StudentAidUnit 2: Quadratic Equations — Exercise 2 7
10th Class Mathematics · Unit 2: Quadratic Equations
2.7.1.A town's population is modeled by P(t) = -2t² + 40t + 800, where t is years since 2020. Find the years when the population will be at least 1000.
Given
P(t)ge1000
-2t2+40t+800ge1000 ⇒ t2-20t+100le0 (div-2)
(t-10)2le0 ⇒ t=10
Result
Year
2020+10=2030
2.7.2.A company models its profit P in thousands of rupees by the equation: P(x) = -5x² + 150x - 1000, where x is the price per item in rupees. Find the price that gives maximum profit.
Given
P(x)=-5x2+150x-1000, a=-5<0
Formula
Vertex
x=-b2a
Working
x=-1502(-5)=15
Result
Price for maximum profit
Rs. 15
2.7.3.A toy car rolls down an incline and covers a distance given by d = t² - 0.5t metres, where t is the time in seconds. Find the time when the car has travelled a distance 12.5 metres.
Given
t2-0.5t=12.5
Multiply by 2
2t2-t-25=0
Formula
Quadratic formula
t=frac{1 ± sqrt{1+200}{4}=frac{1 ± sqrt{201}{4}
Result
t=frac{1+sqrt{201}{4} ≈ 3.79 s (negative root rejected)
2.7.4.A ball's height (in metres) after t seconds is h(t) = -4t² + 24t. For what time interval is the ball at least 20m above the ground?
Given
h(t)ge20
-4t2+24t-20ge0 ⇒ t2-6t+5le0 (div-4)
(t-1)(t-5)le0
Result
Time interval
1le tle5 seconds
2.7.5.A ball is thrown upward with an initial velocity of 40 ms⁻¹. Calculate the maximum height it reaches above ground level.
Given
u=40 ms-1, g=9.8 ms-2
Formula
Height
h(t)=ut-frac12gt2=-4.9t2+40t
Time of max height
t=-402(-4.9) ≈ 4.082 s
Result
Maximum height
≈ 81.65 metres
2.7.6.A freelancer's earnings follow the model E(h) = -2h² + 40h, where E is earning in rupees and h is hours worked per week. What is the maximum number of hours he should work to maximize earnings?
Given
E(h)=-2h2+40h, a=-2<0
Formula
Vertex
h=-b2a
Working
h=-402(-2)=10
Result
Maximum hours to work
10 hours