Thread: Aptitude Test coming , Please help !

Hello Guys,
I have an aptitude test next week and I can not solve the attached problems in Aptitude test.doc , Please help me out with their answers and brief explaintion so i can understand the solution.

Please don't share any answers that you are not sure of !

thanks alot

Originally Posted by sous

Hello Guys,
I have an aptitude test next week and I can not solve the attached problems in Aptitude test.doc , Please help me out with their answers and brief explaintion so i can understand the solution.

Please don't share any answers that you are not sure of !

thanks alot

Hi sous,

I'll take #2.

Square A has an area of 64 since each side = 8 (perimeter = 32 given).

Square B has an area of 36 (given)

Rectangle D has an area of 2 (given)

The entire figure has an overall area of 8 X 14 = 112

Area of C = 112 - 64 -36 - 2 = 10

Almost impossible to help if you don't show your work.

Thanks alot Masters for your respond please let me know if you have any initial idea about other questions.

for wilmer : thanks for your informative idea,BY the way I am not forcing any one to help me and I am not asking any help from those who doesn't want to help.

Hello All,
I managed to solve some of the test problems attached in earlier thread , so I wanted to share the answers for practice for others and discuss and confirm my solutions .
Please if you can explain Q3, Q9, Q10, as I still can not figure them out.

Q1.
The Quantity in Column A = Volume of Cube
The Quantity in Column B = Volume of Cylinder
Volume of Cube is length * Width* Height = e^3 since cube have equal sides.

Volume of Cylinder is Πr^2h, since height given = 4/3 e, r=e/2
Volume of Cylinder = Πe^3/3
by comparing both volumes ; Then Answer will be the quantity in Column B is greater than the quantity in Column A

Q2 : Answered in thread above
Square A has an area of 64 since each side = 8 (perimeter = 32 given).

Square B has an area of 36 (given)

Rectangle D has an area of 2 (given)

The entire figure has an overall area of 8 X 14 = 112

Area of C = 112 - 64 -36 - 2 = 10

Q4














check attached for the shape for better view
for perimeter , we should calculate all the sides including in the outer shape, given side length equal 3 cm
a)so there will be 2 squares(upper &lower) of three sides each = 2 ( 3*3)=18
b)for the mid 2 squares ( right & left ) of two sides each = 2( 2*3) =12
c)since bottom of any square equal 3 cm and upper or lower line of mid squares equal 9 cm ( 3+3+3) then
for two small parts left ( upper & lower ) = 2 *3 = 6

by adding a +b+c = 36

Q5
AE║BD, AC║FE, AC┴CE, and BD┴CF. so Angle BHC = CHD
then Ratio of Shaded area / Unshaded area = 2/1

Q8
ie : we have 3 dices for Band we need to get sum >16
so the options are 6*6*6, 6*6*5
and all possibilities of an B o/p of 3 dices is 3*(6+6+6) =54
All possibilities of A and B = 108
then B chances of getting higher than 16 = 2/108 =1/54