Carry It Over I (Part 1)

by on September 11th, 2010

One of the ways that the GMAT makes quantitative questions more challenging is by forcing you to carry over information from one type of environment to another.  This is most commonly seen in geometry questions, where the side of a triangle might also be a chord on a circle inscribed within a square, for example; because the test is going after your care and attention as much (or more) than it’s going after your math knowledge, carrying information over is a key skill.  This “Carry It Over” series will challenge you to transfer information from one type of problem to another, within the same problem.  Let’s start with this one; I’ll provide answer and explanation in the next installment:

Raj is working on a set of Data Sufficiency problems for his December GMAT: a geometry problem, an algebra problem, and a data interpretation problem. He has determined that statement 1 of the geometry problem is insufficient on its own, that both statement 1 and 2 of the algebra problem are insufficient on their own, and that statement 2 of the data interpretation problem is insufficient on its own. If the probabilities are expressed as percents, approximately how much greater is the probability that all three answers are “C” after Raj figures out that statement 1 of the data interpretation problem is also insufficient on its own?

A) 2.3%
B) 2.8%
C) 3.3%
D) 5.6%
E) 8.3%

So what do you think?  Provide your answers in the comments below and I’ll provide the full explanation in my next article.  Stay tuned!

• Geometry : Statement 1 is insufficient => possible answers are BCE
P(G_C) = 1/3 = probability of C for geometry.

Algebra : both are insufficient on their own => Possible answers are C,E
P(A_C) = 1/2 =probability of C for algebra.

DI :
CASE 1: Statement 1 is insufficient => Possible answers are BCE
P(D_C_1) = 1/3 = probability of C for DI in case 1
CASE 1: Statement 1 is insufficient => Possible answers are CE
P(D_C_2) = 1/2 = probability of C for DI in case 2

Total probability of Case 1 = P(G_C)*P(A_C)*P(D_C_1)
[ Independent events]
= 1/3 * 1/2 * 1/3 = 1/8

Total probability of Case 2 = P(G_C)*P(A_C)*P(D_C_2)
[ Independent events]
= 1/3 * 1/2 * 1/2 = 1/12

Increase in probability = 1/12 - 1/18 = (3-2)/36 = 1/36

in %age ,1/36 = 100/36 % = 25/9 % = approx 2.8

• Geometry : Statement 1 is insufficient => possible answers are BCE
P(G_C) = 1/3 = probability of C for geometry.

Algebra : both are insufficient on their own => Possible answers are C,E
P(A_C) = 1/2 =probability of C for algebra.

DI :
CASE 1: Statement 1 is insufficient => Possible answers are BCE
P(D_C_1) = 1/3 = probability of C for DI in case 1

CASE 1: Statement 1 is insufficient => Possible answers are CE
P(D_C_2) = 1/2 = probability of C for DI in case 2

Total probability of Case 1 = P(G_C)*P(A_C)*P(D_C_1)
[ Independent events]
= 1/3 * 1/2 * 1/3 = 1/8

Total probability of Case 2 = P(G_C)*P(A_C)*P(D_C_2)
[ Independent events]
= 1/3 * 1/2 * 1/2 = 1/12

Increase in probability = 1/12 - 1/18 = (3-2)/36 = 1/36

in %age ,1/36 = 100/36 % = 25/9 % = approx 2.8

• Before:
1/3 (Answer C out of B,C,E in geometry)
x
1/2 (Answer C out of C,E in algebra)
x
1/3 (Answer C out of A,C,E in Data interp.)
is
Equals: ( 1/3 x 1/2 x 1/3 ) = 1/18 = 8.3%

After
1/2 (Answer C out of C,E in Geometry)
x
1/2 (Answer C out of C,E in Algebra)
x
1/3 (Answer C out of A,C,E in Data Interp)
is
Equals: ( 1/2 x 1/2 x 1/3 ) = 1/12 = 5.5%

8.3-5.5= *2.8%*