Question

A and B are 2 numbers. 6 times the square of B is 735 more than the square of A. if the respective ratio between A and B is 3:2, what is the value of B?​

Answer 1

Gɪᴠᴇɴ :-

A and B are 2 numbers. 6 times the square of B is 735 more than the square of A. if the respective ratio between A and B is 3:2.

ᴛᴏ ғɪɴᴅ :-

• Value of A and B

sᴏʟᴜᴛɪᴏɴ :-

Now,

✞ According to 1st condition :-

➠ 6(B)² = A² + 735

➠ 6B² = A² + 735

➠ A² = 6B² - 735. ----(1)

✞ According to 2nd condition :-

• Ratio of A and B = 3/2

$\implies \: \sf \frac{A}{B} = \frac{3}{2} \\ \\ \\ \implies \: \sf A = \frac{3B}{2}$

We get,

➠ A = 3B/2. --(2)

Put A = 3B/2 in (1) , we get,

$\implies \sf \: {A}^{2} = 6 {B}^{2} - 735 \\ \\ \\ \implies \sf \: { \frac{3B}{2} }^{2} = 6 {B}^{2} - 735 \\ \\ \\ \implies \sf \: \frac{9 {B}^{2} }{4} = 6 {B}^{2} - 735 \\ \\ \\ \implies \sf \: 9 {B}^{2} = 4(6 {B}^{2} - 735) \\ \\ \\ \implies \sf \: 9 {B}^{2} = 24 {B}^{2} - 2940 \\ \\ \\ \implies \sf \: 24 {B}^{2} - 9 {B}^{2} = 2940 \\ \\ \\ \implies \sf \: 15 {B}^{2} = 2940 \\ \\ \\ \implies \sf \: {B}^{2} = \frac{2940}{15} \\ \\ \\ \implies \sf \: {B}^{2} = 196 \\ \\ \\ \implies \sf \: B = \sqrt{196} \\ \\ \\ \implies \sf \: B = 14$

We get,

➠ B = 14

Put B = 14 in (2) , we get,

➠ A = 3B/2

➠ A = 3×14/2

➠ A = 42/2

➠ A = 21

Hence,

Value of :-

• A = 21
• B = 14

Answer 2

Given

A and B are 2 numbers. 6 times the square of B is 735 more than the square of A. if the respective ratio between A and B is 3:2.

To find

what is the value of B?

Solution

★ Let the " A " number be 3x and " B " number be 2x

**According to the given condition**

➥ 6(2x)² = 735 + (3x)²

➥ 6 × 4x² = 735 + 9x²

➥ 24x² = 735 + 9x²

➥ 24x² - 9x² = 735

➥ 15x² = 735

➥ x² = 735/15

➥ x² = 49

➥ x = √49 = 7

Hence, the value of B = 2x = 2 × 7 = 14

And the value of A = 3x = 3 × 7 = 21