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Reasoning and General Intelligence  Study Materials:

 Quantitative Aptitude Study Materials:

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Here, we have provided the links which contains the study materials which will help you in studying and preparing for your examinations of the TISS RAT. Referring to the links we’ve provided below and the links which contains the study materials of TISS RAT. PDF Format along with the list of recommended books which we’ve provided below, you will be able to ace your examinations. We have also provided you the further details which will allow you to do well in your exams and learn more. These study materials help you understand the concepts and everything easily and creates a better space for you to work on. These study materials give you the best resources to study from.

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Name of BoardTata Institute of Social Sciences (TISS)
Exam NameResearch Aptitude Test (RAT)
Course NameM.Phil / Ph.D
Status Model Question Paper Released

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Name of the BoardChennai Petroleum Cooperation Limited (CPCL)
Post NameWorkmen & Deputy Company Secretary
Vacancy56
Notification PDFDownload
StatusSyllabus & Exam Pattern Released

Production/ Production TYrainee Syllabus PDF

Instrumentation Discipline Syllabus PDF

Electrical Discipline Syllabus PDF

Production Discipline Syllabus PDF

Quality Control Discipline Syllabus PDF

Mechanical Discipline Syllabus PDF

Purchase & Store Discipline Syllabus PDF

Marketing Discipline Syllabus PDF

Accounts Discipline Syllabus PDF

Nursing Discipline Syllabus PDF

Here, we have provided the links which contains the study materials which will help you in studying and preparing for your examinations of the SBI SO and Clerical Cadre. Referring to the links we’ve provided below and the links which contains the study materials of SBI SO and Clerical Cadre. PDF Format along with the list of recommended books which we’ve provided below, you will be able to ace your examinations. We have also provided you the further details which will allow you to do well in your exams and learn more. These study materials help you understand the concepts and everything easily and creates a better space for you to work on. These study materials give you the best resources to study from.

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Details

Name of the BoardState Bank of India
Post nameSO & Clerical Cadre and other posts
Vacancies106
Last Date to apply12.02.2020
NotificationDownload here
StatusSyllabus & Exam Pattern are available

Reasoning:

Quantitive Aptitude:

English Language:

General Awareness:

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Name of the BoardRajasthan Subordinate and Ministerial Service Selection Board
Post NameJunior Engineer
Vacancy1054
StatusSyllabus Available

Exam Pattern

PartsSubjectMarksDuration
Part AGeneral Knowledge402 Hours
Part BRelated Subjects80

General Knowledge Syllabus:

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Solved Example

  1. In the adjacent figure lines AB and CD intersects at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40° then find∠BOE and reflexive ∠COE                       

Solution: ∠AOC = ∠BOD (Vertically opposite angle)                       

∴ ∠AOC = 40°

According to question, ∠AOC + ∠BOE = 70° or,

40° + ∠BOE = 70°

∴∠BOE = 30°

Now, ∠COD + ∠DOB + ∠BOE = reflexive ∠COE

∴180° + 40° + 30° = reflexive∠COE

∴reflexive ∠COE = 250°

  1. In the figure given below lines XY and MN intersect at O.

If LPOY = 90° and a : b = 2 : 3 then find the measure of c.          

Solution: Given,

Let a = 2k and b = 3k

∵∠POY = 90°                   ∴ ∠POX = 90°

or, ∠a + ∠b = 90°              or, 2k + 3k = 90°

or, 5k = 90°                       or, = 18°

∴∠b = 3k = 3 × 18° = 54°

Now, ZXOM = LYON (Vertically opposite angle)

∴∠YON = 54°

Again, ∠XON + ∠YON = 180° (Linear pair of angles axiom)

or, c + 54° = 180°              ∴c = 126°

  1. Given that ∠XYZ = 64° and line XY is produced to point P. Draw a diagram from the given information. If ray YQ bisects ∠ Then find the measure of ∠XYQ and reflex ∠QYP.

Solution: ∠XYZ + ∠ZYQ + ∠QYP = 180° (Linear pair of angles)

or, 64° + 2∠ZYQ = 180° (∵∠XYZ = 64° and ∠ZYQ = ∠QYP)

or, 2∠ZYQ = 116°

∴∠ZYQ = 58°;

∵ ∠XYQ = ∠XYZ + ∠ZYQ

∴∠XYQ = 64° + 58° = 122°

Now reflex, ∠QYP = ∠PYX + ∠XYQ

= 180° + 122° = 302°                                                              

  1. In the figure given below find x and y and hence prove that AB||CD.

Solution:∠DRS = ∠CRQ (Vertically opposite angle)

∴y = 130°

Also, ∠ASP + ∠ASR = 180°

(Linear pair of angles)

or, 50° + x = 180°

or, x = 130°

∵x = y = 130°

AB||CD (Alternate angle)

  1. In the given figure if PQ||ST, ∠PQR = 110° and ∠RST = 130°, Find ∠QRS

Solution: From point R draw a line RM parallel ST.

∠RST + ∠SRM = 180° 1 (Consecutive interior angle)

or, 130° + ∠SRM = 180° ∴∠SRM = 50° … (i)

Now, ∠QRM = ∠PQR

or, ∠QRM = 110°             or, ∠QRS + ∠SRM = 110°

or, ∠QRS + 50° = 110° (from (i)) ∴∠QRS = 60°

  1. In the given figure if AB||CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

Solution: AB||CD             ∴ LAPR = LPRD (Alternate angle)

or, ∠APQ + ∠QPR = ∠PRD

or, 50° + y = 127° or, y = 77°

and ∠APQ = ∠PQR         (Alternate angle)

or, 50° = x, ∴ x = 50°

  1. In ∠ABC, ∠A = 40°. If bisector of AB and AC meets at O then prove that ∠BOC = 110°

Solution: In∠ABC,

∠A + ∠B + ∠C = 180°

or, 40° + ∠B + ∠C = 180°

or, ∠B + ∠C = 140°

or,                                       … (i)

Now, In ∆BOC, ∠OBC + ∠OCB + ∠BOC = 180°

or,  + ∠BOC = 180°or, 70° + ∠BOC = 180°                                           (from (i))

∠BOC = 110° Proved.

[Shortcut: BOC = 90° += 90° + 20° = 110° Proved.]

  1. If angles of a triangle are is the ratio 2:3:4, then find the least and greatest angle.

Solution: Let angle be 2x°, 3x° and 4x°.

∴2x° + 3x° + 4x° = 180°

or, 9x° = 180°                    or, x° = = 20°

∴least angle = 2x° = 2 × 20° = 40°

greatest angle = 4x° = 4 × 20° = 80°

  1. The exterior angle of a triangle is 110° and one of its interior opposite angle is 30°, find other angles.

Solution: Consider the triangle ABC in which exterior ∠ACD = 110° and ∠A = 30°, we have to find ∠B and ∠C.                                                              

∵∠ABC + ∠BAC = ∠ACD

or, ∠ABC + 30° = 110°

or, ∠ABC = 80°

or, ∠ACB + ∠ACD = 180°

or, ∠ACB + 110° = 180°

∵ ∠ACB = 70°

  1. In triangle PQR, sides QP and RQ respectively produced to point S and T. If ∠SPR = 135° and ∠PQT = 110° find ∠PRQ                                             

Solution:∠SPR + ∠QPR = 180° (linear pair of angles)

or, 135° + ∠QPR = 180°

or, ∠QPR = 45°                                        … (i)

Now, ∠QPR + ∠PRQ = ∠PQT

or, 45° + ∠PRQ = 110° (from(i))

∴∠PRQ = 65°

  1. In the adjacent figure ∠X = 62° and ∠XYZ = 54°. If VO and ZO respectively bisects ∠XYZ and ∠XZY then find ∠OZY and ∠YOZ                                       

Solution: In ∆XYZ, ∠YXZ + ∠XYZ + ∠XZY = 180°

or, 62° + 54° + ∠XZY = 180°

∴ ∠XZY = 64°

From question,

∠OYZ = ∠XYZ = × 54° = 27°

and ∠OZY = ∠XZY = × 64° = 32°

Now, In ∆OYZ,

∠OYZ + ∠OZY + ∠YOZ = 180°

or, 27° + 32° + ∠YOZ = 180°

∴∠YOZ = 121°

  1. In the given figure lines PQ and RS intersect at point T such that ∠PRT = 40°; ∠RPT = 95° and ∠TSQ = 75° Find ∠SOT                                          

Solution: In ∆PRT, ∠RPT + ∠PRT = ∠PTS

or, 95° + 40° = ∠PTS

∠PTS = 135°                                             … (i)

Again in,

∆TSQ, ∠TSQ + ∠SQT = ∠PTS

or, 75° + ∠SQT = 135° (∵∠TSQ = 75° and ∵∠PTS = 135°)

∠SQT = 135° – 75° = 60°.

  1. In the figure given below if PQ⊥PS, PQ||SR, ∠SQR = 28° and ∠QRT = 65° find x and y.

Solution:∵PQ||SR

∠PQR = ∠QRT (Alternate angle)

or, x + 28° = 65°

or, x = 65°-28° = 37°                                 … (i)

Now, In

∆PQS, ∠SPQ + ∠PQS + ∠QSP = 180°

or, 90° + x + y = 180°

or, 90° + 37° + y = 180°                (from (i))

or, y = 180° -127° = 53°

Hence, x = 37° and y = 53°.