Question

solve these questions with explanation. ​

Answer 1

$\orange{ \rm{Solution:- }}$

$\small{ \rm{ \green{4.5 - \frac{1}{2} \: of \: (7.6 - 3.5) + 2.3 \times 4.05}}}$

$\small{ \rm{ \green{ = 4.5 - \frac{1}{2} \: of \: (4.1) + 2.3 \times 4.05}}}$

$\small{ \rm{ \green{ = 4.5 - \frac{1}{2} \times 4.1 + 2.3 \times 4.05}}}$

$\small{ \rm{ \green{ = 4.5 - \frac{4.1}{2} + 2.3 \times 4.05}}}$

$\small{ \rm{ \green{ = 4.5 - \frac{4.1}{2} + 9.315}}}$

$\small{ \rm{ \green{ = 4.5 - 2.05 + 9.315 }}}$

$\small{ \rm{ \green {= 13.815 - 2.05}}}$

$\small{ \rm{ \green{ \underline{ \boxed{ = 11.765}}}}}$

Therefore, By simplifying the question we get 11.765.

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$\orange{ \rm{Solution:- }}$

$\small{ \rm{ \green{ {25}^{n - 1} + 100 = {5}^{2n - 1}}}}$

$\small{ \rm{ \green{ {5}^{2(n - 1)} + 100 = {5}^{2n - 1}}}}$

$\small{ \rm{ \green{ {5}^{2n - 2} + 100 = {5}^{2n - 1}}}}$

$\small{ \rm{ \green{ {5}^{2n - 2} - {5}^{2n - 1} = - 100}}}$

$\small{ \rm{ \green{ {5}^{2n} ( {5}^{ - 2} - {5}^{ - 1}) = - 100}}}$

$\small{ \rm{ \green{ {5}^{2n} ( \frac{1}{25} - \frac{1}{5} ) = - 100}}}$

$\small{ \rm{ \green{ {5}^{2n} \times \frac{ - 4}{25} = - 100}}}$

$\small{ \rm{ \green{{5}^{2n} \times - 4 = - 100 \times 25}}}$

$\small{ \rm{ \green{ {5}^{2n} = \frac{ - 100 \times 25}{( - 4)}}}}$

$\small{ \rm{ \green{ {5}^{2n} = 625}}}$

$\small{ \rm{ \green{ {5}^{2n} = {5}^{4}}}}$

When, bases are same POWERS are equal.

$\small{ \rm{ \green{2n = 4}}}$

$\small{ \rm{ \green{n = \frac{4}{2}}}}$

$\small{ \rm{ \green{ \underline{ \boxed{n = 2}}}}}$

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$\orange{ \rm{Solution:- }}$

$\small{ \rm{ \green{m - \frac{m - 1}{2} = 1 - \frac{m - 2}{3} }}}$

$\small{ \rm{ \green{ \frac{m}{1} - \frac{m - 1}{2} + \frac{m - 2}{3} = 1}}}$

Taking LCM,

$\small{ \rm{ \green{ \frac{m \times \frac{6}{1} - (m - 1) \frac{6}{2} + ( m- 2) \frac{6}{3} }{6} = 1}}}$

$\small{ \rm{ \green{ \frac{6m - (m - 1)3 + (m - 2)2}{6} = 1}}}$

$\small{ \rm{ \green{ \frac{6m - 3m - 3 \times - 1 + 2m - 2 \times 2}{6} = 1}}}$

$\small{ \rm{ \green{6m - 3m - 3 + 2m - 4 = 1 \times 6}}}$

$\small{ \rm{ \green{3m + 2m - 1 = 6}}}$

$\small{ \rm{ \green{5m = 6 + 1}}}$

$\small{ \rm{ \green{5m = 7}}}$

$\small{ \rm{ \green{ \underline{ \boxed{m = \frac{7}{5}}}}}}$

Let's Verify it,

L.H.S,

$\small{ \rm{ \green{m - \frac{m - 1}{2} }}}$

$\small{ \rm{ \green{ = \frac{7}{5} - \frac{ \frac{7 }{5} - 1 }{2} }}}$

$\small{ \rm{ \green{ = \frac{7}{5} - \frac{ \frac{7 - 5}{5} }{2}}}}$

$\small{ \rm{ \green{ = \frac{7}{5} - \frac{2}{10}}}}$

$\small{ \rm{ \green{ = \frac{7}{5} - \frac{1}{5} = \frac{6}{5}}}}$

R.H.S,

$\small{ \rm{ \green{1 - \frac{m - 2}{3}}}}$

$\small{ \rm{ \green{ = 1 - \frac{ \frac{7}{5} - 2}{3}}}}$

$\small{ \rm{ \green{ = 1 - \frac{ \frac{7 - 2(5)}{5} }{3}}}}$

$\small{ \rm{ \green{ = 1 - \frac{ \frac{7 - 10}{5} }{3}}}}$

$\small{ \rm{ \green{ = 1 - \frac{ - \frac{3}{5} }{5} }}}$

$\small{ \rm{ \green{ = 1 + \frac{3}{15} }}}$

$\small{ \rm{ \green{ = 1 + \frac{1}{5}}}}$

$\small{ \rm{ \green = { \green{ \underline { \boxed{ \frac{6}{5}}}}}}}$

∴ L.H.S = R.H.S

Hence, Verified!!

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$\orange{ \rm{Solution:- }}$

This question can be done in Two ways....

First,

Let AB & CD be two lines.

$\small{ \rm{ \green{ \angle1 + \angle4 = 180 \degree(</strong><strong>CD</strong><strong> \: is \: a \: straight \: line)}}}$

$\small{ \rm{ \green{ \angle4 + \angle3 = 180 \degree(</strong><strong>AB</strong><strong> \: is \: a \: straight \: line</strong><strong>)</strong><strong>}}}$

$\small{ \rm{ \green{ \angle1 + \angle 4 = \angle4 + \angle3}}}$

$\small{ \rm{ \green{ \angle1 = \angle3 }}}$

Similarly,

$\small{ \rm{ \green{ \angle2 = \angle4}}}$

Hence, Proved!!

N.B : Refer the fig.1 in attachment.

Second,

Vertically opposite angles are equal. As angles x and y are on a straight line,

$\small{ \rm{ \green{ \therefore{ \angle x + \angle y = 180 \degree}}{\: \: ..... (1)}}}$

$\small{ \rm{ \green{ \therefore{ \angle y + \angle z = 180 \degree}}} \: \: .....(2)}$

Thus, from eq (1) and (2), we get:

$\small{ \rm{ \green{\angle x + \angle y = \angle y + \angle z}}}$

$\small{ \rm{ \green{ \therefore{ \angle x = \angle z}}}}$

Similarly, We can prove:

$\small{ \rm{ \green{ \angle y = \angle w}}}$

N.B :- Refer the fig.2 in the attachment.

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Answer 2

$\sf\green{ QUESTION :- }$

Simplify : 4.5 - ½ of ( 7.6 - 3.5 ) + 2.3 × 4.05

$\sf\blue{ ANSWER :- }$

4.5 - ½ of ( 7.6 - 3.5 ) + 2.3 × 4.05

↝ 4.5 - ½ × ( 7.6 - 3.5 ) + 2.3 × 4.05

↝4.5 - ½ × ( 4.1 ) + 9.315

↝4.5 - 0.5 × ( 4.1 ) + 9.315

↝4.5 - 2.05 + 9.315

↝4.5 +7.265

↝11.765

so, 4.5 - ½ of ( 7.6 - 3.5 ) + 2.3 × 4.05 = 11.765

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$\sf\green{ QUESTION :- }$

if 25 ^n-1 + 100 = 5 ^2n-1 , find the value of n

$\sf\blue{ ANSWER :- }$

↝$\tt{25}^{n - 1} + 100 = {5}^{2n - 1}$

↝$\tt {(5}^{2} ) ^{n - 1} + 100 = {5}^{2n - 1}$

↝$\tt {5}^{2 \times n - 2} + 100 = {5}^{2n - 1}$

↝$\tt \dfrac{ {5}^{2n} }{ {5}^{2} } + 100 = \dfrac{ {5}^{2n} }{ {5} }$

↝$\tt 100 = \dfrac{ {5}^{2n} }{ {5}} - \dfrac{ {5}^{2n} }{ {25} }$

↝$\tt 100 = \dfrac{ {25 - 5}^{2n} - {5.5}^{2n} }{ {25 \times 5}}$

↝$\tt 100 = \dfrac{ { 5}^{2n} (25 - {5}) }{ {25 \times 5}}$

↝$\tt { 5}^{2n} = \dfrac{ 25 \times 5 \times 100 }{ 20} = 5 \times 5 \times 5 \times 5$

↝$\tt { 5}^{2n} = {5}^{4}$

Now , bases are same

so, powers are equal

$\tt2n = 4$

$\tt n = \dfrac{4}{2} = 2$

N = 2

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$\sf\green{ QUESTION :- }$

solve for m : m - m-1/2 = 1 - m-2/3 and verify the result

$\sf\blue{ ANSWER :- }$

↝$\tt m - \dfrac{m - 1}{2} = 1 - \dfrac{m - 2}{3}$

↝$\tt \dfrac{m}{1} - \dfrac{m - 1}{2} + \dfrac{m - 2}{3} = 1$

↝$\tt \dfrac{6m - 3(m - 1) + 2(m - 2)}{6} = 1$

↝$\tt \dfrac{6m - 3m + 3+ 2m - 4}{6} = 1$

↝$\tt \dfrac{5m - 1}{6} = 1$

↝$\tt {5m - 1} = 6$

↝$\tt 5m = 6 + 1$

↝$\tt 5m =7$

↝$\tt m = \dfrac{7}{5}$

To check

↝$\tt \dfrac{7}{5} - \dfrac{ \dfrac{7}{5} - 1 }{2} = 1 - \dfrac{ \dfrac{7}{5} - 2}{3}$

↝$\tt \dfrac{7 }{5} - \dfrac{ \dfrac{7 - 5}{5} }{2} = 1 - \dfrac{ \dfrac{7 - 10}{5} }{3}$

↝$\tt \dfrac{7 }{5} - \dfrac{ 2 }{5 \times 2} = 1 - \dfrac{ - 3 }{5 \times 3}$

↝$\tt \dfrac{7 }{5} - \dfrac{1}{5} = 1 + \dfrac{ 1 }{5 }$

↝$\tt \dfrac{7 - 1 }{5} = \dfrac{5 + 1 }{5 }$

↝$\tt \dfrac{6 }{5} = \dfrac{6 }{5 }$

LHS = RHS

Therefore, it is correct

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$\sf\green{ QUESTION :- }$

prove that if two lines intersect then the vertically opposite angles are equal

$\sf\blue{ ANSWER :- }$

★ Note :- See the attachment

Given -

AB and CD intersect at O

To Prove -

∠ AOC = ∠ BOD

∠ AOD = ∠ BOC

Prove -

∠ AOD + ∠ AOC = 180............(1)

∠ AOD + ∠ BOD = 180............(2)

1 & 2 ;

∠ AOD + ∠ AOC = ∠ AOD + ∠ BOD

★ ∠ AOD will cancelled

∠ AOC = ∠ BOD

NOW,

∠ AOD + ∠ AOC = 180............(3)

∠ AOC + ∠ BOC = 180............(4)

3 & 4 ;

∠ AOD + ∠ AOC = ∠ AOC + ∠ BOC

★ ∠ AOC will cancelled

∠ AOD = ∠ BOC

therefore, vertically opposite angles are equal