# Quantitative Aptitude Quiz for SBI / IBPS RRB 2018

**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.**

**T**wo 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**

**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.

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**

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**

x/21+2*(1/14+1/7) = 1

x = 12

**Q7. SOL: B**

(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

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

X= 17

D = 17*12 =204

**Q10. SOL: A**

First car : 5,12,17

Second Car: 4,3,17

Second Car: 4,3,17

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