Question

A large water tank begins to leak at the rate of 25 litres per hour. The amount if water escaping is 25 litres more than the previous hour till the leak was undetected. When the leak was eventually discovered the rate of leakage was 900 litres per hour. For how long had the tank been leaking? What was the total amount of water lost between the beginning of the leakage and the time of discovery?​

Answer 1

Answer:

Step-by-step explanation:

Starting leakage 25 liter per hour then increases next hour 50 lit similarly .....

25 , 50 ,75 , .......900

a= 25

d= 25

tₙ=a +(n-1)d

900=25 (n-1)25

n=36

36 hrs tank had leakage

now Sn

Sn = n/2 [t₁+tₙ]

S₃₆= 36/2 [25+900]

S₃₆ = 18 × 925

S₃₆ = 16650 liter

Amount of water leakage 16650 liter .......

Answer 2

For a period of 36 hours the water tank leaked. The total amount of water lost between the beginning of the leakage and the time of discovery is 16650 liters.

• From given question, we can arrange the data in AP series, i.e.,
• the water tank began to leak at =25 liters
• The amount of water escaping doubled the previous time = 50 liters
• At last  the rate of leakage was found to be = 900 liters
• now, we have series,
• 25,50,.........,900
• a=25
• d=50-25=25
• $a_n$=900
• we have formula
• $a_n=a+(n-1)d\\\\900=25+(n-1)25\\\\875=(n-1)25\\\\35=n-1\\\\n=36$
• The n implies the number of hours the tank leaked.
• Now,
• the total amount of water lost between the beginning of the leakage and the time of discovery
• we have formula
• $S_n=\dfrac{n}{2}[a+a_n]\\\\=\dfrac{36}{2}[25+900]\\\\=18*925\\\\=16650$liters of water