31 Jan 2024, 17:53 UTC≈1,060 viewsread 6 August 2026 A ship follows certain mileage characteristics in a speed limit from 15 km/hr to 90 km/hr. The ship can travel 12 km in one litre of petrol while traveling at 40 km/hr. From this point onwards in the direction of current every increase 5 km/hr in speed the ship mileage increases by 1 km/litre. Similarly, in the opposite direction of current for every decrease 5 km/hr in speed mileage decrease by 1 km/litre. Harshit h…
Signed KAMAL M
31 Jan 2024, 17:52 UTC≈1,140 viewsread 6 August 2026 First 2 hours=40*2=80km
80/12=6+2/3
15-6-2/3==8+1/3 =25/3ltr left
Increase =(90-40)=50
Increase in mileage=10
12+10==22km/ltr
22*25/3==550/3
(80+550/3)==790/3
Signed KAMAL M
22 Jul 2022, 16:37 UTC≈3,000 viewsread 6 August 2026 5:9
4*2+1 = 9
5:3 = 8 === 9.6
5 === 6
72/6 = 12
42/12 = 3.5hrs
Signed KAMAL M
22 Jul 2022, 16:37 UTC≈2,920 viewsread 6 August 2026 Photo
Photo, posted without a caption
Signed KAMAL M
15 Jul 2022, 14:31 UTC≈2,920 viewsread 6 August 2026 Ua-ub=12----(1)
V=12
5(ub+12)=8(ua-12)
8ua-5ub=156----(2)
So, ua=32
Ub=20
240/20+240/8=42hr ans
Signed KAMAL M
15 Jul 2022, 14:31 UTC≈2,820 viewsread 6 August 2026 Photo
Photo, posted without a caption
Signed KAMAL M
4 Nov 2021, 10:25 UTC≈3,700 viewsread 6 August 2026 P------q-----+++-----r---------s
---x----*-0.5(x+y)-*-----y----
X/a+y/a=12-(x+y)/2a....1st
X+y=8a...2nd
X/a+y/a+0.5(x+y)/(a+5)= 34/3
0.5(x+y)/a-0.5(x+y)/(a+5)=2/3
4a/a-4a/(a+5)=2/3
5/3=2a/(a+5)
A= 25 km/hr
X+y=25*8= 200 km
Qr=200/2= 100 km
Tym=200/(25-10)+100/(25-10)
=300/15= 20 hr
Signed Kamal M
4 Nov 2021, 10:25 UTC≈2,870 viewsread 6 August 2026 Dis = 2a+2b+a+b = 3a+3b
x = a+b/4
2(a+b)/x+(a+b)/x+5 = 34/3
x = 25kmph
Dis = 300km
Req: 300/25-10 = 20h ans
Signed Kamal M
4 Nov 2021, 10:25 UTC≈2,930 viewsread 6 August 2026 4*s=10/3 (s+5)
12s=10s+50
s=25
25*12/15=20
Signed Kamal M
4 Nov 2021, 10:25 UTC≈3,340 viewsread 6 August 2026 B*12==15(b-5)
B==25 kmph
25x==30(x-2/3)
X==4
QR==25*4==100 km
100*3==300 km --> distance
300/(25-10)==20 hrs
Signed Kamal M
4 Nov 2021, 10:25 UTC≈2,770 viewsread 6 August 2026 There are four points P, Q, R and S, in the given order, in a river along the same route from P to S
downstream. A boat can travel from P to S in 12 hours if there is no current in the river. The
distance QR is the arithmetic mean of distance PQ and RS. If there was a current of 5 km/hr from
Q to R only, then the boat would have taken (34/3) hours in going from P to S, and if there was
same current of 5 km/hr lowing
…
Signed Kamal M
17 Sept 2021, 04:17 UTC≈2,960 viewsread 6 August 2026 Photo
Photo, posted without a caption
Signed Kamal M
Showing the 12 most recent of 20 posts we hold for @boat_and_stream. View and reaction counts are the latest single reading for each post, not a live figure, and a recent post is still accumulating both. A view count marked ≈ was rounded by Telegram before we ever saw it — t.me prints views in full below 1,000 and to three significant figures above, so ≈1,200,000 means somewhere between 1,150,000 and 1,249,999. Unmarked counts are exact. Text is reproduced from the public post preview and truncated for length.