# Common Problem Solving Practice Test 3

Question – 1

1. Which of the following is equal to C(10,6)? Indicate the correct option..

• A. 210

• B. 6300

• C. 5040

• D. 250

• E. 360

C(10,6) = 10!/[(10-6)!6!]
= 10!/[4!6!]
= 10*9*8*7/(4*3*2)
= 210
Option A is true.

Question – 2

2. In an experiment, a dice is rolled. If the number on the dice is even, then a coin is tossed. If the number on the coin is odd, then the coin is not tossed. How many elements does the sample space for this experiment contain? Indicate the correct option.

• A. 36

• B. 15

• C. 20

• D. 6

• E. 9

The outcomes of rolling a dice are {1, 2, 3, 4, 5, 6}
The coin is tossed when the number on the dice is even.
The sample space of the experiment is {(1), (2,H), (2,T), (3), (4,H), (4,T), (5), (6,H), (6,T)}
There are 9 events in the sample space.
Option E is true.

Question – 3

3. Two adjacent sides of a parallelogram differ by 10 cm and one diagonal is 150 cm. The area of the parallelogram is 16800 sq.cm. Which of the following is the length of the shorter side? Indicate the correct option.

• A. 130 cm

• B. 140 cm

• C. 150 cm

• D. 170 cm

• E. 180 cm

Let the two sides be x and x+10 and the diagonal is 150
Semi-perimeter, s = (x+x+10+150)/2
= (2x+160)/2 = x+80
Area = 2*Sqrt[s(s-x)(s-x-10)(s-150)]
16800*16800 = 4* (x+80)(x+80-x)(x+80-x-10)(x+80-150)
16800*16800 = 4*(x+80)(80)(70)(x-70)
(x-70)(x+80)=16800*16800/(80*70*4)
= 12600
x^2+10x-5600=12600
x^2+10x-18200=0
x^2+140x-130x-18200=0
x(x+140)-130(x+140)=0
x=130 since it cannot be negative.
The shorter side is 130 cm long.
Option A is correct.

Question – 4

4. The ratio of the areas of two circles 16:81. What is the ratio of their radii? Indicate the correct option.

• A. 4:9

• B. 2:3

• C. 16:81

• D. 256:6561

• E. sqrt(2):sqrt(3)

Let the two areas be 16x^2 and 81x^2
Area of a circle = pi*radius^2
Ration of the radii of the two circles will be Sqrt(16x^2/pi):Sqrt(81x^2/pi)
= 4x/sqrt(pi):9x/sqrt(pi)
= 4:9
Option A is true.
[x^2=x*x].

Question – 5

5. Which of the following is the next number in the sequence 200, 199, 195, 186, 170? Indicate the correct option.

• A. 150

• B. 143

• C. 134

• D. 140

• E. 145

200-1^2=199
199-4=199-2^2=195
195-9=195-3^2=186
186-16=186-4^2=170
170-25=170-5^2=145
Option E is correct.
x^2=x*x].

Question – 6

6. Which of the following equations represents the y-axis? Indicate the correct option.

• A. x=0

• B. y=1

• C. y=0

• D. x+y=0

• E. x-y=0

The equation representing y-axis is x=0.
Option A is correct.

Question – 7

7. The tax on a commodity decreased by 30% and its consumption increased by 20%. Which of the following is the increase or decrease in the expenditure? Indicate the correct option.

• A. 16% decrease

• B. 10% decrease

• C. 15% increase

• D. 10% increase

• E. 10% decrease

Expenditure = tax*consumption
Let the original tax be x and consumption be y
Original expenditure = x*y=xyNew tax = x-30% of x
= x-30x/100
= 70x/100

New consumption = y+20% of y
= y+20y/100
=120y/100

New expenditure = 70x/100*120y/100
= 84xy/100

Percentage decrease in expenditure = xy-84xy/100
=(100xy-84xy)/100
= 16%
Option A is correct.

Question – 8

8. The average of 5 consecutive odd numbers in a sequence of 9 numbers is A and the average of the last 4 numbers is B. Which of the following is the average of the 9 numbers? Indicate the correct option.

• A. A+B

• B. (A+B)/2

• C. (A+B)/9

• D. (4A+5B)/9

• E. (5A+4B)/9

Let the numbers be x1, x2, x3, x4, x5, x6, x7, x8, x9A=(x1+x2+x3+x4+x5)/5
x1+x2+x3+x4+x5 = 5A
x6+x7+x8+x9 = 4B

Average of 9 numbers =(x1+x2+x3+x4+x5+x6+x7+x8+x9)/9
= (5A+4B)/9
Option E is true.

Question – 9

9. Sam bought 100 bottles of juice for \$40 each. He planned to earn a profit of 50% on his purchase and so marked the items accordingly. However, due to a sudden change in weather, he had to sell them at a discount of 20%. How much profit did he finally earn? Indicate the correct option.

• A. \$800

• B. \$600

• C. \$8

• D. \$48

• E. \$60

Marked price = 40+50% of 40
= 40+50*40/100
= 40+20=60
Selling price after 20% discount = 60-20% of 60
= 60 – 20*60/100
= 60 – 12
= 48
The profit earned on each item = 48-40=\$8
Total profit earned = \$8*100=\$800
Option A is correct..

Question – 10

10. The diagonal of a parallelogram AC makes an angle of 32 degrees with side AB. What angle does AC make with the side BC? Indicate the correct option.

• A. 64 degrees

• B. 148 degrees

• C. 32 degrees

• D. 58 degrees

• E. Cannot be determined

The angle depends on the positioning of the sides BC and AD. We cannot determine the angle unless additional information is provided.
Option E is correct.

Question – 11

11. If the following system of equations represents lines that meet at a point, then which of the following is the point?ax+by=c and lx+my=nIndicate the correct option..

• A. ((bn-cm)/(am-bl), (cl-an)/(am-bl))

• B. ((bn-cm), (cl-an))

• C. ((bn-cm)/(am+bl), (cl-an)/(am+bl))

• D. ((bn-cm)/(am-bl), (an-cl)/(am-bl))

• E. ((cl-an)/(am-bl), (bn-cm)/(am-bl))

The solution is given by
x/(bn-cm) = y/(lc-an) = 1/(am-lb)
x=(bn-cm)/(am-lb), y=(lc-an)/(am-lb)
The point is given by
((bn-cm)/(am-bl), (cl-an)/(am-bl))
Option A is correct.

Question – 12

12. Suzan tossed a coin. If a head appeared she would buy two new dresses and if a tail appeared she would buy one new dress. How many outcomes are possible? Indicate the correct option.

• A. 2

• B. 4

• C. 2^3

• D. 4^2

• E. 2*3

When Suzan tossed the coin, either she would get a tail or a head.
If she got a head she would buy two new dresses and if she got a tail she would buy one new dress.
Her purchases depend on her tossing the coin. Only two outcomes are possible.
Option A is true.

Question – 13

13. There are 5 suitable players for chess, 5 for hockey and 5 for football. One player is required for each, the chess team, the hockey team and the football team. In how many ways can the Sports Club Secretary make the choice? Indicate the correct option.

• A. 3

• B. 5

• C. 15

• D. 125

• E. 27

The chess player can be chosen in 5 ways, the hockey player in 5 ways and the football player in 5 ways.
Total ways of choosing the team = 5*5*5=125
Option D is correct…

Question – 14

14. In how many ways can six differently coloured flags be hoisted on 8 buildings, with at most one flag on each building? Indicate the correct option.

• A. 20160

• B. 56

• C. 48

• D. 36

• E. 6

Since we have only six flags and 8 buildings, we have to choose 6 buildings out of 8.
This can be done in P(8,6) ways.
P(8,6) = 8!/(8-6)!
= 8!/2!=8*7*6*5*4*3
= 20160
Option A is true.

Question – 15

15. In how many ways can Peter choose 5 subjects out of 7 if one language subject is compulsory for everyone? Indicate the correct option.

• A. 5

• B. 35

• C. 15

• D. 24

• E. 30