1) 55°

2) cyclic quadrilateral

3) 40 m

4) 6 cm

5) k = 9

6) 2029√cm

7) 13

8) 120°

9) acute

10) 3, 4, 5

Class - X

ANSWER KEYMATHEMITCS SA II MOCK TEST (2010 - 11) - 1

TEST ID : MATHX01

Question 11 :Find the discriminant of the quadratic equation 2x2 − 4x+ 3 = 0, and hence find thenature of its roots.

Suggested MarksSuggested Answers

Part Answer1

The given quadratic equation is 2x2 − 4x+ 3 = 0∴ Discriminant D( ) = −4( )2 − 4 ·2 ·3= 16 − 24= − 8

Part Answer2

Since D = − 8 < 0, so the given quadratic equation has no real roots

Question 12 : Find the sum of the first 100 terms of the following arithmetic progression1 + 3 + 5 + 7 + . ......

Suggested Answers

Part Answer1

Here, first term (a) = 1Common difference (d) = 3 − 1 = 2, n = 100The sum of first n terms is

Sn = n2⎡⎣2a+ n− 1( )d⎤⎦

Question 11 :

Question 12 :

Part Answer2

S100 = 1002⎡⎣2×1 + 100 − 1( )2⎤⎦

= 50[2 + 198]= 10,000∴ The sum of the first 100 terms of the given A.P is 10,000

Question 13 : A card is drawn at random from a pack of 52 cards. Find the probability that carddrawn is red and a king.

Suggested MarksSuggested Answers

Part Answer1

Total number of possible outcomes = 52Number of favourable outcomes = 2

Part Answer2

∴ P(Red and a king card) = 252 = 1

26

Question 14 :A spherical ball of diameter 21 cm is melted and recasted into cubes, each of side 1cm.Find the number of cubes thus formed.

Suggested MarksSuggested Answers

Part Answer1

Diameter of spherical ball = 21cm∴ Radius of spherical ball = 21

2 cm

∴ Number of cubes = = Vol. of spherical ballVol. of one cube

Part Answer2

=43×22

7 ×212 ×21

2 ×212

1×1×1= 4851

∵ Volueme of sph. = 4 ∕ 3πr2

Volume of cube = side( )3

Question 15 : Find a point on the y-axis which is equidistant from the point A (6, 5) and B (-4, 3).Suggested Answers

Question 13 :

Question 14 :

Question 15 :

Suggested MarksPart Answer1

Let P (0, y) be any point on y-axisA (6, 5) and B(-4, 3) are given pointsPA = PB ...(Given)

∴ 0 − 6( )2 + y− 5( )2√ = 0 + 4( )2 + y− 3( )2√

Part Answer2

Squaring both sides, we get36 + y2 + 25 − 10y = 16 + y2 + 9 − 6yy2 − 10y− y2 + 6y = 25 − 61−4y = − 36 ⇒ y = 9∴ Coordinates of P are (0, 9)

Question 16 : Find the value of 'K' for which the points (7, -2), (5, 1) and (3, K) are collinear.

Suggested MarksSuggested Answers

Part Answer1

Let the coordinates of the vertices of Δ ABC be A (7, -2), B(5, 1) and C(3, K)Since the points are collinear∴ Area of Δ ABC = 012⎡⎣7 1 − K( ) + 5 K + 2( ) + 3 −2 − 1( )⎤⎦ = 0

Part Answer2

⇒ 12⎡⎣7 − 7K + 5K + 10 − 9] = 0

⇒ −2K +8 = 0 ⇒ −2 K = −8⇒ K = 4 ∴ Value of K = 4

Question 17 :The length of a tangent from a point A at distance 5 cm from the centre of the circle is4cm.

Find the radius of the circle.

Suggested MarksSuggested Answers

Question 16 :

Question 17 :

Part Answer1

∵ Tangent at any point to a circle is perpendicular to the radius at the point of contact

∴ ∠OTA = 90°In rt Δ OTAOT2 + AT2 = OA2

Part Answer2

OT2 + (4)2 = (5)2

OT2 + 16 = 25OT2 = 25-16 = 9∴ OT (radius) = 9√ = 3 cm

Question 18 :A circle is inscribed in a square of side 14 cm. Find the area of the square not includedin the circle.

Suggested Answers

Part Answer1

r = 142 = 7 cm

Area of shaded region= (side)2 − πr2

Question 18 :

Part Answer2

= 14( )2 − 227 ×7×7

= 196 - 154= 42 cm2

Question 19 :Two pipes running together can fill a more than the other to fill the cistern, find the timein which each pipe would fill the cistern.

Suggested MarksSuggested Answers

Part Answer1

Let the two pipes fill the cistern x and x + 5 minutesThen 1

x + 1x+5 = 1

6

Part Answer2

⇒ 6 x+ 5( ) + 6x = x x+ 5( )⇒ 6x+ 30 + 6x = x2 + 5x⇒ x2 + 5x− 6x− 30 − 6x = 0⇒ x2 − 7x− 30 = 0⇒ x2 − 10x+ 3x− 30 = 0

Part Answer3

⇒ x x− 10( ) + 3 x− 10( ) = 0⇒ x− 10( ) x+ 3( ) = 10⇒ x− 10 or x = − 3Since time cannot be−ve∴ Two pipes would fill the cistern in 10 and 10 + 5 i.e. 15 minutes.

Question 20 : Find the sum of first 40 positive integers divisible by 6. Also find the sum of first 20positive integers divisible by 5 or 6.

Suggested MarksSuggested Answers

Question 19 :

Question 20 :

Part Answer1

(i) Integers divisible by 6 are 6, 12, 18, ...Here a = 6, d = 12 -6 = 6, n = 40

Sn = n2⎡⎣2a+ n− 1( )d⎤⎦

= 402⎡⎣2×6 + 40 − 1( ).6⎤⎦

Part Answer2

= 20[12 + 39×6]= 20[12 + 234]= 20×246 = 4920

Part Answer3

(ii) Numbers which are divisible by 5 or 6 i.e. divisible by 30∴ Integers divisible by 30 are 30, 60, 90, ...

Here a = 30, d = 60-30 = 30, n = 20Using same formula, we get

S20 = 202⎡⎣2×30 + 20 − 1( )30⎤⎦

Part Answer4

= 10 [60 +19 × 30]= 10 [60 + 570]= 10 × 630 = 6300

Question 2 1 : Construct a ΔABC in which CA = 6 CM, AB = 5 cm and ∠BAC = 45°, thenconstruct a triangle similar to the given triangle whose sides are 6

5 of the corresponding sides of the

ΔABC.

Suggested MarksSuggested Answers

Question 2 1 :

Part Answer1

∴ ΔAB′C′ is the required Δ.

Question 22 : Prove that the tangents at the extremities of any chord make equal angle with thechord.

Suggested MarksSuggested Answers

Part Answer1

Given : PA and PB are the tangents to a circle from the point P and AB is a chord.

To prove:∠PAC = ∠PBCConst. : Join OP which meets AB at C.

Part Answer2

Proof : In ΔPAC and ΔPBCPA = PB ...[tangents from an external point are equal]∠APC = ∠BPC

[ ∵ PA and PB are equally inclined to OP]PC = PC [common]∴ ΔPAC ≅ ΔPBC [SAS]∴ ∠PAC = ∠PBC [c.p.c.t]

Question 23 :A square field and an equilateral triangular park have equal perimeters. If the cost ofploughing the field at the rate of Rs. 5/m2 is Rs. 720, find the cost of maintaining the park at the rate

of Rs. 10/m2.Suggested Answers

Question 22 :

Question 23 :

Suggested MarksPart Answer1

Let the side of the square be x mArea of the square = Total Cost

Rate per m2

x2 = 7205 = 144m2

x = 144√ = + 12m ∵ side can not be − ve( )

Part Answer2

Perimeter of square = 4x = 4(12) = 48 mLet side of Δ be y mPerimeter of a Δ = Perimeter of a square ...(Given)3y = 48 ∴ y = 48

3 = 16 m

Area of an equilateral Δ = 3√4 (side)2

= 3√4 y( )2

= 3√4 ×16×16 = 64 3√ m2

Part Answer3

Cost of maintaining the park @ Rs. 10 per m2

= 64 3√ ×10= 640×1.732… [ ∵ 3√ = 1.732]

= Rs. 1108.48

Question 24 :A cylinder of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball isdropped into the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball.

π = 22 ∕7⎛⎝

⎞⎠

Suggested Answers

Part Answer1

Radius of cylinder (r) = 12 cmHeight of water level raised (h) = 6.75 cmVolume of water raised = πr2h= π 12( )2 6.75( ) sq.cm

Question 24 :

Part Answer2

Suppose radius of sphere = r1

Volume of sphere = 43 πr1

3

Now Volume of sphere = Volume of water raised∴ = 4

3 πr3 = π 12( )2 6.75( )

Part Answer3

⇒ r3 = 12( )2 6.75( )×34

⇒ r3 = 144×20.254 = 729

⇒ r( )3 = 7293 √ = 9( )3 ∴ r = 9 cm

Question 25 :The angle of elevation of the top of a tower from a point on the ground, which is 30 maway from the foot of the tower is 30°. Find the height of the tower.

Suggested Answers

Part Answer1

Let AB be the tower of height h meters and C be a point on the ground such that the angle ofelevation of the top A of tower AB is of 30°.In Δ ABC, we are given ∠C = 30° and Base BC = 30 m and we have to find perpendicular AB.

Part Answer2

In ΔABC, taking tangent of ∠C , we have,tanC = AB

BC⇒ tan30° = AB

BC

Question 25 :

Part Answer3

⇒ 13√ = h

30

⇒ h = 303√ metres = 10 3√ metres

Hence, the height of the tower is 10 3√ metres.

Question 26 :Find the co-ordinates of the circumcentre of the triangle, whose vertices are (8,6),(8,-2) and (2,-2). Also find its circum-radius.

Suggested Answers

Part Answer1

Let A (8,6), B(8,-2) and C(2,-2). be the vertices of the given triangle and let P(x,y) be thecircumcentre of the triangle.

Then PA = PB = PC∴ PA2 = PB2 = PC2

Part Answer2

Taking PA2 = PB2

(x-8)2 + (y-6)2 = (x-8)2+ (y +2)2

(y - 6)2 = (y +2)2

y2 + 36 -12y = y2 + 4 + 4y-12y -4y = 4-36−16y = − 32 ⇒ y = 2

Question 26 :

Part Answer3

Again, taking PB2 = PC2

(x - 8)2 + (y + 2)2 = (x - 2)2 + (y + 2)2

(x - 8)2 = (x - 2)2

x2 + 64 - 16x = x2 + 4 -4x-16x + 4x = 4-64−12x = − 60 ⇒ x = 5∴ Coordinates of P are (5,2)

Part Answer4

Now circum-radius = PA = PB = PC ... (Proced)∴ Circum-radius (PA)

= 5 − 8( )2 + 2 − 6( )2√ = −3( )2 + −4( )2√= 9+16√ = 25√ = 5

Question 27 :A lot consists of 144 ball-pens of which 20 are defective and the others are good. Nuriwill buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at

random and gives it to her. What is the probability that(i) She will buy it?

(ii) She will not buy it?

Suggested MarksSuggested Answers

Part Answer1

Total ball-pens = 144Defective ball-pens = 20Good ball-pens = 144 - 20 = 124∴ Total number of events = 144

Part Answer2

P(she will buy a pen) = P (good ball-pens)= 124

144 = 3136

Part Answer3

P(she will not buy a pen) = P (defective ball-pen)= 20

144 = 536

Question 28 :Find the length of medians of Δ whose vertices are A(-1, 3), B(1, -1) and C(5, 1).

Question 27 :

Question 28 :

Suggested MarksSuggested Answers

Part Answer1

Let, D, E, F be the mid-points of BC, CA and AB respectivelyThen, co-ordinates of these points are

1+52 , −1+1

2⎛⎝

⎞⎠ ; 5−1

2 , 1+32

⎛⎝

⎞⎠

and −1+12 , 3−1

2⎛⎝

⎞⎠

i.e., D(3, 0), E(2,2) and F(0, 1) respectively

Part Answer2

∴ Median AD = 3 + 1( )2 + 0 − 3( )2√= 16 + 9√ = 25√ = 5 unitsMedian BE = 2 − 1( )2 + 2 + 1( )2√= 1 + 9√ = 10√ unitsand Median CF = 5 − 0( )2 + 1 − 1( )2√= 25√ = 5 units

Question 29 :The length of tangents drawn from an external point to a circle are equal. Prove it.Using the above, do the following :

PA and PB are tangents from P to the circle with centre O. At point M, a tangent is drawn cuttingPA at K and PB at N. Prove that KN = AK + BN.

Suggested Answers

Part Answer1

Part I: PT and PS are tangents from an external point P to the circle with centre O.

To prove : PT = PSConst. Join O to P, T and S

Question 29 :

Part Answer2

Proof : In ΔOTP and ΔOSPOT = OS ...(radii of same circle)OP = OP ...(common)∠OTP = ∠OSP … each 90°( )∴ ΔOTP ≅ ΔOSP … R.H.S( )∴ PT = PS … c.p.c.t.( )

Part Answer3

Part II :

Since the tangents drawn from an external point to a circle are equal in lengths∴ KA = KM (Form K) ...(i)NB = NM (From N) ...(ii)

Part Answer4

Adding (i) and (ii), we getKA + NB = KM + NM⇒ AK + BN = KN

Question 30 :In Fig. depicts a racing track whose left and right ends are semi-circular.

The distance between the two inner parallel line segments is 60 m and they are each 106 m long. Ifthe track is 10 m wide, find

(i) the distance around the track along its inner edge.(ii) the area of the track.

Suggested MarksSuggested Answers

Question 30 :

Part Answer1

Width of track = 10 mInner radius (r) = 60

2 = 30 cmOuter radius (R) = 10 + 30 = 40 mLength of rectangular part (l) = 106 mWidth of rectangular part (b) = 10 m

Part Answer2

(i) Distance around the track along its inner edge= 2l + 2 × circumference of inner semi-circle= 2 × 106 + 2 × πr

Part Answer3

= 212 + 2×227 ×30

= 212 + 13207

= 1484+13207 = 2804

7 m

Part Answer4

(ii) Area of the track = Area of 2 rectangular parts + Difference between 2 semi-circles

= 2 l×b( ) + 2×12 π R

2 − r2⎛⎝

⎞⎠

Part Answer5

= 2×106×10 + 227⎡⎣ 40( )2 − 30( )2⎤

⎦= 2120 + 22

7 × 1600 − 900( )= 2120 + 22

7 ×700= 2120 + 2200 = 4320m2

Question 31 :A circus tent is made or canvas and is in the form of a right circular cylinder and aright circular cone above it. The diameter and height of the cylindrical part of the tent are 126 m and5 m respectively. The total height of the tent is 21 m. find the total cost of the tent if the canvas used

costs Rs. 12 per square metre.Suggested Answers

Part Answer1

Diameter = 126 m∴ Radius (r) = 63 m

Height of the cylinder (h) = 5mTotal height = 21 m∴ Height of the conical portion, H = 21 - 5 = 16 m

Part Answer2

Curved Surface Area of cylinder = 2πrh= 2×22

7 ×63×5= 1980 sq. m

Question 31 :

Part Answer3

Lateral Surface Area of cone = πrl

where slant height (l) = H2 + r2√∴ Lateral Surface Area of cone

= 227 ×63× 16( )2 + 63( )2√

Part Answer4

= 22×9 256 + 3969√= 198 4225√ = 198×65 = 12,870sq. cm

Part Answer5

Canvas used in the tent= Curved Surface Area of cylinder + lateral surface area of cone= 1,980 + 12,870 = 14,850 sq. mcost of canvas per sq.m = Rs. 12∴ Cost of canvas used = 14,850 × 12

= Rs. 1,78,200

Question 32 :Ram can row a boat 8 km downstream and return in 1 hour 40 minutes. If the speedof the stream is 2 km/hr, find the speed of the boat in still water.Suggested Answers

Part Answer1

Let speed of the boat in still water = x km/hrSpeed of the stream = 2 km/hr ...(Given)Resultant speed in downward direction = (x + 2) km/hrResultant speed in upward direction = (x - 2) km/hr

Part Answer2

Distance = 8 kmTotal time taken in upward and downwarddirections = 1 hr 40 minutes= 1 40

60 = 1 23 = 5

3 hours

Question 32 :

Part Answer3

Now according to the question8x+2 + 8

x−2 = 53

Part Answer4

⇒ 8 x−2( )+8 x+2( )x2 −4

= 53

⇒ 3 (8x - 16 + 8x + 16) = 5(x2 - 4)⇒ 24x + 24x = 5x2 - 20⇒ 48x = 5x2 - 20⇒ 5x2 - 48x - 20 = 0

Part Answer5

⇒ 5x2 - 50x + 2x - 20 = 0⇒ 5x(x - 10) + 2(x - 10) = 0⇒ (x - 10) (5x + 2) = 0⇒ x - 10 = 0 or 5x + 2 = 0

Part Answer6

⇒ x = 10 or 5x = - 2⇒ x = 10 or x = − 2

5Since speed can never be -ve∴ Speed of boat in still water = 10 km/hr

Question 33 :If the mth term of an A.P. is 1n and the nth term is 1

m, show that the sum of mn terms

is 12 mn+ 1( ).

Suggested Answers

Part Answer1

Let first term = aand common difference = d of an A.P.∴ am = 1

n ⇒ a+ m− 1( )d = 1n … i( )

an = 1m ⇒ a+ n− 1( )d = 1

m … ii( )

Question 33 :

Part Answer2

Solving these equations, we geta+ m− 1( )d = 1

n

±a± n− 1( )d = ± 1m

m− 1 − n+ 1( )d = 1n − 1

m …(by subtracting)m− n( )d = m−n

mn ⇒ d = 1mn … iii( )

Part Answer3

Putting the value of d in (i), we geta+ m− 1( ). 1mn = 1

n ⇒ a+ m−1mn = 1

na = 1

n − m−1mn

a = m−m+1mn ⇒ a+ 1

mn … iv( )

Part Answer4

Again, Sn = n2⎡⎣2a+ n− 1( )d⎤⎦

Substituting the values of a and b, we get

Smn = mn2⎡⎣2. 1

mn + mn− 1( ). 1mn⎤⎦

= mn2⎡⎣

2+mn−1mn

⎤⎦ = 1

2 mn+ 1( )

Question 34 :From the top of a building 100m high, the angles of depression of the top and bottomof a tower are observed to be 45° and 60° respectively. Find the height of the tower. Also find the

distance between the foot of the building and the bottom of the tower.

Suggested MarksSuggested Answers

Question 34 :

Part Answer1

In right ΔBACtan60° = AB

AC⇒ 100

AC = tan 60°

⇒ AC = 1003√

m

∴ DE = AC = 1003√

m

Part Answer2

In right ΔBED, BEDE = tan45°

⇒ BEDE = 1 ⇒ BE = DE

∴ BE = 1003√

×

3√3√ = 100 3√

3

= 100×1.7323 = 57.73 m [ ∵ 3√ = 1.732]

Part Answer3

∴ Height of tower (CD) =AE= AB - BE= (100-57.73)m= 42.27 mDistance between the foot of the building and the bottom of the tower (AC) = 57.73 m.