'Buy three, get one free.' What is the percentage of discount being offered here?
25
20
28.56
33.33
Given linear equations, I, II and III, a learner is not able to solve III algebraically. The most likely area of difficulty is that the learner has not understood
that two equations can be solved by method of substitution
the method of solving equations using graphs
that both the equations in III can be altered by multiplying with suitable numbers
that two equations can be added or subtracted to solve then
C.
that both the equations in III can be altered by multiplying with suitable numbers
Equations 1 and 2 are easier for the child to solve as the two have one of the variables which have the same value but the equation 3 has different values of variables. The child when not be able to solve the third indicates that he/she is having a difficulty in the idea that the two equations can be solved by putting as one of the values of variable equal, which can be done by multiplying the variable by a number to make one of the variables equal.
While teaching ratio and proportion, Ms Rama demonstrated some computer operations on the screen - 'copy and paste' and 'copy and enlarge' or 'copy and reduce'. This activity maybe
formative assessment activity
fun activity to pass time
pre-content activity to introduce ratio
post-content activity
The symbol drawn to any size means a + 4 and the symbol drawn to any size means b2, where a and b are numbers. Then, the value of
32
9
75
35
Two positive numbers x and y are inversely proportional. If x increases by 10%, then y decreases by
%
%
10%
%
A's income is 150% more than that of B, than how much per cent is B's income less than that of A?
50%
60%
12.5%
25%