Practice Model question set 3 - non verbal reasoning Online Quiz (set-1) For All Competitive Exams

Horizontal lines = BL, CK, DJ, EI = 4

Slant lines = AE, LF, KG, AI, BH, CG, DF = 7

∴ Total number of straight lines = 4 + 7 = 11

In the given diagram, there are 22 triangles which can be better understood with the help of following diagram.

In the above figures,

Triangles consisting of 1 unit are ΔADG, ΔAGH, ΔAHE, ΔDMF, ΔFME, ΔDIB, ΔIBJ, ΔJBF, ΔCFL, ΔCLK and ΔCKE i.e., 11 triangles.

Triangles consisting of 2 units are ΔADH, ΔAGE, ΔDFE, ΔDBJ, ΔIBF, ΔECL and ΔCKF i.e., 7 triangles.

Triangles consisting of 3 units are ΔADE, ΔBDF and ΔCFE i.e., 3 triangles.

Triangles consisting of whole figure is ΔABC i.e., 1 triangle.

So, the total number of triangles = 11 + 7 + 3 + 1 = 22.

Pentagons in the figure as follow

Smallest pentagons = KLABM, BCDEN, EFGHO, HIJKP = 4

Medium pentagons = KMBEH, BNEHK, EOHKB, HPKBE = 4

Largest pentagons = KMBFG, BNEIJ, EOHLA, HPKCD = 4

∴Total pentagons = 4 + 4 + 4 = 12

The figure, in question may be labelled as shown in following figure.

There are 12 triangles in the figure, namely ΔEOG, ΔAGK, ΔKOR, ΔKGT, ΔTUP, ΔRUS, ΔSPV, ΔSUW, ΔWNQ, ΔVNH, ΔHQJ and ΔHNC.

The figure in question has been labelled as shown in the following figure

Smallest Triangles = ΔAEF, ΔEPR, ΔFSQ, ΔPRD, ΔRSD, ΔSQD = 6

Single Triangles = ΔPBD, ΔQDC, ΔEFD = 3

Triangles formed with two triangles = ΔPSD, ΔRQD, ΔFDQ, ΔEDP = 4

Triangles formed with three or four triangles = ΔAPQ, ΔFDC, ΔBDE, ΔPQD = 4

Largest Triangles = ΔABC = 1

∴ Total Triangles = 6 + 3 + 4 + 4 + 1 =18

We may the label the figure as shown

The Simplest triangles are ADE,BDF, DEF and EFC i.e., 4 in number

There is only one triangle ABC composed of four components.

Thus, there are 4 + 1 = 5 triangles in the given figure.

We may the label the figure as shown

The Simplest triangles are ABG, BCG, CGE, CDE, AGE and AEF i.e., 6 in number.

The triangles composed of two components each are ABE, ABC, BCE and ACE i.e., 4 in number.

Thus, there are 6 + 4 = 10. triangles in the given figure.

There are a total of 19 squares in the figure, which have been marked as shown in the following figure.

There are three suqares in the upper row □EFIH, □FGJI, □GBLJ.

There are five suqares in the middle row namely □TQRS, □QHPR, □HIOP, □IJNO, □JLMN

There are three suqares in the lower row namely □SRUD, □RPVU, □POWV

By combining two upper rows there are three suqares namely □AEPS, □EGNP, □FBMO.

By combining two lower rows there are three suqares namely □ THVD, □ QIWU, □ ILCW

From all the three rows combined together there are two suqres namely □ AFWD and □ EBCV

∴Total number of squares = 3 + 5 + 3 + 3 + 3 + 2 = 19

The figure, in question has been labelled at different points as shown in the following figure

From the above diagram parallelograms are as follow

i) Parallelograms from the outer figure = ▱ABCD, ▱AIJD, ▱GBCH, ▱ADHG, ▱GHJI, ▱IJCB = 6

ii) Parallelograms from the upper half of the figure = ▱AGKE, ▱GILK, ▱IBFL, ▱AILE, ▱GBFK, ▱ABFE = 6

iii) Parallelograms from the lower half of the figure = ▱EKHD, ▱KLJH, ▱LFCJ, ▱KFCH, ▱ELJD, ▱EFCD = 6

∴ there are a total of 18 parallelograms in the figure.

Shortcut Method

Number of parallelograms = (3 + 2 + 1) × (2 + 1)

[∵ C = 3, R = 2]

= 6 × 3 = 18

The figure may be labelled as shown

The Simplest triangles are AHL, LHG, GHM, HMB, GMF, BMF, BIF, CIF, FNC, CNJ, FNE, NEJ, EKJ and JKD i.e.,14 in number.

The triangles composed of two components each are AGH, BHG, HBF, BFG, HFG, BCF, CJF, CJE, JEF, CFE and JED i.e.,11 in number.

The triangles composed of four components each are ABG, CBG, BCE and CED i.e.,4 in number.

∴ Total number of triangles in the given figure = 14 + 11 + 4 = 29.