(Part IX) – common aptitude for competitive exam like IBPS consists of shortcut methods, tips and tricks to solve question from Partnerships, Surds and Indices and Pipes and Cistern. This particular learning section (Part X) includes another two very important sections from quantitative aptitudes including “Volume and Surface Area”, “Races and Games”.
Shortcut methods and tricks to solve questions from ‘Volume and Surface Area’ are given below:
Volume and Surface formula for solids, sphere, prism, cuboid:
Let’s consider the length of the cuboid is p units, breadth is a units and height is s units. Then,
a. Volume = (p x a x s) cubic units
b. Surface Area = 2 (pa + as + ps) square units
c. Diagonal = √(p2 + a2 + s2) units
If each edge of a cube is considered to be of length x, then
a. Volume = x3 cubic units
b. Surface Area = 6x2 square units
c. Diagonal = √3x units.
Let the radius of base = I, height or length = m. Then,
a. Volume = ∏l2m cubic units
b. Curved Surface Area = 2∏lm square units.
c. Total Surface Area = 2∏l (l + m)
Let the radius of base = l and height or length = m. Then,
a. Slant height S = √(l2 + m2) units
b. Volume = 1/3 (∏l2m) cubic units
c. Curved Surface Area = ∏lm square units
d. Total Surface Area = (∏lm + ∏l2) square units
If the radius of the sphere r, Then
a. Volume = 4/3 (∏r3) cubic units
b. Surface Area = (4∏r2) square units
Shortcut methods and tricks to solve questions from ‘Races and Games’ are given below:
Races: A race is a contest of speed in running, riding, driving, sailing, rowing etc over a particular distance.
Race Course: It’s defined as the ground or path on which contests are conducted.
Starting Point: It’s a particular point from which the actual race starts
Winning Point (or Goal): It’s a particular point where a face is finished.
Dead Heat Race: If all the persons contesting the race reach the winning point (goal) exactly at the same time, then that particular race is called Dead Heat Race.
Winner: When a person reaches the winning post first, then he is declared as winner.
Suppose P and Q are two contestants in a race. Then considering the below mathematical interpretations:
P beats Q by t seconds – P finishes the race t seconds before Q
P gives a start of t seconds to Q – P starts t seconds after Q starts from the same starting point.
P gives Q a start of S metres – While P starts from the starting point, Q starts s meters ahead from the same starting point at the same time. To cover a race of 50 metres in this case, A will have to cover 50 metres while B will have to cover only (50 – s) metres.
In a game of 60, P can give Q 40 points – While P scores 60 points, Q scores only 60-40=20 points.
If X is n times as fast as Y and X gives Y a start of s meters, then the length of the race course, so that X and Y reaches the winning post at the same time = s [n / (n-1)] metres.
If X can run s metre race in t1 seconds and Y in t2 seconds, where t1 < t2, then X beats Y by a distance: (s / t2) (t2 – t1) metre
(……to be continued in Part XI)