Relations and Numbers | B.Stat Objective | TOMATO 63
Try this TOMATO problem from I.S.I. B.Stat Objective based on Relations and Numbers.
Relations and Numbers (B.Stat Objective problems)
We consider the relation , "a person x shakes hand with a person y".Obviously if x shakes hand with y, then y shakes hand with x. In a gathering of 99 persons , one of the following statements is always true, considering 0 to be an even number, find which one is it.
there is at least one person who shakes hand with an odd number of persons
there is at least one person who shakes hand with an even number of persons
there are even number of persons who shake hand exactly with an even number of persons
none of these
Key Concepts
Logic
Relations
Numbers
Check the Answer
Answer: there is at least one person who shakes hand with an even number of persons
B.Stat Objective Question 63
Challenges and Thrills of Pre-College Mathematics by University Press
Try with Hints
Let R be handshakes among 99 persons holds
first person may handshake with at most 98(even) other persons, for second person similar arguments hold and this holds with similar arguments for all persons
then there exists at least one person who shakes hand with an even number of persons.
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2008 based on Percentage.
Percentage Problem - AIME I, 2008
Of the students attending a party, 60% of the students are girls, and 40% of the students like to dance. After these students are joined by 20 more boy students, all of whom like to dance, the party is now 58% girls, find number of students now at the party like to dance.
is 107
is 252
is 840
cannot be determined from the given information
Key Concepts
Ratios
Percentage
Numbers
Check the Answer
Answer: is 252.
AIME I, 2008, Question 1
Elementary Number Theory by David Burton
Try with Hints
Let number of girls and boys be 3k and 2k, out of 3k girls, 2k likes to dance and 2k+20(boys) like to dance
here given that \(\frac{3k}{5k+20}\)=\(\frac{58}{100}\) then k=116
Try this beautiful problem from the PRMO, 2019 based on Smallest Positive Integer.
Smallest Positive Integer - PRMO 2019
Find the smallest positive integer n\(\geq\)10 such that n+6 is a prime and 9n+7 is a perfect square.
is 107
is 53
is 840
cannot be determined from the given information
Key Concepts
Integers
Primes
Perfect Square
Check the Answer
Answer: is 53.
PRMO, 2019, Question 14
Elementary Number Theory by David Burton
Try with Hints
Let 9n+7=\(m^{2}\) n+6 prime then n+6 odd then n is odd then n=2k+1 then 9(2k+1)+7=\(m^{2}\) then 18k=\(m^{2}\)-16=(m+4)(m-4) then 18k even m is even then m=2p
18k=(2p+4)(2p-4)=4(p+2)(p-2) then 9k=2(p+2)(p-2)then k even then k=2d then 18d=2(p+2)(p-2) then 9d=(p+2)(p-2) then p of form 9q+2,9q-2
for p=9q-2 then m=2(9q-2) for q=1 then\(m^{2}\)=196then n=21 then n+6=27 non prime, for p=9q+2 then m=2(9q+2) for q=1 \(m^{2}\)=484 then n=53 then n+6=59 prime then n=53.
Circles and Triangles | AIME I, 2012 | Question 13
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on Circles and triangles.
Circles and triangles - AIME I, 2012
Three concentric circles have radii 3,4 and 5. An equilateral triangle with one vertex on each circle has side length s. The largest possible area of the triangle can be written as \(a+\frac{b}{c}d^\frac{1}{2}\) where a,b,c,d are positive integers b and c are relative prime and d is not divisible by the square of any prime, find a+b+c+d.
is 107
is 41
is 840
cannot be determined from the given information
Key Concepts
Angles
Trigonometry
Triangles
Check the Answer
Answer: is 41.
AIME I, 2012, Question 13
Geometry Revisited by Coxeter
Try with Hints
In triangle ABC AO=3, BO=4, CO=5 let AB-BC=CA=s [ABC]=\(\frac{s^{2}3^\frac{1}{2}}{4}\)
\(s^{2}=3^{2}+4^{2}-2(3)(4)cosAOB\)=25-24cosAOB then [ABC]=\(\frac{25(3)^\frac{1}{2}}{4}-6(3)^\frac{1}{2}cosAOB\)
of the required form for angle AOB=150 (in degrees) then [ABC]=\(\frac{25(3)^\frac{1}{2}}{4}+9\) then a+b+c+d=25+3+4+9=41.
Box and ball Probability | B.Stat Objective TOMATO Problem 59
Try this problem from I.S.I. B.Stat TOMATO Objective Problem based on Box and ball Probability.
Box and ball Probability ( B.Stat Objective Problem )
A box contains 100 balls of different colours 28 red 17 blue 21 green 10 white 12 yellow 12 black. The smallest number n such that any n balls drawn from the box will contain at least 15 balls of the same colour is
8
77
11
10
Key Concepts
number Series
Logic
Integers
Check the Answer
Answer: 77
B.Stat Objective Problem
Challenges and Thrills of Pre-College Mathematics by University Press
Try with Hints
28 -15 gives 13 and 17-15 gives 2 and 21 -15 gives 6 rest are less than 15
Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2012 based on Arrangement.
Arrangement - AIME 2012
Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chicken meal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waiter could serve the meal types to the nine people so that exactly one person receives the type of meal ordered by that person.
is 107
is 216
is 840
cannot be determined from the given information
Key Concepts
Arrangements
Algebra
Number Theory
Check the Answer
Answer: is 216.
AIME, 2012, Question 3
Combinatorics by Brualdi
Try with Hints
Here the number of ways to order the string BBBCCCFFF, such that one B is in first three positions
one C is in 4th to 6th positions, and one F is for last three positions. There are (3)(3)(3)=27 ways for first 3. Then for next two, 2 ways.
Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2015 based on Theory of Equations.
Theory of Equations - AIME 2015
Let (f(x)) be a third-degree polynomial with real coefficients satisfying[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.]Find (|f(0)|).
is 107
is 72
is 840
cannot be determined from the given information
Key Concepts
Series
Theory of Equations
Algebra
Check the Answer
Answer: is 72.
AIME, 2015, Question 10.
Polynomials by Barbeau.
Try with Hints
Let (f(x)) = (ax^3+bx^2+cx+d). Since (f(x)) is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up. By drawing a coordinate axis, and two lines representing (12) and (-12), it is easy to see that (f(1)=f(5)=f(6)), and (f(2)=f(3)=f(7)); otherwise more bends would be required in the graph.
Since only the absolute value of f(0) is required, there is no loss of generalization by stating that (f(1)=12), and (f(2)=-12). This provides the following system of equations.[a + b + c + d = 12][8a + 4b + 2c + d = -12][27a + 9b + 3c + d = -12][125a + 25b + 5c + d = 12][216a + 36b + 6c + d = 12][343a + 49b + 7c + d = -12]
Using any four of these functions as a system of equations yields (|f(0)| = \boxed{072})
