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A class has 50 pupils. 20 like Math only, 15 like English only, and 5 like neither subject. Find the number of pupils who like Math.

Mathematics

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Here are the solutions to the questions:

Question 21:

  • Total pupils = 50
  • Pupils who like Mathematics (M) only = 20
  • Pupils who like English (E) only = 15
  • Pupils who do not like any of the two subjects = 5
  • Let KK be the number of pupils who like both subjects.

a) Draw a Venn diagram and represent the above information.
A Venn diagram would consist of a rectangle representing the universal set (50 pupils). Inside this rectangle, there would be two overlapping circles, one labeled 'M' for Mathematics and the other 'E' for English.

  • The region of circle M only (not overlapping with E) would contain the number 20.
  • The region of circle E only (not overlapping with M) would contain the number 15.
  • The region outside both circles (but inside the rectangle) would contain the number 5.
  • The overlapping region (intersection of M and E) would contain the variable KK.

b) Find the number of pupils who like Mathematics.

  1. Calculate the number of pupils who like both subjects (KK).

    The total number of pupils is the sum of pupils in all regions:

    Total=(M only)+(E only)+(Both)+(Neither)\text{Total} = (\text{M only}) + (\text{E only}) + (\text{Both}) + (\text{Neither}) 50=20+15+K+550 = 20 + 15 + K + 5 50=40+K50 = 40 + K K=50−40K = 50 - 40 K=10K = 10
  2. Calculate the number of pupils who like Mathematics.

    Pupils who like Mathematics include those who like Mathematics only and those who like both subjects.

    Pupils who like Mathematics=(M only)+(Both)\text{Pupils who like Mathematics} = (\text{M only}) + (\text{Both}) Pupils who like Mathematics=20+10\text{Pupils who like Mathematics} = 20 + 10 Pupils who like Mathematics=30\text{Pupils who like Mathematics} = 30

    The number of pupils who like Mathematics is 30\boxed{\text{30}}.

    Question 22:
    The figure shows a triangle KLM with angles:

    • ∠M=(2y+20)∘\angle M = (2y + 20)^\circ
    • ∠K=(3y+10)∘\angle K = (3y + 10)^\circ
    • ∠L=(y+30)∘\angle L = (y + 30)^\circ

    a) Find the value of y.

  3. Use the property that the sum of angles in a triangle is 180∘180^\circ.

    ∠M+∠K+∠L=180∘\angle M + \angle K + \angle L = 180^\circ (2y+20)+(3y+10)+(y+30)=180(2y + 20) + (3y + 10) + (y + 30) = 180
  4. Combine like terms.

    (2y+3y+y)+(20+10+30)=180(2y + 3y + y) + (20 + 10 + 30) = 180 6y+60=1806y + 60 = 180
  5. Solve for yy.

    6y=180−606y = 180 - 60 6y=1206y = 120 y=1206y = \frac{120}{6} y=20y = 20

    The value of yy is 20\boxed{20}.

    b) Work out the size of angle KLM.

  6. Substitute the value of yy into the expression for ∠L\angle L (which is ∠KLM\angle KLM).

    ∠KLM=(y+30)∘\angle KLM = (y + 30)^\circ ∠KLM=(20+30)∘\angle KLM = (20 + 30)^\circ ∠KLM=50∘\angle KLM = 50^\circ

    The size of angle KLM is 50∘\boxed{50^\circ}.

    Question 23:

    a) Write the integer represented by the arrow on the number line above.

    i) a = 4\boxed{4}

    ii) b = −3\boxed{-3}

    iii) c = 2\boxed{2}

    b) Write down the mathematical statement shown on the number line above.
    The number line shows a segment from -3 to 2, inclusive.
    The mathematical statement is −3≤x≤2\boxed{-3 \le x \le 2}.

    Question 24:
    The table shows marks scored by some pupils in a Mathematics test.
    Marks: 80, 70, 60, 90
    Number of pupils: (empty row)

    No questions are provided for this table, and the "Number of pupils" row is empty. Therefore, this question cannot be answered with the given information.

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