Simplification Questions with Solutions for Bank PO

Simplification is an important chapter in the mathematics subject in which arithmetic and algebra questions are solved on the basis of formulas or a particular rule. Currently, simplification questions are asked in almost all the competitive exams. There are many students who are confused to solve simplification questions in the competitive exams but if you have the right way to solve simplifications questions with solutions, your confidence level automatically increases in the exam.

So, to build your confidence or increase your performance level, learn how to solve simplification questions with solutions in this post for SSC and banking exams. You can practice also from Simplification questions and answers for your exams because of these Simplification Questions related to SBI Clerk and other bank exams.

Simplification Questions with Solutions for Competitive Exams

Q.1.  + 2.8 +   – 2.32 = 5.33

Solution : Let 8.25 – 4.20 +2.8 +  – 2.32 = 5.33.

Then,  = ( 5.33 + 4.20 + 2.32) – (8.25 + 2.8) = 11.85 – 11.05 = 0.80 ↔  = 5.

Q.2. 2.375 × 5.22 ÷ 0.87 – 1.425 × 0.02 = ?

Solution: Given exp. = 2.375 × – 0.0285 = 2.375 × 6 – 0.0285 = 14.25 – 0.0285 = 14.2215.

Q.3. 0.2+0.2 – 0.2 ÷ 0.2 × (0.2 × 0.2), on simplification, gives:

Solution: Given exp. 0.2 + 0.2 – 1 × 0.04 = 0.4 – 0.04 = 0.36.

Q.4. 24-[2.4 –{.24 x 2 – (.024 - ?)}] = 22.0584

Solution: Let 24 – [2.4 – {.24 x 2 – (.024 – x )}] = 22.0584.

Then, 24 – {.48 - .024 + x}] = 22.0584 ↔ 24 – [2.4 – 0.456 – x ] = 22.0584

↔ 24 – 1.994 + x = 22.0584 ↔ x = 22.0584 – 22.056 = 0.0024

Q.5. 54.27 – [12.84 – {(?).87 – (3.41 × 2 – 1.85)}] = 38.33

Solution: Let 54.27 – [12.84 – {x – (6.82 – 1.85)}] = 38.33.

Then, 54.27 – [12.84 – {x – 4.97}] = 38.33

↔ 54.27 – [12.84 – x + 4.97] = 38.33 ↔ 54.27 – [17.81 – x ] = 38.33

↔ 54.27 – 17.81 + x = 38.33 ↔ x = 38.33 – 36.46 = 1.87.

Q.6. If 2p + 3q = 18 and 2p – q = 2, then 2p + q = ?

Solution: (2p +3q) + (2p – q) = 18 +2

↔ 4p + 2q = 20 ↔ 2(2p +q) = 20

↔ 2p +q = 10.

Q.7. If 2x + y = 5 and 3x – 4y = 2, then the value of 2xy is : 

Solution : 2x + y = 5………(i)       and 3x – 4y = 2

Multiplying (i) by 4 and adding (ii) to it, we get : 11x = 22 or x = 2

Putting x = 2 in (i), we get : y = 1. So, 2xy = 2 x 2 x 1 = 4.

Q.8. If 2x + y = 17; y + 2z = 15 and x + y = 9, then what is the value of 4x + 3y+z ?

Solution : 2x + y = 17 ……….(i);  y + 2z = 15………..(ii) and x+ y = 9.

Subtraction (iii) from (i), we get : x = 8.

Putting x = 8 in (i); we get : y = 1. Putting y = 1 in (ii), we get : 2z = 14 or z = 7

  4x + 3y + z = 4 x 8 + 3 x 1 + 7 = 42.

Q.9. If 3x-4y+z = 7; 2x – z + 3y = 19; x+2y+2z = 24, then what is the value of z ?

Solution : 3x – 4y + z = 7 ……….(i); 2x + 3y – z = 19…………(ii) and x + 2y +2z = 24………….(iii)

Adding (i) and (ii), we get : 5x – y = 26

 Subtraction (i) from (ii) and adding to (iii), we get : 9y = 36 or y = 4

Putting y = 4 in (iv), we get : 5x = 30 or x = 6.

Putting x = 6, y = 4 in (iii), we get : 2z = 10 or z = 5.

Q.10. The value of  is 

Solution: Given exp. =  x 999 = 999000 – 4 = 998996.

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