(x × x² × x³ × x⁴ × x⁵) ÷ x⁸

₌ x^1+2+3+4+5÷x⁸
=x^15-8
=x⁷

দেওয়া আছে, (১০০০) এর ঘাত y/৩ = ১০
বা, (১০³) এর ঘাত y/৩ = ১০ [যেহেতু ১০ × ১০ × ১০ = ১০০০]
বা, ১০ এর ঘাত (৩ × y/৩) = ১০ [শক্তির শক্তি গুণ হয়]
বা, ১০ এর ঘাত y = ১০ এর ঘাত ১
যেহেতু ভিত্তি (১০) উভয় পাশে সমান, তাই ঘাতগুলোও সমান হবে।
সুতরাং, y = ১

প্রদত্ত রাশি: (a²b³ / c²d) ÷ (a³b² / cd³)
= (a²b³ / c²d) × (cd³ / a³b²) [ভাগ চিহ্ন গুণ হলে ভগ্নাংশ উল্টে যায়]
এখন লব ও হরের ঘাতগুলো কাটাকাটি করি:
= (a² / a³) × (b³ / b²) × (c / c²) × (d³ / d)
= (১ / a) × b × (১ / c) × d²
= bd² / ac

(2√7)² = (2√7) × (2√7) 

= 4 × (√7 × √7) 

= 4 × 7 

= 28

So, (√7 + √7)² = 28.

x² + y² = ( x - y) ² + 2xy

= 2 + 2 × 2

= 2 + 4 

= 6

Given,
2x= 3
=> x= 3/2,

and, y/4= 3
=> y=12.

Now, (3+y)/(4+x)

= (3+12)/(4+(3/2))

=15/(8+3)/2)

=15× 2/11
=30/11

a³ + 3a²b + 3ab² + b³ =3√3
or,(a + b)³ = 3 √3
or,a + b = ³√(3√3)
or,a + b =(3√3)^1/3
or,√(a+b)=√{(3√3)^1/3}
=(3√3) ^1/(3×2)
=(3√3)^1/6
={(√3)² × ×3}^1/6
={(√3)³}^1/6
={3^1/2}^1/2
=3^1/4
=⁴√3

Let the quotient after division be x² + ax + b.

⇒ (x² – 3x + n)( x² + ax + b) + q = x⁴ – 11x³ + 31x² – 46x + 20

⇒ x⁴ + (a – 3)x³ + (n – 3a + b)x² + (an – 3b)x + bn + q = x⁴ – 11x³ + 31x² – 46x + 20

Comparing, we get a = -8, n – 3a + b = 31, an – 3b = -46, bn + q = 20.

Put value of a, we get n + b = 7, 8n + 3b = 46

Solving, we get n = 5, b = 2.

Now, bn + q = 20

⇒ q = 20 – 5× 2 = 10

∴ Product of n and q will be 50.

Here, p, m, q and n could attain different values in different cases. For example, if x² + px + q correspond to roots 1 and 2, and x² + mx + n correspond to roots 3 and 4, then p = -3, q = 2, m = -7, n = 12. Here, (p + m)(q + n) = (-10)(14) = -140. And, if x² + px + q correspond to roots 1 and 3, and x² + mx + n correspond to roots 2 and 4, then p = -4, q = 3, m = -6, n = 8. Here, (p + m)(q + n) = (-10)(11) = -110.

We see that unique value of (p + m)(q + n) cannot be determined.

Let my age be M years and my mother’s age be N years.

⇒ Age of my father = (M + N) years

When I was born, the ratio of ages of my mother and father was 5 : 6.

I was born M years back.

⇒ (N - M)/(M + N - M) = 5/6

⇒ 6N - 6M = 5N

⇒ N = 6M

After 8 years, the ratio of my age and my mother’s age will be 3 : 8.

⇒ (M + 8)/(N + 8) = 3/8

⇒ 8M + 64 = 3N + 24

Put N = 6M

⇒ 8M + 40 = 18M

⇒ M = 40/10 = 4

⇒ N = 6M = 6 × 4 = 24

Mother’s age = 24 years

∴ I am 4 years old. When I was 2 years old, it would be 2 years back from now, my mother’s age was 22 years.

Equation 1:

q² – 11q + 30 = 0

⇒ q² – 6q – 5q + 30 = 0

⇒ q (q – 6) – 5(q – 6) = 0

⇒ (q – 5) (q – 6) = 0

⇒ q = 5 or q = 6

Equation 2:

2p² – 7p + 6 = 0

⇒ 2p² – 4p – 3p + 6 = 0

⇒ 2p(p – 2) – 3(p – 2) = 0

⇒ (2p – 3) (p – 2) = 0

⇒ p = 3/2 or p = 2

Thus, p values are less than q. So, p < q

First we will find Quantity A,

Quantity A:

Let the number be ‘x’.

(x – 53) = (175 – x)

⇒ x + x = 175 + 53

⇒ x = 228/2 = 114

∴ x = 114

Now,

Quantity B:

\(p = \frac{{5 - {q^2}}}{{p - q}}\)

⇒ p (p – q) = 5 – q²

⇒ p² – pq + q² = 5

⇒ (p² – pq + q²) = 5

We know that,

a³ + b³ = (a + b)(a² – ab + b²)

Now,

p³ + q³ = (p + q)(p² – pq + q²)

⇒ 45 = (p + q) × 5

⇒ (p + q) = 45/5 = 9

∴ (p+q)^3/2 = 9^3/2 = 27

∴ Quantity A > Quantity B

Notice that in each of the quantities, we can factor 3x out of the  given expressions.

0 < x < 1

Quantity A: (3x³ - 9x) (5x + 7) = 3x(x² - 3) (5x + 7)

Quantity B: (3x² + 9) (5x² + 7x) = 3x(x² + 3) (5x + 7)

Because we know that x is not 0, we can divide away 3x from both quantities.

Also, 0 < x < 1, so (5x + 7) ≠ 0, we can divide away (5x + 7) from both quantities as well.

Quantity A: 3x(x² - 3) (5x + 7) = x² - 3

Quantity B: 3x(x² + 3) (5x + 7) = x² + 3

Since x² will always be positive, (x² + 3) will always be bigger than (x² – 3).

Let the son’s present age be x years.

From the given data, it is known that

Father’s age at the time of son’s birth = Son’s present age

∴ 50 – x =x

⇒ 50 = 2x

⇒ x = 25

∴ son’s age 10 yrs ago = (25 - 10) = 15 yrs.

According to the given equations,

I. 121x² – 17 = 8

⇒ 121x² = 8 + 17 = 25

⇒ x = \(\sqrt {\frac{{25}}{{121}}}\)

∴ x = ± 5/11

II. \(13y + \sqrt 9 = \sqrt {64}\)

⇒ 13y + 3 = 8

⇒ 13y = 8 – 3 = 5

∴ y = 5/13

∵ The value of y is greater than -5/11 but less than 5/11

Hence, no relationship can be established

We will solve both the equations separately.

Equation I:

x² – 19x + 84 = 0

⇒ x² – 12x – 7x + 84 = 0

⇒ (x – 12) (x – 7) = 0

⇒ x = 12 or x = 7

Equation II:

y² – 25y + 126 = 0

⇒ y² – 18y – 7y + 126 = 0

⇒ (y – 18) (y – 7) = 0

⇒ y = 18 or y = 7

So, the relation cannot be determined

We will solve both the equations separately.

Equation I:

\(\frac{{135}}{{\sqrt p }} - 22\sqrt p = 23\sqrt p\)

Multiplying both sides by √p, we get

⇒ 135 – 22p = 23p

⇒ 23p + 22p = 135

⇒ 45p = 135

⇒ p = 135/45

⇒ p = 3

Equation II:

\(\begin{array}{l} \frac{{2 \times 324 \div \sqrt q }}{{108}} - 2\sqrt q = 0\\ \Rightarrow \frac{{2 \times 324}}{{108 \times \sqrt q }} = 2\sqrt q \end{array}\)

Multiplying both sides by √q, we get

\(\Rightarrow q = \;\frac{{2 \times 324}}{{108 \times 2}}\)

⇒ q = 3

Comparing the values of p and q, we notice that p = q

n percent (%) of a number x = \(x \times \frac{n}{{100}}\)

Suppose the original fraction to be x/y.

Then, \(\frac{{x+x \times \frac{{200}}{{100}}}}{{y+y \times \frac{{300}}{{100}}}} = \frac{{12}}{{15}}\)

\(\Rightarrow \frac{x}{y} = \frac{{12 \times 4}}{{15 \times 3}}\)

\(\Rightarrow \frac{x}{y} = \frac{{16}}{{15}}\)

∴ Debit card pin is 1615

9q² – 12q + 4 = 0

⇒ (3q – 2)² = 0

q = 2/3

Also,\(p = \frac{{\sqrt 4 }}{{\sqrt 9 }}\)

Hence p = 2/3

Hence we have p = q

|x| is always positive.

The given equation is x² + 4|x| + 5 = 0

∴ all the terms on the left hand side are positive.

Thus, it will never be equal to zero for any value of x.

∴ Number of roots of the given equation are 0.

We solve both equations separately

Equation I:

(125)^1/3 l – (1331)^1/3 = 4 × (729)^1/6 + (1024)^1/10

⇒ 5l – 11 = 4 × 3 + 2

⇒ 5l = 14 + 11

⇒ l = 25/5

⇒ l = 5

 Equation II:

(m)² – (15625)^1/3 = 0

⇒ (m)² = (15625)^1/3

⇒ m² = 25

⇒ m² – 25 = 0

By using the formula: a² – b² = (a + b) (a – b)

⇒ m = ± 5

Comparing values of l and m, we obtain

l ≥ m

We will solve both the equations separately.

Equation I:

\(\begin{array}{l} {\left( {\frac{{1024 \times 36}}{{900}}} \right)^p} = \frac{{32}}{5}\\ \Rightarrow \;{\left( {\frac{{{2^{10}} \times 1}}{{25}}} \right)^p} = \frac{{{2^{10\; \times \frac{1}{2}}}}}{{{{25}^{\frac{1}{2}}}}}\\ \Rightarrow \;\frac{{{2^{10 \times p}}}}{{{{25}^p}}} = \frac{{{2^{10\; \times \frac{1}{2}}}}}{{{{25}^{\frac{1}{2}}}}} \end{array}\)

On comparing both sides, we get

⇒ p = 1/2

Equation II:

q² + 2 × √2 × √2 × q + 2 × √2 × √2 = 0

⇒ q² + 2 × 2× q + 22 = 0

By using formula: (a + b)²= a² + 2ab + b²

⇒ (q + 2)2 = 0

⇒ q = -2, -2

Comparing the values of p and q we get,

p > q

Let the fraction be a/b

We consider fraction a/b = x

A boy was asked to find 8/9 th of a fraction. He made a mistake of dividing the fraction by 8/9 and so got an answer which exceeds the correct answer by 17/54

\(\Rightarrow \frac{x}{{\frac{8}{9}}}\;-\frac{8}{9} \times x\; = \frac{{17}}{{54}}\)

⇒ 9x/8 – 8x/9 = 17/54

\(\begin{array}{l} \Rightarrow \;\frac{{9x \times 9 - 8x \times 8}}{{72}} = \;\frac{{17}}{{54}}\\ \Rightarrow {\rm{\;}}\frac{{81x - 64x}}{{72}} = \frac{{17}}{{54}} \end{array}\)

⇒ 17x/72 = 17/54

\(\Rightarrow {\rm{\;x\;}} = \frac{{17}}{{54}}\; \times \frac{{72}}{{17}}\)

⇒ x = 4/3

So, our fraction a/b = x = 4/3

We will solve both the equations separately.

Equation I:

9x – 15.45 = 54.55 + 4x

⇒ 9x – 4x = 70

⇒ 5x = 70

⇒ x = 14

Equation II:

\(\begin{array}{l} \sqrt {y + 155} - \sqrt {36} = \sqrt {49} \\ \Rightarrow \sqrt {y + 155} - 6 = 7\\ \Rightarrow \sqrt {y + 155} = 13 \end{array}\)

On squaring both sides,

y + 155 = 169

⇒ y = 169 – 155

⇒ y = 14

Comparing the values of x and y, we get,

x = y

We will solve both the equations separately.

Equation I:

\(\begin{array}{l} \frac{{15 \times 4}}{{{x^{\frac{4}{7}}}}} - \frac{{8 \times 3}}{{{x^{\frac{4}{7}}}}} = {x^{\frac{{10}}{7}}}\\ \Rightarrow \;\frac{{60}}{{{x^{4/7}}}} - \frac{{24}}{{{x^{4/7}}}} = {x^{10/7}}\\ \Rightarrow \;\frac{{36}}{{{x^{4/7}}}} = {x^{10/7}}\\ \Rightarrow \;36 = {x^{\frac{{10}}{7} + \frac{4}{7}}}\\ \Rightarrow \;36 = {x^2} \end{array}\)

⇒ x = ± 6

Equation II:

y³ + 783 = 999

⇒ y³ = 999 – 783

⇒ y³ = 216

⇒ y = 6

Comparing the values of x and y, we get,

x ≤ y

We will solve both the equations separately.

Equation I:

a√25 + 2b = √961

Here √25 can only be evaluated as 5

⇒ 5a + 2b = 31

Equation II:

3a + (2401)^1/4b = 36

⇒ 3a + 7b = 36

Multiplying the simplified values of equation (I) by 3 and that of (II) by 5, we get,

b = 3

Putting this value of y in the simplified value of equation (I), we get

a = 5

∴ a > b

We will solve both the equations separately.

Equation I:

(729)^1/3 l ÷ 423 = 1

\(\Rightarrow {\rm{\;}}\frac{{{{729}^{1/3}}}}{{423}}l\; = \;1\)

⇒ 9l = 423

⇒ l = 423/9 = 47

Equation II:

6 × (3375)^1/3m – (3240000)^1/2 = 36

⇒ 6 × 15m – 1800 = 36

⇒ 90m = 36 + 1800

⇒ m = 1836/90 = 20.4

Comparing the values of l and m we get,

l > m

Equation 1:

2p² = 23p – 63

⇒ 2p2 - 14p – 9p + 63 = 0

⇒ 2p (p- 7) -9 (p - 7) = 0

⇒ (2p - 9) (p – 7) = 0

⇒ 2p – 9 = 0 and p – 7 = 0

⇒ p = 9/2 and 7

Equation 2:

2q (q^-8) = q^-36

⇒ 2q^-7 = q^-36

⇒ 2 q^-7 × q³⁶ = 1

⇒ q²⁹ = ½

⇒ q = (1/2)²⁹

Clearly, q < 1

And hence: p > q

3x² + 7x = 6

⇒ 3x² + 7x - 6 = 0

⇒ 3x² + 9x - 2x - 6 = 0

⇒ 3x (x + 3) - 2 (x + 3) = 0

⇒ (3x - 2) (x + 3) = 0

⇒ 3x - 2 = 0 or x + 3 = 0

⇒ x= 2/3 or -3

Equation 2:

6(2y² + 1) = 17y

⇒ 12y² + 6 = 17y

⇒ 12y² – 17y + 6 = 0

⇒12y² – 9y – 8y + 6 = 0

⇒ 3y (4y- 3) – 2 (4y - 3) = 0

⇒ (3y- 2) (4y - 3) = 0

⇒ 3y – 2 = 0 and 4y - 3 = 0

⇒ y = 2/3 and 3/4

Comparing values of x and y

x ≤ y

3x² - 7x + 2 = 0 

⇒ 3x²- 6x - x + 2 = 0
⇒ 3x (x - 2) - 1 (x - 2) = 0 

⇒ (3x - 1) (x - 2) = 0

⇒ 3x - 1 = 0 
⇒ 3x = 1 
⇒ x = 1/3 
or,
x - 2 = 0 
⇒ x = 2

Equation 2:

2y² - 11y + 15 = 0
⇒ 2y² - 6y - 5y +15 = 0
⇒ 2y (y - 3) - 5 (y - 3) = 0
⇒ (2y - 5) (y - 3) = 0
⇒ (2y - 5) = 0
⇒ y = 5/2
Also (y - 3) = 0 
⇒ y = 3

Now comparing x and y:

x = 1/3 is smaller than 5/2 and 3

x= 2 is smaller than 3 and 5/2

Thus we see that x is smaller than y