AIEEE Questions

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The common greenhouse gas containing only hydrogen and oxygen is H2O, this is water vapor.

The common greenhouse gases are water vapor, carbon dioxide, methane, nitrous oxide, and ozone.

The composition (chemical formula) of those gases are:

Water vapor: H2O
Carbon dioxide: CO2
Methane: CH4
Nitrous oxide: NO
Ozone: O3.

As you see the only one of those that contains only hydrogen and oxygen is H2O, water vapor.

9321 = 3 x 13 x 239 = 31 x 13 1 x 239 1 ; add one to all the powers ie.,

(1+1) x (1+1) x (1+1)= 8 ;

we get eight factors.

The man's brother's father means his father is the only son of woman's grandfather

=> grandfather to the man also

=> Woman and the man are brother and sisters.

Let the sum be Rs. 100. Then,
S.I. for first 6 months = (100 x 8 x 1) / (100 x 2) = Rs. 4
S.I. for last 6 months = (104 x 8 x 1) / (100 x 2) = Rs. 4.16
So, amount at the end of 1 year = (100 + 4 + 4.16) = Rs. 108.16
Effective rate = (108.16 - 100) = 8.16%.

PARAM 8000 was was the first supercomputer developed by Indian scientists launched on July 01, 1991 is considered India's first supercomputer.

• It was indigenously built in 1991 by Centre for Development of Advanced Computing and was replicated and installed at ICAD Moscow in 1991 under Russian collaboration.

Let M and m grams be the mass of boat + rock in it. By law of floatation(M+m) grams of water will be displaced. This has a volume of (M+m)cc. That means the level of water in pool has risen up by (M+m)cc with the boat + rock when it is floating. If the rocks in the boat are thrown not into the pool but outside, the weight left is only M grams the weight of the boat. This will have displaced M cc of water only i.e less water than in the case of loaded boat. But if the rocks are put in the water in the pool and not outside, then the water in the pool will rise not only by M cc on account of empty boat but also by a volume of the stone put into it. This volume will be mdwhere m is the mass of rock and d density of rock.

Now md is  < m. Thus the total rise will be M+mdwhich is < M + m.

Let 'P' be the original population at the beginning

P x (95/100) x (85/100) = 3553

P = 4400.

Numbers of students in section A = x
∴ Numbers of students in section B and C = (100 – x)
∴ x 70 + (100 – x) 87.5 = 84 100
=> 70x + 87.5 100 – 87.5x = 8400
=> 8750 – 17.5x = 8400
=> 17.5x = 8750 – 8400 => x = 20.