Live practice

Formula or theorem used: For grouped data use class marks and frequencies systematically. Apply the direct, assumed-mean, or step-deviation formula and show the working table totals.

Working: Write the relevant mean, median, or mode formula, identify the modal/median class, substitute the table totals, and interpret the numerical result in the question's context.

After substitution and simplification:

Median = 8.05, Mean = 8.32, Modal size = 7.88

Opposite sides of a parallelogram are parallel, so \(AB\parallel CD\) and \(AD\parallel BC\). Since \(F\) lies on \(CD\), \(CF\parallel AB\); since \(E\) lies on the extension of \(AD\), \(AE\parallel BC\).

With \(BE\) as a transversal,

Therefore, \(\triangle ABE\sim\triangle CFB\) by AA similarity.

Formula or theorem used: For an AP use \(a_n=a+(n-1)d\) and \(S_n=\frac n2[2a+(n-1)d]\).

Working: Identify \(a\), \(d\), and the required term or sum; substitute them in the appropriate formula and simplify line by line.

After substitution and simplification:

20th term from the last term is 158.

Two dice give (6\times6=36) equally likely ordered outcomes. The sums (2) to (12) occur in (1,2,3,4,5,6,5,4,3,2,1) ways, respectively.

The eleven sums are not equally likely, so assigning each probability (\frac1{11}) is incorrect.

Formula or theorem used: For grouped data use class marks and frequencies systematically. Apply the direct, assumed-mean, or step-deviation formula and show the working table totals.

Working: Write the relevant mean, median, or mode formula, identify the modal/median class, substitute the table totals, and interpret the numerical result in the question's context.

After substitution and simplification:

Median weight = \(56.67\,\mathrm{kg}\)

\[\begin{aligned}\angle ABC &= \angle AMP\\&= 90^\circ\end{aligned}\]. Also \(AB\) and \(AP\) are along the same straight line, while \(AC\) and \(AM\) are along the same line; hence \(\angle BAC=\angle PAM\).

Therefore, \(\triangle ABC\sim\triangle AMP\) by AA similarity.

The hypotenuse \(CA\) corresponds to \(PA\), and side \(BC\) corresponds to \(MP\). Therefore,

Formula or theorem used: For an AP use \(a_n=a+(n-1)d\) and \(S_n=\frac n2[2a+(n-1)d]\).

Working: Identify \(a\), \(d\), and the required term or sum; substitute them in the appropriate formula and simplify line by line.

After substitution and simplification:

–13, –8, –3

For three tosses, (S=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}). Hanif wins for (HHH) or (TTT), so he loses in (6) outcomes.

Since \(\triangle ABC\sim\triangle FEG\), \(\angle A=\angle F\), \(\angle B=\angle E\), and \(\angle C=\angle G\). The angle bisectors give

Thus \(\angle ACD=\angle FGH\) and \(\angle CAD=\angle GFH\), so \(\triangle ACD\sim\triangle FGH\) by AA. Hence

Further, \(\angle DCB=\angle HGE\) and \(\angle DBC=\angle HEG\), so \(\triangle DCB\sim\triangle HGE\).

Similarly, \(\angle DCA=\angle HGF\) and \(\angle DAC=\angle HFG\), so \(\triangle DCA\sim\triangle HGF\).

Formula or theorem used: For an AP use \(a_n=a+(n-1)d\) and \(S_n=\frac n2[2a+(n-1)d]\).

Working: Identify \(a\), \(d\), and the required term or sum; substitute them in the appropriate formula and simplify line by line.

After substitution and simplification:

11th year