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When a fifth of the class is absent, there are 24 pupils present. What is the total number of pupils in the class?

Since 1/5 of the class is absent, this must mean that 4/5 of the class is present. We are told that 24 pupils are present, so 4/5 of the class must equal 24 pupils. Now we need to find out how many pupils 1/...
AT
Answered by Arisera T. • Maths tutor
9523 Views

Solve algebraically 6a + b = 16 & 5a - 2b = 19

6a + b = 16 Equation 1 5a - 2b = 19 Equation 2 Scale up one equation so both equations have same coeffient of one variable ie 2 times Equ 1 12a + 2b = 32 5a - 2b = 19 Add variables of each equation, hence el...
EM
Answered by Euan M. • Maths tutor
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Find the solutions to the quadratic equation x^2-9x+20=0

By factorising the expression, (x-5)(x-4)=0. This means that either (x-5)=0 or (x-4)=0. By solving these equations we can work out the solutions to be x=4,5.
CM
Answered by Chris M. • Maths tutor
3892 Views

Find the perpendicular bisector passing through the stationary point of the curve y=x^2+2x-7.

First thing to do is to find the stationary point of the curve. This is done by differentiating the function and then equating to zero, as the gradient of the stationary point is zero. Setting dy/dx to zero ...
CM
Answered by Chris M. • Maths tutor
3624 Views

12 + 37 + 47

12 + 37 + 47 = (10 + 30 + 40) + (2 + 7 + 7) = 80 + 16 = 96 It is important to visualize well the operation, being able to separate and put together the numbers that are easy to calculate together. This makes...
MH
Answered by Marie H. • Maths tutor
3867 Views