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

**Explanation :**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.

**Explanation :**. 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\(\frac {2}{5}\)

**Explanation :**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.

**Explanation :**

**Explanation :**

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

**Explanation :**(364.824/x) = (3794.1696 + 36.4824) – 3648.24 = 3830.652 – 3648.24 = 182.412.

x = (364.824/182.412) =2.

**Explanation :**

**Explanation :**

**Explanation :**

**Explanation :**

**Explanation :**(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

**Explanation :**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

**Explanation :**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.

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

**Explanation :**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.

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

**Explanation :**Length of board=7ft. 9 inches=(7*12+9)inches=93 inches.

Length of each part = (93/3) inches = 31 inches = 2ft. 7 inches

**Explanation :**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;

**Explanation :**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

**Explanation :**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

**Explanation :**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

**Explanation :**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

**Explanation :**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

**Explanation :**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

**Explanation :**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

**Explanation :**\((a+b)^2\)=\(a^2\)+\(b^2\)+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

**Explanation :**Given expression= \((75983)^2\)-\((45983)^2\)/ (75983-45983) =\((a-b)^2\)/(a-b)

=(a+b)(a-b)/(a-b)

=(a+b)

=75983+45983

=121966

**Explanation :**Given expression= (\(a^3\)-\(b^3\))/ (\(a^2\)+ab+\(b^2\))

=(a-b)

=(343-113)

.=230

**Explanation :**Let the population of two villages be equal after p years Then,68000-1200p=42000+800p

2000p=26000

p=13

**Explanation :**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

**Explanation :**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

**Explanation :**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

**Explanation :**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.

**Explanation :**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.

**Explanation :**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

**Explanation :**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.

**Explanation :**. 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.