Word Problems on Time and Work Class 10 Concise Exe-6A Selina Solutions Ch-6 Problems on Quadratic Equations. In this article you will learn how to solve simple word problems on quadratic Equations based on Number , Time and Work. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.

Word Problems on Time and Work Class 10 Concise Exe-6A Selina Solutions Ch-6 Problems on Quadratic Equations

Board	ICSE
Publications	Selina
Subject	Maths
Class	10th
Chapter-6	Solving (Simple) Problems (Based on Quadratic Equations)
Writer	R.K. Bansal
Exe-6A	Problems based on Number and Time and Work.
Edition	2025-2026

Word Problems on Time and Work

Class 10 Concise Exe-6A Selina Solutions Ch-6 Problems on Quadratic Equations

Que-1: The product of two consecutive integers is 56. Find the integers.

Sol: Let the two consecutive integers be x and x + 1.
From the given information,
x(x + 1) = 56
x² + x – 56 = 0
(x + 8) (x – 7) = 0
x = -8 or 7
Thus, the required integers are – 8 and -7; 7 and 8.

Que-2: The sum of the squares of two consecutive natural numbers is 41. Find the numbers.

Sol: Let the numbers be x and x + 1.
From the given information,
x² + (x + 1)² = 41
2x² + 2x + 1 – 41 = 0
x² + x – 20 = 0
(x + 5) (x – 4) = 0
x = -5, 4
But, -5 is not a natural number. So, x = 4.
Thus, the numbers are 4 and 5.

Que-3: Find the two natural numbers which differ by 5 and the sum of whose squares is 97.

Sol: Let the two numbers be x and x + 5.
From the given information,
x² + (x + 5)² = 97
2x² + 10x + 25 – 97 = 0
2x² + 10x – 72 = 0
x² + 5x – 36 = 0
(x + 9) (x – 4) = 0
x = -9 or 4
Since, -9 is not a natural number. So, x = 4.
Thus, the numbers are 4 and 9.

Que-4: The sum of a number and its reciprocal is 4.25. Find the number.

Sol:  Let the numbers be x and 1𝑥
Form the given information,
𝑥 + (1/𝑥) = 4.25
(𝑥²+1)/𝑥 = 425/100 = 17/4
4x² – 17x + 4 = 0
4x² – 16x – x + 4 = 0
4x(x – 4) – 1(x – 4) = 0
(x – 4)(4x – 1) = 0
𝑥 = 4, 1/4
𝑥 = 4 ⇒ 1𝑥 = 1/4
𝑥 = 1/4  ⇒ 𝑥 = 4
Thus, the number are 4 and 1/4

Que-5: Two natural numbers differ by 3. Find the numbers, if the sum of their reciprocals is 7/10

Sol: Let the numbers be x and x + 3.
From the given information,
(1/𝑥) + {1/(𝑥+3)} = 7/10
{𝑥+3+𝑥}/{𝑥⁢(𝑥+3)} = 7/10
(2⁢𝑥+3)/(𝑥²+3⁢𝑥) = 7/10
20x + 30 = 7x² + 21x
7x² + x – 30 = 0
7x² – 14x + 15x – 30 = 0
7x(x – 2) + 15(x – 2) = 0
(x – 2)(7x + 15) = 0
𝑥 = 2,  −15/7
Since, x is a natural number, so x = 2.
Thus, the numbers are 2 and 5.

Que-6: Divide 15 into two parts such that the sum of their reciprocals is 3/10

Sol: Let the two parts be x and 15 – x.
(1/𝑥) + {1/(15−𝑥)} = 3/10
(15−𝑥+𝑥)/{𝑥⁢(15−𝑥)} = 3/10
15/(15⁢𝑥−𝑥²) = 3/10
{15/(15⁢𝑥−𝑥²)} × (10/3) = 0
{5/(15⁢𝑥−𝑥²) × 10 = 0
50 =15⁢𝑥 −𝑥²
𝑥² −15⁢𝑥 +50 =0
𝑥² −10⁢𝑥 −5⁢𝑥 +50 =0
(x – 10)(x – 5) = 0
∴ x – 10 = 0
∴ x = 10
∴ x – 5 = 0
∴ x = 5
Thus the required two parts are 5 and 10.

Que-7: The sum of the square of two positive integers is 208. If the square of larger number is 18 times the smaller number, find the numbers.

Sol: Let the two numbers be x and y, y being the bigger number. From the given information,
x² + y² = 208 ….. (i)
y² = 18x ….. (ii)
From (i), we get y²=208 – x². Putting this in (ii), we get,
208 – x² = 18x
⇒ x² + 18x – 208 = 0
⇒ x² + 26X – 8X – 208 = 0
⇒ x(x + 26) – 8(x + 26) = 0
⇒ (x – 8)(x + 26) = 0
⇒ x can’t be a negative number , hence x = 8
⇒ Putting x = 8 in (ii), we get y² = 18 x 8=144
⇒ y = 12, since y is a positive integer
Hence, the two numbers are 8 and 12.

Que-8: The sum of the squares of two consecutive positive even numbers is 52. Find the numbers.

Sol: Let the consecutive positive even numbers be x and x + 2.
From the given information,
x² + (x + 2)² = 52
2x² + 4x + 4 = 52
2x² + 4x – 48 = 0
x² + 2x – 24 = 0
(x + 6) (x – 4) = 0
x = -6, 4
Since, the numbers are positive, so x = 4.
Thus, the numbers are 4 and 6.

Que-9: Find two consecutive positive odd numbers, the sum of whose squares is 74.

Sol: Let the consecutive positive odd numbers be x and x + 2.
From the given information,
x² + (x + 2)² = 74
2x² + 4x + 4 = 74
2x² + 4x – 70 = 0
x² + 2x – 35 = 0
(x + 7)(x – 5) = 0
x = -7, 5
Since, the numbers are positive, so, x = 5.
Thus, the numbers are 5 and 7.

Que-10: The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2.9; find the fraction.

Sol:  Let the required fraction be 𝑥/(2⁢𝑥+1)
From the given information,
{𝑥/(⁢𝑥+1)} + {(2⁢𝑥+1)/𝑥} = 2.9
(𝑥²+4⁢𝑥²+1+4⁢𝑥)/{𝑥⁢(2⁢𝑥+1)} = 29/10
(5⁢𝑥²+1+4⁢𝑥)/(2⁢𝑥²+𝑥) = 29/10
50⁢𝑥² +10 +40⁢𝑥 = 58⁢𝑥² + 29⁢𝑥
8⁢𝑥² −11⁢𝑥 −10 =0
𝑥 = {11±√(121+320)}/16
𝑥 = {11±√441}/16
𝑥 = (11±21)/16
𝑥 = 2, −5/8
Thus, the required fraction is 2/5

Que-11: Three positive numbers are in the ratio 1/2 : 1/3 : 1/4. Find the numbers if the sum of their squares is 244.

Sol: Given, three positive numbers are in the ratio 1/2 : 1/3 : 1/4 = 6 : 4 : 3
Let the numbers be 6x, 4x and 3x.
From the given information,
(6x)² + (4x)² + (3x)² = 244
36x² + 16x² + 9x² = 244
61x² = 244
x² = 4
x = ± 2
Since, the numbers are positive, so x = 2.
Thus, the numbers are 12, 8 and 6.

Que-12: Divide 20 into two parts such that three times the square of one part exceeds the other part by 10.

Sol: Let the two parts be x and y.
From the given information,
x + y = 20 ⇒ y = 20 – x
3x² = (20 – x) + 10
3x² = 30 – x
3x² + x – 30 = 0
3x² – 9x + 10x – 30 = 0
3x(x – 3) + 10(x – 3) = 0
(x – 3) (3x + 10) = 0
x = 3, -10/3
Since, x cannot be equal to -10/3, so, x = 3.
Thus, one part is 3 and other part is 20 – 3 = 17.

Que-13: Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60.
Assume the middle number to be x and form a quadratic equation satisfying the above statement. Hence; find the three numbers.

Sol: Let the numbers be x – 1, x and x + 1.
From the given information,
x² = (x + 1)² – (x – 1)² + 60
x² = x² + 1 + 2x – x² – 1 + 2x + 60
x² = 4x + 60
x² – 4x – 60 = 0
(x – 10) (x + 6) = 0
x = 10, -6
Since, x is a natural number, so x = 10.
Thus, the three numbers are 9, 10 and 11.

Que-14: Out of three consecutive positive integers, the middle number is p. If three times the square of the largest is greater than the sum of the squares of the other two numbers by 67; calculate the value of p.

Sol: Let the numbers be p – 1, p and p + 1.
From the given information,
3(p + 1)² = (p – 1)² + p² + 67
3p² + 6p + 3 = p² + 1 – 2p + p² + 67
p² + 8p – 65 = 0
(p + 13)(p – 5) = 0
p = -13, 5
Since, the numbers are positive so p cannot be equal to -13.
Thus, p = 5.

Que-15: A can do a piece of work in ‘x’ days and B can do the same work in (x + 16) days. If both working together can do it in 15 days; calculate ‘x’.

Sol: Work done by A in one day = 1/𝑥
Work done by B in one day = 1/(𝑥+16)
Together A and B can do the work in 15 days.
Therefore, we have
(1/𝑥) + {1/(𝑥+16)} = 1/15
(𝑥+16+𝑥)/{𝑥⁢(𝑥+16)} = 1/15
(2⁢𝑥+16)/(𝑥²+16⁢𝑥) = 1/15
30⁢𝑥 +240 = x² + 16⁢𝑥
𝑥² −14⁢𝑥 −240 =0
(𝑥−24)⁢(𝑥+10) =0
x = 24, –10
Since, x cannot be negative
Thus, x = 24

Que-16: One pipe can fill a cistern in 3 hours less than the other. The two pipes together can fill the cistern in 6 hours 40 minutes. Find the time that each pipe will take to fill the cistern.

Sol:  Let one pipe fill the cistern in x hours and the other fills it in (x – 3) hours.
Given that the two pipes together can fill the cistern in 6 hours 40 minutes,
i.e., 640/60 hours = 6*(2/3) hour  = 20/3 ⁢hours
(1/𝑥) + {1/(𝑥−3)} = 3/20
(𝑥−3+𝑥)/{𝑥⁢(𝑥−3)} = 3/20
(2⁢𝑥−3)/(𝑥²−3⁢𝑥) = 3/20
40⁢𝑥 −60 = 3⁢𝑥² − 9⁢𝑥
3⁢𝑥² − 49⁢𝑥 + 60 = 0
3⁢𝑥² − 45⁢𝑥 − 4⁢𝑥 + 60 = 0
3⁢𝑥⁢(𝑥−15) −4⁢(𝑥−15) =0
(𝑥−15)⁢(3⁢𝑥−4) =0
𝑥 = 15, 4/3
If 𝑥 = 4/3,
Then 𝑥 −3 = (4/3) − 3
= 4 − (9/3)
= −5/3,
Which is not possible
So, x = 15
Thus, one pipe fill the cistern in 15 hours and other fills in (x – 3) = 15 – 3 = 12 hours.

Que-17: A positive number is divided into two parts such that the sum of the squares of the two parts is 20. The square of the larger part is 8 times the smaller part. Taking x as the smaller part of the two parts, find the number.

Sol: Let the smaller part be x
Then, (larger part)² = 8x
∴ Larger part = √8⁢𝑥
Now, the sum of the squares of both the terms is given to be 20
𝑥² + (√8⁢𝑥)² = 20
⇒𝑥² +8⁢𝑥 =20
⇒𝑥² +8⁢𝑥 −20 =0
⇒𝑥² −2⁢𝑥 +10⁢𝑥 −20 =0
⇒𝑥⁢(𝑥−2) +10⁢(𝑥−2) =0
⇒(𝑥−2)⁢(𝑥+10) =0
⇒𝑥 = 2⁢  or  ⁢𝑥 = −10
x = –10 is rejected as it is negative
∴ x = 2
Smaller part = 2
Larger part = √(8×2) = 4
Thus, the required number = 2 + 4 = 6

— : End of Word Problems on Time and Work Class 10 Concise Exe-6A Selina Solutions Ch-6 Practice Problems. :–

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