## Slot -2 – Quantitative Aptitude – Geometry – Mensuration – The area of a rectangle and the square

The area of a rectangle and the square of its perimeter are in the ratio 1 ∶ 25. Then the lengths of the shorter and longer sides of the rectangle are in the ratio?

a) 1:3

b) 3:8

c) 2:9

d) 1:4

Solution:
Given ratio of areas of rectangle and square = 1:25 = 4∶ 100=(1×4):(10×10)
Thus possible ratio of perimeter =(1+4)/(10+10) = 5/20= 1∶4

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## Slot -2 – Quantitative Aptitude – Geometry – Mensuration – A parallelogram ABCD has area 48 sqcm

A parallelogram ABCD has area 48 sqcm. If the length of CD is 8 cm and that of AD is s cm, then which one of the following is necessarily true?

a) 5≤s≤7

b) s≤6

c) s≥6

d) s≠6

Solution: 1 Solution: As the area of ABCD = 48=s×h ————1)
In right-angled triangle CKD, DK ≤ CD ( CD is hypotenuses )
So h ≤8
Thus s ≥6——-(from eq 1)

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## Slot -1 – Quantitative Aptitude – Geometry – Mensuration – Let ABCD be a rectangle inscribed in a circle

Let ABCD be a rectangle inscribed in a circle of radius 13 cm. Which one of the following pairs can represent, in cm, the possible length and breadth of ABCD?

a) 25, 10

b) 24, 12

c) 25, 9

d) 24, 10

Solution: As ABCD is a rectangle angles A,B,C and D will be 90°. Thus AC will be diameter of circle of length 13*2 = 26

So length , width and 26 will form a Pythagorean triplet . From given options ,

only option d) 24, 10 satisfy this condition as
10^2+24^2=26^2

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## Slot -1 – Quantitative Aptitude – Geometry – Mensuration – Given an equilateral triangle T1 with side 24 cm

Given an equilateral triangle T1 with side 24 cm, a second triangle T2 is formed by joining the midpoints of the sides of T1. Then a third triangle T3 is formed by joining the midpoints of the sides of T2. If this process of forming triangles is continued, the sum of the areas, in sq cm, of infinitely many such triangles T1, T2, T3,… will be?

a) 164√3

b) 188√3

c) 248√3

d) 192√3

Solution: As P,Q and R are mid points of AC,AB and BC so both triangle PQR and CBA will be similar and PQ = BC/2
So if the area of triangle ABC = A then area of triangle PQR = A/4
Thus
Sum of Areas = A+A/4+A/16+A/64+ ——–=A/(1-1/4)=4/3 A=4/3×√3/4×24^2=192√3

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## Slot -1 – Quantitative Aptitude – Geometry – Mensuration – Points E, F, G, H lie on the sides

Points E, F, G, H lie on the sides AB, BC, CD, and DA, respectively, of a square ABCD. If EFGH is also a square whose area is 62.5% of that of ABCD and CG is longer than EB, then the ratio of length of EB to that of CG is?

a) 2 : 5

b) 4 : 9

c) 3 : 8

d) 1 : 3

Solution: in triangle EBF ,EF^2=x^2+(a-x)^2

Ratio of Areas = AB^2:EF^2

1∶5/8=a^2: { x^2+(a-x)^2 }

8/5=a^2/(2x^2+a^2-2x )

16x^2+8a^2-16ax=5a^2
16x^2-16ax+3a^2=0
16x^2-12ax-4ax+3a^2=0

(4x-3a)(4x-a)=0
So 4x=a or 4x=3a

Thus CG=a-x=3x or 1/3 x
As CG > EB so CG = 3x

EB : CG = 1 : 3

Option d) 1:3 is correct.

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## Slot -1 – Quantitative Aptitude – Geometry – Mensuration – A right circular cone of height 12 ft

A right circular cone, of height 12 ft, stands on its base which has diameter 8 ft. The tip of the cone is cut off with a plane which is parallel to the base and 9 ft from the base. With π = 22/7, the volume, in cubic ft, of the remaining part of the cone is (TITA)?

Solution: Volume of remaining part ( A’B’BA) = 1/3 *22/7 { 4^2 * 12 – 1^2 *3 } = 22/21 *189 = 198

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