GATE Questions

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Let the actual distance be t kms

Now, According to the data,

$$\begin{gathered} 5t + 860 = 6t \\ => 560t + 860 = 6t \\ => 60t + 8 = 72t \\ => 8 = 12t \\ => t = 8/12 = 2/3 hrs \end{gathered}$$

Now, the total distance = 6t = 6 x 2/3 = 2 x 2 = 4 kms.

0.46 = 46/100 = 23/50.

Sum of the numerator and denominator is 23 + 50 = 73

The given series follows,

2

2 x 2 = 4

4 x 3 = 12

12 x 4 = 48

48 x 5 = 240.

From the given data,

When the price of rice is decreased by 10%, the value of Rs. 270 would become

270 x 100/90 = 300

Now, the value 270 becomes 300

And given, the girl could buy 1 kg extra for that value

Change in value = 300 - 270 = Rs. 30

Therefore, 1 kg extra rice she can get for Rs. 30

=> The original price of rice = Rs. 30/kg.

As she gets only 9 kgs for Rs. 270 before 10% discount, now after discount she gets 10 kgs instead of 9 kgs for Rs. 270.  

Let 'M' be the maximum marks in the examination.

Therefore, Madhu got 32% of M = 32M/100 = 0.32M
And Kumar got 42% of M = 42M/100 = 0.42M.
In terms of the maximum marks Kumar got 0.42M - 0.32M = 0.1M more than Madhu. ---- (1)

The problem however, states that Kumar got 16 marks more than the cut-off mark and Madhu got 8 marks less than the cut-off mark. Therefore, Kumar has got 16 + 8 = 24 marks more than Madhu. ---- (2)

Now, Equating (1) and (2), we get
0.1M = 24 => M = 24/0.1 = 240
'M' is the maximum mark and is equal to 240 marks.
We know that Madhu got 32% of the maximum marks.
Therefore, Madhu got 32 x 240/100 = 76.8 marks.
We also know that Madhu got 8 marks less than the cut-off marks.

Therefore, the cut-off marks will be 8 marks more than what Madhu got
= 76.8 + 8 = 84.8.

Here by trial and error method, we can obseve that

404404 = 404 x 1001; 415415 = 415 x 1001, etc.

So, any number of this form is divisible by 1001.

Unit digit of this expression is always 1 as the base ends with 1. 

For the tenth place digit we need to multiply the digit in the tenth place of the base and unit digit of the power and take its unit digit.

i.e, tenth place digit in 2151 is 5 and

tenth place digit in power 415 is 1

And the units digit in the product of 5 x 1 = 5

Therefore, last two digits of 2151415 is 51.

The Chipko movement or Chipko Andolan refers to a forest conservation movement or deforestation. In modern india, it began in 1973 in the state of Uttar Pradesh (now Uttarakhand). Sunderlal Bahuguna and 84 villagers risked their lives to protect the forest trees from being felled on the order of the maharaja (king) and went on to become a rallying point for many future environmental movements all over the world it created a precedent for starting of nonviolent protest in India, and its success meant that the world immediately took notice of this non violent movement, which was to inspire in time many such eco-groups by helping to slow down the rapid deforestation, expose vested interests, increase ecological awareness, and demonstrate the viability of people power. Above all, it stirred up the existing civil society in India, which began to address the issues of tribal and marginalized people.

Chipko type movement dates back to 1730 AD when in a khejarli village of Rajasthan, 363 people sacrificed their lives to save khejari trees.