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MATHEMATICS Comprehensive Exam 1 Answer Guide 1. A The dimensions of the prism formed after joining the

three cubes are:

Length = 21 cm; Breadth = 7 cm; Height = 7 cm.

Surface area of the resulting prism ( )( )

2

2 21 7 7 7 7 212 147 49 1472 343 686 cm

= × + × + ×= + +

= × = 2. E The midpoint of the points ( )4,7, 8− and ( )5,3, 4− − is

given by

4− 52

, 7 + 32

, −8− 42

⎛⎝⎜

⎞⎠⎟ = −0.5, 5, − 6( ) .

3. E By multiplication principle, the number of ways is equal to the product of the number of choices in each variety, i.e.

15 18 12 3,240.= × × =

4. D The simple interest per month is given to be 5%. Hence to increase its value by 300%, the money needs to be

invested for 300% 60 months = 5 years.5%

=

5. C This is the case of a 30 60 90− −o o o right triangle. If the length of the shorter leg (opposite the 30o angle) is a, the length of the longer leg (opposite the 60o angle) is given by 3a .

Here 5 3a = .

Therefore length of longer leg is

( )3 3 5 3 15.a= = =

6. B Two movements in the x-y-z coordinate plane <a, b, c> and <x, y, z> are perpendicular to each other when

0ax by cz+ + = .

( ) ( )( ) ( )1 9 3 4 7 027 28 01

xx

x

⇒ − + − − + − =⇒ − + − =⇒ = −

7. B Let AE and BD represent the two poles. The difference between the heights of the two poles is given by

8mAC = . The difference between their feet is given by 15mDE BC= = . We then use Pythagoras theorem to

find the distance AB between their tops.

2 2 28 15AB⇒ = +

17 mAB⇒ = ±

But, AB cannot be negative, therefore 17 mAB = .

8. A The distance d of a point ( )1 1,x y from the line

0ax by c+ + = is given by 1 1

2 2

ax by cd

a b

+ +=

+.

Using this formula we get the distance of ( )4,1 from the line 3 4 9 0x y− − =

( ) ( )( )22

3 4 4 1 9 12 4 9 1 1 0.2.5 59 163 4

− − − − −= = = = =

++ −

9. E The average rate of change between any two points is equal to the slope of the line joining the two given points.

Using the given end points of the interval we find the values of y at those points

( ) ( ) ( )32 2 4 2 7

8 8 7 7

f − = − − − +

= − + + =

( ) ( )34 4 4 4 764 16 7 55

f = − +

= − + =

Thus we need to find the slope between ( )2, 7− and

( )4,55 which is equal to ( )

55 7 48 84 2 6

− = =− −

.

10. D The slope of line containing the points ( )1, 4− and

( )0, 6 is given by ( )

6 4 20 1

− =− −

.

The slope of line containing the points ( )3, y and ( )2, 7

is equal to 7 72 3

y y− = −−

.

Both the lines are parallel. Hence, their slopes are equal

MATHEMATICS COMPREHENSIVE EXAM 1 ANSWER GUIDE PAGE 2 OF 5 DEMIDEC RESOURCES ©2011

7 29

yy

⇒ − =⇒ =

11. C The formula for the number of diagonals in a polygon is

given by ( )32

n n −

( )

( )

327

23 54

9.

n n

n nn

−⇒ =

⇒ − =⇒ =

The sum of interior angles of a 9 sided polygon is equal to

( )9 2 180 7 180 1260 .− = × =o o o

12. D When two chords intersect in a circle, the angle between them is equal to the average of the arc angles formed by the chords.

46 77246 154108

x

xx

+⇒ =

⇒ + =

⇒ = o

13. E Let the base area of pyramid be equal to b. Then, the base area of prism is equal to 2b.

Volume of prism 12b h= ×

Volume of pyramid 213

b h= × ×

We are given that volumes are equal.

1 2

1

2

123

16

b h b h

hh

⇒ × = × ×

⇒ =

or 1 2: 1 : 6h h =

14. A Let ( ),P x y be equidistant from the points ( )7,1A and

( )3,5B .

We are given that AP BP= .

So, 2 2AP BP=

( ) ( ) ( ) ( )2 22 27 1 3 5x y x y⇒ − + − = − + −

2 2

2 2

14 49 2 16 9 10 25

x x y yx x y y

⇒ − + + − + =

− + + − +

2x y⇒ − = , which is the required relation.

15. C The first letter can be chosen in 26 ways. Since all the letters need to be distinct, we are left with 25 choices for the second letter and hence, 24 choices for the third letter.

By multiplication principle, number of words that can be made using three distinct letters 26 25 24 15600.= × × =

16. B Since it is a regular octagon, therefore the length of each side of the octagon is equal to one-eighth of the perimeter of the octagon.

Length of each side 1 24 2 3 28

= × =

The ratio of areas is equal to the square of the ratio of the length of sides as

23 2 9 .

4 8⎛ ⎞

= =⎜ ⎟⎝ ⎠

Hence, the ratio is 9:8.

17. C The volume of sphere 343

rπ= . Since the sphere of radius

r is melted and cast into a solid cone, therefore

Volume of sphere = Volume of cone

3 24 13 3

r R hπ π= , where R is the radius of the base of the

cone and h is its height. We are given that height of cone is equal to r.

3 2

2 2

4 13 342

r R r

r Rr R

π π⇒ =

⇒ =⇒ =

or 2R r= , which is the radius of the base of cone.

18. B The slope of the graph of 2 3y x x= + at 1 7,4 5

⎛ ⎞ ⎜ ⎟⎝ ⎠ is the

same as the slope of the tangent at that point. Slope of tangent is equal to the first derivative of the function, which in this case is

3' 22

y xx

= +

At 14

x =

1 3 1 7' 2 34 2 212

4

y = × + = + =

19. C We have ( )( )( )4 1 2 3n P n n n n= − − − and

( )( )( )1 3 1 2 3n P n n n− = − − − .

4 1 3: 9 :1n nP P− =

( )( )( )( )( )( )

1 2 3 91 2 3 1

9

n n n nn n n

n

− − −⇒ =

− − −⇒ =


20. E All equilateral triangles are similar, therefore ABCΔ and BDEΔ are similar.

D is the mid-point of side BC

2BC BD⇒ =

2BCBD

⇒ =

We know that ratio of areas of similar triangles is equal to the square of the ratio of their sides.

( )( )

2 41

ar ABC BCar BDE BD

Δ ⎛ ⎞⇒ = =⎜ ⎟Δ ⎝ ⎠

Hence, required ratio is 4 :1.

21. D Volume of a coin ( )2 30.75 0.2 cmπ⎡ ⎤= × ×⎣ ⎦

Volume of cylinder ( )2 32.25 10 cmπ⎡ ⎤= × ×⎣ ⎦

Number of coins

( )( )

2

2

Volume of cylinderVolume of coin

2.25 100.75 0.2

2.25 2.25 100.75 0.75 0.23 3 50 450 coins

ππ

=

× ×=

× ×× ×=× ×

= × × =

22. E This question must be solved using the binomial distribution method of probability. Here getting an odd number, i.e. 1, 3, 5, is a success. Hence,

Probability of success, p 3 16 2

= =

Probability of failure, q 3 16 2

= =

Here n = 6. We need to find the probability of at most 5 successes which can be calculation by finding the probability of 6 successes and then subtracting that value from 1.

P (at most 5 successes) ( )6 0

6

1 P 6 successes6

16

1 64 1 631 12 64 64

p q

= −

⎛ ⎞= − ⎜ ⎟

⎝ ⎠−⎛ ⎞= − = =⎜ ⎟⎝ ⎠

23. B From 1 to 25 there are 9 prime numbers, 2, 3, 5, 7, 11, 13, 17, 19 and 23.

Two tickets bearing prime numbers can be selected in 92⎛ ⎞⎜ ⎟⎝ ⎠

ways.

Two tickets out of 25 can be selected in 252

⎛ ⎞⎜ ⎟⎝ ⎠

ways.

Hence, probability that both tickets bear prime numbers

9 9 82 32

25 2425 2522

⎛ ⎞ ×⎜ ⎟⎝ ⎠= = =×⎛ ⎞⎜ ⎟⎝ ⎠

24. D A quick way to do this is by plotting the coordinates and seeing what type of quadrilateral you get by joining them.

Otherwise you can also do it by finding the slopes as:

( )

( )

( )

3 2 14 1 52 3 5

3 43 2 1

2 3 53 2 5

2 1

PQ

QR

RS

SP

−= =− −

− −= =−

− − −= =

− −− −= =

− − −

Since the opposite sides of the quadrilateral are parallel, this figure is a parallelogram.

However, we need to specify what type of parallelogram is this. Since the slopes of adjacent sides are not negative reciprocals of each other, therefore the angle between them is not 90o . Hence, this figure is neither a square nor a rectangle.

We find the slope of diagonals as:

( )2 2 1

3 13 3 12 4

PR

QS

− −= = −− −

− −= =− −

The slopes of diagonals are negative reciprocals of each other, which means the diagonals are perpendicular to each other. Hence, this figure is a rhombus.


25. B When two chords intersect inside a circle, the product of the lengths of the segments of one chord equals the product of the lengths of the segments of the other chord.

7 4 142

RN SN BN AN

AN

⇒ × = ××⇒ = =

Now, 14 1 13MN AN AM= − = − =

13 2 15MB MN BN⇒ = + = + =

When chords PQ and AB intersect we have

1 15 53.

AM MB PM MQz

z

× = ×⇒ × = ×⇒ =

26. A AB bisects the chord MN at O.

4MO NO⇒ = =

Using the rule for intersecting chords, we get

4 4 82

EO BO MO NO

EO

× = ××⇒ = =

Now using the rule for intersecting secants, we get

( ) ( )( ) ( )2 2 10 3 3

3 85.

AE AE EB AF AF FCFC

FCFC

+ = +⇒ + = +⇒ + =⇒ =

27. C Compound interest25 4411

100 40041400

x x x x

x

⎡ ⎤⎛ ⎞= + − = −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

=

Simple interest 5 2100 10

x x× ×⎛ ⎞= =⎜ ⎟⎝ ⎠

Given that the difference between the compound and simple interest is $5.

41 5400 10

5400

$2,000

x x

x

x

⇒ − =

⇒ =

⇒ =

28. B Use your imagination to create a figure starting from point A and add specific number of sides as in the question to create a figure as below. From the figure, we can see that it is a 12-sided polygon.

29. B The figure is as shown below:

Since tangent is perpendicular to the radius, at the point of contact, therefore 90ORD OSD∠ = ∠ = o

It is given that 90D∠ = o . Also, OR OS= . Therefore ORDS is a square.

Since tangents from an exterior point to a circle are equal in length

BP BQ= , CQ CR= and DR DS=

Now, BP BQ= , 27BQ⇒ =

Also 27BC CQ− = , 11CQ CR⇒ = =

11CD DR− = , 14.DR⇒ =

But ORDS is a square, Therefore, 14.OR DR= =

Hence, 14.r =

30. D The given curve is 2 4 1y x x= + +

When 3x = , then 23 4 3 1 22y = + × + = .

∴The point at which tangent is required is ( )3, 22 .

Now, 2 4dy xdx

= +


Slope of tangent to curve at ( )3, 22 is

2 3 4 10.dydx

⎛ ⎞ = × + =⎜ ⎟⎝ ⎠

∴Equation of the tangent at ( )3, 22 is

( )22 10 3y x− = − or 10 8 0x y− − = .

31. D Since AB DCP , therefore the given figure is a trapezoid.

Since the diagonals of a trapezium divide each other proportionally,

( ) ( )( )

( )( )

2

2

3 19 35 3

3 3 19 5 39 57 8 15

17 72 08 9 0

AO BOOC OD

x xx

x x xx x x

x xx x

⇒ =

− −⇒ =−

⇒ − = − −

⇒ − = − +

⇒ − + =⇒ − − =

8x⇒ = or 9x =

Hence the largest value is 9x = .

32. B Three numbers out of 20 can be chosen in 20 20 19 18 20 19 33 1 2 3

⎛ ⎞ × ×= = × ×⎜ ⎟ × ×⎝ ⎠ways

These numbers can be consecutive in 18 ways as they can be

1, 2, 3; 2, 3, 4; 3, 4, 5……….18, 19, 20.

∴Required probability 18 320 19 3 190

= =× ×

33. B The largest possible hexagon inside the given circle will have the length of its diagonals, that pass through the center of the circle, equal to the diameter of the circle.

Each angle in a regular hexagon is equal to 120o . Thus, when a regular hexagon is divided as below, we get six equilateral triangles.

We see that the length of each side of these equilateral triangles is equal to the radius of the circle which is equal to 8.

Area of Hexagon = 6 x area of each equilateral triangle

236 84

96 3

= × ×

=

34. D

( )( )

( )( ) ( )

3 3 3

1 1

2

1

2

12

1 1lim lim1 1

1 1lim

1lim 1

1 1 1 3

x x

x

x

x xx x

x x xx

x x

→− →−

→−

→−

⎛ ⎞ ⎛ ⎞+ +=⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠+ − +

=+

= − +

= − − − + =

35. C Since A is the only letter that comes before E in the dictionary order, therefore the required number of words is equal to the number of words which begin with the letter A.

When A is fixed in the first place we can arrange the remaining 10 letters, of which N is repeated twice and I is

repeated twice, in 10! 907,2002!2!

= = ways.

MATHEMATICS Comprehensive Exam 2 Answer Guide 1. D We have to choose one flag after the other. The upper

flag can be chosen in 5 ways and then the lower flag can be chosen in 4 ways.

Number of signals that can be generated 5 4 20= × =

2. A Since the sun makes the same angle with all the objects, these objects create similar triangles. This means there is a constant ratio between the side lengths, and we can use a proportion to solve for the height of the tower as:

20 1050100 m

xx

=

⇒ =

3. C The number of diagonals in a convex polygon with n

sides is given by ( )3

2n n −

.

Required answer 14 11 772×= =

4. E First the man goes 10 m due west . When he turns right he starts walking towards north for 8 m. He then turns left and again starts walking in the west direction for 5 m. His overall movement is pictured as below.

To find the distance of the man from his starting point, we use Pythagoras theorem as:

Distance 2 215 8 289 17= + = = m

5. D Slope of tangent to the curve 43 4y x x= − is given by 312 4x − .

Thus, slope at 4x = is equal to 312 4 4 764× − = .

6. B Lateral surface area of a cylinder 22 264 mrhπ= =

Volume of cylinder 2 2924 mr hπ= =

Ratio of volume to lateral surface area

2 9242 264

72 2

7 m

r hrh

r

r

ππ

⇒ =

⇒ =

⇒ =

Therefore, diameter 14 m=

Putting the value of r in lateral surface area gives us height 6 m= (approximately).

Hence, ratio of diameter to height 14 76 3

= =

7. B Simple Interest (S.I.) 100

P R T× ×=

. . 100

1,050 100750 2.5

56 months = 4 years 8 months

S ITP R×⇒ =×

×=×

=

8. D Using distance formula, we get the distance between( )cos sin , 0a bθ θ+ and ( )0, sin cosa bθ θ −

( ) ( )

( ) ( )

2 2

2 2 2 2

2 2 2 2

2 2 2 2 2 2

2 2

cos sin 0 0 sin cos

cos sin 2 cos sinsin cos 2 sin cos

cos sin in cos

a b a b

a b aba b ab

a b s

a b

θ θ θ θ

θ θ θ θθ θ θ θ

θ θ θ θ

= + − + − +

+ +=

+ + −

= + + +

= +

9. A The slope of line containing the points ( )0, 7 and

( )4, 6− is given by 6 7 14 0 4− =

− −.

The slope of line containing the points ( ), 4x and

( )3, 1 − is equal to 1 4 53 3x x− − = −− −

.

Both the lines are perpendicular. Hence, their slopes are negative reciprocals of each other.

( )

5 43

4 3 512 4 54 7

74

xxx

x

x

⇒ − = −−

⇒ − =⇒ − =⇒ =

⇒ =

10. D Given 1

2

35

rr

= and 1

2

23

hh

=

Ratio of volumes:


221 1 1 1 1

22 2 2 2 2

9 2 625 3 25

V r h r hV r h r h

ππ

⎛ ⎞ ⎛ ⎞= = = × =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠.

Hence, the ratio of volumes is 6:25

11. C Since each page is equally likely to be chosen the sample space contains 100 events. The sample space favorable to the event contains 10 choices as 9, 18, 27, 36, 45, 54, 63, 72, 81 and 90.

Hence, required probability 10 0.1.100

= =

12. A A goalkeeper can be selected in 3 ways. The captain can be selected out of the remaining 15 players and the assistant-captain can be selected in 14 ways after the captain is selected.

Total combinations possible 15 14 3 630= × × =

13. E Let D be the midpoint of BC, then D has coordinates

equal to ( )1 5 1 1, 3,0 .2 2+ − +⎛ ⎞ =⎜ ⎟⎝ ⎠

Required length AD ( ) ( )2 23 1 0 3 25 5= + + − = =

14. A Let the edge length of a cube be 2x. The largest sphere that can be fit inside it will have a radius equal to half the edge length, i.e. x.

Ratio of volumes ( )3 3

33

2 3 8 64 43

x xxx π ππ

× ×= = =× ×

Required ratio is 6 :π

15. C A quick way to do this is by plotting the coordinates and seeing what type of quadrilateral you get by joining them.

Otherwise you can also do it by finding the slopes as:

( )( )

( )( )

( )( )

4 1 32 4 2

0 4 24 2 33 0 32 4 23 1 22 4 3

PQ

QR

RS

SP

− − −= = −− − −− −

= =− −−= = −−− −

= =− −


However, we need to specify what type of parallelogram is this. Since the slopes of adjacent sides are negative reciprocals of each other, therefore the angle between them is 90o . Hence, this figure is either a square or a rectangle.


( )( )( )( )

0 1 14 4 83 4 72 2 4

PR

QS

− −= =

− −− −

= =− −

The slopes of diagonals are not negative reciprocals of each other, which means the diagonals are not perpendicular to each other. Hence, this figure is a rectangle.

16. B Let us take the three vertices to be ( )2, 1A − − , ( )1,0B ,

( )4,3C . Let the fourth vertex D of the parallelogram be

( ),a b .

The diagonals of a parallelogram bisect each other. Therefore the midpoint of the diagonals is the same.

For the given parallelogram ABCD, we have

Midpoint of AC = Midpoint of BD

2 4 1 3 1 0, ,2 2 2 2

a b− + − + + +⎛ ⎞ ⎛ ⎞⇒ = ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( ) 1 01,1 ,2 2

a b+ +⎛ ⎞⇒ = ⎜ ⎟⎝ ⎠

1 2a⇒ + = and 0 2b+ =

1a⇒ = and 2b =

Thus, coordinates of D are ( )1, 2

17. E Girls can sit in chairs in 4 4 4! 24P = = ways and boys can stand in a row in 3 3 3! 6P = = ways.

By the multiplication rule, required number of ways = 24 6 144× = ways.

18. B This is the case of a 30 60 90− −o o o right triangle. If the length of the hypotenuse is 2a, the length of the shorter leg (opposite the 30o angle) is given by a and the length of the longer leg (opposite the 60oangle) is given by 3a.

Here 2 18a = .

Therefore length of shorter leg = a = 9.

Also the length of longer leg 3 9 3.a= =

Therefore perimeter of triangle

( )9 18 9 3 27 9 3 9 3 3 .= + + = + = +

19. C Midpoint of the given points is

( )2 0 8 0, 1, 4 .2 2

− + +⎛ ⎞ = − ⎜ ⎟⎝ ⎠

The slope of the graph is equal to the value of the first derivative of y at the point ( )1, 4− .


3' 2 6y x x⇒ = − +

At 1x = − , ( ) ( )3' 2 1 6 1 2 6 4y⇒ = − − + − = − = −

Hence, required slope 4.= −

20. D Total length to be fenced is equal to the sum of perimeter of hexagonal field and the length of 3 diagonals.

When a regular hexagon is divided as below, each triangle formed is equilateral. Hence the length of each diagonal is twice the length of the side.

Length of diagonal 2 120 240= × = m

Total length 6 120 3 240 1,440= × + × = m

Total cost of fencing 1440 3.5 $5,040= × =

21. A Since the money is being invested for 4 years and is compounded quarterly, there are a total of 16 compoundings.

Amount after 4 years is given by 1ntrP

n⎛ ⎞+⎜ ⎟⎝ ⎠

.

Here P =$3,800, r = 7.5%, n = 4 and t = 4 years.

⇒ Amount

( )

16

16

0.075$3,800 14

$3,800 1.01875$5,115.23

⎛ ⎞= +⎜ ⎟⎝ ⎠==

Interest earned ( )$ 5,115.23 3,800$1,315.23

= −=

22. D An inscribed angle whose ray passes through the center of the circle has measure equal to one half of the arc angle it intercepts.

⇒ Inscribed 33ABD∠ = o and inscribed 21DBC∠ = o

33 21 54ABC⇒∠ = + =o o o .

The quadrilateral ABCD is a cyclic quadrilateral and as we know that the sum of opposite angles in a cyclic quadrilateral is 180 .o

180 54 126ADC⇒∠ = − =o o o .

23. C The distance from the point ( )1 1 1, ,x y z to the plane

0Ax By Cz D+ + + = is 1 1 1

2 2 2

Ax By Cz D

A B C

+ + +

+ +.

Required distance ( )( )22 2

2 2 1 4 1 9

1 2 4

13 13 21 units2121

− × + × − −

+ − +

−= =

24. D Volume of cone A with radius 1and height 4

( )21 41 43 3π π= × = .

Ratio of volume of cone B to cone A

64 2343

16 2

π

π=

=

Therefore the scale factor between the lengths of the two

cones 3 16 2 2 2= = .

Hence, the radius of cone B is 2 2 .

25. C Since from a point outside the circle, the distances to the two possible points of tangency are equal, we have:

PB PX= , QC QX= and AB AC=

Perimeter of

( )( ) ( )( ) ( )

APQ AP PQ AQAP PX XQ AQAP PX AQ XQAP PB AQ QC

Δ = + += + + += + + += + + +

⇒ Perimeter of 10APQ AB ACΔ = + = cm

26. D Slope of the given line 25

= − . Since the required line is

parallel to the given line, therefore their slopes are same,

i.e. 25

− .

Midpoint of the segment joining the given points is

( )2 4 7 1, 1, 42 2− +⎛ ⎞ = − ⎜ ⎟⎝ ⎠

.

∴Required equation of line is ( )24 15

2 5 18 0

y x

x y

− = − +

⇒ + − =

27. C Let the segment AO be equal to 7x and segment OB equal to 3x.

Using the rule for intersecting chords, we get


2

2

7 3 6 1421 84

42

AO BO PO QOx xx

xx

× = ×⇒ × = ×

⇒ =

⇒ =⇒ =

Length of segment AO = 7x = 14 units.

28. B Three black balls can be selected out of 4 in 43⎛ ⎞⎜ ⎟⎝ ⎠

ways

and 3 red balls can be selected out of 5 in 53⎛ ⎞⎜ ⎟⎝ ⎠

ways.

Hence the number of ways in which a drawing of 6 balls consisting of 3 black and 3 red balls can be made in

4 54 10 40

3 3⎛ ⎞ ⎛ ⎞

× = × =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

ways.

29. A Here success is ‘jumping over a hurdle successfully without knocking it down’ and n = 10.

p = P (a success) 56

=

5 116 6

q⇒ = − =

The athlete must knock down fewer than 2 hurdles, that is, he should successfully jump over 9 or 10 hurdles.

Hence required probability ( ) ( )9 1 10 0

9 10

9 9

9

9 1010 109 10

5 1 510 16 6 6

5 5 5 5 156 3 6 6 6

5 52 6

P P

p q p q

= +

⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + = ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞= ⎜ ⎟⎝ ⎠

30. D Area of prism

= Area of 2 hexagonal bases + Area of 6 rectangular faces

When a regular hexagon is divided as show below, each of the triangles shown below is an equilateral triangle with edge equal to 6 cm.

Area of hexagon 6= × area of each equilateral triangle

2

2

36 64

54 3 cm

= × ×

=

Area of rectangular face 26 5 30 cm= × =

Total area of prism

( ) 2 22 54 3 6 30 cm 108 3 180 cm= × + × = +

31. B The word MALENKOV contains 8 letters which can be arranged in 8 8 8!P = ways.

Since vowels are to occur together, we take all the three vowels as a combination. We are left with 5 consonants and 1 group of vowels. These 6 objects can be rearranged in 6 6 6!P = ways and in any such arrangement, 3 vowels can reshuffle among themselves in 3 3 3!P = ways. Hence number of arrangements in which vowels can occur together are equal to 6! 3! × .

Required probability 6! 3! 6! 6 6 38! 8 7 6! 56 28 × ×= = = =

× ×

32. B Sum of angles of polygon is equal to ( )2 180n− o.

( )2 180 14402 810

nnn

⇒ − =⇒ − =⇒ =

o o

We anticipate using 10 triangles to create the decagon as below


Each vertex of the decagon has an angle of 144o . Therefore, by symmetry each base angle of each triangle will be 72o .

Using trigonometry we get the height of each triangle as

4 tan 72 4 3.08 12.32= × =o sq. units.

Area of each triangle 1 8 12.32=49.282

= × × sq. units

Area of decagon = 10× area of each triangle10 49.28 492.8= × = sq. units

33. E The average rate of change between any two points is equal to the slope of the line joining the two given points.

Using the given end points of the interval we find the values of y at those points.

( ) ( ) ( )23 2 3 7 3 186 63 18 39

f = − += − + = −

( ) ( ) ( )25 2 5 7 5 1810 175 18 147

f = − += − + = −

Thus we need to find the slope between ( )3, 39 − and

( )5, 147 − which is equal to

( )147 39 108 545 3 2

− − −= − = −

−.

34. D Using the rule for intersecting chords, we get

2 10 5.4

CO DO FO BO

BO

× = ××⇒ = =

Now using the rule for intersecting secants, we get

( ) ( )( ) ( )6 6 5 5 9

36 6 706 34

17 .3

AE AE EC AF AF FBxx

x

x

+ = +⇒ + = +⇒ + =⇒ =

⇒ =

35. E

( )

2

21 1

1

1

1 1 1 1 1lim lim1 11

1 1 1lim

1 11 1lim

11 1 1 2 1 2 2 .

21 1 2

x x

x

x

x x x x xx xx

x x

x xx

x

→ →

→

→

− + − + − + −=+ −−

− + +=

+ −+ +=+

+ + + += = =+

MATHEMATICS Comprehensive Exam 3 Answer Guide 1. D Tangent is perpendicular to the radius at the point of

contact, therefore 90ABC ACB∠ = ∠ = o

Sum of angles of quad. ABOC is equal to 360o

90 69 90 360111

BOCBOC

⇒∠ + + + =

⇒∠ =

o o o o

o

2. C There are 14 options when entering the hall. However, when leaving, the person cannot leave through the door he entered. Thus, he can leave in 13 ways.

Total number of ways in which he can enter and leave is equal to 14 13 182× =

3. C This is the case of a right angled triangle whose hypotenuse is the wire of length 24 m. We use Pythagoras theorem to find the distance for which the stake needs to be pulled as:

Required length 2 224 18 252 6 7 m= − = =

4. A We are given that 2AB DE=

12

ABDE

⇒ =

The given triangles are similar, hence their sides are

proportional which gives us the relation, 12

AB BCDE EF

= =

Putting value of 8BC = , we get

8 12

16EF

EF

=

⇒ =

5. B A non-leap year has 365 days which contains 52 full weeks equal to 364 days. The remaining one day can be

either of the seven days of a week. Thus, that day has a 17chance of being a Sunday.

6. A The sum of exterior angles of any polygon is equal to 360o .

7. C It is given that the water is flowing with velocity 10 km/hr. Therefore length of water column formed in 2 hours 20 km 20,000 m= =

Volume of this water column ( ) 3

3

3 1.2 20,000 m72,000 m

= × ×

=

8. B lim

22

x a

x a a ax a a a

a aa a

→

+ +=+ +

= =

9. A The slope of the graph at a given point is the value of the first derivative of the graph at that given point.

Differentiating 2 23y x x= + with respect to x we get:

2 . ' 6 16 1'

2

y y xxy

y

= ++⇒ =

At ( )2, 10− , 12 1 11 11 10'202 10 2 10

y − += = − = −×

10. A Let the base length of the right isosceles triangle be x.

Then length of hypotenuse is given by 2x .

We are given that length of hypotenuse is 7 6

2 7 6

7 3

x

x

⇒ =

⇒ =

Area of right isosceles triangle with base length x is given

by 212

x .

Therefore, required area ( )21 1477 3 73.52 2

= × = = sq.

units

11. E Interest earned in five years ( )$ 140 50 $90= − =

. .100. . 100

90 100 36%50 5

P R TS I

S IRP T

R

× ×=

×⇒ =××⇒ = =×

12. C The ratio of areas of similar triangles is square of the ratio of their corresponding sides. In this case ratio of areas

22 45 25

⎛ ⎞= =⎜ ⎟⎝ ⎠ or 4 : 25 .

13. D The distance between ( )4, p and ( )1,0 is given to be 5.

( ) ( )22

2 2

2

2

4 1 0 5

3 59 25

164

p

pp

pp

⇒ − + − =

⇒ + =

⇒ + =

⇒ =⇒ = ±

But, ( )4, p lies in fourth quadrant. Therefore p must be negative.

Hence 4p = − .

14. D Area of rhombus 12

= ×product of diagonals


Length of diagonal AC

( ) ( )2 22 4 10 8 36 324

360 6 10

= − − + − − = +

= =

Length of diagonal BD

( ) ( )2 24 2 2 0 36 4

40 2 10

= + + − − = +

= =

Area 1 6 10 2 10 602

= × × = sq. units

15. A The volume of a pyramid is given by 13

Bh , where B is

area of base and h is the height.

Area of base 2150 150 22,500 m= × =

Volume of sand in pyramid = 80% of volume of pyramid

Volume of sand 380 1 22,500 80 480,000 m100 3

= × × × = .

16. B Using the theorem for intersecting chords, we get

( )

( )( )

2

2

2

1 3 103030 0

6 5 30 06 5 06 or 5

x xx xx xx x xx x

x x

+ = ×

⇒ + =

⇒ + − =

⇒ + − − =⇒ + − =⇒ = − =

Since x is a length and hence cannot be negative, therefore 5x = is the answer.

17. D When three dice are rolled once, the sample space contains 6 6 6 216× × = equally likely events.

However, only 6 of these are favorable to the event, “sum of numbers is 5”, i.e. ( ) ( ) ( ) ( ) ( )1,1,3 , 1,3,1 , 3,1,1 , 1, 2, 2 , 2,1, 2 and ( )2,2,1 .

Hence, required probability 6 1216 36

= =

18. E Determining the slopes of each side gives us:

4 3 16 4 26 4 25 65 6 13 5 25 3 23 4

AB

BC

CD

DA

−= =−−= = −−−= =−−= = −−


However, we need to specify what type of parallelogram is this. Since the slopes of adjacent sides are negative reciprocals of each other, therefore the angle between them is 90o . Hence, this figure is either a square or a rectangle.


6 3 35 45 4 13 6 3

AC

BD

−= =−−= = −−

The slopes of diagonals are negative reciprocals of each other, which means the diagonals are perpendicular to each other. Hence, this figure is a square.

19. C Volume of sphere 34 3 363

π π= × × =

Volume rise in the vessel 24 16h hπ π= × × = ×

Since the sphere is completely immersed in the vessel, the volume rise in the vessel is equal to the volume of sphere

16 369 cm4

h

h

π π⇒ × =

⇒ =

20. B The perpendicular from the origin cuts the given line A at the point ( )2,9−

Therefore slope of perpendicular from origin 9 0 92 0 2−= = −

− −

The two lines are perpendicular therefore their slopes are negative reciprocals of each other.

Hence required slope 29

=

21. E Two movements in the x-y-z coordinate plane , ,a b c< > and , ,x y z< > are perpendicular to each

other when 0ax by cz+ + = .

Putting the answer choices one by one, we see that only (E) satisfies the equation as:

( ) ( )( )21 1 9 3 3 2 21 27 6 0× + × − + − − = − + = .

22. A The number is three digit, therefore in hundred’s place we cannot have 0. Also, no digit can be 5. Thus, in hundred’s place 0 and 5 are not to be placed. In ten’s and unit’s place, 5 is not to be placed.

We are left with 8 possibilities for hundred’s place and 9 possibilities each for ten’s and unit’s place.

⇒ Required number = 8 9 9 648× × =

23. A Slope of chord joining ( )2, 0 and ( )4, 4 4 0 24 2−= =−


Equation of given curve is ( )22y x= − . Slope of tangent

to the curve is given by ( )2 2dy xdx

= −

For the point at which the tangent is parallel to chord joining given points, we must have

( )2 2 22 13

xxx

− =⇒ − =⇒ =

When 3x = , then ( )23 2 1y = − =

Thus, the required point is ( )3,1

24. C The word ENGINEERING has 11 letters of which 3 are E’s; 3 are N’s; 2 are G’s; 2 are I’s and 1 is R.

Hence the required number of permutations 11! 277,200

3! 3! 2! 2!= =

× × ×

25. C Volume of a barrel ( ) 3 33.14 0.25 0.25 7 cm 1.374 cm= × × × =

Volume of ink in bottle 3200 cm=

Total number of barrels that can be filled from the given

volume of ink 2001.374

=

So, required number of words 200 330 48,0351.374

= × =

26. A The midpoint of the line joining ( )3, 4 and ( ), 7k is

given by 3 4 7,2 2

k+ +⎛ ⎞ ⎜ ⎟⎝ ⎠. We are given that this point is

equal to ( ),x y .

32

kx +⇒ = and 112

y = . Putting these values in

2 2 1 0x y+ + = we get 3 11 1 0k+ + + =

15 0k⇒ + =

15k⇒ = −

27. B Equation of line joining the points ( )1,3 and ( )2,6− is

given by ( )6 33 12 1

y x−− = −− −

or 4 0x y+ − =

The distance d of a point ( )1 1,x y from the line


2 2

ax by cd

a b

+ +=

+.

Using this formula we get the distance of ( )0, 1 − from the line 4 0x y+ − =

( )2 2

0 1 4 5 221 1

+ − −= =

+

28. B The graph of 2 2y x= − + cuts the x-axis when 0y =

2

2

2 02

2

xx

x

⇒ − + =

⇒ =

⇒ = ±

However 2 0 2 20 2 2

− = − = −−

falls on the negative x-

axis. Hence, the required first point is ( )2, 0 .

Also, the graph cuts the y-axis when 0x = , i.e. at ( )0, 2 .

Average rate of change between ( )2, 0 and ( )0, 2 is

equal to 2 0 2 20 2 2

− = − = −−

.

29. A Using the theorem for intersecting secants, we have

( ) ( )8 8 7 7 191828 22.75

822.75 8 14.75

QC

QC

QC

+ = +

⇒ + = =

⇒ = − =

30. A Sum of interior angles in an octagon ( )8 2 180 1080= − × =o o

Sum of three angles whose average is given 3 107 321= × =o o

Sum of remaining five angles 1080 321 759= − =o o o

Average of other five angles 759 151.85

= =o

o

31. D Let principal amount be x and interest rate be r%

Then S.I. 2 50100

2,500

x r

x r

× ×= =

⇒ × =

Also, C.I. 2

1 52100

rx x⎛ ⎞= + − =⎜ ⎟⎝ ⎠

2

2

1 5210,000 50

5210,000 50

1 5210,000 50

r rx x

r rx

rxr

⎛ ⎞⇒ + + − =⎜ ⎟

⎝ ⎠⎛ ⎞

⇒ + =⎜ ⎟⎝ ⎠⎛ ⎞⇒ + =⎜ ⎟⎝ ⎠

Putting 2,500xr = , we get:


12,500 5210,000 50

50 524

8%

r

r

r

⎛ ⎞⇒ + =⎜ ⎟⎝ ⎠

⇒ + =

⇒ =

32. E Each side of the regular hexagon is tangent to the circle with center O. The length of the line joining the center of the regular hexagon is equal to the side of the hexagon since the triangle formed by joining any two vertices and the center form an equilateral triangle.

Also we know that the tangent is perpendicular to radius at point of contact. Thus, using trigonometry we find the length of radius OB of the circle as

sin 60

3 5 352 2

OB OA

OB

=

= × =

o

Thus, area of circle 2

2 5 3 752 2

r ππ π⎛ ⎞

= = × =⎜ ⎟⎝ ⎠

33. D Let the length of side of square and decagon be x.

Then, area of square is equal to 2x .

Cost of painting area 2x is $84.

⇒ Cost of painting unit area 2

84$x

=

Area of decagon ( )

( )

2

2

2

2 2

2 180tan

4 2

8 18010 tan4 20

5 tan 7225 3.08 7.72

nsnn

x

x

x x

⎛ ⎞− ×= × ⎜ ⎟

⎝ ⎠⎛ ⎞×= × ⎜ ⎟⎝ ⎠

=

= × =

o

o

o

Cost of painting the decagon 22

847.7 $646.8xx

= × =

34. D Suppose taking a step forward is considered a success and backward be regarded as a failure.

Then we have 0.3p = , 0.7q = and 9n = .

Gary will be one step away from the starting point if he takes

(i) 5steps forward and 4 steps backwards

(ii) 4 steps forward and 5 steps backwards

Hence, the required probability

( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( )

5 4 4 5

5 4 4 5

4 4

4 4

5 49 9

0.3 0.7 0.3 0.75 4

9 9 9 90.3 0.7 0.3 0.7

5 5 4 5

90.3 0.7 0.3 0.7

59

0.3 0.75

P P= +

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + ∴ =⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦⎛ ⎞

= +⎜ ⎟⎝ ⎠⎛ ⎞

= ⎜ ⎟⎝ ⎠

35. D First 10 letters of the alphabet are a, b, c, d, e, f, g, h, i, and j. These contain 3 vowels and 7 consonants.

One of the letters must be vowel which can be selected in

3 ways. The other 2 letters can be selected in 72⎛ ⎞⎜ ⎟⎝ ⎠

ways

since repetition is not allowed.

These 3 selected letters can then be re-arranged in 3! ways.

Thus total number of words 7 6 3 7 6 63 3! 378

2 2× × × ×= × × == =

ONMATHEMATICS Comprehensive Exam 4 Answer Guide 1. B ABCΔ and DEFΔ are similar triangles. Hence, their

corresponding angles are equal, which gives us 83B E∠ = ∠ = o . Using angle sum property in ABCΔ ,

180 47 83 50C∠ = − − =o o o o

2. E Two tangents of a circle can be parallel only if they touch the end points of the same diameter. Thus, the distance between them is equal to the length of the diameter equal to 12 cm.

3. C The sum of interior angles is given by ( )2 180n− o.

The sum of interior angles of a 29-sided figure is

= 29− 2( )×180 = 27 ×180 = 4860 .

4. C The first person has 5 choices for a hat. Based on his choice, the second person is left with 4 hats to choose from since their purchases need to be different. Hence total number of ways in which the hats can be chosen are 5 4 20× = .

5. E The given sample space can have infinitely many items as it is possible for 6 to never show up.

6. B Area of octagonal base 2Volume of prism 176 16 cm

Height of prism 11= = =

7. C In a right isosceles triangle both the legs are equal. If the length of each leg is x (say), then the area of the triangle is

equal to 212

x

21 492

7 2

x

x

⇒ =

⇒ =.

The length of the hypotenuse in a right isosceles triangle is 2 times the length of the leg.

Length of hypotenuse 2 7 2 2 14x= = × =

8. C DEFΔ is similar to ABCΔ . Hence their sides are proportional which gives us the relation, AB BC CADE EF FD

= =

The proportionality ratio of the sides is given by 2 14 2

BCEF

= = . Using this ratio we find the length of other

sides of triangle DEF as 6 cmDE = and 5 cmFD = .

Perimeter of DEFΔ ( )4 5 6 cm 15 cm= + + = .

9. A Since a particular player is sure to be the captain, we need to select the remaining 10 players of the team out of remaining 13 candidates. This can be done in

13 13 13 12 11 28610 3 3 2 1⎛ ⎞ ⎛ ⎞ × ×= = =⎜ ⎟ ⎜ ⎟ × ×⎝ ⎠ ⎝ ⎠

ways.

10. D Determining the slopes of each side gives us:

( )

( )( )

2 5 76 52 2

07 6

5 2 74 7 3

5 5 04 5

PQ

QR

RS

SP

− −= = −−

− − −= =

− −− −

= =− − −−= =

− −

Since only one pair of opposite sides is parallel, this figure is a trapezoid.

11. E

2161

3

1 11 16 6

1 16 6

116

16

1

11lim lim1 1

1 1lim

1

lim 1 1 1 2

x x

x

x

zz

z z

z z

z

z

→ →

→

→

⎛ ⎞−⎜ ⎟

− ⎝ ⎠=− −

⎛ ⎞⎛ ⎞− +⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠=−

⎛ ⎞= + = + =⎜ ⎟

⎝ ⎠

12. D The direction perpendicular to a plane is given by the coefficients of x, y and z in the equation of the plane.

Thus, 3 2 7 8x y z+ − = is perpendicular to the direction 3,2, 7< − > .

13. A When two secants intersect outside a circle, the angle between them is half of the difference between the angles of the intercepted arcs.

2 2

82372 2

8

a a

a

a

m BC m DEBAC

m DE

m DE

⇒∠ = −

⇒ = −

⇒ =

oo

o

14. D The average rate of change between any two points is equal to the slope of the line joining the two given points.


( ) ( )3

3

2 78 7

f p pp

= −

= −

( ) ( )3

3

2 78 7

f p pp

− = − −

= − −


Thus we need to find the slope between ( )3, 8 7p p −

and ( )3, 8 7p p− − − which is equal to 3 3 3

28 7 8 7 16 82

p p p pp p p

− − − + = =− −

.

15. D The figure can be more precisely drawn as below

Using Pythagoras theorem we can find the height of trapezoid from triangle ADM as:

2 210 5 100 25

75 5 3

DM = − = −

= =

Now, Area of trapezoid 12

= × sum of parallel sides×

height

1 30 5 32

= × ×

75 3= sq. units

16. B The largest sphere that can be cut off from the cylindrical log will have the same radius as the base of the cylinder, i.e. 1cm

Volume of sphere ( )34 413 3π π= =

17. B Let the principal amount be p and rate percent be equal to x.

Rate percent is equal to time, therefore time is also x.2

. .100 100 100

p x x pxP R TS I × ×× ×= = =

Also simple interest is 916

of the sum.

2

2

916 100

900 22516 4

15 7.52

p px

x

x

⇒ =

⇒ = =

⇒ = =

Hence, rate is equal to 7.5% and time is 7.5 years

18. E Distance travelled by the first boat in 1.5 hours 1.5 100 150= × = km

Distance travelled by second boat in 1.5 hours 1.5 120 180= × = km

The boats travel at right angle to each other. Therefore, the distance between them can be calculated using Pythagorean theorem as:

Distance 2 2150 180 54900 234.3= + = = km

19. B The diagonals of a parallelogram bisect each other.

Therefore the midpoint of the diagonals is the same.

For the given parallelogram ABCD, we have

Midpoint of AC = Midpoint of BD

15 7 3 3 1, ,2 2 2 2

15 ,5 2,2 2

15 2 and 52 29 and 9

9 9 0

yx

yx

yx

x yx y

+− + + +⎛ ⎞⎛ ⎞⇒ = ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠+− + ⎛ ⎞⎛ ⎞⇒ = ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

+− +⇒ = =

⇒ = =⇒ − = − =

20. D Let the points ( )0,0A , ( )1,0B and ( )0,1C be the vertices of triangle ABC. Using distance formula we get the length of the sides of the triangle as

1AB = unit

2BC = units

1AC = unit

Thus the perimeter of the triangle ABC

1 2 1 2 2AB BC AC= + + = + + = + units.

21. B Volume of spherical bullet 3

34 5 125 cm3 2 6π π⎛ ⎞= =⎜ ⎟⎝ ⎠

Volume of rectangular block of metal 3110 100 50 cm= × ×

Number of bullets cast Volume of metal block

Volume of spherical bullet110 100 50

1256

110 100 50 6125 3.14

8,408 (approximately)

π

=

× ×=

× × ×=×

=

22. E If 10 more red balls are added, total number of balls become 30.

And, the total number of red balls become 10p + .


New probability of getting a red ball 1030

p + .

We are given that new probability is double that of 20p

10 2

30 2010 3

2 105

p p

p pp

p

+⇒ = ×

⇒ + =⇒ =⇒ =

23. D The given digits are 0, 1, 2, 3 and 4 which are five in number. Since we are to form the numbers that are greater than 20000 and as no digit is to be repeated, every such number must start with either of 2, 3, or 4. Thus, the extreme left place can be filled in 3 ways only.

Now the remaining four places can be filled up with the remaining four digits in 4 4P ways. By using the principle of association, the required number of numbers formed are

4 43 3 4! 72P= × = × = .

24. C Given that ( ) 2 9 8f x x x= + − .

Number of diagonals ( ) ( )27 7 9 7 8 49 63 8 104f= = + − = + − = .

Number of diagonals in n-sided polygon are ( )3

2n n −

( )

( )( )

2

3104

23 208 016 13 016 or 13

n n

n nn n

n n

−⇒ =

⇒ − − =⇒ − + =⇒ = = −

But n cannot be negative, therefore 16n = is the answer.

25. C When the tangent makes an angle of 45owith x-axis, then the inclination of the tangent with positive x-axis is either 45o or 135o i.e., the slope of tangent is either 1 or –1. But, we are given that slope is positive. Hence slope of tangent is 1.

Slope of tangent to the curve 2y x= is given by 2x.

2 1x⇒ = or 12

x =

Putting 12

x = in 2y x= gives us 14

y = .

Thus the required point is 1 1,2 4

⎛ ⎞ ⎜ ⎟⎝ ⎠.

26. E Let the length of segment CD be x.

It is known that perpendicular from center to the chord bisects the chord

5AD BD⇒ = = cm.

In ,AODΔ 5AD = cm, 7AO = cm and 90ADO∠ = o .

Using Pythagorean theorem, we get: 2 2 2

2 2 2

2

7 549 25 24

24 cm

AO AD ODOD

OD

OD

= +

⇒ = +

⇒ = − =

⇒ =

7 24CD⇒ = − cm

27. D Let the slope of line be x. Then it’s perpendicular has

slope equal to 1x

− . Given that the sum of slopes is 5524

2

2

2

1 5524

1 5524

24 24 5524 55 24 0

xx

xxx xx x

⇒ − =

−⇒ =

⇒ − =

⇒ − − =

( ) ( )

224 64 9 24 08 3 8 3 3 8 0

8 3 or 3 8

x x xx x x

x

⇒ − + − =⇒ − + − =

⇒ = −

28. D The given curve is 3y x=

23dy xdx

⇒ =

According to the given condition, we have

( )

2 3

2

33 00 or 3

dy ydx

x xx xx x

=

⇒ =

⇒ − =⇒ = =

When 0x = , then 30 0y = = . Hence the required point is ( )0, 0

When 3x = , then 33 27y = = . Hence the required point is ( )3, 27 .Looking at the options the answer is

( )3, 27


29. B Area of a regular nonagon is given by: ( )

( )

2

2

2

2 180tan

4 2

8 7 1809 tan4 18

144 tan 70144 2.75396 m

nsnn

⎛ ⎞− ×× ⎜ ⎟

⎝ ⎠⎛ ⎞×= × ⎜ ⎟⎝ ⎠

== ×

=

o

o

o

Amount of paper wasted 2708 396 312 m= − =

30. A Given amount is x. Let r be the rate percent per year. When interest is compounded annually for 15 years, the resulting amount becomes eight fold, i.e. 8x

15

1 8100

rx x⎛ ⎞⇒ + =⎜ ⎟⎝ ⎠

15

1 8100

r⎛ ⎞⇒ + =⎜ ⎟⎝ ⎠…..(1)

When interest is compounded for 5 years, the resulting

amount is 5

1100

rx ⎛ ⎞+⎜ ⎟⎝ ⎠

115 3

1100

rx⎡ ⎤⎛ ⎞= +⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

Using the result from equation (1) above we get:

[ ]1

15 3 131 8 2

100rx x x

⎡ ⎤⎛ ⎞+ = =⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

31. C The distance d of a point ( )1 1,x y from the line


2 2

ax by cd

a b

+ +=

+.

Using this formula we get the distance of ( )1,1− from the

line ( ) ( )12 6 5 2x y+ = − or 12 5 82 0x y− + =

( ) ( )( ) ( )2 2

12 1 5 1 82 65 51312 5

− − += = =

+ −

32. C When a right angled triangle with sides 3 cm, 4 cm and 5 cm is revolved about the side of length 4 cm, we get a cone with base radius 3cm, height 4cm and slant height 5 cm.

Surface area of given cone ( )( )

2

3 5 3 24rl r r l rπ π π

π π= + = += × + =

33. B Let the radius of the circle be r. Then length of segment EC is equal to 2r. Using the theorem for intersecting secants, we get:

( ) ( )5 5 2 4 4 1625 10 8010 55

5.5

rr

rr

+ = +⇒ + =⇒ =⇒ =

Now, circumference of circle 2 2 5.5 11rπ π π= = × =

34. E This is a case of Binomial Distribution with 5n = , where success is guessing a correct answer to a multiple choice question with 3 options.

There are three possible answers to each question, therefore the probability of guessing an answer correct

just by guessing is 13

p =

And hence, 1 213 3

q = − =

Required probability =P (four or more correct answers)

( ) ( )4 1 5 0

4 5

5 5 5

4 55 51 2 1 24 53 3 3 35 51 2 1 5 2 1 11 111 03 3 3 3 3 3 243

P P= +

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞ ×⎛ ⎞ ⎛ ⎞= + = + = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

35. D The given Venn-diagram has 8 regions as shown below.

Each of these regions can either be shaded or not. Hence there are 2 choices for each region. Hence total number of choices for 8 regions are

82 2 2 2 2 2 2 2 2 256× × × × × × × = = .

MATHEMATICS Comprehensive Exam 5 Answer Guide 1. C Tangent is perpendicular to radius at the point of contact.

Hence, triangle AOB is a right angled triangle. Using Pythagorean theorem, we get:

2 28 15 17AO = + =

2. D In each place we can either use a letter or a number. Thus total choices for each place are 26 10 36+ = .

Since each digit has to be distinct, therefore total choices are 36 35 34 42,840= × × = .

3. E Rotating a line by 180o has no effect in its orientation, hence its slope remains same.

Thus, rotating a line by 270o is the same as rotating it by 90o . Thus the new slope is negative reciprocal of the

original 94

= − .

4. C Let the length of the shorter leg of triangle be a. Then length of longer leg is 3a . Thus area of the triangle is

21 332 2

a a a× × = .

2

2

3 49 32 8

494

72

a

a

a

⇒ =

⇒ =

⇒ =

Length of hypotenuse in a 30 60 90− −o o o triangle is twice the length of shorter leg. Thus, here length is 2a

72 72

= × =

5. A

( )

2 2

2

1 1 22 2lim lim2 2

1lim2

1 12 2 4

x x

x

xx xx x

x

→− →−

→−

++=

+ +

=

= = −−

6. B Ratio of volume of two spheres

3 31

1

3 22

4343

r rrr

π

π

⎛ ⎞= = ⎜ ⎟

⎝ ⎠

We are given that ratio of volumes is 125:8

Therefore, 1

2

52

rr

⇒ =

Ratio of surface areas 2 22

1 12

2 2

4 5 254 2 4

r rr r

ππ

⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

7. B Slope of tangent to the curve 3 54 2y x x= − is given by

2 412 10dy x xdx

= − .

At ( )2, 32− slope is equal to

( ) ( )2 412 2 10 2 48 160 112dydx

= − = − = − .

Tangent and normal are perpendicular to each other.

Therefore, slope of normal is equal to 1112

.

8. C Let the sum be x and rate be r%. We are given that the sum doubles itself, that is, the interest after 10 years is equal to the sum itself.

10. .100

1010010%

x rS I

x r x

r

× ×⇒ =

× ×⇒ =

⇒ =

Now we need to find the time in which the sum becomes 4x, i.e. the interest generated must be equal to 3x.

10. .100

10 3100

x tS I

x t x

× ×⇒ =

× ×⇒ =

30t⇒ = years

9. D When two chords intersect in a circle, the angle between them is equal to the average of the arc angles formed by the chords.

110 262 2

55 13 68

AOC

AOC

⇒∠ = +

⇒∠ = + =

o o

o o o

Also, 180AOC BOC∠ +∠ = o

180 68 112BOC⇒∠ = − =o o o

10. E Let the other end of the diameter of the circle be ( ),a b .

The center ( )2,1 is the midpoint of the diameter. Therefore, we use the midpoint formula to determine the values of a and b.

By midpoint formula center of circle is 7 1,2 2

a b+ −⎛ ⎞ ⎜ ⎟⎝ ⎠

7 22

a +⇒ = and 1 12

b − =

3a⇒ = − and 3b = .

Thus, other end of diameter is ( )3,3− .


11. A The given curve is 3y x= . Slope of tangent to curve is given by 23x . At 2x = , slope is equal to 23 2 12× = .

Equation of tangent is given by

( )8 12 28 12 2412 16

y xy xy x

− = −⇒ − = −⇒ = −

.

12. A Sum of interior angles of a 15-sided polygon( )15 2 180 2340= − × =o o

13. C Two movements in the x-y-z coordinate plane , ,a b c< > and , ,x y z< > are perpendicular to each

other when 0ax by cz+ + = .

Putting the answer choices one by one, we see that only (C) satisfies the equation as:

( ) ( ) ( )( )1 4 8 1 3 4 4 8 12 0− × + × − + − − = − − + = .

14. D Each question can be marked with any 1 choice out of 5. Hence, there are 5 choices for each question. Similarly for 10 questions the total number of different response sheets is equal to 105 5 5 5 5 5 5 5 5 5 5× × × × × × × × × = .

15. C The longest rod has the length equal to the diagonal of the cube given by 2 2 2length + breadth + height .

Length of diagonal ( )

2 2 27 4 11 49 16 121

186 16.4 m approx

= + + = + +

= =

16. D There are a total of 12 balls. Out of these 2 balls can be

selected in 122

⎛ ⎞⎜ ⎟⎝ ⎠

ways.

The number of ways of drawing one white ball out of 5 is

equal to 51⎛ ⎞⎜ ⎟⎝ ⎠

. Similarly, the number of ways of drawing

one red ball out of 7 is equal to 71⎛ ⎞⎜ ⎟⎝ ⎠

.

Hence, P (one white and one red ball)

5 71 1 5 7 1 2 35

12 1 1 12 11 662

⎛ ⎞ ⎛ ⎞×⎜ ⎟ ⎜ ⎟

×⎝ ⎠ ⎝ ⎠= = × × =×⎛ ⎞

⎜ ⎟⎝ ⎠

17. D The two prisms are similar, hence their proportionality ratio, r is given by the ratio of their base lengths

3 67 72

= =

The ratio of their surfaces areas is equal to square of the proportionality ratio, r.

Hence, ratio of surface areas 2

2 6 367 49

r ⎛ ⎞= = =⎜ ⎟⎝ ⎠

18. E The ratio of the volumes is given by the cube of the proportionality ratio, r i.e. 3r

Hence, ratio of volumes 3

3 6 2167 343

r ⎛ ⎞= = =⎜ ⎟⎝ ⎠

19. A Let the points ( )0,0A , ( )3, 3B and ( )3,C λ be the

vertices of equilateral triangle ABC.

Length 223 3 9 3 12AB = + = + =

Length 2 2 23 9AC λ λ= + = +

As the length of each side of equilateral triangle is same, therefore length AB = length AC

2

2

9 129 12

3

λλ

λ

⇒ + =

⇒ + =

⇒ = ±

But λ cannot be equal to 3 because then the points B and C would coincide and it would no longer be a triangle. Hence 3λ = − .

20. D To form a number between 100 and 1000, we are to fill up three places, the hundred’s place, the ten’s place and the unit’s place.

The hundred’s place can be filled up using any of the non-zero numbers, i.e. in 9 ways. The ten’s place can also be filled up in 9 ways using any of the nine remaining digits . Corresponding to this, the unit’s place can be filled up in 8 ways using any of the remaining eight digits.

Thus, the three places can together be filled up in 9 9 8 648× × = ways.

21. A The base of the largest cone will be the circle inscribed in a face of the cube and its height will be equal to an edge of the cube.

9radius of base of cone cm2

Height of cone 9 cm

r

h

∴ = =

= =

Hence, volume of cone 213

r hπ=

3

3

1 9 9 9 cm3 2 2

243 cm4

π

π

⎛ ⎞= × × × ×⎜ ⎟⎝ ⎠

=

22. E Determining the slopes of each side gives us:


( )

7 3 41 6 52 7 53 1 42 2 4

2 3 52 3 5

2 6 4

AB

BC

CD

DA

−= = −−−= =

− −− −= = −− −

− −= =−

Since the opposite sides of the quadrilateral are parallel, this figure is a parallelogram. Also, the slopes of adjacent sides are negative reciprocals of each other, therefore the angle between them is 90o . Hence, this figure is either a square or a rectangle.


2 3 13 6 92 7 92 1

AC

BD

−= =− −− −= = −

−

The slopes of diagonals are negative reciprocals of each other, which means the diagonals are perpendicular to each other. Hence, this figure is a square.

23. D Total number of ways of arranging the letters of the word

MUMMY are 5!3!

.

When the letter’s at extreme ends are M’s, the remaining 3 letters in between can be arranged in 3! ways.

Hence, required probability 3! 6 6 35! 120 103!

×= = =

24. B Let the side length of heptagon and decagon be x.

Area of a regular heptagon is given by: ( )

( )

2

2

2

2

2 180tan

4 2

5 1807 tan4 14

7 tan 64.343.64

nsnn

x

x

x

⎛ ⎞− ×× ⎜ ⎟

⎝ ⎠⎛ ⎞×= × ⎜ ⎟⎝ ⎠

=

=

o

o

o

Similarly we find the area of a regular decagon which comes out to be equal to 27.7x

Ratio 2

2

3.64 267.7 55

xx

= = .

25. C The diagonals of a rhombus bisect each other and are perpendicular to each other.

Let BD be the given diagonal. It is bisected at the point O.

Therefore BO = 8 cm.

We use Pythagorean theorem in right triangle AOB as: 2 2 2AB AO BO= +

2 210 8 36 6AO⇒ = − = =

Length of diagonal AC = 2× length AO = 12cm

26. B Amount of cloth needed

2 2

2

2

surface area of hemisphere surface area of base amount of cloth wasted

2 33 33 3.14 21 21 34,154.22 3 4,157.22 m

r rr

π ππ

=++

= + +

= += × × × +

= + =

27. C The average rate of change between any two points is equal to the slope of the line joining the two given points.


22 3 3sin3 3 2 4

f π π ⎛ ⎞⎛ ⎞ ⎛ ⎞= = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

223 3 1 1sin4 4 22

f π π ⎛ ⎞⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Thus we need to find the slope between 3,3 4π⎛ ⎞ ⎜ ⎟⎝ ⎠

and

3 1,4 2π⎛ ⎞ ⎜ ⎟⎝ ⎠

which is equal to

1 3 132 4 4

3 5 54 3 12π π π π

−= =

−.

28. A The distance from the point ( )1 1 1, ,x y z to the plane

0Ax By Cz D+ + + = is 1 1 1

2 2 2

Ax By Cz D

A B C

+ + +

+ +.


Required distance ( )

( )22 2

5 8 3 7 4 2 9 40 21 8 9505 3 4

20 4 =2 2 units5 2 2

× − × + − + − − +=

+ − +

= =

29. B Population growth is a type of compound interest problem.

We are given that the population grew from 1 million to 4 million in 20 years or 2 decades. Let the rate growth be r% per decade.

Starting with an initial population of 1 million, the

population after 2 decades is given by 2

1 1100

r⎛ ⎞+⎜ ⎟⎝ ⎠million

However we are given that final population is 4 million 2

1 4100

1 2100

1100

100%

r

r

r

r

⎛ ⎞⇒ + =⎜ ⎟⎝ ⎠⎛ ⎞⇒ + =⎜ ⎟⎝ ⎠

⇒ =

⇒ =

Hence, rate of population growth per decade is 100%.

30. E We have to select one king out of 4, which can be done in 41⎛ ⎞⎜ ⎟⎝ ⎠

ways. The other four other cards are to be selected

out 48 cards. This can be done in 484

⎛ ⎞⎜ ⎟⎝ ⎠

ways.

⇒ Required number of ways 4 481 4

4 194,580778,320

⎛ ⎞ ⎛ ⎞= ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= ×=

31. B Number of diagonals of hexagon, a ( )6 6 3

92−

= =

Number of diagonals of heptagon, b ( )7 7 3

142−

= =

2 2 2 214 9 196 81 115b a⇒ − = − = − =

32. A Using the theorem for intersecting chords, we get

( )( )

( )( )

2

2

2

2

4 3 2 3 512 6 56 6 12 0

2 02 2 02 1 02 or 1

x x xx x x

x xx xx x xx x

x x

− + = ×

⇒ − − =

⇒ + − =

⇒ + − =

⇒ + − − =⇒ + − =⇒ = − =

Since x is a length and hence cannot be negative, therefore 1x = is the answer.

33. A Using the theorem for intersecting secants for AB and AC, we get:

( ) ( )2

2

11 5 5 711 6011 60 011 or 4

AE AEAE AEAE AEAE AE

+ = +

⇒ + =

⇒ + − =⇒ = − =

Since length cannot be negative, therefore 4AE = .

Now using Pythagoras theorem in right triangle AEP,

2 2 2 24 12

16 144 160 4 10

EP AE AP= + = +

= + = =

34. C This is a case of Binomial Distribution with 12n = , where success is ‘a defective item’.

Here 10 1100 10

p = =

And hence, 1 91 110 10

q p= − = − =

Required probability =P (9 items are defective)

( )9 3 9 3

3

12 12 12 11

912 121 9 1 99 310 10 10 10

12 11 10 9 220 729 160,380 16,0381 2 3 10 10 10 10

P=

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠× × ×= × = = =× ×

35. E Let r be the radius and h be the height of the original cone.

Volume of original cone 213

r hπ=

If a cone is cut into two parts through the mid-point of its axis, we get two objects, a smaller cone and a frustum . The smaller cone has half the height and half the radius of the original cone.

Volume of new (smaller) cone 2

21 1 13 2 2 8 3

r h r hπ π⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


Hence, required ratio

2

2

1 118 3

1 83

r h

r h

π

π

⎛ ⎞⎜ ⎟⎝ ⎠ =

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