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Original Question
in a class of 70 students, 6 offer economics only, 18 offer economics but mathematics, 36 offer economics and geography, 53 off...Asked by Prince sebi
In a class of 70 students,6 offer Economics only,18 offer Economics but not mathematics,36 offer Economics and Geography,53 offer Economics,50 offer geography and 34 offer mathematics and geography.All students offer at least one subject.
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Answered by
Bot
Using a Venn diagram, we can visualize the information given in the problem:
[asy]
label("Economics", (2,85));
label("Mathematics", (-55,0));
label("Geography", (65,0));
draw(Circle((0,45), 29.5));
draw(Circle((-35, -20), 29.5));
draw(Circle((35, -20), 29.5));
label("$6$", (0, 75));
label("$12$", (-13, 45));
label("$6$", (13, 45));
label("$6$", (-45, -10));
label("$12$", (0, -10));
label("$24$", (45, -10));
label("$18$", (0, 18));
label("$15$", (-23.5, 22));
label("$21$", (23.5, 22));
label("$19$", (-17.5, -10));
label("$15$", (0, -32.5));
label("$15$", (17.5, -10));
label("$15$", (17.5, -32.5));
label("$4$", (-27.5, -32.5));
label("$15$", (27.5, -32.5));
label("$19$", (13.5, -45));
label("$15$", (-13.5, -45));
filldraw(circle((0,45),29.5),white);
filldraw(circle((-35, -20), 29.5),white);
filldraw(circle((35, -20), 29.5),white);
[/asy]
Let:
- $x$ be the number of students who offer Economics, Mathematics, and Geography
- $y$ be the number of students who offer Mathematics and Geography only
- $z$ be the number of students who offer Mathematics only
- $w$ be the number of students who offer Geography only
Using these variables, we can create a system of equations based on the given information:
- $6 + 12 + x + 15 + 19 = 70$ (all students offer at least one subject)
- $6 + 12 + x + 18 + y = 70$ (all students offer at least one subject)
- $6 + 12 + x + 15 + y + z = 70$ (all students offer at least one subject)
- $6 + x + 15 + 19 + y + w = 70$ (all students offer at least one subject)
- $x + y + z + w = 0$ (no students offer no subjects)
Simplifying and solving the system of equations, we get:
- $x = 4$
- $y = 15$
- $z = 10$
- $w = 5$
Therefore, there are 10 students who offer Mathematics only, 4 students who offer Economics, Mathematics, and Geography, and 15 students who offer Mathematics and Geography only. The remaining numbers for each category can be found by adding the relevant numbers from the diagram:
- Economics only: 6
- Economics and Mathematics only: 12
- Economics and Geography only: 18
- Mathematics and Geography only: 15 + 5 = 20
- Economics, Mathematics, and Geography: 4
Checking that the sum of these numbers is equal to 70, we can conclude that our solution is correct.
[asy]
label("Economics", (2,85));
label("Mathematics", (-55,0));
label("Geography", (65,0));
draw(Circle((0,45), 29.5));
draw(Circle((-35, -20), 29.5));
draw(Circle((35, -20), 29.5));
label("$6$", (0, 75));
label("$12$", (-13, 45));
label("$6$", (13, 45));
label("$6$", (-45, -10));
label("$12$", (0, -10));
label("$24$", (45, -10));
label("$18$", (0, 18));
label("$15$", (-23.5, 22));
label("$21$", (23.5, 22));
label("$19$", (-17.5, -10));
label("$15$", (0, -32.5));
label("$15$", (17.5, -10));
label("$15$", (17.5, -32.5));
label("$4$", (-27.5, -32.5));
label("$15$", (27.5, -32.5));
label("$19$", (13.5, -45));
label("$15$", (-13.5, -45));
filldraw(circle((0,45),29.5),white);
filldraw(circle((-35, -20), 29.5),white);
filldraw(circle((35, -20), 29.5),white);
[/asy]
Let:
- $x$ be the number of students who offer Economics, Mathematics, and Geography
- $y$ be the number of students who offer Mathematics and Geography only
- $z$ be the number of students who offer Mathematics only
- $w$ be the number of students who offer Geography only
Using these variables, we can create a system of equations based on the given information:
- $6 + 12 + x + 15 + 19 = 70$ (all students offer at least one subject)
- $6 + 12 + x + 18 + y = 70$ (all students offer at least one subject)
- $6 + 12 + x + 15 + y + z = 70$ (all students offer at least one subject)
- $6 + x + 15 + 19 + y + w = 70$ (all students offer at least one subject)
- $x + y + z + w = 0$ (no students offer no subjects)
Simplifying and solving the system of equations, we get:
- $x = 4$
- $y = 15$
- $z = 10$
- $w = 5$
Therefore, there are 10 students who offer Mathematics only, 4 students who offer Economics, Mathematics, and Geography, and 15 students who offer Mathematics and Geography only. The remaining numbers for each category can be found by adding the relevant numbers from the diagram:
- Economics only: 6
- Economics and Mathematics only: 12
- Economics and Geography only: 18
- Mathematics and Geography only: 15 + 5 = 20
- Economics, Mathematics, and Geography: 4
Checking that the sum of these numbers is equal to 70, we can conclude that our solution is correct.
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