बीजगणित प्रश्नावली 2

2x + 3y = 78 ...(i)

3x + 2y = 72 ...(ii)

From equations (i) and (ii),

x = 12, y = 18

∴ x + y = 12 + 18 = 30

x + y = 18

⇒ (x + y)² = 18² = 324

⇒ x² + y² + 2xy = 324

⇒ x² + y² = 324 – 2xy

⇒ x² + y² = 324 – 2(72)

⇒ x² + y² = 324 – 144 = 180

3x + 5y = 44 ...(i)

10x – 2y = 16 ...(ii)

By equation (i) × 2 + equation (ii) × 5

we have,

6x + 10y + 50x – 10y = 88 + 80

or 56x = 168

∴ x = = 3

729 = 9³ = 3⁶, Now 4X – 2 = 6 ⇒ X = 2.

2^{x + 4} – 2^{x + 2} = 3

⇒ 2^{x + 2} (2² – 1) = 3

⇒ 2^{x + 2} = 2⁰

∴ x + 2 = 0 ⇒ x = –2

4 × 5 = 20z/x × 20z/y = 20z/x + z/y = 20

i.e. 1 = z/x + z/y

i.e. z = xy/(x + y)

We have,

⇒

⇒

⇒

⇒ x =

x^abc = x^ax^bx^c

⇒ x^abc = x^{a + b + c}

⇒ abc = a + b + c.

This happens for positive integer value of a, b, c, only when a, b, c take value of 1, 2 & 3.

Hence, a² + b² + c² = 1² + 2² + 3² = 1 + 4 + 9 = 14.

∴ Required answer

∴ Number of rows = 124.

=

=

=

=

Hence, the number of prime factors = (2 + 3 + 2) = 7

Let the larger and smaller numbers be x and y respectively.

Then, x – y = 3 ...(i)

and, x² – y² = 63

⇒ (x + y) (x – y) = 63

...(ii)

From equation (i) and (ii), x = 12

= 1040 – 920 = 120

=

= =

= =

= = + 1

a^y = 19683

⇒ 3⁹ = 27³ = (19683)¹ = 19683

As y is a multiple of a, a = 3, y = 9, b^x = 1024

⇒ 2¹⁰ = 4⁵ = 32² = (1024)¹ = 1024

As b is a factor of x, b = 2, x = 10

Now, x^a + y^b

⇒ 10³ + 9² = 1000 + 81 = 1081.

Given, a + n = 12 ⇒ n = 12 – a

∴ Given expression becomes

= 2^{12 – a} . 3^{12 – a}

By comparing, power of 3, we get a + 2 = 12 – a ⇒ a = 5 and n = 7

∴ Required answer = (a² + n²) – (n² – a²) = (49 + 25) – (49 – 25) = 50

mx^m = nxⁿ

∴ x^m =

∴ Given

=

= = .

Check all options one by one.

(a)

(b) + (15.2 × 4) = 100

(c) (15 × 8) – (5 × 4) = 100

(d) (76 × 1.5) – (5.5 × 2.6) = 99.7

All others are equal to 100. But option (c) gives a value equal to 98.

Others equal 5650.

Except it, others are equal to 550.

Other parts are equal to 300.