5005-5000+10=5005-(5000/10)=5005-500=4505.

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.

. Let \(\frac {9}{2}\)+\(\frac {19}{6}\)+ x + \(\frac {7}{3}\)=\(\frac {67}{5}\)
Then x=\(\frac {67}{5}\)-\(\frac {9}{2}\) + \(\frac {19}{6}\)+ \(\frac {7}{3}\)
x=\(\frac {67}{5}\)-\(\frac {(27+19+14)}{6}\)=\(\frac{67}{5}\)-\(\frac {60}{6}\)
x=\(\frac {67}{5}\)-10=\(\frac {17}{5}\)=3\(\frac {2}{5}\)

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.

Given exp. = 12.05*(5.4/0.6) = (12.05*9) = 108.45

(364.824/x) = (3794.1696 + 36.4824) – 3648.24 = 3830.652 – 3648.24 = 182.412.
x = (364.824/182.412) =2.

(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

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

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.

Given expression = (2/3)*(3/4)*(4/5) *…….* (99/100) = 2/100 = 1/50.

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.

Given expression = (100-1/49) * 245=(4899/49) * 245 = 4899 * 5=24495.

Length of board=7ft. 9 inches=(7*12+9)inches=93 inches.
Length of each part = (93/3) inches = 31 inches = 2ft. 7 inches

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;

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

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

Suppose x bottles can fill the tin completely
Then 4/5x - 3/4x =6-4
X/20=2
X=40
Therefore required no of bottles =40

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

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 workers

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

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

\((a+b)^2\)=\(a^2\)+\(b^2\)+2ab = 117+2*24 = 225
a+b=15
\((a-b)^2\)=\(a^2\)+\(b^2\)-2ab = 117-2*54
a-b=3
\(\frac {a+b}{a-b}\)=\(\frac {15}{3}\)=5

Given expression= \((75983)^2\)-\((45983)^2\)/ (75983-45983) =\((a-b)^2\)/(a-b)
=(a+b)(a-b)/(a-b)
=(a+b)
=75983+45983
=121966

Given expression= (\(a^3\)-\(b^3\))/ (\(a^2\)+ab+\(b^2\))
=(a-b)
=(343-113)
.=230

Let the population of two villages be equal after p years Then,68000-1200p=42000+800p
2000p=26000
p=13

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

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

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

Let the total amount be Rs. X the,
\(\frac {x}{14}\) - \(\frac {x}{18}\) = 80
\(\frac {2 x}{126}\) = 80
\(\frac {x}{63}\) =80 = 5040.
Hence the total amount is 5040.

Suppose Mr. Bhaskar is on tour for x days. Then,
\(\frac {360}{x}\) - \(\frac {360}{x+4}\) = 3
\(\frac {1}{x}\) - \(\frac {1}{x+4}\) = \(\frac {1}{120}\)
x(x+4) =4 * 120 =480 \( x^2\) +4x –480 = 0
(x+24) (x-20) = 0
x =20. Hence Mr. Bhaskar is on tour for 20 days.

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

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.

. 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.