Wassce Core Mathematics 1993 Paper 2 Question 21 – Answer

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Wassce Core Mathematics 1993 Paper 2 Question 21 – Answer is an e-learning material we have prepared online. This learning resource has been designed to help students to pass their Core Maths WAEC exams. I believe you are imagining what our website can offer you? If so, then just wait and see. Read more. about our website and what we offer.

 

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Wassce Core Mathematics 1993 Paper 2 Question 21 – Answer is an e-learning material we have prepared online. This learning resource has been designed to help students to pass their Core Maths WEC exams. I must however, make it known that we may not always provide a pdf download of this wassce past question. This means that WASSCE Core Mathematics past questions and answers may not be available for a direct pdf download. So, what does our website offer? Just wait and see. Also, have you had challenges trying to learn, write or pass your WAEC Maths exams? If the answer is yes, then I can assure you that educareguide will help and guide you.  

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Details Of the Wassce Core Mathematics 1993 Question 21

In a Senior Secondary School, there are 174 students in Form 2. Of these, 86 play table tennis, 84 play football, 94 play volleyball; 30 play table tennis and volleyball, 34 play volleyball and football and 42 play table tennis and football. Each student plays at least one of the three games and x students play all three games.

  1. Display these facts in a Venn diagram.
  2. Write down an equation in x and hence find
  3. If a student is chosen at random from Form 2, what is the probability that he plays two games?

The answer To Wassce Core Mathematics 1993 Question 21

n(U) =174 , n(T) = 86, n(F) = 84, n(V) = 94, n(T nV) = 30, n(F nV) = 34, n(T nV) = 42.

a.   The Venn diagram is shown below:

U (174)

 

 

 

 

 

 

b.   From the Venn diagram,

Number of students who play Table tennis only,       a = 86—(30—x+x+42—x) =14+x

Furthermore, the number of students who play Football only,       b=84—(42—x+x+34—x) =8+x

Also, the number of students who play Volleyball only,       c=94—(30—x+x+34—x) =30+x

 

n(T F V)= n(U)

(14+x)+(8+x)+(30+x)+x+(42—x)+ (30—x) + (34—x) =174 158+x=174

x =16

The ans is 16 students

 

(c) Number of students who play two games

= (42—x) + (30—x) + (34—x)

= (42 —16) + (30 —16) + (34 –16)

=26+14+18 =58

 ⸫ Probability that a student plays two games

      58/173 = 1/3

 

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