Problem Solving

sureshbala wrote:Folks, the next question is from geometry.....Have a look at the following figure..


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Here is the solution....

Here is the next one...

N, the set of natural numbers is portioned into subsets
S1={1}
S2={2,3}
S3={4,5,6}
S4={7,8,9,10} and so on.

The sum of the numbers in subset S50 is

(A) 61,250
(B) 65,525
(C) 42,455
(D) 62,525
(E) None of these

The first element of any given set can be determined using the following formula.
Tn = n(n-1)/2 + 1
Using this formula, the first number of S50 will be 1226.
S50 has 50 elements starting with 1226 which is in an arithmetic progression. Hence sum of all the elements = 50 * 1226 + (50*49)/2 = 62525.

v_shiv wrote:The first element of any given set can be determined using the following formula.
Tn = n(n-1)/2 + 1
Using this formula, the first number of S50 will be 1226.
S50 has 50 elements starting with 1226 which is in an arithmetic progression. Hence sum of all the elements = 50 * 1226 + (50*49)/2 = 62525.

Patterning First Term.

S1 1

S2 1+1

S3 1+ 1 + 2

S4 1 + 1 +2 +3

S5 1 + 1 + 2 + 3 + 4

So Sn = 1 + Sum of first n-1 numbers

We know the sum of first n numbers is n(n+1)/2 ( A Must Know Formula!!)

S50 = 1 + (50)*51* 1/2 =1226

According to the observed pattern, The number of terms in each set =n. So S50 gas 50 terms beginning with 1226.

We know the formula for the sum of an arithemetic sequence is

Sn = n/2 (2a+(n-1)d) ( A Must Know Formula!!) where a is the first term and d is arithmetic difference, which according to pattern is 1.

Plugging gives S50 = 50/2 (2*1226 +(50-1)1)=62,525

This approach is more intuitive since it uses well known formulas.

dtweah wrote:

v_shiv wrote:The first element of any given set can be determined using the following formula.
Tn = n(n-1)/2 + 1
Using this formula, the first number of S50 will be 1226.
S50 has 50 elements starting with 1226 which is in an arithmetic progression. Hence sum of all the elements = 50 * 1226 + (50*49)/2 = 62525.

Patterning First Term.

S1 1

S2 1+1

S3 1+ 1 + 2

S4 1 + 1 +2 +3

S5 1 + 1 + 2 + 3 + 4

So Sn = 1 + Sum of first n-1 numbers

We know the sum of first n numbers is n(n+1)/2 ( A Must Know Formula!!)

S50 = 1 + (50)*51* 1/2 =1226

According to the observed pattern, The number of terms in each set =n. So S50 gas 50 terms beginning with 1226.

We know the formula for the sum of an arithemetic sequence is

Sn = n/2 (2a+(n-1)d) ( A Must Know Formula!!) where a is the first term and d is arithmetic difference, which according to pattern is 1.

Plugging gives S50 = 50/2 (2*1226 +(50-1)1)=62,525

This approach is more intuitive since it uses well known formulas.

Error in calculating sum of n-1 terms. S50 = 1 + (49)*50* 1/2 =1226

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

I feel E

Total B+G=x+1 where x is number of chairs
2B( adjusted for one B) +1G= z+2+3 where z is the number of chairs occupied by B or G

I dont think we have enough info to find B if z =0 T=5 but we dont know whether it is 4,1 or 3,2 (B,G)

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

Let X = number of boys r

Y= Number of girls

Then x+y= the total number of people to be seated at any time.
In the first case we are told that one person is left standing, so there is one less seat than there are people to occupy them.

Example:

seated: B B G G G G Unseated: B

Since x+y is total of people, in this case it is 7. But 6 people are seated. So number of people seated (=number of seats occupied) is 1 less than total number of people:
x+y-1= number of seats occupied.

Now in the second case, two boys occupy a single and one girl occupies a single seat and we are told three 3 seats remain unoccupied. If we can find number of seats occupied in this case, we simply take the difference between the 2 cases.

if x is total boys and and 2 occupy one seat then each occupies
x/2 seats. Number of seats occupied in this case is:

x/2 + y

Number of seats occupied in case 1 - Number of seats occupied in case 2 =3
x+y-1 -(x/2+y)=3
x/2=3
x=6.
Choose B.

dtweah wrote:

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

Let X = number of boys r

Y= Number of girls

Then x+y= the total number of people to be seated at any time.
In the first case we are told that one person is left standing, so there is one less seat than there are people to occupy them.

Example:

seated: B B G G G G Unseated: B

Since x+y is total of people, in this case it is 7. But 6 people are seated. So number of people seated (=number of seats occupied) is 1 less than total number of people:
x+y-1= number of seats occupied.

Now in the second case, two boys occupy a single and one girl occupies a single seat and we are told three 3 seats remain unoccupied. If we can find number of seats occupied in this case, we simply take the difference between the 2 cases.

if x is total boys and and 2 occupy one seat then each occupies
x/2 seats. Number of seats occupied in this case is:

x/2 + y

Number of seats occupied in case 1 - Number of seats occupied in case 2 =3
x+y-1 -(x/2+y)=3
x/2=3
x=6.
Choose B.

A perfect example of the carelessness that can happen on the GMAT even when you know how to solve.
x+y-1 -(x/2+y)=3
x/2=4
x=8.
Choose C but I don't get this opportunity on the GMAT! .

As per your answer question should say every 2 boy and each girl ...something of that sort

hemantsood wrote:As per your answer question should say every 2 boy and each girl ...something of that sort

No it need not say. 2 boys sitting on one chair and 1 girl sitting on another chair, with 3 chairs remaining should not be interpreted as there being only 2 boys and 1 girl, 3 people. I considered that too. If this is true then there would be only 5 chairs. Then this would contradict case 1 where it is clear that each was seated in one chair with one boy standing. Then we would have 1 boy sitting 1 girl siting 3 empty seats and 1 boy standing. The problem breaks down logically. More besides, if we assume only 2 boys and one girl then we already know the answer merely from case 2.

The thing to note is that in both the case the girls are occupying the same number of chairs. So the difference in the number of chairs being occupied is due to the changed seating pattern of the boys.

So let us consider that there are X+1 number of boys, in the first situation the boys are occupying X number of chairs, in the second they are occupying (X+1)/2 number of chairs

So X-(X+1)/2 = 3, solve for X and we have X =7 and therefore the number of boys is equal to 8.

Cheers.

sureshbala wrote:Folks, finally here is the solution....Thought of being more clear on this problem, so uploaded the image....Have a look at this....

I think That You should look this