Time and Work / Pipe and Cistern
Square, Cube, Indices & Surds
Geometry
Average
Miscellaneous
Number Series
Algebraic Expressions and Inequalities
LCM, HCF and Fraction
Probability
Simple Interest
Permutation and Combination
Alligation and Mixture
Ratio & Proportion
Compound Interest
Problem Based on Ages
Quantitative Aptitude
Simplification
Distance, Speed and Time
Number System
Profit, Loss & Discount
Mensuration
Simple & Compound Interest
Data Interpretation
Percentage

Socio-Eco-Political Environment of India
Books & Authors
Events/ Organisation/ Summits
Micro Finance & Economics
Awards & Honours
Foreign Trade
History of Banking & its Development
Current Banking
Banking Terminologies
Government Schemes
Banking Product & Services
Country and Capital
Sports & Games
Appointment/ Election/ Resignation
RBI & its Monetary Policies
Science & Technology
General Awareness
Current Affairs
Agriculture
Financial Awareness
Marketing

Reading Comprehension
Direct and Indirect Speech
Deriving conclusion from passages
Correct Usage of Preposition
Sentence Improvement
Homonyms
Sentence Completion
Para Jumbles
Active and Passive Voice
Synonyms & Antonyms
Passage Completion
One Word Substitution
Theme Detection
English Language
Synonyms
Vocabulary Test
Double Synonms
Antonyms
Cloze Test
Similar Substitution
Idioms and Phrases
Spotting the Errors
Spelling
English Language / Hindi Language

Fundamentals of Marketing, Product and Branding
Modern Marketing/ Marketing in Banking Industry
Logic Gates
Computer Awareness
Computer Fundamental/ Binary System/ OS
DBMS (Database management system)
Internet, Networking & Computer Abbreviations
Market Situation, Distribution, Promotion & Advertising
MS Office/ Commands & Shortcut Keys
Market Segmentation, Targeting & Positioning
Computer Architecture
Softwares/ Programming
Professional Knowledge (IT)
Data Structure Compiler

Geometry

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 1. If (5, 1), (x, 7) and (3, -1) are 3 consecutive verticles of a square then x is equal to :

-3

-4

5

6

None of these

For the verticles to form a square, we know that the length of each side of the square should be equal. Therefore,

(x - 5){tex}^2{/tex} + (7 - 1){tex}^2{/tex} = (x - 3){tex}^2{/tex} + (7 + 1){tex}^2{/tex}

[x{tex}^2{/tex} + 5{tex}^2{/tex} - 2 (x) (5) ] + [36] = [x{tex}^2{/tex} + 3{tex}^2{/tex} - 2 (x) (3) ] + [64]

[5{tex}^2{/tex} + 36] - [9 + 64] = (10 - 6) x

x = -{tex}\frac{12}{4}{/tex} = - 3

This gives the side of the square x = - 3.

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 2. What is the area of an obtuse angled triangle whose two sides are 8 and 12 and the angle included between two sides is 150°?

24 sq units

48 sq units

24{tex}\sqrt 3{/tex}

48{tex}\sqrt 3{/tex}

Such a triangle does not exist

If two sides of a triangle and the included angle 'y' is known, then the area of the triangle can be found using the formula

{tex}\frac{1}{2}{/tex}* (product of sides) * sin y

Substituting the values in the formula, we get {tex}\frac{1}{2}{/tex}* 8 * 12 * sin 150 = 24 sq units

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 3. What is the measure of the radius of the circle that circumscribes a triangles whose sides measure 9, 40 and 41?

6

4

24.5

20.5

12.5

From the measure of the length of the sides of the triangle 9, 40 and 41 we can infer that the triangle is a right angled triangle. 9, 40, 41 is a Pythagorean triplet.

In a right angled triangle, the radius of the circle that circumscribes the triangle is half the hypotenuse.

In the given triangle, the hypotenuse = 41

Therefore, the radius of the circle that circumscribes the triangle ={tex}\frac{41}{2}{/tex} = 20.5 units

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 4. Verticles of a quadrilateral ABCD are A (0, 0), B (4, 5), C (9, 9) and D (5, 4). What is the shape of the quadrilateral?

Square

Rectangle but not a square

Rhombus

Parallelogram but not a rhombus

None of these

The lengths of the four sides AB, BC, CD and DA are all equal to {tex}\sqrt{41}{/tex}.

Hence, the given quadrilateral is either a Rhombus or a Square.

Now let us compute the lengths of the two diagonals AC and BD.

The length of AC is {tex}\sqrt{162}{/tex} and the length of BD is {tex}\sqrt{2}{/tex}

As the diagonals are not equal and the sides are equal, the given quadrilateral is a Rhombus.

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 5. If the sum of the interior angles of a regular polygon measures upto 1440 degrees, how many sides does the polygon have?

10 sides

8 sides

12 sides

9 sides

None of these

We know that the sum of an exterior angle and an interior angle of a polygon = 180°

We also know that sum of all the exterior angles of a polygon = 360°

The question states that the sum of all interior angles of the given polygon = 1440°

Therefore, sum of all the interior and exterior angles of the polygon = 1440 + 360 = 1800

If there are ‘n' sides to this polygon, then the sum of all the exterior and interior angles = 180 x n = 10

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 6. What is the radius of the in circle of the triangle whose sides measure 5, 12 and 13 units?

2 units

12 units

6.5 units

6 units

7.5 units

The triangle given is a right angled triangle as its sides are 5, 12 and 13 which is one of the Pythagorean triplets.

Note: In a right angled triangle, the radius of the incircle is given by the following relation

{tex}\frac{sum\ of\ perpendicular\ sides - hypotenuse}{2}{/tex}

As the given triangle is a right angled triangle, radius of its incircle = {tex}\frac{5+12-13}{2}{/tex}= 2 units

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 7. How many diagonals does a 63 sided convex polygon have?

3780

1890

3843

3906

1953

The number of diagonals of an n-sided convex is {tex}\frac{n(n-3)}{2}{/tex}

This polygon has 63 sides. Hence, {tex}n{/tex} = 63

Therefore, number of diagonals = {tex}\frac{63\times60}{2}{/tex} = 1890

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 8. If 10, 12 and ‘x' are sides of an acute angled triangle, how many integer values of ‘x' are possible?

7

12

9

13

11

For any triangle sum of any two sides must be greater than the third side.

The sides are 10, 12 and ‘x'.

From Rule 2, x can take the following values : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21 - A total of 19 values.

When x = 3 or x = 4 or x = 5 or x = 6, the triangle is an OBTUSE angled triangle.

The smallest value of x that satisfies both conditions is 7. (10{tex}^{2}{/tex} + 7{tex}^{2}{/tex} > 12{tex}^{2}{/tex})

The highest value of x that satisfies both conditions is 15. (10{tex}^{2}{/tex} + 12{tex}^{2}{/tex} + 15{tex}^{2}{/tex})

When x = 16 or x = 17 or x = 18 or x = 19 or x = 20 or x = 21, the triangle is an OBTUSE angled triangle.

Hence, the values of x that satisfy both the rules are x = 7, 8, 9, 10, 11, 12, 13, 14, 15. A total of 9 values.

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 9. Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are 6 inches and 8 inches.

10 inches

11 inches

18 inches

20 inches

None of these

Test the ratio of the lengths to see if it fits the 3n : 4n : 5n ratio.

6 : 8 : ?= 3 (2) : 4 (2) : ?

Yes, it is a 3-4-5 triangle for n =

Calculate the third side 5n = 5 x 2 = 10

The length of the hypotenuse is 10 inches.

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 10. Find the length of one side of a right triangle if the length of the hypotenuse is 15 inches and the length of the other side is 12 inches.

8 inches

7 inches

9 inches

13 inches

None of these

Test the ratio of the lengths to see if it fits the 3n : 4n : 5n ratio.

? : 12 : 15 = ? : 4 (3) : 5 (3)

Yes, it is a 3-4-5 triangle for n = 3

Calculate the third side 3n = 3 x 3 = 9

The length of the side is 9 inches.

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 11. Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are both 3 inches.

{tex}5{/tex} inches

{tex}3\sqrt{4}{/tex} inches

{tex}6{/tex} inches

{tex}3\sqrt{2}{/tex} inches

None of these

This is a right triangle with two equal sides so it must be a 45° - 45° - 90° triangle. You are given that the both the sides are 3. If the first and second value of the ratio n : n : n{tex}\sqrt{2}{/tex} is 3 then the length of the third side is 3{tex}\sqrt{2}{/tex}

The length of the hypotenuse is 3{tex}\sqrt{2}{/tex} inches.

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 12. Find the lengths of the other two sides of a right triangle if the length of the hypotenuse is 4{tex}\sqrt{2}{/tex} inches and one of the angles is 45°.

4 inches

9 inches

8 inches

7 inches

None of these

This is a right triangle with a 45° so it must be a 45° - 45° - 90° triangle.

You are given that the hypotenuse is 4{tex}\sqrt{2}{/tex} . If the third value of the ratio n : n : n{tex}\sqrt{2}{/tex} is 4{tex}\sqrt{2}{/tex} then the lengths of the other two sides must 4.

The lengths of the two sides are both 4 inches.

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 13. Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are 4 inches and 4{tex}\sqrt{3}{/tex} inches.

8 inches

9 inches

10 inches

11 inches

None of these

Test the ratio of the lengths to see if it fits the n : n{tex}\sqrt{3}{/tex} : 2n ratio.

4 : 4{tex}\sqrt{3}{/tex} : ? n : n{tex}\sqrt{3}{/tex} : 2n

Yes, it is a 30° - 60° - 90° triangle for n = 4

Calculate the third side

2n = 2 x 4 = 8

The length of the hypotenuse is 8 inches.

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 14. What is the area of the following square, if the length of BD is 2{tex}\sqrt{2}{/tex} ?

1

2

3

4

5

We need to find the length of the side of the square in order to get the area.

The diagonal BD makes two 45° - 45° - 90° triangles with the sides of the square.

Using the 45° - 45° - 90° special triangle ratio n: n : n{tex}\sqrt{2}{/tex}. If the hypotenuse is 2{tex}\sqrt{2}{/tex} then the legs must be 2. So, the length of the side of the square is 2.

Area of square = 5{tex}^{2}{/tex} = 2{tex}^{2}{/tex} = 4

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 15. In the figure below, what is the value of y?

40

50

60

100

120

Vertical angles being equal allows us to fill in two angles in the triangle that y° belongs to.

Sum of angles in a triangle = 180°

So, y° + 40° + 80° = 180°

y° + 120° = 180°

y° = 60°

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 16. Two circles both of radii 6 have exactly one point in common. If A is a point on one circle and B is a point on the other circle, what is the maximum possible length for the line segment AB?

12

15

18

20

24

Sketch the two circles touching at one point.

The furthest that A and B can be would be at the two ends as shown in the above diagram.

If the radius is 6 then the diameter is 2 x 6 = 12 and the distance from A to B would be 2 x 12 = 24

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 17. Note: Figures not drawn to scale.

In the figures above, x = 60, How much more is the perimeter of triangle ABC compared with the triangle DEF.

0

2

4

6

8

Note: Figures not drawn to scale
Since x = 60°, triangle ABC is an equilateral triangle with sides all equal.
The sides are all equal to 8.
Perimeter of triangle ABC = 8 + 8 + 8 = 24
Triangle DEF has two angles equal, so it must be an isosceles triangle.
The two equal sides will be opposite the equal angles

So, the length of DF = length of DE = 10

Perimeter of triangle DEF = 10 + 10 + 4 = 24 Subtract the two perimeters.

24 - 24 = 0

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 18. An equilateral triangle has one side that measures 5 in. What is the size of the angle opposite that side?

55°

70°

110°

60°

None of these

Since it is an equilateral triangle all its angles would be 60°. The size of the angle does not depend on the length of the side.

The size of the angle is 60°.

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 19. An isosceles triangle has one angle of 96°. What are the sizes of the other two angles?

24°

34°

42°

96°

None of these

Since it is an isosceles triangle it will have two equal angles. The given 96° angle cannot be one of the equal pair because a triangle cannot have two obtuse angles.

Let x be one of the two equal angles. The sum of all the angles in any triangle is 180°.

x + x + 96° = 180°

2x = 84°

x = 42°

The sizes of the other two angles are 42° each.

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 20. Find the circumference of the circle with a diameter of 8 inches?

25 inches

25.163 inches

29.45 inches

35.62 inches

None of these

Formula C = {tex}\pi {/tex}d

C = 8{tex}\pi {/tex}

The circumference of the circle is 8{tex}\pi {/tex} = 25.163 inches

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 21. Find the area of the circle with a diameter of 10 inches?

55.78 sq. inches

99.75 sq. inches

92 inches

78.55 sq. inches

None of these

Formula A = {tex}\pi {/tex}r{tex}^{2}{/tex}

Change diameter to radius r = {tex}\frac{1}{2}{/tex}d = {tex}\frac{1}{2}{/tex}x 10 = 5

Plug in the value: A = {tex}\pi {/tex}5{tex}^{2}{/tex} = 25 {tex}\pi {/tex}

The area of the circle is 25{tex}\pi {/tex}

78.55 sq. inches

**Correct Marks**
1

**Incorrectly Marks**
-0.25

Q 22. Find the area of the circle with a radius of 10 inches?

314.2 sq. inches

115 inches

320.29 sq. inches

56.12 sq. inches

None of these

Formula A = {tex}\pi {/tex}r2

Plug in the value A = {tex}\pi {/tex}102 = 100 {tex}\pi {/tex}

The area of the circle is 100 {tex}\pi {/tex}

314.2 sq. inches

Your request has been placed successfully.