Online Test      LOGIN      SIGN UP
Forgot your password?
  • IBPS PO CWE Exam Quantitative Aptitude Study Material

Digitization help student to explore and study their academic courses online, as this gives them flexibility and scheduling their learning at their convenience. Kidsfront has prepared unique course material of Quantitative Aptitude Mensuration for IBPS PO CWE Exam student. This free online Quantitative Aptitude study material for IBPS PO CWE Exam will help students in learning and doing practice on Mensuration topic of IBPS PO CWE Exam Quantitative Aptitude. The study material on Mensuration, help IBPS PO CWE Exam Quantitative Aptitude students to learn every aspect of Mensuration and prepare themselves for exams by doing online test exercise for Mensuration, as their study progresses in class. Kidsfront provide unique pattern of learning Quantitative Aptitude with free online comprehensive study material and loads of IBPS PO CWE Exam Quantitative Aptitude Mensuration exercise prepared by the highly professionals team. Students can understand Mensuration concept easily and consolidate their learning by doing practice test on Mensuration regularly till they excel in Quantitative Aptitude Mensuration.


Mensuration
The total surface area of a regular triangular pyramid with each edge of length 1 cm is

a) √3 cm2
b) 4/3 √3 cm2
c) 4√3 cm2
d) 4 cm2



Answer
Solution
Correct Answer Is : √3 cm2
Solution Is : Total surface area = 4 x √3/4 x (1)2 = √3 sq.cm
The number of paving stones each measuring 2.5 m x 2m required to pave a rectangular courtyard 30m long and 17.5m wide is

a) 105
b) 99
c) 33
d) 80



Answer
Solution
Correct Answer Is : 105
Solution Is : Number of paving stones = Area of courtyard / Area of a stone = 30x 17.5 / 2.5 x 2 = 105
The perimeter of a sheet of paper in the shape of a quadrant of a circle is 75 cm. Its area would be ( π = 22/7)

a) 346.5 cm2
b) 100 cm2
c) 693 cm2
d) 512.25 cm2



Answer
Solution
Correct Answer Is : 346.5 cm2
Solution Is :
The perimeter of two similar triangles are 30 cm and 20 cm respectively. If one side of the first triangles is 9 cm. Determine the corresponding side of the second triangle.

a) 13.5 cm
b) 6 cm
c) 15 cm
d) 5cm



Answer
Solution
Correct Answer Is : 6 cm
Solution Is : If the required side be x cm, then 30/20 = 9/x => 3x = 9 x 2 => x = 9x2/3 = 6 cm
The diagonal of a quadrilateral shaped field is 24m and the perpendiculars dropped on it from the remaining opposite vertices are 8 cm and 13 cm. The area of the field is

a) 252 m2
b) 156 m2
c) 96 m2
d) 1152 m2



Answer
Solution
Correct Answer Is : 252 m2
Solution Is :
If the base right of right prism remains same and the lateral edges are halved, then its volume will be reduced by

a) 33.33%
b) 66%
c) 25%
d) 50%



Answer
Solution
Correct Answer Is : 50%
Solution Is : Volume of prism = Area of base x height Base remains constant. Hence, volume α height
The radius of a right circular cone is 3 cm and its height is 4 cm. The total surface area of the cone is

a) 48.4 sq.cm
b) 64.4 sq.cm
c) 96.4 sq.cm
d) 75.4 sq.cm



Answer
Solution
Correct Answer Is : 75.4 sq.cm
Solution Is :
A wooden box of dimensions 8 metre x 7 metre x6 metre is to carry rectangular boxes of dimensions 8 cm x 7 cm x6 cm. The maximum number of boxes that can be carried in 1 wooden box is

a) 7500000
b) 9800000
c) 1200000
d) 1000000



Answer
Solution
Correct Answer Is : 1000000
Solution Is : Volume of wooden box = (8 * 7 * 6) cu.m. = (8 * 7 * 6 * 100?3) cu.m. Maximum number of boxes = 8 * 7 * 6 * 1003/ 8 * 7 * 6 =1000000
Two circular cylinders of equal volume have their heights in the ratio 1 : 2. Ratio of their radii is

a) 1:04
b) 1 : √2
c) √2 : 1
d) 1:02



Answer
Solution
Correct Answer Is : √2 : 1
Solution Is :
A rectangular piece of paper of dimensions 22 cm by 12 cm is rolled along its length to form a cylinder. The Volume ( in cu. Cm) of the cylinder so formed is

a) 562
b) 412
c) 462
d) 362



Answer
Solution
Correct Answer Is : 462
Solution Is : Paper is folded along the length. Circumference of the base = 22 cm, Height of cylinder = 12 cm 2 π r = 22 => 2 * 22/7 * r = 22 => r = 7/2 cm Volume of cylinder = π r2 h 22/7 * 7/2 * 7/2 * 12 = 462 cu.cm.

Preparation for Exams

script type="text/javascript">