mensuration Model Questions & Answers, Practice Test for ibps clerk prelims 2024

Answer: (a)

Total area of road

= Area of road which parallel to length + Area of road which parallel to breadth – overlapped road

= 70 × 5 + 30 × 5 – 5 × 5

= 350 + 150 – 25

= 500 – 25 = 475 $m^2$

∴ Cost of gravelling the road

= 475 × 4 = Rs 1900

Answer: (c)

Let ΔABC is a right triangle with side AB, BC and AC are 4, 3 and 5 units.

$(AB)^3 = (4)^3$ = 64

$(BC)^3 = (3)^3$ = 27

$(AC)^3 = (5)^3$ = 125.

Here, we see that 125 > 64 + 27

⇒ $(AC)^3 > (AB)^3 + (BC)^3$

Answer: (a)

According to the question

$√3 a = 2r$ ∴ a = ${2r}/√3$

Required ratio = $({2r}/√3)^3 : 4/3 π r^3$

= ${8r^3}/{3√3} : 4/3 π r^3 = 2 : √3 π$

Answer: (b)

Surface area of sphere = 616 $cm^2$

4π $r^2$ = 616

⇒ $r^2 = {616 × 7}/{4 × 22}$

⇒ $r^2$ = 7 × 7

∴ r = 7 cm

∴ Diameter of largest circle which lying on sphere = 2 × r = 14 cm

Answer: (c)

Here, Area of Rectangle lies between 40$cm^2$ and 45$cm^2$

Given that one side = 5cm.

Area of Rectangle = 5 × second side

Now, If Area = 40$cm^2$

then, 40 = 3 × second side

∵ second side = 8cm.

Again, If Area = 45$cm^2$

45 = 3 × second side

∵ Second side = 9cm.

It means that second side varies between 8 cm to 9 cm.

Let diagonal = d

⇒ $√{8^2 + 5^2} < d < √{5^2 + 9^2} $

⇒ $√{64 + 25} < d < √{25 + 81}$

⇒ $√{87} < d < √{106}$

⇒ $√{81} < √{89} < d < √{106} < √{121}$

⇒ $√{81} < d < √{121}$

⇒ 9cm < d < 11cm.