AIEEE Questions

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Total savings = 200 + 200 + 200 + 240 + 280 + ... to n months = 11040

= 120 + 80 + 160 + 40 + 200 + 240 + ....n terms
= 40 + 80 + 120 + 160 + 200 + 240 + ...n terms
= n/2[2a + (n-1)d] + 80 + 40 = 11040
= n/2[240 + (n-1)40] = 10920
= n[6+n-1] = 546
= n[n+5] = 546
n = 21

Here 5, 7 & 9 are repeating after modulation.

The computer that process both analog and digital is called Hybrid computer.

As given in the question, Marked price is 25% more than the Cost price.

=> C.P of the article = 34 x 400 = 300

Now, 

Let the original S.P of the article be Rs. P

Now the new S.P = P + 16.666 +13.333300 x P

=> S.P = 7P6

According to the question,

7P6 - 300 = 2P - 300

=> 5P = 1800

=> P = Rs. 360

Hence, the increased S.P = 360 x 7/6 = Rs. 420. 

Two passengers behind the operator is safe for a PWC.

HCF of 210 and 55 is 5

Now, 210x5 + 55P = 5

=> 1050 + 55P = 5

=> 55P = -1045

=> P = -1045/55

=> P = -19.

A bubble is what lighter than air but a hundred people cannot lift me.

This question concerns a committee’s decision about which five of eight areas of expenditure to reduce. The question requires you to suppose that K and N are among the areas that are to be reduced, and then to determine which pair of areas could not also be among the five areas that are reduced.

The fourth condition given in the passage on which this question is based requires that exactly two of K, N, and J are reduced. Since the question asks us to suppose that both K and N are reduced, we know that J must not be reduced:

Reduced         ::      K, N
Not reduced   ::      J

The second condition requires that if L is reduced, neither N nor O is reduced. So L and N cannot both be reduced. Here, since N is reduced, we know that L cannot be. Thus, adding this to what we’ve determined so far, we know that J and L are a pair of areas that cannot both be reduced if both K and N are reduced:

Reduced        ::      K, N
Not reduced  ::      J, L

Answer choice (B) is therefore the correct answer.