Vardaan Learning Institute
Answer Key: Competency Based Assessment
Class: 8 (CBSE)
Subject: Mathematics
Solutions
SOLUTIONS
Case 1: The Charity Drive
1. Total Fraction Donated = $\frac{1}{3} + \frac{1}{4} = \frac{4+3}{12} = \frac{7}{12}$.
2. Fraction Left = $1 - \frac{7}{12} = \frac{5}{12}$.
3. Donation to Food Bank = $\frac{1}{4}$ of Salary.
Amount = $\frac{1}{4} \times 60000 = \text{₹}15,000$.
2. Fraction Left = $1 - \frac{7}{12} = \frac{5}{12}$.
3. Donation to Food Bank = $\frac{1}{4}$ of Salary.
Amount = $\frac{1}{4} \times 60000 = \text{₹}15,000$.
Case 2: The School Excursion
1. Equation: Bus + Car + Van = Total $\Rightarrow$ $\frac{x}{2} + \frac{x}{4} + 20 = x$.
2. Solving: $\frac{2x+x}{4} + 20 = x \Rightarrow \frac{3x}{4} + 20 = x$.
$20 = x - \frac{3x}{4} = \frac{x}{4} \Rightarrow x = 80$.
3. Students by Bus = $\frac{x}{2} = \frac{80}{2} = 40$.
2. Solving: $\frac{2x+x}{4} + 20 = x \Rightarrow \frac{3x}{4} + 20 = x$.
$20 = x - \frac{3x}{4} = \frac{x}{4} \Rightarrow x = 80$.
3. Students by Bus = $\frac{x}{2} = \frac{80}{2} = 40$.
Case 3: The Farm Land
1. Angle A + Angle B = $180^\circ$ (Adjacent angles in parallelogram are supplementary).
$\angle B = 180^\circ - 70^\circ = 110^\circ$.
2. $\angle C = \angle A = 70^\circ$ (Opposite angles).
$\angle D = \angle B = 110^\circ$ (Opposite angles).
3. Measures of rectangle angles are $90^\circ$. So $\angle A$ would be $90^\circ$.
$\angle B = 180^\circ - 70^\circ = 110^\circ$.
2. $\angle C = \angle A = 70^\circ$ (Opposite angles).
$\angle D = \angle B = 110^\circ$ (Opposite angles).
3. Measures of rectangle angles are $90^\circ$. So $\angle A$ would be $90^\circ$.
Case 4: Sports Selection
1. Fraction Cricket = $\frac{120^\circ}{360^\circ} = \frac{1}{3}$.
2. Angle Basketball = $360^\circ - (120^\circ + 90^\circ) = 360^\circ - 210^\circ = 150^\circ$.
3. Football Fraction = $\frac{90^\circ}{360^\circ} = \frac{1}{4}$.
Students = $\frac{1}{4} \times 720 = 180$.
2. Angle Basketball = $360^\circ - (120^\circ + 90^\circ) = 360^\circ - 210^\circ = 150^\circ$.
3. Football Fraction = $\frac{90^\circ}{360^\circ} = \frac{1}{4}$.
Students = $\frac{1}{4} \times 720 = 180$.
Case 5: The Auditorium Seating
1. Concept used: Square Root.
2. Rows = $\sqrt{1764}$. Factors: $1764 = 2^2 \times 3^2 \times 7^2$.
$\sqrt{1764} = 2 \times 3 \times 7 = 42$. So, 42 rows.
3. Total Chairs = $1764 + 100 = 1864$.
$43^2 = 1849$, $44^2 = 1936$. 1864 is not a perfect square. No.
2. Rows = $\sqrt{1764}$. Factors: $1764 = 2^2 \times 3^2 \times 7^2$.
$\sqrt{1764} = 2 \times 3 \times 7 = 42$. So, 42 rows.
3. Total Chairs = $1764 + 100 = 1864$.
$43^2 = 1849$, $44^2 = 1936$. 1864 is not a perfect square. No.
Case 6: The Festival Sale
1. Discount = $10\%$ of 40,000 = $\text{₹}4,000$.
2. Selling Price (before tax) = $40000 - 4000 = \text{₹}36,000$.
3. GST = $18\%$ of 36,000 = $0.18 \times 36000 = \text{₹}6,480$.
Total Bill = $36000 + 6480 = \text{₹}42,480$.
2. Selling Price (before tax) = $40000 - 4000 = \text{₹}36,000$.
3. GST = $18\%$ of 36,000 = $0.18 \times 36000 = \text{₹}6,480$.
Total Bill = $36000 + 6480 = \text{₹}42,480$.
Case 7: The Rectangular Park
1. Length $l = 2b + 5$.
2. Perimeter $P = 2(l + b) = 2(2b + 5 + b) = 2(3b + 5) = 6b + 10$.
3. If $b=10$, $l = 2(10) + 5 = 25$m.
Area = $l \times b = 25 \times 10 = 250 m^2$.
2. Perimeter $P = 2(l + b) = 2(2b + 5 + b) = 2(3b + 5) = 6b + 10$.
3. If $b=10$, $l = 2(10) + 5 = 25$m.
Area = $l \times b = 25 \times 10 = 250 m^2$.
Case 8: The Water Tank
1. CSA = $2\pi rh = 2 \times \frac{22}{7} \times 7 \times 10 = 440 m^2$.
2. Cost = $440 \times 20 = \text{₹}8,800$.
3. Volume (Capacity) = $\pi r^2 h = \frac{22}{7} \times 7^2 \times 10 = 1540 m^3$.
2. Cost = $440 \times 20 = \text{₹}8,800$.
3. Volume (Capacity) = $\pi r^2 h = \frac{22}{7} \times 7^2 \times 10 = 1540 m^3$.
Case 9: The Micro-Organisms
1. Plant Cell = $1.275 \times 10^{-5}$ m.
2. Red Blood Cell = $7.0 \times 10^{-6}$ m.
3. Total Size = $100 \times 1.275 \times 10^{-5} = 1.275 \times 10^{-3}$ m.
2. Red Blood Cell = $7.0 \times 10^{-6}$ m.
3. Total Size = $100 \times 1.275 \times 10^{-5} = 1.275 \times 10^{-3}$ m.
Case 10: The Delivery Boy
1. Horizontal line indicates the boy was at rest (stopped) during that time interval.
2. Speed = Distance / Time = 10 km / 1 hr = 10 km/hr.
3. A steeper slope indicates greater speed (traveling faster), as more distance is covered in less time.
2. Speed = Distance / Time = 10 km / 1 hr = 10 km/hr.
3. A steeper slope indicates greater speed (traveling faster), as more distance is covered in less time.