3*3^3*5

Hi sureshbala.. waiting for your next question.

sureshbala wrote:Folks, both Naresh and Bluementor gave the correct explanation to this question and I am sure everyone understood them.

I suggest you to look into the solution provided by bluementor because he generalized the value of f(x) which will help us to answer this question even if it is asked in terms of DS. Also the solution provided by Naresh is quick enough.

Anyway, I have simplified a bit of calculation...look at the below image

sorry for digging this out.
i am a new comer...

i just wondered why we have to multiple Equation 2 by 2 and then minus Equation 1?

juyau800 wrote:

sureshbala wrote:Folks, both Naresh and Bluementor gave the correct explanation to this question and I am sure everyone understood them.

I suggest you to look into the solution provided by bluementor because he generalized the value of f(x) which will help us to answer this question even if it is asked in terms of DS. Also the solution provided by Naresh is quick enough.

Anyway, I have simplified a bit of calculation...look at the below image

sorry for digging this out.
i am a new comer...

i just wondered why we have to multiple Equation 2 by 2 and then minus Equation 1?

hi, let me write it in detail. may be that will simplify your thoughts.

f(x) + 2f(2002/x) = 3x

f(2002/x) + 2f(x) = (6006/x)
when i multiply this by 2,
2f(2002/x) + 4f(x) = (12012/x)

in other terms,

2f(2002/x) = (12012/x) - 4f(x)

sub this in the orig equation for 2f(2002/x)
f(x) + (12012/x) - 4f(x) = 3x

simplify this further,

3f(x) = (12012/x) - 3x

for x= 0.5,

substitute x=0.5 in the equation and you will get the answer for f(0.5).

hope i have clarified your doubt.