Practice Set for WBCS/Rail NTPC/Group D/WBP/PSC/SSC Exams | Set - 02

Hello Everyone,

welcome to www.jobguidee.com , A guide for all competitive exams. Today we are providing a very special Math Practice Set - 02 for WBCS/Rail NTPC/Group D/SSC/PSC and many more exams.

First of all see the questions with options given below.Then try to solve the 10 problems in less than 20 mins. Then match it with the answer key. After that for any doubt you can have the solutions given below also. Friends, Math is an integral part of each and every State or Central competitive exams so For that You have to practice hard and also manage the time limit. So just go to below and solve the questions very carefully. You can comment your score out of 10×2 = 20 marks ↡

1. A car covers a distance of 715 km at a constant speed.If the speed of the car would have been 10 km/hr more, then it would have taken 2 hours less to cover the same distance. What is the original speed of the car ?

একটি গাড়ি 715 কিমি দূরত্ব নির্দিষ্ট বেগে যায়। যদি গাড়িটির গতিবেগ 10 কিমি/ঘন্টা বেশি হতো, তবে ওই দূরত্ব যেতে 2 ঘন্টা সময় কম লাগতো। গাড়িটির প্রকৃত গতিবেগ কত ?

[A] 45 km/hr

[B] 50 km/hr

[C] 55 km/hr

[D] 65 km/hr

Q.2. A plane left 30 minutes later than the scheduled time and in order to reach the destination 1500 km away in time, it had to increase the speed by 250 km/hr from the usual speed. Its usual Speed is :

একটি প্লেন নির্ধারিত সময়ের 30 মিনিট পরে যাত্রা শুরু করে, 1500 কিমি দূরত্ব সঠিক সময়ে পৌঁছাতে তাকে গতিবেগ 250 কিমি/ঘন্টা বাড়াতে হয়, তবে প্রকৃত গতিবেগ কত ?

[A] 720 km/hr

[B] 730 km/hr

[C] 740 km/hr

[D] 750 km/hr

Q.3.A train leaves Meerut at 6 A.M and reaches Delhi at 10 A.M. Another train leaves Delhi at 8 A.M. and reaches Meerut at 11:30 A.M .At what time do the two trains cross each other ?

সকাল 6 টায় একটি ট্রেন মিরাট থেকে রওনা দেয়, বেলা 10 টায় দিল্লি পৌঁছায়। আর একটি ট্রেন দিল্লি থেকে সকাল 8 টায় রওনা দিয়ে মিরাট এ বেলা 11:30 টায় পৌঁছায়। কখন দুটো ট্রেন পরস্পর কে অতিক্রম করবে ?

[A] 9:26 AM

[B] 9 AM

[C] 8:36 AM

[D] 8:56 AM

Q.4. A car covers four successive 3 km stretches at 10 km/hr , 20 km/hr , 30 km/hr and 60 km/hr respectively. Its average speed over the distance is :

একটি গাড়ি 4 টে পর পর 3 কিমি দূরত্ব যায় 10 কিমি/ঘণ্টা, 20 কিমি/ঘন্টা, 30 কিমি/ঘন্টা এবং 60 কিমি/ঘন্টা গতিবেগে যথাক্রমে। গড় বেগ কত ?

[A] 10 km/hr

[B] 20 km/hr

[C] 25 km/hr

[D] 30 km/hr

Q.5. A train passes two persons walking in the same direction in which the train is going. These persons are walking at the rate of 3 km/hr and 5 km/hr respectively and the train passes them completely in 10 seconds and 11 seconds respectively. The speed of the train is :

একটি ট্রেন একই দিকে দুটো চলমান ব্যক্তিকে অতিক্রম করে। একজন ব্যক্তির গতিবেগ 3 কিমি/ঘন্টা, অন্যজনের 5 কিমি/ঘন্টা। যদি ট্রেনটি তাদের কে যথাক্রমে 10 সেকেন্ড এবং 11 সেকেন্ড এ অতিক্রম করে যথাক্রমে, তবে ট্রেনটির গতিবেগ কত ?

[A] 24 km/hr

[B] 25 km/hr

[C] 27 km/hr

[D] 28 km/hr

Q.6. A train overtakes two persons walking in the same direction in which the train is going. These persons are walking at the rate of 2 km/hr and 4 km/hr and the train passes them completely in 9 seconds and 10 seconds respectively. The length of the train is -

একটি ট্রেন একই দিকে চলমান দুই ব্যক্তিকে অতিক্রম করে। এক ব্যক্তির গতিবেগ 2 কিমি/ঘন্টা, অন্যজনের 4 কিমি/ঘন্টা । যদি ট্রেনটি তাদের কে 9 সেকেন্ড এবং 10 সেকেন্ড সময়ে সম্পূর্ণ অতিক্রম করে, তবে ট্রেনটির দৈঘ্য কত ?

[A] 72 m

[B] 54 m

[C] 50 m

[D] 45 m

Q.7. In a stream running at 2 km/hr, a motor boat goes 10 km upstream and back again to the starting point in 55 minutes . Find the speed of the motorboat in still water.

2 কিমি/ঘন্টা স্রোতের বেগে একটি নৌকা 10 কিমি প্রতিকূলে যায় এবং আবার আগের জায়গায় ফিরে আসে 55 মিনিটে। স্থির জলে নৌকাটির গতিবেগ কত ?

[A] 20 km/hr

[B] 22 km/hr

[C] 24 km/hr

[D] 16 km/hr

Q.8. The speed of a boat in still water is 15 km/hr, it can go 30 km upstream and return downstream to the original point in 4 hrs 30 min, The speed of the stream is :

স্থির জলে একটি নৌকার গতিবেগ 15 কিমি/ঘন্টা। যদি এটি 30 কিমি প্রতিকূলে যায় এবং আগের জায়গায় ফিরে আসে 4 ঘন্টা 30 মিনিটে, তবে স্রোতের বেগ কত ?

[A] 5 km/hr

[B] 8 km/hr

[C] 10 km/hr

[D] 15 km/hr

Q.9. A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup ?

তরলে ভর্তি একটি পাত্রে, 3 ভাগ জল এবং 5 ভাগ সিরাপ। কত টা মিশ্রণ তুলে নিয়ে সেই জায়গায় জল মেশাতে হবে যাতে ওই মিশ্রনে অর্ধেক জল এবং অর্ধেক সিরাপ হয় ?

[A] 1/3

[B] 1/4

[C] 1/5

[D] 1/7

Q.10. A can contains a mixture of two liquids A and B in the ratio 7:5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7:9. How many litres of liquid A was there in the can initially ?

7:5 অনুপাতে দুটি তরল A এবং B দিয়ে একটি পাত্র পূর্ণ রয়েছে। যখন 9 লিটার মিশ্রণ তুলে নেওয়া হয় এবং B দিয়ে পাত্রটি পূর্ণ করা হয়,তখন A এবং B এর অনুপাত হয় 7:9 । শুরুতে ঐ পাত্রে কত লিটার A তরল ছিল ?

[A] 10 litres

[B] 20 littes

[C] 21 litres

[D] 25 litres

Answer Key :

1) C

2) D

3) D

4) B

5) B

6) C

7) B

8) A

9) C

10) C

Set Solutions :

1) Solution :

Let the constant speed be x km/hr. Then

715/x - 715/(x+10) = 2

=> 1/x - 1/(x+10) = 2/715

=> {(x+10) - x}/ x(x-10) = 2/715

=> x(x+10) = 3575

=> x^2 + 10x - 3575 = 0

=> x^2 + 65x - 55x - 3575 = 0

=> x ( x + 65 ) - 55 ( x + 65) = 0

=> ( x - 55) (x + 65) = 0

=> x = 55

so Original speed of the car is 55 km/hr

2) Solution :

Let the usual speed be x km/hr

1500/x - 1500/(x+250) = 1/2

=> 1/x - 1/(x+250) = 1/3000

=> { (x+250) - x }/x(x+250) = 1/3000

=> x ( x +250 ) = 750000

=> x^2 + 250x - 750000 = 0

=> x^2 + 1000x - 750x - 750000 = 0

=> x ( x + 1000 ) - 750 ( x + 1000) = 0

=> (x+ 1000)(x - 750) = 0

=> x = 750 ( x>0 )

Usual speed = 750 km/hr

3) Solution :

Let the distance between Meerut and Delhi be x km

Average speed of train from Meerut = x/4 km/hr

Average speed of train from Delhi = 2x/7 km/hr

Suppose they meet y hrs after 6 AM , then

(x/4 × y) + 2x/7 × (y - 2) = x

=> y/4 + 2(y-2)/7 = 1

=> 7y + 8(y - 2) = 28

=> 15y = 44

=> y = 44/15 = 2 14/15 hrs = 2 hrs 14/15×60 min

=> y = 2 hrs 56 min

So the trains meet at 8:56 A.M.

4) Solution :

Total Distance = (3×4) km = 12 km

Total time taken = ( 3/10 + 3/20 + 3/30 + 3/60)

=(36+18+12+6)/120 = 72/120 hrs = 3/5 hrs

Average Speed = 12/(3/5) = 12×5/3 = 20 km/hr

5) Solution :

Let the length of the train be x km and its speed be y km/hr

Speed of the train relative ti first man = (y-3) km/hr

Speed of the train relative to second man = (y-5) km/hr

x/(y-3) = 10/(60×60) and x/(y-5) = 11/(60×60)

=> y - 3 = 36x and 11y - 55 = 3600x

=> 10y - 30 = 11y - 55

=> y = 25

Speed of the train = 25 km/hr

6) Solution :

Let the length of the train be x km and its speed be y km/hr

Speed of the train relative to first man = (y-2) km/hr

Speed of the train relative to second man = (y-4) km/hr

x/(y-2) = 9/(60*60) and x/(y-4) = 10/(60×60)

=> y - 2 = 400x and y - 4 = 360x

=> 400x + 2 = 360x + 4

=> 40x = 2

=> x = 1/20 km = 1/20 × 1000= 50 m

Length of the train is 50 m

7) Solution :

Let the speed of the motorboat in still water be x km/hr

Speed downstream = (x+2) km/hr

Speed upstream = (x-2) km/hr

10/(x-2) + 10/(x+2) = 55/60

=> 1/(x-2) + 1/(x+2) = 55/600= 11/120

=> {(x+2) - (x-2)}/(x^2 - 4) = 11/120

=> 11(x^2 - 4) = 240x

=> 11x^2 - 240x - 44 = 0

=> 11x^2 - 242x + 2x - 44 = 0

=> 11x (x-22) + 2 (x-22)

=> (11x + 2) ( x - 22) = 0

=> x = 22

Speed of motorboat in still water = 22 km/hr

8) Solution :

Let the speed of the stream be x km/hr

Speed downstream = (15 +x) km/hr

Speed Upstream = (15-x) km/hr

30/ (15+x) + 30/(15 - x) = 4 30/60

=> 30/(15+x) + 30/(15-x) = 9/2

=> 1/(15+x) + 1/(15-x) = 9/(2×30) = 3/20

=> {(15-x) + (15+x)}/(15+x)(15-x) = 3/2

=> 3 ( 225 - x^2 ) = 600

=> 3 x^2 = 75

=> x^2 = 25

=> x = 5

Speed of the stream = 5 km/hr

9) Solution :

Let the vessel initially contain 8 litres of liquid

Let x litres of this liquid be replaced with water.

Quantity of water in the new mix = ( 3 - 3x/8 + x) litres

Quantity of syrup in the new mix = (5 - 5x/8) litres

3 - 3x/8 + x = 5 - 5x/8

=> 24 - 3x + 8x = 40 - 5x

=> 10x = 16

=> x = 16/10 = 8/5

Part of the mixture replaced = ( 1/8 of 8/5) = 1/5

10) Solution :

Initially let the can hold 7x litres and 5x litres of A and B respectively

Quantity of A in remaining mix = ( 7x - 7/12 ×9) litres = (7x - 21/4) litres

Quantity of B in remaining mix = (5x - 5/12 ×9) = (5x - 15/4) litres

(7x - 21/4)/{(5x - 15/4)+9} = 7/9

=> (28x - 21)/(20x + 21) = 7/9

=> 9 ( 28x - 21) = 7(20x + 21)

=> 252x - 189 = 140x + 147

=> 112x = 336

=> x = 3

Initially, the can contained (7×3) = 21 litres

Thank You.

Practice Set for WBCS/Rail NTPC/Group D/WBP/PSC/SSC Exams | Set - 02