Pioneer Education IOQM-2020-21 SOLUTIONS · 2021. 2. 16. · Questions 1 to 8 carry 2 marks each; questions 9 to 21 carry 3 marks each; questions 22 to 30 carry 5 marks each. 8. All

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IOQM-2020-21

SOLUTIONS

Instructions

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6. The answer you write on OMR sheet is irrelevant. The darkened bubble will be considered as your

final answer.

7. Questions 1 to 8 carry 2 marks each; questions 9 to 21 carry 3 marks each; questions 22 to 30 carry

5 marks each.

8. All questions are compulsory.

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13. You may take away the question paper after the examination.

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1. If a, b, c are real numbers and

(a + b –5)2 + (b + 2c + 3)2 + (c + 3a – 10)2 = 0 find the integer nearest to a3 + b3 + c3.

Ans. 57

Solution:

This is possible only when

a + b – 5 = 0 ….(i)

b + 2c + 3 = 0 …(ii)

c + 3a – 10 = 0 …(iii)

On solving, we get

a = 4, b = 1, c = – 2

a3 + b3 + c3 = 57

2. If ABCD is a rectangle and P is a point inside it such that AP = 33, BP = 16, DP = 63. Find CP.

Ans. 56

Solution:

In 2 2 2PAD, x y 33 ….(i)

In 2 2 2PBD, a x y 16 …(ii)

In 22 2PDF, x b y 63 …(iii)

(ii) + (iii) – (i) 2 2

a x b y

= 2 2 263 16 33

2 22 2 2CP 56 In PCE, CP a x b y

CP 56





3. Sita and Geeta are two sisters. If Sita’s age is written after Geeta’s age a four digit perfect square

(number) is obtained. If the same exercise is repeated after 13 years another four digit perfect

square (number) will be obtained. What is the sum of the present ages of Sita and Geeta?

Ans. 55

Solution:

Let Geeta’s age be be ab 10a b

& Sita’s age be cd 10c d

abcd = m2

2a 1000 b 100 c 10 d m

210a b 100 10c d m …(i)

After 13 years.

210a b 13 100 10c d 13 n ....(ii)

2 2ii i 1313 n m n m

13 101 n m n m

n + m =101 n + m = 1313

n –m = 13 n – m = 1

n = 57, m = 44 n = 657, m = 656 (Rejected)

abcd = m2 = 1936

19 + 36 = 55

4. Let ABC be an isosceles triangle with AB = AC and incentre l. If AI = 3 and the distance

from I to BC is 2, what is the square of the length of BC?

Ans: 80

Solution:

lM = r = 2

rlA

Asin

2

A r 2sin

2 lA 3

A 2 BMtan

2 AM5

a2 2

55

BC = a = 4 5

BC2 = 80





5. Find the number of positive integers n such that the highest power of 7 dividing n! is 8.

Ans. 07

Solution:

Power of 7 in n! = 2

n n

7 7

+ …

n

7 7

Let n = 49

Power of 7 in n! = 7 + 1 = 8

n may equal to 49, 50, 51, 52, 53, 54 55

number of possible values of n = 07

6. Let ABCD be a square with side length 100. A circle with center C and radius CD is drawn.

Another circle of radius r, lying inside ABCD, is drawn to touch this circle externally and

such that the circle also touches AB and AD. If r = m + n k , where m, n are integers and

k is a prime number, find the value of m n

k

.

Ans. 50

Solution:

AQ = 2r

AC = 100 2

AQ + PQ + CP = 100 2

2 r + r + 100 = 100 2

r 2 1 100 2 1

r = 100 2 1

2 1

r = 100 (3 – 2 2 )

r = 300 – 200 2 = m + n k

m = 300, n = – 200, k = 2

m n 300 20050

k 2





7. a, b, c are positive real numbers such that a2 + b2 = c2 and ab = c. Determine the value of

2

(a b c)(a b c)(b c a)(c a b)

c

.

Ans. 04

Solution:

a2 + b2 = c2, ab = c

(a + b) 2 = a2 + b2 + 2ab = c2 + 2c

(a + b)2 – c2 = 2c ...(i)

(a – b) c2 – (a – b) 2 = c2 – 2c 2 = 2c ...(ii)

Given expression is

2 22 2

2 2

a b c c a b 2c 2c4

c c

8. Find the largest 2-digit number N which is divisible by 4, such that all integral powers of

N end with N.

Ans. 76

Solution:

If given condition is satisfied for power 2 (i.e. for square), then it will hold for all (+ve) integral

powers.

Let the number N be "ab".

Units place remains same on squaring b = 0 or 6

If b = 0, then on squaring even ten's place will become 0. So b = 0 is rejected.

b = 6

N is divisible by 4.

Possibilities of N are 16, 36, 56, 76, 96.

Among these only 76 satisfies the required condition

N = 76

9. Find the number of ordered triples (x, y, z) of real numbers that satisfy the system of equation:

x + y + z = 7; x2 + y2 + z2 = 27; xyz = 5.

Ans. 03

Solution:

x + y + z = 7, x2 + y2 + z2 = 27, xyz = 5

xy + yz + zx = 49 27

2

= 11

Cubic whose roots are x, y, z is

t 3 – 7t 2 + 11t – 5 = 0

Its roots are 1, 1, 5

(1, 1, 5) (x, y, z)

So number of triplets =3!

2!=3





10. Let A and B be two finite sets such that there are exactly 144 sets which are subsets of A or subsets

of B. Find the number of elements in AB.

Ans. 08

Solution

2a+b + 2b+c – 2b = 144 = 24 · 9

2b(2a + 2c – 1) = 24 · 9

b = 4 & a = 1, c = 3

n(A B) = 5 + 7 – 4 = 8

11. The prime numbers a, b and c are such that a + b2 = 4c2. Determine the sum of

all possible values of a + b + c.

Ans. 31

Solution:

4c2 – b2 = a

(2c – b) (2c + b) = a

2c – b = 1 and 2c + b = a

4c = a + 1 and 2b = a – 1

If c 5 than c must be of the type 6 ± 1 for N

Let c = 6 1, then a = 24 3 or 24 5

24+ 3 is not a prime, so c = 6 – 1 and

a = 24 5 then b = 24 6

2

= 12 – 3, which is not a prime.

So c cannot be greater than 5.

If c = 3 then a = 7 and b = 5

and If c = 2 then a = 7 and b = 3

Sum of possible values of a + b + c = 31

12. Let A = {m: m an integer and the roots of x2 + mx + 2020 = 0 are positive integers) and

B = (n: n an integer and the roots of x2 + 2020x + n = 0 are negative integers).

Suppose a is the largest element of A and b is the smallest element of B. Find the sum of digits of

a + b.

Ans. 26

Solution:

. 2020 where , are 20 and 101.

So a = –121

Now 2020 where and are negative integers. So n = . is least if , are –1

and –2019.

So b = 2019

a b 1898

Sum of digits = 26





13. The sides of a triangle are x, 2x + 1 and x + 2 for some positive rational number x. If one angle of the

triangle is 60°, what is the perimeter of the triangle?

Ans. 9

Solution:

x < x + 2 < 2x + 1 for existence triangle also triangle is non-equilateral.

Hence cos 600 =

2 22x 2x 1 x 2

2x 2x 1

3

x2

Sides are 3 7

, ,42 2

Perimeter = 9

14. Let ABC be an equilateral triangle with side length 10. A square PQRS is inscribed in it, with

P on AB, Q, R on BC and S on AC. If the area of the square PQRS is m + n k where m, n are integers

and k is a prime number then determine the value of 2

m n.

k

Ans. 10

Solution:

a + 2x = 10

also tan600 = a

x

a

x3

10 3

a2 3

Area of square = a2 = 300 7 4 3

m 2100, n 1200, k 3

2

m n 90010

k 9

15. Ria has 4 green marbles and 8 red marbles. She arranges them in a circle randomly. If the

probability that no green marbles are adjacent is p

q where the positive integers p, q have no

common factors other than 1, what is p + q?

Ans. 40

Solution:

Favourable ways = 7! 8C4.4!

Sample space = 11!

7!8! 4! 7.6.5.4! 7Pr

4!4! 11! 4!.11.10.9 33

p + q = 40





16. If x and y are positive integers such that (x – 4) (x – 10) = 2y, find the maximum possible

value of x + y.

Ans. 16

Solution:

Let y = a + b

(x – 4) (x – 10) = 2a. 2b

a bx 2 4 or 2 10 (vice-versa).

a bx 2 4 2 10.

a b2 2 6.

a 3 and b 1. y 4 and x 12

x y 16.

17. Two sides of a regular polygon with n sides, when extended, meet at an angle of 28°. What is the

smallest possible value of n?

Ans. 45

Solution:

Let the sides extended pass through the vertices which on joining the centre makes angle k 2

n

0n 2 2k28 360

n n

02k 2

152n

0k 1

76n

k 1 45

n19

For minimum k = 20 and n = 45

18. Let D, E, F be points on the sides BC, CA, AB of a triangle ABC, respectively. Suppose AD, BE, CF

are concurrent at P. If PF/PC = 2/3, PE/PB = 2/7 and PD/PA = m/n, where m, n are positive integers

with gcd(m, n) = 1, find m + n.

Ans. 45

Solution:

PF 2

PC 3

1 3PF 2

CF 5

….(i)





Similarly 2 62

9

….(ii)

and 4 5m

m n

….(iv)

Adding (i), (ii) and (iii); we get

2 2 m1

5 9 m n

m 28

1m n 45

m 17

m n 45

28m 17n

m 17

n 28

m n 45

19. A semicircular paper is folded along a chord such that the folded circular arc is tangent to the

diameter of the semicircle. The radius of the semicircle is 4 units and the point of tangency divides

the diameter in the ratio 7: 1. If the length of the crease (the dotted line segment in the figure)

is I then determine l2.

Ans. 39

Solution:

If we complete the circle on arc AOB, we get the centre O2 and radius 4.

Clearly O1O2 = 5

1 2O AO is isosceles, then

2

2 5AD 4

2

= 64 25

2

So AB 39





20. Two people A and B start from the same place at the same time to travel around a circular track of

length 100 m in opposite directions. First B goes more slowly than A until they meet, then by

doubling his rate he next meets A at the starting point. Let d m be the distance travelled by B before

he met A for the first time after leaving the starting point. Find the integer closest to d.

Ans. 41

Solution:

Let speed of A and B are initially vA and vB respectively. Let they started at P and meet first time

at point R.

B A

d 100 d

v v

…(i)

Let they meet sound time at P, then

B A

100 d d

2v v

…(ii)

from equation (i) I equation (ii) we get

2d 100 d

100 d d

222d 100 d

2d 200d 10000 0

200 4000 4000d

2

d 100 2 1

d 41.42





21. Let A = {1, 2, 3, 4, 5, 6, 7, 8}, B = {9, 10, 11, 12, 13, 14, 15, 16} and C = {17, 18, 19, 20, 21, 22, 23, 24}.

Find the number of triples (x, y, z) such that x A, y B, z C and x + y + z = 36.

Ans. 46

Solution:

x,y, z N and

x A, i.e., 1 x 8

and y B, i.e., 9 y 16

and z C, i.e., 17 z 24

Number of solution for x + y + z = 36 is equal to coefficient of x36 in (x + x2 + … + x8).

(x9 + x10 + … + x16). (x17 + x18 + … + x24)

= coefficient of x9 in (1 + x + …. + x7)3

= coefficient of x9 in

38

3

1 x

1 x

= coefficient of x9 in (1 – 3x8 + 3x16 – x24)

23.41 3x x ...

2!

= 11C9 – 3.3

= 55 – 9

= 46

22. Let ABC be a triangle with BAC = 900 and D be the point on the side BC such that ADBC.

Let r, r 1, and r2 be the in radii of triangles ABC, ABD and ACD, respectively. If r, r1, and r2 are

positive integers and one of them is 5, find the largest possible value of r + r1 + r2.

Ans. 30

Solution:

ABD ~ ACD ~ ABC

Any hypotenuse of these triangles follow the Pythagoras theorem.

So 2 2 21 2r r r

If r1, r2, r N and least is 5, then the other two must be 12 and 13.

1 2 maxr r r 30





23. Find the largest positive integer N such that the number of integers in the set {1, 2, 3,...,N} which are

divisible by 3 is equal to the number of integers which are divisible by 5 or 7 (or both).

Ans. 35

Solution:

N N N N

3 5 7 35

N N N N

3 35 5 7

The largest value of N satisfying this equation is 35.

24. Two circles, S1 and S2, of radii 6 units and 3 units respectively, are tangent to each other externally.

Let AC and BD be their direct common tangents with A and B on S1, and C and D on S2. Find the area

of quadrilateral ABDC to the nearest integer.

Ans. 68

Solution:

Refer to diagram.

3 1cos

9 3

Let ar APR

then ar BQS4

Because APR ~ BQS

1AP. PR sin

2

= 1 2 2

6.2. 4 22 3

Area of 1

ABDC 2 72 3 62 4

= 48 2





25. A five digit number n abcde is such that divided respectively by 2, 3, 4, 5, 6 the remainders are

a, b, c, d, e. What is the remainder when n is divided by 100?

Ans. 11

Solution:

a {0, 1}, b {0, 1, 2}, c {0, 1, 2, 3}, d {0, 1, 2, 3, 4}, e {0, 1, 2, 3, 4, 5}

n is a five digit number, so a = 1 and n must be odd.

When n is divided by 5 the remainder is e or e – 5.

So either d = e or (d, e) = (0, 5)

(i) When d = e then (d, e) = (1, 1) or (3, 3)

de c(divisible by 4) b (divisible by 3)

11 3 0, 1, 2

33 1 Not possible

So n = 10311 or 11311 or 12311

Out of which n = 11311 is correct when we check the condition of divided by 6.

(ii) When (d, e) = (0, 5)

then c = 1, b is not possible.

So n = 11311

26. Let a, b, c be three distinct positive integers such that the sum of any two of them is a perfect square

and having minimal sum, a + b + c. Find this sum.

Ans. 55

Solution:

WLOG, a < b < c

Let a + b = x2

a + c = y2

and b + c = z2

So, 2 2 23 x y z

Also N = a + b + c = 2 2 2

2x y zz

2

2 2 2x y z

2 2 2x y z

N2

is an integer so any two out of x, y, z must be odd or all three must be even.

(x, y, z) = (5, 6, z), (4, 6, 8), ……..

for least value of N, (x, y, z) = (5, 6, z)

2 2 21N 5 6 7 55

2

Hence a = 6, b = 19 and c = 30





27. Let ABC be an acute-angled triangle and P be a point in its interior. Let PA, PB, and PC be the images

of P under reflection in the sides BC, CA and AB, respectively. If P is the orthocentre of the triangle

PAPBPC and if the largest angle of the triangle that can be formed by the line segments PA, PB, and PC

is x°; determine the value of x.

Ans. 60

Solution:

AB is bisector PPC & AC is bisector of PPB.

A is point of intersection of bisectors

A is point of intersection of bisectors

A is circumcentre of B CPP P

Circumradius = B CCR PP P PA

Similarly, C ACR PP P PB

A B& CR PP P PC

But CR A B B CPP P CR PP P = C A A B CCR PP P CR P P P

PA = PC Required triangle is equilateral triangle

x = 600

28. For a natural number n, let n’ denote the number obtained by deleting zero digits, if any.

(For example, if n = 260, n’ = 26; if n = 2020, n’ = 22.) Find the number of 3-digit numbers n for

which n’ is a divisor of n, different from n.

Ans. 93

Solution:

n' n so atleast one digit of a must be zero

Let n abc

Case I. If only c = 0 ab|abc

ab|10ab (always true)

Number of numbers n = 9 9 81

Case II. If only b = 0 ac |abc

10a c |100a c

10a c|10a 8c|

So either 10a = 8c or 8c – 10a = 10a c for some natural number .

When 10a 8c 5a 4c a, c 4, 5

When 90

c 8 10a 1 c a 108

for 20

1, c a7

(Not possible)

for 2, c 5a a, c 1, 5





for 3, c 8a a, c 1, 8

for 25

4, c a2

(Not possible)

and so further value of is needed to be checked

Number of numbers n = 3, which are 405, 105 and 108.

Case III. If both b = c = 0, then all values of n are acceptable.

Number of numbers n = 9

So total number of possible values of n = 93.

29. Consider a permutation (a1, a2, a3, a4, a5) of {1, 2, 3, 4, 5}. We say the 5-tuple (a1, a2, a3, a4, a5) is

flawless if for all 1 i j k 5, the sequence(ai , aj , ak) is not an arithmetic progression

(in that order). Find the number of flawless 5-tuples.

Ans. 19

Solution:

There are only four possible three terms AP exist, which are

< 1, 2, 3 >, <2, 3, 4 >, < 3, 4, 5 > and <1, 3, 5>

Consider the sets A = <1, 2, 3> are present in the sequence

B = <2, 3, 4> are present in the sequence

C = <3, 4, 5> are present in the sequence

D = <1, 3, 5> are present in the sequence

53n A n B n C n D C .2. 2 40

54n A B n B C C . 2 10

n A C 1 6 2 13

54n A D C .2 n C D 10

n B D 4. 2 2 16

n A B C 2

n B C D 4

n A C D 2

n A B D 4 and n A B C D 2

Now n A B C D 4 40 69 12 2 101

So number of flawless 5-tupless = 5 11 19

30. Ari chooses 7 balls at random from n balls numbered 1 to n. If the probability that no two of the

drawn balls have consecutive numbers equals the probability of exactly one pair of consecutive

numbers in the chosen balls, find n.

Ans. 54

Solution:





Case I. When no two selected balls have consecutive numbers.

x1 + x2 + x3 + …. + x8 = n – 7

1 2 3 7 8x X X ..... X x n 13

Total ways = n 67C

Case II. When exactly two balls have consecutive numbers

x1 + x2 + x3 + ….x8 = n – 7

Any one out of x2, x3, ….,x7 must be zero

i

1 2 3 7 8

any 5x 's

x X X ..... X x n 12

Total ways = 6 n 61 6C . C

n 6 6 n 67 1 6C C . C

n 12 7 6

n 54


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Pioneer Education IOQM-2020-21 SOLUTIONS · 2021. 2. 16. · Questions 1 to 8 carry 2 marks each; questions 9 to 21 carry 3 marks each; questions 22 to 30 carry 5 marks each. 8. All