General Intelligence and Reasoning

Completion of Figural Pattern
Non-Verbal Classification
Mirror Image and Water Image
Verbal Series
Arrangement of Words in a Logical Order (Alphabetical Order)
General Intelligence and Reasoning
Finding the Missing Number
Cubes and Dices
Verbal Analogy
Deviation of Figure
Blood Relationship
Symbols & Notations
Direction & Distance Test
Logical Venn-Diagram
Scheduled Day/Date/Time
Ranking/Arrangement
Miscellaneous
Non-Verbal Analogy or Similarity
Verbal Classification
Paper Cutting, Folding & Punching
Embedded Figure
Syllogism, Statement & Conclusions
Coding-Decoding
Non-Verbal Series
Word Formation
Arithmetical Problems

General Awareness

Indian Polity and Constitution
Indian Art & Culture
Geography Of India
Economics
History of India
Indian Economy
World History
Physical Geography
World Geography
Honours and Awards
Biology
Chemistry
Miscellaneous
Computer and IT
Physics
Books and Authors
National Events
Discoveries and Inventions
Science and Technology
Sports
Important Decades, Years and Days
UNO, Other International and National Organisations
International Events
Science
Science
Personalities
World Art and Culture

Quantitative Aptitude

LCM and HCF
Simplification
Average
Ratio and Proportion
Quantitative Aptitude
Trigonometry
Boat and Stream
Discount
Percentage
Time and Work
Statistics & Data Interpretation
Pipe and Cistern
Sequence and Series
Miscellaneous
Mensuration
Time and Distance
Profit and Loss
Compound Interest
Geometry
Simple Interest
Algebra
Number System
Power, Indices and Surds
Approximation
Simple and Decimal Fraction
Area and Perimeter
Circles
Triangles
Trigonometric Ratios
Volume and Surface Area
Straight Lines
Factorisation
Set Theory
Height and Distance
Logarithms
Polynomials
Quadratic Equations
Quadrilateral and Parallelogram
Partnership

English Comprehension

Comprehension Test
Transformation of Sentences (Active/Passive)
Arrangement of Sentences
Improvement of Sentences
Idioms/ Phrases
One word substitution
Cloze Test
Miscellaneous
Direct/ Indirect Speech
Fill in the Blanks
Common Errors
Antonyms
Synonyms
English Comprehension
Selection of misspelt word/ correctly spelt word
General English

Q 1.

Correct2

Incorrect-0.5

The sides of a triangle area in the ratio of {tex} \frac{1}{3}:\frac{1}{4}:\frac{1}{5} {/tex} and its perimeter is 94 Cm.Find the length of the smallest side of the triangle.

18 cm

22.5 cm

24 m

27 cm

Given ratio = {tex} \frac{1}{3}:\frac{1}{4}:\frac{1}{5} {/tex}

Let lengths of the sides be 20x,15x and 12x.

Then, according to the question,

20x + 15x + 12x = 94 =>47x = 94

x = {tex} \frac{94}{47} {/tex}=2

Smallest side =12x = {tex} 12\times 2 {/tex} = 24 cm

Q 2.

Correct2

Incorrect-0.5

The ratio of length of each equal side and the third side of an isosceles triangle is 3:4. If the area of the triangle is {tex} 18\sqrt{5} {/tex} sq units, the third side is

{tex} 8\sqrt{2} {/tex}units

12 units

16 units

{tex} 5\sqrt{10} {/tex}units

Let sides of isosceles triangle are 3x,3x and 4x.

Then, half-perimeter (s) = {tex} \frac{a+b+c}{2} {/tex}

= {tex} \frac{3x+3x+4x}{2} {/tex}=5x

Given, area of isosceles triangle

= {tex} 18\sqrt{5} {/tex}sq units

{tex} \sqrt{s(s-a)(s-b)(s-c)}=18\sqrt{5} {/tex}

{tex} \sqrt{5x(5x-3x)(5x-3x)(5x-4x)}=18\sqrt{5} {/tex}

{tex} \sqrt{5x\times 2x\times 2x\times x}=18\sqrt{5} {/tex}

=> {tex} 2\sqrt{5}x^{2}=18\sqrt{5} {/tex}

=> {tex} x^{2}=9 {/tex}

=> x=3

Third side of isosceles triangle

={tex} 4\times 3 {/tex}= 12units

Q 3.

Correct2

Incorrect-0.5

Three sides of a triangular field are of length 15 m, 20 m and 25 m long, respectively. Find the cost of sowing seeds in the field at the rate of RS. 5 per Sq m.

750

150

300

600

Since,{tex} AC^{2}=AB^{2}+BC^{2} {/tex}

=> {tex} (25)^{2}=(15)^{2}+(20)^{2} {/tex}

=> 625=225+400

=> 625 = 625

So, the triangular field is right angled at B

Area of the field = {tex} \frac{1}{2}\times AB\times BC {/tex}

= {tex} \frac{1}{2}\times 15\times 20 {/tex}

={tex} 150m^{2} {/tex}

So, the cost of sowing seed is RS. 5 per sq m

Cost of sowing seed for

{tex} 150m^{2}=150\times 5 {/tex}

= RS. 750

Q 4.

Correct2

Incorrect-0.5

The length of a rectangular field is 100 m and its breadth is 40 m. What will be the area of the field?

{tex} (4\times 10^{2}) {/tex}sq m

{tex} (4\times 10) {/tex}sq m

{tex} (4\times 10^{4}) {/tex}sq m

{tex} (4\times 10^{3}) {/tex}sq m

Requnred area = {tex} Length\times Breadth {/tex}

= {tex} 100\times 40 {/tex}=4000 sq m

= {tex} (4\times 10^{3}) {/tex} sq m

Q 5.

Correct2

Incorrect-0.5

The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is

{tex} 2520 m^{2} {/tex}

{tex} 2480 m^{2} {/tex}

{tex} 2420 m^{2} {/tex}

None of the above

Let the length of rectangle = L m

Breadth of rectangle = B m

using conditions from the question,

L - B = 23 ....(i)

2(L + B) = 206

L + B = 103 ....(ii)

On adding Eqs. (i) and (ii), we get

2L = 126

=> L = 63 m

=> B = 103-63=40m

Then, area of rectangle = {tex} L\times B {/tex}

= {tex} 63\times 40 {/tex}

= {tex} 2520 m^{2} {/tex}

Q 6.

Correct2

Incorrect-0.5

If the length of a rectangle decreases by 5 m and breadth increases by 3 m, then its area reduces 9 sq m. If length and breadth of this rectangle increased by 3 m and 2 m respectively, then its area increased by 67 sq m. What is the length of rectangle?

9 m

15.6 m

17 m

18.5 m

Let length and breadth of a rectangle is x and y. Then, as per first condition,

(x-5)(y+3)=xy-9

=> xy-5y+3x-15=xy-9

=> 3x-5y=6 .....(i)

As per second condition,

(x+3)(y+2)=xy+67

=> xy+3y+2x+6=xy+67

=> 2x + 3y = 61 .....(ii)

On multiplying Eq. (i) by 3 and Eq. (ii) by 5 then adding, we get

9x - 15y = 18

10x + 15y = 305

----------------

=> 19x = 323

=> x = {tex} \frac{323}{19} {/tex}=17

Hence, the length of rectangle is 17 m

Q 7.

Correct2

Incorrect-0.5

The area of a rectangle whose length is 5 more than twice its width is 75 sq units. What is the perimeter of the rectangle?

40 units

30 units

24 units

20 units

Let the width of the rectangle = x units

Length = (2 x + 5) units

According to the question,

Area = x(2x + 5)

=> 75 = {tex} 2x^{2}+5x {/tex}

=> {tex} 2x^{2} +5x-75=0{/tex}

=> {tex} 2x^{2} +15x-10x-75=0{/tex}

=> x(2x+15)-5(2x+15)=0

=> (2x+15)(x-5)=0

=> x = 5 and {tex} \frac{-15}{2} {/tex}

Width cannot be negative.

Width = 5 units

Length=2x+5 ={tex} 2\times 5+5 {/tex}=15 units

Perimeter of the rectangle

= 2(15 + 5) = 40 units

Q 8.

Correct2

Incorrect-0.5

Floor of a square room of side 10 m is to be completely covered with square tiles, each having length 50 cm. The smallest number of tiles needed is

200

300

400

500

Area of square room = {tex} (10)^{2} {/tex} = 100 sq

= {tex} 100\times (100)^{2} {/tex} sq cm

= {tex} 100\times 100\times 100 {/tex} sq cm

Now, area of tile = {tex} (50)^{2}=50\times 50 {/tex} sq cm

Number of tiles needed

= {tex} \frac{Area \ of \ square \ room}{Area \ of \ tile} {/tex}

= {tex} \frac{100\times 100\times 100}{50\times 50} {/tex} =400

Hence, 400 tiles will be needed.

Q 9.

Correct2

Incorrect-0.5

The area of a trapezium is {tex} 384 cm^{2} {/tex}. If its parallel sides are in the ratio 3 : 5 and the perpendicular distance between them is 12 cm, the smaller of the parallel sides is

20 cm

24 cm

30 cm

36 cm

Let the sides of trapezium be 5x and respectively.

According to the question,

{tex} \frac{1}{2}\times(5x+3x)\times 12 {/tex}=384

=> 8x={tex} \frac{384\times 2}{12} {/tex} => x={tex} \frac{64}{8} {/tex}=8 cm

Length of smaller of the parallel sides

= {tex} 8\times 3 {/tex} = 24 cm

Q 10.

Correct2

Incorrect-0.5

In a trapezium, the two non-parallel sides are equal in length, each being of 5 units. The parallel sides are at a distance of 3 units apart. If the smaller side of the parallel sides is of length 2 units, then the sum of the diagonals of the trapezium is

{tex} 10\sqrt{5} {/tex} units

{tex} 6\sqrt{5} {/tex} units

{tex} 5\sqrt{5} {/tex} units

{tex} 3\sqrt{5} {/tex} units

In {tex} \triangle BCF {/tex}, by Pythagoras theorem,

{tex} (5)^{2}=(3)^{2}+(BF)^{2} {/tex}

BF=4 cm

AB=2+4+4=10 cm

Now, in {tex} \triangle ACF {/tex}, {tex} AC^{2}=CF^{2}+FA^{2} {/tex}

=> {tex} AC^{2}=3^{2}+6^{2} {/tex} => AC={tex} \sqrt{45} {/tex}cm

Similarly, BD = {tex} \sqrt{45} {/tex}cm

Sum of diagonals

=AC+BD= {tex} \sqrt{45}+\sqrt{45}=2\sqrt{45}=6\sqrt{5} {/tex}cm

Q 11.

Correct2

Incorrect-0.5

If area of a regular pentagon is {tex} 125\sqrt{3} {/tex} units sq cm, how long is its each side

10 cm

15 cm

16 cm

25 cm

We know that,

Area of regular pentagon = {tex} 5a^{2}\frac{\sqrt{3}}{4} {/tex}

According to the question,

{tex} \frac{5a^{2}\sqrt{3}}{4}=125\sqrt{3} {/tex}=> {tex} a^{2}=\frac{125\sqrt{3}\times 4}{5\sqrt{3}} {/tex}=100

a = 10 cm

Q 12.

Correct2

Incorrect-0.5

The area of a sector of a circle of radius 36 cm is {tex} 72\pi cm^{2} {/tex}. The length of the Corresponding are of the sector 1s

{tex} \pi cm {/tex}

{tex} 2\pi cm {/tex}

{tex} 3\pi cm {/tex}

{tex} 4\pi cm {/tex}

Given, area of sector = {tex} 72\pi cm^{2} {/tex}

or {tex} \frac{\pi r^{2}\theta}{360 degree}=72\pi {/tex}

{tex} \theta=\frac{72\times 360}{36\times 36} {/tex}=> {tex} \theta {/tex}=20 degree

Now,length of arc

={tex} \frac{\pi r\theta}{180 degree}=\frac{\pi\times 36\times 20 degree}{180 degree}=4\pi cm {/tex}

Q 13.

Correct2

Incorrect-0.5

The radius of a circle is so increased that its circumference increased by 5%. The area of the circle, then increases by

12.50%

10.25%

10.50%

11.25%

Increase in Circumference of circle = 5%

Increase in radius is also 5%

Now, increase in area of circle

={tex} \left(2a+\frac{a^{2}}{100} \right)% {/tex}

where,a=increase in radius

={tex} \left(2\times 5+\frac{5\times 5}{100} \right)\% {/tex}=10.25%

Q 14.

Correct2

Incorrect-0.5

What is the total area of three equilateral triangles inscribed in a semi-circle of radius 2cm?

{tex} 12 cm^{2} {/tex}

{tex} \frac{3\sqrt{3}}{4}cm^{2} {/tex}

{tex} \frac{9\sqrt{3}}{4}cm^{2} {/tex}

{tex} 3 \sqrt{3}cm^{2} {/tex}

since, {tex} \triangle's {/tex} AOB,BOC,COD are equilateral.

Sides = 2 cm

Now, total area = {tex} 3\times {/tex}{tex} \frac{\sqrt{3}}{4}(Side)^{2} {/tex}

= {tex} 3\times {/tex}{tex} \frac{\sqrt{3}}{4}\times (Side)^{2}=3\sqrt{3} cm^{2} {/tex}

Q 15.

Correct2

Incorrect-0.5

A circle of radius 10 cm has an equilateral triangle inscribed in it. The length of the perpendicular drawn from the centre to any side of the triangle is

{tex} 2.5\sqrt{3} {/tex}cm

{tex} 5\sqrt{3} {/tex}cm

{tex} 10\sqrt{3} {/tex}cm

None of these

Circumradius ={tex} \frac{2}{3}\times Height {/tex}

Height ={tex} \frac{10\times 3}{2} {/tex}=15 cm

So,length of perpendicular drawn from centre = 15 - 10 = 5 cm

Q 16.

Correct2

Incorrect-0.5

AB and CD are two diameters of a circle of radius r and they are mutually perpendicular. What is the ratio of the area of the circle to the area of the triangle ACD?

{tex} \frac{\pi}{2} {/tex}

{tex} \pi {/tex}

{tex} \frac{\pi}{4} {/tex}

{tex} 2\pi {/tex}

Required ratio ={tex} \frac{Area \ of \ circle}{Area \ of \ \triangle ACD} {/tex}

={tex} \frac{\pi r^{2}}{\frac{1}{2}\times 2r\times r}=\pi {/tex}

Q 17.

Correct2

Incorrect-0.5

The area of a rectangle is 1.8 times the area of a square. The length of the rectangle is 5 times the breadth. The side of the square is 20 cm. What is the perimeter of the rectangle?

145 cm

144 cm

133 cm

135 cm

Area of square ={tex} (Side)^{2}=20^{2} {/tex} = 400 sq cm

Area of rectangle ={tex} 1.8\times 400 {/tex} = 720 sq cm

Let length and breadth of rectangle be 5x and x, respectively.

Then, according to the question.

{tex} 5x\times x {/tex}=720

=> {tex} 5x^{2} {/tex}=720 => {tex} x^{2}=\frac{720}{5} {/tex}=144

={tex} \sqrt{144} {/tex}=12 cm

Perimeter of rectangle = 2 (5x + x) = 12x

={tex} 12\times 12 {/tex} = 144cm

Q 18.

Correct2

Incorrect-0.5

The side of a square is 5 cm which is 13 cm less than the diameter of a circle. What is the approximate area of the circle?

245 sq cm

235 sq cm

265 sq cm

255 sq cm

Diameter of the circle = 13 + 5 = 18cm

Radius ={tex} \frac{Diameter}{2}=\frac{18}{2} {/tex}=9 cm.

Area of the circle = {tex} \pi r^{2}=\frac{22}{7}\times 9^{2} {/tex}

= {tex} \frac{22\times 81}{7}=\frac{1782}{7} {/tex}=254.57 sq cm.

=255 sq cm.

Q 19.

Correct2

Incorrect-0.5

If area of a square is 64 sq cm, then find the area of the circle formed by the same perimeter.

{tex} \frac{215}{\pi} {/tex}sq cm

{tex} \frac{216}{\pi} {/tex}sq cm

{tex} \frac{256}{\pi} {/tex}sq cm

{tex} \frac{318}{\pi} {/tex}sq cm

Area of square = 64 sq cm

{tex} (Side)^{2} {/tex}= 64

Side = {tex} \sqrt{64} {/tex} = 8 cm

According to the question,

=> {tex} 2\pi r=4\times 8 {/tex} => r={tex} \frac{4\times 8}{2\pi}=\frac{16}{\pi} {/tex}

Area of the circle

={tex} \pi \times 16/ \pi\times 16/ \pi=\frac{256}{\pi} {/tex}sq cm.

Q 20.

Correct2

Incorrect-0.5

The perimeter of a square is twice the perimeter of a rectangle. If the perimeter of the square is 72 cm and the length of the rectangle is 12 cm. what is the difference between the breadth of the rectangle and the Side of the square?

9 cm

12 cm

18 cm

3 cm

Perimeter of square = 72 cm

Perimeter of rectangle

={tex} \frac{72}{2} {/tex}=36 cm

=> {tex} 2\times (l+b) {/tex}=36

[l and b are dimension of rectangle]

{tex} 2\times (12+b) {/tex}=36

=> 12 + b = 18

b=18-12=6cm

And side of square = {tex} \frac{72}{4} {/tex}=18

Difference between breadth of rectangle and side of the square

= 18 - 6 = 12 cm.

Q 21.

Correct2

Incorrect-0.5

The area of a square is twice the area of a circle. The area of the circle is 392 sq cm. Find the length of the side of the square.

28 cm

26 cm

24 cm

22 cm

According to the question,

Area of Square ={tex} 2\times 392 {/tex}

=> {tex} a^{2} {/tex}=784

a={tex} \sqrt{784} {/tex}=28 cm.

Q 22.

Correct2

Incorrect-0.5

The area of a rectangle is equal to the area of a circle with circumference equal to 39.6 m. What is the length of the rectangle if its breadth is 4.5 m?

33.52 m

21.63 m

31.77 m

27.72 m

Let the radius of the circle be r.

Perimeter of given circle = 39.6 m

{tex} 2\pi r {/tex}=39.6 => ={tex} \frac{7\times 39.6}{22\times 2} {/tex}=6.3m

Area of circle = {tex} \frac{22}{7}\times 6.3\times 6.3 {/tex}

= {tex} \frac{873.18}{7}=124.74 m^{2} {/tex}

Now, area of a rectangle = Area of a circle

{tex} l\times b {/tex}=124.74 => {tex} l\times 4.5 {/tex}=124.74

l={tex} \frac{124.74}{4.5} {/tex}=27.72 m.

Q 23.

Correct2

Incorrect-0.5

If the area of a circle is equal to the area of square with side {tex} 2\sqrt{\pi} {/tex} units, what is the diameter of the circle ?

1 unit

2 units

4 units

8 units

Area of the circle = Area of the square

={tex} (Side)^{2} {/tex}

{tex} \pi r^{2}=(2\sqrt{\pi})^{2} {/tex}

=> {tex} \pi r^{2}=4\pi {/tex}

=> {tex} r^{2}=\frac{4\pi}{\pi} {/tex}=4

=> r={tex} \sqrt{4} {/tex}=2 units

Diameter of circle (d) = {tex} 2\times r=2\times 2 {/tex}

= 4 units

Q 24.

Correct2

Incorrect-0.5

Consider the following statements.

The area of square is greater than the area of the triangle.

The area of circle is less than the area of triangle.

Which of the statement is/are correct?

Only I

Only II

Both l and II

Neither I nor II

Let the radius of circle is 'r' and a side of a square is 'a', then given condition

{tex} 2\pi r {/tex}=4a => a={tex} \frac{\pi r}{2} {/tex}

Area of square

={tex} \left(\frac{\pi r}{2}\right)^{2}=\frac{\pi^{2}r^{2}}{4}=\frac{9.86 r^{2}}{4}=2.46 r^{2} {/tex}

and area of circle = {tex} \pi r^{2}=3.14 r^{2} {/tex}

and let the side of equilateral triangle is x.

Then, given condition,

3x={tex} 2\pi r {/tex}=> x={tex} \frac{2\pi r}{3} {/tex}

Area of equilateral triangle = {tex} \frac{\sqrt{3}}{4}x^{2} {/tex}

={tex} \frac{\sqrt{3}}{4}{/tex}{tex} \times {/tex}{tex} \frac{4\pi^{2}r^{2}}{9} {/tex}

={tex} \frac{\pi^{2}}{3\sqrt{3}}r^{2}=1.89 r^{2} {/tex}

Hence, Area of circle > Area of square > Area of equilateral triangle.

Q 25.

Correct2

Incorrect-0.5

Which one of the following is a Pythagorean triple in which one side differs from the hypotenuse by two units?

{tex} (2n+1,4n,2n^{2}+2n) {/tex}

{tex} (2n,4n,n^{2}+1) {/tex}

{tex} (2n^{2},2n,2n+1) {/tex}

{tex} (2n,n^{2}-1,n^{2}+1) {/tex}

By hit and trial method.

Put,n = 2 in option (d)

= {tex} (2\times 2),(2)^{2}-1,(2)^{2}+1] {/tex}=(4,3,5)

Which satisfy pythagoras theorem and one side differes from hypnotenuse by 2 units.

Your request has been placed successfully.