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ch quadratic equations class 10
Answers
Sᴏʟᴜᴛɪᴏɴ :-
Let us Assume that, initially ramesh has x chocolates and suresh has y chocolates. ( where x > y).
Than,
→ (x + y) = 25 ----------- Eqn(1).
Now, suresh give 1 chocolate to ramesh .
So,
→ Now suresh have = (y - 1) chocolates.
→ Ramesh have = (x + 1) chocolates .
A/q,
→ (x + 1)² - (y - 1)² = 125
→ (x² + 2x + 1) - (y² - 2y + 1) = 125
→ x² + 2x + 1 - y² + 2y - 1 = 125
→ (x² - y²) + 2(x + y) = 125
Putting values of (x + y) from Eqn (1) in LHS,
→ (x² - y²) + 2*25 = 125
→ (x² - y²) = 125 - 50
→ (x² - y²) = 75
→ (x + y)(x - y) = 75
Again, Putting values of (x + y) from Eqn (1) in LHS,
→ 25 * (x - y) = 75
→ (x - y) = 3 ------------- Eqn(2).
____________
Now,
Adding Both Eqns. we get,
→ (x + y) + (x - y) = 25 + 3
→ x + x + y - y = 28
→ 2x = 28
→ x = 14 (Ans.)
Putting This value in Eqn.(1) Now,
→ (14 + y) = 25
→ y = 25 - 14
→ y = 11 (Ans.)
Hence, initially ramesh has 14 chocolates and suresh has 11 chocolates.
SaTyAmmmmm follow mee
Let us Assume that, initially ramesh has x chocolates and suresh has y chocolates. ( where x > y).
Than,
→ (x + y) = 25 ----------- Eqn(1).
Now, suresh give 1 chocolate to ramesh .
So,
→ Now suresh have = (y - 1) chocolates.
→ Ramesh have = (x + 1) chocolates .
A/q,
→ (x + 1)² - (y - 1)² = 125
→ (x² + 2x + 1) - (y² - 2y + 1) = 125
→ x² + 2x + 1 - y² + 2y - 1 = 125
→ (x² - y²) + 2(x + y) = 125
Putting values of (x + y) from Eqn (1) in LHS,
→ (x² - y²) + 2*25 = 125
→ (x² - y²) = 125 - 50
→ (x² - y²) = 75
→ (x + y)(x - y) = 75
Again, Putting values of (x + y) from Eqn (1) in LHS,
→ 25 * (x - y) = 75
→ (x - y) = 3 ------------- Eqn(2).
____________
Now,
Adding Both Eqns. we get,
→ (x + y) + (x - y) = 25 + 3
→ x + x + y - y = 28
→ 2x = 28
→ x = 14 (Ans.)
Putting This value in Eqn.(1) Now,
→ (14 + y) = 25
→ y = 25 - 14
→ y = 11 (Ans.)
Hence, initially ramesh has 14 chocolates and suresh has 11 chocolates.