Top 10 – SSC CGL maths tricks and shortcuts

SSC CGL maths tricks and shortcuts
A detailed blog on useful SSC CGL maths tricks and shortcuts

The SSC CGL Tier-1 exam consists of 4 sections of 25 questions each, and each question carries two marks. The duration of the exam is 1 hour. Aspirants get less than 0.6 minutes to answer a question. Hence, it is essential to know SSC CGL maths tricks to solve the questions within the time available.

Aspirants can practice SSC CGL mock tests to improve their speed of solving questions and formulate a strategy. Candidates must deploy the SSC CGL maths tricks they learn in the mock tests and must only use the ones they are comfortable with in the actual exam.

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Aspirants must have in mind that shortcuts are a nice add-on to have, but they are never the replacement for sound fundamentals. Hence, candidates must start using SSC CGL maths tricks only after ensuring that they also know the conventional way to solve the question.

SSC CGL maths tricks and shortcuts:

Let us have a look at some questions from the previous year papers of SSC CGL and some specially created examples to get an idea of how to solve certain problems using SSC CGL maths tricks.

Solving using options:

Joan’s age is 42 years, and Kevin’s is 26 years. How many years ago was Kevin’s age half of Joan’s age?
A) 4
B) 10
C) 8
D) 6
Instead of framing equations and solving, try to solve using the options. You can solve the question by substituting options quickly. 10 is the right answer.

Consecutive discounts:

A shopkeeper offers two consecutive discounts of 10% on a piece of cloth worth Rs. 200. The final price of the cloth is:
A) 158
B) 162
C) 160
D) 180
We know that if two consecutive discounts are a% and b%, then the overall discount is a+b-ab/100
So, the overall discount is 19% on Rs. 200 = Rs.38
Final price = 200 – 38 = 162.

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Areas and volumes:

If the surface areas of 2 cubes are in the ratio 4:9, then their volumes will be in the ratio
A) 2:3
B) 9:4
C) 8:27
D) 27:8

Here, we can solve this questions without constructing equations. Since we are dealing with the same type of object (cubes), the difference will be due to the ratio of their sides.

Surface areas will be in the ratio of a^2 which is equal to 4:9.
=> sides will be in the ratio 2:3 and volumes will be in the ratio 8:27

Substitute values:

If $x^{2}+4x+3=0$, then find $\frac{x^{3}}{x^{6}+27x^{3}+27}$.
A) 1
B) -1
C) 1/2
D) -1/2

On looking at the first equation, we can easily guess that x=-1 satisfies it. Therefore x= -1 has to be one of its roots. Now, substitute x = -1 in the second equation. The value of the expression can be easily found out to be -1.

Eliminate options:

A cylinder of radius 2 cm and height 4 cm. It is planned to paint the cylinder along its curved surface area. If the cost of painting a square cm is 2 Rs. , find the cost of painting the cylinder.

A) 32pi
B) 27pi
C) 33pi
D) 39pi

All the parameters involved in the question are even. Hence, the cost should also be an even multiple of pi. Only A satisfies this condition.

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Trigonometry – Plug in values from the table:

$\frac{tan A + tan B}{cot A+ cot B}$ is equal to
A)tan A tan B
B)Sec A cosec B
C)Cot A cot B
D)None of these

Instead of solving the equations, you can substitute some values for A and B in the question and in the options given. The one that matches will be the answer.

Think laterally:

A box of dimensions 8cm, 4cm and 6cm is placed on a table. A cut is made parallel to the side of dimensions 6cm x 8cm. The cut is at an integral distance from the 6cm side. Which of the following can be the volume of one of the cuboids obtained after the cut?

A) 48
B) 56
C) 64
D) 94

The conventional way to find out will be to enumerate the possibilities – 6cm X 8cm X 1cm and 6cm X 8cm X 2cm.

But here, we know that both the final cuboids have the face of dimensions 6cm X 8cm. Therefore, the volume must be a multiple of 6. Only 48 satisfies the given condition.

Use L.C.M to solve problems from unitary methods:

A can complete a piece of work in 10 days. B can 8 days. If A and B work on alternate days starting from A, when will the work get completed?
A) 9 days
B) 5 days
C) 10 days
D) 7 days

L.C.M of 10 and 8 is 40. Let us assume there are 40 units of work to be completed. A will complete 40/10 = 4 units per day. B will complete 5 units per day. If they work on alternate days, they will complete 9 units of work per 2 days. By the end of the eighth day, A and B will complete 36 units of work. A will complete the remaining 4 units of work on the ninth day.

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Try to solve without equations:

The average age of 2 brothers is 9 years. On including their mother, the average increases by 9 years. The age of the mother is
A) 32 years
B) 36 years
C) 30 years
D) 40 years

Usually, students resort to using equations to solve this question. Instead, use intuition to solve.

Had the age of mother been 9 years, the average would have remained the same. Mother’s arrival increases the average by 9.
=> Mother’s age increases both the son’s age and her age by 9 in addition to the 9 years she already has.
=> Age of Mother is (3 x 9) + 9 = 36.

Simplify calculations:

If you have to find the square of 87, it can be done easily by thinking 87 as (80+7) or (90-3) and then by using the$(a+b)^{2}$ or $(a-b)^{2}$ identities.

Try to remember fraction to decimal conversion values for commonly used fractions. This will help to solve the questions fast.

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