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The distance between two stations A and B is 440 km. A train starts at 4 p.m. from A and move towards B at an average speed of 40 km/hr. Another train starts B at 5 p.m. and moves towards A at an average speed of 60 km/hr. How far from A will the two trains meet and at what time?

Option: 1

200,8 p.m.


Option: 2

300,9 p.m.


Option: 3

200,9 p.m.


Option: 4

300,8 p.m.


The sum of the distances covered by both trains should be equal to the total distance between the stations A and B: 40t+60(t−1)=44040t + 60(t - 1) = 44040t+60(t−1)=440

40t+60t−60=440
100t−60=440
100t=500
t=5

So, the trains will meet 5 hours after Train A starts from A.

Train A starts at 4 p.m. so 5 hours from 4 p.m. is 4 p.m. + 5 hours = 9 p.m.

The two trains will meet 200 km from A at 9 p.m.

Option 3: 200, 9 p.m.

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Posted by

Sara P

PQ is a tunnel. A dog sits at the distance of 5/ 11 of PQ from P. The train whistle coming from any end of the tunnel would make the dog run. If a train approaches P and dog runs towards P the train would hit the dog at P. If the dog runs towards Q instead, it would hit the dog at Q. Find ratio of speed of train and dog?

 

Option: 1

5:2


Option: 2

16:5


Option: 3

11:1


Option: 4

34:3


Option 3 - 11:1

########            P____5_____¶______6_____Q

"¶" is the dog located inside the tunnel PQ and #####... is a train approaching to it. 

Case 1 - Dog decides to reach point P 

Distance covered by Dog - 5 parts and train reaches at P

Case 2 - Dog decides to run towards point Q

At the time train reaches to point P, the dog covers 5 parts in direction of Q. Now the left out part covered by dog will be 1 part towards Q .

At the time dog covers 1 part and reaches Q, train covers whole 11 parts of tunnel and reaches from P to Q. 

So, 11 part of train is eual to the 1 part covered by the dog to reach Q. 

Ratio of their speeds - Train:Dog = 11:1  

 

 

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Posted by

Vaibhavi Gupta

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A massless string connects two pulley of masses ' 2 \mathrm{~kg}' and '1 \mathrm{~kg}' respectively as shown in the figure.

The heavier pulley is fixed and free to rotate about its central axis while the other is free to rotate as well as translate. Find the acceleration of the lower pulley if the system was released from the rest. [Given, g=10 \mathrm{~m} / \mathrm{s}^2]

Option: 1

\frac{4}{3} \mathrm{~gm} / \mathrm{s}^2


Option: 2

\frac{3}{2} \mathrm{~gm} / \mathrm{s}^2


Option: 3

\frac{3}{4} \mathrm{~gm} / \mathrm{s}^2


Option: 4

\frac{2}{3} \mathrm{~gm} / \mathrm{s}^2


Not understanding sir 

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Posted by

Raju vittal nandi

The following is the biochemical pathway for purple pigment production in flowers of sweet pea:

Colorless precursor 1 —- Allele A—->  Colorless precursor 2 —--- Allele B—---> Purple pigment

Recessive mutation of either gene A or B leads to the formation of white flowers. A cross is made between two parents with the genotype: AaBb × aabb. Considering that the two genes are not linked, the phenotypes of the expected progenies are:

 

Option: 1

9 purple : 7 white


Option: 2

3 white : 1 purple


Option: 3

1 purple : 1 white


Option: 4

9 purple : 6 light purple : 1 white


3(white):1(purple)

In case of 'duplicate recessive epistasis homozygous recessive gene masks the other gene i.e., aa is epistatic to A or a || bb is epistatic to B or b

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Posted by

Riya

JEE Main high-scoring chapters and topics

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Calculate the acceleration of block m_1 of the following diagram. Assume all surfaces are frictionless . Here m1 = 100kg and m2 = 50kg

 

Option: 1

0.33m/s2


Option: 2

0.66m/s2


Option: 3

1m/s2


Option: 4

1.32m/s2


1.32

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Posted by

balda gayathri

No. of transition state in given figure

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


2

 

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Posted by

Mura

NEET 2024 Most scoring concepts

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When cell has stalled DNA replication fork, which checkpoint should be predominantly activated?

Option: 1

G1/S

 

Option: 2

G2/M

 

 

 

Option: 3

M

 

 

Option: 4

Both G2 M and M

 

G2/M should be activated as the cell has stalled DNA replication fork.

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Posted by

Ajit Kumar Dubey

A random variable X has the following probability distribution:          X: \: \: 1\; \; \: \: 2\; \;\: \: \: \: 3\; \; \: \: \: \: \: 4\: \: \: \; \; \: \: 5 P(X): \;K^{2}\; \; 2K\; \; K\; \; 2K\; \; 5K^{2} Then P(X>2) is equal to: 
Option: 1 \frac{7}{12}
Option: 2 \frac{23}{36}
Option: 3 \frac{1}{36}
Option: 4 \frac{1}{6}
 

Option 2

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Posted by

kpvalli2725@gmail.com

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Let a_{n} be the nth term of a G.P. of positive terms. If \sum_{n=1}^{100}a_{2n+1}=200\: \: and\: \: \sum_{n=1}^{100}a_{2n}=100,\: \: then\: \sum_{n=1}^{200}a_n is equal to :   
Option: 1 300
Option: 2 175
Option: 3 225
Option: 4 150
 

150

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Posted by

Nalla mahalakshmi

If x=\sum_{n=0}^{\infty }(-1)^{n}\tan ^{2n}\theta \: \: and\: \: y=\sum_{n=0}^{\infty }\cos ^{2n}\theta , for 0<\theta < \frac{\pi }{4}, then
Option: 1 y(1+x)=1
Option: 2 x(1-y)=1
Option: 3 y(1-x)=1
Option: 4 x(1+y)=1
 

x=\sum_{n=0}^{\infty}(-1)^{n} \tan ^{2 n} \theta=1-\tan^2\theta+\tan^4\theta..........

y=\sum_{n=0}^{\infty} \cos ^{2 n} \theta=1+\cos^2\theta+\cos^4\theta......

Use \text S_{\infty}=\frac{1}{1-r}

{x=\frac{1}{1+\tan ^{2} \alpha}=\cos ^{2} \theta} \\ {y=\frac{1}{1-\cos ^{2} \theta}=\frac{1}{\sin ^{2} \theta}}

\Rightarrow (1-x)= \sin ^{2} \theta

\Rightarrow y(1-x)=1

Correct Option (3)

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Posted by

avinash.dongre

JEE Main high-scoring chapters and topics

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