Q1.  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 /hr) is :

(a) 0.4            (b) 0.2           (c) 1                (d) 0.6          (e) None

Q2. 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?

(a) 1 hour 358/13 minutes
(b) 1 hour 319/13 minutes
(c) 1 hour 332/13 minutes
(d) 1 hour 345/13 minutes
(e) None

Q3. There are 36 tickets numbered from 1 to 36. If two tickets are drawn at random without replacement one by one than find the probability that both tickets have a number which is multiple of 5?

(a) 1/40         (b) 1/48       (c)1/28         (d)1/30        (e) None

Q4. A is twice efficient as B. A and B together do the same work in as much time as C and D can do together. If the ratio of the number of alone working days of C to D is 2:3 and if B worked 16 days more than C then no of days which A worked alone?

(a) 18 Days    (b) 20 Days   (C) 30 Days    (d) 36 Days     (e) Cannot be determined

Q5. A can do a piece of work in 30 days, B can do in 45 days and C can do same work alone in 60 days. If on the first day A worked alone and on the second day A and B worked together and on the third day A and C worked together. If they repeat the cycle as follows then in how many days total work can be completed?

(a) 21 Days   (b) 21 7/8 Days       (c) 21 5/6 Days      (d) 21 4/9 Days      (e) None

Q6. A can do a piece of work in 21days. B is 50% more efficient than A. C is twice efficient than B. A started the work alone and worked for some days and left the work then B and C joined together and completed the work in 2 days. Then how many days does A worked alone?

(a) 7 Days
(b) 12 Days
(c) 14 Days
(d) 21 Days
(e) None

Q7. A car started from Indore to Bhopal at a certain speed. The Car missed an accident at 40Kms away from Indore, then the driver decided to reduce Car speed to 4/5 of the original speed. Due to this, he reached Bhopal by a late of 1hr 15min.Suppose if he missed an accident at 80Km away from Indore and from then he maintained 4/5 of original speed then he would reach Bhopal by a late of 1hour. Then what is the original speed of the Car?

(a) 20 km/hr    (b) 40 km/hr    (c) 60 km/hr     (d) 80 km/hr   (e) Cannot be determined

Q8. Two places A and B are at a certain distance. Ramu started from A towards B at a speed of 40 kmph. After 2 hours Raju started from B towards A at a speed of 60 kmph. If they meet at a place C then ratio of ratio of time taken by Raju to Ramu to reach Place C is 2:3. Then what is the distance between A and B?

(a)300 Km    (b) 400 Km   (c) 480 Km  (d) 600 Km   (e)Cannot be determined

Q9. Two Cars started at same time, same place and towards same direction. First Car goes at uniform speed of 12Km/hr. Second Car goes at speed of 4 Km/hr in first hour and increases it speed by 1 Km/hr for every hour. Then what is the distance travelled by car B when the both the Cars meet for the first time?

(a) 196 Km   (b) 198 Km   (c) 200 Km   (d) 204 Km   (e) None

Q10. Two Cars started at same time, same place and towards same direction. First Car goes at a speed of 5km/hr in first and 7 Km/hr in second hour and repeats the cycle over the entire journey. Similarly, second Car goes at 4km/hr in first hour and 9km/hr in second hour and repeats the cycle over the entire journey. Then these two Cars for the first time for after how many hours of journey?

(a) 3 Hours   (b) 6 Hours   (c) 9 Hours   (d) 12 Hours            (e) None

Solution:

Q1. SOL:   D

OA distance covered by upstream
Speed in 5 minutes = 5/60 (x - y)
OB = Distance covered by downstream speed in 5minutes
= 5/60 (x + y)
ATQ
OB - OA = AB
5/60 (x + y) – 5/60 (x - y) = 100 / 1000 km
5/60 [x + y – x + y] = 100/ 1000 km
On solving; y = 0.6 km/hr
Thus, Speed of the current = 0.6 km/hr.

Q2. SOL:   D
ATQ
Speed of current ‘y’ = 4 km/h
Distance = 12 km
Speed in upstream= (x + y) km/hr.
Here, ‘x’ is speed of boat in still water, thus, Speed = Distance/Time
x – 4 = 12/5
5x – 20 = 12
5x = - 20 = 12
5x = 32
x = 64 km/hr
Speed in downstream = (x + y) = 6.4 + 4 = 10.4 km/h
Thus, (x + y) = 6.4 + 4 = 10.4 km/h
Time = Distance/ Speed
time = 15/10.4 = 150/104= 1 hour 26 minute (approx.)

Q3. SOL:   D

(7/36) * (6 / 35) = 1/30

Q4. SOL:   A

Assume working days
A = x, B = 2x, C = 2y, D = 3y
1/x+1/2x = 1/2y+1/3y
And 2x-2y = 16
Solving we get x = 18 days
.

Q5. SOL:  C

First day = 1/30
Second day = 1/30+1/45
Third day = 1/30+1/60
3 days’ work = 3/30+1/45+1/60 = 25/180
3*7 = 21 days’ work = 175/180
Now 1/36 work is left which can be completed by A alone
1/36*30 = 5/6
21+5/6 = 21 5/6 Days
x = 100

Q6. SOL:   B

A = 21 B = 14 C =7
x/21+2*(1/14+1/7) = 1
x = 12

Q7. SOL:   B
(x+40/s) + 5/4 = 40/s + 5x/4s — — — 1
(x+40/s) + 1 = 80/s + 5(x-40)/4s — — — 2
s = 40

Q8.SOL:   A

40t 1 +60t 2 = d
T 1 /t 2 = 3/2
T 1 – t 2 = 2
Solving d = 480 Km

Q9. SOL:   D

12*x = x/2(2*4+(x-1)*1)
X= 17
D = 17*12 =204

Q10. SOL:   A

First car : 5,12,17
Second Car: 4,3,17

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