GMAT Problem Solving Discussions

MBA
1863
Hi ,

check these!

MR2 - 4 - c (stem 1) if v (stem 2) same as the above
with stm1 and stm2 --> only one of the odd powered digit is -ve leaving the answer to YES.

5 - D
16. = 8 = .5 -> nonzero
16.1/4=4 =.25 nonzero

8 = 2 .25 nonzero
8.1/8 = 1 .25 .125 nonzero

OA's pls
1864
Hi ,

check these!

MR1

1. A

S+.04(C1-2000) = 3620
S+.04(C2-2000) = 3580

two eqns. solve it and it leads to c1-c2=100

2.
C.

stem 1 => 14.1/2 = 7 or 7./7 or 7.1 =7 ==> not clear
stem 2 => integers can be anything => not clear
combining both, it shud be either 7.1 or 1.7 and since the power is even, answer should be same.

3. x1,x2,x3...xn..

if you look at the given formula, it says, x1=3, x2=5, x3=9..on closely observing..its just equal to 2^n+1...
2^20-2^19 = 2^19.2^1 - 2^19 = 2^19( 2-1) = 2^19

so answer is A

OAs please...
1865

Lines n and p lie in the xy plane. is the slope of line n less than the slope of line p ?

(1) lines n and p intersect at point (5,1)
(2) the y-intercept of line n is greater than the y-intercept of p

can anyone explain this plss..if am wrong..also i am not getting much abt the co-ordinate qns like this..can anyone give me a precise ref material for this..please..

for this qn.
stem.1 -> lines intersecting at a point can be having any slope.more info required
stem.2-> clealry not enough..

from stem1,2-->y intercept of line n > Lp..means line n should be having -ve slope and line P must be having +ve slope..therefore,slope of line p is greater

1866

When a certain tree was first planted,it was 4 ft tall. each year the height increased by a constant amount. at the end of the 6th year, the tree was 1/5 taller than it was at the end of 4th year. by how many feet did the height of the tree increase each year?
1) 3/10
2) 2/5
3) 1/2
4) 2/3
5) 6/5

assuming k as the constant increase..doesnt it lead to 4+5k = 1/5(4+3k) + (4+3k) ?

1867
When a certain tree was first planted,it was 4 ft tall. each year the height increased by a constant amount. at the end of the 6th year, the tree was 1/5 taller than it was at the end of 4th year. by how many feet did the height of the tree increase each year?
1) 3/10
2) 2/5
3) 1/2
4) 2/3
5) 6/5

assuming k as the constant increase..doesnt it lead to 4+5k = 1/5(4+3k) + (4+3k) ?

hi vinod,

since the height increased by a constant amount ....let say k..

so at the end of 1st yr the height will be 4+k

similarly end of 6th yr will be 4+6k and 4th yr will be 4+4k

given that end of 6th yr = 1/5th taller than it was at end of 4th yr.....

==> 4+6k = 4+4k+ 1/5(4+4k)
4+6k = 6/5(4+4k)

solving k = 2/3

...so tree increased by 2/3

correct me if i was wrong

Regards
Madhav
1868

if a,b,c are non zero integers and a=2b/c(people i dnt know how to give power ,here b hav power of three).it a is halved wht will be the change in c...plz give the explaianation along with ur answer!!

1869

A company increases its production by 10% due to rise in demand. Then due to holiday rush, they increase the production by another 20% of the new value. What % should the output now be decreased to return to the initial output?

ANy help? Sorry if I posted in the wrong location...thanks in advance

1870
A company increases its production by 10% due to rise in demand. Then due to holiday rush, they increase the production by another 20% of the new value. What % should the output now be decreased to return to the initial output?

ANy help? Sorry if I posted in the wrong location...thanks in advance


Let the company's initial production be 100.
10% increment -> 110. (100+10)
another 20% to new value -> 132 (110 +22).

To get back to 100 from 132.. it would need approximate 25% reduction.
1871

Hi buddy, thanks for the reply but I just figured it out:
Let the company's initial production be 100.
10% increment -> 110. (100+10)
another 20% to new value -> 132 (110 +22).

So, to find the %decrease, the formula is 32/132*100 = 24% to the nearest %.

Thank you tho!!

1872

hey can anyone solve my problem n tell me
a , b , c are non zero integers. where a=2b^3/c
if a is halved wt is the value of c

1873
hi vinod,

since the height increased by a constant amount ....let say k..

so at the end of 1st yr the height will be 4+k

similarly end of 6th yr will be 4+6k and 4th yr will be 4+4k

given that end of 6th yr = 1/5th taller than it was at end of 4th yr.....

==> 4+6k = 4+4k+ 1/5(4+4k)
4+6k = 6/5(4+4k)

solving k = 2/3

...so tree increased by 2/3

correct me if i was wrong

Regards
Madhav

oa is 6/5..
1874
vinothkumartm Says
oa is 6/5..


Going by OA it should be more than 1/4th taller in the 6th year than it was in the 4th year.

Assume it grows by 1 ft every year, the given ratio of 1/5 would become 2/ (4+4) = 1/4.
Hence the answer cant be more than 1 ft.

I got 2/3.
1875
hey can anyone solve my problem n tell me
a , b , c are non zero integers. where a=2b^3/c
if a is halved wt is the value of c

given a = 2b^3/c or c = 2b^3/a
when a is halved, it becomes c = 8b^3/a
i.e c becomes 4 times its original value.

wats the OA?
thanks
1876
Going by OA it should be more than 1/4th taller in the 6th year than it was in the 4th year.

Assume it grows by 1 ft every year, the given ratio of 1/5 would become 2/ (4+4) = 1/4.
Hence the answer cant be more than 1 ft.

I got 2/3.

Tree is 4ft tall
at the end of 1st yr its height becomes 4+k
hence at the end of 6th yr it becomes 4+6k tall
given 4+6k = 6/5(4+4k)
20+30k = 24+24k => 6k=4 => k=2/3

OA is 6/5 doesn't sound convincing.

frm where hv u picked this prob?is there any solution given there?

thanks
1877
When a certain tree was first planted,it was 4 ft tall. each year the height increased by a constant amount. at the end of the 6th year, the tree was 1/5 taller than it was at the end of 4th year. by how many feet did the height of the tree increase each year?
1) 3/10
2) 2/5
3) 1/2
4) 2/3
5) 6/5

assuming k as the constant increase..doesnt it lead to 4+5k = 1/5(4+3k) + (4+3k) ?


Consider that 1st year the ht of tree is 4x, 2nd year it wl be 4x^2.... 4th year 4x^4 & 6th year 4x^6
it says that the tree was 1/5 taller than it was at the end of 4th year
i.e 4x^6 = 4x^4+1/5(4x^4)
solving this we get x = 6/5..

wats the OA...
1878

Hi friends

I have a trouble with solving work related problems. Please let me know how to handle the work related problems easily.

1879
Hi friends

I have a trouble with solving work related problems. Please let me know how to handle the work related problems easily.


aasan hai yaar..practice karo, is all i can say.
1880

Hi guys,

Can you let me know any good book name on GMAT?

It will be very helpful......

1881
Consider that 1st year the ht of tree is 4x, 2nd year it wl be 4x^2.... 4th year 4x^4 & 6th year 4x^6
it says that the tree was 1/5 taller than it was at the end of 4th year
i.e 4x^6 = 4x^4+1/5(4x^4)
solving this we get x = 6/5..

wats the OA...


as per your solution should it not be sqrt(6/5)........
1882

hey friends
tell me if the question is wrong or right!!!!
coz the answer that i suppose should ,is wrong
tell ur answers so that i clear my doubt as soon as possible
it was a question in cambridge test!!!!

is z an odd integer or not!
i)z/3 is an odd integer
ii)3z is an odd integer