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Olympiad-Classroom Assessment Practice Sheet

O-CAPS-03 : Pre-Regional Mathematics Olympiad (PRMO) (For VIII, IX, X Studying Students)

1. In the plane of given ∆ABC, if P is a point satisfying geometric condition.

[∆PAB] = [∆PAC] where [ . ] represents area enclosed. Then the curve traced by all such positions of P if

divides the plane in m parts then find 5m.

2. If the ratio of areas of circles circumscribing the ∆ABC and the circle circumscribing the ∆ formed by joining

reflection of its orthocentre about the sides respectively is n then find 15n.

3. For x, y, z being sides of ∆. If the least natural ‘n’ satisfies.

( ) ( ) ( )2 2 2x y z y z x z x yn

xyz− + − + −

< then find 10n.

4. Three circles, C1, C2, C3 are drawn by taking the sides of acute ∆ABC as diameters. If the common chords of

C1 and C2, C2 and C3, C1 and C3 meet at a point N inside the triangle and BN extended meets AC at D then

find ∠BDA in degrees.

5. Find the largest angle (in degrees) of a triangle having mid-points of altitudes as collinear.

6. A right triangle D is divided by its altitude into two triangles D1 and D2. If the sum of inradius of D1 and D is

10 units and length of altitude of D is 15 units then find the (inradius of D2) × 4.

7. If sin2θ = cos3θ, then find the value of cot6θ – cot2θ.

8. If sin3x + sin3y + 3sinx siny = 1 for some x, y ∈ R, then find the sum of the fourth power of least and

maximum value of the expression sinx + siny.

9. If the radius of a circle is 8 units and a triangle ABC is constructed by taking three points on it such that the

inradius of triangle is given by r ∈ N then find the sum of square of naturals that r can take.

Topics Covered :

Mathematics : Geometry, Trigonometry

MATHEMATICS

(For VIII, IX, X Studying Students) O-CAPS-03 : Pre-Regional Mathematics Olympiad (PRMO)


10. For –1 ≤ x ≤ 1, find value of (number of roots of the equation 3 63 4 2x x x− = + satisfying the given

interval) + 15.

11. If sin2x – 3sinx + 2 = 0 then find 17sin3x + 2.

12. For 0 ≤ a, b ≤ 3 and the equation x2 + 4 – 3cos(ax + b) – 2x has atleast one solution, then find the value of

99 a b+ π

.

13. If number of ordered pairs (0, x) which satisfy tan2θ + secθ = 6x – 11 – x2 are given by m then find m2018 + 10.

14. In a ∆ABC if D divides BC internally such that 34

BDDC

= and length of AD = x, AB = 8, BC = 5, CA = 4 then

find the value of x2 to nearest integer.

15. For a hexagon inscribed in a circle of radius m, two of its sides have unit length, two have length 2 units, and remaining two as length 3 units, then find the value of 10m3 – 35m.

16. For an equilateral ∆ABC with side length 6 units, if h1, h2 and h3 as lengths of perpendiculars upon sides then find (h1 + h2 + h3)2.

17. In a square ABCD, E is the mid-point of CB, AF is drawn perpendicular to DE. If the side of the square is 29 cm. Find the length of FB in cm.

18. In a scalene triangle ABC, the three altitudes drawn from vertices intersect at a point H. If BC = a, CA = b,

AB = c, AH = x, BH = y and CH = z then find the value of ( )12 xyz a b cabc x y z

+ +

.

19. A semi-circle is drawn outwardly on chord AB of the circle with centre O and radius 8 cm . The

perpendicular from O to AB meets the semi-circle at C. Find the length of chord AB so that the line segment OC has the maximum length.

20. As shown in the figure, triangle ABC is divided into six smaller triangles by lines drawn from the vertices through a common interior point P. The areas of these smaller triangles are as indicated in the figure. Then find the value of x – y.

21. In a scalene triangle ABC, with AB = c, BC = a, CA = b and a2 = b(b + c), ∠A and ∠B is related by 3A = (p + 1)B then find the value of p.

22. Let A be one of the two points of intersection of two circles with radii 2 cm and 3 cm and centres

X and Y, respectively. The angle XAY is 120° and the tangents at A to these two circles meet the circles

again at points B and C. If the square of circumradius of triangle ABC is ( ) c.m.a b c− then value of

a + b + c.



23. Let ABCD be a trapezium with AB || CD such that

(i) Its vertices A, B, C and D lies on a circle with centre O.

(ii) Its diagonals AC and BD intersect at point M and ∠AMD = 60°

(ii) MO = 10, BM = 4 and DM = 6

If the sum of length of AB and CD is a b where ‘a’ and ‘b’ are integers such that ‘b’ is least possible. Find a + b.

24. The diagonals AC and BD of a cyclic quadrilateral ABCD meet at right angles in E. If the radius of the

circumscribing circle is 3 cm, find the value of 2 2 2 2

2EA EB EC ED+ + + in square cm.

25. The trapezium ABCD has area S, AB = b, CD = a with a < b and AB || CD. Two diagonals AC and BC meet

at O. The area of ∆BOC is 29

S. Find (a + b) (where a and b are co-primes).

26. For some positive integer p, there is a quadrilateral ABCD with positive integer side length, perimeter p, right angles at B and C, AB = 2 and CD = AD. How many different values of p < 2018 are possible?

27. ABCD is a square with side length 15 units. E is a point inside the square such that ∠EBC = ∠ECB = 15°. Find the greatest integer value less than or equal to sum of lengths AE and ED.

28. A rectangle ABCD is inscribed in the circle with centre at O. The length of side AB is greater than side BC. The ratio of area of the circle to the rectangle ABCD is : 3π . The line segment DE intersects AB at E such

that ∠ODC = ∠ADE. If the ratio AE : AD is pq

, where gcd(p, q) = 1 then find q – p.

29. In the figure shown, ABCD is a square of side length 85 units. Two semicircles have been drawn on

diameters AB and AD. If the area of the shaded region ispq

, where gcd(p, q) = 1 then find p + q.

30. Consider f (x) = |sinx| + |cosx| and g(x) = |sinx| + |cosx – 1| defined for all real x. Let T1 and T2 be the least positive real numbers such that f (x + T1) = f (x) and g(x + T2) = g(x) for all real x. Find the value of p + q

if T1 + T2 = pqπ

, where gcd(p, q) = 1.


ANSWERS



1. (20) (D)

2. (15) (E)

3. (10) (E)

4. (90) (D)

5. (90) (E)

6. (20) (M)

7. (01) (D)

8. (17) (E)

9. (30) (M)

10. (15) (E)

11. (19) (M)

12. (99) (D)

13. (10) (M)

14. (49) (E)

15. (15) (M)

16. (27) (D)

17. (29) (M)

18. (12) (D)

19. (04) (M)

20. (40) (D)

21. (05) (M)

22. (12) (D)

23. (13) (D)

24. (18) (M)

25. (03) (M)

26. (31) (E)

27. (30) (M)

28. (02) (D)

29. (87) (D)

30. (07) (D)

Question Level Question Number

Easy (E) - 07 2, 3, 5, 8, 10, 14, 26

Moderate (M) - 11 6, 9, 11, 13, 15, 17, 19, 21, 24, 25, 27

Difficult (D) - 12 1, 4, 7, 12, 16, 18, 20, 22, 23, 28, 29, 30


ANSWERS & SOLUTIONS



1. Answer (20)

Area(∆PBD) = Area(∆PCD)

As, BD = CD and height of both the triangles is

same.

⇒ [∆PAB] = [∆PAC]

So P will lie on median AD and extended

Again by a line passing through A and || BC

⇒ For any point P

[∆PAB] = [∆PAC]

As PA is common base between || lines.

⇒ m = 4

2. Answer (15)

As reflections of the orthocentre lies on circumcircle of ∆ABC hence ratio stands as 1.

3. Answer (10) The equation inequality can be written as ( )( )( )x y y z z x

nxyz

− − −<

LHS : |x – y| < z |y – z| < x |z – x| < y ⇒ n = 1 4. Answer (90) If circle is drawn by taking AB as diameter then it

meets AC and BC at D, E such that BD and AE are altitudes.

(For VIII, IX, X Studying Students) O-CAPS-03 : Pre-Regional Mathematics Olympiad (PRMO) - Ans & Sol


The three altitudes of ∆ABC will be common chords of circles hence

∠BDA = 90°

5. Answer (90)

As the mid-points of altitudes lies on the sides of medial ∆, hence they will be collinear if they lie on same side like H1, H2 are endpoints and H3 lies between H1 and H2. The only possibility is right angled ∆.

6. Answer (20)

As for ∆ABC :

Inradius (r) 2

AB BC CA+ −=

Similarly r1 and r2 :

Now, r + r1 + r2 = BM

r = 15 – 10 = 5

7. Answer (01)

8. Answer (17)

sin3x + sin3y + (–1)3 = – 3sinx siny

Hence, either sin x + sin y = 1

as sin x = siny = –1

Therefore sum will be (–2)4 + (–1)4 = 17.

9. Answer (30)

For any such ∆ABC :

R ≥ 2r

Hence, 02Rr< ≤ ⇒ r = 1, 2, 3, 4

10. Answer (15)

Let x = sinθ

⇒ 6sin3 2xθ = +

Therefore LHS ∈ [–1, 1] but RHS > 2

Hence, no root in interval.

11. Answer (19)

(sinx – 1) (sinx – 2) = 0

As sin x = 1 is only valid

Hence, 17sin3x + 2 = 19

12. Answer (99)

x2 – 4 – 2x = –3cos(ax – b)

– 1 ≤ cos(ax + b) ≤ 1

solution only possible.

if a + b = π, 3π, 5π

a + b = π

1a b+

=π

13. Answer (10)

Let tan2θ + secθ = 0

⇒ sec2θ + secθ – 1

⇒ 21 5 –5sec –

2 4 4 θ + ≥

RHS :

–2 –(x – 3)2 ≤ –2

Hence, m2018 + 10 = 10

14. Answer (49)

By theorem on sections using cosine rule :

(8)2 × 4 + (4)2 × 3 = 5(4 × 3 + x2)

O-CAPS-03 : Pre-Regional Mathematics Olympiad (PRMO) - Ans & Sol (For VIII, IX, X Studying Students)


15. Answer (15)

⇒ 31 2 902 2

αα + α= ° −

( )1 2 3cos sin2 2

α + α α=

1 21 1sin ,sin

2 2 2m mα α

= =

3 3sin2 2m

α= , now substituting the value m.

16. Answer (27)

( )1 2 3

1 2 3

1 3As 6 362 4

3 3

h h h

h h h

× + + = ×

+ + =

17. Answer (29)

Construction : Produce DE to cut AB produce at M.

In ∆ECD and ∆EBM

∠1 = ∠2

EC = EB

∠ECD = ∠FBM = 90°

By ASA congruences

∆ECD ≅ ∆EBM

∴ CD = BM

But CD = AB

⇒ BM = AB

⇒ B is the mid-point of hypotenuse AM

⇒ FB = BM = 29

18. Answer (12)

[ABC] = [BHC] + [CHA] + [AHB]

⇒ 14 2abc

R= yz sin(π – A) + 1

2zx sin(π – B) +

12

xy sin(π – C)

= 12

yz sinA + 12

zx sinB + 12

xy sinC

= 12

xyz 2 2 2a b cRx Ry Rz

+ +

⇒ 4 4abc xyz a b c

R R x y z = + +

⇒ 1xyz a b cabc x y z

+ + =

19. Answer (04)

Let OD = a cm

AD = 8 a− cm = BD = DC

OC2 = (OD + DC)2 = ( )28a a+ −

( )8 2 8a a= + −

For OC to be maximum, a = 8 – a

⇒ a = 4

⇒ 8 4 2 cmAD = − =

AB = 4 cm

20. Answer (40) 84 84 35

56 56 40AF xFB y

+ += =

+ +

⇒ 2x – 3y = 50 40 84 56 40

35BDDC y x y

+ += =

+ +

⇒ 7y – 2x = 70

⇒ x = 70, y = 30



21. Answer (05)

CA is produced to D such that AD = AB

x = 2A ...(i)

Also, a2 = b(b + c)

⇒ a b cb a

+=

⇒ BC DCAC BC

=

⇒ ∆ACB ∼ ∆BCD

⇒ ∠CDB = ∠CBA = B ...(ii)

(i) and (ii)

⇒ A = 2B

22. Answer (12)

Perpendicular bisector of side AB will pass

through X and will also be parallel to AY.

Similarly, perpendicular bisector of side AC will

pass through Y and will also be parallel to AX. Hence, for the circumcentre P, AXPY is a

parallelogram.

Circumradius,

( ) ( ) ( )( ) ( )2 2 2 cosAP AX AY AX AY XAY= + +

4 3 2 2 3 cos120= + + × × × °

7 2 3= −

23. Answer (13)

Since ABCD is a cyclic quadrilateral, ∠DCA = ∠DBA. Since, AB || CD, ∠DCA = ∠CAB So, ∆AMB is isosceles Similarly, ∆CMD is isosceles. OA = OB ∴ ∆AMO ≅ ∆BMO

⇒ ∠AMO = ∠BMO

Since, ∠AMB = 90° ∠AMB = 120° ⇒ ∠AMO = ∠BMO = 60° Hence, ∆AMX and ∆BMX are congruent and have

angles 30°, 60° and 90°. Similarly, DMY and CMY are congruents ∆’s 4 3AB =

6 3DC =

24. Answer (18)

Let O be the centre of the circle EA2 + EC2 = (EP + PA)2 + (PC – PE)2 = EP2 + PA2 + PC2 + PE2 + 2EP.PA – 2PC.PE = 2(PA2 + PE2), (as PA = PC) Similarly, EB2 + ED2 = 2(QD2 + EQ2) ∴ EA2 + EB2 + EC2 + ED2 = 2(PA2 + PE2) +

2(QD2 + OQ2) = 2(PA2 + OQ2) + 2(QD2 + OP2) = 2(PA2 + OP2)

+ 2(QD2 + OQ2) = 2(OA2 + OD2) = 2(32 + 32) = 36



25. Answer (03) Let areas of ∆DOC = S1

and ∆AOB = S2

1 22 29 9

S S S S S+ + + =

∴ S1 + S2 = S –49

S

⇒ S1 + S2 = 59

S

and S1S2 22

9S =

On solving,

1 21 4and9 9

S S S S= =

∴ 1

2

12

Sab S

= =

∴ a + b = 3 26. Answer (31) Let BC = x and CD = AD = y By Pythagoras theorem

x2 +(y – 2)2 = y2 x2 + y2 + 4 – 4y = y2 x2 – 4y + 4 = 0 x2 = 4(y – 1) p = 2 + x + 2y 2 2 1 2 2018p y y= + − + <

y = 312 + 1 satisfy ∴ y = n2 + 1 where 1 ≤ n ≤ 31 ⇒ Number of possible values of p are 31.

27. Answer (30)

Draw ∆AFB ≅ ∆BEC.

∠FBE = 90° – (15° + 15°) = 60° and FB = EB

⇒ ∆FBE is equilateral

⇒ ∠FAE = ∠AEF = x(say)

In ∆ABE, sum of angles = 75° + (15° + x)

+ (60° + x) = 180°

⇒ x = 15°

⇒ ∠AEB = 60° + 15° = 75° = ∠ABE

⇒ AE = AB = 15 units.

Similarly, DE = DC = 15 units.

28. Answer (02)

Let AB = a, BC = b

∠ODC = ∠ADE = θ

2

3r

abπ π

=

⇒ 32 2 4a br r

× =

⇒ 2 sinθ cosθ 32

=

⇒ sin2θ 32

=

⇒ θ = 30°

AEAD

= tanθ = tan30° 13

=




1. PQRS is a trapezium with PQ||RS and the diagonals intersect at the point M. The area of ∆PQM is

32 cm2 and the area of ∆RSM is 50 cm2. If area of the trapezium PQRS is λ cm2, then 2λ

is equal to

2. PQR is a right angled triangle with ∠PQR = 90°, PR = 2 units, QR = 1 unit and QS is perpendicular to PR. If the area of rectangle with QS as one of its diagonals is µ, then 512 µ2 is equal to

3. A 20 cm × 20 cm square is divided into n2 congruent squares by equispaced lines parallel to its sides. Circles are inscribed in each of the squares. If the sum of the areas of the circles be πA cm2, then A is equal to

4. AB is a diameter of a circle of radius 24 units. CD is a chord, perpendicular to AB that cuts AB at E. If the arc

CAD is 23

of the circumference of the circle, then the length of AE is equal to

5. In a trapezium ABCD with AB||CD, AB = 20 cm, CD = 3 cm. ∠ABC = 32° and ∠BAD = 58°. The distance from the mid-point of AB to the mid-point of CD is d, then 2d is equal to

6. An isosceles right triangle is removed from each corner (see figure) of a square piece of paper so that a rectangle remains. If the sum of the areas of the cut off pieces is 242 sq. units, then the length of the diagonal of the rectangle is

Topics Covered :

Mathematics : Geometry, Trigonometry

MATHEMATICS

(For VIII, IX, X Studying Students) O-CAPS-04 : Pre-Regional Mathematics Olympiad (PRMO)


7. The quarter circle shown in the figure has centre C and radius 10 units. If the perimeter of the rectangle CPQR is 28 units. If the perimeter of the shaded region is (mπ + n) where m and n are natural numbers and co-prime to each other, then (m + n) is

8. ABCD is a square and P is a point on side AD such that ∆PAB has area 62 sq. units and ∆PCD has area 10 sq. units. The length of the side of the square is

9. P and Q are points on the sides AB and AC respectively of ∆ABC such that BP = CQ = λ, PA = 6 cm, AQ = 20 cm, BC = 25 cm. If ∆PAQ and the quadrilateral BPQC have equal areas, then the value of λ is

10. ∆ABC is isosceles with AB = AC = 10 cm and BC = 6 cm, M is the mid-point of AB. Let l be the line through A parallel to BC. If l intersects the circle through A, C and M at D, then 3(AD) is equal to

11. A sequence of equilateral triangles is drawn. The altitude of each is 3 times the altitude of the preceding

triangle. The difference between the areas of the first triangle and sixth triangle is 968 3 sq. units. The perimeter of the first triangle is

12. The value of tan205 – tan115 sectan245 tan335

x° °= °

° + °, where 0 < x < 99, then x is equal to

13. x1 and x2 are the two values of x in [0, 2π] for which tanx = 100, then 1 2tan tan2 2x x

⋅ is equal to

14. If sin 1 cos 3,sin 2 cos 2

x xy y

= = where x, 0,2

y π ∈

, then the value of sec2(x + y) is

15. The total number of solutions of 1cot cot ,

sinx x

x= + in the interval [0, 3π], is

16. If x, y ∈ [0, 2π], then the total number of ordered pairs (x, y) satisfying the equation sinx ⋅ cosy = 1, is equal to

17. If ABC is a triangle which is not right angled, the value of cos cos cos

sin sin sin sin sin sinA B C

B C C A A B+ + is equal to

18. sinx + sin2x = 1, and the value of sinx is λsin(µ°) λ∈N and 0 < µ < 90, then λµ is equal to

19. If tan3A = 4tanA (tanA ≠ 0), then the value of 30sin3

sinA

A is

20. A quadrilateral inscribed in the circle has side lengths 20, 99, 22 and 97 in the order. Taking

π = 227

, if the area of circle is ab

where a and b are co-prime then find a – 70b.

21. On sides BC and CD of a square ABCD, P and Q are points respectively, such that AP = 8 cm, PQ = 6 cm and

AQ = 10 cm. If the side of the square is given as pq

, where p and q are integers and q is least possible

then find the value of p + q.



22. Triangle ABC has AB = 90, BC = 50 and CA = 70. A circle is drawn with centre P on AB such that CA and CB

are tangents to the circle and 2AP = x. Find 5x

.

23. Two circles with radii 9 units and 16 units touch each other externally. Let r be the radius of the circle that

touches these two circles externally as well as a common tangent to the two circles. If prq

= , where

gcd(p, q) = 1, then find the value of (p + q).

24. The chords ED and AB of the circle AEBD meet at right angle at a point F such that EF = 6, AF = 2 and FD = 4 and the radius of the circle is r, then find r2.

25. Let f (x) = cos sin2x

and T be a real number such that f (x + T) = f (x) for all real values of x. If T = 16kπ

,

then find the smallest positive value of k.

26. In ∆ABC, D is the mid-point of side BC, E is the mid-point of AD, F is the mid-point of BE, and G is the mid-point of FC. If the ratio of ar(∆ABC) to ar(∆EFG) is p : q, where p and q are co-primes, then find p + q.

27. A hexagon is drawn in a plane such that all its interior angles are equal and all its sides have unequal positive integer values. Find the least perimeter of such a hexagon.

28. In a triangle ABC, the lengths of medians through vertices B and C are respectively 9 units and 12 units. If

AG = BC, where G is the centroid of ∆ABC, then find 2

2BC .

29. The three altitudes of a triangle ABC has lengths 3 units, 4 units and 125

units. If the area of the incircle of

the triangle ABC is s square units then find the greatest integer value less than or equal to s.

30. In a triangle ABC, ∠A = 30°, BC2 = 4 ( )10 3 3− and AC : AB = 1 : 3. If the square of the area of the triangle

ABC ispq

, where gcd(p, q) = 1, then find p + q.


ANSWERS



1. (09) (M)

2. (54) (M)

3. (10) (M)

4. (36) (E)

5. (17) (M)

6. (22) (M)

7. (21) (D)

8. (12) (E)

9. (04) (M)

10. (25) (D)

11. (12) (M)

12. (50) (M)

13. (01) (D)

14. (16) (D)

15. (02) (D)

16. (03) (D)

17. (02) (D)

18. (36) (D)

19. (80) (M)

20. (47) (D)

21. (49) (M)

22. (21) (E)

23. (19) (E)

24. (50) (D)

25. (32) (M)

26. (09) (M)

27. (21) (D)

28. (50) (M)

29. (03) (D)

30. (10) (D)

Question Level Question Number

Easy (E) - 04 4, 8, 22, 23

Moderate (M) - 13 1, 2, 3, 5, 6, 9, 11, 12, 19, 21, 25, 26, 28

Difficult (D) - 13 7, 10, 13, 14, 15, 16, 17, 18, 20, 24, 27, 29, 30


ANSWERS & SOLUTIONS



1. Answer (09)

AMB ⊥ SR ∆PQM ~ ∆RSM

⇒ 2 2

22 2

ar( )ar( )

PQM PQ AM kRSM RS MB

∆= = =

∆ (say)

⇒ 2 1625

k =

Now, ar(PQRS) = 12

AB(PQ + RS)

= 12

(AM + MB)(PQ + RS)

= 12

[k(MB) + MB](PQ + RS)

= 12

[(k + 1)MB] [(k + 1)RS]

= 12

(k + 1)2(MB)(RS)

24 1 (50)

5 = +

29 50

5 =

81 5025

= ×

= 162

⇒ 812λ

=

2. Answer (54)

1 3ar( ) 3 12 2

PQR∆ = × × =

Also, 1ar( )2

PQR QS PR∆ = ×

1 22

QS= ×

⇒ 32

QS=

Also, ∆QFS ~ ∆PQR

⇒ SF QS QFQR PR PQ

= =

⇒ 3 3,4 4

SF QF= =

3 3ar( )16

QESF QF SFµ = = × =

3. Answer (10)

The diameter of each of n2 circles is 20n

⇒ Sum of the areas of n2 circles

22

100 100nn

π= × = π

⇒ A = 100



4. Answer (36)

2 (2 )3

CAD r= π

⇒ AC CD DA= =

⇒ ∆ACD is an equilateral triangle. ⇒ O is the centroid AO = 24 ⇒ OE = 12 ⇒ AE = 36 cm 5. Answer (17)

Produce AD and BC to intersect at E. ∠AEB = 90°

PQ = EP – EQ = 12

AB – 12

CD

3 1710 –2 2

= =

6. Answer (22)

2 2 2 21 1 1 1 242

2 2 2 2x y x y+ + + =

⇒ x2 + y2 = 242

Also, ( ) ( )2 222 2x y d+ =

2x2 + 2y2 = d2 ⇒ d = 22

7. Answer (21) CPQR is a rectangle. ⇒ PR = QC = 10 Radius of circle = 10 ⇒ AC + BC = 20 ⇒ AP + PC + CR + RB = 20

⇒ 12

AP + (Perimeter of rectangle) + RB = 20

⇒ AP + 14 + RB = 20 ⇒ AP + RB = 6 ∴ Perimeter of shaded region

AP PR RB AQB= + + +

16 10 (2 10) (5 16)4

= + + π × = π +

8. Answer (12)

1 ( )( ) 622

AP AB =

1 ( )( ) 102

PD CD =

⇒ 1 [ ]( ) 722

AP PD AB+ =

⇒ 21 722

AB =

⇒ AB = 12 9. Answer (04)

ar(∆APQ) = ar( BPQC)

⇒ ar(∆ABC) = 2ar(∆APQ) ⇒ ar( ) 2 ar( )

ar( ) ar( )ABC APQABQ ABQ

∆ × ∆=

∆ ∆

⇒ 20 62 ( 20)( 6) 24020 6

+ λ = ⇒ λ + λ + = + λ

⇒ λ = 4, –30 So, λ = 4



10. Answer (25)

l||BC and AB = AC ⇒ ∠DAC = ∠ACB = ∠MBC ADCM is a cyclic quadrilateral.

∠ADC = 180° – ∠AMC

= ∠BMC ∆ADC ~ ∆BMC

⇒ AD ACBM BC

=

⇒ 253

AD =

11. Answer (12) Let a be the side and h be the altitude of the first

triangle.

32

h a= and 234

A a= [A = area]

⇒ 2 23 2

4 3 3h hA

= =

According to the question,

( )25

2 23 242968 3 –3 3 3

h h h = =

⇒ h2 = 12 ⇒ a = 4 ∴ Perimeter = 12 12. Answer (50)

Given expression 2

2

1 tan 25 sec 501– tan 25

+ °= = °

°

13. Answer (01) tanx = 100

⇒ 2

2 tan2 100

1– tan2

x

x=

⇒ 2100 – 100 tan 2tan2 2x x

=

⇒ 2100 tan 2tan – 100 02 2x x

+ =

⇒ 1 2tan tan –12 2x x

=

14. Answer (16) tan 1

tan 3xy

=

As, tan tantan( )

1– tan tanx yx y

x y+

+ =

2

4 tan1– 3 tan

xx

=

Also, siny = 2sinx, 2cos cos3

y x=

⇒ sin2y + cos2y 2 244sin cos9

x x= +

⇒ 36tan2x + 4 = 9sec2x = 9(1 + tan2x)

⇒ 5tan3 3

x =

⇒ tan( ) 15x y+ =

15. Answer (02) Let cotx > 0

⇒ 1cot cot

sinx x

x= + , which is not possible

Let cotx ≤ 0

∴ 1– cot cot

sinx x

x= +

⇒ 1–2cot

sinx

x=

⇒ 1cos –2

x =

⇒ 2 8,3 3

x π π=

16. Answer (03) sinx = 1, cosy = 1 or sinx = –1, cosy = –1

⇒ 2

x π= , y = 0, 2π or 3

2x π

= , y = π

∴ The solutions are 3, 0 , , 2 , ,

2 2 2π π π π π

17. Answer (02) cos cos( )– (1– cot cot )

sin sin sin sinA B C B C

B C B C+

= =∑ ∑ ∑

3 – cot cotB C= ∑

= 3 – 1 = 2



18. Answer (36) sin2x + sinx – 1 = 0

⇒ –1 5sin2

x ±=

⇒ 5 – 1sin 2sin182

x = = °

19. Answer (80)

2

2

tan3 3 – tan4 4tan 1– 3 tan

A AA A

= ⇒ =

⇒ 2 2 2 13 – tan 4 – 12tan tan11

A A A= ⇒ =

Now, 2

2

sin3 3 – 4sin 43 –sin 1 1 cot

A AA A

= = +

43 –12

=

83

=

20. Answer (47) Let ∠D = θ ∠B =180° – θ

2 20 99 – 2( 20)( 99)cosAC = + θ

2 22 97 – 2( 22)( 97)cos(180 – )AC = + θ

⇒ 2 119 – 2 20 99 cosAC = θ

and 2 119 2 22 97 cosAC = + θ

∴ 119 – 2 20 99 cos 119 2( 22)( 97)cosθ = + θ

Given ( )2cos 22 97 20 99 0θ + =

⇒ cos 0 or 90θ = θ = °

∴ AC2 = 20 + 99 = 119 ∴ (2R)2 = 119

∴ 2 1194

R =

∴ A = πR2 = 22 1197 4

×

= 11 17 1872 2×

=

∴ a = 187 ; b = 2 ⇒ a – 70b = 187 –140 = 47

21. Answer (49)

∠APQ = 2π

AB = 8 cosθ = PB + PC

⇒ 8 cosθ = 8 sinθ + 6 cosθ

⇒ tanθ = 14

⇒ cosθ = 417

AB = 8 cosθ = 3217

22. Answer (21) Let CA touch the circle at R and CB touch the

circle at S.

Since PR = PS Right-angled ∆’s PRC and PSC are congruent.

Hence, CP bisects ∠ACB. By angle bisector theorem 7

5AP ACPB BC

= =

∴ 5AP = 7(PB) = 7(90 – AP) ⇒ 12AP = 630

⇒ 2AP = 105

⇒ 2AP = 105 and x = 105

215x

=



23. Answer (19) PR2 = CK2 = AC2 – AK2 = (AM + MC)2 – (AP – KP)2 = (AM + MC)2 – (AM – MC)2 = 4(AM) (MC)

⇒ PR = 2 9r ; similarly; RQ = 2 16r and

PQ = 2 9 16× but PR + RQ = PQ

⇒ 127

r =

24. Answer (50)

Let O be the centre of circle. FD × FE = FB × FA ⇒ FB = 12

Let K and G be points on AB and ED respectively, such that OB ⊥ AB and GO ⊥ ED.

ED = 6 + 4 = 10

∴ GD = GE = 10 52

=

⇒ OK = GF = GD – FD = 5 – 4 = 1

and AB = AF + FB = 2 + 12 = 14

So, 14 72

AK KB= = =

r = 50

25. Answer (32) As cos(–θ) = cosθ so smallest T should be such

that sin sin sin2 2 2

x T x x+ = − = π +

⇒ 2T = π

⇒ T = 2π

26. Answer (09)

Draw EC

Since altitude of ∆BEC

1 altitude of 2

BAC= × ∆

and 1ar( ) ar ( )2

BEC BAC∆ = ∆

1ar( ) ar( )2

EFC BEC∆ = ∆

1ar( ) ar( )2

EGF EFC∆ = ∆

∴ 1ar( ) ar( )4

EGF BEC∆ = ∆

∴ 1ar( ) ar( )2

BEC ABC∆ = ∆

⇒ 1ar( ) ar( )8

EGF ABC∆ = ∆

∴ 81

pq

=

∴ p + q = 9

27. Answer (21)

Remove from the three corners of an equilateral triangle of side 9 units equilateral triangles of sides 1 unit, 2 units and 3 units respectively. The remaining portion is now an equiangular hexagon ABCDEF with sides 1, 6, 2, 4, 3, 5.



28. Answer (50)

AG = BC

⇒ DG = 12

BC

⇒ Median GD from vertex G is half of opposite side BC in ∆BGC.

⇒ ∠BGC = 90°

⇒ BC2 = BG2 + GC2 = 2 22 29 12

3 3 × + ×

= 100

29. Answer (03)

∆ = 12

× a × r + 12

× b × r + 12

× c × r

= 2

a b c+ +

× r

⇒ 2a

ah∆

=

Similarly, 2b

bh∆

= and 2c

ch∆

=

So, a b cr h h h

∆ ∆ ∆ ∆= + +

⇒ 1 1 1 1

a b cr h h h= + +

⇒ 1 1 1 5 1

3 4 12r= + + =

⇒ r = 1

Area of incircle = π(1)2 = π

30. Answer (10)

cos30° = 2 2 2

296

x x ax

+ −

⇒ ( )2

2 10 3 33 52 3 3x

−= −

⇒ 2

10 3 3 5 3 3 10 3 32 4 4x

− −= − =

⇒ x2 = 4

Area (∆ABC) = 12

x × 3x × sin30° = 23

4x

= 3 44× = 3



29. Answer (87)

Notice that area of region (1) = area of region (3) and area of region (2) = area of region (4). Swap the shaded part 1 with unshaded part 3; and shaded part 2 with unshaded part 4; which doesn’t change the total area of shaded region or that of unshaded region.

Hence, are of shaded region = 12

area of square

852

=

30. Answer (07)

sin cos

2 2 2f x x xπ π π + = + + +

= |sinx| + |cosx| = f (x)

⇒ T1 = 2π

( ) 2 sin cos sin

2 2 2x xg x π

= +

Period of sin2x

is 2π and that of

cos sin2 2x x

+

is π.

Hence, period of g(x) = lcm (π, 2π) = 2π

⇒ T2 = 2π

Edition: 2020-21

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Olympiad-Classroom Assessment Practice Sheet O-CAPS-03 : Pre … · 2020. 6. 9. · Aakash Educational Services Limited - Regd. Office: Aakash Tower, 8, Pusa Road, New Delhi-110005