Olympiad-Classroom Assessment Practice Sheet O-CAPS-05

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Olympiad-Classroom Assessment Practice Sheet

O-CAPS-05 : Pre-Regional Mathematics Olympiad (PRMO) (For VIII, IX, X Studying Students)

1. The numbers 1447, 1005 and 1231 have some properties in common. Each is a four digit number beginning

with 1 that has exactly two identical digits. If K is total such numbers, then find the value of 8K .

2. The number of integers that are both multiples of 20022018 and factors of 20022020 is

3. If the sum of the series 1 1 1 11 .....3 6 10 15

+ + + + + to 2018 terms is mn

where m, n ∈ N and are co-prime to

each other, then the remainder when (m – n) is divided by 100 is

4. The number of common integers for the two A.P’s 1, 8, 15, 22, ....., 2003 and 2, 13, 24, 35, ....., 2004 is

5. If 200100

101 103 105 .... 199 2 ,50!

kC × × × ×= × then k is equal to

6. Find the largest number n such that (2018!)! is divisible by ((n!)!)!

7. The number of factors of 2160 which are multiples of 6 is

8. The number of two digit positive integers having an even number of positive divisors is

9. In an infinite geometric progression whose terms are positive and any term is equal to the sum of next two

terms, if the common ratio is a bc+ where a, b, c are integers, in their lowest form, then (a + b + c) is equal to

10. Sum of the digits of the number (33333333)2 is

11. The number of 3-digit numbers, whose product of digits is an odd number is k, then |k – 50| is equal to

12. The first three elements of a number series are 3, 1, 1 starting with the second element, the product of two neighbours of any element is the same. What is the sum of 49 terms of the series?

13. Four positive integers x1, x2, x3 and x4 are given. There are exactly four distinct ways to choose three of x1, x2, x3 and x4. The mean of each of the four positive triplets is added to the fourth integer. The four sums 29, 23, 21 and 17 are obtained. The sum of the largest of the two numbers x1, x2, x3 and x4 is

Topics Covered :

Mathematics : Combinatorics, Sequence and Series, Probability

MATHEMATICS

(For VIII, IX, X Studying Students) O-CAPS-05 : Pre-Regional Mathematics Olympiad (PRMO)


14. An integer is chosen at random. The probability that, the sum of the digits of its square is 39 is p, then 96p is equal to

15. Two numbers x and y are chosen at random from the set of integers 1, 2, 3, ......, 15. The probability that 2x = 3y is p, then 189p is equal to

16. In order to get at least once a tail with probability greater than or equal to 0.9, the number of times a coin needed to be tossed is

17. If maximum number of handshakes that takes place in a party of 20 people if no two same person

handshake twice is λ, then find 1 .10

λ

18. If all EIGHT digit numbers using each of the digit 1, 2, 3, 4, 5, 6, 7, 8 exactly once are written in increasing order then find the sum of last three digits of 997th number in the list.

19. If λ is the number of ways one can select five natural numbers among first 200 natural numbers so that they

form an AP (for e.g. (1, 3, 5, 7, 9), (3, 5, 7, 9, 11) are distinct ways), find 1 .100

λ

20. Find the number of primes that divides the sum of all possible divisors of 2000.

21. In how many ways 6 persons can be allotted two different rooms each having capacity to accommodate all if no room goes unoccupied?

22. There are 2n – 3 black balls, 2n white balls and 2n + 3 orange balls in a bag. If the number of ways of choosing two balls from the bag having different colours is 1191 then find ‘n’.

23. If all combinations of four digit number, that can be set for a dialer lock with digits 0 – 9 on each dial and all four digits cannot be identical is λ4 – λ, then find λ.

24. If all the roots of equation x5 – 20x4 + λ1x3 + λ2x2 + λ3 = 0 are real and in AP, then find the arithmetic mean of the roots.

25. If there are 12 points in a plane, 6 of which lie on one of the two parallel lines and rest on the other. The number of triangles formed by them is K. Then find sum of digits of K.

26. If 2 2 2 2 21 2 3 4 9na a a a a+ + + …+ = for a1, a2, a3, … an ∈ R and

117,

i n

ii

a=

==∑ then find 1

5× (sum of all possible pairs

formed by taking any two among a1, a2, a3, a4, ... an).

27. Find the smallest odd positive integer ‘m’ >1, for which sum of first m natural numbers is a perfect square.

28. From a group of 10 persons, if the number of ways to form a committee of at least one member is 1000 + λ, then find ‘λ’.

29. If the first term and common ratio of an infinite GP formed by selecting terms from 1 1 1 11, , , ,2 4 8 16

… upto ∞,

having sum 17

are x and y respectively, then find 640 × xy.

30. If the number of ways for 3 persons to stay in room number 1, 2, 3, 4, 5, 6, 7, 8, so that no two persons stay

in consecutive room numbers. {if no person can stay in same room with other person} is N then find .6N


ANSWERS



1. (54)(M)

2. (81) (M)

3. (17) (M)

4. (26) (E)

5. (50) (E)

6. (06) (M)

7. (24) (E)

8. (84) (E)

9. (06) (E)

10. (72) (M)

11. (75) (E)

12. (99) (E)

13. (33) (M)

14. (00) (D)

15. (09) (M)

16. (04) (M)

17. (19) (E)

18. (16) (D)

19. (49) (D)

20. (04) (E)

21. (62) (E)

22. (10) (E)

23. (10) (M)

24. (04) (M)

25. (09) (D)

26. (28) (D)

27. (49) (E)

28. (23) (M)

29. (10) (M)

30. (20) (D)

Question Level Question Number

Easy (E) - 12 4, 5, 7, 8, 9, 11, 12, 17, 20, 21, 22, 27

Moderate (M) - 12 1, 2, 3, 6, 10, 13, 15, 16, 23, 24, 28, 29

Difficult (D) - 06 14, 18, 19, 25, 26, 30


ANSWERS & SOLUTIONS



1. Answer (54) Possible combinations are 11xy, 1x1y and 1xy1

where x ≠ y, x ≠ 1 and y ≠ 1. Hence there 3 × 9 × 8 = 216 numbers of this form.

Suppose the two identical digits are not 1. Possible combinations are 1xxy, 1xyx, 1yxx, where x ≠ y, x ≠ 1 and y ≠ 1. There are 3 × 9 × 8 = 216

∴ K = 216 + 216 = 432

∴ 432 548 8K

= =

2. Answer (81) Required number of integers = Number of factors

of the ratio 2020

22018

2002 20022002

=

= 112 × 132 × 72 × 22 ⇒ Required answer = 34 = 81 3. Answer (17)

Let 1 1 1 11 .....3 6 10 15

S = + + + + + to 2018 terms

1 1 1 1 1 .....2 2 6 12 20 30S

= + + + + + to 2018 terms

⇒ 1 1 1 1 1 1 1 1 11 .....2 2 2 3 3 4 4 5 5 6S

= ⋅ + ⋅ + ⋅ + ⋅ + ⋅ +

1 12018 2019

+ ⋅

1 1 1 1 1 1 11– – – –2 2 3 3 4 4 5

= + + + +

1 1... –2018 2019

+

⇒ 1 20181–2 2019 2019S

= =

⇒ 40362019

S = ⇒ m = 4036, n = 2019

m – n = 2017

4. Answer (26)

The common terms are 57, 134, 211, .....

5. Answer (50)

200100 2 2

200! (100!)(101 102 ..... 200)(100!) (100!)

C × × ×= =

50(101 103 ..... 199)(51 52 .... 100) 2(50!) (51 52 ..... 100)

× × × × × × ⋅=

× × × ×

50(101 103 ..... 199) 250!

× × × ⋅=

6. Answer (06)

For a, b ∈ N, b! is divisible by a! ⇔ b ≥ a.

Thus, (2018!)! is divisible by ((n!)!)!

⇔ 2018! ≥ (n!)!

⇔ 2018 ≥ n!

So, the largest value of n is 6.

7. Answer (24)

2160 = 24 × 33 × 5 = 6 × 23 × 32 × 5

Number of multiples of 6 = (3 + 1)(2 + 1)(1 + 1)

= 24

O-CAPS-05 : Pre-Regional Mathematics Olympiad (PRMO) - Ans & Sol (For VIII, IX, X Studying Students)


8. Answer (84) Only the two-digit perfect squares have odd

number of divisors. i.e. {42, 52, 62, 72, 82, 92} ∴ Required numbers of positive integers

= 90 – 6 = 84 9. Answer (06) Tr = Tr + 1 + Tr + 2 ⇒ arn–1 = arn + arn+1

⇒ 1 = r + r2 ⇒ 5 – 12

r =

10. Answer (72) 32 = 9 332 = 1089 ⇒ Sum of digits = 2 × 9

3332 = 110889 ⇒ Sum of digits = 3 × 9

∴ Sum of digits = 8 × 9 = 72

11. Answer (75) All digits would be odd. So required number of numbers = 5 × 5 × 5 = 125 12. Answer (99) The series is 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, ..... S49 = 12 × 8 + 3 = 99 13. Answer (33) x1 + x2 + x3 + 3x4 = 87 x1 + x2 + x4 + 3x3 = 69 x2 + x3 + x4 + 3x1 = 63 x1 + x3 + x4 + 3x2 = 51 ⇒ x1 + x2 + x3 + x4 = 45

⇒ 21, 12, 9, 3 are the required numbers.

14. Answer (00) Let the selected integer be α, then the sum of the

digits of α2 be 39 which is a multiple of 3.

⇒ 3 digits α2 ⇒ 3 divides α

It follows that 9 divides α2 which is not the case as 39 is not a multiple of 9.

Hence, p = 0 15. Answer (09) (x, y) can be (3, 2), (6, 4), (9, 6), (12, 8), (15, 10)

⇒ Required probability 152

5 121C

= =

16. Answer (04) Since probability of getting atleast one T in n

tosses = 11–2

n

As 1 11– 0.9 0.12 2

n n ≥ ⇒ ≤

⇒ 2n ≥ 10 ⇒ n ≥ 4 17. Answer (19) For any person number of handshakes = 19

hence 20 × 19 but each handshake is repeated

twice hence 20 19 1902×

=

18. Answer (16) 1 2 3 4 8 7 6 5 Total cases

WAYS

=

⇓

Numbers starting with 1 = 7 × 6 × 5 × 4 × 3 × 2 = 5040 Hence, number starting with 1 and then 2

⇒

19. Answer (49) Selecting 5 number as table : Common difference No. of ways

1 1962 1923 1804 ;5 ;; ;; ;; ;

49 449 4900 ways(196 4)2

=+

20. Answer (04) 2000 = 24 × 53 Sum of divisors = (20 + 21 + 22 + 23 + 24) (50 + 51

+ 52 + 53) 5 42 – 1 5 – 1 31 624

2 – 1 5 – 1 4

= = ×

Prime divisors of 13 × 22 × 3 × 31 are 13, 2, 3, 31

(For VIII, IX, X Studying Students) O-CAPS-05 : Pre-Regional Mathematics Olympiad (PRMO) - Ans & Sol


21. Answer (62) Number of ways 26 – 2 = 62 ways 22. Answer (10) Number of ways = (2n – 3) 2n + 2n (2n + 3)

+ (2n – 3) (2n + 3) ⇒ 12n2 – 9 = 1191

⇒ n = 10

23. Answer (10)

10 10 10 10ways ways ways ways

= 104 – (All Identical) = 104 – 10 24. Answer (04) Let (x – α) (x – β) (x – γ) (x – δ) (x – θ)

= x5 – 20x4 + λ1x3 + λ2x2 + λ3x + λ4

On comparison : ⇒ α + β + γ + δ + θ = 20

(a – 2d) + (a – d) + a + (a + d) + (a + 2d) = 20 a = 4 A.M. = a = 4 25. Answer (09)

Number of triangles = 6 6 6 62 1 1 2 180C C C C× + × =

= k 26. Answer (28) a1

2 + a22 + a3

2 … + an2 = 9

a1 + a2 + a3 … + an = 17

Hence,

2 21 2 1 2

–1( ) – 2

i n

n ii

a a a a a a=

+ … + = Σ ∑

1 2

289 – 9 1402

a a= =∑

27. Answer (49)

As required ( 1)2

m m += Perfect square m = 49

28. Answer (23) P1 P2 P10

⇓ ⇓ ⇓

2 ways 2 ways 2 ways = 210 1023 = 210 – 1

29. Answer (10)

1 1 11, , ,2 4 8

…

Let first term 12a and common ratio 1

2b=

Hence, –

12 12Sum

1 72 – 11–2

b aa

b

b

= = =

Hence, 3 31 1; .

2 2x y= =

30. Answer (20) 1 2 3 4 5 6 7 8 x1 ⇓ x2 ⇓ x3 ⇓ x4

1st person 2nd person 3rd person x1 = Number of unoccupied rooms left to 1st

person from left x2 = Number of unoccupied rooms left between

1st and 2nd person x3 = Number of unoccupied rooms left between

2nd and 3rd person x4 = right to 3rd person Hence, x1 + x2 + x3 + x4 = 8 – 3 = 5 Where, x2 and x3 must be natural number but x1

and x4 may be whole number. Hence, x1 + 1 + x2 + 1 + x3 + x4 = 5

1 2 3 4

All whole number

3x x x x+ + + =

1 2 3 4 Sum

0 0 0 3 30 0 3 0 33 0 0 0 30 3 0 0 31 2 0 0 30 2 1 0 31 0 0 2 32 1 0 0 30 1 2 0 3

Total 20 ways

x x x x

− − − − −− − − − −− − − − −

Now, persons can stay in 20 × 31 ways.




1. You are given 100 cards numbered 100, 99, 98, ......, 3, 2, 1 in that order. You want to put them in the order

1, 2, 3, ......, 99, 100 switching only two adjacent cards at a time. If the minimum number of switches necessary is n, then |n – 5000| is equal to

2. The number of integers between 0 and 2002 that are relatively prime to 2002 are m, then 10m

is equal to

3. The number of subsets S of X = {1, 2, 3, 4, ...., 10} with the property : “There exist integers a < b < c with a ∈ S, b ∉ S, c ∈ S” is µ, then the sum of the digits of µ is equal to

4. If a1, a2 and a3 be positive real numbers such that a1a2 + a3 = (a1 + a3)(a2 + a3) and the maximum value of

a1a2a3 is k, then 2k

is equal to

5. Let N(x, y, z) denote the number of multiples of x that are less than z and greater than y, then N(93, 94, 96) = 100a + 10b + c where a, b, c are single digit natural numbers, then (a + b + c) is equal to

6. The value of –11 2r

r

r∞

=∑ is

7. The nth triangular number is defined to be the sum of the first n positive integers. For example, the 5th triangular number is 1 + 2 + 3 + 4 + 5 = 15. In the first 100 terms of the sequence 1, 3, 6, 10, 15, 21, 28, 36, ..... of triangular numbers, how many are divisible by 7?

8. If the sum to infinity of the series 1 1 1 1 1 1 11– – – – – – .......2 4 8 16 32 64 128

ab

+ + = where a and b are natural

numbers in their lowest form. Then (a2 + b2) is equal to

9. A sum of money is rounded off to the nearest rupee. The probability that the rounded off error is atleast ten paise is p, then 100p is equal to

Topics Covered :

Mathematics : Combinatorics, Sequence and Series, Probability

MATHEMATICS

(For VIII, IX, X Studying Students) O-CAPS-06 : Pre-Regional Mathematics Olympiad (PRMO)


10. The probability of India winning a test match against South Africa is 12

. Assuming independence from match

to match, if the probability of India winning second time in the third test, in a five-match series is p, then 168p is equal to

11. Out of the 13 applicants for a job, there are 5 women and 8 men. It is desired to select 2 persons for the job. The probability that at least one of the selected persons is a woman is p, then 39p is equal to

12. A man alternately tosses a coin and throws a die. If the probability of getting a head on the coin before he gets a ‘6’ on the die is p, then 42p is equal to

13. For the number N = 19202122…93, how many natural numbers less than (a + 11) exist such that N = 3a m, if m ∈ N but not a multiple of 3?

14. Natural numbers are written on all the faces of a cube one on each. At each corner (vertex) of the cube the product of numbers, on the faces that meet at the corner is written. The sum of all corner values is 228. If the sum of all possible values of T where T denotes the sum of numbers on all the faces is µ, then find sum of digits of µ.

15. If ‘a’ is among first 9 whole numbers and hence the probability that the equation x4 – 2ax2 + x + a2 – a = 0 has all four real roots is λ then find value of 90λ.

16. Let set X = {1, 2, 3… 9, 10} and A and B disjoint subsets of X such that A ∪ B = X and Prod (A) is divisible

by Prod (B) {where Prod (A) = Product of elements of a set A}. Then find minimum value of Prod ( )Prod ( )

AB

.

17. An online seller has to deliver five different orders to five different addresses. By mistake the delivery boy delivers two correct parcels and three incorrect parcels. In how many distinct ways he can do it?

18. If a point is selected in the plane of ∆ABC such that it is equidistant from the lines containing segments AB, BC and CA. Find (the probability of point lying inside the triangle) × 40.

19. For a rectangle ABCD, with 47, 17.AB BC= = If the probability that PA + PB + PC + PD ≥ 16,

for a point P lying inside the rectangle is µ, then find value of 10µ.

20. If T1, T2, T3… Tn, form a sequence of real numbers such that T1 = T2 = 1 and Tn + 1 = Tn–1Tn + 1(for n ≥ 1)

{For example n = 2 ⇒ T3 = T1T2 + 1 ⇒ 1 × 1 + 1 = 2}, then find (number of terms in first 50 terms of this sequence which are multiple of 4) + 10.

21. From a box containing ‘n’ different balls a ball is selected at random and placed back. Again a ball at random

is selected. If the probability of getting same ball both the times is 1

40, then find ‘n’.

22. If the sum of first ‘20’ terms of series 1 3 7 152 4 8 16

+ + + … is 2–λ + 19 then find λ.

23. How many divisors of 12600 are multiple of 4 but not multiple of 3?

24. How many subsets having 4 elements each can be formed from a set having six distinct elements?

25. If n(S) = 5 and S is a subset of P = {a1, a2, a3,…, a7} and the probability of a1 such that a1 ∈ S is K, then find 7K.

26. Find the maximum possible number of intersection point of diagonals lying inside a hexagon.



27. Let there exist λ different value of natural number n such that 1 1 1 1

1 2 3 2 1n n n n+ + …+

+ + + + is greater than

unity, then find λ + 10.

28. Start with the numbers 1, 2, 3, 4, 5, 6. Replace two of these numbers say x and y by xy

x y+. Repeat this

until there is only one number left, say p. Find the number of different values of p.

29. Let A be the sum of first 20 terms and B be the sum of first 40 terms of series 12 + 2.22 + 32 + 2.42 + 52

+ 2.62 + .... If B – 2A = 100K, then find the unit digit of K.

30. Let 3 5 7 ...

1.2.3 2.3.4 3.4.5+ + + ∞ terms

pq

= , where p and q are integers such that gcd (p, q) = 1. Then, find

the value of p + q.


ANSWERS



1. (50) (M)

2. (72) (M)

3. (23) (D)

4. (54) (M)

5. (17) (M)

6. (04) (M)

7. (28) (M)

8. (53) (E)

9. (81) (M)

10. (42) (M)

11. (25) (E)

12. (36) (M)

13. (11) (M)

14. (13) (D)

15. (80) (D)

16. (07) (M)

17. (20) (M)

18. (10) (M)

19. (10) (M)

20. (10) (E)

21. (40) (E)

22. (20) (E)

23. (12) (M)

24. (15) (E)

25. (05) (M)

26. (15) (M)

27. (10) (M)

28. (01) (D)

29. (08) (D)

30. (09) (D)

Question Level Question Number

Easy (E) - 06 8, 11, 20, 21, 22, 24

Moderate (M) - 18 1, 2, 4, 5, 6, 7, 9, 10, 12, 13, 16, 17, 18, 19, 23, 25, 26, 27

Difficult (D) - 06 3, 14, 15, 28, 29, 30


ANSWERS & SOLUTIONS



1. Answer (50) Initial order of the cards : 100, 99, 98, ....., 3, 2, 1. Required order is 1, 2, 3, ....., 99, 98. To get the required order, the minimum number

of switches needed between two adjacent cards at a time is 99 + 98 + 97 + ..... + 2 + 1

99 100 49502

×= =

2. Answer (72)

Let φ(n) be the number of integers less then n, that are relatively prime to n.

If n is prime, then φ(n) = n – 1

If k1 and k2 are relatively prime, then

φ(k1 k2) = φ(k1) φ(k2)

⇒ φ(2002) = φ(2 × 7 × 11 × 13)

= φ(2) φ(7) φ(11) φ(13)

= 1 × 6 × 10 × 12

⇒ 7210m

=

3. Answer (23) Number of subsets of X = 210 = 1024. Any subsets without the specified property must

be either empty set or a block of consecutive integers.

To specify a block of consecutive integers, we either have just one element (10 choices) on a pair of distinct end points (10C2 = 45 choices).

The number of sets with the required property = 1024 – (1 + 10 + 45) = 968

4. Answer (54) a1a2 + a3 = (a1 + a3)(a2 + a3)

⇒ a32 + (a1 + a2 – 1)a3 = 0

⇒ a3(a3 + a1 + a2 – 1) = 0

⇒ a3 = 0 or a3 = 1 – a1 – a2

Taking a3 = 1 – a1 – a2

⇒ a1 + a2 + a3 = 1

Now, AM ≥ GM ⇒ 1

1 2 3 31 2 3( )

3a a a

a a a+ +

≥

⇒ 127

k = ⇒ 13

1 2 31 ( )3

a a a≥

2 54k

= 1 2 31

27a a a ≤

5. Answer (17) Multiples of 93 which are greater than 94 and less

than 96 are 10 × 93, 11 × 93, 12 × 93, ..... (93 – 1)93 i.e., 719 6. Answer (04)

2 3

2 3 41 .....2 2 2

S = + + + +

2 3 4

1 1 2 3 4 .....2 2 2 2 2

S = + + + +

2 3

1 1 1 1– 1 .....2 2 2 2

S S = + + + +

1 112 1–2

S = ⇒ S = 4



7. Answer (28) Every 6th and 7th term is divisible by 7. So required number of numbers = 16 + 14 – 2 = 28 8. Answer (53) Given series

1 1 1 1 11 .... – ....8 64 2 16 128

= + + + + + +

1 1 1– ....4 32 256

+ + +

1 11 2 4– –

1 1 11– 1– 1–8 8 8

=

8 1 1 8 1 21– –7 2 4 7 4 7

= = ⋅ =

9. Answer (81) The sample space is S = {–0.50, –0.49, –0.48, ..... –0.01, 0.00, 0.01,

...., 0.49} Let E be the event that the rounded off error is

atleast 10 paise, then EC is the event that the rounded off error is atmost 9 paise.

⇒ EC = {–0.09, – 0.08, ...., –0.01, 0.00, ...., 0.09} n(EC) = 19, n(S) = 100 19 81( ) 1–

100 100P E = =

10. Answer (42) Required probability p = (Probability of winning a

test in first two matches) × (Probability of winning in the third match)

⇒ 2–1

21

1 1 1 1 168 422 2 2 4

p C p = ⋅ = ⇒ =

11. Answer (25)

Required probability 5 8 5

1 1 213 13

2 2

2539

C C CC C×

= + =

12. Answer (36) 1 1( ) , (6)

2 6P H P= =

Required probability 2 2 31 5 1 5 1 .....

2 6 2 6 2 = + + ⋅ +

6 42 367

p p= ⇒ =

13. Answer (11) If we add 1 + 9 + 2 + 0 + 2 + 1… + 3 = 717 which

is a multiple of 3 but not of 9 hence a = 1 only 14. Answer (13)

Face number let ABCD ⇒ a BQRC ⇒ b PQRS ⇒ c APSD ⇒ d ABQP ⇒ e CRSD ⇒ f Then the product written at corners are : ade, abe, abf … cdf. The sum of these 8 numbers

is: given to be 228 = 22.3.19 Hence, (a + c) (b + d) (e + f) = 4.3.19 = 2.6.19 = 2.3.38 = 2.2.57 Hence, T ⇒ 4 + 3 + 19 = 26 2 + 6 + 19 = 27 2 + 3 + 38 = 43 2 + 2 + 57 = 61 15. Answer (80) The equation is quadratic in a a2 – a (2x2 + 1) + x4 + x = 0 2 2 2 42 1 (2 1) – 4( )

2 1x x x x

a+ ± + +

=×

2 4 2 42 1 4 4 1– 4 – 42

x x x x xa + ± + +=

22 1 | 2 – 1|2

x xa + ±=

Hence for (+ and – sign) a = x2 + x a = x2 – x + 1 a2 – a (2x2 + 1) + x4 + x = (a – x2 – x) (a – x2

+ x – 1) = 0

O-CAPS-06 : Pre-Regional Mathematics Olympiad (PRMO) - Ans & Sol (For VIII, IX, X Studying Students)


⇒ x2 + x – a = 0, x2 – x + 1 – a = 0

–1 1 4 1 4 – 3,2 2

a ax x± + ±= =

All roots are real iff 34

a ≥

a = 1, 2, 3, 4,…8

Probability 89

=

16. Answer (07)

As by given condition A must contain 7 and B should not, hence minimum value of Prod ( ) 7Prod ( )

AB

=

17. Answer (20)

As the correctly delivered is 2 among them, hence selecting 2

5 4 102×

=

Now, remaining, 3 are to be rearranged :

3 4 3 5

4 5 4 3

5 3 5 4

2 waysP A P AP A P AP A P A

→ →→ → ⇒→ →

Total 20 ways.

18. Answer (10)

Such 4 points exists hence probability = 14

19. Answer (10)

AC2 = 64

AC = 8

BD = 8

For minimum P must be AC ∩ BD

min (PA + PB + PC + PD) = 16

20. Answer (10)

By simple observation

No multiple of 4 will exist.

21. Answer (40)

Let first ball be selected

Probability = 1

For same ball draw probability 1n

=

so, 1 11

40n× =

22. Answer (20)

201 1 1 21– 1– 1– 1–2 4 8 2

+ + …

⇒ 201 1 1 1 120 – ......2 4 8 16 2

+ + + +

⇒ 20

–20

1 1 – 12 220 – 20 – 1 2

1 – 12

⇒ +



23. Answer (12) 12600 = 23 × 7 × 32 × 52 Divisors being multiple of 4 but not of ‘3’ … 22.7x = 0, 1

5y = 0, 1, 2 or 23 7x = 0,1 5y = 0,1,2

Hence, 6 + 6 = 12 Divisors 24. Answer (15) S = {a1, a2 …, a6} To select 4 = To reject 2 6 5 15

2×

= =

25. Answer (05) Selecting 5 out of 7 reject 2 out of 7 7 6 21(total)

2×

= =

Now as a1 is already there, hence selecting 4 from remaining;

6 5 15 ways2×

=

26. Answer (15) Selecting any 4 points, a quadrilateral is formed

and hence 1 intersection point of diagonals. Selecting 4 out of 6 and rejecting 2 out of 6

6 5 152×

= =

27. Answer (10) For any natural n ∈ Nt 1 1

1 1n n=

+ +

1 12 1n n

<+ +

1 13 1n n

<+ +

1 1( 1) 1n n n

<+ + +

Adding All : 1 1 1 1 1

1 2 2 1 1n

n n n n+

+ …+ < =+ + + +

⇒ For any n given sum is always less than unity

No such natural number exists for which sum is more than 1.

28. Answer (01) The sum of the reciprocals of all the numbers

remain constant, no matter which numbers are

chosen, because 1 1 1xyx y

x y

+ =

+

∴ 1

1 1 1 1...1 2 3 6

p =+ + +

29. Answer (08) Sum of first 20 terms = (12 + 22 + ... + 202) + 4 (12 + 22 + ...+ 102) 20.21.41 10.11.214

6 6= +

A = 4410 Sum of first 40 terms B = (12 + 22 + ... + 402) + 4 (12 + 22 + ... + 202) = 33620 B – 2A = 33620 – 8820 = 24800 = 248 × 100 = 100K Then, K = 248 30. Answer (09)

( )( )( )

( )( )12 1

1 2 1 2nn nnt

n n n n n n+ ++

= =+ + + +

( )( ) ( )1 1

1 2 2n n n n= +

+ + +

1 1 1 1 11 2 2 2n n n n

= − + − + + +

1

1 1 1 1 12 3 2 1 3

t = − + −

2

1 1 1 1 13 4 2 2 4

t = − + −

3

1 1 1 1 14 5 2 3 5

t = − + −

4

1 1 1 1 15 6 2 4 6

t = − + −

1 1 1 10 0 02 2 1 2

S∞ = − + + − −

1 1 3 5 (given)2 2 2 4

ab

= + × = =

Edition: 2020-21

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Olympiad-Classroom Assessment Practice Sheet O-CAPS-05