mensuration Model Questions & Answers, Practice Test for ibps clerk prelims 2024
ibps clerk prelims 2024 SYLLABUS WISE SUBJECTS MCQs
Ratio & Proportion
Percentage
Time & Work
Time & Distance
Mensuration
Trigonometric Ratios & Identities
A rectangular lawn 70 m × 30 m has two roads each 5 metres wide, running in the middle of it, one parallel to the length and the other parallel to the breadth. Find the cost of gravelling the road at the rate of Rs 4 per square metre.
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
If ABC is a right-angled triangle with AC as its hypotenuse, then which one of the following is correct?
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
⇒ $(AC)^3 > (AB)^3 + (BC)^3$
A cube of maximum volume (each corner touching the surface from inside) is cut from a sphere. What is the ratio of the volume of the cube to that of the sphere?
Answer: (a)
According to the question
$√3 a = 2r$ ∴ a = ${2r}/√3$
Required ratio = $({2r}/√3)^3 : 4/3 π r^3$
= ${8r^3}/{3√3} : 4/3 π r^3 = 2 : √3 π$
What is the diameter of the largest circle lying on the surface of a sphere of surface area 616 sq cm?
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
The area of a rectangle lies between 40 $cm^2$ and 45 $cm^2$. If one of the sides is 5 cm, then its diagonal lies between
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.
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