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(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

Let the population of two villages be equal after p years 

Then,68000-1200p = 42000 + 800p 

2000p = 26000 

p=13

Given expression=(75983)² -(45983)² /(75983-45983) 

=(a-b)² /(a-b) 

=(a+b)(a-b)/(a-b) 

=(a+b) 

=75983+45983 

=121966

(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

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. 

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

(ii)18800 + 470 + 20 = (18800/470) + 20 = 40/20 = 2.

Given :- 

5m ÷ 5-3 = 5⁵

Concept :-

a^m ÷ aⁿ = a^{m - n}

a^m = aⁿ , then (m = n)

Calculation :- 

⇒ 5m ÷ 5-3 = 5⁵

⇒ 5^{(m - (-3))}  = 5⁵

⇒ 5^{(m + 3)} = 5⁵

Now from above condition

⇒ m + 3 = 5 

⇒ m = 5 - 3 = 2

⇒ m = 2

∴ m = 2

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

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

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.

Concept used:
In this type of question, we simplify the equation by taking two values as a and b and the remaining two which are their twice become 2a and 2b.

Calculation:
\(\sqrt[3]{0.3\space × \space0.3\space ×\space 0.3 \space+\space 0.06\space × \space0.06\space × \space0.06\over0.6\space ×\space 0.6\space ×\space 0.6 \space+ \space0.12\space ×\space 0.12\space×\space 0.12 }\)
Let 0.3 and 0.06 be a and b respectively.
⇒ 0.6 = 2 × 0.3 = 2a
⇒ 0.12 = 2 × 0.06 = 2b
⇒ (a³ + b³)/(2a × 2a × 2a + 2b × 2b × 2b)
⇒ (a3 + b3)/(8a³ + 8b³)
⇒ (a3 + b3)/[8(a3 + b3)]
⇒ 1/8
⇒ \(\sqrt[3]{0.3\space × \space0.3\space ×\space 0.3 \space+\space 0.06\space × \space0.06\space × \space0.06\over0.6\space ×\space 0.6\space ×\space 0.6 \space+ \space0.12\space ×\space 0.12\space×\space 0.12 }=\sqrt[3]{1\over8}\)
⇒ 1/2
⇒ 0.5
∴ The value is 0.5.

Given:

x/y = 4/5

Calculations:

(y - x)   divide each term with y = [(y/y) - (x/y)]

⇒ (1 - x/y)

(y + x)  divide each term with y = [(y/y) + (x/y)]

⇒ (1 + x/y)

(5/8) + (y - x)/(y + x)  = 5/8[(1 - x/y) ÷ (1 + x/y)]

⇒ (5/8) + [(1 - 4/5) ÷ (1 + 4/5)] = (5/8) + {[(5 - 4)/5] ÷ [(5 + 4)/5]}

⇒ (5/8) + {[(5 - 4)/5] ÷  [(5 + 4)/5]} = (5/8) + [(1/5) ÷ (9/5)]

⇒ 5/8 + 1/9 = (5× 9 + 1 × 8)/(8 × 9)

⇒ (45 + 8)/72 = 53/72

∴ The required value is 53/72

Given:

 3 * 4 = 5

 2 * 5 = √29

Calculations:

(3² + 4²)^1/2  = (9 + 16)^1/2

⇒ (25)^1/2 = 5 

(22 + 52)1/2  = (4 + 25)1/2

⇒ (29)1/2 = √29

In the same way

(62 + 82)1/2  = (36 + 64)1/2

⇒ (100)1/2 = 10 

 ∴ The required value is 10

It is in the form of Pythagoras theorem

a² + b² = c² 

\(\sqrt{a^2 + b^2} = c \)

Given:

In a classroom, the number of boys is three times the number of girls

Let us assume that the number of girls is = G students

Now, The number of boys will be = 3G students

Calculation:

Now the Total number of students in a classroom will be = sum of the boys and the girls = 3G + G = 4G

⇒ Check from the option which is not the multiple of 4

∴ Only 42 is not the multiple of 4

Given :- 

133 + 253 

Calculation :- 

⇒ 13³ = 13 × 13 × 13 = 2197

⇒ 25³ = 25 × 25 × 25 = 15625

⇒ 24² = 576

Now,

⇒ 133 + 253 + 576 = ? 

⇒ 2197 + 15625 + 576 = 18398

∴ 133 + 253 + 24 ² = 18,398 = ?

Given∶

(p/q)^{3x – 4} = (q/p)^{2x – 5}

Formula Used∶

(1/a)^m = (a/1)^m, when x^a = x^b, then a = b

Calculation∶

(p/q)^{3x – 4} = (q/p)^{2x – 5}

⇒ (p/q)^{3x – 4} = (p/q)^{-(2x – 5)}

⇒ (p/q)^{3x – 4} = (p/q)^5-2x

Or, 3x – 4 =5 – 2x

⇒ 3x + 2x = 5 + 4

⇒ 5x = 9

⇒ x = 9/5

∴ The required value of x is 9/5

Calculation:

⇒ \(4\dfrac{2}{3}+ 3\dfrac{1}{2}-1\dfrac{2}{3}\)

⇒ \(\dfrac{14}{3}+ \dfrac{7}{2}-\dfrac{5}{3}\)

⇒ \(\dfrac{28 + 21 - 10}{6}\)

⇒ \(\dfrac{49 - 10}{6}\)

⇒ \(\dfrac{39}{6}\) = \(\dfrac{13}{2}\) = \(6\dfrac{1}{2}\)  

∴ Solution for \(4\dfrac{2}{3}+ 3\dfrac{1}{2}-1\dfrac{2}{3}\) is \(6\dfrac{1}{2}\).

Given∶

13√13 × 13³ ÷ 13^-3/2 = 13^{q + 4}

Formula Used∶

^m√a = (a)^1/m, a^m × aⁿ = a^{m + n}

When x^a = x^b, then a = b

Calculation∶

[13. (13)^1/2 × 13³]/(13)^-3/2 = 13^{q + 4}

⇒ 13^{1 + (1/2) + 3}. 13^3/2 = 13^q+4

⇒ 13^{1 + (1/2) + 3 + (3/2)} = 13^{q + 4}

⇒ 13⁶ =13^{q + 4}

⇒ q + 4 = 6

⇒ q = 6 – 4 = 2

Calculation:

⇒ 2 (14 + 6 - 5) 

⇒ 2 × 15 = 30

∴ Solution is 30.

Given∶

(√243)^x ÷ 3^y+1 = 1, also 81^{(4) – (x/2)} – 27^y = 0

Formula Used∶

√a = (a)^1/m, (a^m)ⁿ = a^{m × n}, a^m ÷ aⁿ = a^{m – n}

Calculation∶

(√243)^x ÷ 3^y+1 = 1

{(3⁵)^1/2}^x ÷ 3^y+1 = 1

(3^5/2)^x ÷ 3^y+1 = 1

3^5x/2 ÷ 3^y+1 = 1

3^{(5x/2) – (y + 1)} = 3⁰

(5x/2) – (y + 1) = 0

(5x/2) – y – 1 = 0

5x – 2y – 2 = 0      ----(1)

Also, 81^{4 – (x/2)} – 27^y = 0

(3^{4)[4 - (x/2)]} – (3³)^y = 0

3^{16 – (4x/2)} – 3^3y = 0

3^16 – 2x = 3^3y

16 – 2x = 3y

2x + 3y – 16 = 0      ----(2)

Multiply equation (1) by 3 and equation (2) by 2

15x – 6y – 6 = 0

Or, 15x – 6y = 6      ----(3)

4x + 6y – 32 = 0      ----(4)

Add equation (3) and (4)

(15x – 6y = 6) + (4x + 6y = 32)

⇒ 19x = 38

⇒ x = 2

Put x = 2 in equation (1)

5x – 2y – 2 = 0

⇒ 5(2) – 2y – 2 = 0

⇒ 10 – 2y – 2 = 0

⇒ 2y = 8

⇒ y = 8/2 = 4

⇒ x = 2, y = 4

∴ The value of x and y is 2 and 4.

Given

81 : 810

Calculation

81 × 10 = 810

In the same way, 1 × 10 =10

⇒ 1 : 10

Given:

\(1 - \frac{1}{{1 - \frac{1}{{1 - \frac{1}{x}}}}}\)

Calculations:

\(1 - \frac{1}{{1 - \frac{1}{{1 - \frac{1}{x}}}}}\)

⇒ \(1 - \frac{1}{{1 - \frac{1}{{\frac{x - 1}{x}}}}}\)

⇒ \(1 - \frac{1}{{1 - \frac{x}{{{x - 1}{}}}}}\)

⇒ \(1 - \frac{1}{{\frac{x - 1 - x}{{{x - 1 }{}}}}}\)

⇒ \(1 - \frac{x - 1}{{{ -1}{{{}}}}}\)

⇒ 1 + x - 1

⇒ x

∴ The required value of the given expression is x

Calculation:

Required value = (-7)²

⇒ 49

∴ The required value is 49

Calculation:

65² = 4225

∴ The value of 65² is 4225

Given:

Number is (15)2

Identity used:

(a + b)² = a² + b² + 2ab

Calculation:

(15)² = (10 + 5)²

⇒ (10)²  + 5² + 2 × 10 × 5 = 100 + 25 + 100

∴ 225

Or we can simply calculate the vaue of 15 × 15 i.e.225

Given:

The total ticket sold = 420

Calculation:

According to the question,

Half of the ticket sold at the rate of Rs. 5 each

420 × (1/2) × 5 = Rs. 1,050

one-third at the rate of Rs. 3 each

420 × (1/3) × 3 = Rs. 420

Remaining tickets 

⇒ 420 - {420 × (1/2) + 420 × (1/3)}

⇒ 420 - (210 + 140) = 70

According to the question,

The remaining tickets are are sold at rate Rs. 2 each

⇒ 70 × 2 = Rs. 140

Total amount recieved = 1,050 + 420 + 140 = Rs. 1,610

∴ The total amount recived is Rs. 1,610

Concept

BODMAS rule will be apply.

Calculation

16 - [14 - {2 + 6 × 1}

⇒ 16 - [14 - { 2 + 6}

⇒ 16 - [14 - 8]

⇒ 16 - 6

⇒ 10

\(\frac{7}{{\frac{8}{9}}}\;\) - \(\frac{8}{{\frac{9}{7}}}\;\) 

= 63/8 - 56/9

= (63 × 9 - 56 × 8)/72

= (567 - 448)/72

= 119/72

Given:

If two thirds of a number is 2016

Calculation:

Let be the number be x

According to the question,

⇒ (2/3)x = 2016

⇒ x = 2016 × (3/2)

⇒ x = 3024

1/18 of the x 

⇒ (1/18)× 3024

⇒ 168

∴ 1/18 of that number will be 168.

Given:

22  × 22000 × 2

Formula used:

a^m × aⁿ  = a^{m +n}  

Calculation:

2^2+2000+1   = 2²⁰⁰³   

∴ The required  answer is  22003 . 

4^(-3/2) = ?

2(2 × -3/2) = ?

2(-3) = ?

1/23= ?

1/8= ?

Calculation:

(1/7) – (3/4) + (5/7) = ?

⇒ (4 – 21 + 20)/28 = ?

⇒ 3/28 = ?

∴ The value of ? is 3/28

Calculation:

(1/7) – (3/4) + (5/7) = ?

⇒ (4 – 21 + 20)/28 = ?

∴ 3/28 = ?

∴ The value of ? is 3/28

Calculation:

20% of X = 400

⇒ X/5 = 400

⇒ X = 400 × 5

⇒ 2000

∴ The value of X is 2000

Calculation: 

Calculate this continued fraction from bottom

\( ⇒ 3 + \frac{4}{{5 + \frac{7}{{2 + \frac{6}{{8 + \frac{7}{{2 + \frac{1}{3}}}}}}}}}\)

\( ⇒ 3 + \frac{4}{{5 + \frac{7}{{2 + \frac{6}{{8 + \frac{{21}}{7}}}}}}}\)

\( ⇒ 3 + \frac{4}{{5 + \frac{7}{{2 + \frac{{6}}{{11}}}}}}\)

\( ⇒ 3 + \frac{4}{{5 + \frac{{77}}{{28}}}}\)

\( ⇒ 3 + \frac{{112}}{{217}}\)

\( ⇒ \frac{{763}}{{217}}\)

⇒ 109/31

∴ The value(approx.) of 4.

Note:

109/31 = 3.51, According to the question and given Option 4 is the correct answer, 4 is the nearest approx value.

Given:

N = 0.738738738...

M = 0.531531531...

Concept:

Non-terminating repeating decimal or recurring decimal:-

A decimal fraction in which a figure or group of figures is repeated indefinitely, as in 0.7777 or as in 1.445445445.

Ex- 0.333... = 0.3̅  = 3/9

Calculation:

N = 0.738738738... = \(0.\overline {738} = \frac{{738}}{{999}}\)

M = 0.531531531... = \(0.\overline {531} = \frac{{531}}{{999}}\)

⇒ N = 738/999 = 82/111

⇒ M = 531/999 = 59/111

Now,

(1/N) + (1/M) = (111/82) + (111/59)

\( ⇒ \frac{{111\: ×\: 59\:+ \:111\: × \:82}}{{82\; ×\: 59}}\)

\(⇒ \frac{{111\left( {59\; + \;82} \right)}}{{82\; ×\: 59}} = \;\frac{{111\; × \;141}}{{82\; × \:59}}\)

⇒ 15651/4838

Shortcut:

Given:

N = 0.738738738...

M = 0.531531531...

Calculation:

N = 0.738738738... = \(0.\overline {738} = \frac{{738}}{{999}}\)

M = 0.531531531... = \(0.\overline {531} = \frac{{531}}{{999}}\)

⇒ N = 738/999 = 82/111

⇒ M = 531/999 = 59/111

(1/N) + (1/M) = (111/82) + (111/59)

unit digit of dominator = 2 and 9 = 2 × 9 = 18

∴ we check option which unit digit of dominator is 8, that is my answer. In

option only option b which unit digit is 8 so that is my answer.

Given:

a2/49 = 4

Calculations:

a2/49 = 4

⇒ a² = 4 × 49

⇒ a = √4 × √49

⇒ a = 2 × 7

⇒ a = 14

Given:

(3,375)^1/3 ÷ b = 5

Concept used:

(15)³ = 3,375

Calculation:

(3,375)^1/3 ÷ b = 5

⇒15 ÷ b = 5

⇒ 15 / b = 5

⇒ b = 15 / 5

⇒ b = 3

⇒ b³ = 3³

⇒ b³ = 27

Calculation:

((3 × 10 + 14) ÷ 4) – 2 = 6 

Option 1:

Multiplication and addition

((3 + 10 × 14) ÷ 4) – 2 = 33.75

Option 2:

Division and multiplication

((3 ÷ 10 × 14) × 4) – 2 = 55.2

Option 3:

Addition and subtraction

((3 × 10 – 14) ÷ 4) + 2 = 16/4 + 2 = 6

Option 4:

Subtraction and division

((3 × 10 + 14) – 4) ÷ 2 = (44 – 4) ÷ 2 = 20

∴ Option 3 is correct

Given:

The one-hundredth of centimeter when written in terms of kilometers is as shown below.

Calculation:

As 1 km = 1000 m ( where km = kilometer, m = meter )

And 1 m = 100 cm ( where cm = centimeter )

Now, as given in the question one hundredth ( 1/100 ) of 1 cm

⇒ As, we can see 1 km = 105 cm

⇒ ( 1 cm )/ ( 100× 105 )

∴ 10-7  = 0.0000001 (The one-hundredth of centimeter)