1. The sum of 4 times a number and 3 times its reciprocal is 7 find the no.
2. 2YRS AGO MY AGE WAS 41/2 TIMES THE AGE OF MY SON 6YRS AGO MY AGE WAS TWICE THE SQUARE OF THE AGE OF SON WHAT IS THE PRESENT AGE OF SON
3. IN A TWO DIGIT NO. THE DIGIT IN THE UNITS PLACE IS EQUAL TO THE SQUARE OF THE DIGIT AT TENSE PLACE IF 54 IS ADDED TO THE NO THE DIGIT GET INTERCHANGED FIND THE NO.
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Answers
Step-by-step explanation:
1.
Let the number be x.
Reciprocal = (1/x)
Given,
Four times a number and 3 time its reciprocal is 7.
=> 4x + (3/x) = 7
=> 4x² + 3 = 7x
⇒ 4x² - 7x + 3 = 0
⇒ 4x² - 4x - 3x + 3 = 0
⇒ 4x(x - 1) - 3(x - 1) = 0
⇒ (4x - 3)(x - 1) = 0
⇒ x = 3/4, 1
Therefore, the number are : (3/4), 1
2.
Let the son's age two years ago be x.
Let his father's age two years ago be = (9/2)x
Six years ago, :
Son's age = (x-6)
Father' age = (9/2)x - 6
According to question, his age was twice the square of his son's age.
⇒ (9/2)x - 6 = (x - 6)²
⇒ 9x - 12 = 6x² - 72x + 216
⇒ 2x² - 27x + 76 = 0
⇒ 2x² - 8x - 19x + 76 = 0
⇒ 2x(x - 4) - 19(x - 4) = 0
⇒ (2x - 19)(x - 4) = 0
⇒ x = 19/2, 4
⇒ x = 4{Age cant be in fraction}
So,
Son's present age = 4 + 2 = 6.
Therefore,
Present age of son = 6 years.
3.
Let the tens digit = x, and the units digit =y.
Then y = x^2 .
The original number is then 10x + x^2.
When 54 is added the new number is 10y + x (digits reversed).
Therefore 10x + x^2 + 54 = 10x^2 + x
10x^2 + x - x^2 - 10x - 54 = 0
9x^2 - 9x - 54 = 0
x^2 - x - 6 = 0 (divide throughout by 9)
(x - 3)(x + 2) = 0
Therefore x-3 = 0 or x + 2 = 0.
Therefore x = 3
Therefore y = x^2 = 9
So, the original number = 10x + y = 39.
Therefore,
Number = 39
Hope it helps!