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Learn Exercise 4.3 with Free Lessons & Tips

The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

Let the length of the shorter side be x metres. 

The length of the diagonal= 60+x metres

The length of the longer side =30+x metres

Applying Pythagoras theorem, 

Diagonal²=longer side²+shorter side²

(60+x) ²= (30+x) ² + x²

3600+120x+x²=900+60x+x²+x²

2700+60x-x²=0

2700+90x-30x-x²=0

90(30+x)-x(30+x) =0

X=90, 

Shorter side is 90m,  longer side is 90+30=120m

 

Comments

Sum of the areas of two squares is 468 m2 . If the difference of their perimeters is 24 m, find the sides of the two squares.

Let the sides of first and second square be X and Y .


Area of first square = (X)²

And,

Area of second square = (Y)²

According to question,

(X)² + (Y)² = 468 m² ------------(1).

Perimeter of first square = 4 × X

and,

Perimeter of second square = 4 × Y

According to question,

4X - 4Y = 24 -----------(2)

From equation (2) we get,

4X - 4Y = 24

4(X-Y) = 24

X - Y = 24/4 

X - Y = 6

X = 6+Y ---------(3)

Putting the value of X in equation (1)

(X)² + (Y)² = 468

(6+Y)² + (Y)² = 468

(6)² + (Y)² + 2 × 6 × Y + (Y)² = 468

36 + Y² + 12Y + Y² = 468

2Y² + 12Y - 468 +36 = 0

2Y² + 12Y -432 = 0

2( Y² + 6Y - 216) = 0

Y² + 6Y - 216 = 0

Y² + 18Y - 12Y -216 = 0

Y(Y+18) - 12(Y+18) = 0

(Y+18) (Y-12) = 0

(Y+18) = 0 Or (Y-12) = 0

Y = -18 OR Y = 12

Putting Y = 12 in EQUATION (3)

X = 6+Y = 6+12 = 18

Side of first square = X = 18 m

and,

Side of second square = Y = 12 m.

Comments

The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers

Supoose two numbers are a and b.

........(1)

And   ...........(2)

Put the value of equation (2) in equation (1) ,

,

,

(a+10) (a-18) =0,

a= -10 is not possible because  cannot be negative , so 

a= 18 and b= 12.

Therefore the larger number is 18 and the smaller number is 12.

Comments

Two water taps together can fill a tank in  hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Let the time taken by the smaller diameter tap be A hours

Let the time taken by the larger diameter tap be A-10 hours

Total time taken with both Taps together= 9  3/8 = 75 /8 hours

Amount filled in one hour by smaller diameter tap = 1/A [use concept of proportions, in A hours it fills 1 complete unit then in 1 hour it will fill 1/A units]

and by larger diamter tap = 1/(A-10) units

As it takes 75/8 hours to fill complete unit .... in 1 hour it will fill 1/(75/8) = 8/75

[1/A] + [1/(A-10)] = 8/75

Take LCM

(A-10+A)/[A*(A-10)] = 8/75

(2A-10)/(A²-10A) = 8/75

8(A²-10A) = 75 ×(2A-10){cross multiply}

8/2 (A²-10A) = 75 (A-5) {taking 2 common and dividing}

4A²-40A = 75A-375

4A² -40A-75A +375 = 0

4A²-115A + 375 = 0

4A²-100A-15A +375 = 0

4A(A-25)-15(A-25)=0

(A-25)(4A-15)

A= 25 hours

or

A= 15/4 hours

If A= 25 hours

then A-10 = 25-10 = 15 hours

if A = 15/4 hours

then A-10 = 15/4 - 10 = 15-40/4 = -25/4 hours which is not possible

since time cannot be negative therefore A = 25 hours

 

Comments

Find the roots of the following equations:

i)

Multiply by x

this resembles a Quadratic equation.

a=1,b=-3,c=-1

ii) Multiplying both sides by 

 

x(x-1)-2(x-1)=0

(x-1)(x-2)=0

x=1 or x=2

Comments

The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1/3 Find his present age.

Let Rehman's present age be x

So before 3 years age=(x-3) years

His age after 5 years= (x+5) years

So a/c to given data

 

 

 

 

x(x-7)+3(x-7)=0

(x-7)(x+3)=0

So, x=7 or -3

Age can not be negative soRehman's present age is 7 years.

 

Comments

In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

Let the marks in Maths be x so the marks in English = (30 - x).

ATQ, (x + 2) (27 - x) = 210

or 27x - x² + 54 - 2x = 0

or - x² + 25x - 156 = 0

or x² - 25x + 156 = 0

or x² - 13x - 12x + 156 = 0

or x(x - 13) - 12(x - 13) = 0

or (x - 12) (x - 13) = 0

Therefore, x = 12 or x = 13

Hence, marks in Maths is 12 or 13, therefore, marks in English is either 17 or 18.

Comments

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11km/h more than that of the passenger train, find the average speed of the two trains.

Let the average speed of the passenger train be 'x' km/hr and the average speed of the express train be (x + 11) km/hr
Distance between Mysore and Bangalore = 132 km
It is given that the time taken by the express train to cover the distance of 132 km is 1 hour less than the passenger train to cover the same distance.
So, time taken by passenger train = 132/x hr
The time taken by the express train = {(132)/x+11} hr
Now, according to the question
{132/(x+11)} = 132/x + 1
After taking L.C.M. of 132/x +1 and then solving it we get (132+x)/x.
Now,
{132/(x+11)} = (132+x)/x
By cross multiplying, we get
132x = x²+132x+11x+1452
x² + 11x - 1452 = 0
x² + 44x - 33x - 1452 = 0
x(x+44) - 33(x+44) = 0
(x+33) (x+44) = 0
x - 44 or x = 33
As the speed cannot be in negative therefore, x = 33 or the speed of the passenger train = 33 km/hr and the speed is 33 + 11 = 44 km/hr.

Comments

Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

The answers are in the form of 

On using the Quadratic formula, 

substituting the values in the Quadratic formula, 

we get 

substituting the quadratic formula,

we get, 

substituting the values in the quadratic formula, we get

However, the square of a number cannot be negative, hence it is not a real root.

Comments

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