AIEEE Questions

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Let the radii of the smaller and the larger circles be 's' m and 'l' m respectively.  

 2πs = 264 and 2πl = 352  

 s = 2642π and l = 3522π   

Difference between the areas =πl2-πs2   

  = π1762π2-1322π2 

  = 1762π-1322π  

  = (176 - 132)(176 + 132) / π  

  = (44 x 308) / (22/7) = 4312 sq.m

Let’s assume John bought the laptop for Rs. 100
He then sells it to Mark at a loss of 20% => Rs. 80,
Who consequently sells it to Kevin at a profit of 25% =>  Rs. 100
Then, Kevin sells it back to Mark for Rs. 90 (Rs.4.50 for every Rs. 5 that he had paid initially), so he suffers a loss of Rs. 10
Therefore, for Rs. 100 there is a loss of Rs. 10,
Then for Rs. 1,75,000 there will be a loss of Rs. 17,500.

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.

We know that, a folder holds documents in the same way an envelope holds letters.

Hence, Folder : Document :: Envelope : Letter.

Let marked price = Rs. 100.
Then, C.P. = RS. 54,
S.P. = Rs. 85
Gain % = 31/64 x 100 = 48.4%.

Here the letters follow the rule (AxZ)+(A+Z)

The number corresponding to the letters are used in the rule

A=1, Z=26 => 26x1 + 1 + 26 = 26+27 = 53 =  053

(LxP)+(L+P) = (12 x16) + 12 + 16 = 192+28 = 220

Similarly (11x12) + 11 + 12 => 132 + 23 = 155 

It is useful to have a property of a material that is independent of the size of the sample and can be use to compare its elastic properties with another - this is called the Modulus of Elasticity for the material.

The modulus of elasticity is defined as:

Modulus of elasticity = Stress / Strain

There are three types of moduli. They are :

(a) the Young Modulus - tensile and compression longitudinal stress

(b) the shear modulus - a shearing stress

(c) the bulk modulus - volume changes of the specimen.

Area of a square SxS = 1024 => S = 32 cm
Length of the rectangle = 32x2 cm = 64 cm
Breadth of the rectangle = 32 - 12 = 20 cm
Required ratio = 64 : 20 = 16 : 5