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.
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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
Then
x=(67/5)-(9/2+19/6+7/3)óx=(67/5)-((27+19+14)/6)=((67/5)-(60/6)
óx=((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=8ó4/21x-8/45x=8
ó(4/21-8/45)x=8ó(60-56)/315x=8ó4/315x=8
óx=(8*315)/4=630ó1/2x=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 ó x =
(17.28/1.44) = (1728/14) = 12.
(ii) (364.824/x) = (3794.1696 + 36.4824)
– 3648.24 = 3830.652 – 3648.24 = 182.412.
ó x = (364.824/182.412) =2.
(iii)
8.5-{5.5-(7.5+(2.8/x))}*(4.25/0.04)
= 306
ó 8.5-{5.5-{(7.5x+2.8)/x)}*(425/4) = 306
ó 8.5-{(5.5x-7.5x-2.8)/x}*(425/4) = 306
ó 8.5-{(-2x-2.8)/x}*106.25 =
306
ó 8.5-{(-212.5x-297.5)/x} = 306
ó (306-221)x = 297.5 ó x
=(297.5/85) = 3.5.
Ex. 10. If
(x/y)=(6/5), find the value (x2+y2)/(x2-y2)
Sol. (x2+y2)/(x2-y2)
= ( x2 /y2+
1)/ ( x2 /y2-1)
= [(6/5)2+1] / [(6/5)2-1]
= [(36/25)+1] / [(36/25)-1] = (61*25)/(25*11) =
61/11
Ex. 11. Find the value of 4 - _____5_________
1 + ___1___ __
3 + __1___
2 + _1_
4
Sol. Given exp. = 4 - _____5_______ = 4 -
____5________ = 4 - ____5_____
1 + ___1__ 1 + _____1____ 1 + ___1__
3 + __1___ 3 + __4__ (31/9)
2 + _1_
9
4
= 4 - __5____ = 4 -
__5___ = 4 – (5*31)/ 40 = 4 – (31/8) = 1/8
1 + _9_ (40/31)
31
Ex. 12.
If _____2x______ = 1 ., then find the value of x .
1 + ___1___ __
1+ __x__
1 -
x
Sol. We have : _____2x______
_ = 1 ó _____2x_____ = 1 ó __2x____ = 1
1 + ___1_____ 1 + ___1____ 1+ (1 – x)
_(1 – x) – x [1/(1- x)]
1 - x
ó 2x =
2-x ó 3x = 2 ó x = (2/3).
Ex.13.(i)If
a/b=3/4 and 8a+5b=22,then find the value of a.
(ii)if x/4-x-3/6=1,then find the value of x.
Sol. (i) (a/b)=3/4 Þb=(4/3)
a.
\ 8a+5b=22 Þ
8a+5*(4/3) a=22 Þ 8a+(20/3) a=22
Þ44a = 66 Þ
a=(66/44)=3/2
(ii) (x
/4)-((x-3)/6)=1Û (3x-2(x-3) )/12 = 1 Û 3x-2x+6=12 Û x=6.
Ex.14.If
2x+3y=34 and ((x + y)/y)=13/8,then find the value of 5y+7x.
Sol. The given equations are:
2x+3y=34 …(i)
and, ((x + y) /y)=13/8 Þ
8x+8y=13y Þ 8x-5y=0 …(ii)
Multiplying (i)
by 5,(ii) by 3 and adding, we get : 34x=170 or x=5.
Putting x=5 in
(i), we get: y=8.
\ 5y+7x=((5*8)+(7*5))=40+35=75
Ex.15.If
2x+3y+z=55,x-y=4 and y - x + z=12,then what are the values of x , y and z?
Sol. The given equations are:
2x+3y+z=55 …(i);
x + z - y=4 …(ii); y -x + z =12 …(iii)
Subtracting (ii)
from (i), we get: x+4y=51 …(iv)
Subtracting
(iii) from (i), we get: 3x+2y=43 …(v)
Multiplying (v) by 2 and subtracting (iv) from
it, we get: 5x=35 or x=7.
Putting x=7 in
(iv), we get: 4y=44 or y=11.
Putting x=7,y=11
in (i), we get: z=8.
Ex.16.Find
the value of (1-(1/3))(1-(1/4))(1-(1/5))….(1-(1/100)).
Sol. Given expression = (2/3)*(3/4)*(4/5) *…….* (99/100) = 2/100 = 1/50.
Ex.17. Find
the value of (1/(2*3))+(1/(3*4))+(1/(4*5))+(1/(5*6))+…..+ ((1/(9*10)).
Sol. Given expression=((1/2)-(1/3))+((1/3)-(1/4))+((1/4)-(1/5))+
((1/5)-(1/6))+….+
((1/9)-(1/10))
=((1/2)-(1/10))=4/10
= 2/5.
Ex.18.Simplify:
9948/49 * 245.
Sol. Given expression = (100-1/49) * 245=(4899/49) * 245 = 4899 *
5=24495.
Ex.19.A board 7ft. 9 inches
long is divided into 3 equal parts . What is the length of each part?
Sol. Length of board=7ft. 9 inches=(7*12+9)inches=93 inches.
\Length of each part = (93/3) inches = 31 inches = 2ft. 7 inches
20.A man divides Rs. Among 5 sons,4daughters and 2
nephews .If each daughter receives four times as much as each nephews and each
son receives five times as much as each nephews ,how much does each daughter receive?
Let the share of
each nephews be Rs.x.
Then,share of
each daughter=rs4x;share of each son=Rs.5x;
So,5*5x+4*4x+2*x=8600
25x+16x+2x=8600
=43x=8600
x=200;
21.A man spends 2/5 of his
salary on house rent,3/10 of his salary on food and 1/8 of his salary on
conveyence.if he has Rs.1400 left with him,find his expenditure on food and
conveyence.
Part of salary
left=1-(2/5+3/10+1/8)
Let the monthly
salary be Rs.x
Then, 7/40 of
x=1400
X=(1400*40/7)
=8600
Expenditure on
food=Rs.(3/10*800)=Rs.2400
Expenditure on
conveyence=Rs.(1/8*8000)=Rs.1000
22.A third of
Arun’s marks in mathematics exeeds a half of his marks in english by 80.if he
got 240 marks In two subjects together how many marks did he got inh english?
Let Arun’s marks
in mathematics and english be x and y
Then
1/3x-1/2y=30
2x-3y=180……>(1)
x+y=240…….>(2)
solving (1) and
(2)
x=180
and y=60
23.A tin of oil was 4/5full.when 6 bottles of oil were
taken out and four bottles of oil were poured into it, it was ¾ full. how many
bottles of oil can the tin contain?
Suppose x
bottles can fill the tin completely
Then4/5x-3/4x=6-4
X/20=2
X=40
Therefore
required no of bottles =40
24.if 1/8 of
a pencil is black ½ of the remaining is white and the remaining 3 ½ is blue find the total length of the pencil?
Let the total length be xm
Then black part
=x/8cm
The remaining
part=(x-x/8)cm=7x/8cm
White part=(1/2
*7x/8)=7x/16 cm
Remaining
part=(7x/8-7x/16)=7x/16cm
7x/16=7/2
x=8cm
25.in a certain office 1/3 of the workers are women ½
of the women are married and 1/3 of the married women have children if ¾ of the
men are married and 2/3 of the married men have children what part of workers
are without children?
Let the total no of workers be x
No of women =x/3
No of men
=x-(x/3)=2x/3
No of women
having children =1/3 of ½ ofx/3=x/18
No of men having
children=2/3 of ¾ of2x/3=x/3
No of workers
having children = x/8 +x/3=7x/18
Workers having
no children=x-7x/18=11x/18=11/18 of all wprkers
26.a crate of
mangoes contains one bruised mango for every thirty mango in the crate. If
three out of every four bruised mango are considerably unsaleble and there are
12 unsaleable mangoes in the crate then how msny mango are there in the crate?
Let the total no
of mangoes in the crate be x
Then the no of
bruised mango = 1/30 x
Let the no of
unsalable mangoes =3/4 (1/30 x)
1/40 x =12
x=480
27. a train
starts full of passengers at the first station it drops 1/3 of the passengers
and takes 280more at the second station it drops one half the new total and
takes twelve more .on arriving at the third station it is found to have 248
passengers. Find the no of passengers in the beginning?
Let no of
passengers in the beginning be x
After first
station no passengers=(x-x/3)+280=2x/3 +280
After second
station no passengers =1/2(2x/3+280)+12
½(2x/3+280)+12=248
2x/3+280=2*236
2x/3=192
x=288
28.if a2+b2=177and
ab=54 then find the value of a+b/a-b?
(a+b)2=a2+b2+2ab=117+2*24=225
a+b=15
(a-b)2=a2+b2-2ab=117-2*54
a-b=3
a+b/a-b=15/3=5
29.find the
value of (75983*75983- 45983*45983/30000)
Given
expression=(75983)2-(45983)2/(75983-45983)
=(a-b)2/(a-b)
=(a+b)(a-b)/(a-b)
=(a+b)
=75983+45983
=121966
30.find the
value of 343*343*343-113*113*113
343*343+343*113+113*113
Given
expression= (a3-b3)
a2+ab+b2
=(a-b)
=(343-113)
.=230
31.Village X
has a population of 68000,which is decreasing at the rate of 1200 per
year.VillagyY has a population of
42000,which is increasing
at the rate
of 800 per year .in how many years will the population of the two villages be
equal?
Let the
population of two villages be equal after
p years
Then,68000-1200p=42000+800p
2000p=26000
p=13
32.From a
group of boys and girls,15 girls leave.There are then left 2 boys for each
girl.After this,45 boys leave.There are then 5 girls for each boy.Find
the number of
girls in the beginning?
Let at present
there be x boys.
Then,no of girls
at present=5x
Before the boys
had left:no of boys=x+45
And no of
girls=5x
X+45=2*5x
9x=45
x=5
no of girls in
the beginning=25+15=40
33.An employer pays Rs.20 for each day a worker works
and for feits Rs.3 for each day is ideal at the end of sixty days a worker gets
Rs.280 . for how many days did the worker remain ideal?
Suppose a worker
remained ideal for x days then he worked for 60-x days
20*(60-x)-3x=280
1200-23x=280
23x=920
x=40
Ex 34.kiran had 85 currency notes in all , some of
which were of Rs.100 denaomination and the remaining of Rs.50 denomination the
total amount of all these currency note was Rs.5000.how much amount did she
have in the denomination of Rs.50?
Let the no of fifty rupee notes be x
Then,no of 100
rupee notes =(85-x)
50x+100(85-x)=5000
x+2(85-x)=100
x=70
so,,required
amount=Rs.(50*70)= Rs.3500
Ex. 35.
When an amount was distributed among 14 boys, each of them got rs 80 more than
the amount received by each boy when the same amount is distributed equally
among 18 boys. What was the amount?
Sol. Let
the total amount be Rs. X the,
x
- x = 80 ó 2
x = 80 ó x =63 x 80 = 5040. 14 18 126 63
Hence the total amount is 5040.
Ex. 36.
Mr. Bhaskar is on tour and he has Rs. 360 for his expenses. If he exceeds his
tour by 4 days, he must cut down his daily expenses by Rs. 3. for how many days
is Mr. Bhaskar on tour?
Sol.
Suppose Mr. Bhaskar is on tour for x days. Then,
360
- 360 = 3 ó 1 -
1 = 1
ó
x(x+4) =4 x 120 =480
x
x+4 x x+4 120
ó x2 +4x –480 = 0 ó (x+24) (x-20) = 0 ó x =20.
Hence Mr. Bhaskar is on tour for 20
days.
Ex. 37. Two pens and three pencils cost Rs 86.
four Pens and a pencil cost Rs. 112. find the cost of a pen and that of a
pencil.
Sol. Let the cost of a pen ands a pencil be Rs. X and Rs. Y
respectively.
Then, 2x + 3y = 86 ….(i) and 4x +
y =112.
Solving (i) and (ii), we get: x = 25
and y = 12.
Cost of a pen =Rs. 25 and the cost of a
pencil =Rs. 12.
Ex. 38.
Arjun and Sajal are friends . each has some money. If Arun gives Rs. 30 to
Sajal, the Sajal will have twice the money left with Arjun. But, if Sajal gives
Rs. 10 to Arjun, the Arjun will have thrice as much as is left with Sajal. How
much money does each have?
Sol.
Suppose Arun has Rs. X and Sjal has Rs. Y. then,
2(x-30)= y+30 => 2x-y =90 …(i)
and x +10 =3(y-10) => x-3y = - 40 …(ii)
Solving (i) and (ii),
we get x =62 and y =34.
Arun has Rs. 62 and
Sajal has Rs. 34.
Ex. 39.
In a caravan, in addition to 50 hens there are 45 goats and 8 camels with some
keepers. If the total number of feet be 224 more than the number of heads, find
the number of keepers.
Sol. Let the number of keepers be x then,
Total number of heads =(50 + 45 + 8 +
x)= (103 + x).
Total number of feet = (45 + 8) x 4 +
(50 + x) x 2 =(312 +2x).
(312 + 2x)-(103 + x)
=224ó x =15.
Hence, number of keepers =15.
