GMAT Challenge Series (700+): Q28

Chalkboard-Stats.jpg

Question:

In a set of 5 numbers, the largest number is 4 greater than the median. Is the mean greater than the median?

(1) The largest number plus the median is 34.
(2) The median minus the smallest number is 10.

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. 
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

MBA Wisdom's Answer:

$$\text{Let the 5 numbers be: } \textit{a} \leq \textit{b} \leq \textit{c} \leq \textit{d} \leq \textit{e}$$ $$\text{Median} = \textit{c}$$ $$\text{Mean} = {{\textit{a} + \textit{b} + \textit{c} + \textit{d} + \textit{e}} \over {5}} $$ $$$$ $$\text{We know that: } \textit{e} = {\textit{c} + {4}}$$ $$\text{Mean} = {{\textit{a} + \textit{b} + \textit{c} + \textit{d} + \left(\textit{c} + {4}\right)} \over {5}} $$ $$\text{Mean} = {{\textit{a} + \textit{b} + {2}\textit{c} + \textit{d} + {4}} \over {5}} $$ $$$$ $$\text{We need to work out if:}$$ $${{\textit{a} + \textit{b} + {2}\textit{c} + \textit{d} + {4}} \over {5}} > \textit{c}$$ $${\textit{a} + \textit{b} + {2}\textit{c} + \textit{d} + {4}} > {5}\textit{c}$$ $${\textit{a} + \textit{b} + \textit{d} + {4}} > {3}\textit{c}$$ $$$$ $$\text{Statement 1:}$$ $$\textit{e} = {\textit{c} + {4}}$$ $$\textit{e} + \textit{c} = {34}$$ $$\textit{c} = {15} \text{ and }\textit{e} = {19}$$ $$$$ $$\text{We need to work out if:}$$ $${\textit{a} + \textit{b} + \textit{d} + {4}} > {3}\textit{c}$$ $${\textit{a} + \textit{b} + \textit{d} + {4}} > {45}$$ $${\textit{a} + \textit{b} + \textit{d}} > {41}$$ $$$$ $$\text{Test at maximum values: } \textit{a} = {15} \textit{ b} = {15} \textit{ d} = {19}$$ $${{15} + {15} + {19}} > {41}$$ $${49} > {41} \text{ TRUE}$$ $$$$ $$\text{Test at non-maximum values: } \textit{a} = {0} \textit{ b} = {0} \textit{ d} = {15}$$ $${{0} + {0} + {15}} > {41}$$ $${15} > {41} \text{ FALSE}$$ $$$$ $$\text{We cannot determine whether the mean is greater than the median}$$ $$\text{Statement 1 is NOT SUFFICIENT}$$ $$$$ $$\text{Statement 2:}$$ $$\textit{a} = {\textit{c} - {10}}$$ $$$$ $$\text{We need to work out if:}$$ $${\textit{a} + \textit{b} + \textit{d} + {4}} > {3}\textit{c}$$ $${\left(\textit{c} - {10}\right) + \textit{b} + \textit{d} + {4}} > {3}\textit{c}$$ $${\textit{b} + \textit{d}} > {{2}\textit{c} + {6}}$$ $$$$ $$\text{Test at maximum values: } \textit{b} = \textit{c} \text{ and}\textit{ d} = \textit{e} = \textit{c} + {4}$$ $${\textit{c} + \left(\textit{c} + {4}\right)} > {{2}\textit{c} + {6}}$$ $${{2}\textit{c} + {4}} > {{2}\textit{c} + {6}} \text{ FALSE}$$ $$$$ $$\text{LHS will never be greater than RHS}$$ $$\text{Therefore, the mean is NOT greater than the median}$$ $$\text{Statement 2 is SUFFICIENT}$$

(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. 

GMAT Challenge Series (700+): Q27

TWINS.jpg

Question:

Abe, Beth, Carl and Duncan are four siblings, among which Abe and Carl are twins and Beth and Duncan are also twins. When the present ages of the four siblings are multiplied, the product is 900. If Beth is older than Abe, what is the age of Duncan? Assume the ages of all siblings to be integers.

(1) The difference between Beth’s age and Abe’s age is a prime number.
(2) If Carl had been born four years earlier, the difference between Duncan’s age and Carl’s age would have been a prime number.

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. 
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

MBA Wisdom's Answer:

$$\text{Let the age of Abe and Carl be: } \textit{x}$$ $$\text{Let the age of Beth and Duncan be: } \textit{y}$$ $$\text{We know that:}$$ $$\textit{y} > \textit{x}$$ $$\textit{x}^2 \textit{y}^2 = {900}$$ $$\textit{xy} = {30}$$ $$\text{Therefore, possible values for (} \textit{x} \text{, } \textit{y} \text{) are:}$$ $$\text{(1, 30)}$$ $$\text{(2, 15)}$$ $$\text{(3, 10)}$$ $$\text{(5, 6)}$$ $$\text{To answer the question we need to know which combination is correct}$$ $$$$ $$\text{Statement 1:}$$ $$\text{We are told that } \left| \textit{y} - \textit{x} \right| \text{ is a prime number}$$ $$\text{(1, 30): } \left| \textit{y} - \textit{x} \right| = {29} \text{ (PRIME)}$$ $$\text{(2, 15): } \left| \textit{y} - \textit{x} \right| = {13} \text{ (PRIME)}$$ $$\text{(3, 10): } \left| \textit{y} - \textit{x} \right| = {7} \text{ (PRIME)}$$ $$\text{(5, 6): } \left| \textit{y} - \textit{x} \right| = {1} \text{ (NOT PRIME)}$$ $$$$ $$\text{There are 3 different values for Duncan's age}$$ $$\text{Statement 1 is NOT SUFFICIENT}$$ $$$$ $$\text{Statement 2:}$$ $$\text{We are told that } \left| {\textit{y} - \left( {\textit{x} + {4}} \right)} \right| \text{ is a prime number}$$ $$\text{(1, 30): } \left| {\textit{y} - \left( {\textit{x} + {4}} \right)} \right|= {25} \text{ (NOT PRIME)}$$ $$\text{(2, 15): } \left| {\textit{y} - \left( {\textit{x} + {4}} \right)} \right|= {9} \text{ (NOT PRIME)}$$ $$\text{(3, 10): } \left| {\textit{y} - \left( {\textit{x} + {4}} \right)} \right|= {3} \text{ (PRIME)}$$ $$\text{(5, 6): } \left| {\textit{y} - \left( {\textit{x} + {4}} \right)} \right| = {3} \text{ (PRIME)}$$ $$$$ $$\text{There are 2 different values for Duncan's age}$$ $$\text{Statement 2 is NOT SUFFICIENT}$$ $$$$ $$\text{Statements 1 & 2 together:}$$ $$\text{(3, 10) is the only combination which holds true for statements 1 and 2}$$ $$\text{Statements 1 & 2 together are SUFFICIENT}$$

(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

GMAT Challenge Series (700+): Q26

Cars.jpg

Question:

In a group of 80 college students, how many own a car? 

(1) Of the students who do not own a car, 14 are male. 
(2) Of the students who own a car, 42% are female.

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. 
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

MBA Wisdom's Answer:

$$\text{Statement 1:}$$ $$\text{Clearly not sufficent as we do not know how many female students do not own a car}$$ $$\text{Statement 1 is NOT SUFFICIENT}$$ $$$$ $$\text{Statement 2:}$$ $$\text{Let the number of students who own a car be: } \textit{x}$$ $$\text{Female students who own a car} = {0.42}\textit{x} = \frac{21}{50} \textit{x} \text{ (in lowest terms)}$$ $$\text{We know that } \frac{21}{50} \textit{x} \text{ is an integer}$$ $$\text{Therefore, } \textit{x} \text{ must be a multiple of 50: 50, 100, 150...}$$ $$\text{As } \textit{x} \leq {80} \textit{, x} \text{ must equal 50}$$ $$\text{Statement 2 is SUFFICIENT}$$

(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. 

GMAT Challenge Series (700+): Q25

Cleaner.jpg

Question:

Three employees, A, B and C, clean a certain conference room each day. Working together, A and B can clean the conference room in 3 hours, whereas A and C together can do it in 2 and 1/2 hours. Can A, B and C working together clean the conference room in less than 2 hours?

1) B cleans faster than A.
2) Working alone, C can clean the conference room in less than 5 hours.

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. 
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

MBA Wisdom's Answer:

$$\textit{rate} = {\textit{work} \over \textit{time}}$$ $$$$ $$\text{We know that:}$$ $$\text{A} + \text{B} = \frac{1}{3}$$ $$\text{A} + \text{C} = \frac{2}{5}$$ $$$$ $$\text{We need to figure out if:}$$ $$\text{A} + \text{B} + \text{C} > \frac{1}{2}$$ $$\text{or}$$ $$\text{B} > \frac{1}{10}$$ $$\text{or}$$ $$\text{C} > \frac{1}{6}$$ $$$$ $$\text{Statement 1:}$$ $$\text{If:}$$ $$\text{A} + \text{B} = \frac{1}{3}$$ $$\text{and}$$ $$\text{B} > \text{A}$$ $$\text{then:}$$ $$\text{B} > \frac{1}{6} > \frac{1}{10}$$ $$\text{Statement 1 is SUFFICIENT}$$ $$$$ $$\text{Statement 2:}$$ $$\text{C} > \frac{1}{5} > \frac{1}{6}$$ $$\text{Statement 2 is SUFFICIENT}$$

(D) Each statement ALONE is sufficient.

GMAT Challenge Series (700+): Q24

Salesman.jpg

Question:

A certain salesman's yearly income is determined by a base salary plus a commission on the sales he makes during the year. Did the salesman's base salary account for more than half of the salesman's yearly income last year?

(1) If the amount of the commission had been 30 percent higher, the salesman's income would have been 10 percent higher last year.
(2) The difference between the amount of the salesman's base salary and the amount of the commission was equal to 50 percent of the salesman's base salary last year.

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. 
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

MBA Wisdom's Answer:

$$\text{Income} = \text{Salary} + \text{Commission}$$ $$\text{The question is effectively asking if: } \text{Salary} > \text{Commission?}$$ $$$$ $$\text{Statement 1:}$$ $${1.1}{\left( {\text{Salary} + \text{Commission}} \right)} = \text{Salary} + {1.3}{\left( \text{Commission} \right)}$$ $${0.1}\left( \text{Salary} \right) = {0.2}\left( \text{Commission} \right)$$ $$\text{Salary} = {2}\left( \text{Commission} \right)$$ $$\text{Therfore, } \text{Salary} > \text{Commission}$$ $$\text{Statement 1 is SUFFICIENT}$$ $$$$ $$\text{Statement 2:}$$ $$\left|\text{Salary} - \text{Commission} \right| = {0.5}\left( \text{Salary} \right)$$ $$$$ $$\text{Case 1: } {\text{Salary} - \text{Commission}} > {0}$$ $$\text{Salary} - \text{Commission} = {0.5}\left( \text{Salary} \right)$$ $${0.5}\left( \text{Salary} \right) = \text{Commission}$$ $$\text{Salary} = {2}\left( \text{Commission} \right)$$ $$\text{In this case, } \text{Salary} > \text{Commission}$$ $$$$ $$\text{Case 2: } \text{Salary} - \text{Commission} < {0}$$ $$\text{Commission} - \text{Salary} = {0.5}\left( \text{Salary} \right)$$ $$\text{Commission} = {1.5}\left( \text{Salary} \right)$$ $$\text{In this case, } \text{Salary} < \text{Commission}$$ $$$$ $$\text{Different answers in each case}$$ $$\text{Statement 2 is NOT SUFFICIENT}$$

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

GMAT Challenge Series (700+): Q23

ABC Blocks.jpg

Question:

If a, b, and c are integers such that 0 < a < b < c < 10, is the product abc divisible by 3?

(1) If a/1000 + b/100 + c/10 is expressed as a single fraction reduced to lowest terms, the denominator is 200.
(2) c – b < b – a

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. 
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

MBA Wisdom's Answer:

$$\textit{abc} \text{ will be divisible by 3 if } \textit{a} \text{, } \textit{b} \text{, or} \textit{c} \text{ equals 3, 6, or 9}$$ $$$$ $$\text{Statement 1:}$$ $${\textit{a} \over {1000}} + {\textit{b} \over {100}} + {\textit{c} \over {10}} = {{\textit{a} + {10}\textit{b} + {100}\textit{c}} \over {1000}} = {{{\textit{a} + {10}\textit{b} + {100}\textit{c}} \over {5}} \over {200}}$$ $$\textit{a} + {10}\textit{b} + {100}\textit{c} \text{ must be divisible by 5}$$ $$\text{This implies that } \textit{a} \text{ must be divisible by 5}$$ $$\text{Since } {0} < \textit{a} < {10}, \textit{a} = {5}$$ $$\textit{abc} \text{ wont be divisible by 3 if, and only if, } \textit{b} \text{ and } \textit{c} \text{ are 7 and 8 respectively}$$ $$\text{In this case:}$$ $${{\textit{a} + {10}\textit{b} + {100}\textit{c}} \over {1000}} = \frac{875}{1000} = \frac{7}{8}$$ $$\text{Reduced to the lowest terms the denominator is 8 and not 200}$$ $$\text{Therfore, } \textit{b} \text{ must be 6 or } \textit{c} \text{ must be 9}$$ $$\text{Statement 1 is SUFFICIENT}$$ $$$$ $$\text{Statement 2:}$$ $$\textit{c} - \textit{b} < \textit{b} - \textit{a}$$ $$\textit{a} + \textit{c} < {2}\textit{b}$$ $$\text{Expression valid for: } \textit{a} = {1} \textit{ b} = {5} \textit{ c} = {7} \text{ where} \textit{ abc} \text{ is not divisible by 3}$$ $$\text{Expression valid for: } \textit{a} = {1} \textit{ b} = {6} \textit{ c} = {7} \text{ where} \textit{ abc} \text{ is divisible by 3}$$ $$\text{Statement 2 is NOT SUFFICIENT}$$

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

GMAT Challenge Series (700+): Q22

Dinner Party.png

Question:

At a restaurant, a group of friends ordered four main dishes and three side dishes at a total cost of $89. The prices of the seven items, in dollars, were all different integers, and every main dish cost more than every side dish. What was the price, in dollars, of the most expensive side dish?

(1) The most expensive main dish cost $16.
(2) The least expensive side dish cost $9.

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. 
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

MBA Wisdom's Answer:

$$\text{Let the price of each main dish be: MD1, MD2, MD3 and MD4 (most expensive to least expesive)}$$ $$\text{Let the price of each side dish be: SD1, SD2, and SD3 (most expensive to least expesive)}$$ $$\text{Let the bill be: (MD1, MD2, MD3, MD4, SD1, SD2, SD3)}$$ $$$$ $$\text{Statement 1:}$$ $$\text{Most expensive bill: (16, 15, 14, 13, 12, 11, 10), Total cost: 91}$$ $$\text{We need to reduce the total cost by 2}$$ $$\text{Options:}$$ $$\text{Reduce SD3 by 2: (16, 15, 14, 13, 12, 11, 8), Total cost: 89}$$ $$\text{Reduce SD2 and SD3 by 1: (16, 15, 14, 13, 12, 10, 9), Total cost: 89}$$ $$\text{We cannot reduce SD1 as the total cost will be < 89}$$ $$\text{SD1 can only be 12}$$ $$\text{Statement 1 is SUFFICIENT}$$ $$$$ $$\text{Statement 2:}$$ $$\text{Least expensive bill: (15, 14, 13, 12, 11, 10, 9), Total cost: 84}$$ $$\text{We need to increase the total cost by 5}$$ $$\text{Options:}$$ $$\text{Increase MD1 by 5: (20, 14, 13, 12, 11, 10, 9), Total cost: 89}$$ $$\text{Increase MD1-4 and SD1 by 1: (16, 15, 14, 13, 12, 10, 9), Total cost: 89}$$ $$\text{SD1 can take on different values}$$ $$\text{Statement 2 is NOT SUFFICIENT}$$

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

GMAT Challenge Series (700+): Q21

Amtrak.jpeg

Question:

Train A leaves New York for Boston at 3 PM and travels at the constant speed of 100 mph. An hour later, it passes Train B, which is making the trip from Boston to New York at a constant speed. If Train B left Boston at 3:50 PM and if the combined travel time of the two trains is 2 hours, what time did Train B arrive in New York? 

(1) Train B arrived in New York before Train A arrived in Boston.
(2) The distance between New York and Boston is greater than 140 miles. 

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. 
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

 MBA Wisdom's Answer:

$$\text{Train A:}$$ $$\text{At 4:00 PM Train A has traveled 100 miles}$$ $$\text{Let the time that it takes Train A to get to Boston be: } \textit{A} \text{ hours}$$ $$$$ $$\text{Train B:}$$ $$\text{Let the distance that Train B has traveled from 3:50 PM to 4:00 PM be: } \textit{x} \text{ miles}$$ $$\text{Given this, the speed of Train B is: } {6}\textit{x} \text{ mph}$$ $$\text{Let the time that it takes Train B to get to NY be: } \textit{B} \text{ hours}$$ $$$$ $$\text{Situation at 4:00 PM:}$$ $$\text{NY} \text{ ------------- 100 miles ------------- A > < B ------- x miles ------- }\text{Boston}$$ $$\text{Total distance between NY and Boston is: } {100} + \textit{x} \text{ miles}$$ $$$$ $$\textit{time} = {\textit{distance} \over \textit{speed}}$$ $$\textit{A} = {{{100} + \textit{x}} \over {100}}$$ $$\textit{B} = {{{100} + \textit{x}} \over {6}\textit{x}}$$ $$\text{To work out when Train B arrived in NY we need to know } \textit{x}$$ $$$$ $$\textit{A} + \textit{B} = {2}$$ $${{{100} + \textit{x}} \over {100}} + {{{100} + \textit{x}} \over {6}\textit{x}} = {2}$$ $${600}\textit{x} + {6}\textit{x}^2 + {10,000} + {100}\textit{x} = {1,200}\textit{x}$$ $${6}\textit{x}^2 - {500}\textit{x} + {10,000} = {0}$$ $${3}\textit{x}^2 - {250}\textit{x} + {5,000} = {0}$$ $$\left( \textit{x} - {50} \right) \left( {3}\textit{x} - {100} \right)$$ $$\textit{x} = {50} \textit{ or } {33}\frac{1}{3}$$ $$$$ $$\text{To work out when Train B arrived in NY we need to know if } \textit{x} \text{ is } {50} \text{ or } {33}\frac{1}{3}$$ $$$$ $$\text{Statement 1:}$$ $$\text{If } \textit{x} \text{ is } {50} \text{: } \textit{A} = {1.5} \text{, }\textit{B} = {0.5} \text{, } \textit{A - B} = {1}$$ $$\text{If } \textit{x} \text{ is } {33}\frac{1}{3} \text{: } \textit{A} = {1}\frac{1}{3} \text{, }\textit{B} = \frac{2}{3} \text{, } \textit{A - B} = \frac{2}{3}$$ $$\text{If Train B arrived before Train A, } \textit{A - B} \text{ must be } > \frac{5}{6}$$ $$\text{Thus, } \textit{x} \text{ must be } {50}$$ $$\text{Statement 1 is SUFFICIENT}$$ $$$$ $$\text{Statement 2:}$$ $${100} + \textit{x} > {140}$$ $$\textit{x} > {40}$$ $$\text{Thus, } \textit{x} \text{ must be } {50}$$ $$\text{Statement 2 is SUFFICIENT}$$ $$$$

(D) Each statement ALONE is sufficient.

GMAT Challenge Series (700+): Q20

Cyclist.jpg

Question:

A cyclist travels the length of a bike path that is 225 miles long, rounded to the nearest mile. If the trip took him 5 hrs, rounded to the nearest hour, then his average speed must be between:

A. 38 and 50 mph

B. 40 and 50 mph

C. 40 and 51 mph

D. 41 and 50 mph

E. 41 and 51 mph

MBA Wisdom's Answer:

$$\text{The length of the bike path is 225 miles, rounded to the nearest mile, therefore:}$$ $${224.5} \geq \textit{distance} < {225.5}$$ $$\text{The trip took 5 hrs, rounded to to the nearest hour, therefore:}$$ $${4.5} \geq \textit{time} < {5.5}$$ $$$$ $$\text{Lowest average speed occurs with the lowest distance and the highest time}$$ $$\textit{Lowest average speed} \approx {{224.5} \over {5.5}} \approx {40.8}$$ $$$$ $$\text{Highest average speed occurs with the highest distance and the lowest time}$$ $$\textit{Highest average speed} \approx {{225.5} \over {4.5}} \approx {50.1}$$ $$$$ $${40.1} < \textit{average speed} < {50.1}$$ $${40} < {40.1} < \textit{average speed} < {50.1} < {51}$$ $$$$ $$\text{Range: } {40} < \textit{average speed} < {51}$$

C. 40 and 51 mph

GMAT Challenge Series (700+): Q19

Loan.jpg

Question:

Louie takes out a three-month loan of $1000. The lender charges him 10% interest per month compounded monthly. The terms of the loan state that Louie must repay the loan in three equal monthly payments. To the nearest dollar, how much does Louie have to pay each month?

A. $333

B. $383

C. $402

D. $433

E. $483

MBA Wisdom's Answer:

$$\text{Let the monthly repayment be: } \textit{x}$$ $$$$ $$\text{Month 0:}$$ $$\textit{Unpaid Balance} = {1,000}$$ $$$$ $$\text{Month 1:}$$ $$\textit{Unpaid Balance} = \left({{1} + \textit{Interest Rate}}\right) \left(\textit{Unpaid Balance Month 0}\right) - \textit{Monthly Repayment}$$ $$\textit{Unpaid Balance} = \left({1.1}\right) \left({1,000}\right) - \textit{x}$$ $$\textit{Unpaid Balance} = {1,100} - \textit{x}$$ $$$$ $$\text{Month 2:}$$ $$\textit{Unpaid Balance} = \left({{1} + \textit{Interest Rate}}\right) \left(\textit{Unpaid Balance Month 1}\right) - \textit{Monthly Repayment}$$ $$\textit{Unpaid Balance} = \left({1.1}\right) \left({{1,100} - \textit{x}}\right) - \textit{x}$$ $$\textit{Unpaid Balance} = {1,210} - {2.1}\textit{x}$$ $$$$ $$\text{Month 3:}$$ $$\textit{Unpaid Balance} = \left({{1} + \textit{Interest Rate}}\right) \left(\textit{Unpaid Balance Month 2}\right) - \textit{Monthly Repayment}$$ $$\textit{Unpaid Balance} = \left({1.1}\right) \left({{1,210} - {2.1}\textit{x}}\right) - \textit{x}$$ $$\textit{Unpaid Balance} = {1,331} - {3.31}\textit{x}$$ $$$$ $$\text{At the end of Month 3 the loan is repaid, therefore:}$$ $${0} = {1,331} - {3.31}\textit{x}$$ $$\textit{x} \approx {402}$$

C. $402

GMAT Challenge Series (700+): Q18

Disease Test.jpg

Question:

A terrible disease sweeps around the world, luckily only affecting 1 in 10,000, but for that one, the disease is lethal. Shortly after the disease was discovered, scientists developed a test that is 99% accurate regardless of whether you have the disease. In other words, the test yields the correct positive or correct negative result 99% of the time. You take the test and a week later, you receive the lab report. The outcome of the test is positive. What is the probability you have the disease?

A. 99/1,000,000

B. 1/102

C. 1/100

D. 10,099/1,000,000

E. 99/100

MBA Wisdom's Answer:

$$\text{Probability(Disease)} = {{1} \over {10,000}}$$ $$\text{Probability(Correct Positive)} = {{99} \over {100}}$$ $$\text{Probability(Incorrect Positive)} = {{1} \over {100}}$$ $$$$ $$\text{Probability(Disease & Correct Positive)} = \left({{1} \over {10,000}}\right) \left({{99} \over {100}}\right) = {{99} \over {1,000,000}}$$ $$\text{Probability(No Disease & Incorrect Positive)} = \left({{9,999} \over {10,000}}\right) \left({{1} \over {100}}\right) = {{9,999} \over {1,000,000}}$$ $$$$ $$\text{Probability(Positive)} = \text{Probability(Disease & Correct Positive)} + \text{Probability(No Disease & Incorrect Positive)}$$ $$\text{Probability(Positive)} = {{{99} \over {1,000,000}} + {{9,999} \over {1,000,000}}} = {{10,098} \over {1,000,000}}$$ $$$$ $$\text{Probability(Disease if Postive)} = {\text{Probability(Disease & Correct Positive)} \over \text{Probability(Positive)}}$$ $$\text{Probability(Disease if Postive)} = {{{99} \over {1,000,000}} \over {{10,098} \over {1,000,000}}} = {{99} \over {10,098}} = {{1} \over {102}}$$ $$$$

B. 1/102

GMAT Challenge Series (700+): Q17

Cloud Sun.jpg

Question:

On a partly cloudy day, Derek decides to walk back from work. When it is sunny, he walks at a speed of s miles/hr (s is an integer) and when it gets cloudy, he increases his speed to (s + 1) miles/hr. If his average speed for the entire distance is 2.8 miles/hr, what fraction of the total distance did he cover while the sun was shining on him?

A. 1/4

B. 4/5

C. 1/5

D. 1/6

E. 1/7

MBA Wisdom's Answer:

$$\textit{sunny speed} = \textit{s}$$ $$\textit{cloudy speed} = \textit{s} + {1}$$ $$\textit{average speed} = {2.8} \textit{ miles/hr}$$ $$$$ $$\text{The average speed must be between } \textit{s} \text{ and } \textit{s} + {1}$$ $$\text{As } \textit{s} \text{ is an integer and the average speed is 2.8 } \textit{s} \text{ must equal 2}$$ $$\text{This implies that the ratio of time sunny:cloudy is 1:4}$$ $$$$ $$\text{Let the sunny time be: } \textit{t}$$ $$\text{Then the cloudy time will be: } {4}\textit{t}$$ $$$$ $$\textit{distance} = \left(\textit{speed}\right) \left(\textit{time}\right)$$ $$\textit{sunny distance} = {{2} \textit{t}}$$ $$\textit{cloudy distance} = {\left({3}\right) \left({4}\textit{t}\right)} = {{12} \textit{t}}$$ $$\textit{total distance} = \textit{sunny distance} + \textit{cloudy distance}$$ $$\textit{total distance} = {{{2} \textit{t}} + {{12} \textit{t}}} = {{14} \textit{t}}$$ $$$$ $${\textit{sunny distance} \over \textit{total distance}} = {{{2} \textit{t}} \over {{14} \textit{t}}} = {{1} \over {7}}$$

E. 1/7

GMAT Challenge Series (700+): Q16

Runner.jpg

Question:

Brenda and Sally run in opposite direction on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?
 

A. 250

B. 300

C. 350

D. 400

E. 500

MBA Wisdom's Answer:

$$\text{Meeting points:}$$ $$\textit{1st time: Brenda and Sally, together, run half of the circumference}$$ $$\textit{2nd time: Brenda and Sally, together, run the full circumference}$$ $$$$ $$\text{Brenda:}$$ $$\textit{1st time: Brenda runs 100m}$$ $$\textit{2nd time: Brenda runs 200m (twice as far as the 1st time)}$$ $$$$ $$\text{Sally:}$$ $$\textit{2nd time: Sally runs 150m}$$ $$$$ $$\textit{Length of the track} = {200} + {150} = {350}$$

C. 350

GMAT Challenge Series (700+): Q15

 This question is dedicated to my friend's venture Otto's Burger:  www.ottosburger.com  - congrats Dan!

This question is dedicated to my friend's venture Otto's Burger: www.ottosburger.com - congrats Dan!

Question:

3 cooks have to make 80 burgers. Working together they can make 20 burgers every minute. The 1st cook began working alone and made 20 burgers having worked for sometime more than 3 minutes. The remaining part of the work was done by 2nd and 3rd cook working together. Working in this fashion, it took the 3 cooks a total of 8 minutes to make the 80 burgers. How many minutes would it take the 1st cook alone to cook 160 burgers?

A. 16 minutes

B. 24 minutes

C. 32 minutes

D. 40 minutes

E. 30 minutes

MBA Wisdom's Answer:

$$\text{Let the work rate per minute of the 1st cook be: } \textit{A}$$ $$\text{Let the work rate per minute of the 2nd cook be: } \textit{B}$$ $$\text{Let the work rate per minute of the 3rd cook be: } \textit{C}$$ $$$$ $$\textit{work} = \left(\textit{rate}\right) \left(\textit{time}\right)$$ $$\text{Working together the cooks can make 20 burgers in a minute, therefore:}$$ $$\textit{A} + \textit{B} + \textit{C} = {20}$$ $$$$ $$\text{Let the time that the 1st cook works be: } \textit{t} \text{ where } \textit{ t} > {3}$$ $$\text{The time that the 2nd and 3rd cook work together is: } {8} - \textit{t}$$ $$$$ $$\textit{work of 1st cook} = \left(\textit{A}\right) \left(\textit{t}\right) = {20}$$ $$\textit{work of 2nd cook} = \left(\textit{B}\right) \left({8} - \textit{t}\right)$$ $$\textit{work of 3rd cook} = \left(\textit{C}\right) \left({8} - \textit{t}\right)$$ $$$$ $$\textit{work of 1st cook} + \textit{work of 2nd cook} + \textit{work of 3rd cook} = {80}$$ $$\left(\textit{A}\right) \left(\textit{t}\right) + \left(\textit{B}\right) \left({8} - \textit{t}\right) + \left(\textit{C}\right) \left({8} - \textit{t}\right) = {80}$$ $${4}\textit{A} + {4}\textit{B} + {4}\textit{C} = {80}$$ $$\left(\textit{A}\right) \left(\textit{t}\right) + \left(\textit{B}\right) \left({8} - \textit{t}\right) + \left(\textit{C}\right) \left({8} - \textit{t}\right) = {4}\textit{A} + {4}\textit{B} + {4}\textit{C}$$ $$\textit{t} \text{ must equal 4}$$ $$$$ $$\text{As a result, we know that the 1st cook can make 20 burgers in 4 minutes}$$ $$\text{Multiply this by 8 and we get 160 burgers in 32 minutes}$$

C. 32 minutes

GMAT Challenge Series (700+): Q14

Mobile Phone.jpg

Question:

A customer using a certain telephone calling plan pays a fee of $25 per month, and then receives a discount of 40% on the regular charge for all calls made to country A. If calls to country A are regularly charged at $1.60 per minute for the first 3 minutes, and $0.80 per minute for each minute thereafter, what is the maximum the customer could have saved over regular prices if he was charged for 1 hour of calls made to country A in a certain month?

A. $8.75

B. $12.00

C. $13.40

D. $17.40

E. $24.40

MBA Wisdom's Answer:

$$\text{The maximum saving occurs when the bill is the highest}$$ $$\text{The bill is the highest when every minute is charged at \$1.60}$$ $$$$ $$\textit{max cost without plan} = \left(\textit{minutes}\right) \left(\textit{rate}\right)$$ $$\textit{max cost without plan} = {\left({60}\right) \left({1.6}\right)} = {96}$$ $$$$ $$\textit{max cost with plan} = \textit{plan fee} + \left(\textit{minutes}\right) \left(\textit{rate}\right) \left(\textit{1 - discount}\right)$$ $$\textit{max cost with plan} = {25} + {\left({60}\right) \left({1.6}\right) \left({0.6}\right)} = {25} + {57.6} = {82.6}$$ $$$$ $$\textit{max saving} = \textit{max cost without plan} - \textit{max cost with plan}$$ $$\textit{max saving} = {96} - {82.6}$$ $$\textit{max saving} = {13.4}$$

C. $13.40

GMAT Challenge Series (700+): Q13

Bus.jpg

Question:

A bus from city M is traveling to city N at a constant speed while another bus is making the same journey in the opposite direction at the same constant speed. They meet in point P after driving for 2 hours. The following day the buses do the return trip at the same constant speed. One bus is delayed 24 minutes and the other leaves 36 minutes earlier. If they meet 24 miles from point P, what is the distance between the two cities?

A. 48

B. 72

C. 96

D. 120

E. 192

MBA Wisdom's Answer:

$$\text{Let x equal the speed of the buses}$$ $$\text{Let 2P equal distance between M and N, with P being the midpoint}$$ $$$$ $$\text{Day 1:}$$ $$\text{M ---------------------------> P <--------------------------- N}$$ $$\text{Day 2:}$$ $$\text{M ---------> P - 24 | P + 24 <------------------------------- N}$$ $$$$ $$\textit{speed} = {\textit{distance} \over \textit{time}}$$ $$\text{Late bus:}$$ $${x} = {\text{P - 24} \over \text{1.6}}$$ $$\text{Early bus:}$$ $${x} = {\text{P + 24} \over \text{2.6}}$$ $$\text{Solve:}$$ $${\text{P - 24} \over \text{1.6}} = {\text{P + 24} \over \text{2.6}}$$ $$\text{2.6P} - {2.6}\left({24}\right) = \text{1.6P} + {1.6}\left({24}\right)$$ $$\text{P} = {4}\left({24}\right)$$ $$\text{P} = {96}$$ $$\text{2P} = {192}$$

E. 192

GMAT Challenge Series (700+): Q12

Envelopes.png

Question:

In a certain mathematical activity, we have five cards with five different prime numbers on them. We will distribute these five cards among three envelope: all could go in any envelope, or they could be broken up in any way among the envelopes. Then in each envelop, we find the product of all the cards in that envelope: that is the “number” of the envelope. An envelope containing no cards has the number 1. We then put the three envelope numbers in order, from lowest to highest, and that is our set. How many different sets can be produced by this process?

A. 41

B. 89

C. 125

D. 243

E. 512

MBA Wisdom's Answer:

$$\text{Let's consider the different cases:}$$ $$\text{Case 1: 5-0-0}$$ $$\text{Possibilities} = \text{5C5} = {1}$$ $$\text{Case 2: 4-1-0}$$ $$\text{Possibilities} = \text{5C4} = {5}$$ $$\text{Case 3: 3-2-0}$$ $$\text{Possibilities} = \text{5C3} = {10}$$ $$\text{Case 4: 3-1-1}$$ $$\text{Possibilities} = \text{5C3} = {10}$$ $$\text{Case 5: 2-2-1}$$ $$\text{Possibilities} = {{\text{5C2} \times \text{3C2}} \over {2!}} = {{\left({10}\right) \left({3}\right)} \over {2}} = {15}$$ $$$$ $$\text{Total numbet of sets:} = {1} + {5} + {10} + {10} + {15} = {41}$$

A. 41

GMAT Challenge Series (700+): Q11

Vodka Tonic.jpg

Question:

Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching a profit of 10% and 20% respectively. If the vodkas are mixed in and equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit of

A. 18%

B. 20%

C. 21%

D. 23%

E. Cannot be determined

MBA Wisdom's Answer:

$$\text{Let the profit on vodka 1 = x%}$$ $$\text{Let the profit on vodka 2 = y%}$$ $$$$ $$\text{When the vodkas are mixed in the ratio 1:2 (total of 3 parts) the average profit is 10%:}$$ $${\left( \text{x} + \text{2y}\right) \over {3}} = {10}$$ $$\text{When the vodkas are mixed in the ratio 2:1 (total of 3 parts) the average profit is 20%:}$$ $${\left( \text{2x} + \text{y}\right) \over {3}} = {20}$$ $$$$ $$\text{Solve for x and y:}$$ $$\text{(1): } \text{x} + \text{2y} = {30}$$ $$\text{(2): } \text{2x} + \text{y} = {60}$$ $$$$ $${\text{x} + {2}\left({60}-\text{2x}\right)} = {30}$$ $$\text{x} + {120} - \text{4x} = {30}$$ $$\text{-3x} = {-90}$$ $$\text{x} = {30}$$ $$$$ $${30} + \text{2y} = {30}$$ $$\text{2y} = {0}$$ $$\text{y} = {0}$$ $$$$ $$\text{Solving gives: x = 30% and y = 0%}$$ $$$$ $$\text{After the profit of vodka 1 is increased by 4/3 the profit becomes 40%}$$ $$\text{After the profit of vodka 2 is increased by 5/3 the profit becomes 0%}$$ $$\text{If the vodkas are mixed in an equal ratio of 1:1, then the mixture will fetch a profit of: } {\left({40} + {0}\right) \over {2}} = \text{20%}$$

B. 20%

GMAT Challenge Series (700+): Q10

Waiting for Elevator.jpg

Question:

The elevator in an eleven-story office building travels at the rate of one floor per 1/4 minute, which allows time for picking up and discharging passengers. At the main floor and at the top floor, the operator stops for 1 minute. How many complete trips will an operator make during a 7-hour period?

A. 88

B. 56

C. 42

D. 60

E. 64

MBA Wisdom's Answer:

$$\text{A complete trip comprises of:}$$ $$\text{Stop at first floor: 1 min}$$ $$\text{10 floors up: 2.5 min}$$ $$\text{Stop at top floor: 1 min}$$ $$\text{10 floors down: 2.5 min}$$ $$\text{Total trip time: 7 min}$$ $$$$ $$\text{7 hours} = {{7} \times {60}} \text{ min} = {420} \text{ min}$$ $${trips} = {\text{time} \over \text{trip time}}$$ $${trips} = \frac{420}{7} = {60}$$

D. 60

GMAT Challenge Series (700+): Q9

Trail Mix.jpeg

Question:

A trail mix company keeps costs down by employing the peanuts:cashews:almonds ratio of 10:4:1 in each bag of up to 75 total nuts. What is the maximum percentage by which the company could decrease its number of peanuts per bag and still have peanuts constitute more than half the total amount of nuts? 

A. 40%

B. 48%

C. 49%

D. 50%

E. 58%

MBA Wisdom's Answer:

$$\text{peanuts : cashews : almonds}$$ $$\text{10 :4 :1}$$ $$\text{50 : 20 : 5}$$ $$\text{Peanuts have to be greater than the cashews and almonds}$$ $$\text{peanuts} > \text{cashews} + \text{almonds}$$ $$\text{peanuts} > {20} + {5}$$ $$\text{peanuts} > {25}$$ $$\text{As peanuts have to be an integer the lowest number is 26}$$ $$\text{decrease} = {\text{reduction} \over \text{original number}}$$ $$\text{decrease} = \frac{50 - 26}{50} = \frac{24}{50}$$ $$\text{decrease} = \text{48%}$$