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Q1. . A furniture dealer purchased a desk for $150 and then set the selling price
equal to the purchase price plus a markup that was 40% of the selling price. If the
dealer sold the desk at the selling price, what was the amount of the dealer's gross
profit from the purchase and the sale of the desk?
a) $40
b) $60
c) $80
d) $90
e) $100
Sp = 150 + .4 sp
0.6sp = 150
Sp = 150/0.6 = 250
Profit = 250-150= 100
Answer = E
Q2. . When a certain tree was first planted, it was 4 feet tall, and the heigth of the
tree increased by a constant amount each year for the next 6 years. At the end of the
6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many
feet did the height of the tree increased each year?
A. 3/10
B. 2/5
C. 1/2
D. 2/3
E. 6/5
We have to read the question carefully to make sure we see that the end of the 6th
year is being compared to the end of the 4th year. Also note that the tree grows by a
constant amount each year. This is different than the same % each year.
Let x = the constant annual growth in feet.
4ft is the original height of the tree, so after the 6th year, we have
4 + x + x + x + x + x + x….aslo 4 + 6x
After the 4th year we have 4 + x + x + x+ x or 4 + 4x
Lets compare the two values algebraically as the problem does:
6th year is 1/5 taller than the 4th year.
The difference in growth from the 4th to the 6th year will be 1/5 (or 0.2) of the height
after the 4th year.
So…(4+6x)-(4+4x) = 1/5(4+4x)…now solve for x
4 + 6x -4 – 4x = 4/5 + 4/5x
2x = 4/5 + 4/5x
Subtract 4/5x from both sides. I’ve changed 2x into 10/5x to easily do the subtraction
of fractions.
10/5x – 4/5x = 4/5
6/5x = 4/5
Divide both sides by 6/5, which is the same as multiplying by 5/6
X = 4/5 * 5/6. (Reduce the 5’s or multiply out and) you have 20 / 30 or x = 2/3.
Answer = D
Q3. Five pieces of wood have an average arithmatic mean length of 124 cms and a
median length of 140 cms.What can be the maximum possible length in cms of the
shortest piece of wood?
(A)90
(B) 100
(C)110
(D)130
(E) 140
Mean of the length of five pieces = 124 So total length = 124*5 = 620.
Median = 140, so the length of rest of the 4 pieces = 620 - 140 = 480
Assume the 5 pieces in ascending order be X1 X2 140 X3 X4 where X1 is the
shortest.
For X1 to be maximum, X3 and X4 has to be minimum but keep in mind the median
has to be 140.
The minimum possible values of X3 and X4 could be 140, hence the X1+X2 = 620 420 = 200. Distribute 200 among X1 and X2, the maximum possible value of X1 =
100 So the length of five pieces would be
100, 100, 140, 140, 140
Answer = B
Q4. Working alone at its own constant rate, A machine seals k cartons in 8 hours,
and working alone at its own constant rate, a second machine seals k cartons in 4
hours. if the two machines, working at its own constant rate and for the same period
of time, together sealed a certain number of cartons S, what percentage of the
cartons were sealed by the machine working at the faster rate?
(A)25%
(B) 33.3%
(C.) 50%
( D) 66 and 2/3%
(E) 75%
Let A = 8 cartoons in 8 hours= 1 cartoon per hour
B= 8 cartoons in 4 hours = 2 cartoon per hour
Therefore in 1 hour = 3 cartoons . faster one seals 2 cartons .
Therefore the percentage of cartoons sealed by faster one = (2/3)X 100 = 66 (2/ 3)
%
Answer D
Q5. At a certain committee meeting only associate professors and assistant
professors are present. Each associate professor has brought 2 pencils and 1 chart
to the meeting, while each assistant professor has brought 1 pencil and 2 charts. If a
total of 10 pencils and 11 charts have been brought to the meeting, how many
people are present?
A. 6
B. 7
C. 8
D. 9
E. 10
Say there are 'A' associate professors. So we have 2A pencils and A charts.
Say there are 'B' assistant professors. So we have B pencils and 2B charts.
Total pencils are 10 so 2A + B = 10
Total charts are 11 so A + 2B = 11
Add both: 3A + 3B = 21 so A + B = 7
Total number of people = 7
Answer = B