Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.7 | Set 2

Question 11: The sum of a number and its square is 63/4 , find the numbers.

Solution:

Let the number is = x

so it's square is = x²

Now according to condition-

⇒(number)+(number)²=63/4

⇒ x+x² = 63/4

⇒ x²+x-63/4=0

Multiplying by 4-

⇒ 4x²+4x-63=0

⇒ 4x² +(18-14)x - 63 =0    [because 63*4=252

                                               so 18*14=252  & 18-14=4]

⇒ 4x²+18x-14x-63=0

⇒2x(2x+9)-7(2x+9)=0

⇒(2x+9)(2x-7)=0

either 2x+9=0      or           2x-7=0

            x=-9/2       or                x=7/2

So number is -9/2 or 7/2.  

Question 12: There are three consecutive integers such that the square of the first increased by the product of the other two gives 154. What are the integers?

Solution:

Let the first number is = x

so second number is = x+1

and third number is = x+2

now according to the given  condition-

⇒ (first number)² + (second number)*(third number)=154

⇒ x² + (x+1)(x+2)=154

⇒ x² + x² + 3x + 2 = 154

⇒ 2x²+3x-152=0

⇒ 2x² +(19-16)x-152=0

⇒ 2x² +19x-16x-152=0

⇒ x(2x+19)-8(2x+19)=0

⇒ (2x+19)(x-8)=0

Either  2x+19=0    or       x-8=0

           x=-19/2      or         x=8

but in question it is said number should be integer 

so discard x=-19/2

when x=8

First number is =8

Second number is =9

And Third number is =10

Question 13: The product of two successive integral multiple of 5 is 300. Determine the multiplies.

Solution:

Let the first integral multiple of 5 is =5x

So next is = 5x+5

Now according to the condition-

 ⇒ (first integral multiple of 5)*(next integral multiple of 5)=300

⇒ 5x(5x+5)=300

Dividing by 5-

 ⇒ x(5x+5)=60

again dividing by 5-

⇒ x(x+1)=12

⇒ x² + x -12=0

⇒ x²+(4-3)x-12=0

⇒ x²+4x-3x-12=0

⇒ x(x+4)-3(x+4)=0

⇒ (x+4)(x-3)=0

Either x+4=0    or       x-3=0

              x=-4     or          x=3

when x=-4

first integral multiple of 5 is = 5*-4 = -20

and next integral multiple of 5 is = 5*-4+5= -15

when x=3

first integral multiple of 5 is = 5*3 = 15

and next integral multiple of 5 is = 5*3+5 = 20

Question 14:The sum of the squares of two numbers is 233 and one of the numbers is 3 less than twice the other number. Find the numbers.

Solution:

Let first number is= x

So second number = 2*(first number)-3

                              = 2x-3

Now according to given condition-

⇒(first number)²+(second number)² = 233

⇒x²+(2x-3)²=233

⇒x²+4x²-2*2x*3+9=233

⇒5x²-12x+9-233=0

⇒5x²-12x-224=0

⇒5x²-(40-28)x-224=0

⇒5x²-40x+28x-224=0

⇒5x(x-8)+28(x-8)=0

⇒(x-8)(5x+28)=0

Either   x-8=0      or        5x+28=0

               x=8       or             x=-28/5

but when x=-28/3, it doesn't satisfy given condition.

so on taking x=8

first number is=x= 8

and second number is=2x-3=13 

Question 15:Find two consecutive even integers whose squares have the sum 340.

Solution:

Let first even integer=2x

so second even integer=2x+2

According to given condition-

⇒(first integer)²+(second integer)²=340

⇒(2x)²+(2x+2)²=340

⇒4x²+4x²+2*2x*2+4=340

⇒8x²+8x-336=0

Dividing by 8-

⇒x²+x-42=0

⇒x²+(7-6)x-42=0

⇒x²+7x-6x-42=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

When x=-7

then first integer=2*x= -14

and second integer=2x+2=-12

when x=6

then first integer=2*x= 12

and second integer=2x+2=14

Question 16:The difference of two numbers is 4. If the difference of their reciprocals is 4/21, find the numbers. 

Solution:

Let first number is=x

So second number is=x-4

Reciprocal of first number is=1/x

and reciprocal of second number is=1/x-4

According to given condition-

⇒(reciprocal of first number)-(reciprocal of second number)=4/21

⇒(1/(x-4))-(1/x)=4/21

on taking LCM-

⇒(x-x+4)/(x(x-4))=4/21

⇒21(4)=4(x(x-4)

⇒21=(x²-4x)

⇒x²-4x-21=0

⇒x²-(7-3)-21=0

⇒x²-7x+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

When x=7

numbers are= 3,7

and when x=-3

numbers are=-7,-3

Question 17: Find two natural numbers that differ by 3 and whose squared have the sum 117.

Solution:

Let the first number is=x

So second number is=x-3

Now according to given condition-

⇒(first number)²+(second number)²=117

⇒x²+(x-3)²=117

⇒x²+x²-2*x*3+9=117

⇒2x²-6x+9-117=0

⇒2x²-6x-108=0

Dividing by 2-

⇒x²-3x-54=0

⇒x²-(9-6)x-54=0

⇒x²-9x+6x-54=0

⇒x(x-9)+6(x-9)=0

⇒(x-9)(x+6)=0

Either   x-9=0      or     x+6=0

              x=9        or        x=-6

But x=-6 is not a natural number.

So when x=9

⇒first number is=9

⇒second number is=9-3=6

Question 18: The sum of squares of three consecutive natural numbers is 149. Find the numbers.

Solution:

Let first number is=x

so second is=x+1

and third is=x+2

Now according to given condition-

⇒(first number)²+(second number)²+(third number)²=149

⇒x²+(x+1)²+(x+2)²=149

⇒x²+x²+2*x*1+1+x²+2*x*2+4=149

⇒3x²+6x+5-149=0

⇒3x²+6x-144=0

Dividing by 3-

⇒x²+2x-48=0

⇒x²+(8-6)x-48=0

⇒x²+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

But x=-8 is not a natural number.

so when x=6

⇒first number is=x=6

⇒second number is=x+1=7

⇒third number is=x+2=8 

Question 19:The sum of two numbers is 16. The sum of their reciprocals is 1/3. Find the numbers.

Solution:

Let first number is=x

So second number is=16-x

Reciprocal of first number is=1/x

and Reciprocal of second number is=1/(16-x)

Now according to given condition-

⇒(Reciprocal of first number)+(Reciprocal of second number)=1/3

⇒(1/x)+(1/(16-x))=1/3

on taking LCM-

⇒(16-x+x)/(x(16-x))=1/3

⇒16*3=x(16-x)

⇒48=16x-x²

⇒x²-16x+48=0

⇒x²-(12+4)x+48=0

⇒x²-12x-4x+48=0

⇒x(x-12)-4(x-12)=0

⇒(x-12)(x-4)=0

Either    x-12=0         or        x-4=0

                x=12         or          x=4

so when x=12

numbers are = 12,4

when x=4

            numbers are=4,12

means required numbers are=4,12

Question 20: Determine two consecutive multiples of 3 whose product is 270.

Solution:

Let first multiple of 3 is=3x

so second is=3x+3

Now according to given condition-

⇒ (first multiple of 3)*(second multiple of 3)=270

⇒3x(3x+3)=270

⇒9x²+9x=270

on dividing by 9-

⇒x²+x=30

⇒x²+x-30=0

⇒x²+(6-5)x-30=0

⇒x²+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

when x=-6

         first multiple is=3x=-18

         second multiple is=3x+3=-15

when  x=5

         first multiple is=3x=15

         second multiple is=3x+3=18

Question 21. The sum of a number and its reciprocal is 17/4. Find the number.

Solution:

Let the number be x.

Then from the question, we have

x + 1/x = 17/4

(x² + 1)/x = 17/4

⇒ 4(x²+1) = 17x

⇒ 4x² + 4 – 17x = 0

⇒ 4x² + 4 – 16x – x = 0

⇒ 4x(x – 4) – 1(x – 4) = 0

⇒ (4x – 1)(x – 4) = 0

Now, either x – 4 = 0 ⇒ x = 4

Or, 4x – 1 = 0 ⇒ x = 1/4

Thus, the value of x is 4.