Olympiad Test Level 1: Quantitative Aptitude & Reasoning- 2 Class 8 GK

Radius of each of semi-circle
= 42/2 = 21m
The required area =2 x area of semi-circle EFA + area of rectangle ABDE

Clearly, the playground is of rectangular shape.
So the required length of the wire
= Perimeter of the rectangle
= 2 (length + breadth)
= 2 (250+20) = 540 m.

Obtuse angled triangle as ∠ABC = 120°.

Let C.P. = ₹ x
C.P. + 20% of C.P. = S.P.

= 240 x 5 = ₹ 1200

A : B : C = 12 : 15 :25 
Total parts = 52
One part = 312/52 = 6
∴ A = 12 x 6 = 72, B = 15 x 6 = 90, C = 25 x 6 = 150

Let father’s age = x years and son’s age be y years 
According to question

Also; xy = 1000 .... (ii)
Put the value of y from Eqn. (i) in Eqn. (ii) we get
xy = 1000 ⇒ x × 2x/5 =1000 ⇒ 2x² = 1000 × 5 
⇒ x² = 2500 ⇒ x = 50
So, after 10 years father’s age will be = x + 10
= 50+ 10 = 60 years.

Let x, y and z is the angles, where x = y + z
x + y + z = 180° (angle sum property)
(y + z) + y + z = 180°
2(y + z)= 180°

x = 90°, hence triangle is right angled.

As a > 1/6 (90°) => a > 15°
But a + b = 180° (linear Pair) 
∴ b < 165°

⇒ 3x = y + z 
⇒ 3 x - z = y

Total number of votes = 7500 
Given that 20% of Percentage votes were invalid
⇒ Valid votes = 80%
Total valid votes = 7500*(80/100) 
1st candidate got 55% of the total valid votes. 
Hence the 2nd candidate should have got 45% of the total valid votes 
⇒ Valid votes that 2nd candidate got = total valid votes x (45/100) 
7500*(80/100)*(45/100) = 2700