Question #f982f

2 Answers
Nov 4, 2017

#7#

Explanation:

let the number of tests taken in the first place be#" " i#

then the sum of these test is #sumx_i#

from the data

#((sumx_i)+73)/(i+1)=87#

#:. ((sumx_i)+73)=87(i+1)---(1)#

also

#((sumx_i)+97)/(i+1)=90#

#:.((sumx_i)+97)=90(i+1)---(2)#

#(2) - (1)#

the #sumx_is# cancel

#24=3(i+1)#

#:.8=i+1#

#i=7#

Nov 4, 2017

Andrew has taken 7 tests.

Explanation:

Summary: You need to find the equations of both cases and then simultaneously solve it.

Detailed step-by-step method & explanation:
Firstly, you need to know that the formula to find the mean is #(sumx)/n#. Since we don't know the #sumx# in this case, we can use the variable #y# to represent it. So, the mean for the all the tests before is #y/n#.

Now, we need to find the equation for Andrew's marks if he gets 73 in his next test. That will be #(y+73)/(n+1)=87#. The new mark is 73, so it's added to the total #y# and the total number of tests #n# is increased by one. Rearrange the equation to get #y+73=87n+87#. This will be equation one.

Then, we need to find the equation for Andrew's marks if he gets 97. This will be #(y+97)/(n+1)=90#. Rearrange to get #y+97=90n+90#. This will be equation two.

Now, we need to simultaneously solve it. This can be done in either be done by substitution or elimination. (I will show both methods but you really only need to do one, depending on which you find easier)

  • Substitution: Rearrange equation one to make #y# the subject and you should get #y=87n+14#. Substitute this value for #y# in equation two, which gives
    #(87n+14) +97=90n+90#
    #14+97-90=90n-87n#
    #21=3n#
    #n=7#

  • Elimination: Subtract the equation two from equation one (or vice versa; it doesn't matter). So,
    #( y+73=87n+87)#
    #-(y+97=90n+90)#
    which when you solve gives
    #-24=-3n-3#
    #-24+3=-3n#
    #(-21)/-3=n#
    #n=7#

Hope this helped!