Live practice

Formula or theorem used: In a right triangle use \(\sin\theta=\frac{P}{H}\), \(\cos\theta=\frac{B}{H}\), and \(\tan\theta=\frac{P}{B}\), together with the standard-angle values and identities.

Working: Write the relevant ratio or identity, substitute the known value, simplify one equality at a time, and state restrictions when a denominator could be zero.

After substitution and simplification:

(i) 1 (ii) 2 (iii) 3 2 6
8
− (iv) 43 24 3
11
− (v) 67
12

Formula or theorem used: Use \(PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) or the section formula \(\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\right)\).

Working: Substitute the coordinates before simplifying. For a shape, compare the required side lengths or diagonals and state the conclusion.

After substitution and simplification:

\(\left(2,-\frac53\right)\) and \(\left(0,-\frac73\right)\)

Consider \(\triangle AOB\) and \(\triangle COD\).

\(\angle AOB=\angle COD\) because they are vertically opposite angles. Also, \(\angle ABO=\angle CDO\) because \(AB\parallel DC\) and \(BD\) is a transversal.

Therefore, \(\triangle AOB\sim\triangle COD\) by AA similarity. Corresponding sides are proportional:

Cross-multiplying gives \(AO\cdot DO=BO\cdot CO\). Dividing by \(BO\cdot DO\),

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:

64

Formula or theorem used: Write the equation in the form \(ax^2+bx+c=0\). Factorise where possible; otherwise use \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\).

Working: Display the standard form, the factorisation or discriminant calculation, both roots, and reject any value that is impossible in the stated context.

After substitution and simplification:

No

Formula or theorem used: Translate the information into two linear equations. Use the method named in the exercise, keeping equivalent equations aligned.

Working: Eliminate or substitute one variable, obtain the other variable, substitute back, and verify the ordered pair in both original equations.

After substitution and simplification:

(i) x – y + 2 = 0, 2x – y – 1 = 0, where x and y are the numerator and denominator of the
fraction;
3
5 ⋅
(ii) x – 3y + 10 = 0, x – 2y – 10 = 0, where x and y are the ages (in years) of Nuri and Sonu
respectively. Age of Nuri (x) = 50, Age of Sonu (y) = 20.
(iii) x + y = 9, 8x – y = 0, where x and y are respectively the tens and units digits of the
number; 18.
(iv) x + 2y = 40, x + y = 25, where x and y are respectively the number of ₹ 50 and ₹ 100
notes; x = 10, y = 15.
(v) x + 4y = 27, x + 2y = 21, where x is the fixed charge (in ₹) and y is the additional
charge (in ₹) per day; x = 15, y = 3.

For one die, (S=\{1,2,3,4,5,6\}).

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:

Mode : 30.6, Mean = 29.2. Most states/U.T. have a student teacher ratio of 30.6 and on
an average, this ratio is 29.2.

The cuboid volume is

The volume removed by four cones is

Therefore the wood volume is \(525-1.47\approx523.53\text{ cm}^3\).

For an inscribed regular hexagon, each side equals the radius, \(28\,\mathrm{cm}\). The circle area is

The hexagon consists of six equilateral triangles:

Decorated area \(=2464-1999.2=464.8\text{ cm}^2\). Cost \(=464.8\times0.35=₹162.68\).