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বিষয় ভিত্তিক প্রস্তুতি : গণিত এবং গানিতিক যুক্তি ও দক্ষতা
By apple
#1912
MQ: 01
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?

Solution:

(x+40/s) + 5/4 = 40/s + 5x/4s — — — 1

(x+40/s) + 1 = 80/s + 5(x-40)/4s — — — 2

s = 40

Alternative solution :

Let,
Original Speed =5x km/hr
Reduced Speed =4x km/hr
V=D/T
=> T=D/V
=> (80-40)/4x -(80-40)/5x =15/60
=> 40/4x -40/5x =1/4
=> 40/20x =1/4
=> X=8
Original Speed =5x km/hr
=40 km/hr
Reduced Speed =32 km/hr

MQ: 02
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?

Solution:

40t 1 +60t 2 = d

T 1 /t 2 = 3/2

T 1 – t 2 = 2 Solving d = 480 Km

MQ: 03
Ramu started from A towards B at a speed of 20Km/hr and Raju started from B towards A. They crossed each other after one hour. Raju reached his destination 5/6 hour earlier than Ramu reached his destination.Then what is the distance between A and B?

Solution:

1 = x/20 = D-x/s 2

5/6 = D(1/20 – 1/s 2 )

D =50

MQ: 04
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 traveled by car B when the both the Cars meet for the first time?

Solution:

12*x = x/2(2*4+(x-1)*1)

X= 17

D = 17*12 =204

MQ: 05
A man traveled 100 km by Bike in 2 hours. He then traveled in Bus for 8 hrs and then Train in 9 hrs. Ratio of Speeds of Bus to Train is 4:5. If speed of train is 4/5 of Bike speed then the entire journey covered by him in Km is?

Solution:

Speed of train = 50

Bus = 32

Train = 40

Distance = 100+32*8+40*9 = 716

MQ: 06
A Tank is filled with the mixture of Milk and Water in the ratio of 3:2 up to 2/5 of its capacity. The tank has two inlet pipes i.e., Milk and Water inlets. Milk and Water pipe can fill an empty tank in 12 and 18 hours respectively. Now both pipes are opened simultaneously and closed after the Tank is completely filled, then what is the ratio of Milk and Water in the full Tank if it can accommodate 250Litre?

Solution:

Initial Milk = 2/5*250*3/5 = 60 L

Water = 2/5*250*2/5 = 40 L

Rest of Tank =150 L

Pipes are opened then can fill rest of tank in 108/25 hours

H/W = constant

then (108/25)/12/x = (108/25)/18(150-x)

X = 90 = Milk, Water = 60

Final ratio = 3:2

MQ: 07
An Inlet pipe can fill a tank in 5 hours and an Outlet pipe can empty 4/7 of the same Tank in 4 hours. In the first hour only Inlet pipe is opened and in the second hour, only outlet pipe is opened. They have opened alternately every hour until the Tank is filled. Then in how many hours does the Tank gets filled?

Solution:

2 hours work = 1/5-1/7 = 2/35

34 hours work = 34/35

remaining work = 1/35

Now its inlet pipe turn = 1/35*5 = 1/7

= 34 hours + 60/7 min

MQ: 08
A Tank is already filled up to X% of its capacity. An Inlet pipe can fill Full Tank in 30 minutes and an Outlet pipe can empty Full Tank in 20 Minutes. Now both pipes are opened then the Tank is emptied in 24 Minutes. Then initially up to what % of its capacity is Tank filled?

Solution:

1/30 – 1/20 = -1/60

Full Tank can be emptied 60 Minutes

In 24 minutes 40% of Tank can be emptied.

MQ: 09
Two Inlet Pipes A and B together can fill a Tank in ‘X’ minutes. If A and B take 81 minutes and 49 minutes more than ‘X’ minutes respectively, to fill the Tank. Then They can fill the 5/7 of that Tank in how many minutes?

Solution:

Time taken by two pipes to fill full Tank is = √ab min = 63 min

5/7 Tank = 63*5/7 = 45 min

MQ: 10
Pipe A can fill a Tank in 18 Hours, Pipe B can empty a Tank in 12 Hours, Pipe C can fill Tank in 6 Hours. The Tank is already filled up to 1/6 of its capacity. Now Pipe A is opened in the First Hour alone, Pipe B is opened in the Second Hour alone and Pipe C is opened in the Third Hour alone. This cycle is repeated until the Tank gets filled. Then in How many Hours does the rest of Tank gets filled?

Solution:

In First Hour Tank filled = 1/6+1/18

Second Hour = 1/6+1/18-1/12

Third Hour = 1/6+1/18-1/12+1/6 = 11/36 is filled

25/36 is left

From then 3 hours work = 1/18-1/12+1/6 = 5/36

5*3 Hours = 5*5/36 = 25/36

Total = 5*3+3 = 18 Hours

Collected
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