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Aptitude-4-SIMPLIFICATION
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
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=84/21x-8/45x=8
(4/21-8/45)x=8(60-56)/315x=84/315x=8
x=(8*315)/4=6301/2x=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}]=[13/41/24]=[(13/4)*24]=78
Ex. 6. Simplify: 10836of 1+2*3 1
4 5 4
Sol. Given exp.= 1089+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.
(iii) 8.5-{5.5-(7.5+(2.8/x))}*(4.25/0.04) = 306
8.5-{5.5-{(7.5x+2.8)/x)}*(425/4) = 306
8.5-{(5.5x-7.5x-2.8)/x}*(425/4) = 306
8.5-{(-2x-2.8)/x}*106.25 = 306
8.5-{(-212.5x-297.5)/x} = 306
(306-221)x = 297.5 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 _____2x_____ = 1 __2x____ = 1
1 + ___1_____ 1 + ___1____ 1+ (1 – x)
_(1 – x) – x [1/(1- x)]
1 - x
2x = 2-x 3x = 2 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 b=(4/3) a.
8a+5b=22 8a+5*(4/3) a=22 8a+(20/3) a=22
44a = 66 a=(66/44)=3/2
(ii) (x /4)-((x-3)/6)=1 (3x-2(x-3) )/12 = 1 3x-2x+6=12 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 8x+8y=13y 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 2 x = 80 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 1 - 1 = 1 x(x+4) =4 x 120 =480
x x+4 x x+4 120
x2 +4x –480 = 0 (x+24) (x-20) = 0 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) =224 x =15.
Hence, number of keepers =15.
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