SSC CGL Reasoning · Arithmetic Reasoning

Calculation and Number Facts

Plain arithmetic set as a reasoning question: work the brackets first, know the small squares and cubes, and use LCM and HCF for questions about remainders.

All Arithmetic Reasoning notes

Why this matters

Nineteen questions in the bank, all EASY and all from the 2025 shifts. Twelve are a sum, a bill or a bracket to work out, four test powers, cubes or the digit-sum test, and three test LCM and HCF. Each should take well under the 36 seconds a Tier I question allows.

Concept 1 of 3: Order of operations and quick totals

These questions are pure arithmetic. The only way to lose a mark is to do the steps in the wrong order or to slip on a multiplication, and the options often let you check the last digit before you finish.

Definition

Order: brackets first, then powers and roots, then multiply and divide, then add and subtract.
"The sum of a and b multiplied by c" means $(a + b) \times c$ : add first, then multiply.
A bill: for each item, quantity times price; then add the items.
Fast multiplying: $\times 9$ is $\times 10$ minus the number, and $\times 99$ is $\times 100$ minus the number; $\times 5$ is $\times 10$ then halve; $\times 4$ is doubling twice.
Last-digit check: work out only the last digit of the answer and drop every option that ends differently. It is safe only when the options end in different digits; if two options share a last digit, finish the sum.
Zero: any product that has 0 as a factor is 0.

Worked example

If the sum of 27 and 18 is multiplied by 6, and 15 is then taken away, what is the result?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 13 Sep 2025, Shift 3 · Q25Easy

Example 1 · Arithmetic Reasoning · Calculation and Number Facts

If the sum of 41 and 36 is multiplied by 9, what is the result?

Multiplying before adding the sum

"The sum of 12 and 8 multiplied by 3" is

(12 + 8) \times 3 = 60

. Doing

12 + 8 \times 3 = 36

answers a different question, and SSC prints that wrong total as an option.

Squaring each number inside a bracket

(7 - 4)^2

3^2 = 9

. Squaring each number first gives

49 - 16 = 33

, which is wrong: work out the bracket, then square.

Drill 11 more on order of operations and quick totals

Concept 2 of 3: Squares, cubes and the digit-sum test

A few number facts, known by heart, turn these questions into a glance: the small squares and cubes, how a cube ends, and the rule that a number and its digit sum leave the same remainder on division by 9.

Definition

Cubes 1 to 10: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000.
Squares 11 to 20: 121, 144, 169, 196, 225, 256, 289, 324, 361, 400.
Last digit of a cube: a cube ends in the same digit as its root for 0, 1, 4, 5, 6 and 9; the pairs 2 and 8, and 3 and 7, swap. So a cube ending in 8 has a root ending in 2.
To test a cube: find the two cubes it sits between. If it equals neither, it is not a cube.
Digit-sum test: a number leaves the same remainder on division by 9 as the sum of its digits. The same holds for 3. It does not work for 7, 11 or any other divisor.
Geometric progression: each term is the one before times the same ratio $r$ , so term $n$ is $a r^{n-1}$ . To find $r$ , divide the given term by $a$ and take the root of power $n - 1$ .

Term n of a geometric progression

T_n = a\,r^{\,n-1}

athe first term
rthe common ratio
nthe position of the term

Worked example

Which of these is a perfect cube: 512, 576, 625, 676?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 26 Sep 2025, Shift 3 · Q6Easy

Example 2 · Arithmetic Reasoning · Calculation and Number Facts

Which roll number is a perfect cube?

Taking a square or a fourth power for a cube

625 is

25^2

and

5^4

, but it is not a cube:

8^3 = 512

and

9^3 = 729

, and nothing between them is a cube.

Using the digit sum for 7 or 11

The digit-sum test gives the remainder for 9 and for 3 only. For 7 or 11 it gives a wrong remainder, so divide.

Raising the ratio to the power n

The 6th term of a geometric progression is

a r^5

, not

a r^6

: the first term has had no multiplication yet.

Drill 3 more on squares, cubes and the digit-sum test

Concept 3 of 3: LCM, HCF and remainders

The LCM is the first number every divisor goes into exactly. So a number that leaves the same remainder for every divisor is a multiple of the LCM with that remainder added on.

Definition

LCM: write each number as a product of primes and take the highest power of every prime that appears.
HCF: take the lowest power of each prime that appears in all of them.
Two numbers: HCF $\times$ LCM $=$ the product of the two numbers. This is true for two numbers only; for three it fails.
Same remainder r for every divisor: the number is $k \times$ LCM $+ r$ . The smallest one above r is LCM $+ r$ .
Fast route: with four options, divide each by the divisors and keep the one that leaves the right remainder every time.

HCF, LCM and remainders

\text{HCF} \times \text{LCM} = a \times b \qquad N = k \times \text{LCM} + r

a, bthe two numbers
rthe remainder left by every divisor
ka whole number, 1 for the smallest such number

Worked example

Find the LCM of 16, 24 and 36.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 24 Sep 2025, Shift 3 · Q5Easy

Example 3 · Arithmetic Reasoning · Calculation and Number Facts

Find the smallest number which, when divided by 6, 8, and 10, leaves a remainder of 5 in each case.

Using HCF times LCM for three numbers

For 2, 4 and 6 the HCF is 2 and the LCM is 12, so HCF times LCM is 24, while the product is 48. The rule holds only for two numbers.

Forgetting to add the remainder

The LCM itself leaves remainder 0. A number leaving remainder r for every divisor is the LCM plus r.

Drill 2 more on lcm, hcf and remainders

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 (2)

Squares, cubes and the digit-sum test
Term n of a geometric progression
$T_n = a\,r^{\,n-1}$
LCM, HCF and remainders
HCF, LCM and remainders
$\text{HCF} \times \text{LCM} = a \times b \qquad N = k \times \text{LCM} + r$

Watch out for (7)

Multiplying before adding the sum
→ Order of operations and quick totals
Squaring each number inside a bracket
→ Order of operations and quick totals
Taking a square or a fourth power for a cube
→ Squares, cubes and the digit-sum test
Using the digit sum for 7 or 11
→ Squares, cubes and the digit-sum test
Raising the ratio to the power n
→ Squares, cubes and the digit-sum test
Using HCF times LCM for three numbers
→ LCM, HCF and remainders
Forgetting to add the remainder
→ LCM, HCF and remainders

Test yourself on Arithmetic Reasoning

20 past SSC CGL questions from this chapter, timed at 12 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Arithmetic Reasoning chapter test Drill all 19 questions