Selina Concise Class 10 Math Chapter 6 Solving (simple) problems (Based on Quadratic Equations) Exercise 6D Solutions
Solving (simple) problems
(Based on Quadratic Equations)
Exercise 6D
(Q1) The sum S of n successive odd numbers starting from 3 is given by the relation: S = n (n + 2).
Determine n, if the sum is 168.
Solution:
The sum S of n successive odd numbers starting from 3 is given by the relation
S = n (n + 2) —— (1)
Also, given that, S = 168 —— (2)
∴ From (1) and (2), we get
n (n + 2) = 168
n2 + 2n = 168
n2 + 2n – 168 = 0
n2 + 14n – 12n – 168 = 0
n (n + 14) – 12 (n + 14) = 0
(n + 14) (n – 12) = 0
n + 14 = 0 or n – 12 = 0
n = -14 or n = 12
∴ The natural numbers cannot be negative.
∴ n = -14 is not a solution.
∴ n = 12 is the only solution.
(Q2) A stone is thrown vertically downwards and the formula d = 16 t2 + 4t gives the distance, d meters, that it falls in t seconds, how long does it take to fall 420 meters?
Solution:
A stone is thrown vertically downwards and the formula,
d = 16t2 + 4t —– (i)
d = 16t2 + 4
d = 420 m
Also, given that,
d = 420 m, (d = distance) —— (ii)
From (i) and (ii), we get 16t2 + 4t = 420
16t2 + 4t – 420 = 0
Dividing 4 on both sides
16t2/4 + 4t/4 – 420/4 = 0
4+2 + t – 105 = 0
4t2 + 21t – 20t – 105 = 0
t (4t + 21) – 5 (4t + 21) = 0
(4t + 21) (t – 5) = 0
(4t + 21) = 0 or t – 5 = 0
4t = -21 or t = 5
t = -21/4
But time cannot negative
∴ t = 5 is the only solution
∴ t = 5 seconds
(Q3) The product of the digits of a two digit number is 24. If its unit’s digits exceeds twice its ten’s digit by 2; find the number.
Solution:
Let unit digit be x and Let ten digit be x
If it’s unit’s digit exceeds twice its ten’s digit by 2.
∴ x = 2y + 2 ——– (1)
Also give that, the product of the digit of a two digit number is 24.
xy = 24 —— (2)
From equation (1), we get
(2y + 2) y = 24
2y2 + 2y = 24
2y2 + 2y – 24 = 0
divide by ‘2’ on both sides
2y2/2 + 2y/2 – 24/2 = 0
y2 + y – 12 = 0
y2 + 4y – 3 – 12 = 0
y (y + 4) – 3 (y + 4) = 0
(y + 4) (y – 3) = 0
y + 4 = 0 or y – 3 = 0
y = -4 or y = 3
If y = -4 then x = 2×(-4) + 2
= -8+ 2
x = -6
If y = 3 then x = 2 (3) + 2
= 6 + 2
x = 8
∴ Number = 10y + x
= 10×3 + 8 (1tens = 10)
= 30 + 8
= 38
∴ The required number is 38.
(Q4) The ages of two sisters are 11 years and 14 years. In how many years time will the product of their ages be 304?
Solution:
After x years,
The age of first sister is, (11 + x) and the age of second sister is (14 + x)
Also given the product of their ages be 304
(11 + x) (14 + x) = 304
11 (14 + x) + x (14 +x) = 304
154 + 11x + 14x + x2 = 304
154 + 25x + x2 = 304
x2 + 25x + 154 – 304 = 0
x2 + 25x – 150 = 0
x2 + 30x – 5x – 150 = 0
x (x + 30) – 5 (x + 30) = 0
(x + 30) (x – 5) = 0
x + 30 = 0 or x – 5 = 0
x = -30 or x = 5
But the years is not negative,
∴ x = 5 is the only solution.
∴ The age of first sister is (11 + x) = (11 + 5) = 16 years
and the age of second sister is (14 + x) = (14 + 5) = 19 years
∴ After 5 years the product of their ages be 304.
(Q5) One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s. Find their present ages.
Solution:
Let present age of son is x years and the present age of father is y years
One year ago, a man was 8 times as old as his son.
In one year ago,
The son’s age = (x – 1) years and the father age = (y – 1) years
∴ From question,
(y – 1) = 8 (x – 1) —– (1)
∴ Now, his father age is equal to the square of his son’s age
∴ y = x2 —– (2)
From equation (1) & (2), we get
(x2 – 1) = 8x – 8
x2 – 1 = 8x – 8
x2 – 8x – 1 + 8 = 0
x2 – 8x + 7 = 0
x2 – 7x – 1x + 7 = 0
x (x – 7) – 1 (x – 7) = 0
(x – 7 (x – 1) = 0
x – 7 = 0 or x – 1 = 0
x = 7 or x = 1
∴ The son’s age is x = 7 year and the father age is
y = x2 = (7)2 = 49
y = 49 years
(Q6) The age of the father is twice the square of the age of his son. Eight years hence, the age of the father will be 4 years more than three times the age of the son. Find their present ages.
Solution:
Let the age of father is y years and the age of son is x years
Given that, the age of the father is twice the square of the age of his son.
∴ y = 2x2 —- (1)
Eight years hence, the age of the father will be 4 years more than three times the age of the son.
∴ (y + 8) = 3 (x + 8) + 4 —– (2)
From equation (1) & (2), we get,
(2x2 + 8) = 3 (x + 8) + 4
2x2 + 8 = 3x + 24 + 4
2x2 +8 = 3x + 28
2x2 – 3x + 8 – 28 = 0
2x2 – 3x – 20 = 0
2x2 – 8x + 5 – 20 = 0
2x (x – 4) + 5 (x – 4) = 0
(x – 4) (2x + 5) = 0
x – 4 = 0 or 2x + 5 = 0
2x = -5
x = 4
x = -5/2
But the age cannot be negative
∴ x = 4 is the only solution.
The son’s age = 4 years
and the father age = y = 2x2
y = 2 × (4)2
= 2 × 16
y = 32 year
Here is your solution of Selina Concise Class 10 Math Chapter 6 Solving (simple) problems Exercise 6D (Based on Quadratic Equations)
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