Counting Figures Model Question Set 1 Practice Questions Answers Test With Solutions & More Shortcuts
COUNTING FIGURES PRACTICE TEST [3 - EXERCISES]
Counting Figures Model Question Set 1
Counting Figures Model Question Set 2
Counting Figures Model Question Set 3
Question : 1
a) a
b) b
c) c
d) d
Answer »Answer: (c)
We may the label the figure as shown
The Simplest triangles are AGH, GFO, LFO, DJK, EKP, PEL and IMN i.e., 7 in number.
The triangles composed of two components each are GFL, KEL, AMO, NDP, BHN, CMJ, NEJ and HFM i.e., 8 in number.
The triangles composed of three components each are IOE, IFP, BIF and CEI i.e., 4 in number.
The triangles composed of four components each are ANE and DMF i.e., 2 in number.
The triangles composed of five components each are FCK, BGE and ADL i.e., 3 in number.
The triangles composed of six components each are BPF, COE, DHF and AJE i.e., 4 in number.
Thus, there are 7 + 8 + 4 + 2 + 3 + 4 = 28. triangles in the given figure.
Question : 3
a) a
b) b
c) c
d) d
Answer »Answer: (a)
The figure may be labelled as shown
The Simplest triangles are AHG, AIG, AIB, JFE, CJE and CED i.e., 6 in number.
The triangles composed of two components each are ABG, CFE, ACJ and EGI i.e., 4 in number.
The triangles composed of three components each are ACE, AGE and CFD i.e., 3 in number.
There is only one triangle i.e., AHE composed of four components.
Thus, there are 6 + 4 + 3 + 1 = 14 triangles in the given figure.
Question : 4
a) a
b) b
c) c
d) d
Answer »Answer: (a)
Smallest Triangles = ΔBOC, ΔCOD, ΔFOG, ΔGOH, ΔHOA, ΔAOB = 6
Triangles formed with two small triangles = ΔAOG, ΔAOC = 2
Largest Triangles = ΔACG, ΔEGC = 2
∴ Total Triangles = 6 + 2 + 2 = 10
Question : 5
a) a
b) b
c) c
d) d
Answer »Answer: (b)
The figure may be labelled as shown
Triangles
The Simplest triangles are KJN, KJO, CNB, OEF, JIL, JIM, BLA and MFG i.e.,8 in number.
The triangles composed of two components each are CDJ, EDJ, NKO, JLM, JAH and JGH i.e.,6 in number.
The triangles composed of three components each are BKI, FKI, CJA and EJG i.e.,4 in number.
The triangles composed of four components each are CDE and AJG i.e.,2 in number.
The only triangle composed of six components is BKF.
Thus, there are 8 + 6 + 4 + 2 + 1 = 21. triangles in the given figure.
Parallelograms:
The Simplest ||gms are NJLB and JOFM i.e.,2 in number.
The ||gms composed of two components each are CDKB, DEFK, BIHA and IFGH i.e.,4 in number.
The ||gms composed of three components each are BKJA, KFGJ, CJIB and JEFI i.e.,4 in number.
There is only one ||gm i.e., BFGA composed of four components.
The ||gms composed of five components each are CDJA, DEGJ, CJHA and JEGH i.e.,4 in number.
The only ||gm composed of six components is CEFB.
The only ||gm composed of ten components is CEGA.
Thus, there are 2 + 4 + 4 + 1+ 4 + 1 + 1 = 17. ||gms in the given figure.
IMPORTANT NON VERBAL REASONING EXERCISES
Counting Figures Model Question Set 1 Online Quiz
Click to Start..NON VERBAL REASONING CATEGORIES
RECENT BLOGS
Recently Added Subject & Categories For All Competitive Exams
Direction and Distance Sense Test for IBPS PO MAINS 2025
Direction and Distance Sense Test MCQ questions and easy explanation, download PDF for IBPS PO MAINS 2025. Reasoning model exercises all Competitive Exams
Continue Reading »
Top 100 Ratio Proportion MCQ, PDF for IBPS PO MAINS 2025
Get answers to the Ratio Proportion MCQ questions and easy explanation, download PDF for IBPS PO MAINS 2025. Attitude model exercises all Competitive Exams
Continue Reading »
Important 100+ Synonyms MCQ, PDF for IBPS PO MAINS 2025
Get answers to the Synonyms MCQ questions and easy explanation, download PDF for IBPS PO MAINS 2025. General English model exercises all Competitive Exams
Continue Reading »
IBPS PO MAINS 2025: Blood Relations Reasoning MCQ, PDF
Get answers to the Blood Relations MCQ questions and easy explanation, download PDF for IBPS PO MAINS 2025. Reasoning model exercises all Competitive Exams
Continue Reading »