PSAT Problem Solving Practice Test 2

Question – 1

1. The first day of a leap year is a Sunday. How many Mondays does the year contain? Indicate the correct option.

  • A. 50

  • B. 51

  • C. 52

  • D. 53

  • E. 54

  • Answer:D

  • Answer Explanation:Every seventh day of the year will be a Sunday.A leap year has 365+1=366 daysThe year contains 52*7+2 daysThe second day is a Monday and hence the 365th day will be a Sunday and the 366th day will be a Monday.There are 53 Mondays in the year.


Question – 2

(Note:[x^2=x*x] )

  • A. 2

  • B. 4

  • C. 6

  • D. 7

  • E. 9

  • Answer:D

  • Answer Explanation:Put x = 2 in the given polynomial6x^3-2ax^2+ax-6 = 6*(2^3) -2a*(2^2)+a*2-6=048-8a+2a-6=06a=42a=7Option D is correct.


Question – 3

3. The volume of a cylinder is 550 cc and its radius is 5cm. What is the height per unit radius of the cylinder? Indicate the correct option.

  • A. 1

  • B. 5

  • C. 1.4

  • D. 0.71

  • E. 7

  • Answer:C

  • Answer Explanation:Volume of cylinder = pi*r^2*h, where r and h are radius and height of the cylinder.550=pi*5^2*hh=550*7/(22*5*5)= 7Height of the cylinder = 7 cmHeight/radius = 7/5 = 1.4Option C is correct.


Question – 4

(Note:[x^2=x*x] ).

  • A. 1/3

  • B. 1

  • C. 1/9

  • D. 3

  • E. 27

  • Answer:C

  • Answer Explanation:[81*3^(n+1)-9*3^n]/[81*3^(n+3)-3*3^(n+3)]= [3^(n+5)-3^(n+2)]/[3^(n+7)-3^(n+4)]= [3^(n+5)-3^(n+2)]/{(3^2)*[3^(n+5)-3^(n+2)]}=1/3^2= 1/9Option C is true.


Question – 5

(Note:[pi=22/7] ).

  • A. 169 m

  • B. 10 m

  • C. 26 m

  • D. 13pi m

  • E. 26pi m

  • Answer:C

  • Answer Explanation:The shortest path will be along the diameter of the field.Area = 169pi = pi*r^2, where r is the radius of the circler^2=169r=13The length of the shortest path is 2*13=26m.Option C is true.


Question – 6

6. A cylinder’s diameter is reduced to one-third. The volume remains the same. How many times will the length become? Indicate the correct option.

  • A. 3

  • B. 6

  • C. 5

  • D. 12

  • E. 9

  • Answer:E

  • Answer Explanation:Let the initial length be l and new length be l’.Let the original diameter be 6r and radius be 6r/2=3r.The new diameter will be 6r/3=2r and radius will be 2r/2=rVolume=pi*(3r)^2*l = pi*r^2*l’9r^2*l=r^2*l’9l=l’The length becomes 9 times itself.Option E is true.[x^2=x*x][pi=22/7]


Question – 7

(Note:[x^2=x*x] ).

  • A. 16

  • B. 4

  • C. 18

  • D. 12

  • E. 14

  • Answer:E

  • Answer Explanation:x+1/x=4(x+1/x)^2=x^2+1/x^2+2*x*1/xx^2+1/x^2=(x+1/x)^2-2=4^2-2=16-2=14Option E is correct.


Question – 8

8. Triangle ABC is right angled at A. AL is drawn perpendicular to BC. Which of the following is true? Indicate the correct option.

  • A. Angle ABC = Angle ACB

  • B. Angle BAL = Angle ALC

  • C. Angle ACL = Angle ABL

  • D. Angle BAL = Angle ACB

  • E. Angle ALB = Angle ACB

  • Answer:D

  • Answer Explanation:Options A and C need not be true and hence they are incorrect.Option B is false as angle ALC is a right angle while angle BAL is an acute angle.Similarly option E is false.We check option D now.In triangle ABL,angle BAL+angle ALB+angle B=180angle BAL+90+angle B=180angle BAL+angle B=90angle BAL=90-angle B …(1)In triangle ABC,angle A + angle B + angle C = 18090 + angle B +angle C = 180angle B + angle C = 180-90=90angle ACB=90-angle B …(2)From (1) and (2), we conclude that option D is true.


Question – 9

9. Pam has a field in the shape of a polygon with 10 sides. Into how many triangular fields can she cut her field so that she can access each field from the same vertex? Indicate the correct option.

  • A. 3

  • B. 5

  • C. 7

  • D. 8

  • E. 9

  • Answer:D

  • Answer Explanation:A polygon with n sides can be cut into (n-2) triangles with a common vertex.The bigger field with 10 sides can be cut into (10-2)=8 triangular fields with a common vertex.Option D is true.


Question – 10

10. What is the area enclosed by the points (1,4), (4,-2) and (9,-12)? Indicate the correct option.

  • A. 0

  • B. 24

  • C. 38

  • D. 26

  • E. 36

  • Answer:A

  • Answer Explanation:(1,4), (4,-2), (9,-12) are the three points.Area = 1/2[1*(-2)-4*4+4*(-12)-(-2)*9+9*4-1*(-12)]=1/2[-2-16-48+18+36+12]=1/2(0)=0Option A is true..


Question – 11

11. If 20% of Adam’s share is 10% of Eve’s share then how much percent of Adam’s share is 4% of Eve’s share? Indicate the correct option.

  • A. 2

  • B. 4

  • C. 2/25

  • D. 8

  • E. 0.08

  • Answer:D

  • Answer Explanation:Let Adam’s and Eve’s shares be x and y respectively.20%*x=10%*y20/100*x = 10/100*y2x=y4% of y = 4/100*y= y/25 = 2x/25= 2/25*100% of x= 8% of x.


Question – 12

12. A company gave a bonus of $500 dollars per employee on an average. The bonus packets of five employees worth $300 each and two employees worth $500 each got lost before being distributed. As a result, the company had to make new bonus packets for them. The company spent an added 10% on the bonus per employee as a result of this added expense. How many employees does the company have? Indicate the correct option.

  • A. 10

  • B. 28

  • C. 50

  • D. 55

  • E. 32

  • Answer:C

  • Answer Explanation:Let the number of employees be xAverage = Sum of bonus/number of employees500 = Sum/xSum=500xBonus lost = 5*$300+2*$500= 1500+1000= 2500New average bonus = 500+500*10/100= 500+50=550New average = (500x+2500)/x = 550500x+2500=550x50x=2500x=2500/50=50The company has 50 employees.Option C is true.


Question – 13

13. A water tank can hold 5760 litres of water. The inlet fills water at the rate of 4 litres per minute and the outlet can empty the tank in 6 hours. How many hours would it take to empty the tank if the inlet and outlet are both open? Indicate the correct option

  • A. 4

  • B. 5

  • C. 6

  • D. 7

  • E. 8

  • Answer:E

  • Answer Explanation:Rate of water in = 4 litres/minRate of water out = 5760/(6*60)= 16 litres/minRelative rate at which water goes out when the inlet and outlet are both open = 16-4=12 litres/minTime taken to empty the tank = 5760/12= 480 minutes= 480/60 hours = 8 hoursOption E is true.


Question – 14

14. A and B have to cover a distance of 500m. Their speeds are in ratio 3:4 respectively. How many meters should A be before B at the time of starting so that they reach the finish line together? Indicate the correct option.

  • A. 0

  • B. 125

  • C. 30

  • D. 50

  • E. 150

  • Answer:B

  • Answer Explanation:Let the speed of A be 3x m/s and that of B be 4x m/sLet the distance covered by A be (500-y) meters.Time taken by them is the same,(500-y)/3x=500/4×4*(500-y)=3*5002000-4y=15004y=2000-1500=500y=500/4=125A should be ahead of B by 125 metersOption B is correct.


Question – 15

15. Which of the following is equal to P(15,5)? Indicate the correct option.

  • A. 360306

  • B. 300600

  • C. 306306

  • D. 360360

  • E. 330660

  • Answer:D

  • Answer Explanation:P(15,5) = 15!/(15-5)!= 15!/10!= 15*14*13*12*11= 360360Option D is true.


Score: 0/10

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