Geometry is an important topic that is always covered in MBA exams. Geometry is important because MBA aspirants need to have the diagnostic skill to understand shapes and figures. Some of the topics that are being covered under geometry are circles, triangles, parallel lines. Congruency and Similarity of triangles is a chapter that is being intensively concentrated on geometry.
Below, the details about Geometry under QA for different competitive MBA exams are given: –
CAT :
Geometry | ||||
Year | No of Questions | Good attempt | Difficulty | |
2019 | Slot 1 | 6 | 4 | Moderate |
Slot 2 | 6 | 3 to 4 | Moderate | |
2018 | Slot 1 | 6 to 7 | 4 to 5 | Easy |
Slot 2 | 4 | 2 to 3 | Moderate | |
2017 | Slot 1 | 7 to 8 | 4 to 5 | Moderate |
Slot 2 | 7 | 5 to 6 | Easy |
XAT :
Geometry | |||
Year | No of Questions | Good attempt | Difficulty |
2020
| 6
| 4 to 5
| Moderate
|
2019
| 4
| 2 to 3
| Moderate-Difficult
|
2018
| 6
| 3
| Moderate-Difficult
|
SNAP :
Geometry | |||
Year | No of Questions | Good attempt | Difficulty |
2019 | 6 | 4 to 5 | Easy |
2018 | 7 | 3 to 4 | Moderate |
2017 | 7 | 3 to 4 | Moderate to Difficult |
The list of concepts that are covered in the Geometry chapter is as follows: –
Altitude
Orthocentre
Perpendicular Bisector
Circumcentre
Median
Some of the important CAT questions on Geometry that appeared in the previous papers are: –
1. Let A and B be two regular polygons having a and b sides, respectively. If b = 2a and each interior angle of B is 3/2 times each interior angle of A, then each interior angle, in degrees, of a regular polygon with a + b sides is? (CAT 2019 – SLOT 2)
Each interior angle of n sided polygon = (n-2)*180/n
(b-2)*180/b = 3/2 (a-2)*180/a and b = 2a
So, 2a-2/2 = (a-2)*3/2
2a – 2 = 3a – 6
a = 4 and b = 8
So, a + b = 12, a 12-sided polygon
Each interior angle = (12-2)*180/12 = 150 degrees
2. Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm. If AP is perpendicular to BC, then the maximum possible length of AP, in cm, is? (CAT 2019 – SLOT 2)
Let AB = a and AC = b
A^2 + b^2 = 400 by pythagoras
Let AP = x and BP = y, so CP= 20-y
So, by Pythagoras
X^2 + y^2 = a^2 and b^2 = x^2 + (20-y)^2
Maximum possible value of AP occurs when a = b, so a= b=10 root 2
On solving, we get AP = 10 units
3. In a triangle, ABC, medians AD and BE are perpendicular to each other and have lengths 12 cm and 9 cm, respectively. Then, the area of triangle ABC, in sq. cm is? (CAT 2019 – SLOT 2)
4. Let ABCD be a rectangle inscribed in a circle of a radius 13 cm. Which one of the following pairs can represent, in cm, the possible length and breadth of ABCD? (CAT 2018 – SLOT 1)
10^2+24^2=26^2
5. In a circle, two parallel chords on the same side of diameter have lengths 4 cm and 6 cm. If the distance between these chords is 1 cm, then the radius of the circle, in cm, is? (CAT 2018 – SLOT 1)
Level 1
Level 2
Level 3
Inspiring Education… Assuring Success!!
Ⓒ 2020 – All Rights Are Reserved