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March 3, 2025Simplification and Approximation: One of the simpler topics in the numerical ability area is simplification and approximation, and the only way to complete questions from this section quickly is by efficient calculation. The major goal of the questions on the approximation and simplification topic is to assess a candidate’s aptitude for working with numbers and doing simple computations.
Long decimal numbers and calculations are used in the questions to confound the applicants, however this is one of the areas where a candidate can perform best without making mistakes and yet receive a high score. Continue reading to learn the tips and tricks to solve simplification and approximation problems.
There are several fundamental strategies and techniques that can be used to answer the approximation and simplification questions in order to ensure that the candidate does not lose marks in this topic. Refer to the following tips:
BODMAS is an acronym for a math rule that represents the order of operations. Any simplification or approximation questions need to be solved in the following order:
There are two ways to phrase the questions about simplification:
Missing numbers: Candidates must complete an equation that is either on the left-hand side (LHS) or the right-hand side (RHS) of the paper (Right-hand side). For instance, 18% of 125 x 9 % of 25 = _____ + 100. Candidates must complete the blanks.
Making the equation simpler: The direct method of providing an equation and solving it to obtain the solution is the second way that the issues about simplification may be posed. As an example, 500 + 2000 ÷ 40 x 50 = ? Candidates must respond to these questions by choosing what appears in the place of the question mark (?).
Refer to some examples of simplification and approximation questions:
Example 1
(26.912)2 × 6.001 ÷ 6.12 + (7.03)3 + 40.02 = ? – 210.75
A. 1685
B. 1158
C. 1323
D. 1925
E. 1485
Solution :
(26.912)2 × 6.001 ÷ 6.12 + (7.03)3 + 40.02 = ? – 210.75
Or, (27)2 × 6 × 1/6 + (7)3 + 40 = ? – 211
or, ? = 729 + 343 + 40 + 211 = 1323
Hence, option C is correct.
Example 2
1164 × 128 ÷ 8.008 + 969.007 = ?
A. 18800
B. 19000
C. 19600
D. 19200
E. 18600
Solution : By taking the approximate value for the expression, we have
1164 × 128 ÷ 8.008 + 969.007 = ?
1164 × 128 × 1/8.008 + 969.007 = ?
? = 1164 × 128 × 1/8 + 969
? = 1164 × 16 + 969
? = 18624 + 969 = 19593 ≈ 19600
Hence, option C is correct.
Example 3
(25.03)2 + ( 16.93)2 – (13.06)2 = ? + 72 ÷ 9
A. 730
B. 737
C. 749
D. 762
E. 771
Solution :
(25.03)2 + ( 16.93)2 – (13.06)2 = ? + 72 ÷ 9
≈ 625 + 289 – 169 – 8 = ?
? = 914 – 177
? = 737
Hence, option B is correct.
We hope this detailed article on Approximation and Simplification was truly helpful to you. For more interesting topics, stay tuned to Embibe.
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