SSC CGL Reasoning · Arithmetic Reasoning

Counting, Chance and Shapes

Count pairs with n times (n minus 1) over 2, find a probability as favourable over total, and use the standard volume and diagonal facts for shapes.

All Arithmetic Reasoning notes

Why this matters

Seven questions in the bank, all EASY and all from the 2025 shifts. Four count pairs (handshakes, matches, the diagonals of a polygon), one asks a simple probability, and two use a fact about a shape.

Concept 1 of 3: Counting pairs

A handshake, a match or a diagonal joins two things. Each of n people can pair with n minus 1 others, but that counts every pair twice, once from each end, so halve it.

Definition

Every pair once (handshakes, single matches): $\frac{n(n - 1)}{2}$ .
Every pair twice (home and away matches): $n(n - 1)$ .
Each of n people takes part in exactly k handshakes: $\frac{nk}{2}$ , because each handshake has two people in it.
Diagonals of a polygon with n sides: $\frac{n(n - 3)}{2}$ . Each corner joins every other corner except itself and its two neighbours.
Check: $1 + 2 + \dots + (n - 1)$ gives the same count as $\frac{n(n - 1)}{2}$ .

Pairs and diagonals

\text{Pairs} = \frac{n(n - 1)}{2} \qquad \text{Diagonals} = \frac{n(n - 3)}{2}

nthe number of people, teams or sides

Worked example

In a league of 9 teams, each team plays every other team once. How many matches are played?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 21 Sep 2025, Shift 1 · Q24Easy

Example 1 · Arithmetic Reasoning · Counting, Chance and Shapes

In a room of 6 people, how many different handshakes are possible if each person shakes hands once with every other?

Counting each pair from both ends

n(n - 1)

counts the handshake between A and B once for A and once for B. For single handshakes or single matches, halve it.

Counting the sides as diagonals

\frac{n(n - 1)}{2}

counts every line joining two corners, and that includes the n sides. A diagonal joins corners that are not next to each other, so take the sides away:

\frac{n(n - 3)}{2}

Drill 3 more on counting pairs

Concept 2 of 3: Probability as a fraction

When every outcome is equally likely, a probability is just a count: the outcomes you want over all the outcomes there are.

Definition

Probability $=$ favourable outcomes $\div$ all outcomes, when every outcome is equally likely.
Not happening: $P(\text{not } E) = 1 - P(E)$ . Use it when the event you want is the larger group.
A probability is always between 0 and 1. Options may be fractions or decimals: $\frac{1}{4} = 0.25$ , $\frac{3}{10} = 0.3$ .

Probability and its complement

P(E) = \frac{\text{favourable}}{\text{total}} \qquad P(\text{not } E) = 1 - P(E)

Worked example

A bag has 4 white, 6 black and 2 yellow balls. One ball is drawn at random. What is the probability that it is not black?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 14 Sep 2025, Shift 2 · Q14Easy

Example 2 · Arithmetic Reasoning · Counting, Chance and Shapes

A box has 5 red, 3 blue, 2 green balls. Probability of drawing a non-green ball?

Dividing by the wrong total

With 3 red and 9 other balls, the chance of red is

\frac{3}{12}

, not

\frac{3}{9}

. Divide by every ball in the bag, including the ones you want.

Concept 3 of 3: Shape facts: volumes and diagonals

A few shape facts carry these questions. A cone holds a third of the cylinder with the same base and height, and the diagonal of a rectangle is the long side of a right triangle.

Definition

Cylinder: $\pi r^2 h$ . Cone: $\frac{1}{3}\pi r^2 h$ , a third of the cylinder with the same base and height.
Diagonal of a rectangle $= \sqrt{l^2 + b^2}$ ; of a square $=$ side $\times \sqrt{2}$ .
Triples to know: 3, 4, 5; 5, 12, 13; 8, 15, 17; 7, 24, 25. Multiples work too (6, 8, 10).
When the square root is not a whole number, find the two squares it lies between, and say so: the answer is then approximate.

Cylinder, cone and diagonal

V_{\text{cyl}} = \pi r^2 h \qquad V_{\text{cone}} = \tfrac{1}{3}\pi r^2 h \qquad d = \sqrt{l^2 + b^2}

r, hradius of the base and height
l, blength and breadth of the rectangle

Worked example

A rectangle is 24 cm long and 7 cm wide. What is the length of its diagonal?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 24 Sep 2025, Shift 2 · Q18Easy

Example 3 · Arithmetic Reasoning · Counting, Chance and Shapes

A cone and a cylinder have the same base radius and the same height. What is the ratio of the volume of the cone to the volume of the cylinder?

Taking the cone as half the cylinder

With the same base and height, a cone holds one third of the cylinder, not one half.

Treating an approximate diagonal as exact

\sqrt{l^2 + b^2}

is a whole number only for triples such as 3, 4, 5. Otherwise it lies between two whole numbers, and the nearest option is an approximation.

Drill 1 more on shape facts: volumes and diagonals

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Counting pairs
Pairs and diagonals
$\text{Pairs} = \frac{n(n - 1)}{2} \qquad \text{Diagonals} = \frac{n(n - 3)}{2}$
Probability as a fraction
Probability and its complement
$P(E) = \frac{\text{favourable}}{\text{total}} \qquad P(\text{not } E) = 1 - P(E)$
Shape facts: volumes and diagonals
Cylinder, cone and diagonal
$V_{\text{cyl}} = \pi r^2 h \qquad V_{\text{cone}} = \tfrac{1}{3}\pi r^2 h \qquad d = \sqrt{l^2 + b^2}$

Watch out for (5)

Counting each pair from both ends
→ Counting pairs
Counting the sides as diagonals
→ Counting pairs
Dividing by the wrong total
→ Probability as a fraction
Taking the cone as half the cylinder
→ Shape facts: volumes and diagonals
Treating an approximate diagonal as exact
→ Shape facts: volumes and diagonals

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