A certain airline's fleet consisted of 60 Type A planes at the

Algebraic approach:
Consider x as number of years when both planes A & B will reach towards their break-even and then type B just exceeds type A planes.

i.e. in x years, The number of Type B planes with annual rate of 4 will be just greater than (existing inventory of 60 planes minus Type A inventory reduced at annual rate 3 in x years)
i.e. \(4x\geq{60 - 3x}\)
i.e. \(7x\geq{60}\)
i.e. \(x\geq{8.57}\)
Round to next integer \(x = 9 years\)

Hence choice(D) is the correct answer.

PS: This problem is similar to water tank problem - one pipe is filling the tank and another one draining at different rate.

We are given that the fleet started with 60 type A planes. We are also given that the airline retired 3 of the type A planes and acquired 4 new type B planes each year. We need to determine how many years it took for the number of type A planes left in the airline's fleet to be less than 50 percent of the fleet. We can let n = the number of years it will take for this to occur.

The number of type A planes in the fleet can be expressed as 60 - 3n (because the fleet loses 3 type A planes each year). The total number of planes in the fleet is 60 - 3n + 4n, which takes into account the loss of 3 type A planes and addition of 4 type B planes each year.

We are interested in the number of years it will take until the number of type A planes is less than ½ the total planes in the fleet, as illustrated in the following equation:

60 - 3n < (60 - 3n + 4n) x ½

Multiplying the entire equation by 2, we have:

120 - 6n < 60 + n

-7n < -60

n > -60/-7

n > 8 4/7

Since n > 8 4/7, the smallest integer value n can be is 9.

Answer: D
_________________

60 - 3*n = 4*n , n = 60/7 = 8.5

So, in 8.5 years, there would be equal number of planes A & B(which is of course a hypothetical situation).

So, in 9 years the number of B would be more.

Answer is D.

It should by TYPE B, isnt it?

Thanks. You are right. Its a typo. I have updated the thread.

Let x = number of years. Each year we lose 3 Type A planes and gain 4 Type B planes.
Since we start off with 60 type A planes and 0 Type B planes, the following equation would determine the point in time where Type A planes = Type B planes
(60-3x) = 4x....this comes to 8 4/7 years. Since we are looking for the number of years (rounded to the nearest whole number) where <50% of the planes are of Type A, this must be > 8 years. So 9 years.

Very nice problem fozzy, congrats!
Now this is basically an algebraic translation and that's all we need to do so let's go and nail this

60-3x < 1/2(60+x)

Answer is D

Cheers!
J

I think the best and fastest approach here is to back solve
start with C as usual

No of Years Type A Type B
8 60-8*3= 36 8*4 = 32 --> We need more years
9 60-9*3= 33 9*4 = 36 --> We found answer

Ans: D

Hi

I actually just plugged in the numbers and got 8. Can someone please clarify?

End of year 1980 -- 57 A and 4 B.
End of year 1981 -- 54 A and 8 B... and so on

End up year 1988 -- 33 A and 36 B

Since that's 8 years, I picked C. It tips over at the end of year 8, so it's still in year 8. Why do we choose 9?

From the beginning of 1980 to the end of 1988 there are 8 years and 364 days so 9 years.
_________________

So, I also created a table, since for me this was faster and safer than venturing an equation. True, in many cases an equation is the only way to reach an answer in good time, but there is no reason to torture yourself with finding one if you have another option and you cannot readily come up with the correct equation! Right??

______A:60 - 3 per year____B: 0 + 4 per year
Year1______57_______________4
Year2______54_______________8
Year3______51_______________12
Year4______48_______________16
Year5______45_______________20
Year6______42_______________24
Year7______39_______________28
Year8______36_______________32
Year9______33_______________36

I used an equality and back-solving:
y=years

60-3y = 4y

and then plugging in answer choices starting with E)10
60-30< 40 Could Work since 30 is less than half of 70

So trying C)
60-24 > 32
36 > 24 Nope too high but close

So without even doing the math I picked 9 Ans: D

Hi All,

When dealing with questions that can be solved with simple arithmetic and a bit or brute force, it's often best/fastest to just put pen-to-pad and 'map out' the solution. The prompt mentions that planes retired and acquired at the END of each year. You can create a quick table to track what you're looking for:

Start: 60 A and 0 B
Yr 1: 57 A and 4 B
Yr 2: 54 A and 8 B
Yr 3: 51 A and 12 B
Yr 4: 48 A and 16 B
Yr 5: 45 A and 20 B
Yr 6: 42 A and 24 B
Yr 7: 39 A and 28 B
Yr 8: 36 A and 32 B
Yr 9: 33 A and 36 B

After 9 years, the type A planes were less than 50% of the total.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

pacifist85 and Rich demonstrate the utility (and power) of simply creating a table of values.
For more on using tables to solve quantitative questions, watch the following free video - https://www.gmatprepnow.com/module/gmat- ... /video/929

Cheers,
Brent

I go the answer as D - 9,, however, I used the crude way of solving, where I reduced A by 3 every year and simultaneously added 4 for B.

After seeing some posts I can see there are some short cut ways of solving this.

See some similar posts in the way I solved.

Bunuel,

I did the problem using arithmetic progression:

ana = anb
4+(n-1)*4 = 60+(n-1)*(-3)
n=9

Is that solution right? or it was just luck?

TKS!!!

Hi Bunuel.

The prompt question asks "How many years did it take BEFORE the number of type A planes". I get confused in choosing between 8 and 9 because of the word "BEFORE". I was wondering if it was a trap and actually took 8 years before the planes A become half of the Fleet.

"Before" in this context means "until"

How many years did it take until the number of type A planes left in the airline's fleet was less than 50 percent of the fleet?
_________________

60-3x+4y no of planes in the fleet at any point of year

no of type A = 60-3x
no of type B =4y
to find the solution we can form an equation

1/2*(60-3y+4x)>= 60-3y

X=Y no of years starting from 1980
7x>=60 least integer is 9

so answer is 9