Practice Counting figures - non verbal reasoning Online Quiz (set-1) For All Competitive Exams
Q-1)
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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.
Q-2)
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(d)
Q-3)
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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.
Q-4)
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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
Q-5)
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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.
Q-6)
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The figure in question has been labelled as shown in the following figure
Smallest Triangles = ΔEAP, ΔEDP, ΔAGB, ΔHFG, ΔGBF, ΔPHD = 6
Triangles formed with two triangles = ΔEAD, ΔAGD, ΔBHF, ΔEHD, ΔAPF, ΔABF= 6
Largest Triangles = ΔBDC, ΔAEF, ΔABD = 3
∴ Total Triangles = 6 + 6 + 3 = 15
Q-7)
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Total 28 triangles are as follow
| ΔABG, | ΔAFG, | ΔCHD, | ΔHDE, | ΔBGI, |
| ΔBCI, | ΔHCI, | ΔHGI, | ΔGHJ, | ΔHEJ, |
| ΔEFJ, | ΔGFJ, | ΔABF, | ΔCDE, | ΔBCG, |
| ΔBCH, | ΔHCG, | ΔBHG, | ΔGHE, | ΔHEF, |
| ΔGFE, | ΔGHF, | ΔABH, | ΔAFH, | ΔCDG, |
| ΔGDE, | ΔBHF and | ΔCGE |
Q-8)
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There are total 27 triangles in the given figure.
From this figure it is clear that,
Triangles formed by single units = ΔABC, ΔBDE, ΔBCE, ΔCEF, ΔDGH, ΔDHE, ΔHEI, ΔEIF, ΔFIJ, ΔGKL, ΔGLH, ΔLHM, ΔHMI, ΔMIN, ΔINJ, Δ NJO i.e., 16 triangles.
Triangles formed by Combining 4 triangles are ΔADF, ΔBGI, ΔCHJ, ΔDKM, ΔLEN, ΔMFO and ΔDFM i.e., 7 triangles.
Triangles formed by Combining 9 triangles are ΔAGJ, ΔBKN and ΔCLO i.e., 3 triangles.
One Triangle ΔAKO formed by complete figure
∴ Total number of Triangles = 16 + 7 + 3 + 1
Q-9)
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Smallest hexagons = MBCQPN, OCDSRP, TNPVIJ, UPRWHI = 4
Medium hexagons = KLBCPJ, JNCDRI, IPDEFH, GFDCPH, HRCBNI, IPBALJ, LBCRIJ, NCDFHI, MBDSRN, TNRWHJ = 10
Largest hexagons = LBDFHJ = 1
∴Total hexagons = 4 + 10 + 1 = 15
Q-10)
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The figure in the question may be labelled as shown in the following figure.
There are nine triangles in the upper half of the figure = ΔAEI, ΔAIG, ΔAEG, ΔGEO, ΔGBJ, ΔBFJ, ΔGBF, ΔGOF and ΔGEF.
Similarly, there are nine triangles in the lower half figure i.e., EFCD.
There are two more triangles ΔEGH and ΔFGH.
Hence, there are atotal of 20 triangles.
Q-11)
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Q-12)
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The figure in question looks like as shown below after labelling it.
Smallest Triangles = ΔGAH, ΔHFL, ΔLCI, Δ IDJ ΔJBK, ΔKEG = 6
Largest Triangles = ΔABC, ΔEFD = 2
∴ Total Triangles = 6 + 2 = 8
Q-13)
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The figure may be labelled as shown
The Simplest triangles are ABF, BFG, BCG, CGH, GHD, GED, EFG and AFE i.e., 8 in number.
The triangles composed of two components each are ABG, BGE, AGE, ABE and GCD i.e., 5 in number.
The triangles composed of three components each are BCD, CDE, BED and BCE i.e., 4 in number.
Thus, there are 8 + 5 + 4 = 17 triangles in the given figure.
Q-14)
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We shall join the centres of all the circles by horizontal and vertical lines and then label the resulting figure as shown.
The simplest squares are ABED, BCFE, DEHG, EFIH, GHKJ and HILK i.e., 6 in number.
The squares composed of four simple squares are ACIG and DFLJ i.e., 2 in number.
Thus, 6 + 2 = 8 squares will be formed.
Q-15)
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Given figure can be labelled as follows
In the above figure, triangles formed by single units are ΔABE, ΔAEO, ΔAOF, ΔAFC, ΔDEB, ΔDOE, ΔDFO and ΔDCF i.e., 8 triangles.
Triangles formed by Combining 2 units are ΔABO, ΔAEF, ΔAOC, ΔBOD, ΔDEF, ΔDOC, ΔAED and ΔADF i.e., 8 triangles.
Triangles formed by Combining 3 units are ΔABF, ΔAEC, ΔBDF and ΔEDC i.e., 4 triangles.
Triangles formed by Combining 4 units are ΔABC, ΔBCD, ΔABD and ΔACD i.e., 4 triangles.
∴ Total number of Triangles = 8 + 8 + 4 + 4 = 24
Q-16)
(a)
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The figure, in question may be labelled as shown in following figure.
There are 14 squares in the figure, namely
| □ABCD, | □DCFE, | □CGHF, | □EFPR, |
| □FHQP, | □HISQ, | □RPNO, | □PQMN, |
| □MQSL, | □SJKL, | □DGQR, | □EHMO, |
| □FILN and | □FQNR. |
Q-17)
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The figure may be labelled as shown
The horizontal lines are IK, AB, HG and DC i.e., 4 in number
The vertical lines are AD, EH, JM, FG and BC i.e., 5 in number
The slanting lines are IE, JE, JF, KF, DE, DH, FC and GC i.e., 8 in number
Thus, there are 4 + 5 + 8 = 17 straight lines in the figure.
Q-18)
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We shall label the figure as shown
The Simplest rectangles are ABJI, BCKJ, IJFG and JKEF i.e.,4 in number.
The rectangles composed of two components each are ACKI, BCEF, IKEG and ABFG i.e.,4 in number.
The only rectangle composed of four components is ACEG.
Thus, there are 4 +4 + 1 = 9. rectangles in the given figure.
Q-19)
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The figure may be labelled as shown
The Simplest triangles are EFH, BIC, GHJ, GIJ, EKD and CKD i.e.,6 in number.
The triangles composed of two components each are ABJ, AFJ, GCK, GEK, CED and GHI i.e.,6 in number.
The triangles composed of three components each are GCD, GED, DJB and DJF i.e.,4 in number.
The triangles composed of four components each are ABF and GCE i.e.,2 in number.
The triangles composed of five components each are ABD and AFD i.e.,2 in number.
There is only one triangle i.e., FBD composed of six components.
Thus, there are 6 + 6 + 4 + 2 + 2 + 1 = 21 triangles in the given figure.
Q-20)
(a)
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The figure may be labelled as shown
Triangles
The Simplest triangles are JBO, BKO, KDO, DFO, FGO, GHO, HIO, IJO, ABJ, BCK, CKD and DEF i.e.,12 in number.
The triangles composed of two components each are IBO, BDO, DGO, GIO, ABO, CBD and DEO i.e.,7 in number.
The triangles composed of four components each are IBD, BDG, DGI, GIB, ACO and COE i.e.,6 in number.
There is only one triangle ACE which is composed of eight components.
Thus, there are 12 + 7 + 6 + 1 = 26 triangles in the given figure.
Squares
The squares composed of two components each are BKOJ, KDFO, OFGH and JOHI i.e.,4 in number.
There is only one square i.e., CDOB composed of four components.
There is only one square i.e., BDGI composed of eight components.
Thus, there are 4 + 1 + 1 = 6 squares in the given figure.