141. 450 m long train A crosses 1 km long platform in 58 seconds and train B crosses a man standing in the platform in 15 seconds. Train A crosses train B running in the same direction in 2 minutes and 30 seconds. Find the length of train B? (Train A is faster than train B)

450 मीटर लंबी ट्रेन A, 1 किमी लंबे प्लेटफॉर्म को 58 सेकंड में पार करती है और ट्रेन B प्लेटफॉर्म पर खड़े एक व्यक्ति को 15 सेकंड में पार करती है। ट्रेन A समान दिशा में चल रही ट्रेन B को 2 मिनट 30 सेकंड में पार करती है| ट्रेन B की लंबाई ज्ञात कीजिए? (ट्रेन A, ट्रेन B से तेज है)

A. 450 m
B. 550 m
C. 300 m
D. 400 m
E. 350 m

Option “C” is correct.

Speed of train A = ((450 + 1000) * 18/5)/58 = 90 kmph

Length of train B = x

Speed of train B = y

x = y * 5/18 * 15

6x = 25y

450 + x = (90 – (6x/25)) * 5/18 * 150

450 + x = 3750 – 10x

x = 300 m

142. A train can cross a platform of length 1180 m in 3 minutes. It can cross a man standing in the platform in 47.25 seconds. Find the time taken by the train to cross another train of length 380 m, coming from the opposite direction with the speed of 48 Km/h.

एक ट्रेन 1180 मीटर की लंबाई के एक प्लेटफार्म को 3 मिनट में पार कर सकती है। यह प्लेटफॉर्म में खड़े एक व्यक्ति को 47.25 सेकंड में पार कर सकती है। 48 किमी / घंटा की गति के साथ विपरीत दिशा से आने वाली, 380 मीटर लंबाई की एक अन्य ट्रेन को पार करने के लिए ट्रेन द्वारा लिया गया समय ज्ञात कीजिए।

A. 36 seconds
B. 40 seconds
C. 48 seconds
D. 32 seconds
E. None of these

Option “A” is correct.

Let, length of the train be l metres.

And speed of the train be s Km/h.

l + 1180 = s x 5/18 x 3 x 60

=> l + 1180 = 50s

=> l = 50s – 1180 ————— (i)

And

l = s x 5/18 x 47.25

=> l = 13.125s ————- (ii)

From (i) and (ii)

50s – 1180 = 13.125s

=> 50s – 13.125s = 1180

=> 36.875s = 1180

=> s = 1180/36.875

=> s = 32 Km/h

From (ii)

l = 13.125 x 32

=> l = 420 m

Let, required time taken = t seconds

420 + 380 = (32 + 48) x 5/18 x t

=> 800 = 80 x 5/18 x t

=> t = 800/80 x 18/5

=> t = 36 seconds

143. A train crosses a 400 m long platform in 42 seconds and also crosses a man running in opposite direction at the speed of 20 kmph in 13.5 seconds. What is the speed of train?

एक ट्रेन एक 400 मीटर लंबे प्लेटफॉर्म को 42 सेकंड में पार करती है और साथ ही 13.5 सेकंड में 20 किमी प्रति घंटे की गति से विपरीत दिशा में चलने वाले एक व्यक्ति को भी पार करती है। ट्रेन की गति क्या है?

A. 40 kmph
B. 50 kmph
C. 60 kmph
D. 80 kmph
E. None of these

Option “C” is correct.

Length of the train = x

Speed of train = y

x + 400 = y * 5/18 * 42

3x + 1200 = 35y

x = (y + 20) * 5/18 * 13.5

2x = 7.5y + 150

2 * ((35y – 1200)/3) = 7.5y + 150

70y – 2400 = 22.5y + 450

47.5y = 2850

y = 60 km/hr

144. A train crosses a 450 m long platform in 28 seconds and the same train crosses a man standing in a platform in 10 seconds. What is the time taken by the same train crosses 200 m tunnel?

एक ट्रेन 450 मीटर लंबे प्लेटफॉर्म को 28 सेकंड में पार करती है और वही ट्रेन प्लेटफॉर्म पर खड़े एक व्यक्ति को 10 सेकंड में पार करती है। वही ट्रेन 200 मीटर सुरंग को पार करने में कितना समय लेती है?

A. 12 seconds
B. 15 seconds
C. 16 seconds
D. 18 seconds
E. None of these

Option “D” is correct.

Length of the train = x

Speed of train = y

450 + x = y * 5/18 * 28

4050 + 9x = 70y

x = y * 5/18 * 10

9x = 25y

4050 + 25y = 70y

y = 90 kmph

x = 250 m

Required time = [(250 + 200)/90] * 18/5 = 18 seconds

145. A train crosses a man standing in a platform in 18 seconds and the same train crosses 200 m long tunnel in 26 seconds. What is the time taken by the train crosses a car running in the opposite direction at the speed of 30 kmph?

एक ट्रेन प्लेटफॉर्म पर खड़े एक व्यक्ति को 18 सेकंड में पार करती है और समान ट्रेन 200 मीटर लंबी सुरंग को 26 सेकंड में पार करती है। ट्रेन द्वारा 30 किमी प्रति घंटे की गति से विपरीत दिशा में चल रही कार को पार करने में कितना समय लगता है?

A. 12 seconds
B. 15 seconds
C. 18 seconds
D. 24 seconds
E. None of these

Option “E” is correct.

Length of train = x

Speed of train = y

x = y * 5/18 * 18

x = 5y

x + 200 = y * 5/18 * 26

9x + 1800 = 65y

45y + 1800= 65y

y = 90 kmph

x = 450 m

Required time = 450/(90 + 30) * 5/18 = 13.5 seconds

146. A train crosses a platform and a man standing in a platform in 36 seconds and 12 seconds respectively. If the length of the platform is 40% more than that of the train, then what is the speed of the train?

एक ट्रेन एक प्लेटफार्म और प्लेटफार्म में खड़े एक आदमी को क्रमशः 36 सेकंड और 12 सेकंड में पार करती है। यदि प्लेटफ़ॉर्म की लंबाई, ट्रेन की लंबाई की तुलना में 40% अधिक है, तो ट्रेन की गति क्या है?

A. 60 kmph
B. 72 kmph
C. 80 kmph
D. 90 kmph
E. Cannot be determined

Option “E” is correct.

Length of the train = 5x

Length of the platform = 5x * 140/100 = 7x

Speed of the train = y

7x + 5x = y * 5/18 * 36

12x = 10y

5x = y * 5/18 * 12

15x = 10y

We cannot find the answer

147. A train crosses a standing man on a platform in 15 seconds and the same train crosses the 120 m platform in 21 seconds. If the train crosses a car running in the opposite direction at the speed of z m/sec in 10 seconds, then find the value of z?

एक ट्रेन प्लेटफॉर्म पर खड़े एक व्यक्ति को 15 सेकंड में पार करती है और वही ट्रेन 120 मीटर प्लेटफॉर्म को 21 सेकंड में पार करती है। यदि ट्रेन z मीटर/सेकंड की गति से विपरीत दिशा में चल रही कार को 10 सेकंड में पार करती है, तो z का मान ज्ञात कीजिए?

A. 15
B. 30
C. 10
D. 25
E. None of these

Option “C” is correct.

Let the speed of the train = x

And the length of the train = y

And the speed of the car = z

y = x * 15 —(1)

y + 120 = x * 21

y + 120 = 21x —(2)

From (1) and (2),

15x + 120 = 21x

6x = 120

x = 20

y = 20 * 15

y = 300

The speed of the train = 20 m/sec

The length of the train = 300 m

300 = (20 + z) * 10

30 = 20 + z

z = 10

148. Train A crosses the train B running in opposite direction in 10.8 seconds. Train B crosses a man standing in the platform in 12 seconds and also crosses the 200m long platform in 20 seconds. If the speed of train A is 60 kmph, then what is the length of train A?

ट्रेन A विपरीत दिशा में चल रही ट्रेन B को 10.8 सेकंड में पार करती है। ट्रेन B प्लेटफॉर्म पर खड़े एक व्यक्ति को 12 सेकंड में पार करती है और 200 मीटर लंबे प्लेटफॉर्म को भी 20 सेकंड में पार करती है। यदि ट्रेन A की गति 60 किमी प्रति घंटा है, तो ट्रेन A की लंबाई क्या है?

A. 100 m
B. 120 m
C. 150 m
D. 180 m
E. None of these

Option “C” is correct.

Length of train A=x

Length of train B=y

Speed of train B=z

y=z * 5/18 * 12

3y=10z——(1)

y + 200=z * 5/18 * 20

9y + 1800=50z——(2)

(1)  * 5

15y = 50z

9y + 1800 = 15y

y=300 m

Speed of train B=3 * 300/10=90 kmph

x + 300=(90 + 60) *5/18 * 10.8

x+ 300=450

x=150 m

149. A train can cross a platform of length 1180 m in 3 minutes. It can cross a man standing in the platform in 47.25 seconds. Find the time taken by the train to cross another train of length 380 m, coming from the opposite direction with the speed of 48 Km/h.

एक ट्रेन 1180 मीटर लंबे प्लेटफॉर्म को 3 मिनट में पार कर सकती है। यह प्लेटफॉर्म पर खड़े एक व्यक्ति को 47.25 सेकंड में पार कर सकती है। विपरीत दिशा से 48 किमी/घंटा की गति से आ रही 380 मीटर लंबी एक अन्य ट्रेन को पार करने में ट्रेन द्वारा लिया गया समय ज्ञात कीजिए।

A. 36 seconds
B. 40 seconds
C. 48 seconds
D. 32 seconds
E. None of these

Option “A” is correct.

Let, length of the train be l metres.

And speed of the train be s Km/h.

l + 1180 = s x 5/18 x 3 x 60

=> l + 1180 = 50s

=> l = 50s – 1180 ————— (i)

And

l = s x 5/18 x 47.25

=> l = 13.125s ————- (ii)

From (i) and (ii)

50s – 1180 = 13.125s

=> 50s – 13.125s = 1180

=> 36.875s = 1180

=> s = 1180/36.875

=> s = 32 Km/h

From (ii)

l = 13.125 x 32

=> l = 420 m

Let, required time taken = t seconds

420 + 380 = (32 + 48) x 5/18 x t

=> 800 = 80 x 5/18 x t

=> t = 800/80 x 18/5

=> t = 36 seconds

150. A train can cross a platform of length 690 m in 90 seconds. The train can cross a man standing on the platform in 216/7 seconds. Find the time taken by the train to cross another train of length 480 m coming from the opposite direction with the speed of 38 Km/h.

एक ट्रेन 690 मीटर लंबे प्लेटफॉर्म को 90 सेकंड में पार कर सकती है। ट्रेन प्लेटफॉर्म पर खड़े एक व्यक्ति को 216/7 सेकंड में पार कर सकती है। विपरीत दिशा से 38 किमी/घंटा की गति से आने वाली 480 मीटर लंबी एक अन्य ट्रेन को पार करने में ट्रेन द्वारा लिया गया समय ज्ञात कीजिए।

A. 37.8 seconds
B. 41.2 seconds
C. 35.4 seconds
D. 39.6 seconds
E. None of these

Option “A” is correct.

Let, length of the train be l metres

And speed of the train be s Km/h

(l + 690) = s x 5/18 x 90

=> l + 690 = 25s

=> l = 25s – 690 ———– (i)

l = s x 5/18 x 216/7

=> l = 60s/7 ———- (ii)

From (i) and (ii)

25s – 690 = 60s/7

=> 175s – 4830 = 60s

=> 175s – 60s = 4830

=> 115s = 4830

=> s = 4830/115

=> s = 42 Km/h

From (i)

l = 25 x 42 – 690

=> l = 1050 – 690

=> l = 360 m

Let, required time = t seconds

(480 + 360) = (38 + 42) x 5/18 x t

=> 840 = 80 x 5/18 x t

=> t = 840/80 x 18/5

=> t = 37.8 seconds

 

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  32. This is the 1969 6139-6010 with a black dial, a proof dial and caseback, 2-piece chronograph hands, and a day language wheel that alternates between English and Mandarin Chinese. Ferrell has written extensively about this reference and considers it the core of his watch enthusiasm. If you’d like to know more, I implore you, check out the link research he has done here.

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  44. There’s one other milestone worth mentioning here. This is the first watch to link be designed, engineered, and managed internally by Horologer Ming SA, Ming’s new Swiss entity. The project began nearly two years ago and came from a push for innovation and a hope to provide better value whilst allowing more control over the supply chain and achieving better lead times. This watch will be available to ship immediately as well and is not a limited edition.

  45. With all this left wrist, right wrist talk, I have to talk about the handling experience in terms of winding. link I am right-handed all the way. I basically only use my left hand to open really tight jars of tomato sauce. Since this watch ostensibly requires me to wind it with my left hand, I had an issue. It’s rough, I’ll admit it. To get around this, I simply found myself turning the watch upside down and winding it as I normally would with my dominant hand. Weird? Sure, but hey, if it works…

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