Q1 the difference of two numbers is 5 and the difference of their reciprocals is 1/10 find the - Maths - Quadratic Equations

Question

Q1 the difference of two numbers is 5 and the difference of
their reciprocals is 1/10 find the numbers
Q2 by increasing the list price of a book by RS 10 a person can
buy 10 less books for RS 1200 find the original list price of the
book
Q3 a passenger train takes three hours less for a journey of 360
km, if its speed is increased by 10 km/hr what is the usual
speed?
Q4 the sum of the squares of two consecutive odd numbers is 394
find the numbers
Q5 the numerator of a fraction is one less than its denominator
if three is added to each of the numerator and denominator, the
fraction is increased by 3/28 find the fraction

Chetna Khanna · Accepted Answer

(i) Let the first number be x So, the next number will be x + 5 According to the given condition, $\frac{1}{x} - \frac{1}{x - 5} = \frac{1}{10} \Rightarrow \frac{x - 5 - x}{x (x - 5)} = \frac{1}{10} \Rightarrow \frac{- 5}{x (x - 5)} = \frac{1}{10} \Rightarrow - 50 = x (x - 5) \Rightarrow - 50 = x^{2} - 5 x \Rightarrow x^{2} - 5 x + 50 = 0 \Rightarrow x = \frac{5 \pm \sqrt{25 - 200}}{2} \Rightarrow x = \frac{5 \pm \sqrt{- 175}}{2}$ Roots of this equation are imaginary, so the given question is incorrect Please recheck your query (ii) Let the list price of the book be Rs x Number of books purchased = $\frac{1200}{x}$ When the price of books is increased by Rs 10 Number of books purchased = $\frac{1200}{x + 10}$ According to the given condition, $\frac{1200}{x} - \frac{1200}{x + 10} = 10 \Rightarrow 1200 (x + 10 - x) = 10 x (x + 10) \Rightarrow 1200 = x^{2} + 10 x \Rightarrow x^{2} + 10 x - 1200 = 0 \Rightarrow x^{2} + 40 x - 30 x - 1200 = 0 \Rightarrow x (x + 40) - 30 (x + 40) = 0 \Rightarrow (x - 30) (x + 40) = 0 \Rightarrow x = 30 o r - 40$ Thus, the list price of the book is Rs 30 Try to solve the other questions on your own in the similar manner and feel free to ask us if you face any problem in any step