**RRB JE Mensuration Question & Answers PDF**

Download Top 20 RRB JE Mensuration Questions and Answers PDF. RRB JE Maths questions based on asked questions in previous exam papers very important for the Railway JE exam

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**Question 1: **$\triangle$ ABC is an equilateral triangle. D is a point on AB such that $\angle BCD = 25^\circ$. E is a point outside the triangle extending BC towards right. Then find the sum of $\angle ACE$ and $\angle CDB$

a) $195^\circ$

b) $215^\circ$

c) $205^\circ$

d) $180^\circ$

**Question 2: **$\triangle$ ABC is a right angled triangle right angled at B. If AB = BC, then, find $\angle$ ACB

a) $60^\circ$

b) $35^\circ$

c) $45^\circ$

d) $30^\circ$

**Question 3: **$\triangle$ ABC is an equilateral triangle. Then find $\angle$ ABD.

a) $105^\circ$

b) $120^\circ$

c) $90^\circ$

d) $60^\circ$

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**Question 4: **Find the total surface area of cuboid 40mmx30mmx15mm ?

a) 4000 sq.mm

b) 4500 sq.mm

c) 5000 sq.mm

d) 5500 sq.mm

**Question 5: **Find the volume( in $cm^3$) of cuboid, if its length is 2 times more than breadth and breadth is 12cm which is twice its height ?

a) 2295

b) 2952

c) 2592

d) can not be determined

**Question 6: **Find the total surface area of the cylinder of height 12 cm and diameter 14 cm ?

a) 823

b) 845

c) 856

d) 836

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**Question 7: **Find the volume of the cylinder of radius 7 cm and height 12 cm ?

a) 1123

b) 2348

c) 1845

d) 1848

**Question 8: **Find the volume of a sphere of radius 7 cm ?

a) 1477.33

b) 1437.33

c) 1433.33

d) 1473.33

**Question 9: **Find the ratio of the volumes of a sphere of radius 6cm and a cone of height 6cm and radius 4cm ?

a) 9:1

b) 1:9

c) 3:1

d) 1:3

**Question 10: **Find the area of trapezium ABCD, AB=7 cm, CD=5 cm & height of trapezium is 4 cm ? It is known that AB is parallel to CD

a) 20 sq.cm

b) 24 sq.cm

c) 28 sq.cm

d) 32 sq.cm

**Question 11: **How many cubes of side 2cm can be formed by melting a cuboid of length 6 cm, breadth 4 cm and height 4 cm. ?

a) 12

b) 14

c) 16

d) 18

**Question 12: **Find the total surface area of the solid hemisphere of radius 8 cm?

a) $192\pi cm^2$

b) $256\pi cm^2$

c) $64\pi cm^2$

d) $128\pi cm^2$

**Question 13: **In a rectangular measuring 6 m * 10 m, a path of width 1 m is constructed inside the garden on all the sides. Find the area of the path?

a) 13

b) 16

c) 28

d) 22

**Question 14: **In a rhombus, the length of the two diagonals are 8cm and 10cm respectively. Find the area of the rhombus?

a) 30

b) 20

c) 40

d) 50

**Question 15: **The length of one pair of opposite sides of the square is increased by 20% and the other pair of opposite sides is reduced by 20%. The area of new rectangle as compared to the area of square is

a) Same

b) 4% greater

c) 4% less

d) 1% greater

**Question 16: **The area of rectangle decreases by 60 sqm if its breadth decreases by 3m and length increases by 1m. The area decreases by 40 sq m if the breadth increases by 1m and length decreases by 4m. The length of the rectangle is

a) 22m

b) 26m

c) 24m

d) 16m

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**Question 17: **If the length, breadth and height of the four walls of the room are 20 m, 30 m, and 40 m respectively then what will be the area of four walls of the room?

a) 1800 sq.m.

b) 3600 sq.m.

c) 4000 sq.m.

d) 4500 sq.m.

**Question 18: **The diagonal of a rectangle is $\sqrt{130}$ units and its area is 63 sq units. Find the perimeter of the rectangle.

a) 16

b) 20

c) 32

d) 40

**Question 19: **If the length is increased by 2 cm and the breadth is decreased by 4 cm, the area of the rectangle decreases by 40 square cms. If the length is decreased by 2 cm and the breadth is increased by 2cm, the area increases by 4 square cms. Find the area of the original rectangle.

a) 90

b) 95

c) 96

d) 100

**Question 20: **There is a path 1 m wide outside a rectangular field of 16 m length and 11m breadth then the total area of the path is

a) 58 m

b) 58 sq.m

c) 36 sq.m

d) 28 sq.m

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**Answers & Solutions:**

**1) Answer (B)**

In an equilateral triangle, each angle equals to $60^\circ$

Given, $\angle BCD = 25^\circ$

Here, $\angle ACB = \angle BCD + \angle ACD$

$\angle ACB = 60^\circ$

$\angle BCD = 25^\circ$

Then, $\angle ACD = 35^\circ$

Angle in a straight line equals to $180^\circ$

Here, $\angle ACB + \angle ACE = 180^\circ$

Therefore, $\angle ACE = 180^\circ – 60^\circ = 120^\circ$

$\angle BDC = 180^\circ – 25^\circ-60^\circ$

=$95^\circ$

sum=95+120=215

**2) Answer (C)**

Given $\angle ABC = 90^\circ$

In a triangle, if two sides are equal, then, opposite angles are equal.

Here, AB = BC ⇒ $\angle A = \angle C$

Let $\angle A = \angle C = x^\circ$

Sum of angles in a triangle is equal to $180^\circ$

⇒ $90^\circ + 2x = 180^\circ$

⇒ $ 2x = 90^\circ$

⇒ $ x = 45^\circ$

Hence, $\angle ACB = 45^\circ$

**3) Answer (B)**

In an equilateral triangle, each angle will be equal to $60^\circ$

Here, $\angle ABC + \angle ABD = 180^\circ$

⇒ $\angle ABD = 180^\circ – 60^\circ = 120^\circ$

**4) Answer (B)**

Total surface area of cuboid = 2(lb+bh+hl) = 2[(40)(30)+(30)(15)+(15)(40)] = 2[1200+450+600] = 2[2250] = 4500 sq.mm

So the answer is option B.

**5) Answer (C)**

Breadth = b = 12 cm

Length = 2b+b = 3b = 36 cm

Height = b/2 = 12/2 = 6 cm

Volume of cuboid = lbh = (36)(12)(6) = 2592 $cm^3$

So the answer is option C.

**6) Answer (D)**

Radius = 14/2 = 7 cm

Total surface area = $2\pi r (h+r) = 2(22/7)(7)(12+7) = 44(19) = 836 sq.cm$

So the answer is option D.

**7) Answer (D)**

Volume of cylinder = $\pi r^2h = (22/7)(7)^2(12) = 1848 sq.cm$

So the answer is option D.

**8) Answer (B)**

Volume of sphere = $\frac{4}{3} \pi r^3$ = $\frac{4}{3} \times \frac{22}{7}(7)^3 = 1437.33$

So the answer is option B.

**9) Answer (A)**

Volume of sphere = $\frac{4}{3}\pi r^3 = \frac{4}{3} \pi (6)^3 = 288\pi$

Volume of cone = $\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (4)^2(6) = 32\pi$

Ratio = $\frac{288\pi}{32\pi} = 9/1$

So the answer is option A.

**10) Answer (B)**

Area of trapezium = $\frac{1}{2}(a+b)h = \frac{1}{2}(7+5)(4) = 24sq.cm$

So the answer is option B.

**11) Answer (A)**

no of cubes = (vol.of cuboid)/(vol.of cube) = (6*4*4)/(2*2*2) = 96/8 = 12

So the answer is option A.

**12) Answer (A)**

Total surface area of the solid hemisphere =$3\pi r^2$ =$(3*8*8)\pi = 192\pi cm^2$

**13) Answer (C)**

Since the dimensions of the reactangle is 6m and 10m and the length of the path is 1m

Dimensions of the rectangle PQRS = (6-2) and (10-2) = 4 and 8

Area of rectangle PQRS = 4*8 = 32

Area of rectangle ABCD = 6*10 = 60

Area of path = (60-32) = 28

**14) Answer (C)**

Area of the rhombus = 1/2 *8 *10 = 40

**15) Answer (C)**

Let the side of the original square be x.

Area of the original square =$x^2$

Length of one pair of sides = 1.2x

Length of another pair of sides = 0.8x

Area of the new rectangle =$1.2x*0.8x = 0.96 x^2$

Area of the rectangle is 4% less as compared to that of square.

**16) Answer (C)**

Let l and b be the length and breadth respectively. Hence, from the information we get

(l+1)(b-3)=lb-60 => 3l-b=57 —(1)

(l-4)(b+1)=lb-49 => l-4b=-36 —(2)

Multiplying (1) by 4 and subtracting (2) from it we get

11l=228+36=264 => l=24.

**17) Answer (C)**

As we know that Area of the four walls of a room = 2*height (length + breadth)

Hence, area = 2*40(50) = 4000 sq.m.

**18) Answer (C)**

Let length and breadth of the rectangle be l and b respectively.

lb = 63

$\sqrt{l^2+b^2}$ = $\sqrt{130}$

=> $l^2+b^2$ = 130

$(l+b)^2$ = $l^2+b^2+2lb$

=> $(l+b)^2$ = 256

=> l+b = 16

Perimeter = 2(l+b) = 2*16 = 32

**19) Answer (C)**

Let l and b be length and breadth of a rectangle respectively.

(l+2)(b-4) = lb-40 => 2l-b=16

(l-2)(b+2) = lb+4 => l-b = 4

=> l = 12 and b = 80

=> Area = 96

**20) Answer (B)**

area of rectangular field=16 x 11=176m$^{2}$

area of field including area of path=(16+2)x(11+2)=234$^{2}$

∴area of path=234-176=58m$^{2}$

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