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
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