Answer:
24
Step-by-step explanation:
A + B = 40.
And at Rs 7 a kg for 40 kg, you want a total of Rs.280.
So the second equation is 9A + 4B = 280
From the first equation: B = 40 – A
and sub into the second equation:
9A + 4(40-A) = 280
9A + 160 – 4A = 280
5A = 120
A = 24.
And you should check: B should equal 40-24 = 16. Check with the final equation: 9*24 + 4*16 = 216 + 64 = 280. So it works.
Your answer, of course, is A = 24
DISTANCE=TIME *SPEED
D=?
48*(5/18)=13.33m/s
D=9*13.333=120m
D also increases
D & T are directly proportional
According to the question:
⇒(x+5)(y−10)=300
⇒xy+5y−10x−50=xy
⇒5×300−10x−50=0
⇒−150+x2+5x=0
⇒x2+15x−10x−150=0
⇒x(x+15)−10(x+15)=0
⇒x=10 or −15
Price cannot be negative
So, As x>0,x=10.
114
Let d = 7r. And use distance is = rate × time
7r= ( r+12) 5
7r= 5r + 60
Subtract 5r from both sides
2r = 60
Divide out 2
Rate = 30 km/h
Original question 30 km/h × 7 = 210
30 +12 = 42× 5 = 210
consider the tank capacity as 90 litres.
to fill the tank in 3 hrs first pipe must flow at a rate of 30 litres/hr
2nd pipe has to flow at a rate of 45 litres/hr
if two pipes are opened at a time the flow rate will become 75 litres/hr
90/75= 6/5 = 1.2= (i.e) 1hr and 12 minutes 1 1/5 hr
Answer: 8:10, 7:10
Explanation:
The bus b1, which started at P, reached S at 10:40, passing through the intermediary cities Q and R.
The time taken to travel from P to S
= 3 * 40 + 2 * 15 = 150minute
(journey)+(stoppage) = 2 hrs 30 minutes.
Hence, b1 started at 10:40 – 2:30 = 8:10 at P.
b2 reached Q, starting at U, through the city T, S and R.
The time taken by it to reach S = 4 * 40 + 3 * 15 = 205 minutes = 3 hr 25 minutes.
Hence, b2 started at, 10:35 – 3:25 = 7:10, at U.
4/5
37 SEATS….
because if passengers will occupy 3 seats, per head amount
payable will be 26.89.80.67 can not be divisible other than
3 and 26.89. so the,
Number of occupied seats will be 3
Number of unoccupied seats will be 40-3=37
1.6
– While the train is moving, the jogger will also be running in the same direction.
– for the head(engine) of the train to get to the current position of the jogger 240m away, it will take:
45km/hr => 12.5m/s => 240/12.5 = 19.2 seconds.
– But in the same period of time, the jogger will still be running and will have moved to a new location by: 9km/hr => 2.5m/s => 2.5 * 19.2 = 48m
To get to the new location at the speed of 12.5m/s will take the train:
48/12.5 = 3.84sec
In this additional time, the jogger will move forward by:
3.84 * 2.5 = 9.6m
at a speed of 12.5m/s, it will take the train less than a second to cover the additional 9.6m
If we add the distance the jogger will cover in 1 second to 9.6, it is still less than what the train can cover per second. let us see (9.6 + 2.5 = 12.1)
Therefore, the head of the train will pass the runner at approximately: 19.2 + 3.84 + 1 => 24.04 seconds.
For the train to completely pass the runner, it will need its whole length of 120m to be in front of the runner.
This will take an additional (9.6 + 2) seconds.
Therefore for the length of the train to be ahead of the runner it will take approx. 35.65 (24.04 + 9.6 + 2) seconds
47
16
1+1=2
2+2=4
3+4-7
4+7=11
5+11=16