ML Aggarwal Quadratic Equations Exe-5.5 Class 10 ICSE Maths Solutions . We Provide Step by Step Answer of Exe-5.5 ( Word Problems ) Questions for Quadratic Equations in One Variable as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10.

## ML Aggarwal Ch-5 Quadratic Equations in one Variable Exercise- 5.5 (Word Problems) Class 10 ICSE Maths Solutions

Board | ICSE |

Subject | Maths |

Class | 9th |

Chapter-5 | Quadratic Equations |

Writer / Book | Understanding |

Topics | Solutions of Exe-5.5 |

Academic Session | 2024-2025 |

**Problems on Quadratic Equations in One Variable Exe-5.5**

(Class 10 ICSE ML Aggarwal Maths Solutions)

**Question- 1**

**(i) Find two consecutive natural numbers such that the sum of their squares is 61.**

**(ii) Find two consecutive integers such that the sum of their squares is 61.**

**Answer -1**

Let the first natural number = x

then second natural number = x + 1

According to the condition, (x)² + (x + 1)² = 61

on simplify

x^{2} + x^{2} + 1^{2} + 2x – 61 = 0

2x^{2} + 2x – 60 = 0

x^{2} + x – 30 = 0

on factorize,

x^{2} + 6x – 5x – 30 = 0

x(x + 6) – 5 (x + 6) = 0

(x + 6) (x – 5) = 0

either (x + 6) = 0 or (x – 5) = 0

x = -6 or x = 5

∴ x = 5 [Since -6 is not a positive number]

So the first natural number = 5

and then Second natural number = 5 + 1 = 6

##### (ii) Let us consider first integer number be ‘x’

Second integer number be ‘x + 1’

So according to the question,

x^{2} + (x + 1)^{2} = 61

on simplify

x^{2} + x^{2} + 1^{2} + 2x – 61 = 0

2x^{2} + 2x – 60 = 0

x^{2} + x – 30 = 0

on factorize,

x^{2} + 6x – 5x – 30 = 0

x(x + 6) – 5 (x + 6) = 0

(x + 6) (x – 5) = 0

(x + 6) = 0 or (x – 5) = 0

x = -6 or x = 5

If x = -6, then

First integer number = -6

Second integer number = -6 + 1 = -5

and If x = 5, then

First integer number = 5

Second integer number = 5 + 1 = 6

**Question- 2 ****(i) If the product of two positive consecutive even integers is 288, find the integers.**

**(ii) If the product of two consecutive even integers is 224, find the integers.**

**(iii) Find two consecutive even natural numbers such that the sum of their squares is 340.**

**(iv) Find two consecutive odd integers such that the sum of their squares is 394.**

**Answer -2**

**(i) **Let first positive even integer number= ‘2x’

then Second even integer number be ‘2x + 2’

according to condition

2x × (2x + 2) = 288

4x^{2} + 4x – 288 = 0

x^{2} + x – 72 = 0

on factorize,

x^{2} + 9x – 8x – 72 = 0

x(x + 9) – 8(x + 9) = 0

(x + 9) (x – 8) = 0

either

(x + 9) = 0 or (x – 8) = 0

so x = -9 or x = 8

∴ Value of x = 8 [since, -9 is not positive]

First even integer = 2x = 2 (8) = 16

Second even integer = 2x + 2 = 2(8) + 2 = 18

**(ii) **Let first positive even integer number = ‘2x’

then Second even integer number = ‘2x + 2’

according to condition

2x × (2x + 2) = 224

4x^{2} + 4x – 224 = 0

x^{2} + x – 56 = 0

on factorize,

x^{2} + 8x – 7x – 56 = 0

x(x + 8) – 7(x + 8) = 0

(x + 8) (x – 7) = 0

either (x + 8) = 0 or (x – 7) = 0

so x = -8 or x = 7

∴ Value of x = 7 [since, -8 is not positive]

First even integer = 2x = 2 (7) = 14

Second even integer = 2x + 2 = 2(7) + 2 = 16

**(iii) **Let first positive even natural number = ‘2x’

Second even number= ‘2x + 2’

according to condition

(2x)^{2} + (2x + 2)^{2} = 340

4x^{2} + 4x^{2} + 8x + 4 – 340 = 0

8x^{2} + 8x – 336 = 0

x^{2} + x – 42 = 0

on factorize,

x^{2} + 7x – 6x – 56 = 0

x(x + 7) – 6(x + 7) = 0

(x + 7) (x – 6) = 0

either (x + 7) = 0 or (x – 6) = 0

x = -7 or x = 6

∴ Value of x = 6 [since, -7 is not positive]

First even natural number = 2x = 2 (6) = 12

Second even natural number = 2x + 2 = 2(6) + 2 = 14

**(iv) **Let first odd integer number = ‘2x + 1’

Second odd integer number = ‘2x + 3’

according to condition

(2x + 1)^{2} + (2x + 3)^{2} = 394

4x^{2} + 4x + 1 + 4x^{2} + 12x + 9 – 394 = 0

8x^{2} + 16x – 384 = 0

x^{2} + 2x – 48 = 0

on factorize,

x^{2} + 8x – 6x – 48 = 0

x(x + 8) – 6(x + 8) = 0

(x + 8) (x – 6) = 0

either (x + 8) = 0 or (x – 6) = 0

x = -8 or x = 6

When x = -8, then

First odd integer = 2x + 1 = 2 (-8) + 1 = -16 + 1 = -15

Second odd integer = 2x + 3 = 2(-8) + 3 = -16 + 3 = -13

and When x = 6, then

First odd integer = 2x + 1 = 2 (6) + 1 = 12 + 1 = 13

Second odd integer = 2x + 3 = 2(6) + 3 = 12 + 3 = 15

Hence The required odd integers are -15, -13, 13, 15.

**Question -3 ****The sum of two numbers is 9 and the sum of their squares is 41. Taking one number as x, form ail equation in x and solve it to find the numbers.**

**Answer- 3**

Sum of two numbers = 9

Let first number = x

then second number = 9 – x

Now according to the condition,

(x)^{2} + (9 – x)^{2} = 41

x^{2} + 81 – 18x + x^{2} – 41 = 0

2x^{2} – 18x + 40 = 0

Divide by 2, we get

x^{2} – 9x + 20 = 0

Let us factorize,

x^{2} – 4x – 5x + 20 = 0

x(x – 4) – 5(x – 4) = 0

(x – 4) (x – 5) = 0

either (x – 4) = 0 or (x – 5) = 0

x = 4 or x = 5

When x = 4,

then First number = x = 4

Second number = 9 – x = 9 – 4 = 5

and When x = 5,

then First number = x = 5

Second number = 9 – x = 9 – 5 = 4

Hence The required numbers are 4 and 5.

**Question- 4 ****Five times a certain whole number is equal to three less than twice the square of the number. Find the number.**

**Answer- 4**

Let number = x

Now according to the condition,

5x = 2x² – 3

2x^{2} – 3 – 5x = 0

2x^{2} – 5x – 3 = 0

on factorize,

2x^{2} – 6x + x – 3 = 0

2x(x – 3) + 1(x – 3) = 0

(x – 3) (2x + 1) = 0

either (x – 3) = 0 or (2x + 1) = 0

x = 3 or 2x = -1

x = 3 or x = -1/2

**Question- 5 ****Sum of two natural numbers is 8 and the difference of their reciprocal is 2/15. Find the numbers.**

**Answer -5**

Let two numbers = ‘x’ and ‘y’

according to condition

1/x – 1/y = 2/15 ….. (i)

given that, x + y = 8

So, y = 8 – x … (ii)

substitute the value of y in equation (i),

1/x – 1/(8 – x) 2/15

[8 – x – x] / x(8 – x) = 2/15

(8 – 2x) / x(8 – x) = 2/15

15(8 – 2x) = 2x(8 – x)

120 – 30x = 16x – 2x^{2}

120 – 30x – 16x + 2x^{2} = 0

2x^{2} – 46x + 120 = 0

x^{2} – 23x + 60 = 0

on factorize,

x^{2} – 20x – 3x + 60 = 0

x(x – 20) – 3 (x – 20) = 0

(x – 20) (x – 3) = 0

either

(x – 20) = 0 or (x – 3) = 0

x = 20 or x = 3

Now,

Sum of two natural numbers, y = 8 – x = 8 – 20 = -12, which is a negative value.

So value of x = 3, y = 8 – x = 8 – 3 = 5

∴ The value of x and y are 3 and 5.

## Word Problems of Quadratic Equations in one Variable

**Question -6 ****The difference between the squares of two numbers is 45. The square of the smaller number is 4 times the larger number. Determine the numbers.**

**Answer- 6**

Let the larger number = x

then smaller number = y

Now according to the condition,

x^{2} – y^{2} = 45 … (i)

y^{2} = 4x … (ii)

substitute the value of y in equation (i),

x^{2} – 4x = 45

x^{2} – 4x – 45 = 0

on factorize,

x^{2} – 9x + 5x – 45 = 0

x(x – 9) + 5 (x – 9) = 0

(x – 9) (x + 5) = 0

either (x – 9) = 0 or (x + 5) = 0

so x = 9 or x = -5

When x = 9, then

The larger number = x = 9

and Smaller number = y => y^{2} = 4x

y = **√**4x = **√**4(9) = **√**36 = 6

When x = -5, then

The larger number = x = -5

Smaller number = y => y^{2} = 4x

y = **√**4x = **√**4(-5) = **√**-20 (which is not possible)

hence The value of x and y are 9, 6.

**Question -7 ****There are three consecutive positive integers such that the sum of the square of the first and the product of other two is 154. What are the integers?**

**Answer -7**

Let the first integer = x

then second integer = x + 1

and third integer = x + 2

Now according to the condition,

x^{2} + (x + 1) (x + 2) = 154

on simplify,

x^{2} + x^{2} + 3x + 2 – 154 = 0

2x^{2} + 3x – 152 = 0

on factorize,

2x^{2} + 19x – 16x – 152 = 0

x(2x + 19) – 8 (2x + 19) = 0

(2x + 19) (x – 8) = 0

either (2x + 19) = 0 or (x – 8) = 0

2x = -19 or x = 8

x = -19/2 or x = 8

∴ The value of x = 8 [since -19/2 is a negative value]

then

First integer = x = 8

Second integer = x + 1 = 8 + 1 = 9

Third integer = x + 2 = 8 + 2 = 10

∴ The numbers are 8, 9, 10.

**Question -8 ****(i) Find three successive even natural numbers, the sum of whose squares is 308.**

**(ii) Find three consecutive odd integers, the sum of whose squares is 83.**

**Answer -8**

(i) Let first even number = 2x

second even number = 2x + 2

third even number = 2x + 4

Now according to the condition,

(2x)^{2} + (2x + 2)^{2} + (2x + 4)^{2} = 308

4x^{2} + 4x^{2} + 8x + 4 + 4x^{2} + 16x + 16 – 308 = 0

12x^{2} + 24x – 288 = 0

x^{2} + 2x – 24 = 0

on factorize,

x^{2} + 6x – 4x – 24 = 0

x(x + 6) – 4(x + 6) = 0

(x + 6) (x – 4) = 0

either (x + 6) = 0 or (x – 4) = 0

x = -6 or x = 4

∴ Value of x = 4 [since, -6 is not positive]

First even natural number = 2x = 2 (4) = 8

Second even natural number = 2x + 2 = 2(4) + 2 = 10

Third even natural number = 2x + 4 = 2(4) + 4 = 12

∴ The numbers are 8, 10, 12.

**(ii) **Let three odd numbers = ‘x’, ‘x + 2’, ‘x + 4’

according to condition

(x)^{2} + (x + 2)^{2} + (x + 4)^{2} = 83

x^{2} + x^{2} + 4x + 4 + x^{2} + 8x + 16 – 83 = 0

3x^{2} + 12x – 63 = 0

x^{2} + 4x – 21 = 0

on factorize,

x^{2} + 7x – 3x – 21 = 0

x(x + 7) – 3 (x + 7) = 0

(x + 7) (x – 3) = 0

either (x + 7) = 0 or (x – 3) = 0

x = -7 or x = 3

hence

The numbers

= x, x+2, x+4

=> -7, -7+2, -7+4

=> -7, -5, -3

Or the numbers = x, x+2, x+4

=> 3, 3+2, 3+4,

=> 3, 5, 7

**Question -9 ****In a certain positive fraction, the denominator is greater than the numerator by 3. If 1 is subtracted from both the numerator and denominator, the fraction is decreased by 1/14. Find the fraction?**

**Answer-9 **Let the numerator be ‘ $x$ ‘

Denominator be ‘x+3’

So the fraction is x / ($x+ $

According to the question,

⇒ (x – 1)(14x + 42) = (13x – 3)(x + 2)

⇒ 14x^{2} + 42x – 14x – 42 = 13x^{2} + 26x – 3x 6

⇒ 14x^{2} + 28x 42 – 13x^{2} – 23x + 6 = 0

⇒ x^{2} + 5x – 36 = 0

⇒ x^{2} + 9x – 4x – 36 = 0

x(x + 9) -4(x + 9) = 0

⇒ (x + 9)(x – 4) = 0

Either x + 9 = 0,

then x = -9,

but it is not possible as the fraction is positive.

or

x – 4 = 0,

then x = 4

∴ Fraction = x/(x+3)=4/(4+3)=4/7.

**Question -10 ****The sum of the numerator and denominator of a certain positive fraction is 8. If 2 is added to both the numerator and denominator, the fraction is increased by 4/35. Find the fraction.**

**Answer- 10**

## ML Aggarwal Exercise- 5.5 Class 10 ICSE Maths Solutions

**Question- 11 ****A two digit number contains the bigger at ten’s place. The product of the digits is 27 and the difference between two digits is 6. Find the number.**

**Answer -11 **Let unit’s digit = x

then tens digit = x + 6

Number = x + 10(x + 6)

= x + 10x + 60

= 11x + 60

according the condition

x(x + 6) = 27

x^{2} + 6x – 27 = 0

on factorize,

x^{2} + 9x – 3x – 27 = 0

x(x + 9) – 3 (x + 9) = 0

(x + 9) (x – 3) = 0

either

(x + 9) = 0 or (x – 3) = 0

x = -9 or x = 3

so, value of x = 3

[since, -9 is a negative number]

∴ The number = 11x + 60

= 11(3) + 60

= 33 + 60

= 93

**Question- 12 ****A two digit positive number is such that the product of its digits is 6. If 9 is added to the number, the digits interchange their places. Find the number. (2014)**

**Answer- 12**

Let 2-digit number = xy = 10x + y

Reversed digits = yx = 10y + x

According to first condition

10x + y + 9 = 10y + x

second condition

xy = 6

y = 6/x

substituting the value

10x + 6/x + 9 = 10(6/x) + x

10x^{2} + 6 + 9x = 60 + x^{2}

10x^{2} + 6 + 9x – 60 – x^{2} = 0

9x^{2} + 9x – 54 = 0

x^{2} + x – 6 = 0

on factorize,

x^{2} + 3x – 2x – 6 = 0

x(x + 3) – 2 (x + 3) = 0

(x + 3) (x – 2) = 0

either (x + 3) = 0 or (x – 2) = 0

x = -3 or x = 2

Value of x = 2

[because, -3 is a negative value]

substitute the value of x in y = 6/x,

y = 6/2 = 3

Hence2-digit number = 10x + y = 10(2) + 3 = 23

**Question -13 ****A rectangle of area 105 cm² has its length equal to x cm. Write down its breadth in terms of x. Given that the perimeter is 44 cm, write down an equation in x and solve it to determine the dimensions of the rectangle.**

**Answer -13**

Perimeter of rectangle = 44 cm

length + breadth = = 22 cm

Let length = x

then breadth = 22 – x

According to the condition,

x(22 – x) = 105

22x – x^{2} – 105 = 0

x^{2} – 22x + 105 = 0

on factorize,

x^{2} – 15x – 7x + 105 = 0

x(x – 15) – 7(x – 15) = 0

(x – 15) (x – 7) = 0

either (x – 15) = 0 or (x – 7) = 0

x = 15 or x = 7

length > breadth,

x = 7 is not permittable

so Length = 15cm

then Breadth = 22 – x

= 22 – 15

= 7cm

**Question -14 ****A rectangular garden 10 m by 16 m is to be surrounded by a concrete walk of uniform width. Given that the area of the walk is 120 square metres, assuming the width of the walk to be x, form an equation in x and solve it to find the value of x. **

**Answer 14**

Length of garden = 16 m

and width = 10 m

Let the width of walk = x m

Outer length = 16 + 2x

and outer width = 10 + 2x

Now according to the condition,

(16 + 2x) (10 + 2x) – 16(10) = 120

160 + 32x + 20x + 4x^{2} – 160 – 120 = 0

4x^{2} + 52x – 120 = 0

x^{2} + 13x – 30 = 0

x^{2} + 15x – 2x – 30 = 0

x(x + 15) – 2 (x + 15) = 0

(x + 15) (x – 2) = 0

either (x + 15) = 0 or (x – 2) = 0

so x = -15 or x = 2

hence Value of x is 2 [Since, -15 is a negative value]

page-90

**Question -15 **** The length of a rectangle exceeds its breadth by 5 m. If the breadth were doubled and the length reduced by 9 m, the area of the rectangle would have increased by 140 m². Find its dimensions.**

**Answer- 15**

Let length of the rectangle = ‘x’

Width = (x – 5) meter (first condition)

Area = l b

= x(x – 5) sq.m

**condition second**

if Length = (x – 9) meter

if Width = 2 (x – 5) meter

then Area = (x – 9) 2(x – 5) = 2(x – 9) (x – 5) sq.m

but new area increased 140

2(x – 9) (x – 5) = x(x – 5) + 140

2(x^{2} – 14x + 45) = x^{2} – 5x + 140

2x^{2} – 28x + 90 – x^{2} + 5x – 140 = 0

x^{2} – 23x – 50 = 0

on factorize,

x^{2} – 25x + 2x – 50 = 0

x(x – 25) + 2 (x – 25) = 0

(x – 25) (x + 2) = 0

either (x – 25) = 0 or (x + 2) = 0

x = 25 or x = -2

So Length of the first rectangle = 25meters.

[because, -2 is a negative value]

then Width = x – 5

= 25 – 5

= 20meters

Area = l b

= 25 × 20

= 500 m^{2}

### ML Aggarwal Ch-5 Quadratic Equations in one Variable Exercise- 5.5 (Word Problems) Class 10 ICSE Maths Solutions

**Question- 16 ****The perimeter of a rectangular plot is 180 m and its area is 1800 m². Take the length of the plot as x m. Use the perimeter 180 m to write the value of the breadth in terms of x. Use the values of length, breadth and the area to write an equation in x. Solve the equation to calculate the length and breadth of the plot. **

**Answer -16 **

#### The perimeter of a rectangular field = 180 m

and area = 1800 m²

Let length = x m

Perimeter of rectangular field

= 2 (length + breadth)

So, (length + breadth) = perimeter/ 2

x + breadth = 180/2

⇒ breadth = 90 – x

Length × breadth = 1800

x × (90 – x) = 1800

90x – x^{2} = 1800

x^{2} – 90x + 1800 = 0

on factorization,

x^{2} – 60x – 30x + 1800 = 0

x(x – 60) – 30(x – 60) = 0

(x – 30) (x – 60) = 0

either x – 30 = 0 or x – 60 = 0

x = 30 or x = 60

length > breadth,

So, Length = 60 m

and breadth = (90 – 60)

= 30 m

**Question- 17 ****The lengths of the parallel sides of a trapezium are (x + 9) cm and (2x – 3) cm and the distance between them is (x + 4) cm. If its area is 540 cm², find x.**

**Answer -17**

Area of a trapezium = 1/2

(sum of parallel sides) x height

Lengths of parallel sides are (x + 9) and (2x – 3)

and height = (x + 4)

According to the condition,

½ × (x + 9 + 2x – 3) × (x + 4) = 540

(3x + 6) (x + 4) = 540 × 2

3x^{2} + 12x + 6x + 24 = 1080

3x^{2} + 18x – 1056 = 0

x^{2} + 6x – 352 = 0

on factorize

x^{2} + 22x – 16x – 352 = 0

x(x + 22) – 16(x + 22) = 0

(x – 16) (x + 22) = 0

either x – 16 = 0 or x + 22 = 0

x = 16 or x = -22

Any measurements cannot be negative

So x = -22 impossible

Hence, x = 16

**Question -18 ****If the perimeter of a rectangular plot is 68 m and the length of its diagonal is 26 m, find its area.**

**Answer -18**

Perimeter = 68 m and diagonal = 26 m

Length + breadth = Perimeter/2

= 68/2

= 34 m

Let the length = ‘x’ m

Then, breadth = (34 – x) m

the diagonal of the rectangular

length^{2} + breadth^{2} = diagonal^{2}

[By Pythagoras Theorem]

x^{2} + (34 – x)^{2} = 26^{2}

x^{2} + 1156 + x^{2} – 68x = 676

2x^{2} – 68x + 1156 – 676 = 0

2x^{2} – 68x + 480 = 0

x^{2} – 34x + 240 = 0 [Dividing by 2]

on factorization

x^{2} – 24x – 10x + 240 = 0

x(x – 24) – 10(x – 24) = 0

(x – 10) (x – 24) = 0

either x – 10 = 0 or x – 24 = 0

x = 10 or x = 24

length > breadth,

So length = 24 m

and breadth = (34 – 24) m

= 10 m

, area of the rectangular plot = 24 × 10

= 240 m^{2}.

**Question -19 ****If the sum of two smaller sides of a right – angled triangle is 17cm and the perimeter is 30 cm, then find the area of the triangle.**

**Answer -19**

The perimeter of the triangle = 30cm.

Let one of the two small sides = x

then, other side = 17 – x

∴ Length of hypotenuse

= perimeter – sum of other two sides

= 30cm – 17cm

= 13cm.

x^{2} + (17 – x)^{2} = (13)^{2} …(Pythagoras theorem)

⇒ x^{2} + 289 + x^{2} – 34x = 169

⇒ 2x^{2} – 34x + 289 – 169 = 0

⇒ 2x^{2} – 34x + 120 = 0

⇒ x^{2} – 17x + 60 = 0 …(Dividing by 2)

⇒ x^{2} – 12x – 5x + 60 = 0

⇒ x(x – 12) – 5(x – 12) = 0

⇒ (x – 12)(x – 5) = 0

Either x – 12 = 0,

then x = 12

or

x – 5 = 0,

then x = 5

(i) when x = 12, then first side = 12cm

and second side = 17 – 12 = 5cm

(ii) When x = 5, then first side = 5

and second side = 17 – 5 = 12

∴ Sides are 5cm. 12cm

Now,

Area of the triangle = ½ (5 × 12)

= 60/2

= 30 cm^{2}

**Question- 20 ****The hypotenuse of grassy land in the shape of a right triangle is 1 metre more than twice the shortest side. If the third side is 7 metres more than the shortest side, find the sides of the grassy land.**

**Answer- 20**

##### Let the length of the shortest side be x metres.

## Ch-5 Quadratic Equations in one Variable Exercise- 5.5 (Word Problems)

**Question -21 ****Mohini wishes to fit three rods together in the shape of a right triangle. If the hypotenuse is 2 cm longer than the base and 4 cm longer than the shortest side, find the lengths of the rods.**

**Answer- 21**

Let the length of hypotenuse = x cm

then base = (x – 2) cm

and shortest side = x – 4

According to the condition,

(x)^{2} = (x – 2)^{2} + (x – 4)^{2}

⇒ x^{2 }= x^{2} – 4x + 4 + x^{2 }– 8x + 16

⇒ x^{2} = 2x^{2} – 12x + 20

⇒ 2x^{2} – 12x + 20 – x2 = 0

⇒ x^{2} – 12x + 20 = 0

⇒ x^{2 }– 10x – 2x + 20 = 0

⇒ x(x – 10) -2(x – 10) = 0

⇒ (x – 10)(x – 2) = 0

Either x – 10 = 0,

then x = 10

or

x – 2 = 0,

then x = 2,

but it is not possible as the hypotenuse is the longest side.

∴ Hypotenuse = 10cm

Base = 10 – 2 = 8cm

and shortest side = 10 – 4 = 6cm.

**Question- 22 ****In a P.T. display, 480 students are arranged in rows and columns. If there are 4 more students in each row than the number of rows, find the number of students in each row.**

**Answer- 22**

Total number of students = 480

Let the number of students in each row = x

then number of row = 480/x

According condition

x = (480/x) + 4

⇒ x^{2} = 480 + 4

⇒ x^{2} – 4x – 480 = 0

⇒ x^{2} – 24x + 20x – 480 = 0

⇒ x(x – 24) + 20(x – 24) = 0

⇒ (x – 24)(x + 20) = 0

Either x – 24 = 0

or

x + 20 = 0

⇒ x = 24

or

x = -20

which is not possible as it is negative

∴ Number of students in each row = 24.

**Question- 23 ****In an auditorium, the number of rows are equal to the number of seats in each row. If the number of rows is doubled and number of seats in each row is reduced by 5, then the total number of seats is increased by 375. How many rows were there?**

**Answer -23**

Let the number of rows = x

then no. of seats in each row = x

and total number of seats = x × x = x^{2}

According to the condition,

2x x (x – 5) = x^{2} + 375

⇒ 2x^{2 }– 10x = x^{2} + 375

⇒ 2x^{2} – 10x – x^{2} – 375 = 0

⇒ x^{2} – 10x – 375 = 0

⇒ x^{2} – 25x + 15x – 375 = 0

⇒ x(x – 25) + 15(x – 25) = 0

⇒ (x – 25)(x + 15) = 0

Either x – 25 = 0,

then x = 25

or

x + 15 = 0,

then x = -15,

but it is not possible as it is negative.

∴ Number of rows = 25.

**Question -24 ****At an annual function of a school, each student gives the gift to every other student. If the number of gifts is 1980, find the number of students.**

**Answer -24**

Let the number of students = x

then the number of gifts given = x – 1

Total number of gifts = x (x – 1)

According to the condition,

x (x – 1) = 1980

⇒ x^{2} – x – 1980 = 0

⇒ x^{2} – 45x + 44x – 1980 = 0

⇒ x(x – 45) + 44(x – 45) = 0

⇒ (x – 45)(x + 44) = 0

Either x – 45 = 0,

then x = 45

or

x + 44 = 0,

then x = -44,

but it is not possible as it is negative.

Hence number of students = 45.

**Question- 25 ****A bus covers a distance of 240 km at a uniform speed. Due to heavy rain, its speed gets reduced by 10 km/h and as such it takes two hours longer to cover the total distance. Assuming the uniform speed to be ‘x’ km/h, form an equation and solve it to evaluate x. (2016)**

**Answer -25**

##### Uniform speed $=xkm/h$

distance $=240km$

### Quadratic Equations ,Word Problems Questions and Solutions

**Question- 26 ****The speed of an express train is x km/hr and the speed of an ordinary train is 12 km/hr less than that of the express train. If the ordinary train takes one hour longer than the express train to cover a distance of 240 km, find the speed of the express train. (2009)**

**Answer -26**

Let the speed of express train = x km

Then speed of the ordinary train = (x – 12) km

Time is taken to cover 240 km by the express

ime require to cover for each train is 240/ x and 240/x-12 respectively.

According to question

240x-12-240x=1

240x-240(x-12)(x-12)(x)=1

240x – 240 (x – 12) = x (x – 12)

x^{2} – 12x – 2880 = 0

(x – 60) (x + 48) = 0

∴ x = 60 km/hr.

Speed of the express train is 60 km/hr.

**Question -27 ****A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car. (2012)**

**Answer -27**

page-91

**Question- 28 ****An aero plane travelled a distance of 400 km at an average speed of x km/hr. On the return journey, the speed was increased by 40 km/hr. Write down an expression for the time taken for**

**(i)the onward journey,**

**(ii) the return journey.**

**If the return journey took 30 minutes less than the onward journey, write down an equation in x and find its value.** (2002)

**Answer- 28**

Distance = 400 km

Average speed of the aero plane = x km

x(x + 200) – 160(x + 200) = 0

(x + 200)(x – 160) = 0

X = − 200, 160

Since, the speed cannot be negative. Thus, x = 160

**Question- 29 ****The distance by road between two towns A and B, is 216 km, and by rail it is 208 km. A car travels at a speed of x km/hr, and the train travels at a speed which is 16 km/hr faster than the car. Calculate :**

**(i) The time taken by the car, to reach town B from A, in terms of x ;**

**(ii) The time taken by the train, to reach town B from A, in terms of x ;**

**(iii) If the train takes 2 hours less than the car, to reach town B, obtain an equation in x and solve it.**

**(iv) Hence find the speed of the train.**

**Answer -29**

Speed of car = x km/hr

Speed of train = (x + 16) km/hr

(i) we know: Time = distance/speed

Time taken by the car to reach town B from A = 216/x hrs

(ii) Time taken by the train to reach town B from A = 208/(x+16) hrs

**Question- 30 ****An aero plane flying with a wind of 30 km/hr takes 40 minutes less to fly 3600 km, than what it would have taken to fly against the same wind. Find the planes speed of flying in still air.**

**Answer -30**

Let the speed of the plane in still air = x km/hr

Speed of wind = 30 km/hr

Distance = 3600 km

⇒ 2x^{2} – 1800 = 3 x 3600 x 60

⇒ 2x^{2} – 1800 = 648000

⇒ 2x^{2} – 1800 – 648000 = 0

⇒ 2x^{2} – 649800 = 0

⇒ x^{2} – 324900 = 0 ..(Dividing by 2)

⇒ x^{2} – (570)2 = 0

⇒ (x + 570)(x – 570) = 0

Either x + 570 = 0,

then x = -570

which is not possible as it is negative

or

x – 570 = 0,

then x = 570

Hence speed of plane in still air = 570km/hr.

### ML Aggarwal Class 10 ICSE Maths Solutions

Ch-5 Quadratic Equations in one Variable Exercise- 5.5 (Word Problems)

**Question -31 ****A school bus transported an excursion party to a picnic spot 150 km away. While returning, it was raining and the bus had to reduce its speed by 5 km/hr, and it took one hour longer to make the return trip. Find the time taken to return.**

**Answer- 31**

**Solution:**

Distance = 150 km

Let the speed of bus = x km/hr

**Question -32 ****A boat can cover 10 km up the stream and 5 km down the stream in 6 hours. If the speed of the stream is 1.5 km/hr. find the speed of the boat in still water.**

**Answer -32**

Distance up stream = 10 km

and down stream = 5 km

Total time is taken = 6 hours

Speed of stream = 1.5 km/hr

Let the speed of a boat in still water = x km/hr

According to the condition,

10/ (x-1.5) + 5/ (x+1.5) = 6

⇒ 10x + 15 + 5x + 5x – 7.5 = 6(x – 15)(x + 15)

⇒ 15x + 7.5 = 6(x^{2} – 2.25)

⇒ 15x + 7.5 = 6x^{2} – 13.5

⇒ 6x^{2} – 15x – 13.5 – 7.5

⇒ 6x^{2} – 15x – 21 = 0

⇒ 2x^{2} – 5x – 7 = 0 …(Dividing by 3)

⇒ 2x^{2} – 7x + 2x – 7 = 0 …

⇒ x(2x – 7) + 1(2x – 7) = 0

⇒ (2x – 7)(x + 1) = 0

Either 2x – 7 = 0,

then 2x = 7

⇒ x = 72

or

x + 1 = 0,

then x = -1

But it is not possible being negative

∴ x = 72 = 3.5

∴Speed of boat = 3.5km/hr

**Question- 33 ****Two pipes running together can fill a tank in minutes. If one pipe takes 5 minutes more than the other to fill the tank, find the time in which each pipe would/fill the tank.**

**Answer- 33**

Let time taken by pipe A be x minutes. and time taken by pipe B be x + 5 minutes.

Hence pipe A will fill the tank in 20 minutes and pipe B will fill it in 25 minutes.

**Question- 34 ****(i) Rs. 480 is divided equally among ‘x’ children. If the number of children was 20 more then each would have got Rs. 12 less. Find ‘x’.**

**(ii) Rs. 7500 is divided equally among a certain number of children. Had there been 20 less children, each would have receive Rs 100 more. Find the original ****number of Children.**

**Answer -34**

(i) Share of each child = Rs 480/ x

According to the question

(ii)

Let the original number of person be x, then 7500 divided equally between x person,

each one gets = 7500/x

7500 divided equally between x – 20 children

each one gets 75 =7500/ (x-20)

According to the question

7500x = (x-20)(7500+100x)

75x = (x – 20)(75 + x)

75x = 75x + x^{2} – 1500 – 20x

x^{2} – 20x – 1500 = 0

x = 50 or x = -30 (not possible)

∴ original number of children = 50

**Question- 35 ****2x articles cost Rs. (5x + 54) and (x + 2) similar articles cost Rs. (10x – 4), find x.**

**Answer -35**

If 2x articles cost 5x+54 then the cost of one article=5x+54/2x

If x+2 articles cost 10x-4 then the cost of one article=10x-4/x+2

Now equating the price of one article since they both are same articles, we get-

5x+54/2x=10x-4/x+2

By cross multiplying, we get-

15x²-72x-108=0

So x=6,-36/30

Since articles cannot be negative,

Therefore** x=6**

### Ch-5 Quadratic Equations in one Variable Exercise- 5.5 (Word Problems)

**Question -36 ****A trader buys x articles for a total cost of Rs. 600.**

**(i) Write down the cost of one article in terms of x. If the cost per article were Rs. 5 more, the number of articles that can be bought for Rs. 600 would be four less.**

**(ii) Write down the equation in x for the above situation and solve it to find x. **

**Answer -36**

Total cost = Rs. 600,

No. of articles = x

Cost of one article = Rs 600/ x

according condition

600/ (x-4) – 600/x =5

on simplify

x²-4x-480=0

x²-24x+20x-480=0

x(x – 24) + 20(x – 24) = 0

(x – 24)(x + 20) = 0

x = 24, -20

So x = 24.

Since number of article can not be negative

page-92

**Question -37 ****A shopkeeper buys a certain number of books for Rs 960. If the cost per book was Rs 8 less, the number of books that could be bought for Rs 960 would be 4 more. Taking the original cost of each book to be Rs x, write an equation in x and solve it to find the original cost of each book.**

**Answer -37**

original cost of each book= x

number of books=960/x

if cost of each books=x – 8

than no of books for 960=960/x -8

according to give condition

as cost is not in negative so

x =48

**Question -38 ****A piece of cloth costs Rs. 300. If the piece was 5 metres longer and each metre of cloth costs Rs. 2 less, the cost of the piece would have remained unchanged. How long is the original piece of cloth and what is the rate per metre?**

**Answer -38**

The total cost of cloth piece = Rs. 300

Let the length of the piece of cloth in the beginning = x m

Then cost of 1 metre = Rs. 300/x

In second case, length of cloth = (x + 5)

Cost of 1 metre = Rs. 300 / (x+5)

According to the condition,

Either x + 30 = 0,

then x = -30

which is not possible being negative

or

x – 25 = 0,

then x = 25

∴ Length of cloth piece in the begining = 25 metres

and rate per metre = Rs. 300/25 = Rs. 12.

**Question -39 ****The hotel bill for a number of people for an overnight stay is Rs. 4800. If there were 4 more, the bill each person had to pay would have reduced by Rs. 200. Find the number of people staying overnight. (2000)**

**Answer- 39**

Let the number of people = x

Amount of bill = Rs. 4800

Then bill for each person = Rs. 4800/x

In second case,

the number of people = x + 4

then bill of each person = 4800/ (x+4)

According to the condition,

on simplify

⇒ 19200 = 200x^{2} + 800x

⇒ 200x^{2} + 800x – 19200 = 0

⇒ x^{2} + 4x – 96 = 0 …(Dividing by 200)

⇒ x^{2} + 12x – 8x – 96 = 0

⇒ x(x + 12) -8(x + 12) = 0

⇒ (x + 12)(x + 8) = 0

Either x + 12 = 0,

then x = -12,

but it is not possible as it is in negative

or

x – 8 = 0,

then x = 8

∴ No. of people = 8.

**Question -40 ****A person was given Rs. 3000 for a tour. If he extends his tour programme by 5 days, he must cut down his daily expenses by Rs. 20. Find the number of days of his tour programmed.**

**Answer- 40**

Let the number of days of tour programme = x

Amount = Rs. 3000

Express for each day = 3000/x

In second case, no of days = x + 5

then expenses of each day = 3000/(x+5)

Now according to the condition,

⇒ 3000 x 5 = 20x^{2 }+ 100x

⇒ 20x^{2} + 100x – 15000 = 0

⇒ x^{2} + 5x – 750 = 0 …(DIviding by 20)

⇒ x^{2} – 25x + 30x – 750 = 0

⇒ x(x – 25)(x + 30) = 0

⇒ (x – 25)(x + 30) = 0

EIther x – 25 = 0,

then x = 25

or

x + 30 = 0,

then x = -30,

but it is not possible as it is in negative.

∴ Number days = 25.

**Question- 41 ****Ritu bought a saree for Rs. 60 x and sold it for Rs. (500 + 4x) at a loss of x%. Find the cost price.**

**Answer -41**

According to. Question :-

The cost price of saree = Rs. 60x

and selling price = Rs. (500 + 4x)

Loss = x%

Now according to the condition

The cost price of saree = Rs. 60x

and selling price = Rs. (500 + 4x)

Loss = x%

Now according to the condition

⇒ 50000 + 400x = 6000x – 60x^{2}

⇒ 60x^{2 }– 6000x + 400x + 50000 = 0

⇒ 60x^{2} – 5600x + 50000 = 0

⇒ 3x^{2 }– 280x + 2500 = 0 …(Dividing by 20)

⇒ 3x^{2} – 30x – 250x + 2500 = 0

⇒ 3x(x – 10) – 250(x – 10) = 0

⇒ (x – 10)(3x – 250) = 0

Either x – 10 = 0,

then x = 10

or

3x – 250 = 0,

then 3x = 250

⇒ x = 250/3

But it is not possible

∴ Loss = 10%

Cost price = 60x

= 60 x 10

= Rs. 600.

**Question- 42 ****(i) The sum of the ages of Vivek and his younger brother Amit is 47 years. The product of their ages in years is 550. Find their ages. (2017)**

**(ii) Paul is x years old and his father’s age is twice the square of Paul’s age. Ten years hence, the father’s age will be four times Paul’s age. Find their present ages.**

**Answer -42**

(i) Let Vivek’s present age be x years.

His brother’s age = (47 – x) years

According to question,

x(47 – x) = 550

⇒ 47x – x² = 550

⇒ x² – 47x + 550 = 0

or ⇒ x² – 25x – 22x + 550 = 0

⇒ x(x – 25) – 22(x – 25) – 0

⇒ (x – 25)(x – 22) = 0

⇒ x – 25 = 0 or x – 22 = 0

⇒ x = 25 or x = 22

When x = 25, then 47 – x = 47 – 25 = 22

When x = 22, then 47 – x = 47 – 22 = 25 …(does not satisfy the given condition)

∴ Vivek’s age = x = 25 years.

His younger brother’s age = 22 years.

(ii)

Age of Paul = x years

Father’s age = 2x^{2}

10 years hence,

Age of Paul = x + 10

and father’s age = 2x^{2} + 10

According to the conditions,

2x^{2} + 10 = 4(x + 10)

⇒ 2x^{2 }+ 10 = 4x + 40

⇒ 2x^{2} + 10 – 4x – 40 = 0

⇒ 2x^{2} – 4x – 30 = 0

⇒ x^{2 }– 2x – 15 = 0 …(Dividing by 2)

⇒ x^{2 }– 5x + 3x – 15 = 0

⇒ x(x – 5) + 3(x – 5) = 0

⇒ (x – 5)(x + 3) = 0

Either x – 5 = 0,

then x = 5

or

x + 3 = 0,

then x = -3,

but it is not possible as it is in negative.

∴ Age of Paul = 5 years.

and his father’s age

= 2x^{2}

= 2(5)^{2}

= 2 x 25

= 50 years.

**Question -43 ****The age of a man is twice the square of the age of his son. Eight years hence, the age of the man will be 4 years more than three times the age of his son. Find the present age.**

**Answer -43**

Let the present age of the son be x years.

∴ Present age of father = 2x^{2} years

Eight years hence,

Son’s age = (x + 8) years

Father’s age = (2x^{2} + 8) years

2x^{2} + 8 = 3(x + 8) + 4

⇒ 2x^{2} + 8 = 3x + 24 + 4

⇒ 2x^{2} – 3x – 24 – 4 + 8 = 0

⇒ 2x^{2 }– 3x – 20 = 0

⇒ 2x^{2 }– 8x + 5x – 20 = 0

⇒ 2x(x – 4) +5(x – 4) = 0

⇒ (x – 4)(2x + 5) = 0

Either x – 4 = 0,

then x = 4

or

2x + 5 = 0,

then 2x = -5

⇒ x = -5/2

But, it is not possible.

hence Present age of the son = 4 years

and present age of the man = 2x^{2}

= 2(4)^{2} years

= 32 years.

**Question -44 ****Two years ago, a man’s age was three times the square of his daughter’s age. Three years hence, his age will be four times his daughter’s age. Find their present ages.**

**Answer -44**

2 years ago,

Let the age of daughter = x

age of man = 3x²

then present age of daughter = x + 2

and mean = 3x² + 2

and 3 years hence , the age of

the daughter = x + 2 + 3 = x + 5

and man = 3x^{2} + 2 + 3 = 3x^{2} + 5

According to the condition.

3x^{2} + 5 = 4(x + 5)

⇒ 3x^{2} + 5 = 4x + 20

⇒ 3x^{2} – 4x + 5 – 20 = 0

⇒ 3x^{2} – 4x – 15 = 0

⇒ 3x^{2} – 9x + 5x – 15 = 0

⇒ 3x(x – 3) + 5(x – 3) = 0

⇒ (x – 3)(3x + 5) = 0

EIther x – 3 = 0,

then x = 3

or

3x + 5 = 0,

then 3x = -5

⇒ x = -53

Which is not possible, as age can’t be negative

If x = 3, then

Present age of man

= 3x^{2} + 2

= 3(3)^{2} + 2

= 27 + 2

= 29 years

and age of daughter

= x + 2

= 3 + 2

= 5 years.

**Question -45 ****The length (in cm) of the hypotenuse of a right-angled triangle exceeds the length of one side by 2 cm and exceeds twice the length of another side by 1 cm. Find the length of each side. Also, find the perimeter and the area of the triangle.**

**Answer- 45**

Let the length of one side = x cm

and other side = y cm.

then hypotenuse = x + 2, and 2y + 1

∴ x + 2 = 2y + 1

⇒ x – 2y 1 – 2

⇒ x – 2y = -1

⇒ x = 2y – 1 …(i)

and using Pythagorous theorem,

x^{2} + y^{2} = (2y + 1)^{2}

⇒ x^{2} + y^{2} = 4y^{2} + 4y + 1

⇒ (2y – 1)^{2} + y^{2} = 4y^{2} + 4y + 1 …[From (i)] ⇒ 4y^{2} – 4y + 1 + y^{2} = 4y^{2} + 4y + 1

⇒ 4y^{2} – 4y + 1 + y^{2} – 4y^{2} – 4y – 1 = 0

⇒ y^{2} – 8y = 0

⇒ y(y – 8) = 0

Either y = 0,

but it is not possible

or

y – 8 = 0,

then y = 8

Substituting the value of y in (i)

x = 2(8) – 1

= 16 – 1

= 15

∴ Length of one side = 15 cm

and length of other side = 8 cm

and hypotenuse

= x + 2

= 15 + 2

= 17

∴ Perimeter

= 15 + 8 + 17

= 40cm

and Area

= 1/2 x one side x other side

= (1/2 )×15×8

= 60cm^{2}.

**Question- 46 ****If twice the area of a smaller square is subtracted from the area of a larger square, the result is 14 cm². However, if twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm². Determine the sides of the two squares.**

**Answer- 46**

Let the side of the smaller square be x cm and side of the larger square be y cm. Then,

According to the given conditions, we have

y^{2} – 2x^{2} = 14 …(i)

and 2y^{2} + 3x^{2} = 203 …(ii)

Substituting y^{2} = 14 + 2x^{2
}From (i) in (ii), we get

2(14 + 2x^{2}) + 3x^{2}= 203

⇒ 28 + 4x^{2} + 3x^{2} = 203

⇒ 7x^{2} = 175

⇒ x^{2 =} 25

⇒ x = ±5

∴ x = 5 [x = –5 is rejected]

From (i), y^{2} –2x^{2} = 14

⇒ y^{2 }– 2(5)^{2} = 14

⇒ y^{2} – 50 = 14

⇒ y^{2} = 64

⇒ y = ±8

⇒ y = 8 [y = –8 is rejected]

Hence, side of smaller square = 5 cm

and side of larger square = 8 cm.

— : End of ML Aggarwal Quadratic Equations Exe-5.5 Class 10 ICSE Maths Solutions : –

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