MGMAT- VICS PS

[bblast]

Mr. and Mrs. Wiley have a child every J years. Their oldest child is now Tyears old. If
they have a child 2 years from now, how many children will they-have in total?
(A) {(T+2)/J} +1 (B)JT+1 (C)J/T + T/J (D)TJ-1 (E) (J+T)/J

anyone with algebraic approach for this ?


oa[spoiler]=a[/spoiler]

[manpsingh87]

Mr. and Mrs. Wiley have a child every J years. Their oldest child is now Tyears old. If
they have a child 2 years from now, how many children will they-have in total?
(A) {(T+2)/J} +1 (B)JT+1 (C)J/T + T/J (D)TJ-1 (E) (J+T)/J

anyone with algebraic approach for this ?


oa[spoiler]=a[/spoiler]

here j is the difference in years after which mr. and mrs. wiley will have kid;
lets assume that difference is 2; and after 6 years, no. of kids born will be (6/2)+1; to understand why i add "1" consider the following diagram..!!
k1--2yr---k2--2yr---k3--2yr---k4; kid born in the 0th yr{i.e. the year from which counting starts,(reference year for other kids)} was not covered when we divide 6/2 hence we will have to add 1 in the expression,

now consider the scenario given in the question;

age of the eldest son is T years; after 2 years; gap of j years will be fulfilled and the couple wil have a new child; therefore age of eldest son after two year will be (T+2) and total no. of kids will be {T+2/j}+1; hence A

[GMATGuruNY]

Mr. and Mrs. Wiley have a child every J years. Their oldest child is now Tyears old. If
they have a child 2 years from now, how many children will they-have in total?
(A) {(T+2)/J} +1 (B)JT+1 (C)J/T + T/J (D)TJ-1 (E) (J+T)/J

anyone with algebraic approach for this ?


oa[spoiler]=a[/spoiler]

I would plug in values.

Let J=2.
Let 2000 = year in which the first child is born.
Let now = 2006.
Since a child is born every 2 years, there are now 4 children -- born in 2000, 2002, 2004 and 2006 -- and the oldest age is T=6.
When a child is born 2 years from now, the number of children = 5. This is our target.

Now we plug J=2 and T=6 into the answers to see which yields our target of 5.

Only answer choice A works:
(T+2)/J + 1 = (6+2)/2 + 1 = 5.

The correct answer is A.

To solve algebraically:

Since the ages are all J years apart, they are evenly spaced integers.

Number of evenly spaced integers = (biggest - smallest)/(distance between each successive pair) + 1.

Biggest = T+2 (the age of the oldest child 2 years from now)
Smallest = 0 (the age of the last child at its birth 2 years from now)
Distance between each successive pair = J.

Number of children = (T+2 - 0)/J + 1 = (T+2)/J + 1.

[Whitney Garner]

Mr. and Mrs. Wiley have a child every J years. Their oldest child is now Tyears old. If
they have a child 2 years from now, how many children will they-have in total?
(A) {(T+2)/J} +1 (B)JT+1 (C)J/T + T/J (D)TJ-1 (E) (J+T)/J

anyone with algebraic approach for this ?


oa[spoiler]=a[/spoiler]

If you are interested in the "Algebraic Approach" I would definitely recommend utilizing a "Hybrid" approach: pick numbers just to understand the relationship, and then convert back to the algebra so you do not need to test all the answer choices.

Let's pretend that the Wiley's oldest is 10. That means that they have been having kids for 10 years. They will have 1 more child in 2 years - total number of years having kids = 10 + 2 = (T+2) = 12.

Ok, if we now say they have had kids every 4 years, they would have had the 12 year old, an 8 year old, a 4 year old and then the newest kid. The intuition would be to divide 12/3, but we forget that we are starting and ending with a kid (0, 4, 8, 12) so it is similar to finding the number of terms in a list (Most-Least + ONE). So this is like taking the total number of years the Wiley's had kids, dividing by the interval of the kids, then adding one for the newest arrival!

(T+2)/J + 1


Whit

[bblast]

all three approaches to this good gmat problem covered. thank you experts.