# 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 = x

^{2}Now according to condition-

⇒(number)+(number)

^{2}=63/4⇒ x+x

^{2}= 63/4⇒ x

^{2}+x-63/4=0Multiplying by 4-

⇒ 4x

^{2}+4x-63=0⇒ 4x

^{2}+(18-14)x – 63 =0 [because 63*4=252so 18*14=252 & 18-14=4]

⇒ 4x

^{2}+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⇒ x

^{2}+ (x+1)(x+2)=154⇒ x

^{2}+ x^{2}+ 3x + 2 = 154⇒ 2x

^{2}+3x-152=0⇒ 2x

^{2}+(19-16)x-152=0⇒ 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

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

⇒ x

^{2}+ x -12=0⇒ x

^{2}+(4-3)x-12=0⇒ x

^{2}+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⇒x

^{2}+(2x-3)^{2}=233⇒x

^{2}+4x^{2}-2*2x*3+9=233⇒5x

^{2}-12x+9-233=0⇒5x

^{2}-12x-224=0⇒5x

^{2}-(40-28)x-224=0⇒5x

^{2}-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⇒4x

^{2}+4x^{2}+2*2x*2+4=340⇒8x

^{2}+8x-336=0Dividing by 8-

⇒x

^{2}+x-42=0⇒x

^{2}+(7-6)x-42=0⇒x

^{2}+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=(x

^{2}-4x)⇒x

^{2}-4x-21=0⇒x

^{2}-(7-3)-21=0⇒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

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⇒x

^{2}+(x-3)^{2}=117⇒x

^{2}+x^{2}-2*x*3+9=117⇒2x

^{2}-6x+9-117=0⇒2x

^{2}-6x-108=0Dividing by 2-

⇒x

^{2}-3x-54=0⇒x

^{2}-(9-6)x-54=0⇒x

^{2}-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⇒x

^{2}+(x+1)^{2}+(x+2)^{2}=149⇒x

^{2}+x^{2}+2*x*1+1+x^{2}+2*x*2+4=149⇒3x

^{2}+6x+5-149=0⇒3x

^{2}+6x-144=0Dividing by 3-

⇒x

^{2}+2x-48=0⇒x

^{2}+(8-6)x-48=0⇒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

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-x

^{2}⇒x

^{2}-16x+48=0⇒x

^{2}-(12+4)x+48=0⇒x

^{2}-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

⇒9x

^{2}+9x=270on dividing by 9-

⇒x

^{2}+x=30⇒x

^{2}+x-30=0⇒x

^{2}+(6-5)x-30=0⇒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

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

(x

^{2}+ 1)/x = 17/4⇒ 4(x

^{2}+1) = 17x⇒ 4x

^{2 }+ 4 – 17x = 0⇒ 4x

^{2 }+ 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.

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