19. boats and streams c&add [www.itmona.com]

BOATS AND STREAMS 【629】
BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE BANK MATH BIBLE
awi, w¯’i cvwb‡Z (in still water) †bŠKvi ev gvwSi †eM = u
†¯ªv‡Zi †eM = v
Boat u
w¯’i cvwb ZvB GLv‡b †¯ªv‡Zi †eM k~b¨
I. AbyK‚‡j †eM (Down Stream Speed)
Boat u
v
†¯ªv‡Zi AbyK‚‡j †bŠKvi †eM = u + v
[u I v GKBgyLx ZvB u I v †hvM n‡”Q]
cÖwZK‚‡j †eM (upstream speed)
Boat u
v
†¯ªv‡Zi cÖwZK‚‡j †bŠKvi †eM = u – v
[u I v wecixZgyLx Ges u > v ZvB u n‡Z v we‡qvM n‡”Q]
Ab¨fv‡e g‡b ivL‡Z cv‡ib  †bŠKvq DVvi AwfÁZv i‡q‡Q, Zviv wbðq †Lqvj K‡i‡Qb †¯ªvZ †hw`‡K hvq †bŠKv †mw`‡K `ªæZ P‡j| A_©vr ZLb
AbyK‚‡j †eM †ewk, ZvB u I v †hvM n‡e, Avevi †bŠKv hLb †¯ªv‡Zi wecixZ w`‡K hvq ZLb †bŠKv A‡bK ax‡i P‡j, A_©vr †eM Kg ZvB, ZLb u I v
we‡qvM n‡e|
II. †¯ªv‡Zi AbyK‚‡j †bŠKvi †eM = a Ges †¯ªv‡Zi cÖwZK‚‡j †bŠKvi †eM = b n‡j,
w¯’i cvwb‡Z †bŠKvi †eM, u =



a + b
2
Ges †¯ªv‡Zi †eM, v =
a  b
2
III. GKRb †jvK w¯’i cvwb‡Z u †e‡M muvZvi KvU‡Z cv‡i| awi, †m u †e‡M OB
c‡_ hvÎv ïiæ Kij| †¯ªv‡Zi †eM v Gi Rb¨ †m OB c‡_ †h‡Z cvij bv, †m
OA c‡_ b`x cvi n‡jv| b`x cvi n‡Z Zvi mgq jvM‡e =
x
u
, †hLv‡b x = b`xi cÖ¯’
wP‡Î, OB eivei †jv‡Ki †eM = u, OD eivei †¯ªv‡Zi †eM = v
IV. GLb g‡b Kwi, †jvKwU OB c‡_B j¤^fv‡e (perpendicular) b`x cvi n‡e|
Zvn‡j Zv‡K cv‡ki wP‡Îi b¨vq OE c‡_ Ggb fv‡e hvÎv Ki‡Z n‡e †hb,
†¯ªv‡Zi †eM v Gi Kvi‡Y †m OB c‡_ j¤^ fv‡e b`x cvi n‡Z cv‡i| OE Gi
mgvb I mgvšÍivj †iLv BD AsKb Kwi, GB BD †iLvB †jvKwUi †eM u †K
wb‡`©k K‡i| OB eivei †jvKwUi cÖK…Z ev jwä †eM = R
OBD mg‡KvYx wÎfz‡R,
u2
= v2
+ R2
..........................(i)
 R = u2
 v2
 j¤^fv‡e b`x cvi n‡Z mgq jvM‡e, T =
b`xi cÖ¯’ (x)
jwä †eM (R)
 T =
x
u2
 v2
GLb b`x‡Z †¯ªvZ bv _vK‡j (in still water) x wgUvi †h‡Z t mgq jv‡M wKš‘ hw` †¯ªvZ _v‡K Zvn‡j, H `~iZ¡ †h‡Z t mgq jv‡M G‡ÿ‡Î,
u =
x
t
Ges R =
x
t
(i) bs mgxKiY n‡Z, v2
= u2
 R2
v = u2
 R2
=



x
t
2




x
t
2
= x2



1
t2 
1
t2
 †¯ªv‡Zi †eM, v = x
1
t2 
1
t2
GB Aa¨v‡qi ¸iæZ¡c~Y© Z_¨ I m~Î
19 Boats and Streams

【630】 BANK MATH BIBLE
UvBc bs UvBc Gi bvg cÖkœ b¤î
1 †bŠKvi †eM, †¯ªv‡Zi †eM m¤úwK©Z mgm¨v 1, 2, 3, 4, 5, 6, 8, 9, 12, 13, 14
2 mgq I †eM m¤úwK©Z mgm¨v 7, 15, 16, 17, 18, 30, 35, 36
3 mgxKiY m¤úwK©Z mgm¨v 10, 11, 19, 21, 22, 23, 24, 27, 28, 33
4 we‡kl mgm¨v 20, 25, 26, 29, 31, 34, 37
cÖkœ b¤î:
Average Speed = Mo MwZ/‡eM
Current = †¯ªvZ
Downstream = †¯ªv‡Zi AbyK‚‡j
Equidistant = mg`~iZ¡
Motorboat = BwÄbPvwjZ †bŠKv
Perpendicularly = j¤^fv‡e
Velocity = †eM
Stream = †¯ªvZ
Speed = MwZ/†eM
Stationary = w¯’i
Row = ùvo †e‡q Pjv
Ratio = AbycvZ
Upstream = †¯ªv‡Zi cÖwZK‚‡j
Journey = ågY/hvÎv
1. *A boat goes 8 km in one hour along the stream and 2
km in one hour against the stream. The speed in km/hr
of the stream is (GKwU †bŠKv †¯ªv‡Zi AbyK‚‡j N›Uvq 8 wK.wg.
hvq Ges †¯ªv‡Zi cÖwZK‚‡j N›Uvq 2 wK.wg. hvq| †¯ªv‡Zi †eM N›Uvq KZ
wK.wg.?) [Pubali Bank (TAJO Cash) – 19 + + www.brainly.in]
a 2 b 3 c 4 d 5 b
 mgvavb : awi, †bŠKvi cÖK…Z †eM (w¯’i cvwb‡Z) = u
kZ©g‡Z, u + v = 8 ....................... (i)
u – v = 2 ...................... (ii)
(–) (+) (–)
(i) – (ii) Kwi, 2v = 8 – 2
 v =
8 – 2
2
= 3 km/hr
 †¯ªv‡Zi †eM = 3 km/hr
GB ai‡bi mgm¨vi †ÿ‡Î weKí m~Î,
†¯ªv‡Zi †eM =
AbyK‚‡j †eM – cÖwZK‚‡j †eM
2
=
a – b
2
=
8 – 2
2
=
6
2
= 3
2. *In one hour, a boat goes 11 km along the stream and 5
km against the stream. The speed of the boat in still
water (in km/hr) is (GK N›Uvq GKwU †bŠKv †¯ªv‡Zi AbyK‚‡j 11
wK.wg. hvq Ges †¯ªv‡Zi cÖwZK‚‡j 5 wK.wg. hvq| w¯’i cvwb‡Z †bŠKvi
†eM N›Uvq KZ wK.wg.?) [www.examveda.com; www.indiabix.com]
a 3 b 5 c 8 d 9 c
 mgvavb : w¯’i cvwb‡Z †bŠKvi †eM = u
†¯ªv‡Zi AbyK‚‡j †bŠKvi †eM, u + v = 11 ..................... (i)
†¯ªv‡Zi cÖwZK‚‡j u – v = 5 ....................... (ii)
(i) + (ii) Kwi, 2u = 16  u = 8 km/hr
G‡ÿ‡Î weKí mswÿß m~Î,
w¯’i cvwb‡Z †bŠKvi †eM =
AbyK‚‡j †eM + cÖwZK‚‡j †eM
2
=
a + b
2
=
11 + 5
2
= 8
3. A man rows downstream 32 km and 14 km upstream.
If he takes 6 hours to cover each distance, then the
velocity (in kmph) of the current is (GKRb †jvK †¯ªv‡Zi
AbyK‚‡j 32 wK.wg. Ges cÖwZK~‡j 14 wK.wg. `vo †e‡q P‡j| hw`
cÖwZwU `~iZ¡ AwZµg Ki‡Z Zvi 6 N›Uv jv‡M, Zvn‡j †mªv‡Zi †eM
N›Uvq KZ wK.wg.?)
a
1
2
b 1 c 1
1
2
d 2 c
 mgvavb : †¯ªv‡Zi AbyK‚‡j †bŠKvi †eM, u + v =
32
6
......... (i)
†¯ªv‡Zi cÖwZK‚‡j u – v =
14
6
........ (ii)
(–) (+) (–)
(i) – (ii) Kwi, 2v =
32
6
–
14
6
 2v =
32 – 14
6
=
18
6
= 3
 v =
3
2
= 1
1
2
wK.wg./N›Uv
4. *A boatman rows 1 km in 5 minutes, along the stream and
6 km in 1 hour against the stream. The speed of the stream
is (GKRb gvwS †¯ªv‡Zi AbyK‚‡j 5 wgwb‡U 1 wK‡jvwgUvi Ges †¯ªv‡Zi
cÖwZK‚‡j 1 N›Uvq 6 wK.wg hvq| †¯ªv‡Zi †eM KZ?) [www.competoid.com]
a 3 kmph b 6 kmph c 10 kmph d 12 kmph a
 mgvavb : †¯ªv‡Zi AbyK‚‡j 5 wgwb‡U hvq 1 wK.wg.
1 NÈvq ev 60
1  60
5
= 12 wK.wg.
AZGe, †¯ªv‡Zi AbyK‚‡j †bŠKvi †eM, u + v = 12 km/hr ....... (i)
Ges †¯ªv‡Zi cÖwZK‚‡j u – v = 6 km/hr ........ (ii)
(i) + (ii) Kwi, 2u = 18  u = 9 km/hr
u Gi gvb (i) bs G ewm‡q cvB, 9 + v = 12
 v = 12 – 9 = 3 km/hr
Dr. R.S. AGGARWAL m¨v‡ii eB‡qi c~Y©v½ evsjv mgvavb
kãfvÐvi GB Aa¨v‡qi AvÛvijvBb Kiv k‡ãi A_© GLv‡b †`Lyb 
wiwfkb e· cieZx©‡Z †h cÖkœ¸‡jv Avcbvi wiwfkb Kiv cÖ‡qvRbÑ †m¸‡jvi b¤î wj‡L ivLyb 
GKB wbq‡gi AsK¸‡jv GK mv‡_ Abykxjb Ki‡Z

5. *A boat takes half time in moving a certain distance
downstream than upstream. What is the ratio between
the rate in still water and the rate of current? (GKwU
†bŠKv wbw`©ó `~iZ¡ AwZµg Ki‡Z, †¯ªv‡Zi cÖwZK‚‡ji †_‡K AbyK‚‡j
A‡a©K mgq †bq| w¯’i cvwb‡Z †bŠKvi †eM I †¯ªv‡Zi †e‡Mi
AbycvZ KZ?) [www.examveda.com; www.competoid.com]
a 1 : 2 b 2 : 1 c 1 : 3 d 3 : 1 d
 mgvavb : mg‡qi AbycvZ (cÖwZK‚j : AbyK‚j) = 1 :
1
2
= 2 : 1
†h‡nZz, †eM =
`~iZ¡
mgq A_©vr †eM I mgq ci¯úi e¨¯ÍvbycvwZK ev
wecixZgyLx
ZvB, †e‡Mi AbycvZ (cÖwZK‚j : AbyK‚j)
 u – v : u + v =
1
2
:
1
1
= 1 : 2
AbyK‚‡j †eM u + v
G cÖwZK‚‡j †eM u – v

u – v
u + v
=
1
2

u – v + u + v
u – v – u – v
=
1 + 2
1 – 2
[†hvRb we‡qvRb K‡i]

2u
– 2v
=
3
– 1

u
v
=
3
1
6. If a man goes 18 km downstream in 4 hours and return
against the stream in 12 hours, then the speed of the
stream in km/hr is (hw` GKRb †jvK †¯ªv‡Zi AbyK‚‡j 4 N›Uvq
18 wK.wg. hvq Ges †¯ªv‡Zi cÖwZK‚‡j 12 N›Uvq wd‡i Av‡m, Zvnv‡j
†¯ªv‡Zi †eM wK.wg./ NÈv‡Z KZ?)
a 1 b 1.5 c 1.75 d 3 b
 mgvavb : †¯ªv‡Zi AbyK‚‡j 4 N›Uvq hvq 18 wK.wg.
 1 =
18
4
= 4.5 wK.wg.
†¯ªv‡Zi cÖwZK‚‡j 12 N›Uvq wd‡i Av‡m 18 wK.wg.
 1 =
18
12
= 1.5 wK.wg.
 †¯ªv‡Zi AbyK‚‡j †eM, u + v = 4 .5 ....................... (i)
†¯ªv‡Zi cÖwZK‚‡j u – v = 1.5 ........................ (ii)
(–) (+) (–)
(i) – (ii) Kwi, ev, 2v = 3 wK.wg./N›Uv
 v =
3
2
= 1.5 wK.wg./N›Uv
7. *A boatman goes 2 km against the current of the stream
in 1 hour and goes 1 km along the current in 10
minutes. How long will it take to go 5 km in stationary
water ? (GKRb gvwS N›Uvq †¯ªv‡Zi cÖwZK‚‡j 2 wK.wg. hvq Ges
†¯ªv‡Zi AbyK‚‡j 10 wgwb‡U 1 wK.wg. hvq | w¯’i cvwb‡Z 5 wK.wg.
†h‡Z KZ mgq jvM‡e?) [www.Examveda.com; www.indiabix.com]
a 40 minutes b 1 hour
c 1 hr 15 min d 1 hr 30 min c
 mgvavb : †¯ªv‡Zi cÖwZK‚‡j †eM, u – v = 2 (wK.wg./N›Uv) ......... (i)
†¯ªv‡Zi AbyK‚‡j 10 wgwb‡U hvq 1 wK.wg.
 60 =
60  1
10
= 6 wK.wg.
 †¯ªv‡Zi AbyK‚‡j †eM, u + v = 6 (wK.wg./N›Uv) .................. (ii)
(i) + (ii) K‡i, 2u = 6 + 2  u =
8
2
= 4 wK.wg./N›Uv
w¯’i cvwb‡Z †bŠKvi †eM = 4 wK.wg./N›Uv
w¯’i cvwb‡Z 5 wK.wg. †h‡Z mgq jv‡M =
`~iZ¡
†eM =
5
4
= 1
1
4
1 N›Uv +
1
4
× 60 wgwbU = 1 N›Uv 15 wgwbU
8. *A man can row
3
4
of a km against the stream in 11
1
4
minutes and returns in 7
1
2
minutes. Find the speed of
the man in still water. (GKRb gvwS †¯ªv‡Zi cÖwZK‚‡j
3
4
wK.wg.
`~iZ¡ 11
1
4
wgwb‡U hvq Ges wd‡i Av‡m 7
1
2
wgwb‡U| w¯’i cvwb‡Z
gvwSi †eM KZ?)
a 3 km/hr b 4 km/hr c 5 km/hr d 6 km/hr c
 mgvavb : cÖwZK‚‡j 11
1
4
ev
45
4
wgwb‡U hvq =
3
4
wK.wg.
1 =
3
4
÷
45
4
 60 =



3
4

4
45
 60
= 4 wK.wg.
AbyK‚‡j 7
1
2
ev
15
2
wgwb‡U hvq =
3
4
wK.wg.
 60 =



3
4
÷
15
2
 60
=



3
4

2
15
 60 = 6 wK.wg.
awi, u = †bŠKv ev gvwSi †eM, v = †¯ªv‡Zi †eM
†¯ªv‡Zi AbyK‚‡j †eM, u + v = 6 .................................. (i)
†¯ªv‡Zi cÖwZK‚‡j u – v = 4 ................................. (ii)
(i) + (ii) Kwi, ev, 2u = 10 wK.wg./N›Uv
 u = 5 wK.wg./N›Uv
9. *A boat, while going downstream in a river covered a
distance of 50 miles at an average speed of 60 miles per
hour. While returning, because of the water resistance,
it took 1 hour 15 minutes to cover the same distance.
What was the average speed during the whole journey?
(GKwU †bŠKv †¯ªv‡Zi AbyK‚‡j N›Uvq 60 gvBj Mo †e‡M 50 gvBj
`~iZ¡ AwZµg K‡i| †diZ Avmvi mgq †¯ªv‡Zi evavi Kvi‡Y GKB
`~iZ¡ 1 N›Uv 15 wgwb‡U AwZµg K‡i| cy‡iv hvÎvq †bŠKvi Mo
MwZ‡eM KZ wQj?) [www.examveda.com]
a 40 mph b 48 mph c 50 mph d 55 mph b
 mgvavb : †¯ªv‡Zi AbyK‚‡j 50 gvBj `~iZ¡ 60 gvBj/N›Uv †e‡M †h‡Z
mgq jv‡M =
50
60
=
5
6
hr



 mgq =
`~iZ¡
†eM
†¯ªv‡Zi cÖwZK‚‡j 50 gvBj †h‡Z mgq jv‡M = 1 hr 15 min
= 1 +
15
60
= 1 +
1
4
=
5
4
hr
 Mo MwZ‡eM =
†gvU `~iZ¡
†gvU mgq =
50 + 50
5
6
+
5
4
=
100
10 + 15
12
[hvIqv + Avmvq, †gvU `~iZ¡ = (50 + 50) = 100]
= 100 
12
25
= 48 mph
10. *A man swimming in a stream which flows 1
1
2
km/hr
finds that in a given time he can swim twice as far with
the stream as he can against it. At what rate does he
swim? (GKwU b`x‡Z †¯ªv‡Zi MwZ †eM 1
1
2
wK.wg./N›Uv| GKRb
e¨w³ H b`x‡Z mvZvi KvU‡Z wM‡q †`L‡jv, †m †¯ªv‡Zi cÖwZK‚‡ji
†P‡q AbyKy‡j wØMyY MwZ‡Z mvZvi KvU‡Z cv‡i| Zvi mvZv‡ii
MwZ‡eM KZ?)
a 4
1
2
km/hr b 5
1
2
km/hr
c 7
1
2
km/hr d None of these a

 mgvavb : †¯ªv‡Zi MwZ, v = 1
1
2
=
3
2
wK.wg./N›Uv
kZ©g‡Z, AbyK‚‡j †eM = 2  cÖwZK‚‡j †eM
 (u + v) = 2(u – v) 



u +
3
2
= 2



u –
3
2
 u +
3
2
= 2u – 3  2u –u = 3 +
3
2
=
6 +3
2
 u =
9
2
= 4
1
2
km/hr
11. *A boat running upstream takes 8 hours 48 minutes to
cover a certain distance, while it takes 4 hours to cover the
same distance running downstream. What is the ratio
between the speed of the boat and speed of the water
current respectively? (GKwU †bŠKv GKwU wbw`©ó `~iZ¡ †¯ªv‡Zi
cÖwZK‚‡j 8 N›Uv 48 wgwb‡U AwZµg K‡i| GKB `~iZ¡ †¯ªv‡Zi AbyK‚‡j
4 N›Uvq AwZµg K‡i| †bŠKvi †eM I †¯ªv‡Zi MwZ‡e‡Mi AbycvZ
KZ?) [www.examveda.com; www.indiabix.com; www.competoid.com]
a 2 : 1 b 3 : 2
c 8 : 3 d Cannot be determined
e None of these c
 mgvavb : 8 N›Uv 48 wgwbU = 8 +
48
60
= 8 +
4
5
=
40 + 4
5
=
44
5
N›Uv
Dfq †ÿ‡Î wbw`©ó `~iZ¡ = (mgq  †eM)AbK~‡j
= (mgq  †eM)cÖwZK‚‡j
 4  (u + v) =
44
5
(u – v)
 (u + v) =
11
5
(u – v) [Dfq cÿ‡K 4 Øviv fvM K‡i]
 u + v =
11
5
u –
11
5
v
 u –
11
5
u = – v –
11
5
v  u



1 –
11
5
= – v



1 +
11
5
 u 
– 6
5
= – v 
16
5

u
v
=
16
5

5
6

u
v
=
8
3
 u : v = 8 : 3
12. If a boat goes 7 km upstream in 42 minutes and the
speed of the stream is 3 kmph, then the speed of the
boat in still water is : (hw` GKwU †bŠKv †¯ªv‡Zi cÖwZK‚‡j 42
wgwb‡U 7 wK.wg hvq Ges †¯ªv‡Zi MwZ 3 wK.wg./N›Uv nq, Zvn‡j w¯’i
cvwb‡Z †bŠKvi †eM KZ?)
a 4.2 km/hr b 9 km/hr
c 13 km/hr d 21 km/hr c
 mgvavb : 42 wgwb‡U †¯ªv‡Zi cªwZK‚‡j hvq = 7 wK.wg.
 =
7
42
 =
7  60
42
= 10 wK.wg.
†`Iqv Av‡Q, †¯ªv‡Zi MwZ, v = 3
awi, w¯’i cvwb‡Z †bŠKvi †eM = u
 †¯ªv‡Zi cÖwZK‚‡j †eM, u – 3 = 10
 u = 10 + 3 = 13 wK.wg./N›Uv
13. *A man’s speed with the current is 15 km/hr and the
speed of the current is 2.5 km/hr. The man’s speed
against the current is : (GKRb †jv‡Ki †eM †¯ªv‡Zi AbyK‚‡j
15 wK.wg./N›Uv Ges †¯ªv‡Zi †eM 2.5 wK.wg./N›Uv| †¯ªv‡Zi cÖwZK‚‡j
†jvKwUi MwZ KZ?) [www.examveda.com; www.indiabix.com]
a 8.5 km/hr b 9 km/hr
c 10 km/hr d 12.5 km/hr c
 mgvavb : AbyK‚‡j †jv‡Ki MwZ, u + v = 15
Ges †¯ªv‡Zi MwZ, v = 2.5
 u = 15 – v = 15 – 2.5 = 12.5
cÖwZK‚‡j †jv‡Ki MwZ = u – v = 12.5 – 2.5 = 10 km/hr
14. *If a man rows at the rate of 5 kmph in still water and
his rate against the current is 3.5 kmph, then the man’s
rate along the current is : (hw` GKRb †jvK N›Uvq 5 wK.wg.
†e‡M w¯’i cvwb‡Z P‡j Ges †¯ªv‡Zi cÖwZK‚‡j Zvi †eM NÈvq 3.5
wK.wg. nq| Zvn‡j †¯ªv‡Zi AbyK‚‡j †jvKwUi MwZ KZ?)
a 4.25 kmph b 6 kmph
c 6.5 kmph d 8.5 kmph c
 mgvavb : w¯’i cvwb‡Z †jv‡Ki MwZ, u = 5 (wK.wg./N›Uv)
cÖwZK~‡j MwZ, u – v = 3.5  5 – v = 3.5  v = 5 – 3.5 = 1.5
 AbyK‚‡j †jvKwUi MwZ = u + v = 5 + 1.5 = 6.5 km/hr
15. A motorboat in still water travels at a speed of 36
km/hr. It goes 56 km upstream in 1 hour 45 minutes.
The time taken by it to cover the same distance down
the stream will be (GKwU BwÄb PvwjZ †bŠKv w¯’i cvwb‡Z N›Uvq
36 wK.wg. †e‡M ågY K‡i| GwU †¯ªv‡Zi cÖwZK‚‡j 56 wK.wg. hvq 1
N›Uv 45 wgwb‡U| Zvn‡j †¯ªv‡Zi AbyK‚‡j GKB `~iZ¡ AwZµg Ki‡Z
KZ mgq jvM‡e?) [www.examveda.com]
a 1 hour 24 minutes b 2 hour 21 minutes
c 2 hour 25 minutes d 3 hour a
 mgvavb : w¯’i cvwb‡Z †bŠKvi †eM, u = 36 (wK.wg./N›Uv)
Avevi, 1 N›Uv 45 wgwbU ev 1 +
45
60
= 1 +
3
4
=
7
4
N›Uv
cÖwZK‚‡j,
7
4
N›Uvq hvq 56 wK.wg.
 1
56  4
7
= 32 N›Uv
cÖwZK‚‡j †eM, u – v = 32
 36 – v = 32  v = 36 – 32 = 4 (wK.wg./N›Uv)
AbyK‚‡j †eM, u + v = 36 + 4 = 40 (wK.wg./N›Uv)
AbyK‚‡j mgq =
`~iZ¡
AbyK‚‡j †eM =
56
40
=
7
5
= 1
2
5
GLb, 1
2
5
N›Uv = 1 N›Uv +
2
5
 60 wgwbU = 1 N›Uv 24 wgwbU
16. *Speed of boat in standing water is 9 kmph and the
speed of the stream is 1.5 kmph. A man rows to a place
at a distance of 105 km and comes back to the starting
point. The total time taken by him is : (w¯’i cvwb‡Z GKwU
†bŠKvi †eM N›Uvq 9 wK.wg. Ges †¯ªv‡Zi †eM N›Uvq 1.5 wK.wg.|
Zvn‡j GKRb e¨w³i 105 wK.wg. †h‡Z Ges wd‡i Avm‡Z †gvU KZ
mgq jvM‡e?) [Exam Taker AUST : P.K.B. (E.O. Cash-2019);
www.examveda.com; www.indiabix.com]
a 16 hours b 18 hours
c 20 hours d 24 hours d
 mgvavb : AbyK‚‡j †bŠKvi †eM = u + v = 9 + 1.5 = 10.5
cÖwZK‚‡j = u – v = 9 – 1.5 = 7.5
†gvU mgq = AbyK‚‡j mgq + cÖwZK‚‡j mgq
=



105
10.5
+
105
7.5 


 mgq =
`~iZ¡
†eM
= 10 + 14 = 24 N›Uv
17. *The speed of a boat in still water is 15 km/hr and the
rate of current is 3 km/hr. The distance travelled
downstream in 12 minutes is : (w¯’i cvwb‡Z GKwU †bŠKvi
†eM 15 wK.wg.| N›Uv Ges †¯ªv‡Zi †eM 3 wK.wg./N›Uv| †¯ªv‡Zi
AbyK‚‡j 12 wgwb‡U †bŠKvwU KZUzKz `~iZ¡ AwZµg Ki‡e?)
[Exam Taker AUST : Combined 8 Banks (S.O.-2019); Combined 6 Bank’s & 2 Fin.
Inst. (Senior Officer) – 19; www.indiabix.com; www.doubtnut.com;
www.examveda.com; www.careerbless.com; www.brainly.in]
a 1.2 km b 1.8 km
c 2.4 km d 3.6 km d

 mgvavb : w¯’i cvwb‡Z †bŠKvi †eM, u = 15 (wK.wg./N›Uv)
†¯ªv‡Zi †eM, v = 3 (wK.wg./N›Uv)
†¯ªv‡Zi AbyK‚‡j †bŠKvi †eM = u + v = 15 + 3 = 18 (wK.wg./N›Uv)
 1 N›Uvq ev 60 wgwb‡U †¯ªv‡Zi AbyK‚‡j hvq = 18 wK.wg.
 1 =
18
60
wK.wg.
 12 =
12  18
60
= 3.6 wK.wg.
18. *A man can row at 5 kmph in still water. If the velocity
of current is 1 kmph and it takes him 1 hour to row to
a place and come back, how far is the place? (GKRb
e¨w³ w¯’i cvwb‡Z N›Uvq 5 wK.wg. †h‡Z cv‡i| hw` †¯ªv‡Zi †eM 1
wK.wg./N›Uv nq Zvn‡j GKwU wbw`©ó ¯’v‡b wM‡q wd‡i Avm‡Z 1 N›Uv
mgq jv‡M| ¯’vbwUi `~iZ¡ KZ?) [www.examveda.com; www.indiabix.com]
a 2.4 km b 2.5 km
c 3 km d 3.6 kmwwww a
 mgvavb :
x
†¯ªv‡Zi AbyK‚‡j †bŠKvi †eM = u + v = 5 + 1 = 6 (wK.wg./N›Uv)
†¯ªv‡Zi cÖwZK‚‡j = u – v = 5 – 1 = 4 (wK.wg./N›Uv)
awi, `~iZ¡ = x wK.wg.
kZ©g‡Z, AbyK‚‡j mgq + cÖwZK‚‡j mgq = †gvU mgq

x
6
+
x
4
= 1



 mgq =
`~iZ¡
†eM

2x + 3x
12
= 1 
5x
12
= 1  x =
12
5
= 2.4 wK.wg.
19. *A boat takes 19 hours for travelling downstream from
point A to point B and coming back to a point C
midway between A and B. If the velocity of the stream
is 4 kmph and the speed of the boat in still water is 14
kmph, What is the distance between A and B? (†¯ªv‡Zi
AbyK‚‡j GKwU †bŠKv ¯’vb A †_‡K B ¯’v‡b wM‡q D³ `yB ¯’v‡bi
ga¨eZ©x ¯’vb C †Z wd‡i Avm‡Z 19 N›Uv mgq jv‡M| hw` †¯ªv‡Zi
†eM 4 wK.wg./N›Uv nq Ges w¯’i cvwb‡Z †bŠKvi †eM 14 wK.wg./ N›Uv
nq| Zvn‡j A †_‡K B-Gi `~iZ¡ KZ?)
[Exam Taker AUST : Aggarwal-19; Sonali Bank (Officer Cash FF-2019);
Sonali Bank (Officer Cash FF) – 19; www.brainly.in; www.sawaal.com]
a 160 km b 180 km c 200 km d 220 km b
 mgvavb :
x
A C B
x/2
awi, †gvU `~iZ¡ = x wK.wg.
†¯ªv‡Zi AbyK‚‡j †eM = u + v = 14 + 4 = 18 (wK.wg./N›Uv)
cÖwZK‚‡j = u – v = 14 – 4 = 10 (wK.wg./N›Uv)
kZ©g‡Z,
A n‡Z B AbyK‚‡j †h‡Z mgq + B n‡Z C †Z cÖwZK‚‡j †h‡Z mgq = †gvU mgq

x
18
+
x
2
10
= 19 
x
18
+
x
20
= 19

10x + 9x
180
= 19



 mgq =
`~iZ¡
†eM

19x
180
= 19  x =
19 
19
= 180 wK.wg.
20. *P, Q and R are three towns on a river which flows
uniformly. Q is equidistant from P and R. I row from P
to Q and back in 10 hours and I can row from P to R in
4 hours. Compare the speed of my boat in still water
with that of the river. (P, Q Ges R Gi GKwU b`xi cvk^©eZ©x
wZbwU kni| Q, P Ges R Gi †_‡K mg-`~i‡Z¡ Aew¯’Z| P †_‡K
Q-†Z wM‡q wd‡i Avm‡Z Avgvi 10 N›Uv mgq jv‡M Ges P †_‡K R
G †h‡Z Avgvi mgq jv‡M 4 N›Uv| Zvn‡j w¯’i cvwb‡Z Avgvi
†bŠKvi †eM Ges †¯ªv‡Zi †e‡Mi AbycvZ KZ?) [www.examveda.com]
a 4 : 3 b 5 : 3 c 6 : 5 d 7 : 3 b
 mgvavb : P n‡Z Q †Z wM‡q Avevi
Q n‡Z P †Z wd‡i Avm‡Z 10 hr
mgq jv‡M,
PQ = QR = x (awi)
x
4 hr (hvIqv)
10 hr
cÖwZK‚jAbyK‚j
x RP Q

x
u + v
+
x
u – v
= 10



AbyK‚‡j †eM = u + v
cÖwZK‚‡j †eM = u – v
 x



1
u + v
+
1
u–v
= 10  x 
u – v + u +v
(u + v) (u –v)
= 10
 x 
2u
(u + v) (u – v)
= 10  x =
5(u + v) (u – v)
u
Avevi, †¯ªv‡Zi AbyK‚‡j P n‡Z R †Z †h‡Z mgq jv‡M 4 N›Uv
kZ©g‡Z,
2x
u + v
= 4 [ †gvU `~iZ¡ = x + x = 2x, AbyK‚‡j †eM = u + v]

2
u + v
 x = 4

2
(u + v)

5(u + v) (u – v)
u
= 4

10 (u – v)
u
= 4

u – v
u
=
4
10
=
2
5
[x Gi gvb ewm‡q, jÿ¨ Kiæb
Avgv‡`i
u
v
Gi gvb cÖ‡qvRb,
GLv‡b x Sv‡gjv Ki‡Q ZvB x
Gi gvb u I v Gi gva¨‡g
GLv‡b emv‡bv n‡q‡Q]

u
u
–
v
u
=
2
5
 1 –
v
u
=
2
5

v
u
= 1 –
2
5
=
5 – 2
5
=
3
5

u
v
=
5
3
weKí mgvavb : g‡b Kwi, †¯ªv‡Zi AbyK‚‡j †eM = a
†¯ªv‡Zi cÖwZK‚‡j †eM = b
GKs PQ = QR = x.
P RQ
x x
Zvn‡j, †bŠKvi †eM =
a + b
2
; †¯ªv‡Zi †eM =
a – b
2
kZ©g‡Z,
x
a
+
x
b
= 10 ...... (i) [P n‡Z Q †h‡q Avevi wd‡i Av‡m 10 NÈvq]
2x
a
= 4 [P n‡Z R ch©šÍ †h‡Z mgq †bq 4 NÈv]
 x =
4  a
2
= 2a
mgxKiY (i) G x = 2a ewm‡q cvB,
2a
a
+
2a
b
= 10
 2 +
2a
b
= 10 
2a
b
= 10 – 2 = 8 
a
b
= 4

a + b
a – b
=
4 + 1
4 – 1
[†hvRbÑwe‡qvRb]

a + b
2
a – b
2
=
5
3
[je I ni‡K 2 Øviv fvM]

†bŠKvi †eM
†¯ªv‡Zi †eM =
5
3
21. A man can row 9
1
3
kmph in still water and finds that it
takes him thrice as much time to row up than as to row
down the same distance in the river. The speed of the
current is : (GKRb †jvK w¯’i cvwb‡Z N›Uvq 9
1
3
wK.wg †h‡Z cv‡i
Ges †¯ªv‡Zi AbyK‚‡j GKwU wbw`©ó `~iZ¡ AwZµg Ki‡Z †h mgq
jv‡M †¯ªv‡Zi cÖwZK‚‡j H `~iZ¡ AwZµg Ki‡Z wZb¸Y mgq jv‡M|
†¯ªv‡Zi †eM KZ?
a 3
1
3
km/hr b 3
1
9
km/hr c 4
2
3
km/hr d 4
1
2
km/hr c

 mgvavb : w¯’i cvwb‡Z †jv‡Ki †eM, u = 9
1
3
=
28
3
(wK.wg./N›Uv)
awi, wbw`©ó `~iZ¡ = x
kZ©g‡Z, AbyK‚‡j mgq = 3  cÖwZK‚‡j mgq

x
u + v
=
3x
u – v

1
u + v
=
3
u – v





 mgq =
`~iZ¡
†eM
AbyK‚‡j †eM = u + v
cÖwZK‚‡j †eM = u – v
 3u + 3v = u – v  3v + v = u – 3u  4v = – 2u
 v = –
u
2
=
–28
3
2
=
–28
6
= – 4
2
3
(wK.wg./N›Uv) 


 u =
28
3
[†h‡nZz †bŠKvi †eM I †¯ªv‡Zi †eM wecixZgyLx ZvB v Gi gvb
FbvZ¡K G‡m‡Q ]
22. *A boat takes 8 hours to cover a distance while
travelling upstream, whereas while travelling
downstream it takes 6 hours. If the speed of the current
is 4 kmph, what is the speed of the boat in still water?
(GKwU †bŠKv †¯ªv‡Zi cÖwZK‚‡j 8 N›Uvq GKwU wbw`©ó `~iZ¡ AwZµg
K‡i| †hLv‡b AbyK‚‡j mgq jv‡M 6 N›Uv| †¯ªv‡Zi MwZ N›Uvq 4
wK.wg. n‡j w¯’i cvwb‡Z †bŠKvi †eM KZ?)
a 12 kmph b 16 kmph
c 28 kmph d Cannot be determined
e None of these c
 mgvavb : Avgiv Rvwb, wbw`©ó `~iZ¡
= AbyK‚‡j †eM  AbyK‚‡j mgq = cÖwZK‚‡j †eM  cÖwZK‚‡j mgq
 (u + v)  6 = (u – v)  8
 (u + 4)  6 = (u – 4)  8 [ †¯ªv‡Zi MwZ, v = 4]
 6u + 24 = 8u – 32
 2u = 24 + 32 = 56  u =
56
2
= 28 (wK.wg./N›Uv)
23. A motor boat can travel at 10 km/hr in still water. It
travelled 91 km downstream in a river a then returned
taking altogether 20 hours. Find the rate of flow of the
river. (GKwU BwÄbPvwjZ †bŠKv w¯’i cvwb‡Z N›Uvq 10 wK.wg. ågY
Ki‡Z cv‡i| †¯ªv‡Zi AbyK‚‡j 91 wK.wg. `~iZ¡ AwZµg K‡i Avevi
wd‡i Avm‡Z †gvU 20 N›Uv mgq jv‡M| †¯ªv‡Zi †eM KZ?)
a 3 km/hr b 5 km/hr c 6 km/hr d 8 km/hr a
 mgvavb : w¯’i cvwb‡Z †bŠKvi †eM = 10 wK.wg./N›Uv
awi, †¯ªv‡Zi †eM = v wK.wg./N›Uv
 AbyK‚‡j †bŠKvi †eM = 10 + v wK.wg./N›Uv
cÖwZK‚‡j †bŠKvi †eM = 10 – v wK.wg./N›Uv

91
10 + v
+
91
10 – v
= 20

91(10 – v + 10 + v)
(10 + v) (10 – v)
= 20 
91  20
102
– v2 = 20
 91 = 100 – v2
 v2
= 9  v = 3 (wK.wg./N›Uv)
24. *The speed of a boat in still water is 10 km/hr. If it can
travel 26 km downstream and 14 km upstream in the
same time, the speed of the stream is : (w¯’i cvwb‡Z †bŠKvi
MwZ‡eM N›Uvq 10 wK.wg.| hw` †bŠKvwU †¯ªv‡Zi AbyK‚‡j 26 wK.wg.
Ges †¯ªv‡Zi cÖwZK‚‡j 14 wK.wg. åg‡Y mgvb mgq jv‡M, Z‡e
†¯ªv‡Zi MwZ‡eM KZ?) [www.examveda.com]
a 2 km/hr b 2.5 km/hr
c 3 km/hr d 4 km/hr c
 mgvavb : w¯’i cvwb‡Z †bŠKvi MwZ‡eM = 10 wK.wg./N›Uv
†¯ªv‡Zi MwZ‡eM = v
 AbyK‚‡j †eM = 10 + v
cÖwZK‚‡j †eM = 10 – v
kZ©g‡Z, AbyK‚‡j mgq = cÖwZK‚‡j mgq

26
10 + v
=
14
10 – v 


 mgq =
`~iZ¡
†eM

13
10 + v
=
7
10 – v
 70 + 7v = 130 – 13v  20v = 60  v = 3 (wK.wg./N›Uv)
25. *A boat takes 90 minutes less to travel 36 miles
downstream than to travel the same distance upstream.
If the speed of the boat in still water is 10 mph, the
speed of the stream is : (GKwU †bŠKvi †¯ªv‡Zi cÖwZK‚‡j 36
gvBj `~iZ¡ AwZµg Ki‡Z †h mgq jv‡M †¯ªv‡Zi AbyK‚‡j H GKB
`~iZ¡ AwZµg Ki‡Z †bŠKvwUi 90 wgwbU Kg mgq jv‡M| hw` w¯’i
cvwb‡Z †bŠKvi MwZ‡eM N›Uvq 10 gvBj nq| Zvn‡j †¯ªv‡Zi †eMÑ)
[www.examveda.com; www.indiabix.com]
a 2 mph b 2.5 mph c 3 mph d 4 mph a
 mgvavb : w¯’i cvwb‡Z †bŠKvi †eM = 10 gvBj/N›Uv
AbyK‚‡j MwZ = 10 + v
cÖwZK‚‡j MwZ = 10 – v
†h‡nZz, cÖwZK‚‡j †bŠKvi mgq †ewk jv‡M|
ZvB kZ©g‡Z, cÖwZK‚‡j mgq – AbyK~‡j mgq = mgq e¨eavb

36
10 – v
–
36
10 + v
=
90
60
=
3
2

1
10 – v
–
1
10 + v
=
3
2  36
[Dfq cÿ‡K 36 Øviv fvM K‡i]

10 + v – 10 + v
(10 – v) (10 + v)
=
1
24

2v
100 – v2 =
1
24
 48v = 100 – v2
 v2
+ 48v – 100 = 0
 v2
+ 50v – 2v – 100 = 0  v (v + 50) – 2 (v + 50) = 0
 (v + 50) (v – 2) = 0  v = – 50, v = 2
†h‡nZz †¯ªv‡Zi †eM †bŠKvi †eM n‡Z †ewk n‡Z cv‡i bv| ZvB v =
– 50 MÖnY‡hvM¨ bq|  †¯ªv‡Zi †eM = 2 (gvBj/N›Uv)
26. *A man rows to a place 48 km distant and back in 14
hours. He finds that he can row 4 km with the stream in
the same time as 3 km against the stream. The rate of the
stream is : (GKRb †jvK `uvo †e‡q 48 wK.wg. `~i‡Z¡i GK RvqMvq
wM‡q Avevi wd‡i Avm‡Z mgq jv‡M 14 NÈv| †m jÿ Kij †h, Zvi
†¯ªv‡Zi AbyK‚‡j 4 wK.wg. Ges †¯ªv‡Zi cÖwZK‚‡j 3 wK.wg. †h‡Z GKB
mgq jv‡M| †¯ªv‡Zi †eM KZ?) [www.examveda.com; www.indiabix.com]
c 1.8 km/hr d 3.5 km/hr a
 mgvavb : awi, H wbw`©ó mgq = t N›Uv
AbyK‚‡j t N›Uvq hvq = 4 wK.wg.
 1 =
4
t 48 km
14 hr
 AbyK‚j †eM =
4
t
wK.wg./ N›Uv
Abyiƒcfv‡e, cÖwZK‚‡j †eM =
3
t
wK.wg./ N›Uv
kZ©g‡Z, AbyK‚‡j mgq + cÖwZK~‡j mgq = †gvU mgq

48
4
t
+
48
3
t
= 14 
48  t
4
+
48  t
3
= 14
 12t + 16t = 14  28t = 14  t =
14
28
=
1
2
(N›Uv)
 †¯ªv‡Zi AbyK‚‡j †eM, u + v =
4
1/2
= 8
†¯ªv‡Zi cÖwZK‚‡j †eM, u – v =
3
1/2
= 6
(–) (+) (–)
we‡qvM K‡i cvB, 2v = 2  v = 2/2 = 1(wK.wg./N›Uv)

27. *A boat covers 24 km upstream and 36 km downstream
in 6 hours while it covers 36 km upstream and 24 km
downstream in 6
1
2
hours. The velocity of the current is
(GKwU †bŠKv 6 N›Uvq †¯ªv‡Zi cÖwZK‚‡j 24 wK.wg. Ges †¯ªv‡Zi
AbyK‚‡j 36 wK.wg. †h‡Z cv‡i Avevi GwU 6
1
2
N›Uvq †¯ªv‡Zi
cÖwZK‚‡j 36 wK.wg. Ges AbyK‚‡j 24 wK.wg. `~iZ¡ AwZµg K‡i|
†¯ªv‡Zi MwZ‡eMÑ) [www.competoid.com]
c 2 km/hr d 2.5 km/hr c
 mgvavb : awi, w¯’i cvwb‡Z †bŠKvi †eM = u Ges †¯ªv‡Zi †eM = v
1g kZ©g‡Z, cÖwZK‚‡j mgq + AbyK‚‡j mgq = †gvU mgq

24
u – v
+
36
u + v
= 6



 mgq =
`~iZ¡
†eM
Avevi 2q kZ©g‡Z, cÖwZK‚‡j mgq + AbyK‚‡j mgq = †gvU mgq

36
u – v
+
24
u + v
= 6
1
2
=
13
2
Simple Kivi Rb¨ awi, u – v = x Ges u + v = y

24
x
+
36
y
= 6 ................................................. (i)
36
x
+
24
y
=
13
2
.............................................. (ii)
(i) bs n‡Z cvB,
4
x
+
6
y
= 1 [Dfq cÿ‡K 6 Øviv fvM K‡i]

4
x
= 1–
6
y

1
x
=
1
4 


1 –
6
y
......................... (iii)
1
x
Gi gvb (ii) bs G ewm‡q cvB-
36 
1
4 


1 –
6
y
+
24
y
=
13
2
 9



1 –
6
y
+
24
y
=
13
2
 9 –
54
y
+
24
y
=
13
2

– 54 + 24
y
=
13
2
– 9 
– 30
y
=
13 – 18
2
=
– 5
2

y
30
=
2
5
 y =
2
5
 30 = 12 wK.wg/N›Uv
y Gi gvb (iii) bs G ewm‡q cvB-
1
x
=
1
4 


1 –
6
12
=
1
4




1 –
1
2
=
1
4

1
2
=
1
8
 x = 8
 u + v = 12 (y Gi gvb)
Ges u – v = 8 (x Gi gvb)
(–) (+) (–)
2v = 4 [we‡qvM K‡i cvB]
 v = 2  †¯ªv‡Zi MwZ‡eM = 2 wK.wg./N›Uv
28. A boat goes 30 km upstream and 44 km downstream in
10 hours. In 13 hours, it can go 40 km upstream and 55
km downstream. The speed of the boat in still water is
(GKwU †bŠKv 10 N›Uvq †¯ªv‡Zi cÖwZK‚‡j 30 wK.wg. Ges †¯ªv‡Zi
AbyK‚‡j 44 wK.wg hvq| †bŠKvwU 13 N›Uvq †¯ªv‡Zi cÖwZK‚‡j 40 wK.wg.
Ges AbyK‚‡j 55 wK.wg. †h‡Z cv‡i| w¯’i cvwb‡Z †bŠKvi MwZ‡eMÑ)
a 3 km/hr b 4 km/hr
c 8 km/hr d None of these c
 mgvavb : 1g kZ©g‡Z,
cÖwZK‚‡j mgq + AbyK‚‡j mgq = †gvU mgq

30
u – v
+
44
u + v
= 10



 mgq =
`~iZ¡
†eM
awi, cÖwZK‚‡j †eM = u – v = x
AbyK‚‡j †eM = u + v = y

30
x
+
44
y
= 10 ... ... ... (i)
2q kZ©g‡Z,
40
u – v
+
55
u + v
= 13

40
x
+
55
y
= 13 ... ... ... (ii)
(i) bs n‡Z cvB,
30
x
+
44
y
= 10

30
x
= 10 –
44
y

1
x
=
1
30 


10 –
44
y
.......... (iii)
1
x
Gi gvb (ii) bs G ewm‡q cvB-
40 
1
30 


10 –
44
y
+
55
y
= 13

4
3 


10 –
44
y
+
55
y
= 13

40
3
–
44  4
3y
+
55
y
= 13 
– 44  4 + 55 3
3y
= 13 –
40
3

– 11
3y
=
39 – 40
3

3y
11
=
3
1
 y = 11
y Gi gvb (iii) bs G ewm‡q cvB-
1
x
=
1
30 


10 –
44
11
=
1
30
  x = 5
 u + v = 11
u – v = 5
2u = 16 [†hvM K‡i]
 u = 8
29. *At his usual rowing rate, Rahul can travel 12 miles
downstream in a certain river in 6 hours less than it
takes him to travel the same distance upstream. But if
he could double his usual rowing rate for his 24 mile
round trip, the downstream 12 miles would then take
only on hour less than the upstream 12 miles. What is
the speed of the current in miles per hour? (ivûj GKwU
wbw`©ó MwZ‡Z 12 gvBj wM‡q wd‡i Avmvi mgq jÿ¨ Ki‡jv †¯ªv‡Zi
cÖwZK‚‡j hZÿY mgq jv‡M †¯ªv‡Zi AbyK~‡j Zvi †P‡q 6 N›Uv mgq
Kg jv‡M| hw` Zvi MwZ wØ¸b n‡Zv Zvn‡j GB 24 gvBj `~i‡Z¡i
†ÿ‡Î AbyK‚‡ji 12 gvBj †h‡Z cÖwZK‚‡ji 12 gvBj hvIqvi mg‡qi
PvB‡Z 1 N›Uv Kg mgq jvM‡Zv| †¯ªv‡Zi †eM KZ gvBj/N›Uv?)
a 1
1
3
b 1
2
3
c 2
1
3
d 2
2
3
d
 mgvavb : awi, w¯’i cvwb‡Z ivû‡ji MwZ‡eM = u (gvBj/N›Uv)
†¯ªv‡Zi MwZ‡eM = v (gvBj/N›Uv)
`~iZ¡ = 12 (gvBj)
†h‡nZz cÖwZK‚‡j †ewk mgq jv‡M,
ZvB 1g kZ©g‡Z, cÖwZK‚‡j mgq – AbyK‚‡j mgq = mgq e¨eavb

12
u – v
–
12
u + v
= 6



 mgq =
`~iZ¡
†eM

2
u – v
–
2
u + v
= 1 [Dfq cÿ‡K 6 Øviv fvM K‡i]

2 (u + v) – 2(u – v)
(u – v) (u + v)
= 1 
2u + 2v – 2u + 2v
u2
– v2 = 1
 4v = u2
– v2
................................................. (i)
Avevi, ivû‡ji MwZ‡eM = 2u n‡j,
 †¯ªv‡Zi cÖwZK‚‡j Zvi †eM = 2u – v
†¯ªv‡Zi AbyK‚‡j Zvi †eM = 2u + v
2q kZ©g‡Z, cÖwZK‚‡j mgq – AbyK‚‡j mgq = mgq e¨eavb

12
2u – v
–
12
2u + v
= 1

12{(2u + v) – (2u – v)}
(2u – v) (2u + v)
= 1 
12{2u + v – 2u + v}
4u2
– v2 = 1
 24v = 4u2
– v2
............................................ (ii)

u2
†K ev` †`qvi Rb¨ (i)  4 – (ii) Kwi
16v = 4u2
– 4v2
24v = 4u2
– v2
(–) (–) (+)
– 8v = – 3v2
 3v = 8  v =
8
3
 †¯ªv‡Zi †eM =
8
3
= 2
2
3
30. A man can swim in still water at a rate of 4 km/hr. The
width of the river is 1 km. How long will he take to
cross the river straight, if the speed of the current is 3
km/hr? (GKRb †jvK w¯’i cvwb‡Z N›Uvq 4 wK.wg. †e‡M muvZvi
KvU‡Z cv‡i| b`xi cÖ¯’ 1 wK.wg. n‡j Ges †¯ªv‡Zi †eM 3 wK.wg./
N›Uv n‡j, b`xwU cvi n‡Z KZÿY mgq jvM‡e?)
a 10 min b 15 min c 18 min d 20 min b
 mgvavb : b`xi cÖ¯’ OA eivei †jv‡Ki †eM = 4 km/hr
b`xi cÖ¯’ ev `~iZ¡ = 1 km
b`xi cÖ¯’ OA eivei muvZvi KvU‡j
me©wb¤œ mg‡q B we›`y‡Z †cŠQv‡e
 mgq =
`~iZ¡
†eM =
1
4
hr =
1
4
 60 = 15 min
4km/hr1 km
A
O
B
31. A man wishes to cross a river perpendicularly. In still
water he takes 4 minutes to cross the river, but in flowing
river he takes 5 minutes. If the river is 100 metres
wide, the velocity of the following water of the river is
(GKRb e¨w³ j¤^fv‡e GKwU b`x cvi n‡Z Pvq| w¯’i cvwb‡Z b`xwU
cvi n‡Z Zvi 4 wgwbU jv‡M wKš‘ cÖevngvb b`x‡Z Zvi 5 wgwbU
jv‡M| hw` b`xwU 100 wg. cÖk¯’ nq, Zvn‡j †¯ªv‡Zi †eM KZ?)
a 10 m/min b 15 m/min c 20 m/min d 30 m/min b
 mgvavb : w¯’i cvwb‡Z †jvKwUi †eM, u =
`~iZ¡
mgq =
100 wgUvi
4 wgwbU
= 25 wgUvi/wgwbU
†¯ªvZ we‡ePbv Ki‡j, cv‡ki wP‡Î,
†jvKwU u †e‡M OA eivei Pj‡Z †Póv
Ki‡Q| †¯ªv‡Zi †eM V Gi Rb¨ †m
OA c‡_ bv wM‡q cÖK…Zc‡ÿ OD
eivei j¤^fv‡e R †e‡M b`x cvi n‡Z
†c‡i‡Q| OA Gi mgvb I mgvšÍivj
mij‡iLv BD A¼b Kwi| Zvn‡j BD
†iLv †jvKwUi MwZ‡eM u wb‡`©k K‡i|
D
O BV
A
R = 20
u = 25
u = 25
jwä‡eM, R =
100
5
= 20 wgUvi/wgwbU [‹†¯ªvZ _vK‡j mgq = 5 wgwbU]
OBD wÎf‚‡R, u2
= R2
+ v2
 v = u2
– R2
= 252
– 202
= 225 = 15 wgUvi/wgwbU
32. A man can row upstream at 10 kmph and downstream
at 18 kmph. Find the man’s rate in still water? (GKRb
†jvK †¯ªv‡Zi cÖwZK‚‡j 10 wK.wg./N›Uv Ges AbyK‚‡j 18 wK.wg./N›Uv
†e‡M †h‡Z cv‡i| Zvn‡j w¯’i cvwb‡Z Zvi †eM KZ?)
a 14 kmph b 4 kmph c 12 kmph d 10 kmph a
 mgvavb : cÖwZK‚‡j †jvKwUi †eM, u – v = 10
AbyK‚‡j †jvKwUi †eM, u + v = 18
†hvM K‡i, 2u = 28  u = 14
33. *A man takes 2.2 times as long to row a distance upstream
as to row the same distance downstream. If he can row 55
km downstream in 2 hours 30 minutes, what is the speed
of the boat in still water? (GKRb †jvK †¯ªv‡Zi AbyK‚‡j GKwU
wbw`©ó `~iZ¡ hZ mg‡q AwZµg Ki‡Z cv‡i, †¯ªv‡Zi cÖwZK‚‡j 2.2 ¸b
mgq †ewk jv‡M| hw` wZwb †¯ªv‡Zi AbyK‚‡j 2 N›Uv 30 wgwb‡U 55
wK.wg. †h‡Z cv‡i, w¯’i cvwb‡Z Zvi †eM KZ?) [www.examveda.com]
a 40 km/h b 8 km/h c 16 km/h d 24 km/hr c
 mgvavb : 2 N›Uv 30 wgwbU = 2 +
30
60
= 2 +
1
2
=
5
2
N›Uv
†¯ªv‡Zi AbyK‚‡j
5
2
N›Uvq hvq = 55 wK.wg.
 =
55
5
2
= 55 
2
5
= 22 wK.wg.
 AbyK‚‡j †¯ªv‡Zi †eM, u + v = 22 ... ... ... (i)
awi, `~iZ¡ = x
1g kZ©, cÖwZK‚‡j mgq = 2.2  AbyK‚‡j mgq

x
u – v
= 2.2 
x
u + v

1
u – v
=
2.2
u + v 


 mgq =
`~iZ¡
†eM

1
u – v
=
2.2
22
=
1
10
 u – v = 10 ... ... ... (ii)
(i) I (ii) †hvM Kwi, 2u = 22 + 10 = 32
 u =
32
2
= 16 wK.wg./N›Uv
34. Boat A travels downstream from Point X to Point Y in
3 hours less than the time taken by Boat B to travel
upstream from Point Y to Point Z. The distance
between X and Y is 20 km, which is half of the distance
between Y and Z. The speed of Boat B in still water is
10 km/h and the speed of Boat A in still water is equal
to the speed of Boat B upstream. What is the speed of
Boat A in still water? (Consider the speed of the current
to be the same.) (B †bŠKvwU †¯ªv‡Zi cÖwZK‚‡j Y c‡q›U n‡Z Z
c‡q‡›U †h‡Z hZ mgq jv‡M, A †bŠKvwU †¯ªv‡Zi AbyK‚‡j X c‡q›U
n‡Z Y c‡q‡›U †h‡Z Zvi †_‡K 3 N›Uv mgq Kg jv‡M| X I Y Gi
`~iZ¡ 20 wK.wg. hv Y n‡Z Z Gi `~i‡Z¡i A‡a©K| B †bŠKvwUi w¯’i
cvwb‡Z †eM 10 wK.wg./N›Uv Ges A †bŠKvwUi w¯’i cvwb‡Z †eM, B
†bŠKvwUi cÖwZK‚‡ji †e‡Mi mgvb| w¯’i cvwb‡Z A †bŠKvwUi †eM
KZ? (†¯ªv‡Zi †eM Dfq †ÿ‡Î GKB)
a 10 km/h b 16 km/h c 12 km/h d 8 km/h d
 mgvavb :
Y
A
X Z40 km20 km
cÖwZK‚jAbyK‚j
B
awi, w¯’i cvwb‡Z, A †bŠKvi †eM = u Ges †¯ªv‡Zi †eM = v
w¯’i cvwb‡Z, B †bŠKvi †eM = 10 wK.wg./N›Uv (†`Iqv Av‡Q)
eY©bv g‡Z, XY = 20 wK.wg. Ges YZ = 2 × 20 = 40 wK.wg.
1g kZ©, B †bŠKvi cÖwZK‚‡j mgq – A †bŠKvi AbyK‚‡j mgq = 3 N›Uv

40
10 – v
–
20
u + v
= 3 ........................ (i)
2q kZ©, A †bŠKvi w¯’i cvwb‡Z †eM = B †bŠKvi cÖwZK‚‡j †eM|
 u = 10 – v  v = 10 – u
v Gi gvb (i) bs G emvB-
40
10 – 10 + u
–
20
u + 10 – u
= 3

40
u
–
20
10
= 3 
40
u
– 2 = 3 
40
u
= 5
 u =
40
5
= 8 wK.wg./N›Uv
35. *The speed of the boat in still water is 5 times that of the
current, it takes 1.1 hours to row to point B form point
A downstream. The distance between point A and
point B is 13.2 km. How much distance (in km) will it
cover in 312 minutes upstream? (w¯’i cvwb‡Z GKwU †bŠKvi
†eM †¯ªv‡Zi †e‡Mi cuvP¸Y Ges AbyK‚‡j A †_‡K B-†Z hvIqvi
Rb¨ 1.1 N›Uv mgq jv‡M| A †_‡K B Gi `~iZ¡ 13.2 wK.wg.|
Zvn‡j, †¯ªv‡Zi cÖwZK‚‡j 312 wgwb‡U †bŠKvwU KZ`~i †h‡Z cvi‡e?)
[www.examveda.com]
a 43.2 b 48 c 41.6 d 44.8 c

 mgvavb : awi, †¯ªv‡Zi †eM = v wK.wg./N›Uv
w¯’i cvwb‡Z †bŠKvi †eM, u = 5v wK.wg./N›Uv
kZ©g‡Z, AbyK‚‡j A n‡Z B †h‡Z mgq jv‡M = 1.1 N›Uv

13.2
u + v
= 1.1 
13.2
5v + v
= 1.1
 1.1  6v = 13.2
 v =
13.2
1.1  6
= 2 wK.wg./N›Uv
w¯’i cvwb‡Z †bŠKvi †eM, u = 5v = 5  2 = 10 wK.wg./N›Uv
 cÖwZK‚‡j †bŠKvi †eM = u – v = 10 – 2 = 8 wK.wg./N›Uv
GLv‡b, 312 wgwbU =
312
60
=
26
5
N›Uv
cÖwZK‚‡j 1 N›Uvq hvq = 8 wK.wg.

26
5
=
26
5
 8 = 41.6 wK.wg.
36. *A boat can travel 36 km upstream in 5 hours. If the
speed of the stream is 2.4 kmph, how much time will the
boat take to cover a distance of 78 km downstream? (in
hours) (GKwU †bŠKv †¯ªv‡Zi cÖwZK‚‡j 5 N›Uvq 36 wK.wg. †h‡Z
cv‡i| hw` †¯ªv‡Zi †eM 2.4 wK.wg./N›Uv nq, †¯ªv‡Zi AbyK‚‡j 78
wK.wg. †h‡Z KZ NÈv mgq jvM‡e?) [www.examveda.com]
a 5 b 6.5
c 5.5 d 8 b
 mgvavb : awi, w¯’i cvwb‡Z †bŠKvi †eM = u
†`Iqv Av‡Q, †¯ªv‡Zi †eM, v = 2.4 wK.wg./N›Uv
†¯ªv‡Zi cÖwZK‚‡j †eM =
`~iZ¡
mgq
 u – v =
36
5
 u – 2.4 = 7.2
 u = 7.2 + 2.4 = 9.6 wK.wg./N›Uv
†¯ªv‡Zi AbyK‚‡j †eM = u + v = 9.6 + 2.4 = 12 wK.wg./N›Uv
A_©vr, †¯ªv‡Zi AbyK‚‡j 12 wK.wg. hvq = 1 N›Uvq
1 =
1
12
78 =
78
12
= 6.5 N›Uv
Direction (Question No. 37): The following question is
followed by two statements number I and II are given. You
have to read both the statements and then give the answer.
[wb‡Pi cÖkœ¸wj, 37bs cÖ‡kœi (i) I (ii)bs wee„wZi Av‡jv‡K| wee„wZ `ywU
fvj K‡i co–b Ges wb‡Pi cÖkœ¸wji DËi w`b|]
a. If the data given in statement I alone are sufficient to
answer the question whereas the data given in statement
II alone are not sufficient to answer the questions. [ïay
(i)bs kZ© †_‡K DËi wbY©q Kiv hvq wKš‘ (ii)bs kZ© †_‡K hvq bv|]
b. If the data given in statement II alone are sufficient to
answer the question I alone are not sufficient to answer
the question. [ïay (ii) bs kZ© †_‡K DËi wbY©q Kiv hvq, (i)
bs kZ© †_‡K hvq bv|]
c. If the data in either statement I alone or in statement II
alone are sufficient to answer the question. [(i) I (ii)
Dfq kZ© †_‡KB Avjv`vfv‡e DËi wbY©q Kiv hvq|]
d. If the data in both the statement I and II are not
sufficient to answer the question. [(i) I (ii) Dfq kZ©
†_‡KB Avjv`vfv‡e DËi wbY©q Kiv hv‡e bv|]
e. If the data given in both the statements I and II are
necessary to answer the question. [(i) I (ii) Dfq kZ©
wgwj‡q DËi wbY©q Kiv m¤¢e|]
37. What is the speed of the boat in still water? (in km/hr)
(w¯’i cvwb‡Z †bŠKvi †eM KZ?)
I. The boat takes total time of 4h to travel 14 km
upstream and 35 km downstream together. (†¯ªv‡Zi
cÖwZK‚‡j 14 wK.wg. Ges †¯ªv‡Zi AbyK‚‡j 35 wK.wg. †h‡Z
†bŠKvwUi †gvU 4 N›Uv mgq jv‡M|)
II. The boat takes total time of 5h travel 29 km upstream
and 24 km downstream together. (†¯ªv‡Zi cÖwZK‚‡j 29
wK.wg. Ges †¯ªv‡Zi AbyK‚‡j 24 wK.wg. †h‡Z †bŠKvwUi †gvU 5
N›Uv mgq jv‡M| Dc‡ii kZ© †_‡K DËi `vI|)
 mgvavb : awi, w¯’i cvwb‡Z †bŠKvi †eM = u wK.wg./N›Uv
Ges †¯ªv‡Zi †eM = v wK.wg./N›Uv
I. Gi kZ©g‡Z,
14
u – v
+
35
u + v
= 4 ..................................... (i)
II. Gi kZ©g‡Z,
29
u – v
+
24
u + v
= 5 .................................... (ii)
GLv‡b ïay (i)bs †_‡K u I v †ei Kiv m¤¢e bq| KviY †h KqwU
AÁvZ ivwk mgxKi‡Y _v‡K Zv wbY©q Kivi Rb¨ ZZwU ¯^vaxb mgxKiY
cÖ‡qvRb| Abyiƒcfv‡e ïay (ii)bs †_‡KI u I v wbY©q Kiv m¤¢e bq|
Z‡e Dfq (i) I (ii) `ywU kZ© †_‡K u I v wbY©q Kiv hv‡e|
38. The speed of a boat when travelling downstream is 32
km/hr, whereas when travelling upstream it is 28
km/hr, what is the speed of the boat in still water and
the speed of the stream? (†¯ªv‡Zi AbyK‚‡j GKwU †bŠKv 32
wK.wg./N›Uv MwZ‡e‡M hvÎv K‡i Ges †¯ªv‡Zi cÖwZK‚‡j 28 wK.wg/N›Uv
MwZ‡e‡M hvÎv K‡i| w¯’iR‡j †bŠKvi MwZ‡eM KZ Ges †¯ªv‡Zi
MwZ‡eM KZ?)
 mgvavb: awi, ïay †bŠKvi †eM x wK.wg./N›Uv
Ges †¯ªv‡Zi †eM y wK.wg./N›Uv
cÖkœg‡Z, x + y = 32 ........(i)
Ges x  y = 28 .......(ii)
(i) Ges (ii) †hvM K‡i cvB
2x = 60
 x = 30
(i) †_‡K (ii) we‡qvM K‡i cvB
2y = 4
 y = 2
†bŠKvi †eM 30 wK.wg./N›Uv
Ges †¯ªv‡Zi †eM 2 wK.wg./N›Uv
39. The speed of a motor boat is that of the current of
water as 36 : 5. The boat goes along with the current in
5 hours 10 minutes. How much time will it take to come
back? (GKwU †gvUi †evU Ges R‡ji †¯ªv‡Zi MwZ‡e‡Mi AbycvZ
36 : 5| †bŠKvwU †¯ªv‡Zi AbyK‚‡j 5 N›Uv 10 wgwb‡U hvq| †bŠKvwU
KZ mg‡q wd‡i Avm‡e?)
 mgvavb: awi, †bŠKvi †eM 36x wK.wg./N›Uv
†¯ªv‡Zi †eM 5x wK.wg./N›Uv
Ges †gvU `~iZ¡ d wK.wg.
cÖkœg‡Z,
d
36x + 5x
=
5  60 + 10
60

d
41x
=
310
60
 d =
310x  41
60
=
1271x
6
Zvn‡j, †¯ªv‡Zi cÖwZK‚‡j mgq jvM‡e

d
36x  5x
=



1271x
6

1
31x
N›Uv
=
41
6
N›Uv = 6 N›Uv 50 wgwbU
Dr. R.S. AGGARWAL m¨v‡ii eB‡qi D`vniY

40. A man can row 6 km/hr in still water. It takes him
twice as long to row up as to row down the river. Find
the rate of stream. (GK e¨w³ w¯’i R‡j 6 wK.wg./N›Uv †e‡M
†h‡Z cv‡i| †¯ªv‡Zi AbyK‚‡j †h‡Z Zvi hZ mgq jv‡M, †¯ªv‡Zi
cÖwZK‚‡j †h‡Z Zvi wØ¸Y mgq jv‡M, †¯ªv‡Zi MwZ‡eM KZ?)
 mgvavb: awi, †¯ªv‡Zi MwZ‡eM x wK.wg./N›Uv
Ges †gvU `~iZ¡ d wK.wg.
†¯ªv‡Zi AbyK‚‡j †eM 6 + x wK.wg./N›Uv
†¯ªv‡Zi wecix‡Z †eM 6  x wK.wg/N›Uv
cÖkœg‡Z,
2d
6 + x
=
d
6  x
[ 2  AbyK‚‡j mgq = cÖwZK‚‡j mgq]
 12  2x = 6 + x  6 = 3x
 x = 2
†¯ªv‡Zi †eM 2 wK.wg./N›Uv|
41. A man can row 7
1
2
kmph in still water. If in a river
running at 1.5 km an hour, it takes him 50 minutes to
row to a place and back, how far off is the place? (GK
e¨w³ w¯’iR‡j 7
1
2
wK.wg./N›Uv MwZ‡Z †h‡Z cv‡i| hw` 1.5
wK.wg./N›Uv MwZ‡e‡M eB‡q hvIqv †Kv‡bv b`x w`‡q GKwU ¯’v‡b †h‡Z
Ges wd‡i Avm‡Z 50 wgwbU jv‡M, Zvn‡j ¯’vbwUi `~iZ¡ KZ?)
 mgvavb: awi, †gvU `~iZ¡ x wK.wg.
†¯ªv‡Zi AbyK‚‡j †eM 7.5 + 1.5 ev 9 wK.wg./N›Uv
†¯ªv‡Zi cÖwZK‚‡j †eM 7.5  1.5 ev 6 wK.wg./NÈv
cÖkœg‡Z,
x
9
+
x
6
=
50
60
 2x + 3x =



5
6
 18  5x = 15
 x = 3
 †gvU `~iZ¡ 3 wK.wg.|
42. A boat goes 8 km upstream and then returns. Total
time taken is 4 hrs 16 minutes. If the velocity of current
is 1 km/hr, find the actual velocity of the boat. (GKwU
†bŠKv †¯ªv‡Zi cÖwZK‚‡j 8 wK.wg. hvq Ges wd‡i Av‡m| †gvU mgq
jv‡M 4 N›Uv 16 wgwbU| hw` †¯ªv‡Zi †eM 1 wK.wg./N›Uv nq, Zvn‡j
†bŠKvi cÖK…Z †eM KZ?)
 mgvavb: awi, †bŠKvi †eM x wK.wg./N›Uv
†¯ªv‡Zi AbyK‚‡j †eM x + 1 wK.wg./N›Uv
†¯ªv‡Zi wecix‡Z †eM x  1 wK.wg./N›Uv
cÖkœg‡Z,
8
x + 1
+
8
x  1
=
4  60 + 16
60

x + 1 + x  1
(x + 1) (x  1)
=
8
15
 30x = 8 (x2
 1)  8x2
 30x  8 = 0
 4x2
 15x  4 = 0  4x2
 16x + x  4 = 0
 4x (x  4) + (x  4) = 0  (x  4) (4x + 1) = 0
GLb, x  
1
4
†h‡nZz †eM FbvZ¥K n‡Z cv‡i bv
 x = 4
 †bŠKvi †eM 4 wK.wg./N›Uv|
43. A boatman rows to a place 45 km distant and back in
20 hours. He finds that he can row 12 km with the
stream in the same time as 4 km against the stream.
Find the speed of the stream. (GKRb gvwS 45 wK.wg. `~‡i
Aew¯’Z †Kv‡bv ¯’v‡b †bŠKvq †Mj Ges wd‡i Avmj 20 N›Uvq| †m
†`L‡Z †cj, †¯ªv‡Zi AbyK‚‡j 12 wK.wg. c_ †h‡Z hZ mgq jv‡M,
GKB mg‡q †¯ªv‡Zi cÖwZK‚‡j 4 wK.wg. †h‡Z cv‡i| †¯ªv‡Zi MwZ‡eM KZ?)
 mgvavb: awi, †m †¯ªv‡Zi AbyK‚‡j 12 wK.wg. hvq x N›Uvq
 †¯ªv‡Zi AbyK‚‡j †eM
12
x
, cÖwZK‚‡j †eM
4
x
cÖkœg‡Z,
45
12
x
+
45
4
x
= 20
x
12
+
x
4
=
20
4x
 x =
4
3
N›Uv
AbyK‚‡j †eM 


12 
3
4
wK.wg./N›Uv = 9 wK.wg./N›Uv
cÖwZK‚‡j †eM = 4 
3
4
wK.wg./N›Uv = 3 wK.wg./N›Uv
 †¯ªv‡Zi †eM
1
2
(9  3) ev 3 wK.wg./N›Uv [ v =
1
2
(a – b)]
44. *A man rows 12 km in 5 hours against the stream and the
speed of current being 4 kmph. What time will be taken by
him to row 15 km with the stream?[www.examveda.com;www.competoid.com]
a 1 hour 27
7
13
minutes b 1 hour 24
7
13
minutes
c 1 hour 25
7
13
minutes d 1 hour 26
7
13
minutes d
 mgvavb: Given, Speed of current, v = 4 km/hr
Speed of man against the current, u  v =
12
5 


 Speed =
Distance
Time
 u  4 =
12
5
[‹ v = 4]
 u =
12
5
+ 4  u =
12 + 20
5
 u =
32
5
Speed of man with the stream = u + v =
32
5
+ 4
=
32 + 20
5
=
52
5
km/hr
Time taken in downstream =
Distance
downstream speed
=
15
52
5
km/hr
=
15  5
52
hr = 1 hr 26
7
13
min
45. †bŠKv I †¯ªv‡Zi †eM N›Uvq h_vµ‡g 10 wKwg I 5 wKwg| b`x c‡_ 45
wKwg `xN© c_ GKevi AwZµg K‡i wd‡i Avm‡Z KZ N›Uv mgq
jvM‡e? [Exam Taker AUST : Sonali Bank (Sub-Asst. Engr. Civil-2016)]
a 8 N›Uv b 10 N›Uv c 12 N›Uv d 14 N›Uv c
 mgvavb : †bŠKv AwZµg K‡i Avevi wd‡i Avm‡j †h †Kvb GK mgq
†¯ªv‡Zi AbyK‚‡j Ges Ab¨ mgq †¯ªv‡Zi cÖwZK‚‡j hvq|
 †¯ªv‡Zi AbyK‚‡j †eM = (10 + 5) wK.wg./N›Uv = 15 wK.wg/N›Uv
Ges †¯ªv‡Zi cÖwZK‚‡j †eM = (10 – 5) wK.wg./N›Uv = 5 wK.wg./N›Uv
†¯ªv‡Zi AbyK‚‡j cÖ‡qvRbxq mgq t1 n‡j, t1 =
45
15
= 3 N›Uv
†¯ªv‡Zi cÖwZK‚‡j cÖ‡qvRbxq mgq t2 n‡j, t2 =
45
5
= 9 N›Uv
 †gvU mgq = 3 + 9 = 12 N›Uv
46. A boat against the current of water goes 9 km/hr and
in the direction of the current 12 km/hr. The boat takes
4 hours and 12 minutes to move upwared and
downward direction from A to B. What is the distance
between A and B? [www.competoid.com]
a 21.6 km b 21.0 km c 22 km d 30 km a
 mgvavb: Let, the distance between A and B is x km
Time to go from A to B =
x
9
hrs
wewfbœ I‡qemvBU Ges weMZ eQ‡ii cÖkœmg~‡ni mgvavb

Time to go from B to A =
x
12
hrs
According to question,
x
9
+
x
12
= 4 +
12
60

4x + 3x
36
= 4 +
1
5

7x
36
=
21
5
 x =
21
5

36
7
 x = 21.6 km
47. A boat can travel form point A to point B and return
back to point a in 9 hours. Speed of the boat in still
water is 8 km/h and the speed of the stream is 4 km/h.
Find the distance between A and B.
[Combined 4 Bank’s (Officer General) – 19]
a 18 km b 27 km c 36 km d 45 km b
 mgvavb: In a round up travel one of the travel is with the
stream and other is against the stream.
Given, Speed of the boat u = 8 km/h
and Speed of the stream v = 4 km/h
Let, the distance is s
So, in upstream,
s = (u + v) × t1
 t1 =
s
12
.................... (i)
and in downstream,
s = (u  v) × t2
 t2 =
s
4
..................... (ii)
(i) + (ii) 
t1 + t2 = s



1
12
+
1
4
 9 = s



1 + 3
12
[Total time 9 hours]
 s =
12 × 9
4
= 27 km
48. A boat can travel from point A to point B and return
back to point A in 9 hours. Speed of the boat in still
water is 8 km/h and the speed of the stream is 4 km/h.
Find the distance between A and B. (GKwU †bŠKv A †_‡K
B †Z †h‡Z Ges cybivq A †Z wd‡i Avm‡Z 9 N›Uv mgq jv‡M| w¯’i
cvwb‡Z †bŠKvwUi †eM 8 km/h Ges †¯ªv‡Zi †eM 4 km/hr| A Ges B
Gi ga¨eZ©x `~iZ¡Ñ) [Exam Taker AUST : Combined 4 Banks (Officer-2019)]
a 18 km b 27 km c 36 km d 45 km b
 mgvavb : A †_‡K B †Z wM‡q cybivq A †Z wd‡i Avmv gv‡b
†h‡Kv‡bv GK mgq †bŠKv †¯ªv‡Zi AbyK‚‡j wM‡q‡Q Ges Ab¨ mgq
†¯ªv‡Zi cÖwZK‚‡j wM‡q‡Q|
†¯ªv‡Zi AbyK‚‡j †eM = w¯’i cvwb‡Z †bŠKvi †eM + †¯ªv‡Zi †eM = 8
+ 4 = 12 km/h
Ges †¯ªv‡Zi cÖwZK‚‡j †eM = w¯’i cvwb‡Z †bŠKvi †eM – †¯ªv‡Zi
†eM = 8 – 4 = 4 km/h
GLb, `~iZ¡ = †eM  mgq
ev, s = 12  t1 [awi, A †_‡K B Gi `~iZ¡ s wK.wg. Ges †¯ªv‡Zi
AbyK‚‡j A †_‡K B †Z †h‡Z t1 N›Uv mgq jv‡M]
 t1 =
s
12
†¯ªv‡Zi cÖwZK‚‡j, s = 4  t2 [awi, †¯ªv‡Zi cÖwZK‚‡j t2 N›Uv mgq jv‡M]
 t2 =
s
4
GLb, t1 + t2 = 9 [†h‡nZz hvIqv Avmvq †gvU 9 N›Uv mgq jv‡M]

s
12
+
s
4
= 9 
s + 3s
12
= 9
 4s = 108  s = 27 km
 A Ges B Gi ga¨eZx© `~iZ¡ = 27 km
49. A boat can travel with a speed of 13 km/hr in still water.
If the speed of the stream is 4 km/hr, find the time
taken by the boat to go 68 km downstream.
[Combined 5 Bank’s (Officer Cash) – 19 + www.indiabix.com
+ www.examveda.com + www.competoid.com + www.brainly.in]
a 4 hours b 3 hours c 5 hours d 2 hours a
 mgvavb: Given, speed of boat, u = 13 km/hr
and speed of stream, v = 4 km/hr
speed at downstream = u + v = 13 + 4 = 17 km/hr
 Time taken =
68
17
= 4 hours
50. A boat can travel with a speed of 13 km/hr in still water.
If the speed of the stream is 4 km/hr, find the time taken
by the boat to go 68 km downstream. (GKwU †bŠKvi w¯’i
cvwb‡Z †eM 13 km/hr| †¯ªv‡Zi †eM 4 km/hr n‡j †¯ªv‡Zi AbyK‚‡j
68 km †h‡Z KZ mgq jv‡MÑ) [Combined 5 Banks (Officer Cash-2019);
Exam Taker AUST : Janata Bank (A.E.O.-2019);
a 4 hours b 3 hours c 5 hours d 2 hours a
 mgvavb : †¯ªv‡Zi AbyK‚‡j †eM = w¯’i cvwb‡Z †bŠKvi †eM +
†¯ªv‡Zi †eM = 13 + 4 = 17 km/hr
GLb, Avgiv Rvwb, `~iZ¡ = mgq  †eM
 68 = mgq  17  mgq =
68
17
= 4 N›Uv
51. A boat covers a certain distance downstream in 1 hour,
while it comes back in 1.5 hour. If the speed of the
stream be 3 km/hr, what is the speed of the boat in still
water? (GKwU †bŠKv †¯ªv‡Zi AbyK‚‡j 1 N›Uvq GKwU wbw`©ó `~iZ¡
AwZµg K‡i Ges 1.5 N›Uvq wd‡i Av‡m| †¯ªv‡Zi †eM 3 km/hr
n‡j, w¯’i cvwb‡Z †bŠKvi †eM?)
[Exam Taker AUST : Janata Bank (A.E.O.-2019); Sonali Bank (Officer FF-2019);
a 12 km/hr b 15 km/hr c 13 km/hr d 14 km/hr b
 mgvavb : g‡b Kwi, wbw`©ó `~iZ¡ d km
†bŠKvi †eM u km/hr
†¯ªv‡Zi AbyK‚‡j †eM u + 3 =
d
1 


†eM =
`~iZ¡
mgq
 d = u + 3 ....................... (i)
†¯ªv‡Zi cÖwZK‚‡j †eM u  3 =
d
1.5
 d = 1.5 (u  3) ............. (ii)
(i) Ges (ii) n‡Z cvB, u + 3 = 1.5(u  3)
 u + 3 = 1.5u  4.5  7  5 = 0.54
 u =
7.5
0.5
= 15 km/hr
52. A boat goes 20 km upstream in 2 hours and
downstream in 1 hour. How much time this boat will
take to travel 30 km in all still water? [www.competoid.com]
a 1 hr b 2 hrs c 1.5 hrs d 2.5 hrs b
 mgvavb : Let, Speed of boat in still water = u
” ” stream = v
Upstream velocity, u  v =
20
2





‹ velocity
=
Distance
Time
 u  v = 10 ..........(i)
Downstream velocity, u + v =
20
1
 u + v = 20 ..........(ii)
(i) + (ii)  u  v + u + v = 10 + 20  2u = 30
 u = 15 km/hr
Again, Required time =
Given distance
Speed of boat in still water
=
30 km
15 km/hr
= 2 hr

53. A boat goes 6 km an hour in still water, it takes thrice
as much time in going the same distance against the
current comparison to direction of current. The speed
of the current (in km/ hour) is: [www.competoid.com]
a 4 b 5 c 3 d 2 c
 mgvavb : A boat goes 6 km an hour which means  Speed
of boat in still water, u = 6 km/hr
Speed of stream = v (Let)
Time taken in upstream = 3  Time taken downstream

x
u  v
= 3 
x
u + v
[Here, Same distance = x]

1
6  v
=
3
6 + v
[Dfq cÿ †_‡K x ev` w`‡q]
 6 + v = 18  3v  4v = 12  v = 3 km/hr
54. A boat moves downstream at the rate of 1 km in 7
1
2
minutes and upstream at the rate of 5 km an hour.
What is the speed of the boat in the still water?
[www.examveda.com; www.competoid.com]
a 8 km/hour b 6
1
2
km/hourc 4 km/hour d 3
1
2
km/hour b
 mgvavb : 7
1
2
min =
15
2
min =
15
2  60
hr =
1
8
hr
Downstream velocity, u + v =
1
1
8





Here Distance = 1 km
Time =
1
8
hr
 u + v = 8 ........(i)
Upstream velocity, u  v =
5
1 


Here Distance = 5 km
Time = 1 hr
 u  v = 5 .........(ii)
(i) + (ii) 
u + v + u  v = 8 + 5
 2u = 13  u =
13
2
= 6
1
2
km/hr
55. A boat running upstream takes 8 hours 48 minutes to
cover a certain distance, while it takes 4 hours to cover
the same distance running downstream. What is the ratio
between the speed of the boat and the speed of the
water current respectively? (GKwU †bŠKv †¯ªv‡Zi cÖwZK‚‡j
GKwU wbw`©ó `~iZ¡ AwZµg Ki‡Z 8 NÈv 48 wgwbU mgq jv‡M|
Avevi, †¯ªv‡Zi AbyK‚‡j H GKB `~iZ¡ AwZµg Ki‡Z 4 NÈv jv‡M|
Zvn‡j †bŠKvi †eM I †¯ªv‡Zi †e‡Mi AbycvZ KZ?)
[Exam Taker Arts : P.K.B. (EO-cash)-2019; Bangladesh House Building
Finance Corporation (SO) Written-2017]
 mgvavb :
8 hr 48 min = 8 +
48
60
= 8 +
4
5
=
40 + 4
5
=
44
5
hr
Let, the velocity of boat in still water = u
,, ,, ,, stream = v
for same distance,
(velocity  time)uspstream = (velocity  time)downstream
 (u  v) 
44
5
= (u + v)  4
 (u  v)
11
5
= (u + v)
[devided by 4 in both side]
 11u  11v = 5u + 5v
[Applying cross multidicaton]
 11u  5u = 11v + 5v
 6u = 16v 
u
v
=
16
6
=
8
3
 u : v = 8 : 3
56. A boat runs at 22 km per hour along the stream and 10
km per hour against the stream. Find the ratio of the speed
of the boat in still water to that of the speed of the stream.
[Exam Taker AUST : P.K.B. (E.O. General-2019)]
a 2 : 3 b 5 : 3 c 7 : 3 d 8 : 3 d
 mgvavb : awi, †bŠKvi †eM = u kmph
†¯ªv‡Zi ” = v kmph
cÖkœg‡Z, u + v = 22 .................... (i)
u – v = 10 ................... (ii)
(i) ÷ (ii) 
u + v
u – v
=
22
10

u + v
u – v
=
11
5
 11u – 11v = 5u + 5v  11u – 5u = 5v + 11v
 6u = 16v 
u
v
=
16
6
=
8
3
 u : v = 8 : 3
57. A boat sailing against a stream of river takes 6 hours to
travel 24 kms, while sailing with the stream it takes 4
hours to travel the same distance. What is the speed of
the stream? (†¯ª‡Zi cÖwZK‚‡j 24 km ågY Ki‡Z GKwU †bŠKvi
6 N›Uv mgq jv‡M Ges †¯ªv‡Zi AbyK‚‡j H `~iZ¡ cvwo w`‡Z 4 N›Uv
mgq jv‡M| †¯ªv‡Zi †eM KZ?)
[Exam Taker AUST : Basic Bank (Asst. Manager-2018)]
a 2.5 km/hr b 1.5 km/hr c 1 km/hr d 0.5 km/hr c
 mgvavb : awi, †bŠKvi †eM (w¯’i cvwb‡Z) = u
 †¯ªv‡Zi AbyK‚‡j †bŠKvi †eM, a = u + v
cÖwZK‚‡j , b = u – v
a =
`~iZ¡
†¯ªv‡Zi AbyK‚‡j cvwo w`‡Z mgq
 u + v =
24 km
4 hours
 u + v = 6 km/hr
b =
`~iZ¡
†¯ªv‡Zi cÖwZK‚‡j cvwo w`‡Z mgq
u – v =
24
6
km/hr
 u – v = 4 km/hr
u + v = 6 ......(i)
u – v = 4 .........(ii)
(i) – (ii) 
u + v = 6
u – v = 4
(–) (+) (–)
2v = 2
v =
2
2
 v = 1 km/hr
58. A boat takes 4 hours to cover a certain distance running
downstream, while it requires 8 hours 48 minutes to
cover the same distance running upstream. Find the
ratio between speed of stream and speed of the boat?
(GKwU †bŠKv †¯ªv‡Zi AbyK‚‡j wbw`©ó `~iZ¡ †h‡Z 4 N›Uv mgq jv‡M|
Avevi, †¯ªv‡Zi cÖwZK‚‡j mgvb `~iZ¡ †h‡Z 8 N›Uv 48 wgwbU mgq
jv‡M| †¯ªv‡Zi †eM I †bŠKvi †e‡Mi AbycvZÑ)
[Exam Taker AUST : Sonali Bank (Officer Cash FF-2019);
Sonali Bank (Officer Cash FF) – 19 + www.careerbless.com
+ www.examveda.com + www.indiabix.com + www.brainly.in]
a 1 : 2 b 3 : 8 c 2 : 3 d 4 : 3 b
 mgvavb : g‡b Kwi, †bŠKvi †eM = u
 †¯ªv‡Zi AbyKz‡j A_©vr (u + v) †e‡M 4 NÈvq hvq
= (u + v)  4
 †¯ªv‡Zi cÖwZK‚‡j A_©vr (u – v) †e‡M 8 NÈv 48 wgwbU
ev
44
5
NÈvq
Upstream
velocity = u  v
Downstream
velocity = u + v

wd‡i Av‡m = (u – v) 
44
5
cÖkœg‡Z, (u + v)  4 = (u – v) 
44
5
[†h‡nZz hvIqvi mgq †gvU c_ = Avmvi mgq †gvU c_]
 5(u + v) = 11(u – v)
 5u + 5v = 11u – 11v  16v = 6u

v
u
=
6
16
=
3
8
 †¯ªv‡Zi †eM : †bŠKvi †eM = 3 : 8
59. A boat takes total 16 hours for traveling downstream
from point A to point B and coming back point C
which is somewhere between A and B. If the speed of
the Boat in still water is 9 km/hr and rate of stream is 6
km/hr, then what is the distance between A and C?
[www.affairscloud.com]
a 30 km b 60 km c 90 km d 100 km
e Cannot be determined
 mgvavb: Speed of the boat in downstream = (9 + 6) km/hr
= 15 km/hr
Speed of the boat in upstream = (9 – 6) km/hr = 3 km/hr
Let, the distance between A and B is x
the distance between A and C is y
So, the distance between B and C is (x – y)
x
15
+
x – y
3
= 16

x + 5x – 5y
15
= 16  6x – 5y = 240
To solve y, we should know the value of x or a relation
between x and y.
So, the distance can not be determined.
60. A boat travel with a speed of 10 km/hr in still water. If
the speed of the stream is 3 km/hr then find time taken
by boat to travel 52 km downstream. (w¯’i cvwb‡Z GKwU
†bŠKvi †eM 10 km/hr| †¯ªv‡Zi †eM 3 km/hr n‡j †¯ªv‡Zi AbyK‚‡j
52 km †h‡Z †bŠKvwUi KZ mgq jvM‡e?)
[Exam Taker AUST : Combined 8 Banks (S.O.-2018); Combined 2 Banks
(Officer 2018); Exam Taker AUST : Bangladesh Bank (AD)-2019]
a 2 hrs b 4 hrs c 6 hrs d 9 hrs b
 mgvavb : †bŠKvi †eM, u = 10 km/hr
†¯ªv‡Zi †eM, v = 3 km/hr
 †¯ªv‡Zi AbyK‚‡j †eM, a = u + v = (10 + 3) = 13 km/hr
 52 km †h‡Z mgq =
`~iZ¡
†eM =
52
13
= 4 hrs
61. A boat travels upstream from B to A and down stream
from A to B in 3 hours. If the speed of boat in still
water is 9 km/hr. and the speed of current is 3 km/hr
the distance between A and B (km) is [www.competoid.com]
a 4 b 6 c 8 d 12 d
 mgvavb : Let, Distance = x
Given Condition  Time taken in upstream + Time taken
downstream = Total time

x
u  v
+
x
u + v
= 3

x
9  3
+
x
9 + 3
= 3 


u = 9 km/hr
v = 3 km/hr

x
6
+
x
12
= 3 
2x + x
12
= 3 
3x
12
= 3 
x
4
= 3
 x = 12 km
62. A boatman takes twice as long to row a distance
against the stream as to row the same distance with the
stream. Find the ratio of speeds of the boat in still
water and the stream. [www.competoid.com]
a 2 : 1 b 3 : 1 c 1 : 2 d 1 : 3 b
 mgvavb : Let, Speed of Boat in still water = u
” ” Stream = v
Given condition,
Time taken against the stream = 2  time taken with the stream

x
u  v
= 2 
x
u + v





Same distance = x
Time =
Distance
Speed

1
u  v
=
2
u + v
 2u  2v = u + v  2u  u = 2v + v
 u = 3v 
u
v
= 3
 u : v = 3 : 1
63. A certain river has current of 4 miles per hour. A boat
takes twice as a long to travel upstream between two
points as it does to travel downstream between the same
two points. What is the speed of the boat in still water?
(GKwU b`xi †¯ªv‡Zi †eM 4 mph `ywU wbw`©ó ¯’v‡bi g‡a¨ hvZvqv‡Zi
mgq †Kvb †bŠKvi †¯ªv‡Zi cÖwZK‚‡j †h‡Z †h mgq jv‡M Zv †¯ªv‡Zi
AbyKz‡j †h‡Z cÖ‡qvRbxq mg‡qi wØ¸Y| w¯’i cvwb‡Z †bŠKvi †eM KZ?)
[Exam Taker IBA : IFIC Bank Ltd. (TAO-2018)]
a 6 miles per hour b 8 miles per hour
c 12 miles per hour d cannot be determined c
 mgvavb : awi, w¯’i cvwb‡Z †bŠKvi †eM = v mph
 †¯ªv‡Zi AbyK‚‡j †bŠKvi †eM = (v + 4) mph
†¯ªv‡Zi cÖwZK‚‡j †bŠKvi †eM = (v – 4) mph
H ¯’vbØ‡qi ga¨eZ©x `~iZ¡ x n‡j,
†¯ªv‡Zi AbyK‚‡j †h‡Z mgq =
x
v + 4
†¯ªv‡Zi cÖwZK‚‡j †h‡Z mgq =
x
v – 4
cÖkœg‡Z,
x
v – 4
= 2



x
v + 4
 2v – 8 = v + 4
 2v – v = 4 + 8  v = 12 mph
64. A man can row 30 km upstream and 44 km
downstream in 10 hrs. It is also known that he can row
40 km upstream and 55 km downstream in 13 hrs. Find
the speed of the man in still water. (GKwU †jvK `uvo †e‡q
10 NÈvq †¯ªv‡Zi cÖwZK‚‡j 30 wK.wg. Ges †¯ªv‡Zi AbyK‚‡j 44 wK.wg.
†h‡Z cv‡i| Av‡iv Rvbv hvq †h, †m 13 NÈvq †¯ªv‡Zi cÖwZK‚‡j 40
wK.wg. Ges †¯ªv‡Zi AbyK‚‡j 55 wK.wg. hvq| w¯’i cvwb‡Z †bŠKvi †eM
KZ?) [Exam Taker AUST : Combined 3 Banks (Officer-Cash)-2018]
 mgvavb :
G ai‡bi A¼¸‡jv‡Z u, v bv a‡i †¯ªv‡Zi AbyK‚‡j †bŠKvi †eM a
I cÖwZK‚‡j †bŠKvi †eM b aiv †ewk myweavRbK|
Avgiv Rvwb, u =
a + b
2
Ges v =
a  b
2
Let, downstream velocity = a
upstream velocity = b
1st condition :
30
b
+
44
a
= 10... (i)
Similarly,
2nd condition:
40
b
+
55
a
= 13... (ii)
(i)  4  (ii)  3 
[b †K ev` †`qvi Rb¨ (i) bs †K 4 Øviv I (ii) bs †K 3 Øviv ¸Y K‡i
we‡qvM Ki‡Z n‡e|]
120
b
+
176
a
= 40
120
b
+
165
a
= 39
() () ()
1
a
(176  165) = 1
AbyK‚‡j ev cÖwZK‚‡j mgq = `~iZ¡/AbyK‚‡j ev cÖwZK‚‡j †eM
AbyK‚‡j mgq + cÖwZK‚‡j
mgq = †gvU mgq
Ges mgq =
`~iZ¡
‡eM


1
a
 11 = 1  a = 11 (Ans.)
Putting the value of a in equation ... (i)
30
b
+
44
11
= 10

30
b
+ 4 = 10 
30
b
= 6  b =
30
6
= 5
we know, u =
a + b
2
=
11 + 5
2
=
16
2
= 8
 speed of man in still water is 8 km/hr
w¯’i cvwb‡Z †bŠKvi †eM = u
AbyK‚‡j †bŠKvi †eM, a = u + v
cÖwZK‚‡j †bŠKvi †eM, b = u  v
[†hvM K‡i] a + b = 2u
 u =
a + b
2
we‡qvM Ki‡j Avgiv cvB, v =
a  b
2
65. A man can row 5 km per hour in still water. If the river
is flowing at 1km per hour, it takes him 75 minutes to
row to a place and back. How far is the place?
[www.competoid.com]
a 3 km b 2.5 km c 4 km d none of these a
 mgvavb : Let, Distance = x
Man's speed u = 5 km/hr
Stream's speed, v = 1 km/hr;
A B
Time taken (A  B) + Time taken (B  A) = Total Time

x
u + v
+
x
u  v
=
75
60
=
5
4

x
5 + 1
+
x
5  1
=
5
4

x
6
+
x
4
=
5
4

2x + 3x
12
=
5
4

5x
12
=
5
4
 x =
5  12
4  5
 x = 3 km
66. A man can row 6 km/hr in still water. If the speed of
the current is 2 km/hr, it takes 3 hrs more in upstream
than in the downstream for the same distance. The
distance is– (GKRb e¨w³ w¯’i cvwb‡Z 6 km/hr †e‡M `uvo †e‡q
Pj‡Z cv‡i| †¯ªv‡Zi †eM 2 km//hr n‡j, †¯ªv‡Zi cÖwZK‚‡j GKwU
wbw`©ó `~iZ¡ cvwo w`‡Z Zvi †h mgq jv‡M, Zv †¯ªv‡Zi AbyK‚‡j H
GKB `~iZ¡ AwZµg Ki‡Z †h mgq Zvi Zzjbvq 3 N›Uv †ewk, H
`~iZ¡ KZ?) [Exam Taker AUST : Combined 3 Banks (Officer Cash-2018);
Combined 5 Banks (Asst. Engr. IT-2018); www.competoid.com]
a 30 km b 24 km c 20 km d 32 km b
 mgvavb : g‡b Kwi, wbw`©ó `~iZ¡ x km
w¯’i cvwb‡Z †bŠKvi †eM, u = 6 km/hr
†¯ªv‡Zi †eM, v = 2 km/hr
 †¯ªv‡Zi AbyK‚‡j †bŠKvi †eM, a = u + v = (6 + 2) km/hr
= 8 km/hr
†¯ªv‡Zi cÖwZK‚‡j †bŠKvi †eM, b = (6 – 2) km/hr = 4 km/hr
myZivs, x km AwZµg‡Y
†¯ªv‡Zi AbyK‚‡j cÖ‡qvRbxq mgq =
x
a
†¯ªv‡Zi cÖwZK‚‡j cÖ‡qvRbxq mgq =
x
b
cÖkœg‡Z, †¯ªv‡Zi cÖwZK‚‡j †h‡Z mgq
– †¯ªv‡Zi AbyK‚‡j †h‡Z mgq = 3 N›Uv

x
b
–
x
a
= 3  x



1
b

1
a
= 3  x =
3
1
b

1
a
 x =
3
1
4

1
8
km  x =
3
2  1
8
= 24  x = 24 km
67. A man can row upstream a distance of
2
3
km in 10
minutes and returns the same distance downstream in
5 minutes. Ratio of man’s speed in still water and that
of the stream will be? [www.competoid.com]
a 3 : 1 b 1 : 3 c 2 : 3 d 3 : 2 a
 mgvavb: Let, the speed of the man in still water and the speed
of stream is v1 and v2 respectively.
Speed of downstream = v1 + v2
Speed of upstream = v1 – v2
Speed =
Distance
Time
; Distance =
2
3
km
According to condition-1, v1 – v2 =
2
3
10
 v1 – v2 =
1
15
................... (i)
According to condition-2, v1 + v2 =
2
3
5
 v1 + v2 =
2
15
= 2 
1
15
 v1 + v2 = 2(v1 – v2) [From (i)]
 2v1 – 2v2 = v1 + v2  2v1 – v1 = v2 + 2v2
 v1 = 3v2 
v1
v2
=
3
1
 v1 : v2 = 3 : 1
68. A man can swim at the rate of 4 km/hr in still water. If
the speed of the water is 2 km/ hr, then the time taken
by him to swim 10 km upstream is: [www.competoid.com]
a 25 hrs b 35 hrs c 5 hrs d 4 hrs c
 mgvavb : Speed of man, u = 4 km/hr
” ” water, v = 2 km
Upstream speed = u  v = 4  2 = 2 km/hr
Time taken in upstream =
Distance
Upstream speed
=
10 km
2 km/hr
= 5 hr
69. A man rows a certain distance along the stream and
against the stream in 1 hour and 1.5 hours respectively.
If the velocity of the current is 3 km/hr. what is the
speed of a man in still water?
[Sonali Bank (Officer FF) – 19 + www.careerbless.com]
a 12 km/hr b 13 km/hr c 11 km/hr d 15 km/hr d
 mgvavb: Let, the speed of the man in still water = x km/hr
So, Downstream speed (x + 3) km/hr
Upstream speed (x  3) km/hr
According to the condition
(x +3)  1 = (x  3)  1.5
x + 3 = 1.5x  4.5
 1.5x  x = 7.5  0.5x = 7.5
 x =
7.5
0.5
= 15
70. A man rows to a place 35km in distance and back in 10
hours 30 mintues. He found that he can row 5 km with
the stream in the same time as he can row 4 km against
the stream. Find the rate of flow of the stream.
[www.competoid.com]
a 1 km/hr b 0.75km/hr c 1.33 km/hr d 1.5 km/hr b
 mgvavb: Let, Speed of man is still water = u
Speed of stream = v
2nd condition : Time taken downstream = Time taken upstream

5
u + v
=
4
u  v
 5u  5v = 4u + 4v  5u  4u = 4v + 5v
 u = 9v ..........(i)

1st Condition: Time taken downstream + Time taken
upstream = Total time

35
u + v
+
35
u  v
= 10
1
2
=
21
2 


10 h 30 min = 10
1
2
hr

5
u + v
+
5
u  v
=
3
2
[dividing both side by 7]

5
9v + v
+
5
9v  v
=
3
2
[From equation (i)]

5
10v
+
5
8v
=
3
2

5
5v
+
5
4v
= 3 
1
v
+
5
4v
= 3

4 + 5
4v
= 3 
9
4v
= 3  v =
9
4  3
=
3
4
= 0.75
71. A man rows to a place 40 km distant and comes back in
a total of 18 hours. He finds that he can row 5 km with
the stream in the same time as 4 km against the stream.
What is the speed of boat in still water? (GKRb †jvK `uvo
†e‡q 40 wK.wg. `~i‡Z¡i GK ¯’v‡b wM‡q Avevi wd‡i Avm‡Z †gvU 18
NÈv mgq jv‡M| †m jÿ¨ Kij †h, Zvi †¯ªv‡Zi AbyK‚‡j 5 wK.wg.
Ges †¯ªv‡Zi cÖwZK‚‡j 4 wK.wg. †h‡Z GKB mgq jv‡M| w¯’i cvwb‡Z
cvwb‡Z †bŠKvi †eM KZ?) [ExamTakerAUST :Combined8Banks(SO)-2018]
 mgvavb : Let, Speed of the boat = u km/hr
Speed of the stream = v km/hr
 down stream speed would be = (u + v) km/hr
 up stream speed would be = (u – v) km/hr
According to 1st
situation,
Time needed to travel 40 km at the speed (u + v) + time
needed to travel 40 km at the speed (u – v) = 18 hrs

40
u + v
+
40
u – v
= 18

1
u + v
+
1
u – v
=
18
40
................. (i)
According to 2nd
situation,
Time needed to travel 5 km at the speed (u + v) = time
needed to travel 4 km at the speed (u – v)

5
u + v
=
4
u – v
 5u – 5v = 4u + 4v  u = 9v ........... (ii)
from (i) and (ii) we get,
1
9v + v
+
1
9v – v
=
18
40

1
10v
+
1
8v
=
9
20

1
v




1
10
+
1
8
=
9
20
 v =



1
10
+
1
8

20
9
=
1
2
 u = 9v = 9 
1
2
= 4.5 km/hr
 speed of the boat = 4.5 km/hr
72. A man takes 3 hours 45 minutes to row a boat 22.5 km
downstream of a river and 2 hours 30 minutes to cover a
distance of 10 km up stream. Find the speed of the river
current in km/hr. (†¯ªv‡Zi AbyK‚‡j 22.5 km `uvo †e‡q †h‡Z
GKRb e¨w³i 3 N›Uv 45 wgwbU mgq jv‡M| †¯ªv‡Zi cÖwZK‚‡j 10
km †h‡Z 2 N›Uv 30 wgwbU mgq jv‡M| †¯ªv‡Zi †eM KZ?)
[Exam Taker AUST : P.K.B (A.P.-2019); Combined 3 Banks (Officer Cash-2018)]
 mgvavb : awi, †bŠKvi †eM = u
 †¯ªv‡Zi AbyK‚‡j †eM = u + v
†¯ªv‡Zi cÖwZK‚‡j †eM = u – v
†¯ªv‡Zi AbyK‚‡j 22.5 km †h‡Z mgq jv‡M 3 N›Uv 45 wgwbU =



3 +
45
60
N›Uv =
15
4
N›Uv

15
4
(u + v) = 22.5 [mgq × †eM = `~iZ¡]
 u + v =
4 × 22.5
15
u + v = 6 ......... (i)
†¯ªv‡Zi cÖwZK‚‡j 10 km †h‡Z mgq jv‡M 2 N›Uv 30 wgwbU =



2 +
30
60
N›Uv =
5
2
N›Uv

5
2
(u – v) = 10 [mgq × †eM = `~iZ¡]
u – v = 4 .........(ii)
(i) – (ii) 
u + v = 6
u – v = 4
(–) (+) (–)
2v = 2
 v = 1
73. A man takes 3 hours and 45 minutes to boat 15 km
with the current in a river and 2 hours 30 minutes to
cover a distance of 5 km against the current. Speed of
the boat in still water and speed of the current
respectively will be [www.competoid.com]
a 3 km/hr, 1 km/hr b 1 km/hr, 3 km/hr
c 2 km/hr, 4 km/hr d none of these a
 mgvavb : 3hr 45 min =



3 +
45
60
hr = 3 +
3
4
=
15
4
hr
2hr 30 min =



2 +
30
60
hr = 2 +
1
2
=
5
2
hr
Let, Speed of boat in still water = u
” ” current = v
Downstream speed (with the current), u + v =
15 km
15
4
hr
 u + v = 4 .......... (i)
Upstream speed (Against the current), u  v =
5
5
2
 u  v = 2 ......... (ii)
(i) + (ii)  u + v + u  v = 4 + 2  2u = 6
 u = 3
Putting the value of u in equation (i)
3 + v = 4
 v = 4  3 = 1
74. A man takes twice as long to row a distance against the
stream as to row the same distance in favor of the
stream. The ratio of the speed of the boat (in still
water) and the stream is : (GKRb e¨w³i †Kv‡bv wbw`©ó `~iZ¡
†¯ªv‡Zi AbyK‚‡j †h‡Z †h mgq jv‡M †¯ªv‡Zi cÖwZK‚‡j †h‡Z Zvi
wØ¸Y mgq jv‡M| †bŠKvi †eM I †¯ªv‡Zi †e‡Mi AbycvZ?)
[Exam Taker AUST : Janata Bank Ltd. (A.E.O Teller-2019); P.K.B.
(E.O. Cash-2019); www.examveda.com; www.indiabix.com]
a 2 : 1 b 3 : 2 c 4 : 3 d 3 : 1 d
 mgvavb : g‡b Kwi, †bŠKvi †eM u
Ges †¯ªv‡Zi †eM v
wbw`©ó `~iZ¡ S Ges †¯ªv‡Zi AbyK‚‡j cÖ‡qvRbxq mgq t n‡j,
S = (u + v)  t
Ges S = (u  v)  2t [‹ wØ¸Y mgq jv‡M]
GLb, (u + v)  t = (u  v)  2t

u + v
u  v
= 2 
u + v + u  v
u + v  u + v
=
2 + 1
2  1
[†hvRb-we‡qvRb K‡i]

2u
2v
=
3
1
 u : v = 3 : 1

75. A man went downstream for 28 km in a motor boat and
immediately returned. It took the man twice as long to
make the return trip. If the speed of the river flow were
twice as high, the trip downstream and back would take
672 minutes. Find the speed of the boat in still water
and the speed of the river flow. (GKRb †jvK †¯ªv‡Zi
AbyK‚‡j gUi-PvwjZ †bŠKvq †P‡c 28 wK.wg. †Mj, Gi ciciB wd‡i
Avmj| wd‡i Avmvi Rb¨ Av‡Mi Zzjbvq G‡Z wØ¸Y mgq jvMj|
hw` †¯ªv‡Zi †eM wØ¸Y n‡Zv Zvn‡j hvIqv-Avmvq †gvU 672 wgwbU
mgq jvMZ| w¯’i cvwb‡Z †bŠKvi †eM I †¯ªv‡Zi †eM KZ?)
[Exam Taker AUST : Joint Recruitment for 4 Banks (Office)-2019;
Combined 2 Banks (Officer)-2018]
 mgvavb :
1g †ÿ‡Î, †h‡nZz hvIqvi Zzjbvq wd‡i Avm‡Z †ewk mgq (wØ¸Y)
jv‡M| ZvB †¯ªv‡Zi AbyK‚‡j hvq Ges †¯ªv‡Zi cÖwZK‚‡j wd‡i Av‡m|
A
28 km
B
backstart
Let, Speed of Boat and stream u and v respectively
 Time taken in downstream (A  B) =
28
u + v
Again, Time taken in upstream (B  A) =
28
u  v
According to Question,
28
u  v
= 2 
28
u + v
 cÖwZK‚‡j mgq = 2  AbyK‚‡j mgq

1
u  v
=
2
u + v
Dfqc‡ÿ 28 ev` w`‡q
 2u  2v = u + v
 2u  u = v + 2v  u = 3v ... ... (i)
If the speed of stream becomes 2v then,
†¯ªv‡Zi †eM v = 2v n‡j
 AbyK‚‡j mgq + cÖwZK‚‡j mgq = †gvU mgq Avevi 672
wgwbU‡K NÈvq wb‡Z 60 Øviv fvM Kiv n‡q‡Q|
28
u + 2v
+
28
u  2v
=
672
60

28
3v + 2v
+
28
3v  2v
=
56
5
(i) bs †_‡K u = 3v

1
5v
+
1
v
=
2
5
Dfqcÿ‡K 28 Øviv fvM

1
v 


1
5
+ 1 =
2
5

1
v

5 + 1
5
=
2
5

1
v

6
5
=
2
5

1
v
=
2
5

5
6

1
v
=
1
3
 v = 3
Putting the value of 'v' in equation ... (i)
 u = 3v = 3  3 = 9
 speed of boat in still water is 9 km/hr
and speed of stream 3 km/hr
76. A motorboat, whose speed in 15 km/hr in still water
goes 30 km downstream and comes back in a total of 4
hours 30 minutes. The speed of the stream (in km/hr)
is: [www.examveda.com; www.indiabix.co]
a 4 km/hr b 5 km/hr c 6 km/hr d 10 km/hr b
 mgvavb : Speed of Motorboat, u = 15 km/hr
” ” Stream = v (let)
4 hr 30 min =



4 +
30
60
hr = 4 +
1
2
=
9
2
hr
Time taken in downstream + Time taken in upstream =
Total time

30
u + v
+
30
u  v
=
9
2

10
u + v
+
10
u  v
=
3
2
[dividing both side by 3]
 10





1
15 + v
+
1
15  v
=
3
2
 10





15  v + 15 + v
(15 + v) (15  v)
=
3
2
 10 
30
152
 v2 =
3
2
 10 
10
225  v2 =
1
2
 225  v2
= 200  v2
= 225  200
 v2
= 25  v = 25 = 5
77. A river is flowing at a speed of 5 km/h in a particular
direction. A man, who can swim at a speed of 20 km/h
in still water, starts swimming along the direction of
flow of the river from point A and reaches another
point B which is at a distance of 30 km from the
starting point A. On reaching point B, the man turns
back and starts swimming against the direction of flow
of the river and stops after reaching point A. This total
time taken by the man to complete his journey is?
(GKwU b`x NÈvq 5 wK‡jvwgUvi †e‡M GKwU wbw`©ó w`‡K e‡q P‡j|
GKRb e¨w³, whwb w¯’i cvwb‡Z 20 wK.wg./NÈv †e‡M mvZvi KvU‡Z
cv‡i, wZwb b`x cÖev‡ni w`‡K A we›`y n‡Z mvZvi KvU‡Z ïiæ K‡i
30 wK.wg. `~‡i B we›`y‡Z †cuŠQv‡jb| B we›`y‡Z †cuŠQv‡bv gvÎB
Avevi †¯ªvZ cÖev‡ni wecixZ w`‡K muvZvi †K‡U A we›`y‡Z G‡m
_vg‡jb| m¤ú~Y© hvÎv †kl Ki‡Z e¨w³wUi KZ mgq jv‡M|)
[Exam Taker AUST : Sonali Bank Ltd. (Officer-Cash)-2019; www.competoid.com]
 mgvavb :
†h‡nZz, e¨w³wU A n‡Z hvÎv ïiæ K‡i B †Z †cuŠQv‡bv gvÎB Avevi
wecixZ w`‡K hvÎv K‡i ZvB AbyK‚‡j I cÖwZK‚‡ji mgq‡K †hvM
Ki‡j †gvU mgq cvIqv hv‡e|
A 30 km B
Given data, Speed of the man, u = 20 km/hr
Speed of stream, v = 5 km/hr
 Distance = 30 km
down stream speed = u + v = 20 + 5 = 25 km/hr
upstream speed = u  v = 20  5 = 15 km/hr
time taken in downstream =
Distance
Downstream speed
=
30
25
=
6
5
hr
time taken in upstream =
Distance
upstream speed
=
30
15
= 2 hr
 Total time =



6
5
+ 2 hr =



6 + 10
5
hr =
16
5
hr
=
16
5
= 3
1
5
hr = 3 hr +
1
5
 60 min = 3 hr 12 min
78. A river is flowing with a steady speed of 4 km/h. One
rows his boat downstream in the river and then returns
by rowing upstream in the same river. When he
returns to the starting point, the total distance covered
by him is 42 km. If the return journey takes 2 h more
than his outward journey, then the speed of his rowing
in still water must be [www.competoid.com]
a 12 km/h b 10 km/h c 9 km/h d 8 km/h b
 mgvavb : Speed of rowing = u (Let)
Speed of current, v = 4 km/h
Total distance covered including
depart and return = 42
21 km
A B
 AB =
42
2
= 21
16
15
1
3
1
5

Departure journey  downstream
Return journy  upstream
Given condition:
Time taken in upstream  Time taken downstream = 2

21
u  v

21
u + v
= 2  21.





1
u  4

1
u + 4
= 2 [‹ v = 4]
 21





u + 4  u + 4
(u  4) (u + 4)
= 2  21 
8
u2
 42 = 2
 21 
4
u2
 16
= 1  u2
 16 = 84
 u2
= 84 + 16  u2
= 100
 u = 10 km/hr
79. A steamer goes downstream from one port to another
in 4 h. It covers the same distance upstream in 5 h. If
the speed of the stream is 2 km/h, then find the distance
between the two ports. [www.competoid.com]
a 50 km b 60 km c 70 km d 80 km d
 mgvavb : Let, The distance, AB = x
Speed of stream, v = 2 km/hr
For same distance we can write, X
A B
x = (speed  time)downstream = (speed  time)upstream,
 (u + v) 4 = (u  v)  5
 (u + 2).4 = (u  2)  5  4u + 8 = 5u  10
 5u  4u = 8 + 10  u = 18
Distance, x = (u + v)  4
 x = (18 + 2)  4  x = 20  4
 x = 80 km
80. A swimmer swims from a point A against a current for
5 minutes and then swims backwards in favour of the
current for next 5 minutes and comes to the point B. If
AB is 100 metres, the speed of the current (in km per
hour) is : [www.competoid.com]
a 0.4 b 0.2 c 1 d 0.6 d
 mgvavb : The swimmer swims from A to C in upstream,
C to A in downstream and A to B in downstream
B A C
x100m
So, The swimmer swims x distance in upsream for 5 min.
And, From C to B the swimmer swims (x + 100) distance
in downstream for next 5 min.
Upstream speed, u + v =
x
5
60
= 12x ..........(i)
Downstream speed, u  v =
x + 100
5
60
= 12 (x + 100)
 u  L = 12x + 1200 ...........(ii)
(i)  (ii)
u + v  u + v = 12x  12x  1200
 2L =  1200  v = 600 m/hr =
600
1000
km/hr = 0.6 km/hr
81. In still water, a boat can travel at 5 km/hr. It takes 1
hour to row to a place and come back. It the velocity of
the stream is 1 km/hr, how far is the place? (w¯’i cvwb‡Z
GKwU †bŠKvi †eM 5 km/hr| †bŠKvwU 1 N›Uvq GKwU ¯’v‡b †h‡q
Avevi wd‡i Av‡m| †¯ªv‡Zi †eM 1 km/hr n‡j, ¯’vbwUi `~iZ¡ KZ?)
[ExamTakerAUST :RupaliBankLtd. (S.O.-2019);SonaliBank(S.O.FF-2019)]
a 2.4 km b 3.5 km c 2.6 km d Noneofthese a
 mgvavb : awi, wbw`©ó ¯’v‡bi `~iZ¡ x km
†¯ªv‡Zi AbyK‚‡j †eM = 5 + 1 = 6 km/hr
†¯ªv‡Zi cÖwZK‚‡j †eM = 5 – 1 = 4 km/hr
cÖkœg‡Z,
x
6
+
x
4
= 1

2x + 3x
12
= 1  x =
12
5
= 2.4 km
82. Ishwar is rowing a boat. He takes half time in moving a
certain distance downstream than upstream. What is
the ratio of the rate of boat in still water to the rate of
current? [www.competoid.com]
a 2 : 1 b 5 : 1 c 7 : 1 d 3 : 1 d
 mgvavb : Let, Speed of the boat u and speed of current v
We have to find out u : v = ?
If x distance is covered then the time for downstream is =
x
u + v
and time for upstream is =
x
u  v
According the question,
x
u + v
=
1
2
x
u  v

u + v
u  v
= 2  u + v = 2u  2v
 u  2u =  2v  v   u =  3v 
u
v
=
3
1
 u : v = 3 : 1
83. On a river, Q is the midpoint between two points P and
R on the same bank of the river. A boat can go from P
to Q and back in 12 hours, and from P to R in 16 hours
40 minutes. How long would it take to go from R to P?
(GKwU b`x‡Z, P Ges R Gi ga¨we›`y Q| GKwU †bŠKv P †_‡K Q
G †h‡q wd‡i Avm‡Z 12 NÈv mgq jv‡M Ges P †_‡K R G †h‡Z
16 NÈv 40 wgwbU mgq jv‡M| R †_‡K P †Z †h‡Z KZ mgq
jvM‡e?) [Exam Taker AUST : Rupali Bank Ltd. (S.O.-2019);
www.examveda.com; www.competoid.com]
a 3
3
7
hr b 5 hr c 6
2
3
hr d 7
1
3
hr d
 mgvavb : P R
Q
Here, Q is midpoint so, PQ = QR
and PQ =
1
2
PR
 To travel from P to R the boat takes 16 hr 40 min or



16 +
40
60
or



16 +
2
3
or
50
3
hours
As, PQ =
1
2
PR so To travel from P to Q the boat takes



50
3
 2 =
25
3
hours
Now, To travel from Q to P the boat takes



12 –
25
3
=
11
3
hours
As, PR = 2 PQ so, To travel from R to P the boat takes



2 ×
11
3
or
22
3
hours
= 7 hours +
1
3
hours = 7 hours +



1
3
× 60 minutes
= 7 hours 20 min
Alternative Solution:
P Q R
x
2x
Clearly here, forward speed of the boat and backward
speed of the boat are not same.
Let,
Forward speed of boat = a [P → Q or P → R or Q → R]
Backward speed of boat = b [ Q → P or R → P or R → Q]
And distance PQ = x, so distance PR = 2x
According to 2nd
condition,
2x
a
= 16
2
3
or
50
3

x
a
=
25
3
.................... (i)

According to 1st
condition,
x
a
+
x
b
= 12 
x
b
= 12 –
x
a
= 12 –
25
3
=
11
3
 Time to go from R to P =
2x
b
= 2 
x
b
= 2 
11
3 



x
b
=
11
3
=
22
3
= 7 hours 20 minutes.
84. Rahima can row 16 km/hr in still water. It takes her thrice
as long to row up as to row down the river. Find the
difference between her speed in still water and that of
the stream. (w¯’i cvwb‡Z iwngv 16 km/hr †e‡M `uvo †e‡q †h‡Z
cv‡i| †¯ªv‡Zi AbyK‚‡j †h‡Z †h mgq jv‡M, H GKB `~iZ¡ †¯ªv‡Zi
cÖwZK‚‡j †h‡Z Zvi wZb¸Y mgq jv‡M| w¯’i cvwb‡Z iwngvi `vuo
†U‡b Pjvi †eM I †¯ªv‡Zi †e‡Mi cv_©K¨ wbY©q Kiæb|)[Exam Taker
AUST : Combined 8 Banks (S.O.-2018)]
 mgvavb : w¯’i cvwb‡Z †eM, u = 16 km/hr
awi, †¯ªv‡Zi †eM = v
 †¯ªv‡Zi AbyK‚‡j †eM = u + v = 16 + v
†¯ªv‡Zi cÖwZK‚‡j †eM = 16 – v
†¯ªv‡Zi cÖwZK‚‡j †h‡Z mgq = 3 × †¯ªv‡Zi AbyK‚‡j †h‡Z mgq
 †¯ªv‡Zi cÖwZK‚‡j †eM =
1
3
× †¯ªv‡Zi AbyK‚‡j †eM
†¯ªv‡Zi AbyK‚‡j x `~iZ¡ AwZµg‡Y mgq =
x
a
†¯ªv‡Zi cÖwZK‚‡j x `~iZ¡ AwZµg‡Y mgq =
x
b

x
b
= 3
x
a
 b =
1
3
a
 16 – v =
1
3
(16 + v)  16 + v = 48 – 3v
v = 32  v = 8 km/hr
 wb‡Y©q cv_©K¨ = (16 – 8) = 8 km/hr
85. Sameer can row a certain distance downstream in 24 h
and can come back covering the same distance in 36 h.
If the stream flows at the rate of 12 km/h, find the
speed of Sameer in still water. [www.competoid.com]
a 30 km/h b 15 km/h c 40 km/h d 60 km/h d
 mgvavb : Let, The speed of sameer is u
The speed of the boat v = 12 km/h
We know, Speed =
distance
time
If the covered distance is x,
Then according to the question
For downstream, u + v =
x
24
 x = 24 (u + v)
For upstream, u  v =
x
36
 x = 36 (u  v)
 24 (u + v) = 36 (u  v)
 2 (u + v) = 3 (u  v)  2u + 2v = 3u  3v
 3u  2u = 3v + 2v  u = 5v = 5  12 km/h = 60 km/h
86. The effective speed of a boat travelling downstream is
20 kmph, whereas it is 10 kmph upstream. What is the
speed of the stream current in kmph? (†¯ªv‡Zi AbyK‚‡j
†bŠKvi †eM 20 kmph Ges †¯ªv‡Zi cÖwZK‚‡j †bŠKvi †eM 10 kmph|
†¯ªv‡Zi †eM KZ?) [Exam Taker IBA : Dutch-Bangla Bank Ltd. (PO-2015)]
a 3 b 4 c 5 d None c
 mgvavb : awi, †bŠKvi †eM = x kmph
Ges †¯ªv‡Zi †eM = y kmph
cÖkœg‡Z, x + y = 20 .......(i)
Ges x  y = 10
(–) K‡i, 2y = 10
 y =
10
2
= 5
87. The ratio of speed of a motorboat to that of the current
of water is 36 : 5. The boat goes along with the current
in 5 hours 10 minutes. It will come back in :
[www.competoid.com]
a 5 hr 50 min b 6 hr
c 6 hr 50 min d 12 hr 10min c
 mgvavb : The ratio of the speed of motorboat and current u :
v = 36 : 5
For downstream, time t1 = 5 hours 10 min
= 5  60 + 10 = 310 min
We know, time =
distance
velocity
If the covered distance is x then
For downstream, Time t1 =
x
u + v
or, x = t1 (u + v)
For upstream, Time t2 =
x
u  v
or, x = t2 (u  v)
 t1 (u + v) = t2 (u  v)

u + v
u  v
=
t2
t1

u + v + u  v
u + v  u + v
=
t2 + t1
t2  t1
[Addition Subtruction]

2u
2v
=
t2 + t1
t2  t1

36
5
=
t2 + 310
t2  310
 36 t2  310  36 = 5t2 + 5  310
 36t2  5t2 = 36  310 + 5  310
 31t2 = 41  310
 t2 =
41  310
31
= 410 min =
410
60
hour = 6 hour 50 min
88. The ratio of the speed of boat in still water to the speed
of stream is 16 : 5. A boat goes 16.5 km in 45 minute
upstream, find the time taken by boat to cover the distance
of 17.5 km downstream. (w¯’i cvwb‡Z †bŠKvi †eM I †¯ªv‡Zi
†e‡Mi AbycvZ 16 : 5| GKwU †bŠKv †¯ªv‡Zi cÖwZK‚‡j 45 wgwb‡U
16.5 km hvq| †¯ªv‡Zi AbyK‚‡j 17.5 km †h‡Z cÖ‡qvRbxq mgq
KZ?) [Exam Taker AUST : Combined 2 Banks (Officer-2018)]
a 30 minutes b 25 minutes
c 50 minutes d 45 minutes b
 mgvavb : awi, †bŠKvi †eM 16x I †¯ªv‡Zi †eM 5x|
 †¯ªv‡Zi AbyK‚‡j †eM, a = 16x + 5x = 21x
†¯ªv‡Zi cÖwZK‚‡j †eM, b = 16x – 5x = 11x
†¯ªv‡Zi cÖwZK‚‡j AwZµvšÍ `~iZ¡ = †¯ªv‡Zi cÖwZK‚‡j †eM × mgq
= b ×
45
60
km =
3
4
b km
cÖkœg‡Z,
3
4
b = 16.5  b =
4 × 16.5
3
 b = 22  11x = 22  x = 2
 †¯ªv‡Zi AbyK‚‡j †eM, a = 21 × 2 km/hr
 a = 42 km/hr
†¯ªv‡Zi AbyK‚‡j 17.5 km †h‡Z mgq =
`~iZ¡
†eM
=
17.5
42
hr =
17.5
42
× 60 minutes = 25 minutes
89. The speed of a boat in still water is 15 km/hr and the
rate of current is 3 km/hr. The distance travelled
downstream in 12 minutes is
a 1.2 km b 3.6 km c 1.8 km d 2.4 km b

 mgvavb: Given, Speed of boat u = 15 km/hr
Speed of current v = 3 km/hr
in downstream, speed = u + v = 15 + 3 = 18 km/hr
Given time = 12 minutes =
12
60
hour
Distance = Speed × times = 18 ×
12
60
= 3.6 km
90. The speed of the boat in still water is 24 kmph and the
speed of the stream is 4 km/hr. The time taken by the
boat to travel from A to B downstream is 36 minutes
less than the time taken by the same boat to travel
from B to C upstream. If the distance between A and B
is 4 km more than the distance between B and C, what
is the distance between A and B? (w¯’i cvwb‡Z †bŠKvi †eM
24 wK.wg./NÈv Ges †¯ªv‡Zi †eM 4 wK.wg./NÈv| †¯ªv‡Zi cÖwZK‚‡j
†bŠKvwUi B n‡Z C-†Z †h‡Z †h mgq jv‡M, †¯ªv‡Zi AbyK‚‡j A n‡Z
B †Z †h‡Z Zvi †_‡K 36 wgwbU mgq Kg jv‡M| hw` A I B Gi
`~iZ¡, B I C Gi `~iZ¡ A‡cÿv 4 wK.wg. †ewk nq, Zvn‡j A I B
Gi `~iZ¡ KZ?) [Exam Taker AUST : Sonali Bank Ltd. (SO)-2019]
 mgvavb : Let, Distance between B and C, BC = x km
 Distance between A and B, AB = (x + 4) km
Here, Speed of the boat, u = 24 km/hr
Speed of the stream, v = 4 km/hr
So, Needed time to travel BC distance
= Needed time to travel x km at the speed
(u – v) upstream.
=
x
u – v
=
x
24 – 4
=
x
20
Needed time to travel AB distance
= Needed time to travel (x + 4)km distance
at the speed (u + v) downstream
=
x + 4
u + v
=
x + 4
24 + 4
=
x + 4
28
Accordingly,
x
20
–
x + 4
28
=
36
60

7x – 5x – 20
140
=
3
5
 2x – 20 = 84  2x = 104  x = 52
 AB distance = (x + 4) = (52 + 4) = 56 km
91. The time taken by a man to travel 36 miles downstream
is 90 min less than to go the same distance upstream.
The speed of the man in still water is 10 mph. Find the
speed of the stream. (†¯ªv‡Zi AbyK‚‡j 36 gvBj c_ AwZµg
Ki‡Z †Kvb e¨w³i †h mgq jv‡M Zv †¯ªv‡Zi cÖwZK‚‡j H `~iZ¡
AwZµg Ki‡Z cÖ‡qvRbxq mg‡qi Zzjbvq 90 wgwbU Kg| w¯’i cvwb‡Z
H e¨w³i †eM 10 mph| †¯ªv‡Zi †eM wbY©q Kiæb|)
[Exam Taker AUST : P.K.B. (E.O. General-2019)]
a 2 mph b 2.5 mph c 3.5 mphd 5 mph a
 mgvavb : awi, †¯ªv‡Zi †eM = V mph
 †¯ªv‡Zi AbyK‚‡j †eM = (10 + V) mph
” cÖwZK‚‡j ” = (10 – V) mph
†¯ªv‡Zi AbyK‚‡j †h‡Z mgq =
36
10 + V
h
” cÖwZK‚‡j ” ” =
36
10 – V
h
90 wgwbU =
90
60
h =
3
2
h
cÖkœg‡Z,
36
10 – V
–
36
10 + V
=
3
2



1
10 – V
–
1
10 + V
=
3
2
 36 
10 + V – 10 + V
(10 – V) (10 + V)
=
3
2

36  2V
102
– V2 =
3
2
 144V = 300 – 3V2
 3V2
+ 144V – 300 = 0
 V2
+ 48V – 100 = 0  V2
+ 50V – 2V – 100 = 0
 V(V + 50) – 2(V + 50) = 0
 (V + 50) (V – 2) = 0
V  – 50 |  V = 2
 †¯ªv‡Zi †eM 2 mph
92. The water in a river is flowing at the rate of 4 km/hr. If
the width and depth of the river is 8m and 4m
respectively, then how much water will enter the sea in
15 minutes. [www.competoid.com]
a 60000 m3
b 18000 m3
c 28800 m3
d 32000 m3
d
 mgvavb : The speed of current is 4 km/hr
=
4  1000
60
m/min
=
200
3
m/min
We know, distance = speed  time
 The length of the entering water in
The river in 15 minutes is, L =
200
3
 15 m = 1000 m
Given, width of the river, B = 8 m
Depth of the river, D = 4 m
 The volume of water = LBD = 1000  8  4 = 32000 m3
93. Two boats A and B start towards each other from two
places, 108 km apart. Speed of the boat A and B in still
water are 12 km/ hr and 15 km/ hr respectively. If A
proceeds down and B up the stream, they will meet
after. [www.competoid.com]
a 4.5 hours b 4 hours c 5.4 hours d 6 hours b
 mgvavb : Let, The speed of current V
The speed of boat A, UA = 12 km/hr
The speed of boat B, UB = 15 km/hr
Given distance x = 108 km
We know, Distance = Velocity  time
If they meet after time t then
Distance covered by Boat – A in down-stream, x1 = (uA + v) t
Distance covered by Boat – B in upstream, x2 = (uB – v) t
Now, x = x1 + x2
 108 = (UA + V) t + (UB  V) t
 108 = UA t + V t + UB t  V t
 108 = (UA + UB) t
 t =
108
UA + UB
=
108
12 + 15
=
108
27
km/hr
= 4 hours
94. Two boats on opposite banks of a river start moving
towards each other. They first pass each other 1,400
meters from one bank. They each continue to the
opposite bank, immediately turn around and start back
to the other bank. When they pass each other a second
time, they are 600 meters from the other bank. We
assume that each boat travels at a constant speed all
along the journey. Find the wih of the river? (`yBwU †bŠKv
wecixZ `yBK‚j n‡Z G‡K Ac‡ii w`‡K hvÎv ïiæ K‡i| Zviv hLb
GK K‚j n‡Z 1,400 wgUvi `~‡i ZLb cÖ_gevi ci¯úi‡K AwZµg
K‡i| Gfv‡e Zviv wecixZK‚‡ji w`‡K hvB‡Z _v‡K Ges K‚‡j
†cuŠQv‡bv gvÎB Avevi Av‡Mi K‚‡j wdi‡Z ïiæ K‡i| Gevi Zviv
hLb Ab¨ K‚j n‡Z 600 wgUvi `~‡i _v‡K ZLb wØZxqevi ci¯úi‡K
AwZµg K‡i| Avgiv a‡i wbw”Q †h, mgMÖ c‡_ cÖwZwU †bŠKv aªæe
(constant) MwZ‡Z P‡j| b`xi cÖ¯’ KZ?)
[Exam Taker AUST : Combined 3 Banks (SO)-2018]
 mgvavb : Let, the width of the river is x m speed of the boats
is v1 and v2.
They meet t1 seconds after leaving their corresponding banks.

19. boats and streams c&add [www.itmona.com]

19. boats and streams c&add [www.itmona.com]

Recommended

Recommended

More Related Content

What's hot

What's hot (16)

Similar to 19. boats and streams c&add [www.itmona.com]

Similar to 19. boats and streams c&add [www.itmona.com] (20)

More from Itmona

More from Itmona (20)

Recently uploaded

Recently uploaded (20)