Table of contents  
Introduction  
Basic Rules of Simplification  
Tips to Crack Approximation 
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Here is a short study guide to help you crack questions on “Simplification and Approximation“
B = Bracket,
O = Order (Powers, Square Roots, etc.)
D = Division
M = Multiplication
A = Addition
S = Subtraction
Example 1: Solve 12 + 22 ÷ 11 × (18 ÷ 3)^2  10
Solution:
= 12 + 22 ÷ 11 × 6^2  10 (Brackets first)
= 12 + 22 ÷ 11 × 36  10 (Exponents)
= 12 + 2 × 36  10 = 12 + 72  10 (Division and multiplication, left to right)
= 84  10 = 74 (Addition and Subtraction, left to right)
Example 2: Solve 4 + 10  3 × 6 / 3 + 4
Solution:
= 4 + 10  18/3 + 4 = 4 + 10  6 + 4 (Division and multiplication, left to right)
= 14  6 + 4 = 8 + 4 = 12 (Addition and Subtraction, left to right)
x= x {if x ≥ 0} and −x {if x < 0}
Exmaple: Solve 8
Solution: 8 = 8 = 8
Example1: Solve 4433.764  2211.993  1133.667 + 3377.442
Solution: Here,
4433.764 = 4434
2211.993 = 2212
1133.667 = 1134
3377.442 = 3377Now simplify, 4434  2212  1134 + 3377 = 4466
Example 2: Solve 530 x 20.3% + 225 x 16.8%
Solution: Here, 20.3% becomes 20% and 16.8% becomes 17%
Now, simplify 530 x 20% + 225 x 17%
= 106 + 38.25 = 144.25
NOTE: Check that the outside number squared times the inside number should equal the original number inside the square root.
Following are some solved examples to help you prepare well for the upcoming exams:
Ques1. A man has Rs. 480 in the denominations of onerupee notes, fiverupee notes, and tenrupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has?
A. 45
B. 60
C. 75
D. 90
Answer: Option D
Explanation:
Let the number of notes of each denomination be x.
Then x + 5x + 10x = 480
=> 16x = 480
Therefore, x = 30.
Hence, total number of notes = 3x = 90.
Ques2. There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:
A. 20
B. 80
C. 100
D. 200
Answer: Option C
Explanation:
Let the number of students in rooms A and B be x and y respectively.
Then, x  10 = y + 10 x  y = 20 .... (i)
and x + 20 = 2(y  20) x  2y = 60 .... (ii)
Solving (i) and (ii) we get: x = 100 , y = 80.
Therefore, The required answer A = 100.
Ques3. The price of 10 chairs is equal to that of 4 tables. The price of 15 chairs and 2 tables together is Rs. 4000. The total price of 12 chairs and 3 tables is:
A. Rs. 3500
B. Rs. 3750
C. Rs. 3840
D. Rs. 3900
Answer: Option D
Explanation:
Let the cost of a chair and that of a table be Rs. x and Rs. y respectively.
Then, 10x = 4y or y = 5/2 x.
Therefore, 15x + 2y = 4000=> 15x + 2*(5/2)x = 4000
=> 20x = 4000
Therefore, x = 200.
So, y = (5/2) * 200 = 500.
Hence, the cost of 12 chairs and 3 tables = 12x + 3y
= Rs. (2400 + 1500)
= Rs. 3900.
Ques4. If a  b = 3 and a^{2} + b^{2} = 29, find the value of ab.
A. 10
B. 12
C. 15
D. 18
Answer: Option A
Explanation:
2ab = (a^{2} + b^{2})  (a  b)^{2}
= 29  9 = 20
=> ab = 10.
Ques5. The price of 2 sarees and 4 shirts is Rs. 1600. With the same money one can buy 1 saree and 6 shirts. If one wants to buy 12 shirts, how much shall he have to pay ?
A. Rs. 1200
B. Rs. 2400
C. Rs. 4800
D. Cannot be determined
E. None of these
Answer: Option B
Explanation:
Let the price of a saree and a shirt be Rs. x and Rs. y respectively.
Then, 2x + 4y = 1600 .... (i)
and x + 6y = 1600 .... (ii)
Divide equation (i) by 2, we get the below equation.=> x + 2y = 800.  (iii)
Now subtract (iii) from (ii)
x + 6y = 1600 ()
x + 2y = 800

4y = 800
Therefore, y = 200.
Now apply value of y in (iii)
=> x + 2 x 200 = 800
=> x + 400 = 800
Therefore x = 400
Solving (i) and (ii) we get x = 400, y = 200.
Therefore, Cost of 12 shirts = Rs. (12 x 200) = Rs. 2400.
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