Let the filling capacity of the pump = x m3/ min.
Then the emptying capacity of the pump = (x + 10) m3/ min.
Time required for filling the tank = 2400/x minutes
Time required for emptying the tank = 2400/x+10 minutes
Pump needs 8 minutes lesser to empty the tank than it needs to fill it
⇒2400/x−2400/x+10=8
⇒300/x−300/x+10=1
⇒300/(x+10)−300/x=x(x+10)
⇒3000=x2+10x
⇒x2+10x−3000=0
(x+60)(x−50)=0
x = 50 or -60
Since x can not be negative, x=50
i.e.,filling capacity of the pump = 50 m3/min.
If A covers 400m, B covers 395 m
If B covers 400m, C covers 396 m
If D covers 400m, C covers 384 m
Now if B covers 395 m, then C will cover 396/400×395=391.05m
If C covers 391.05 m, then D will cover 400/384×391.05=407.24
If A and D run over 400 m, then D win by 7.2 m (approx.)
Here as per condition let us consider third number = x,
then 2nd number = 2x and 3rd number = 4x,
Given that (4x+2x+x)/3 = 77,
7x = 3×77, x = 3 x 11 = 33,
first number = 4x = 4 x 33 = 132.
Let the speeds of the two trains be s and 2s m/s respectively.
Also, suppose that the lengths of the two trains are P and Q metres respectively.
Then,
P+Q/2s−s=60------(1)
and
P/2s−s=40------(2)
On dividing these two equation we get:
P+Q/P=60/40
P:Q = 2 : 1.
Average speed of train leaving Delhi = 2004=50 km/hr
Average speed of train leaving Ambala cantt. = 200×27=4007
By the time the other train starts from Ambala cantt, the first train had travelled 100 km
Therefore, the trains meet after:
=200−100/(50+400/7)=14/15hr
=14/15×60=56minutes
Hence they meet at 8:56 am.
Distance = Speed × time
Here time = 2hr 45 min = 114 hr
Distance = 4×114=11 km
New Speed =16.5 kmph
Therefore time = DS=1116.5=40 min.
Let the total distance covered be S km.
Total time taken= S (2×40) + S (2×60) = 5S 240 hr
Average speed =S× 240 5S = 48 km/hr.
Given Exp= log75/16 – 2 log5/9 + log32/343
= log [(25 x 3) / (4 x 4)] – log (25/81) + log [(16 x 2) / (81 x 3)]
= log(25 x 3) – log ( 4 x 4 ) – log(25) + log81 + log(16 x 2) -log (81 x 3)
= log 25 + log 3 – log 16 – log 25 + log 81 + log 16 + log 2 – log 81 – log 3
= log 2.
Find the total profit made in the first 15 days:
Mean daily profit for the first 15 days = Ra 275
Total profit for the first 15 days = 275 x 15 = Rs 4125.
Find the total profit for the last 15 days:
10,500 - 4125 = Rs 6375.
Find the mean profit for the last 15 days:
Mean profit = 6375 ÷ 15 = Rs 425.
log2x = 10 ⇒ x = 210.
∴ logx y = 100
⇒ y = x100
⇒ y = (210)100 [put value of x]
⇒ y = 21000.
Let the speed of goods train be x km/hr.
Then,
(50+x)×( 5 18 )= 187.5 9
⇒ x= 25 km/hr.
A can give B a 100 m start and C a 150m.
Start me when A runs 1000m,
B runs 900m and C runs 850m.
When B runs 1000m,
C will run 1000 x (850/900) m (i.e. 8500/9 m) Thus,
B can give C a start of - 1000 - (8500/9), i.e. 500/9 m.
log2(164)=xlog2(164)=x
Rewrite the equation as x=log2(164)x=log2(164).
x=log2(164)x=log2(164) Logarithm base 22 of 164164 is −6-6.
x=−6.
Time = 50/60=5/6 hr
Speed = 48 mph
Distance = S×T=48×5/6=40 km
Time = 40/60 hr
New speed = 40×3/2=60 kmph.
Distance = 110 m
Relative speed = 60 + 6 = 66 kmph
(Since both the train and the man are in moving in opposite direction) = (66*5/18) m/sec = 55/3 m/sec.
Let x be the number of passengers and y be the fare taken from passengers.
3xy + 50xy = 1325 => xy = 25
Amount collected from II class passengers = 25 × 50 = Rs. 1250.
et both of them meet after T min
4500 m are covered by Suresh in 60 m.
In T min he will cover 4500T60
Likewise, In T min Suresh's wife will cover 3750T60
Given,
4500T60+3750T60=726
T=5.28 minutes.
Let the normal speed be 's' km/hr
Then new speed = (s+5) km/hr
300/s−2=300/s+5
On solving this equation we get:
s = 25 km/hr.
We need to find out the HCF for given length.
15 meter 75 cm = 1575 cm.
11 meter 25 cm = 1125 cm 7 meter 65 cm = 765 cm.
1575 = 5 *5 *3 *3 *7 1125 = 5 *5 *5 *3 *3 765.
= 5 *3 *3 *17 HCF of 1575, 1125 and 765 is 45 (5 *3*3).
Let Boys in class = B
Girls in class = 20
Now, (20B+15*20)/(B+20) = 18
⇒ B = 30.
Let d be the average daily expenditure
Original expenditure = 35 × d
New expenditure = 35 × d + 42
New average expenditure will be :
(35 × d + 42)/42 = d - 1
On solving, we get d = 12
Therefore original expenditure = 35 × 12 = 420.
Let the cost price be 100%. It is sold at 10% loss.
So it is sold at 90% of the cost price.
90 % of the cost price + 450 = 108% of the Cost price 18% of the cost price = Rs 450 Cost price of the book = 450/18 x 100 = Rs 2500.
L.C.M. of 21, 36, 66 = 2772 Now,
2772 = 2 x 2 x 3 x 3 x 7 x 11 To make it a perfect square,
it must be multiplied by 7 x @So,
required number = 2 x 2 x 3 x 3 x 7 x 7 x 11 x 11 = 213444.
1525-1300= 225 for 1.5 yrs (3.5-2)
so for one yr 225/1.5= 150
then for 2 yrs interest is 150+150=300
Then principal 1300-300=1000.
Now 150/1000*100= 15%.