Question

Hello guys!


i challenging in social science can anybody answer this question


correct answer should be appreciated ​

Answer 1

Answer:

Let the usual speed be x km/hr.

Then, the speed taken =(x+100) km/hr

Distance =1500 km

Usual time = distance/ speed

= 1500 / x hrs

Time taken = 1500÷ (x+100)hrs

It is given that Time taken=Usual time−0.5 hrs

⇒ x1500 − x+1001500 = 21

⇒ x(x+100)1500(x+100)−1500x = 21

⇒x 2 +100x=1500×100×2

⇒x2+100x−300000=0

⇒x= 2−100± 100 2 +4×300000

= 2−100±1100

=500 or−600

Since speed cannot be negative, x=500 km/hr

∴ The usual speed of the plane is 500 km/hr.

Answer 2

Answer:

hey here is your solution

pls mark it as brainliest

Step-by-step explanation:

$so \: let \: the \: usual \: speed \: of \: plane \: be \: x.km \div hour$

$now \: we \: know \: that \\ speed = distance \div time \\ ie \: time = distance \div speed$

$so \: distance \: covered \: by \: plane = 1500 \: km$

$now \: original \: time \: taken \: by \: plane(t1) = 1500 \div x \: hours \\ but \: when \: speed \: is \: increased \: by \: 100 \: km \div hr \\ time \: taken(t2) = 1500 \div x + 100 \: hours$

$so \: we \: know \: that \\ half \: hour = 30 \div 60 \: th \: hour \: ie \: 30 \: mins$

$so \: here \: t2 > t1 \\ according \: to \: given \: condition \\ (1500 \div x + 100) = (1500 \div x) - 30 \div 60 \\ ie \: (1500 \div x) - (1500 \div x + 100) = 30 \div 60 \\ \\ ie \: 1500 \times (1 \div x - 1 \div x + 100) = 1 \div 2 \\ 1500 \times (x + 100 - x) \div x(x + 100) = 1 \div 2$

$so \: hence \: \\ 1500 \times 2 \times 100 = x(x + 100) \\ ie \: x.square + 100x = 300000 \\ x.square + 10x - 300000 = 0$

$ie \: apply \: here \: formula \: method \: for \: better \: calculation \: rather \: than \: factorising \: and \: such \: headache \: and \: time - wasting \: \\ so \: calculation \: part \: is \: in \: above \: pic$

$so \: hence \: the \: usual \: speed \: of \: plane \: was \: 500.km \div hr$