**Circle:**

- Diameter, D = 2R
- Area = πR
^{2}sq. units - Circumference = 2πR units

**Square:**

- Area = a
^{2}sq. units - Perimeter = 4a units
- Diagonal, d = √2 a units

**Rectangle:**

- Area = L*B sq. units
- Perimeter = 2(L+B) units
- Diagonal, d = √L
^{2}+B^{2}units

**Right Angled Triangle:**

- Area = (½)bxh sq. units
- Perimeter = b + h + hypotenuse
- Hypotenuse = √b
^{2}+h^{2}units

**Equilateral Triangle:**

- Area = √4 a
^{2}sq. units - Perimeter = 3a units, where a = side of the triangle

**Scalene Triangle:**

- Area: √s(s-a)(s-b)(s-c) sq. units; s = (a+b+c)/2
- Perimeter = (a+b+c) units

**Isosceles Triangle:**

- Area = b/4 √4a
^{2}-B^{2}sq units - Perimeter = 2a + b units, where b = base length; a = equal side length

**Cube:**

- Volume = a
^{3}cubic units - Lateral Surface Area (LSA) = 4a
^{2}sq. units - Total surface area (TSA) = 6a
^{2}sq. units - Length of diagonal = a√3 units

**Cuboid:**

- Volume = (Cross section area * height) = L * B * H cubic units
- Lateral Surface Area (LSA) = 2[(L+B)H] sq. units
- Total surface area (TSA) = 2(LB+BH+HL) sq. units
- Length of the diagonals = √L
^{2}+B^{2}+H^{2}units

**Sphere:**

- Volume = (4/3) πR
^{3}cubic units - Surface Area = 4πR
^{2}sq. units - If R and r are the external and internal radii of a spherical shell, then its Volume
= (4/3) [R
^{3}-r^{3}] cubic units

**Hemisphere:**

- Volume = (2/3) πR
^{3}cubic units - TSA = 3πR
^{2}sq. units

**Cylinder:**

- Volume = πr
^{2}h cubic units - Curved surface Area (CSA) (excludes the areas of the top and bottom circular regions) = 2πRh sq. units
- TSA = Curved Surface Area + Areas of the top and bottom circular regions = 2πRh + 2πR
^{2}= 2πR[R+h] sq. units

**Cone:**

- Volume = (1/3) πR
^{2}h cubic units - Slant Height of cone, L = √R
^{2}+H^{2}units - CSA = πRL sq. units

- A right circular cone is placed over a cylinder of the same radius. Now the combined structure is painted on all sides. Then they are separated now the ratio of area painted on Cylinder to Cone is 3:1. What is the height of Cylinder if the height of Cone is 4 m and radius is 3 m?

A. 5m

B. 6m

C. 8m

D. 10m - The diameter of Road Roller is 84 cm and its length is 150 cm. It takes 600 revolutions to level once on a particular road. Then what is the area of that road in m²?

A. 2376

B. 2476

C. 2496

D. 2516 - A smaller triangle is having three sides. Another big triangle is having sides exactly double the sides of the smaller triangle. Then what is the ratio of Area of Smaller triangle to Area of the bigger triangle?

A. 1:2

B. 2:1

C. 1:4

D. 4:1 - ABCD is a square of 20 m. What is the area of the least-sized square that can be inscribed in it with its vertices on the sides of ABCD?

A. 100 m²

B. 120 m²

C. 200 m²

D. 250 m² - A hemispherical bowl of diameter 16cm is full of ice cream. Each student in a class is served exactly 4 scoops of ice cream. If the hemispherical scoop is having a radius of 2cm, then ice cream is served to how many students?

A. 16

B. 32

C. 64

D. 128 - A hollow cylindrical tube is made of plastic is 4 cm thick. If the external diameter is 18 cm and length of the tube is 59cm, then find the volume of the plastic?

A. 10380 cm³

B. 10384 cm³

C. 10440 cm³

D. 10444 cm³ - What is the radius of the circle whose area is equal to the sum of the areas of two circles whose radii are 20 cm and 21 cm?

A. 27m

B. 28m

C. 29m

D. 30m - A well with 14 m diameter is dug up to 49 m deep. Now the soil taken out during dug is made into cubical blocks of 3.5m side each. Then how many such blocks were made?

A. 22

B. 44

C. 88

D. 176 - If the ratio of radius two Cylinders A and B are in the ratio of 2:1 and their heights are in the ratio of 2:1 respectively. The ratio of their total surface areas of Cylinder A to B is?

A. 1:2

B. 2:1

C. 1:4

D. 4:1 - The area of the Circular garden is 88704 m². Outside the garden a road of 7m width laid around it. What would be the cost of laying road at Rs. 2/m².

A. Rs. 7,546

B. Rs. 10,036

C. Rs. 11,092

D. Rs. 15,092

Cylinder painted area = 2πrh+πr²

Cone painted area = πrl

2h+r/√ (r² +h1²) = 3:1

h = 6

Area: 600*2*22/7*42/100*150/100 = 2376

Smaller triangle sides = a, b, c

Area = √s(s-a) (s-b) (s-c);

s = a+b+c/2

= √(a+b+c)(b+c-a)(a+c- b)(a+b-c)/4

Bigger triangle =2a, 2b, 2c

Area = √(a+b+c)(b+c-a)(a+c- b)(a+b-c)

Ratio = 1:4

It touches on midpoints on the sides of the square ABCD

Side = √ (10² +10²) = √200

Area = 200 m²

2/3*π*8³ = n*4*2/3*π*2³

n = 16

R = 9, r = 5

V = 22/7*59(9

= 22/7*59(81 - 25)

= 10384

πR² = πr1² + πr2²

πR² = π(r1² + r2²)

R² = (400 + 441)

R = 29

22/7*7²*49 = n*(7/2)³

n = 176

Cylinder A: 2πr1 (r1 + h1)

Cylinder B: 2πr2 (r2 + h2)

r1/r2 = 2:1; h1/h2 = 2:1

TA/TB = 2πr1 (r1 + h1)/2πr2 (r2 + h2)

88704 = 22/7*r

r = 168

Outer radius = 168+7 = 175

Outer area = 22/7*175

Road area = 96250 – 88704 = 7546

Cost = 7546*2 = 15092

Quantitative Aptitude Made Simpler

Quantitative Aptitude Made Simpler

Quantitative Aptitude Made Simpler

Quantitative Aptitude Made Simpler

Quantitative Aptitude Made Simpler

Quantitative Aptitude Made Simpler

Quantitative Aptitude Made Simpler

Quantitative Aptitude Made Simpler

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