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 operatio ns
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
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
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4
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5
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4
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Sol. Given exp.=
108¸9+2*13 =108 +13
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=12+13
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=133
= 13 3
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5
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4
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9
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10
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10
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10
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10
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Ex.7 Simplify:
(7/2)¸(5/2)*(3/2)
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¸
5.25
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(7/2)¸(5/2)of (3/2)
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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.
(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_________
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1 + ___1___ __
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3 + __1___
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2 + _1_
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4
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Sol. Given exp. = 4 -
_____5_______ = 4 -
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____5________ = 4 - ____5_____
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1 + ___1__
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1 + _____1____
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1 + ___1__
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3 + __1___
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3 + __4__
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(31/9)
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2 + _1_
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9
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4
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= 4 - __5____
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=
4 - __5___ = 4 – (5*31)/ 40 = 4
– (31/8) = 1/8
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1 + _9_
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(40/31)
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31
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Ex. 12. If _____2x______ = 1 ., then find the value of x .
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1
+ ___1___ __
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1+ __x__
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1 - x
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Sol.
We have : _____2x______ _ = 1 ó _____2x_____
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= 1 ó __2x____ = 1
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1 + ___1_____
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1 + ___1____
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1+ (1 – x)
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_(1 – x) – x
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[1/(1- x)]
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1 - x
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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 nephe ws .If each daughter
receives four times as much as each nephe ws 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
=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 we re taken out and four bottles of oil we are 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?
No
of women =x/3
No of men
=x-(x/3)=2x/3
No
of women having children =1/3 of ½ of x/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 workers
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 unsalable and
there are 12 unsalable mangoes in the crate then how many 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 ne w 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 feats 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 curre ncy notes in all , some of which were of Rs.100
denaomination and the re maining of Rs.50 denomination the total amount of all
these curre ncy 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?
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Sol. Let the total
amount be Rs. X the,
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x -
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x
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=
80 ó 2 x = 80 ó x =63 x 80 =
5040.
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14
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18
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126
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63
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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 -
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1
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= 1 ó x(x+4) =4 x 120 =480
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x
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x+4
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x
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x+4
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120
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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 numbe r of
feet be 224 more than the number of heads, find the numbe r 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.
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