AMC 8 Math Olympiad USA Math Olympiad

Time & Work Problem | PRMO-2017 | Problem 3

Try this beautiful problem from PRMO, 2017 from Arithmetic based on Time & Work.

Time & Work Problem | PRMO | Problem 3

A contractor has two teams of workers: team A and team B. Team A can complete a job in 12 days and team B can do the same job in 36 days. Team A starts working on the job and team B joins A after four days. Team A withdraws after two more days. For how many more days should team B work to complete the job?

  • \(24\)
  • \(16\)
  • \(22\)
  • \(18\)

Key Concepts


Unitary process

Work done

Check the Answer


PRMO-2017, Problem 3

Pre College Mathematics

Try with Hints

At first we have to find out A’s 1 days work and B’s 1 days find out A and B both together 1 day’s work .

Can you now finish the problem ……….

Team A completes job in 12 days and Team B completes job in 36 days

1 day work of team A =\(\frac{1}{12}\)

1 day work of team B=\(\frac{1}{36}\)

1 day work of team A and team B (when they both work together \(\frac{1}{12} +\frac{1}{36}\)=\(\frac{1}{9}\)

Now according to question,
Let more number of days should team B works to complete the job be x days

\(4 \times \frac{1}{12} +2 \times \frac{1}{9} + x \times \frac{1}{36}=1\)

\(\Rightarrow x=16\)

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AMC 10 USA Math Olympiad

Arithmetic Sequence | AMC 10B, 2003 | Problem No. 24

Problem – Arithmetic Sequence (AMC 10)

The first four terms in an arithmetic sequence are  \( x+y\) , \( x-y\) xy , \(\frac {x}{y}\) in that order. What is the fifth term?

  • \(\frac{12}{110}\)
  • \(\frac {123}{40}\)
  • \(\frac {16}{17}\)
  • 20

Key Concepts

Arithmetic Sequence

Series and Sequence


Check the Answer

Answer: \(\frac {123}{40}\)

American Mathematics Competition

Challenges and Thrills – Pre – College Mathematics

Try with Hints

Here is the first hint to start this sum:

There is a very easy method to do this sum

At first we can try to find the difference between two consecutive terms which is

\( (x-y) – (x+y) = -2y \)

So after that we can understand the third and forth terms in terms of x and y.

They can be : \( ( x-3y ) \) and \( ( x – 5y )\)

Now try to do rest of the sum………………………………..

If you got stuck after the first hint you can use this :

Though we from our solution we find the other two terms to be \((x-3y)\) and \((x-5y)\)

but from the question we find that the other two terms are \(xy\) and \(\frac {x}{y}\)

So both are equal.Thus ,

\(xy = x – 3y\)

\( xy – x = – 3y \)

\( x (y – 1) = -3y \)

\( x = \frac {-3y}{ y – 1} \) ……………………………(1)

Again , similarly

\(\frac {x}{y} = x -5y \)

Now considering the equation (1) we can take the value of \(\frac {x}{y}\)

\(\frac {-3}{y – 1}= \frac {-3y}{y -1} – 5y \) …………………….(2)

\( -3 = -3y – 5y(y-1) \)

\( 0 = 5y^2 – 2y – 3\)

\( 0 = ( 5y +3)(y-1)\)

\( y = – \frac {3}{5} , 1\)

We are almost there with the answer. Try to find the answer…..

Now from the last hint we find the value of \( y = – \frac {3}{5} , 1\)

But we cannot consider the value of y to be 1 as the 1st and 2 nd terms would be \(x+1\) and \(x-1)\) but last two terms will be equal to x .

So the value of y be \(- \frac {3}{5}\) and substituting the value of y in either \(eq^n\) (1) or \(eq^n\) (2) we get x = -\(\frac {9}{8}\)

so , \(\frac {x}{y} -2y = \frac {9.5}{8.3} + \frac {6}{5} \)

= \(\frac {123}{40} \) (Answer )

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AMC 10 Australian Math Competition USA Math Olympiad

Statistics Problem – Australian Mathematics Competition, 2014

Problem No – 21 – Australian Mathematics Competition – 2014

Here is a problem based on Statistics from Australian Mathematics Competition, 2014.

In a competition between four people, Sally scored twice as many points as Brian and 30 points more than Corrie. Donna scored 50 points more than Brian. Which of the following statements is definitely true?

  • Sally won the competition.
  • Brian came last in the competition.
  • Donna won the competition.
  • Corrie beat Brian.
  • Sally and Donna together scored more than Brian and Corrie.

Key Concepts

Mathematical Analysis



Check the Answer

Answer: Sally and Donna together scored more than Brian and Corrie.

Australian Mathematics Competition – Upper Primary Division – 2014 – UP 21

Statistics 10th Edition – Robert S. Witte and John S. Witte

Try with Hints

For first hint we can use a table form with some possibilities :

statistics problem - solution

So in the previous hint from the table its clear that the 1st , 2nd and 4th options are not correct. Now if we go with the higher numbers like 90,100 etc then there will be a change with the values.Try to find it out using a table.

I guess you have noticed the differences in values let try to do that……

statistics problem - solution

So can understand from this table that the 3rd option is also not correct .

We are told that Sally scored 30 points more than Corrie and Donna scored 50 points more than Brian, and so together Sally and Donna always scored 80 points more than Corrie and Brian

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AMC 10 USA Math Olympiad

Missing Numbers – Australian Mathematics Competition

Problem – Hidden Faces of Dice (AMC Middle Primary)

Let’s solve a problem based on missing numbers from the hidden faces of dice.

Three standard dice are sitting next to each other as shown in the diagram. There are 7 faces visible. How many dots are hidden on the other 11 sides?

Missing numbers - dice
  • 26
  • 36
  • 41
  • 63

Key Concepts

Dice Problem

Missing Values


Check the Answer

Answer: 41

Australian Mathematics Competition – Middle Primary Division

Basic Arithmetic by Robert Moon

Try with Hints

This one is very easy but if you really need any hints at first take a dice and try to understand the number of dots in each sides .

So for first dice : we can see 4,5 and 1. So the no of dots which are not seen 6,2 and 3

For Second Dice : The visible dots are 2 and 6 . So the dots in sides which are not visible are 4,3,1 and 5.

For Third Dice : The Visible dots are 3 and 1 .So the dots which are not visible 2,4,5 and 6.

I think you have already got the answer but if you really need this last step :

Now add all the number of dots from the invisible sides:

2+4+5+6+4+3+1+5+6+2+3 = 41 .

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