There, we explained solving of
the Quadratic Equations by Factoring
and by Quadratic formula respectively
with a number of solved Examples.
Here, a number of math word problems are worked out.
Several math word problems are given
for practice in exercise with answers.
Practice makes one perfect.
This is especially true for learning
of solving Math Word Problems.
So, take the exercises seriously and
practice solving the Math Word Problems.
Example 1 of Math Word Problems
Solve the Following Math Word Problem.
The product of two consecutive odd integers is 899. Find them.
Solution to Example 1 of Math Word Problems :
We know, two consecutive odd integers differ by 2.
Let the two consecutive odd integers be (x - 2) and x.
By data, their product = 899 ⇒ (x - 2)x = 899 ⇒ x2 - 2x - 899 = 0
Comparing this equation with ax2 + bx + c = 0, we get a = 1, b = -2 and c = -899
We know by Quadratic Formula, x = {-b ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get x = {-b ± √(b2 - 4ac)}⁄2a
= [-(-2) ± √{ (-2)2 - 4(1)( -899)}]⁄2(1)
= [2 ± √{ 4 + 4(1)(899)}]⁄2 = [2 ± √{4(1 + 899)}]⁄2
= [2 ± √{4(900)}]⁄2 = [2 ± 2(30)}]⁄2 = 1±30 = 1+30 or 1-30 = 31 or -29
The two consecutive odd integers are (x - 2) and x.
When x. = 31, the required integers are 31 - 2 and 31 i.e. 29 and 31. Ans.
When x. = -29, the required integers are -29 - 2 and -29 i.e. -31 and -29. Ans.
Example 2 of Math Word Problems
Solve the Following Math Word Problem.
Find the number which is less than its square by 182.
Solution to Example 2 of Math Word Problems :
Let the required number be x.
Then, by data, it is less than its square by 182.
⇒ x = x2 - 182 ⇒ x2 - x - 182 = 0
Comparing this equation with ax2 + bx + c = 0, we get a = 1, b = -1 and c = -182
We know by Quadratic Formula, x = {-b ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get x = {-b ± √(b2 - 4ac)}⁄2a
= [-(-1) ± √{ (-1)2 - 4(1)( -182)}]⁄2(1)
= [1 ± √{ 1 + 4(1)(182)}]⁄2 = [1 ± √{(1 + 728)}]⁄2
= [1 ± √{729}]⁄2 = [1 ± 27]⁄2 = (1 + 27)⁄2, (1 - 27)⁄2 = (28)⁄2, (- 26)⁄2 = 14, -13
Thus, the number which is less than its square by 182 is 14 or -13. Ans.
Example 3 of Math Word Problems
Solve the Following Math Word Problem.
The sum of the squares of two odd integers is 202. Find them.
Solution to Example 3 of Math Word Problems :
We know, two consecutive odd integers differ by 2.
Let the two consecutive odd integers be (x - 2) and x.
By data, the sum of their squares = 202 ⇒ (x - 2)2 + x2 = 202
⇒ x2 - 4x + 4 + x2 = 202 ⇒ 2x2 - 4x + 4 = 202
⇒ x2 - 2x + 2 = 101 ⇒ x2 - 2x - 99 = 0
Comparing this equation with ax2 + bx + c = 0, we get a = 1, b = -2 and c = -99
We know by Quadratic Formula, x = {-b ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get x = {-b ± √(b2 - 4ac)}⁄2a
= [-(-2) ± √{ (-2)2 - 4(1)( -99)}]⁄2(1)
= [2 ± √{ 4 + 4(1)(99)}]⁄2 = [2 ± √{4(1 + 99)}]⁄2
= [2 ± √{4(100)}]⁄2 = [2 ± 2(10)}]⁄2 = 1±10 = 1+10 or 1-10 = 11 or -9
The two consecutive odd integers are (x - 2) and x.
When x. = 11, the required integers are 11 - 2 and 11 i.e. 9 and 11. Ans.
When x. = -9, the required integers are -9 - 2 and -9 i.e. -11 and -9. Ans.
Example 4 of Math Word Problems
Solve the Following Math Word Problem.
The length and breadth of a rectangle differ by 17 m.
The length and diagonal differ by 1 m. Find the length and breadth.
Solution to Example 4 of Math Word Problems :
Let l be the length of the rectangle. Then, by data it breadth = (l - 17).
Its diagonal = √{(l)2 + (l - 17)2}
By data, the length and diagonal differ by 1 m.
⇒ √{(l)2 + (l - 17)2} = l + 1
Squaring both sides, we get
(l)2 + (l - 17)2 = (l + 1)2
⇒ l2 + l2 - 34l + 289 = l2 + 2l + 1
⇒ l2 - 36l + 288 = 0
Comparing this equation with al2 + bl + c = 0, we get a = 1, b = -36 and c = 288
We know by Quadratic Formula, l = {-b ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get l = {-b ± √(b2 - 4ac)}⁄2a
= [-(-36) ± √{ (-36)2 - 4(1)( 288)}]⁄2(1)
= [36 ± √{ 36 x 36 - 4 x 8 x 36]⁄2 = [36 ± √{36(36 - 32)}]⁄2
= [36 ± √{36(4)}]⁄2 = [36 ± 6(2)}]⁄2 = 18 ± 6 = 18+6 or 18-6 = 24 or 12
Length can not be 12m,( because breadth = length - 17 = -5 is not possible.) ∴ l ≠ 12 ; ∴ l = 24.
Length = 24 m; Breadth = 24 - 17 = 7m. Ans.
Example 5 of Math Word Problems
Solve the Following Math Word Problem.
The length and breadth of a rectangle differ by 10 m.
Its Area in square meters is 92 more than it perimeter in meters.
Find the length and breadth.
Solution to Example 5 of Math Word Problems :
Let l be the length of the rectangle. Then, by data, it breadth = (l - 10).
Its area = length x breadth = l(l - 10)
By data, Its Area in square meters is 92 more than it perimeter in meters.
⇒ l(l - 10) = 2 {l + (l - 10) } + 92 ⇒ l2 - 10l = 2(2l - 10) + 92 = 4l - 20 + 92 = 4l + 72
⇒ l2 - 14l - 72 = 0
Comparing this equation with al2 + bl + c = 0, we get a = 1, b = -14 and c = -72
We know by Quadratic Formula, l = {-b ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get l = {-b ± √(b2 - 4ac)}⁄2a
= [-(-14) ± √{(-14)2 - 4(1)(-72)}]⁄2(1)
= [14 ± √{196 + 288}]⁄2 = [14 ± √{4(49 + 72)}]⁄2
= [14 ± √{4(121)}]⁄2 = [14 ± 2(11)}]⁄2 = 7 ± 11 = 7+11 or 7-11 = 18 or -4
Length can not be negative. ∴ l ≠ -4 ; ∴ l = 18.
Length = 18 m; Breadth = 18 - 10 = 8m. Ans.
Example 6 of Math Word Problems
Solve the Following Math Word Problem.
The denominator of a fraction exceeds the numerator by 4
and the fraction formed by squaring both numerator and denominator
is equal to 4⁄9. Find the fraction.
Solution to Example 6 of Math Word Problems :
Let the denominator of the fraction be x.
Then its numerator as per data is (x - 4)
Square of the fraction = {(x - 4) ⁄x}2 = 4⁄9 [By data ]
(x - 4)2⁄x2 = 4⁄9
crossmultiplying, we get
9(x - 4)2 = 4x2 ⇒ 9(x2 - 8x + 16) = 4x2
⇒ 9x2 - 72x + 144 = 4x2 ⇒ 9x2 - 4x2 - 72x + 144 = 0
⇒ 5x2 - 72x + 144 = 0
Comparing this equation with ax2 + bx + c = 0, we get a = 5, b = -72 and c = 144
We know by Quadratic Formula, x = {-b ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get x = {-b ± √(b2 - 4ac)}⁄2a
= [-(-72) ± √{ (-72)2 - 4(5)( 144)}]⁄2(5)
= [72 ± √{ 72 x 72 - 4(5)(2 x 72)}]⁄10 = [72 ± √{72 x 8(9 - 5)}]⁄10
= [72 ± √{8 x 9 x 8 x 4}]⁄10 = [72 ± 8 x 3 x 2)}]⁄10 = (72±48)⁄10 = (72+48)⁄10 or (72-48)⁄10 = 12 or 2.4
Taking the integral value, x = denominator = 12. Then Numerator = 12 - 4 = 8.
The required fraction = 8⁄12. Ans.
Example 7 of Math Word Problems
Solve the Following Math Word Problem.
A stream flows from A to B, a distance of 15 km. A man who can
row in still water at 4 kmph can row up and down in 8 hours.
Find the rate of the stream.
Solution to Example 7 of Math Word Problems :
Let the rate of the stream be xkmph. xkmph;
rowing speed of the man in the opposite direction of stream
= rowing speed in still water - speed of the stream. = 4 kmph - xkmph;
We know, Time = Distance⁄Speed
Time in hours for rowing up and down = (15 km)⁄( 4 + x)kmph + (15 km)⁄(4 - x)kmph
By data this is equal to 8 hours.
∴ 15⁄(4 + x) + 15⁄(4 - x) = 8
Multiplying both sides with (4 + x)(4 + x), we get
15(4 - x) + 15(4 + x) = 8(4 + x)(4 + x)
⇒ 15(4 - x + 4 + x) = 8(42 - x2)
⇒ 15(8) = 8(16 - x2) ⇒ 15 = 16 - x2
⇒ x2 = 16 - 15 = 1 ⇒ x = ± 1 = +1, -1
But x can not be negative. ∴ x = 1.
Thus, the rate of the stream = 1 kmph. Ans.
Example 8 of Math Word Problems
Solve the Following Math Word Problem.
A cyclist covers a distance of 60 km in a given time. If he increases
his speed by 2 kmph, he will cover the distance one hour before.
Find the original speed of the cyclist.
Solution to Example 8 of Math Word Problems :
Let the original speed of the cyclist be x kmph.
Then, the time cyclist takes to cover a distance of 60 km = 60⁄x
If he increases his speed by 2 kmph, the time taken = 60⁄(x + 2)
By data, the second time is less than the first by 1 hour.
∴ 60⁄(x + 2) = 60⁄x - 1
Multiplying both sides with (x + 2)x, we get
60x = 60(x + 2) - 1(x + 2)x = 60x + 120 - x2 - 2x
⇒ x2 + 2x - 120 = 0
Comparing this equation with ax2 + bx + c = 0, we get a = 1, b = 2 and c = -120
We know by Quadratic Formula, x = {-b ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get x = {-b ± √(b2 - 4ac)}⁄2a
= [-2 ± √{ (2)2 - 4(1)( -120)}]⁄2(1)
= [-2 ± √{ 4 + 4(1)(120)}]⁄2 = [-2 ± √{4(1 + 120)}]⁄2
= [-2 ± √{4(121)}]⁄2 = [-2 ± 2(11)}]⁄2 = -1±11 = -1+11 or -1-11 = 10 or -12
But x can not be negative. ∴ x = 10.
∴ The original speed of the cyclist = x kmph. = 10 kmph. Ans.
Example 9 of Math Word Problems
Solve the Following Math Word Problem.
Two years ago, a man's age was three times the square of his son's age.
In three years time, his age will be four times his son's age.
Find their present ages.
Solution to Example 9 of Math Word Problems :
Let the present age of the son be x years.
Two years ago, the son's age = x - 2
By data, Two years ago, a man's age was three times
the square of his son's age.
⇒ Two years ago, the man's age = 3(x - 2)2
⇒ Present age of man = 3(x - 2)2 + 2
In three years time, man's age = {3(x - 2)2 + 2} + 3
In three years time, son's age = x + 3
By data, In three years time, man's age will be four times his son's age.
⇒ {3(x - 2)2 + 2} + 3 = 4(x + 3)
⇒ 3(x2 - 4x + 4) + 2 + 3 = 4x + 12
⇒ 3x2 - 12x + 12 + 5 = 4x + 12
⇒ 3x2 - 16x + 5 = 0
Comparing this equation with ax2 + bx + c = 0, we get a = 3, b = -16 and c = 5
We know by Quadratic Formula, x = {-b ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get x = {-b ± √(b2 - 4ac)}⁄2a
= [-(-16) ± √{ (-16)2 - 4(3)(5)}]⁄2(3)
= [16 ± √{ 256 - 60}]⁄6 = [16 ± √{196}]⁄6
= [16 ± 14]⁄6 = (8 ± 7)⁄3
= (8+7)⁄3, (8-7)⁄3 = 15⁄3, 1⁄3 = 5, 1⁄3
But x can not be 1⁄3. ∴ x = 5.
Present age of son = 5 years. Ans.
Present age of man = 3(x - 2)2 + 2 = 3(5 - 2)2 + 2 = 3 x 9 + 2 = 29 years. Ans.
Example 10 of Math Word Problems
Solve the Following Math Word Problem.
A number consists of two digits whose product is 18.
If 27 is added to the number, the digits interchange their places.
Find the number.
Solution to Example 10 of Math Word Problems :
Let the unit's digit of the original number be x.
By data, the product of the digits 18
∴ The ten's digit of the original number = 18⁄x
Original number's value = 1x unit's digit + 10 x ten's digit = 1 x x + 10 x 18⁄x = x + 180⁄x
Digits reversed number value = 1 x 18⁄x + 10 x x = 18⁄x + 10x
By data, Original number + 27 = Digits reversed number
⇒ x + 180⁄x + 27 = 18⁄x + 10x
Multiplying both sides with x, we get x2 + 180 + 27x = 18 + 10x2
⇒ -9x2 + 27x + 162 = 0
Dividing by -9, we get x2 - 3x - 18 = 0
Comparing this equation with ax2 + bx + c = 0, we get a = 1, b = -3 and c = -18
We know by Quadratic Formula, x = {-b ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get x = {-b ± √(b2 - 4ac)}⁄2a
= [-(-3) ± √{ (-3)2 - 4(1)(-18)}]⁄2(1)
= [3 ± √{ 9 + 72}]⁄2 = [3 ± √{81}]⁄2
= [3 ± 9]⁄2 = (3 + 9)⁄2, (3 - 9)⁄2 = 6, -3
But x can not be negative. ∴ x = 6.
unit's digit of the original number = x = 6
The ten's digit of the original number = 18⁄x = 18⁄6 = 3
∴ the required number is 36. Ans.
Exercie : Math Word Problems
The product of two cosecutive integers is 812. Find them.
Thedifference of two numbers is 6 and their product is 391. Find them.
Twice the square of a number exceeds 4 times the number by 240. Find it.
The length and breadth of a rectangle differ by 1 m.
The length of the diagonal is 29 m. Find the length and breadth.
The length and breadth of a rectangle differ by 7 m.
The area of the rectangle is 144 m2.Find the length and breadth.
A number is of two digits, one of which is the square of the other.
The number formed by reversing the digits exceeds
twice the number by 15. Find the number.
A stream flows from A to B, a distance of 30 km at 2 kmph. A man can row up
and down in 8 hours. Find the rate of the man in still water.
A rectangle is 100 m by 60 m. It has grass for certain width all around it
on the outside. Area of grass is 3⁄5 of the area of the rectangle.
Find the width of the grass.
The sum of the numerator and denominator of a certain fraction is 10.
If 1 is subtracted from both the numerator and denominator, the fraction
is decreased by 2⁄21. Find the fraction.
A man walks a distance of 48 kms. in a given time. If he walks
2 kmph faster, he will perform the journey 4 hours before. Find his
normal rate of walking.
