In class 6T, when only one girl stands up and the rest of the children are sitting down, the number of boys sitting down is 12\frac{1}{2} the number of girls sitting down. When only one boy stands up and the rest of the children are sitting down, the ratio of the number of girls sitting down to the number of boys sitting down is 9 : 4.

  1. (a)

    What is the total number of children in class 6T?

  2. (b)

    After an equal number of girls and boys left the class for competition, the ratio of the number of girls to the number of boys in the class became 9 : 2. Find the total number of children who left the class for competition.

Work through it with AI →

It won't just hand you the answer — it asks where you're stuck.How the checker works

Worked solution

14(a)

1. Formulate equations based on the two scenarios given

Let the total number of boys be BB and the total number of girls be GG.
When one girl stands up, the number of girls sitting is G−1G - 1 and the number of boys sitting is BB.
B=12(G−1)  ⟹  G−1=2B  ⟹  G=2B+1B = \frac{1}{2}(G - 1) \implies G - 1 = 2B \implies G = 2B + 1

When one boy stands up, the number of boys sitting is B−1B - 1 and the number of girls sitting is GG.
GB−1=94  ⟹  4G=9(B−1)\frac{G}{B - 1} = \frac{9}{4} \implies 4G = 9(B - 1)

2. Solve the simultaneous equations to find the number of boys and girls

Substitute G=2B+1G = 2B + 1 into the second equation:
4(2B+1)=9(B−1)4(2B + 1) = 9(B - 1)
8B+4=9B−98B + 4 = 9B - 9
9B−8B=4+99B - 8B = 4 + 9
B=13B = 13

Substitute B=13B = 13 back to find GG:
G=2(13)+1=27G = 2(13) + 1 = 27

3. Calculate the total number of children

Total number of children = B+G=13+27=40B + G = 13 + 27 = 40

Answer

4040

14(b)

1. Determine the initial number of boys and girls from part (a)

From part (a), let the number of boys be BB and girls be GG.
When 1 girl stands: B=12(G−1)  ⟹  2B=G−1  ⟹  G=2B+1B = \frac{1}{2}(G - 1) \implies 2B = G - 1 \implies G = 2B + 1.
When 1 boy stands: GB−1=94  ⟹  4G=9(B−1)=9B−9\frac{G}{B - 1} = \frac{9}{4} \implies 4G = 9(B - 1) = 9B - 9.
Substitute GG: 4(2B+1)=9B−9  ⟹  8B+4=9B−9  ⟹  B=134(2B + 1) = 9B - 9 \implies 8B + 4 = 9B - 9 \implies B = 13.
G=2(13)+1=27G = 2(13) + 1 = 27.
Initial boys = 13, Initial girls = 27.

2. Use the constant difference concept since an equal number of boys and girls left

Initial difference between girls and boys = 27−13=1427 - 13 = 14.
Since an equal number of girls and boys left, the difference remains constant.
New ratio of girls to boys = 9:29 : 2.
Difference in ratio units = 9−2=79 - 2 = 7 units.
So, 7 units=14  ⟹  1 unit=27 \text{ units} = 14 \implies 1 \text{ unit} = 2.

3. Calculate the number of children who left

New number of girls = 9×2=189 \times 2 = 18.
Number of girls who left = 27−18=927 - 18 = 9.
Number of boys who left = 13−(2×2)=13−4=913 - (2 \times 2) = 13 - 4 = 9.
Total number of children who left = 9+9=189 + 9 = 18.

Answer

1414

AI analysis — may contain mistakes.

Worked this one out on paper? Snap your steps and Gigle's AI math checker will circle the step where it went wrong.

More questions on this topic

  1. 1Containers X and Y are rectangular containers. The base area of Container X is 30 cm^2 and the base area of Container Y…
  2. 2Andy had $5x. After buying apples at 60c each, he had $2x left. How many apples did Andy buy? Answer: ___
  3. 3n is a positive integer. Explain whether, for all n, (7n+2)2−(7n−1)2(7n+2)^{2}-(7n-1)^{2} is a multiple of 3. Answer ___
  4. 4A bag contains 4 black coins, 7 red coins and 11 white coins. Two coins are drawn from the bag at random, one after anot…
  5. 5给出下列判断:①单项式−12px2-\dfrac{1}{2}px^{2}的系数为−12-\dfrac{1}{2};②整式x−2xy+1x-2xy+1是二次三项式;③多项式3x2y3−px5+12y3x^{2}y^{3}-px^{5}+\dfrac{1}{2}y的次数…