Selina Concise Class 10 Math Chapter 6 Solving (Simple) problems (Based on Quadratic Equations) Exercise 6A Solutions
Solving (Simple) problems
(Based on Quadratic Equations)
Exercise 6A
(Q1) The product of two consecutive integers is 56. Find the integers.
Solution:
Let us consider two consecutive integers be x and x+1.
The product of two consecutive integers is 56.
x (x+1) = 56
x2 + x = 56
x2 + x – 56 = 0
x2 + 8x – 7x – 56 = 0
x (x + 8) – 7 (x + 8) = 0
(x + 8) (8 – 7) = 0
x + 8 = 0 or x – 7 = 0
x + 1 = 7 + 1 = 8
∴ The required integers are (-8, 7) or (7, 8)
(Q2) The sum of the squares of two consecutive natural numbers is 41 find the numbers.
Solutions:
Let us consider the two consecutive natural numbers are x and x+1
The sum of the squares of two consecutive natural numbers is 41.
x2 + (x + 1)2= 41
x2 + [x2 + 2x + 1] = 41 [∵ (a + b)2 = a2 + 2ab + b2]
x2 + x2 + 2x + 1 = 41
2x2 + 2x + 1 – 41 = 0
2x2 + 2x – 40 = 0
x2 + x – 20 = 0
x2 + 5x – 4x – 20 = 0
x (x + 5) – 4x – 20 = 0
x (x + 5) – 4 (x + 5) = 0
(x + 5) (x – 4) = 0
x + 5 = 0 or x – 4 = 0
x = -5 or x = 4
∴ As x = -5 is not a natural number.
∴ x = 4 is the only solution x + 1 = 4+1 = 5
∴ The two consecutive numbers are 4 and 5.
(Q3) Find the two natural number which differ by 5 and the sum of whose squares is 97.
Solution:
Let us assume that the two natural numbers x and x + 5.
We have to assume x + 5 because given that they differ by 5.
The sum of whose squares is 97
x2 (x + 5)2 = 97
x2 + x2 + 2 × x × 5 + 25 = 97
{(a + b)2 = a2 = a2 + 2ab + b2}
x2 + x2 + 10x + 25 – 97 = 0
2x2 + 10x – 72 = 0
2x2/2 + 10x/2 – 72/2 = 0
x2 + 5x – 36 = 0
x2 + 9x – 4x – 36 = 0
x (x + 9) – 4 (x + 9) = 0
(x + 9) (x – 4) = 0
x + 9 = 0 or x – 4 = 0
x = -9 or x = 4
But x = -9 is not a natural number.
∴ x = 4 is the only solution x + 5 = 4 + 5 = 9
∴ The two natural numbers are 4 and 9.
(Q4) The sum of a number and it’s reciprocal is 4.25. Find the number.
Solution:
Let the number be x and it’s reciprocal is 1/x
The sum of a number and it’s reciprocal is 4.25.
x + 1/x = 4.25
x2 + 1/x = 4.25
x2 + 1/x = 425/100
x2 + 1/x = 17/4
4 (x2 + 1) = 17x
4x2+ 4 = 17x
4x2 – 17x + 4 = 0
4x2 – 16x – x + 4 = 0
4x (x – 4) – 1 (x – 4) = 0
(x – 4) (4x – 1) = 0
x – 4 = 0 or 4x – 1 = 0
x = 4 or 4x = 1
x = ¼
∴ The required number is x = 4, ¼
(Q5) Two natural numbers differ by 3. Find the numbers, if the sum of their reciprocals is 7/10.
Solution:
Let us consider the two natural numbers x and x + 3
We have to assume x + 3 because of they differ by 3
The sum of their reciprocals is 7/10
1/x + 1/x + 3 = 7/10
x+3+x/x(x+3) = 7/10
2x + 3/x2 + 3x = 7/10
10 (2x + 3) = 7 (x2 + 3x)
20x + 30 = 7x2 + 21x
7x2 + 21x – 20x – 30 = 0
7x2 + x – 30 = 0
7x2 + x – 30 = 0
7x2 – 14x + 15x – 30 = 0 7× (-30) = 210
7x (x – 2) + 15 (x – 2) = 0
(x – 2) = 0 or 7x + 15 = 0
(x – 2) = 0 or 7x + 15 = 0
7x = -15
x = 2 or x = -15/7
But x = -15/7 is not natural number
∴ x = 2 is the only solution.
x + 3 = 2+3 = 5
Therefore, the two natural numbers are 2 and 5.
(Q6) Divide 15 into two parts such that the sum of their reciprocals is 3/10.
Solution: –
Let us consider or assume the two parts x and 15 – x
We have assume 15 – x because they divide 15 into two parts.
The sum of their reciprocals is 3/10
1/x + 1/15-x = 3/10
15-x+x/x(15-x) = 3/10
15/15x – x2 = 3/10
10×15 = 3(15x – x2)
150 = 45x – 3x2
3x2/3 – 45x/3 + 150/3 = 0
x2 – 15x + 50 = 0
x2 – 15x + 50 = 0
x2 – 10x – 5 + 50 = 0
x (x – 10) – 5 (x – 10) = 0
(x – 10) (x – 5) = 0
x – 10 = 0 or x – 5 = 0
x = 10 or x = 5
∴ The two parts are 5 and 10
(Q7) The sum of the squares of two positive integers is 208. If the square of larger number is 18 times the smaller number, find the numbers.
Solution:
Let the smaller number be x and the larger number be y.
The sum of the squares of two positive integers is 208.
x2 + y2 = 208 —– (i)
and if the square of larger number is 18.
y2 = 18x —– (ii)
From equation (i), we get y2 = 208 – x2 —– (ii)
Put y2 = 208 – x2 in equation (ii)
208 – x2 = 18x
– x2 – 18x + 208 = 0
x2 + 18x – 208 = 0
x2 + 26x – 8x – 208 = 0
x (x + 26) – 8 (x + 26) = 0
(x + 26) (x – 8) = 0
x + 26 = 0 or x – 8 = 0
x = -26 or x = 8
But x cannot be a negative integer. so, x = -26 is not a solution.
∴ x = 8 is the only solution.
∴ x = 8 put in equation (i), we get.
y2 = 18×8
y2 = 144
∴ y = 12 is also a positive integer
∴ The two numbers are 8 and 12
(Q8) The sum of the squares of two consecutive positive even numbers is 52. Find the numbers.
Solution:
Let us consider two positive even numbers be x and x + 2
The sum of the squares of two consecutive positive even numbers is 52
x2 + (x + 2)2 = 52
x2 + x2 + 4x + 4 = 52 [(a + b)2 = a2 + 2ab + b2]
2x2 + 4x + 4 – 52 = 0
2x2 + 4x – 48 = 0
2x2/2 + 4x/2 – 48/2 = 0
x2 + 2x – 24 = 0
x2 – 6x + 4x – 24 = 0
x (x – 6) + 4 (x – 6) = 0
(x – 6) (x + 4) = 0
(x – 6) or x + 4 = 0
x = 6 or x = -4
∴ But x is a positive integers only.
∴ x = -4 is not a solution
x = 6 is the only solution
Therefore, the numbers are 4 and 6
Here is your solution of Selina Concise Class 10 Math Chapter 6 Solving (Simple) problems (Based on Quadratic Equations)
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