AIEEE Questions

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Let p = Rs. 100.

Then, S.I is Rs. 50 and time = 5 years. 

∴ R = 100×50100×5= 10% p.a.

Now, p = Rs. 12,000 , T = 3 years and R = 10% p.a.

C.I. = Rs. 12000×1+101003-1= Rs. 3972

First is the type of houses most commonly found in the second.

4 x 27 x 3125 = 22 × 33 × 55 ;

8 x 9 x 25 x 7 = 23 × 32 × 52 × 7

16 x 81 x 5 x 11 x 49 = 24 × 34 × 5 × 11 × 72

H.C.F = 22 × 32 × 5 = 180.

100C10 + 90C20 + 70C30 + 40C40 = 100!10!x20!x30!x40!

Except Rock color, all other Graded bedding, Ripple marks and Mud cracks are sedimentary structures.

Let the number of students in classes A, B and C be P, Q and R respectively.
Then, total score of A = 83P, total score of B = 76Q, total score of C = 85R.
Also given that,
(83P + 76Q) / (P + Q) = 79
=>4P = 3Q.
(76Q + 85R)/(Q + R) = 81
=>4R = 5Q,
=>Q = 4P/3 and R = 5P/3
Therefore, average score of A, B, C = ( 83P + 76Q + 85R ) / (P + Q + R) = 978/12 = 81.5

Sony and her brother's age sum is 31
Let Sony's mother's age be 'x'.
Now the conditions given for both of their ages related to their mom is
1/5(x-15) + 3/5(x-10) = 31
x-15 + 3x-30 = 155
4x-45 = 155
4x = 200
x = 50
So, Mother's age = 50.

Cross verification : 15 years ago age = 35 => Sony age = 7
10 years ago age = 40 => bro age = 24

Sum = 24 + 7 = 31.

Volume of cuboid = (24 x 9 x 8) cm = 1728 cu.cm
Volume of small cube = (3 x 3 x 3) cm = 27 cu.cm

So, No. of small cubes formed = 1728/27 = 64.