# RRB JE Mensuration Question & Answers PDF

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## 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

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) $60^\circ$

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

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

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

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

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, $\angle ACB = 45^\circ$

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

Here, $\angle ABC + \angle ABD = 180^\circ$
⇒ $\angle ABD = 180^\circ – 60^\circ = 120^\circ$

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

So the answer is option B.

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.

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.

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

So the answer is option D.

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.

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.

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

So the answer is option B.

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

So the answer is option A.

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

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

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

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.

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.

As we know that Area of the four walls of a room = 2*height (length + breadth)
Hence, area = 2*40(50) = 4000 sq.m.

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

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

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$^{2}$