Barisal Board · SSC Mathematics · 2022
Chapter Coverage
| Q | Chapter | Topic | Marks |
|---|---|---|---|
| 1 | Ch 1 – Real Numbers | Rational vs irrational tide measurements | 10 |
| 2 | Ch 4 – Logarithms | Ship cargo weight, log calculations | 10 |
| 3 | Ch 6 – Quadratic Equations | River ferry ticket revenue | 10 |
| 4 | Ch 9 – Geometric Series | Bacterial growth in river water | 10 |
| 5 | Ch 10 – Triangles & Congruence | River crossing triangle proof | 10 |
| 6 | Ch 12 – Similarity | River map scale problem | 10 |
| 7 | Ch 14 – Trigonometry | Two boats, angle of depression from cliff | 10 |
| 8 | Ch 15 – Statistics | Boat passenger count, ogive and median | 10 |
Tide measurements give values p = 3 + √5 and q = 3 − √5.
(a) [2m] Find p + q and p − q. Classify each as rational or irrational.
(b) [4m] Find p × q and p/q in simplified form.
(c) [4m] Show that p² + q² = 38 and p² − q² = 12√5.
p − q = (3+√5)−(3−√5) = 2√5 (irrational)
p/q = (3+√5)/(3−√5) × (3+√5)/(3+√5) = (3+√5)²/4 = (9+6√5+5)/4 = (14+6√5)/4 = (7+3√5)/2
q² = (3−√5)² = 9−6√5+5 = 14−6√5
p²+q² = 28 = 38? Let me recalculate: 14+6√5+14−6√5 = 28.
p²+q² = 28 (not 38 — problem statement has error, correct answer is 28)
p²−q² = (14+6√5)−(14−6√5) = 12√5 ✓
A ship carries cargo weighing 2500 tonnes. (log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990)
(a) [2m] Find log 2500 using the given values.
(b) [4m] If cargo increases to 7500 tonnes, find log 7500.
(c) [4m] Prove: log(2500/7500) = log 2500 − log 7500 and evaluate.
log 2500 = 2·log 5 + 2 = 2(0.6990)+2 = 1.3980+2 = 3.3980
log 7500 = log 3 + 2·log 5 + 2 = 0.4771+1.3980+2 = 3.8751
log(2500/7500) = log(1/3) = −log 3 = −0.4771
Also: 3.3980 − 3.8751 = −0.4771 ✓
A river ferry sells tickets. If the fare is reduced by Tk 5, 20 more passengers travel. Currently 100 passengers travel at Tk 50 each.
(a) [2m] Let fare reduction be x. Write expressions for new fare and new passenger count.
(b) [4m] Write and solve a quadratic equation so that new revenue equals old revenue (Tk 5000).
(c) [4m] Find the fare reduction that maximises revenue. (Let revenue R = (50−x)(100+4x) and find its maximum.)
For every Tk 5 reduction, 20 more passengers: ratio = 20/5 = 4 per Tk 1 reduction.
New passengers = 100 + 4x
5000+200x−100x−4x² = 5000
100x−4x² = 0
4x(25−x) = 0
x = 0 or x = 25
So fare = 50−25 = 25, passengers = 100+100 = 200, revenue = 25×200 = 5000 ✓
dR/dx = 100−8x = 0 → x = 12.5
Max fare = 50−12.5 = 37.5; Passengers = 100+50 = 150
Max R = 37.5×150 = Tk 5,625
Bacteria in river water double every hour. Initially there are 500 bacteria per mL.
(a) [2m] Write the GP for bacterial count at hours 0, 1, 2, 3.
(b) [4m] Find the count after 8 hours.
(c) [4m] After how many hours does the count first exceed 100,000? (log 2 = 0.3010)
500, 1000, 2000, 4000
2ⁿ > 200
n·log2 > log200 = log(2×100) = log2+2 = 2.3010
n × 0.3010 > 2.3010
n > 7.65
n = 8 hours (count = 128,000 > 100,000 ✓)
Two boats leave from opposite banks of a river, crossing at right angles. △ABC has AB = AC = 10 m (isosceles), and BC = 12 m (river width).
(a) [2m] Show that the altitude from A to BC bisects BC.
(b) [4m] Find the length of the altitude AD from A to BC.
(c) [4m] Find angle BAC and angle ABD using trigonometric ratios. (sin⁻¹ of values needed)
10² = AD² + 6²
AD² = 100 − 36 = 64
AD = 8 m
∠ABD = ∠ABC = 53.13° (since △ABD ≅ △ACD)
∠BAC = 180° − 2×53.13° = 73.74°
On a map of the Barisal river system, a river stretch of 4 cm represents an actual distance of 20 km. A lake has map area of 6 cm².
(a) [2m] Find the map scale ratio.
(b) [4m] Find the actual area of the lake in km².
(c) [4m] Two similar river bends have corresponding lengths 5 cm and 8 cm on the map. If the smaller bend's actual perimeter is 30 km, find the larger's actual perimeter.
Scale = 4 : 2,000,000 = 1 : 500,000
Area scale = k² = (500,000)² = 2.5 × 10¹¹ cm²/cm²
Map area = 6 cm² → Actual = 6 × (500,000)² cm² = 6 × 2.5×10¹¹ cm²
= 1.5×10¹² cm² = 1.5×10¹² / 10¹⁰ km² = 150 km²
Perimeter ratio = 5 : 8 (same as length ratio for similar figures)
Smaller actual perimeter = 30 km → Larger = 30 × (8/5) = 48 km
From the top of a cliff 50 m high above a river, two boats are seen. Boat A's angle of depression is 30° and Boat B's (on the same side) angle of depression is 60°.
(a) [2m] Sketch the diagram showing both boats, the cliff, and angles.
(b) [4m] Find the distance of each boat from the base of the cliff. (tan 30° = 1/√3, tan 60° = √3)
(c) [4m] Find the distance between the two boats.
For Boat B (depression = 60°): tan 60° = 50/d_b → d_b = 50/√3 = 50√3/3 ≈ 28.87 m
Distance AB = dₐ − d_b = 50√3 − 50/√3 = 50√3 − 50√3/3 = 50√3(1−1/3) = 50√3 × 2/3 = 100√3/3 ≈ 57.74 m
Daily passenger count on a river ferry over 50 trips: 0–20: 8 trips, 20–40: 14, 40–60: 16, 60–80: 9, 80–100: 3.
(a) [2m] Find the modal class and calculate the mean using midpoints.
(b) [4m] Construct cumulative frequency and find the median.
(c) [4m] Find the percentage of trips with more than 60 passengers.
Mean = Σfx/Σf: midpoints 10,30,50,70,90
Σfx = 8×10+14×30+16×50+9×70+3×90 = 80+420+800+630+270 = 2200
Mean = 2200/50 = 44 passengers
Median at 25th value → in class 40–60 (CF reaches 38 here, previous was 22)
Median = 40 + [(25−22)/16] × 20 = 40 + (3/16)×20 = 40 + 3.75 = 43.75 passengers
Percentage = (12/50) × 100 = 24%