AIEEE Questions

---

Given that the diameters of the three pipes are 2 cm, 3 cm and 4 cm

From the given data,

Amount of water from three pipes is 4 units, 9 units and 16 units.

Let the capacity of cistern be 'p' units.

∴ p/58 = 16

⇒ p = 928 units.

In 1 minute, quantity to be filled by 3 pipes = 29 units

∴ Total time required = 928/29 = 32 minutes.

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

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

30 + (.0825 x 30) + (15% x 30)

Now

=> 0.0825 x 30 = 2.475 -----(1)

=> 15% of 30 = 450/100 = 4.5 -----(2) 

Hence, from (1) & (2)

30 + (.0825 x 30) + (15% x 30)

= 30 + 2.475 + 4.5

30 + (.0825 x 30) + (15% x 30) = 36.975.

Dena Bank Ltd was an Indian public sector bank with the tagline " Your Trusted Family Bank ". It was headquartered in Mumbai. The bank was founded in 1938 and the Indian government nationalized it in 1969.

Here 5, 7 & 9 are repeating after modulation.

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. 

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.