Gmat Prep Growth, exp problem

initial height = 4
height at year x = 4+cx where c is the constant growth rate/year
height at year 6 = 4+c6
height at year 4 = 4+c4

(4+c6)=(6/5)(4+c4)

4+6c=24/5+24/5*c
6/5c=4/5
c=4/6=2/3

thanks

khurram

Hi Khurram,
I have tried to solve the questions given by you.

Q1. n=p^2q and nis multiple of 5

it means q is also a multiple of 5 and thus, p^2q^2 will be multiple of 25

Q2. Lets say there were x no. of employees before July1 last year

so, on July1 last year no. of wmployees=0.9x

Also, Ma=1.1Mb
where Ma= mean after July1 and Mb=mean before july1

Mb=totalSb/x or, totalSb=Mb*x
Ma=totalSa/0.9x or, totalSa=Ma*0.9x

plug the value of Ma=1.1Mb
totalSa=Mb*1.1*0.9x=0.99x*Mb

hence, totalSa/totalSb=0.99 =99% ans

Q3. mean=(33+x)/6 and Medn=(6+8)/2=7
medn=6*Mean/7, mean=7*7/6
so, 49/6=33+x/6
x=49-33=16

Q4. 2^x(1-1/4)=3*2^x-2
x-2=13 so, x=15

Q5. solving the line equation we get y=x^2+(a+b)x+ab
Stem 1 gives value of a+b=-1 NS
Stem2 St line passes through (0,-6) so, ab=-6 NS
combining both stmt Sufficient

Q6.No. of ways to select 2 apples from 5= 5C2=10
probability that two apples including rotten to be selected 2/5.

Q7. T:R:N:M::8:7:3:2
Stem 1. 8x=18750+3x
5x=18750, x=3750 Sufficient

Stem2. 2x=7500, x=3750 sufficient

Hence D

Q8. Lets say karen's Salary in 1995 was $x so, in 1998 it is $x(1+p)

Stem1. x=2000+J in 1995. Not Suff
Stem2. x(1+p)=2440+J in 1998. Not Suff

Combining 1&2, we get x(1+p-1)=440 so, p can be calculated Suff
Hence, ans C

Q9. lets say every year the tree grew by x
1st yr--4+x
2nd yr--4+2x
4th yr--4+4x
6th yr--4+6x

Now, (4+6x)-(4+4x)=1/5*(4+4x)
6x=4
x=2/3

Correct me if I am wrong.

The question I was asking was ksh answered as q9.

Thanks
Khurram