4 CHAP

# 4 CHAP - 4 SIMPLIFICATION IMPORTANT CONCEPTS I BODMASRule...

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4. SIMPLIFICATION IMPORTANT CONCEPTS I. ’BODMAS’Rule: This rule depicts the correct sequence in which the operations are to be executed,so as to find out the value of a given expression. Here, ‘B’ stands for ’bracket’ ,’O’for ‘of’ , ‘D’ for’ division’ and ‘M’ for ‘multiplication’, ‘A’ for ‘addition’ and ‘S’ for ‘subtraction’. Thus, in simplifying an expression, first of all the brackets must be removed, strictly in the order(), {} and []. After removing the brackets, we must use the following operations strictly in the order: (1)of (2)division (3) multiplication (4)addition (5)subtraction. II. Modulus of a real number : Modulus of a real number a is defined as |a| ={a, if a>0 -a, if a<0 Thus, |5|=5 and |-5|=-(-5)=5. III. Virnaculum (or bar): When an expression contains Virnaculum, before applying the ‘BODMAS’ rule, we simplify the expression under the Virnaculum. SOLVED EXAMPLES Ex. 1. Simplify: (i)5005-5000+10 (ii) 18800+470+20 Sol. (i)5005-5000+10=5005-(5000/10)=5005-500=4505. (ii)18800+470+20=(18800/470)+20=40/20=2. Ex. 2. Simplify: b-[b-(a+b)-{b-(b-a-b)}+2a] Sol. Given expression=b-[b-(a+b)-{b-(b-a+b)}+2a] =b-[b-a-b-{b-2b+a}+2a] =b-[-a-{b-2b+a+2a}] =b-[-a-{-b+3a}]=b-[-a+b-3a] =b-[-4a+b]=b+4a-b=4a. Ex. 3. What value will replace the question mark in the following equation? 4 1 +3 1 +?+2 1 =13 2 . 2 6 3 5 Sol. Let 9/2+19/6+x+7/3=67/5

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Then x=(67/5)-(9/2+19/6+7/3) x=(67/5)-((27+19+14)/6)=((67/5)-(60/6) x=((67/5)-10)=17/5=3 2 5 Hence, missing fractions =3 2 5 Ex.4. 4/15 of 5/7 of a number is greater than 4/9 of 2/5 of the same number by 8. What is half of that number? Sol. Let the number be x. then 4/15 of 5/7 of x-4/9 of 2/5 of x=8 4/21x-8/45x=8 (4/21-8/45)x=8 (60-56)/315x=8 4/315x=8 x=(8*315)/4=630 1/2x=315 Hence required number = 315. Ex. 5. Simplify: 3 1 ÷ {1 1 -1/2(2 1 -1/4-1/6)}] 4 4 2 Sol. Given exp. =[13/4 ÷ {5/4-1/2(5/2-(3-2)/12)}]=[13/4 ÷ {5/4-1/2(5/2-1/12)}] =[13/4 ÷ {5/4-1/2((30-1)/12)}]=[13/4 ÷ {5/4-29/24}] =[13/4 ÷ {(30-39)/24}]=[13/4 ÷ 1/24]=[(13/4)*24]=78 Ex. 6. Simplify: 108 ÷ 36of 1 +2 *3 1 4 5 4 Sol. Given exp.= 108 ÷ 9+2 *13 =108 +13 =12+13 =133 = 13 3 5 4 9 10 10 10 10 Ex.7 Simplify: (7/2) ÷ (5/2)*(3/2) ÷ 5.25 (7/2) ÷ (5/2)of (3/2) sol. Given exp. (7/2) × (2/5) × (3/2) ÷ 5.25=(21/10) ÷ (525/100)=(21/10) × (15/14) (7/2) ÷ (15/4) Ex. 8. Simplify: (i) 12.05*5.4+0.6 (ii) 0.6*0.6+0.6*0.6 ( Bank P.O 2003) Sol. (i) Given exp. = 12.05*(5.4/0.6) = (12.05*9) = 108.45 (ii) Given exp. = 0.6*0.6+(0.6*6) = 0.36+0.1 = 0.46 Ex. 9. Find the value of x in each of the following equation: (i) [(17.28/x) / (3.6*0.2)] = 2 (ii) 3648.24 + 364.824 + x – 36.4824 = 3794.1696 (iii) 8.5 – { 5 ½ – [7 ½ + 2.8]/x}*4.25/(0.2) 2 = 306 (Hotel Management,1997) Sol. (i) (17.28/x) = 2*3.6*0.2 x = (17.28/1.44) = (1728/14) = 12.
(ii) (364.824/x) = (3794.1696 + 36.4824) – 3648.24 = 3830.652 – 3648.24 = 182.412. x = (364.824/182.412) =2.

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## This note was uploaded on 08/13/2011 for the course FSD 011 taught by Professor Vinh during the Spring '11 term at Beacon FL.

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4 CHAP - 4 SIMPLIFICATION IMPORTANT CONCEPTS I BODMASRule...

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