Problem Solving

Hello people,
I am starting this new thread under each sub-forum, where I will be posting new catch-points/tips from my notes.

Let this be one point stop to refresh your concepts.
If you find some discrepency in the tips posted, please copy-paste the same in the posts outside this thread, so that we can discuss and I would update the correct one in this thread.

This is just to have flow of the tips maintained.
Happy reading!!!

Points to remember about Square, Parallelogram, Rhombus and Rectangle.

These figures have simiarities. Although there are differences worth noting. This can be helpful to you in your PS as well as DS sections.

Square: 1. Diagonals of square bisect and are perpendicular to each other.
2. Diagonas bisect each pair of opposit angles.
3. Diagonals form four congruent triangle.

Rectangle : 1. Diagonals of rectangle bisect each other.
2. Diagonals form four congruent triangle.

Paralelogram: 1. opposite sides are equal and parallel to each other.
2. Opposite angles are equal.
3. Diagonals bisect each other.

Rhombus: 1. All sides are equal.
2. Diagonals of rhombus (just like square's) bisect and are perpendicular to each other.
3. Opposite angles are equal.
4. Diagonals bisect 4 interior angles.

Polygon Properties you should know for GMAT

Let the number of sides be n

1. Exterior angle = 360/n
2. Interior ange = 180-Exterior angle.
3. Sum of all interior angles = (n-2)*180
4. Sum of al Exterior angles = 4*90
5. Number of Diagonals = [n(n-1)/2] - n

Make sure that you remember atleast property 3 and 5. I have seen extensive use of property 3.
Any question on Polygon, do not think beyond these 5 properties.

Given a/b = c/d, then,

1. b/a = d/c
2. (a+b)/b = (c+d)/d
3. (a+b)/(a-b) = (c+d)/(c-d)
4. a/b = (a+c)/(b+d) = (a-c)/(b-d)

# Given that A/B = r/s and B/C = t/u; then A:B:C = ?
Approach is
A:B:C = (Product of all numerators):(first denominator * 2nd numerator): (Product of all denominator)

= (r*t ): (s*t) : (s:u)

You can also generalize this formula
Let A/B = n1/d1 and B/C = n2/d2, C/D = n3/d3; then A : B : C : D = ?
A:B:C : D = (n1*n2*n3) : (d1*n2*n3) : (d1*d2*n3) : (d1:d2:d3)

Great thread!

Question on your last thread, A/B = B/C. In your last generalization example i notice you have 4 variables, A/B, B/C and C/D to represent A:B:C. What role does D play. I don't see it in your stimulus.

thanks

Point noted and I have edited the same.
Actually, colon and round brackets together made a smiley, so did colon and D together...
I have edited the same and thanks for pointing out.

In the following discussion, letters written in () represent base.

change of base:
log(a)N = log(b)N * log(a)b

log(a^k)N = [1/k] log(a)N

Compare the number A=log(2) 3 and B=log(6) 9
Approach is:
log(2) 2 <log(2) 3 < log(2) 4
=> 1<log(2) 3< 2
=> 1<A<2............................................................1

Now, B = log(6) 9
=> log(6) 6<log (6) 9<log(6) 36

(if bases are same, just the numerator value)
=> 1<b<2............................................................2

Assume A>3/2
=> log(2)3 >3/2
log 3>3/2 * log 2
3>2^(3/2)
9>8

=> A>3/2

similarly try for B and it comes out B<3/2.
hence A>B

"change of base:
log(a)N = log(b)N * log(a)b"


Is this same as log(a)N=log(b)N/log(b)a..I thought this is the correct one.

See,
You are correct that way also, as

log(b)N = log N / log b (generally)
thus,
log(b)N * log(a)b =[log N/log b] * [log b / log a]
Which leads me to your conclusion, that is

log(b)N * log(a)b = [log N/log b] / [log a/ log b].

Type 1: [ax + b]/[px +q] <0
The whole idea is to keep the given LHS less than 0
=> either ax +b >=0 and px + q<0
or
ax +b <=0 and px + q>0

Type 2: In the Equation with Modulus, best way to solve MOD is to square the either sides and proceed.

Type 3: [ax +b]/ [px^2 + qx + z] < y

"Please do not cross multiply as in the denominator, we are not sure if its positive or negative." Hence, best way is to work in a way that we do not change the given inequality sign.
that is,
[ax+b]/[px^2 + qx + z] - y< 0 (take y to other side, as y is constant and we know its sign).
Take the lcm and proceed.

Type 4: [ax+b]/[px+q] >0

As I said above, do not cross multiply until u know the sign of denominator. Hence make the denominator postive by squaring just the denominator.
that is,
[ax+b] * [px+q] / [px+q]^2 > 0
and now proceed as usual.

pahwa wrote:Type 1: [ax + b]/[px +q] <0> either ax +b >=0 and px + q<0
or
ax +b <0>0

Type 2: In the Equation with Modulus, best way to solve MOD is to square the either sides and proceed.

Type 3: [ax +b]/ [px^2 + qx + z] < y

"Please do not cross multiply as in the denominator, we are not sure if its positive or negative." Hence, best way is to work in a way that we do not change the given inequality sign.
that is,
[ax+b]/[px^2 + qx + z] - y<0>0

As I said above, do not cross multiply until u know the sign of denominator. Hence make the denominator postive by squaring just the denominator.
that is,
[ax+b] * [px+q] / [px+q]^2 > 0
and now proceed as usual.

Can you give links to any previous forum posts or PS Set problem numbers(if you remember them) that are of the above type. Quite surprisingly I haven't come across questions of these types so far.An example for each of the type would be great and help us understand better.

Sure, will be done shortly

SIMPLE INTEREST
Basic Facts: The sum that a person lends is called Principle ( P ) and what he gets back is Amount.
Amount = P +Interest

1. Interest = p*Rate*Time / 100
2. Amount = P+ Interest = p[1+(R*T/100)]

pahwa wrote:SIMPLE INTEREST
Basic Facts: The sum that a person lends is called Principle ( P ) and what he gets back is Amount.
Amount = P +Interest

1. Interest = p*Rate*Time / 100
2. Amount = P+ Interest = p[1+(R*T/100)]

Phawa is the above a typo.
Isn't Amount at Compound Interest = P + P*[(1+r/100)]^T

I am talking about SIMPLE INETEREST.

To make it more clear,
Amount = P +Interest
= P + (P*R*T/100)
= P[ 1+ (R*T/100)] taking P common.

Compund Interest will be my next post.