SELINA Solution Class 9 Simultaneous (Linear) Equation (Including problems ) Exercise 6E

Question 1

The ratio of two numbers is ${\displaystyle \frac{2}{3}}$. If 2 is subtracted from the first and 8 from the second, the ratio becomes the reciprocal of the original ratio. Find the numbers.

Sol:

Let the two numbers be x and y.
According to the question,
${\displaystyle \frac{x}{y}=\frac{2}{3}}$
3x - 2y = 0                        .....(1)

Also,${\displaystyle \frac{x-2}{y-8}=\frac{3}{2}}$

2x - 3y = - 20                   .....(2)

Multiplying equations no. (1) by 2 and (2) by 3
6x - 4y = 0                       .....(3)
6x - 9y = - 60                  ......(4)
Subtracting equation (4) from (3)
      6x - 4y = 0
-    6x - 9y = - 60   
      -     +     +        
             5y = 60
               y = 12

From (1), we get
3x - 2(12) = 0
x = ${\displaystyle \frac{24}{3}}$
x = 8
Thus, the numbers are 8 and 12.

Question 2

Two numbers are in the ratio 4 : 7. If thrice the larger be added to twice the smaller, the sum is 59. Find the numbers.

Sol:

Let the smaller number be x and the larger number be y.
According to the question,
${\displaystyle \frac{x}{y}=\frac{4}{7}}$
7x = 4y
7x - 4y = 0                 ......(1)
and, 3y + 2x = 59      ......(2)

Multiplying equation no. (1) by 3 and (2) by 4.
21x - 12y = 0             ......(3)
8x + 12y = 236          .......(4)
Adding equation (3) and (4),
        21x - 12y = 0
+      8x + 12y = 236  
        29x = 236
           x = ${\displaystyle \frac{236}{29}}$
From (1)
${\displaystyle 7\left(\frac{236}{29}\right)=4y}$

${\displaystyle y=7\left(\frac{59}{29}\right)}$   

${\displaystyle y=\left(\frac{413}{29}\right)}$
Hence, the numbers are ${\displaystyle \frac{236}{29}\text{and}\frac{413}{29}}$.         

Question 3

When the greater of the two numbers increased by 1divides the sum of the numbers, the result is ${\displaystyle \frac{3}{2}}$. When the difference of these numbers is divided by the smaller, the result ${\displaystyle \frac{1}{2}}$. Find the numbers.

Sol:

Let the two numbers be a and b respectively such that b > a.
According to given condition,
${\displaystyle \frac{a+b}{b+1}=\frac{3}{2}}$
⇒ 2a + 2b = 3b + 3
⇒ 2a - b = 3                           ......(1)
Also,
${\displaystyle \frac{b-a}{a}=\frac{1}{2}}$ 
⇒ 2b - 2a = a
⇒ 2b - 3a = 0                         .......(2)
Multiplying (1) by 2, We get
4a - 2b = 6                            .......(3)
Adding (2) and (3),We get
      4a - 2b = 6  
+  - 3a + 2b = 0  
        a = 6
Substituting a = 6 in equation (1), We get
2(6) - b = 3
⇒ 12 - b = 3
⇒ b = 9
Thus, the two numbers are 6 and 9 respectively.       

Question 4

The sum of two positive numbers x and y is 50 and the difference of their squares is 720. Find the numbers.

Sol:

Two numbers are x and y such that x > y.
Now,
x + y = 50                          ….(i)
And,
y² - x² = 720
⇒ (y - x)(y + x) = 720
⇒ (y - x)(50) = 720
⇒ y - x = 14.4                     ….(ii)

Adding (i) and (ii), we get
2y = 64.4
⇒ y = 32.2

Substituting the value of y in (i), we have
x + 32.2 = 50
⇒ x = 17.8
Thus, the two numbers are 17.8 and 32.2 respectively.

Question 5

The sum of two numbers is 8 and the sum of their reciprocals is ${\displaystyle \frac{8}{15}}$. Find the numbers.

Sol:

Let the two numbers be x and y respectively.
Then,
x + y = 8                           ….(i)
⇒ x = 8 - y
And,
${\displaystyle \frac{1}{x}+\frac{1}{y}=\frac{8}{15}}$

⇒ ${\displaystyle \frac{y+x}{xy}=\frac{8}{15}}$

⇒ ${\displaystyle \frac{8}{xy}=\frac{8}{15}}$                .....[ From(1) ]
⇒ xy = 15
⇒ ( 8 - y )y = 15
⇒ 8y - y² = 15
⇒ y² - 8y + 15 = 0
⇒ y² - 3y - 5y + 15 = 0
⇒ y( y - 3 ) - 5( y - 3 ) = 0
⇒ ( y - 3 )( y - 5 ) = 0
⇒ y = 3 or y = 5
⇒ x = 5 or x = 3
Thus, the two numbers are 3 and 5 respectively.

Question 6

The difference between two positive numbers x and y ( x > y ) is 4 and the difference between their reciprocals is ${\displaystyle \frac{4}{21}}$. Find the numbers.

Sol:

Two numbers are x and y respectively such that x > y.
Then,
x - y = 4                                          ….(i)
⇒ x = 4 + y
And,
${\displaystyle \frac{1}{y}-\frac{1}{x}=\frac{4}{21}}$
⇒ ${\displaystyle \frac{x-y}{xy}=\frac{4}{21}}$

⇒ ${\displaystyle \frac{4}{xy}=\frac{4}{21}}$                      ......[From(1)]
⇒ xy = 21
⇒ ( 4 + y )y = 21
⇒ 4y + y2 = 21
⇒ y2 + 4y - 21 = 0
⇒ y2 + 7y - 3y - 21 = 0
⇒ y( y + 7 ) - 3( y + 7 ) = 0
⇒ ( y - 3 )( y + 7 ) = 0
⇒ y = 3 and y = -7
We reject y = -7 since y is positive.
⇒ y = 3
⇒ x = 4 + y = 4 + 3 = 7
Thus, the two numbers are 7 and 3 respectively.

Question 7

Two numbers are in the ratio 4:5. If 30 is subtracted from each of the numbers, the ratio becomes 1:2. Find the numbers.

Sol:

Let the common multiple between the numbers be x.
So, the numbers are 4x and 5x.
According to the question,
${\displaystyle \frac{4x-30}{5x-30}=\frac{1}{2}}$
⇒ 8x - 60 = 5x - 30
⇒ 3x = 30
⇒ x = 10
So, 4x = 4(10) = 40 and 5x = 5(10) = 50
Thus, the numbers are 40 and 50.

Question 8

If the numerator of a fraction is increased by 2 and denominator is decreased by 1, it becomes ${\displaystyle \frac{2}{3}}$. If the numerator is increased by 1 and denominator is increased by 2, it becomes ${\displaystyle \frac{1}{3}}$. Find the fraction.

Sol:

Let the numerator and denominator a fraction be x and y respectively .
According to the question,
${\displaystyle \frac{x+2}{y-1}=\frac{2}{3}}$
3x + 6 = 2y - 2
3x - 2y = - 8                               ........(1)
And,
${\displaystyle \frac{x+1}{y+2}=\frac{1}{3}}$
3x + 3 = y + 2
3x - y = - 1                                  .......(2)
Now,
Subtracting equation (1) from (2),
     3x - y = - 1
-   3x - 2y = - 8 
    -     +      +   
            y = 7

From (1) ,
3x - 2 (7) = - 8
3x = - 8 + 14
x = 2
Required fraction = ${\displaystyle \frac{2}{7}}$

Question 9

The sum of the numerator and the denominator of a fraction is equal to 7. Four times the numerator is 8 less than 5 times the denominator. Find the fraction.

Sol:

Let the numerator and denominator of a fraction be x and y respectively.

let the fraction will be ${\displaystyle \frac{x}{y}}$.
According to the question,
x + y = 7                                    ...(1)
5y - 4x = 8                                 ...(2)

Multiplying equation no. (1) by 4
4x + 4y = 28                              ...(3)
Adding equation (3) and (2)
     4x + 4y = 28
+ - 4x + 5y = 8  
              9y = 36
               y = 4

From (1),
x + 4 = 7
x = 3
Required fraction = ${\displaystyle \frac{3}{4}}$.

Question 10

If the numerator of a fraction is multiplied by 2 and its denominator is increased by 1, it becomes 1. However, if  the numerator is increased but 4 and denominator is multiplied by 2, the fraction becomes ${\displaystyle \frac{1}{2}}$. Find the fraction.

Sol:

Let the numerator of the fraction be x and the denominator be y.
So, the Fraction is ${\displaystyle \frac{x}{y}}$
According to the question,
${\displaystyle \frac{2x}{y+1}=1}$
⇒ 2x = y + 1
⇒ 2x - y = 1                       .....(1)
and,
${\displaystyle \frac{x+4}{2y}=\frac{1}{2}}$
⇒ 2x + 8 = 2y
⇒ 2x - 2y = - 8                  ......(2)

Solving equations (1) and (2), We get
y = 9
Putting the value of y in (1), we get
2x - 9 = 1
⇒ 2x = 1 + 9
⇒ x = 5
So, the fraction is ${\displaystyle \frac{5}{9}}$.

Question 11

A Fraction becomes ${\displaystyle \frac{1}{2}}$ if 5 subtracted from its numerator and 3 is subtracted from its denominator. If the denominator of this fraction is 5 more than its numerator. Find the fraction.

Sol:

Let the numerator of the fraction be x and denominator of the fraction be y.
Then, the fraction =${\displaystyle \frac{x}{y}}$
According to given condition, we have
${\displaystyle \frac{x-5}{y-3}=\frac{1}{2}}$
⇒ 2x - 10 = y - 3
⇒ 2x - y = 7                               ...(1)
And,
x + 5 = y
⇒ x - y = - 5                             ...(2)
Subtracting (2) from (1), We get
x = 12
⇒ y = x + 5 
⇒ y = 12 + 5 = 17
Hence, the fraction is ${\displaystyle \frac{12}{17}}$. 

Question 12

The sum of the digits of the digits of two digit number is 5. If the digits are reversed, the number is reduced by 27. Find the number.

Sol:

Let the digit at unit’s place be x and the digit at ten’s place y.
Required no. = 10y + x
If the digit’s are reversed,
Reversed no. = 10x + y

According to the question,
x + y = 5                             ...(1)
and,
(10y + x) - (10x + y) = 27
9y - 9x = 27
y - x = 3                             ...(2)
Now,
Adding equation (1) and (2),
         y - x = 3                  ....(2)
   +   y + x = 5                 ....(1)
        2y = 8
         y = 4

From (1)
x + 4 = 5
x = 1
Require no is
10 (4) + 1 = 41

Question 13

The sum of the digits of a two digit number is 7. If the digits are reversed, the new number decreased by 2, equals twice the original number. Find the number.

Sol:

Let the digit at unit’s place be x and the digit at ten’s place be y.
Required no. = 10y + x
If the digit’s are reversed
Reversed no. = 10x + y

According to the question,
x + y = 7                        ...(1)
and,
10x + y - 2 = 2(10y + x).
8x - 19y = 2                  ...(2)

Multiplying equation no. (1) by 19.
19x + 19y = 133           ...(3)

Now adding equation(2) and (3)
       19x + 19y = 133                
 +      8x - 19y = 2     
                 27x = 135
                     x = 5
From (1)
5 + y = 7
y = 2

Required number is = 10(2) + 5 = 25.

Question 14

The ten’s digit of a two digit number is three times the unit digit. The sum of the number and the unit digit is 32. Find the number.

Sol:

Let the digit at unit’s place be x and the digit at ten’s place be y.
Required no. = 10y + x

According to the question,
y = 3x 
⇒ 3x - y = 0                       ...(1)
and,
10y + x + x = 32
10y + 2x = 32                    ...(2)

Multiplying equation no. (1) by 10.
30x - 10y = 0                    ....(3)
Now,
Adding equation (3) and (2)
          30x - 10y = 0
     +   2x + 10y = 32  
          32x = 32
             x = 1

From (1), we get
y = 3(1) = 3
Required no is = 10(3) + 1 = 31.

Question 15

A two-digit number is such that the ten’s digit exceeds twice the unit’s digit by 2 and the number obtained by inter-changing the digits is 5 more than the the sum of the digits. Find the two digit number.

Sol:

Let the digit a unit’s place be x and the digit at ten’s place be y.
Required no. = 10y + x.

According to the question,
y - 2x = 2
-2x + y = 2                   ...(1)
and,
(10x + y) -3 (y + x) = 5
7x - 2y = 5                   ...(2)

Multiplying equation no. (1) by 2.
 - 4x + 2y = 4              ...(3)

Now adding (2) and (3),
        - 4x + 2y = 4
   +      7x - 2y = 5  
           3x = 9
            x = 3

From (1) ,we get
-2(3) + y = 2
y = 8

Required number is 10(8) + 3 = 83.

Question 16

Four times a certain two digit number is seven times the number obtained on interchanging its digits. If the difference between the digits is 4; find the number.

Sol:

Let x be the number at the ten's place.
and y be the number at the unit's place.
So, the number is 10x + y.

Four times a certain two-digit number is seven times
the number obtained on interchanging its digits.
⇒ 4( 10x + y ) = 7( 10y + x )
⇒ 40x + 4y = 70y + 7x
⇒ 33x - 66y = 0
⇒ x - 2y = 0                  ....(1)

If the difference between the digits is 4, then
⇒ x - y = 4                     ...(2)
Subtracting equation (1) from equation (2), we get :
         x - y = 4
    -   x - 2y = 0  
       -   +      -     
              y = 4
Subtracting y = 4 in equation (1), We get
x - 2(4) = 0
⇒ x = 8
∴ The number is 10x + y = 10(8) + 4 = 84.

Question 17

The sum of two digit number and the number obtained by interchanging the digits of the number is 121. If the digits of the number differ by 3, find the number.

Sol:

Let the tens digit of the number be x and the units digit be y.
So, the number is 10x + y.
The number obtained by interchanging the digits will be 10y + x.
According to question, we have
10x + y + 10y + x = 121
⇒ 11x  + 11y = 121
⇒  11( x + y ) = 121
⇒  x + y = 11                       ...(1)
And, 
x - y = 3                               ...(2)

Adding (1) and (2), We get
∴ 2x = 14
x = 7

∴ y = 11 - x = 11 - 7 = 4
Hence, the number is 74.

Question 18

A two digit number is obtained by multiplying the sum of the digits by 8. Also, it is obtained by multiplying the difference of the digits by 14 and adding 2. Find the number.

Sol:

Let the tens digit of the number be x and the units digit be y.
So, the number is 10x + y.
According to the question,
10x + y = 8( x + y ) 
⇒ 10x + y = 8x + 8y
⇒ 2x = 7y                         ....(1)

and 10x + y = 14( x - y ) + 2 or 10x + y = 14( y - x ) + 2
⇒ 4x - 15y = - 2       ...(2)   or 24x - 13y = 2       ....(3)

Solving (1) and (2), We get
y = 2 and x = 7

Solving (1) and (3), We get
y = ${\displaystyle \frac{2}{71}}$

This is not possible, since y is a digit and cannot be in fraction form.
So, the number is 72.

S.chand class 6 Mathematics Chapter 6 Exercise 6E

 Exercise 6E

Question 1

1. Find:

(i) $8.46 \div 10$

(ii) $68+10$

(iii) $49.5+10$

(iv) $0.082 \div 10$

(v) $584 \div 100$

(vi) $7.6 \div 100$

(vii) $83 \div 1000$

(viii) $90+10000$

2. Find the value of

(i) $34.4 \div 8$

(ii) $0.015 \div 3$

(iii) $3.984+4$

(iv) $0.00081 \div 9$

(v) $20.202+74$

(vi) $11.2575 \div 114$

(vii) $1.296 \div 54$

(viii) $0.0000729 \div 0.09$

3. Divide:

(i) $490.432$ by $7.9$

(ii) $31.083$ by $3.25$

(iii) $0.622454$ by $3.46$

(iv) $490.0456$ by $4.78$

(v) $0.32594$ by $0.086$

(vi) $17.1$ by $0.00019$

(vii) $0.000944$ by $0.008$

(viii) $0.0211788$ by $0.333$

4. Bamboo is one of the fastest growing plants on Earth. In Japan a bamboo plant grew $45.6$ inches in 1 day. What was the average growth of this plant per hour?

5. Hazel is training for a marathon. If she runs at an average speed of $9.25 \mathrm{~km}$ per hour, how long will it take her to run $46.25$ kilometres?

Multiple Choice Questions (MCQs)

Tick (✔) the Correct Option.

6. Pradeep used mental maths to find the answer to a problem. If his answer is 60 , which of the following could be the problem?

(a) $0.24 \div 0.04$

(b) $1.8 \div 0.03$

(c) $0.42 \div 0.7$

(d) $8.4 \div 1.4$

7. Kavita earned $₹ 844.725$ for working $10.5$ hours at her part time job after college. Kavita is paid per hour. How much did she earn per hour?

(a) ₹ $8.45$

(b) ₹ $80.45$

(c) ₹ $84.50$

(d) ₹ $80.50$

RS Aggarwal solution class 8 chapter 6 Operations on Algebraic Expressions Exercise 6E

Exercise 6E

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Question 1:

Tick (✓) the correct answer:
The sum of (6a + 4b − c + 3), (2b − 3c + 4), (11b − 7a + 2c − 1) and (2c − 5a − 6) is
(a) (4a − 6b + 2)
(b) (−3a + 14b − 3c + 2)
(d) (−6a + 17b)
(d) (−6a + 6b + c −4)

Answer 1:

(c) (−6a + 17b)

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Question 2:

Tick (✓) the correct answer:
(3q + 7p² − 2r³ + 4) − (4p² − 2q + 7r³ − 3) = ?
(a) (p² + 2q + 5r³ + 1)
(b) (11p² + q + 5r³ + 1)
(c) (−3p² − 5q + 9r³ − 7)
(d) (3p² + 5q − 9r³ +7)

Answer 2:

(d) (3p² + 5q − 9r³ +7)

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Question 3:

Tick (✓) the correct answer:
(x + 5)(x − 3) = ?
(a) x² + 5x − 15
(b) x² − 3x − 15
(c) x² + 2x + 15
(d) x² + 2x − 15

Answer 3:

(d) x² + 2x − 15

$$\begin{gathered} \left(x+5\right)\left(x-3\right) \\ \Rightarrow \left(x\right)\left(x-3\right)+\left(5\right)\left(x-3\right) \\ \Rightarrow x^2-3x +5x -15 \\ \Rightarrow x^2+2x-15 \end{gathered}$$

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Question 4:

Tick (✓) the correct answer:
(2x + 3)(3x − 1) = ?
(a) (6x² + 8x − 3)
(b) (6x² + 7x − 3)
(c) 6x² − 7x − 3
(d) (6x² − 7x + 3)

Answer 4:

(b) (6x² + 7x − 3)

$$\begin{gathered} \left(2x+3\right)\left(3x-1\right) \\ \Rightarrow \left(2x\right)\left(3x-1\right)+\left(3\right)\left(3x-1\right) \\ \Rightarrow 6x^2-2x +9x-3 \\ \Rightarrow 6x^2+7x-3 \end{gathered}$$

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Question 5:

Tick (✓) the correct answer:
(x + 4)(x + 4) = ?
(a) (x² + 16)
(b) (x² + 4x + 16)
(c) (x² + 8x + 16)
(d) (x² + 16x)

Answer 5:

(c) (x² + 8x + 16)

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Question 6:

Tick (✓) the correct answer:
(x − 6)(x − 6) = ?
(a) (x² − 36)
(b) (x² + 36)
(c) (x² − 6x + 36)
(d) (x² − 12x + 36)

Answer 6:

(d) (x² − 12x + 36)

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Question 7:

Tick (✓) the correct answer:
(2x + 5)(2x − 5) = ?
(a) (4x² + 25)
(b) (4x² − 25)
(c) (4x² − 10x + 25)
(d) (4x² + 10x − 25)

Answer 7:

(b) (4x² − 25)

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Question 8:

Tick (✓) the correct answer:
8a²b³ ÷ (−2ab) = ?
(a) 4ab²
(b) 4a²b
(c) −4ab²
(d) −4a²b

Answer 8:

(c) −4ab²

$$\begin{gathered} 8a^2{b}^{3 }÷ \left(-2ab\right) \\ \Rightarrow \left(\frac{8}{-2}\right)\left(a^{2-1}\right)\left(b^{3-1}\right) \\ \Rightarrow -4a{b}^2 \end{gathered}$$

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Question 9:

Tick (✓) the correct answer:
(2x² + 3x + 1) ÷ (x + 1) = ?
(a) (x + 1)
(b) (2x + 1)
(c) (x + 3)
(d) (2x + 3)

Answer 9:

(b) (2x + 1)

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Question 10:

Tick (✓) the correct answer:
(x² − 4x + 4) ÷ (x − 2) = ?
(a) (x − 2)
(b) (x + 2)
(c) (2 − x)
(d) (2 + x + x²)

Answer 10:

(a) (x − 2)

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Question 11:

Tick (✓) the correct answer:
(a + 1)(a − 1)(a² + 1) = ?
(a) (a⁴ − 2a² − 1)
(b) (a⁴ − a² − 1)
(c) (a⁴ − 1)
(d) (a⁴ + 1)

Answer 11:

(c) (a⁴ − 1)

Q12 | Ex-6E | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions | myhelper

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Question 12:

Tick (✓) the correct answer:
$\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x}-\frac{1}{y}\right)=?$
(a) $\left(\frac{1}{x^2}-\frac{1}{y^2}\right)$
(b) $\left(\frac{1}{x^2}+\frac{1}{y^2}\right)$
(c) $\left(\frac{1}{x^2}+\frac{1}{y^2}-\frac{1}{xy}\right)$
(d) $\left(\frac{1}{x^2}-\frac{1}{y^2}+\frac{1}{xy}\right)$

Answer 12:

a) $\left(\frac{1}{x^2}-\frac{1}{y^2}\right)$

$$\begin{gathered} \left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x}-\frac{1}{y}\right) \\ \Rightarrow According to the formula \left(a+b\right)\left(a-b\right)={\left(a\right)}^2-{\left(b\right)}^2: \\ \Rightarrow \left(\frac{1}{x^2}-\frac{1}{y^2}\right) \end{gathered}$$(1x2−1y2)

Q13 | Ex-6E | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions | myhelper

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Question 13:

Tick (✓) the correct answer:
If $\left(x+\frac{1}{x}\right)=5$, then $\left(x^2+\frac{1}{x^2}\right)=?$
(a) 25
(b) 27
(c) 23
(d) $25\frac{1}{25}$

Answer 13:

(c) 23

$$\begin{gathered} \left(x+\frac{1}{x}\right)= 5 \\ \Rightarrow Squaring both the sides: \\ \Rightarrow {\left(x+\frac{1}{x}\right)}^2= {\left(5\right)}^2 \\ \Rightarrow \left(x^2+\frac{1}{x^2}+2\left(x\right)\left(\frac{1}{x}\right)\right)= 25 \\ \Rightarrow \left(x^2+\frac{1}{x^2}\right)+2 =25 \\ \Rightarrow \left(x^2+\frac{1}{x^2}\right)=25-2 \\ \Rightarrow \left(x^2+\frac{1}{x^2}\right)= 23 \end{gathered}$$

Q14 | Ex-6E | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions | myhelper

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Question 14:

Tick (✓) the correct answer:
If $\left(x-\frac{1}{x}\right)=6$, then $\left(x^2+\frac{1}{x^2}\right)=?$
(a) 36
(b) 38
(c) 32
(d) $36\frac{1}{36}$

Answer 14:

(b) 38

$$\begin{gathered} \left(x-\frac{1}{x}\right)= 6 \\ \Rightarrow Squaring both the sides: \\ \Rightarrow {\left(x-\frac{1}{x}\right)}^2= {\left(6\right)}^2 \\ \Rightarrow \left(x^2+\frac{1}{x^2}-2\left(x\right)\left(\frac{1}{x}\right)\right)= 36 \\ \Rightarrow \left(x^2+\frac{1}{x^2}\right)-2 =36 \\ \Rightarrow \left(x^2+\frac{1}{x^2}\right)=36+2 \\ \Rightarrow \left(x^2+\frac{1}{x^2}\right)= 38 \end{gathered}$$

Q15 | Ex-6E | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions | myhelper

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Question 15:

Tick (✓) the correct answer:
(82)² − (18)² = ?
(a) 8218
(b) 6418
(c) 6400
(d) 7204

Answer 15:

(c) 6400

$$\begin{gathered} {\left(82\right)}^2-{\left(18\right)}^2 \\ =\left(82 + 18\right)\left(82 - 18\right) \\ =\left(100\right)\left(64\right) \\ =6400 \end{gathered}$$                           [using the identity (a-b)(a+b)=a² -b²]

Q16 | Ex-6E | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions | myhelper

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Question 16:

Tick (✓) the correct answer:
(197 × 203) = ?
(a) 39991
(b) 39999
(c) 40009
(d) 40001

Answer 16:

(a) 39991

$$\begin{gathered} \left(197\right)\times \left(203\right) \\ \Rightarrow \left(200-3\right)\left(200+3\right) \\ \Rightarrow {\left(200\right)}^2-{\left(3\right)}^2 \\ \Rightarrow 40000-9 \\ \Rightarrow 39991 \end{gathered}$$                   [using the identity (a+b) (a-b) = a²  -b²]

Q17 | Ex-6E | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions | myhelper

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Question 17:

Tick (✓) the correct answer:
If (a + b) = 12 and ab = 14, then (a² + b²) = ?
(a) 172
(b) 116
(c) 165
(d) 126

Answer 17:

(b) 116

$$\begin{gathered} \left(a+b\right) =12 \\ \Rightarrow Squaring both the sides: \\ \Rightarrow {\left(a+b\right)}^2 = {\left(12\right)}^2 \\ \Rightarrow \left(a^2+ b^2 +2ab\right) = 144 \\ \Rightarrow \left(a^2+ b^2\right) = 144-2ab \\ \Rightarrow \left(a^2+ b^2\right) = 144 -2\left(14\right) \\ \Rightarrow \left(a^2+ b^2\right) = 144 -28 \\ \Rightarrow \left(a^2+ b^2\right) = 116 \end{gathered}$$

Q18 | Ex-6E | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions | myhelper

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Question 18:

Tick (✓) the correct answer:
If (a − b) = 7 and ab = 9, then (a² + b²) = ?
(a) 67
(b) 31
(c) 40
(d) 58

Answer 18:

(a) 67

Q19 | Ex-6E | Class 8 | RS AGGARWAL | chapter 6 | Operations on Algebraic Expressions | myhelper

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Question 19:

Tick (✓) the correct answer:
If x = 10, then the value of (4x² + 20x + 25) = ?
(a) 256
(b) 425
(c) 625
(d) 575

Answer 19:

(c) 625

$$\begin{gathered} \left(4x^2+20x+25\right) \\ \Rightarrow {\left(2x\right)}^2+2\left(2x\right)\left(5\right)+{\left(5\right)}^2 \\ \Rightarrow {\left(2x + 5\right)}^2 \\ \Rightarrow {\left(2\left(10\right)+5\right)}^2 \\ \Rightarrow {\left(20+5\right)}^2 \\ \Rightarrow {\left(25\right)}^2 \\ \Rightarrow 625 \end{gathered}$$