Integers - Class 7 Maths - Extra Questions

Find the missing number:
$$\_\_\_\_\_\_\_\times (-3)=(-21)$$

Let the missing number be $$x$$

$$x\times (-3)=(-21)$$

$$x=\dfrac{(-21)}{(-3)}$$

$$x=7$$


What will be the missing number in the following equation $$(-8)  \times \_\_\_\_\_\_=(+32) \ \ ?$$

Let the missing number be $$x$$

$$(-8)\times x=(+32)$$

$$x=\dfrac{(+32)}{(-8)}$$

$$x=-4$$


Calculate the following.
$$(-24)\div(-6)$$

$$(-24)\div(-6)$$

$$=\dfrac{(-24)}{(-6)}$$

$$=4$$


Find of the following product.
$$\displaystyle (-18)\times (-10)\times 9$$

$$\displaystyle (-18)\times (-10)\times 9\\ $$
$$=180\times 9=1620$$


Find the following products:
$$9\times (-3)\times (-6)$$

$$(9) \times{(-3)}\times {(-6)}$$
As we know that
$$(-) \times{ (-)} = + $$
Therefore, on simplifying, we get
$$  9\times{3}\times{ 6} $$ 
$$ = ( 27 \times 6 ) $$
$$ =  162 $$


Verify the following:
$$18\times [7+(-3)]=[18\times 7]+[18\times (-3)]$$

To    Verify  the  following 

$$ 18\times [ 7+ (-3)] = [18\times  7]  + [18\times (-3)] $$

Taking   LHS   we   have  

$$ 18\times [ 7 -3 ] $$

$$ =  18\times 4 $$

$$ =  72 $$

Now ,  Taking   RHS   we   have  

$$ [18\times  7]  + [18\times (-3)] $$

$$ =  126  -  54 $$

$$ =  72  $$

Hence ,   LHS  =   RHS  


Verify the following:
$$(-21)\times [(-4)+(-6)]=[(-21)\times (-4)+[(-21)\times (-6)]$$

To    Verify  the  following 

$$ (-21) \times [ (-4)+ (-6)] = [(-21)\times  (-4)]  + [(-21) \times (-6)] $$

Taking   LHS   we   have  

$$ (-21) \times  [ (-4)+ (-6)]  $$

$$ =  (-21) \times [-4  - 6 ]  $$

$$ = (-21)\times (-10) $$

$$ =  210  $$

Now ,  Taking   RHS   we   have  

$$ [(-21)\times  (-4)]  + [(-21) \times (-6)] $$

$$ =  84   +126  $$

$$ =  210   $$

Hence ,   LHS  =   RHS


What is distributive property?

The distributive property tells us how to solve expressions in the form of $$a(b + c)=ab+ac.$$  The distributive property is sometimes called the distributive law of multiplication and division.

Normally when we see an expression like this $$3(2+4)=3\times{2}+3\times{4}$$ (According to Distributive Law)


What will be the value of $$\left (\dfrac {18} {-6}\right ) \   ?$$

Given,
 
 $$\dfrac {18} {-6}$$
 
$$=\dfrac { +18 }{ -6 }$$
 
$$=-3$$


Calculate the following.
$$(-18)\div(-6)$$

$$(-18)\div(-6)$$

$$=\dfrac{(-18)}{(-6)}$$

$$=3$$



Find the additive identity of the integers.

$${\textbf{Step-1: Assume unknown as x and use definition of additive identity property}}$$ 

                 $${\text{The additive identity property says that if you add an integer to zero or add zero to an}}$$                                                    $${\text{integer, then you get the same integer back. The number zero is known as the identity}}$$                                                 $${\text{element, or the additive identity. For example, for any integer a,}}$$

                 $${\text{a + 0 = 0 + a = a}}$$

$${\textbf{Hence, the additive identity for integers is 0.}}$$


Calculate the following.
$$|(-25)\div(+5)|$$

$$(-25)\div(+5)$$
$$=\dfrac{(-25)}{(+5)}$$
$$=-5$$


Find the missing number:
$$(-7)  \times \_\_\_\_\_=(-28)$$

Let the missing number be $$x$$

$$(-7)\times x=(-28)$$

$$x=\dfrac{(-28)}{(-7)}$$

$$x=4$$


Find the missing number:
$$\_\_\_\_\_\times (-5)=(-20)$$

Let the missing number be $$x$$

$$x\times (-5)=(-20)$$

$$x=\dfrac{(-20)}{(-5)}$$

$$x=4$$


Find the missing number:
$$(+6)  \times\_\_\_\_\_\_\_=(-42)$$

Let the missing number be $$x$$

$$(+6)\times x=(-42)$$

$$x=\dfrac{(-42)}{(+6)}$$

$$x=-7$$


The point which divides internally , the line joining $$(-1, 2)$$ to $$(4, -5)$$ in the ratio $$2:3$$ is


1330929_1075369_ans_123a3cbd95e14e43b3627e86bb729dca.jpg


Find the product of $$(-1)\times(-2)\times(3)\times4$$.

The required problem is $$\left( { - 1} \right) \times \left( { - 2} \right) \times \left( {  3} \right) \times 4$$
$$ = 2 \times \left( {  12} \right)$$
Hence the result is $$ =  24$$


Using distributive property, solve :
$$122\times 201-112$$

Distributive property is $$a\times (b + c) = a\times b + a\times c$$

$$\therefore 122\times 201 - 112  $$

$$= 122\times ( 200 + 1 ) - 112$$

$$= 122\times 200 + 122\times 1 - 112 $$

$$ = 24400 + 122 - 112$$

$$ = 24400 + 6$$

$$122\times 201 - 112  = 24406$$ 


Using distributive property, solve :
$$23 \times (-10)+23\times 10$$

Distributive property is $$a\times (b+c) = a\times b + a\times c$$

$$\therefore 23\times (-10) + 23\times 10 $$

$$= 23\times ( - 10 + 10)$$

$$= 23\times 0 $$

$$23\times (-10) + 23\times 10 = 0$$ 


Using distributive property, solve :
$$136\times 15+(-10) \times 136+136\times 10$$

Distributive property is $$a\times (b+c) = a\times b + a\times c$$

$$\therefore 136\times 15 + (-10)\times 136 + 136\times 10 $$

$$= 136\times (15 - 10 + 10 )$$

$$= 136\times 15 $$

$$ 136\times 15 + (-10)\times 136 + 136\times 10 = 2040$$ 


Using distributive property, solve :
$$20 \times [(30)+10]$$

Distributive property is $$a\times (b+c) = a\times b + a\times c$$

$$\therefore 20\times [(30) + 10] $$

$$= 20\times (30) + 20\times 10$$

$$=  600 + 200 $$

$$ = 800$$ (Ans)


Multiply the following integers:
$$6$$ by $$80$$

 $$80$$
$$\times 6$$
_________
$$480$$

or
$$80\times 6=10\times 8\times 6$$
             $$=10\times 48=480$$

1188306_1344268_ans_79c9cd19428543a1858083f901895bb7.jpg


( 25) ( 2) = _____.

( 25) ( 2) = _50_


_____ is the multiplicative identity for integers.

1 is the multiplicative identity for integers.
$$\therefore a \times 1=1 \times a=a$$


3 ( 1 ) ( 15) = _____.

3 ( 1 ) ( 15) = _45_


$$[(-8)+$$________$$]+$$_________ $$=$$ _________$$+[(-3)+$$ _________$$]=-3$$

$$[(-8)+$$________$$]+$$_________ $$=$$ _________$$+[(-3)+$$ _________$$]=-3$$
-3, 8, -8, 8 are to be filled.


$$(– 5) × ( – 6) × ( – 7) =$$ _____.

( 5) ( 6) ( 7) = _$$-210$$_


Observe the following patterns and fill in the blanks to make the statements true:
$$ 5 \times 4 = 20$$
5 3 = 15 = 20 ( 5)
5 2 = _______ = 15 ( 5)
5 1 = _______ = _______
5 0 = 0 = _______
5 1 = 5 = _______
5 2 = _______ = _______

5 4 = 20
5 3 = 15 = 20 ( 5)
5 2 = _-10_ = 15 ( 5)
5 1 = _-5_ = _10 - (-5)_
5 0 = 0 = _-5 - (-5)_
5 1 = 5 = _0 - (-5)_
5 2 = _10_ = _5 - (-5)_


From given integers in Column I match an integer of Column II so that their product lies between 19 and 6:

$$\begin{array}{l}-5 \rightarrow 3=-15 \\6 \rightarrow-2=-12\end{array}$$
$$\begin{array}{l}-7 \rightarrow 1=-7 \\8 \rightarrow-1=-8\end{array}$$
The value of the product lies between -19 and -6.


$$286 + 0 = ..$$

$$286 + 0 = 286$$


Find the value of $$8\times 12$$ by using the distributive property.

We need to use the distributive property to find the value of $$8\times 12$$

Distributive property: $$a(b+c)=ab+ac$$

Consider $$8\times 12$$

$$8\times 12=8(10+2)$$

             $$=8(10)+8(2)$$

             $$=80+16$$

             $$=96$$

$$\therefore 8\times 12=96$$


Find three consecutive integers such that their sum is $$78$$

Sum of three consecutive integers$$=78$$
Let first number$$=x$$
The second number$$=x+1$$
and third number$$=x+2$$
Then $$\,x+x+1+x+2=78$$
$$\Rightarrow\,3x+3=78$$
$$\Rightarrow\,3x=78-3=75$$
$$\Rightarrow\,x=\dfrac{75}{3}=25$$
$$\therefore\,$$first number$$=x=25$$
The second number$$=x+1=25+1=26$$
and third number$$=x+2=25+2=27$$
Then the three required numbers are $$25,\,16,\,27$$


Determine each of the following products:
$$(-25) \times (-9)$$ 


1590634_1675664_ans_7ca3486d03da4200be19f03b5ad7ceda.jpg


Determine the following product:
$$125 \times (-8)$$ 


1498812_1675666_ans_098f9b7c55354dba99adb71615c2b8bc.jpg


Multiply:
-35 by 0

Using the rule for Multiplication of integers.
The above question can be written as,
$$=-35 \times 0 $$ 
$$=0$$
($$\because$$ Any integer it may positive or negative multiplied by or to zero the result will be zero itself)


Multiply:
0 by -23

Using the rule for Multiplication of integers.
The above question can be written as,
=0 $$\times$$ -23 
=0
($$\because$$ Any integer it may positive or negative multiplied by or to zero the result will be zero itself)


Find each of the following products:
$$(-5)\times (-8)\times (-3)$$

Using the rule for Multiplication of integers.
In this question we have to multiply the first two integers and then from the answer that we get is multiplied with other integer,
={(-5$$\times$$ -8) $$\times$$ (-3)}          ....[$$\because$$(-$$\times$$ - = +)]
={40 $$\times$$ (-3)}                              ...[$$\because$$(+$$\times$$ - = -)]                  
= -120
(In multiplication we have to multiply the sign of the integers also and then assign the obtained sign to the answer)


Multiply:
-28 by -1

Using the rule for Multiplication of integers.
The above question can be written as,
= -28 ×× -1  (- x - = +)
=28
(In multiplication we have to multiply the sign of the integers also and then assign the obtained sign to the answer)


Multiply:
-105 by -8

Using the rule for Multiplication of integers.
The above question can be written as,
=-105 $$\times$$ -8                ....[$$\because$$(-$$\times$$- = +)]
=840
(In multiplication we have to multiply the sign of the integers also and then assign the obtained sign to the answer)


Find each of the following products:
$$3\times 4\times (-5)$$

Using the rule for Multiplication of integers.
In this question first we have to multiply the integers with same sign and then from the answer that we get is multiplied with other integer of different sign,
$$=(3\times 4) \times (-5) $$
$$=(12) \times (-5) $$                   $$....[\because(+\times- = -)]$$
$$= -60$$
(In multiplication we have to multiply the sign of the integers also and then assign the obtained sign to the answer)


Multiply:
-36 by -50

Using the rule for Multiplication of integers.
The above question can be written as,
=-36 $$\times$$ -50                ....[$$\because$$(-$$\times$$- = +)]
=1800
(In multiplication we have to multiply the sign of the integers also and then assign the obtained sign to the answer)


Multiply:
25 by -11

Using the rule for Multiplication of integers.
The above question can be written as,
=25 $$\times$$ -11                ....[$$\because$$(+$$\times$$- = -)]
$$=-275$$
(In multiplication we have to multiply the sign of the integers also and then assign the obtained sign to the answer)


Find the product:
$$2\times (-5)\times (-6)$$

Using the rule for Multiplication of integers.
In this question first we have to multiply the integers with same sign and then from the answer that we get is multiplied with other integer of different sign,
={2$$\times$$(-5$$\times$$ -6)}          ....[$$\because$$(-$$\times$$ - = +)]
={2 $$\times$$ 30}                    
= 60
(In multiplication we have to multiply the sign of the integers also and then assign the obtained sign to the answer)


Multiply:
-16 by -12

Using the rule for multiplication of integers.
The above question can be written as,
=-16 ××- 12= +(16 x 12)]
=192
(In multiplication we have to multiply the sign of the integers also and then assign the obtained sign to the answer)


Find each of the following products:
$$(-6)\times (-5)\times (-7)\times (-2)\times (-3)$$

Using the rule for Multiplication of integers.
In this question first we have four integers with same sign, so multiply two integers once and then from the answers of both set of integers again multiplied and get the final answer. 
= {$$(-6 \times -5) \times (-7 \times 2) \times (-3)$$}           ....[$$\because (- \times - = +)$$]
= {$$(30) \times (14) \times (-3)$$}
= {$$(30 \times 14) \times (-3)$$}                  
= {$$(420 \times -3)$$}                           ....[$$\because (+ \times - = - )$$]
= 1260


Find each of the following products:
$$(-4)\times (-5)\times (-8)\times (-10)$$

Using the rule for Multiplication of integers.
In this question first we have four integers with same sign, so multiply two integers once and then from the answers of both set of integers again multiplied and get the final answer. 
$$={(-4 \times -5) \times (-8 \times-10)} $$       ....[$$\because$$(-$$\times$$ - = +)]   
$$={(20) \times (80)}         $$      
$$= 1600$$


Find each of the following products:
$$(-30)\times (-20)\times (-5)$$

Using the rule for Multiplication of integers.
In this question first we have 3 integers with same sign, so multiply two integers once and then from the answers of both set of integers again multiplied and get the final answer. 
={(-30 $$\times$$ -20) $$\times$$ (-5)}        ....[$$\because$$  (-$$\times$$ - = +)]   
={(600) $$\times$$ (-5)}               
= -3000                          ...[$$\because$$  (+$$\times$$ - = -)] 


Simplify:
$$(-8) \times 9 + (-8) \times 7$$

Using the Distributive Law of Multiplication $$[(a \times b) + (a \times c) = a (b + c)]$$
Let,
a = -8, b = 9, c = 7
Then, 
=$$(-8) \times 9 + (-8) \times 7$$
The above equation can be written as,
=$$(-8) \times (9 +7)$$            .....$$[\because (a \times b) + (a \times c) = a \times (b + c)]$$
=$$(-8) \times (16)$$                ..... $$[\because (- \times + = -]$$
$$= -128$$


Find each of the following products:
$$(-6)\times 6\times (-10)$$

Using the rule for Multiplication of integers.
In this question first we have to multiply the integers with same sign and then from the answer that we get is multiplied with other integer of different sign,
={(-6$$\times$$ -10) $$\times$$ 6}          ....[$$\because$$(-$$\times$$ - = +)]
={60 $$\times$$ 6}                    
= 360
(In multiplication we have to multiply the sign of the integers also and then assign the obtained sign to the answer)



Simplify:
$$20 \times (-16) + 20 \times 14$$

Using the Distributive Law of Multiplication $$[(a \times b) + (a \times c) = a (b + c)]$$
Let,
$$a = 20, b = -16, c = 14$$
Then, 
=$$20 \times (-16)+ 20 \times 14$$
The above equation can be written as,
=$$20 \times \{(-16) + 14\}$$            .....$$[\because (a \times b) + (a \times c) = a \times (b + c)]$$
=$$20 \times \{-2\}$$                ..... $$[\because (- \times + = -]$$
$$= -40$$


Find each of the following products:
$$(-7)\times (-3)\times 4$$

Using the rule for Multiplication of integers.
In this question first we have to multiply the integers with same sign and then from the answer that we get is multiplied with other integer of different sign,
={(-7 $$\times$$ -3) $$\times$$ 4}        ....[$$\because$$(-$$\times$$ - = +)]       
={21 $$\times$$ 4}               
= 84
(In multiplication we have to multiply the sign of the integers also and then assign the obtained sign to the answer)


Find each of the following products:
$$7\times (-8)\times 3$$

Using the rule for Multiplication of integers.
In this question first we have to multiply the integers with same sign and then from the answer that we get is multiplied with other integer of different sign,
={(7 $$\times$$ 3) $$\times$$ (-8)}          
={21 $$\times$$ -8}        ....[$$\because$$(+$$\times$$ - = -)]            
= -168
(In multiplication we have to multiply the sign of the integers also and then assign the obtained sign to the answer)


Find each of the following products:
$$(-60)\times (-10)\times (-5)\times (-1)$$

Using the rule for Multiplication of integers.
In this question first we have four integers with same sign, so multiply two integers once and then from the answers of both set of integers again multiplied and get the final answer. 
={(-60 $$\times$$ -10) $$\times$$ (-5 $$\times$$-1)}        ....[$$\because$$(-$$\times$$ - = +)]  ={(600) $$\times$$ (5)}               
= 3000


Simplify:
$$9 \times (-13)+ 9 \times (-7)$$

Using the Distributive Law of Multiplication $$(a \times b) + (a \times c) = a (b + c).$$

Let, $$a = 9,\ b = -13,\ c = -7$$

Then, by using the above mentioned property,
$$9 \times (-13) + 9 \times (-7)=9 \times \{(-13) +(-7)\}$$
                                         $$=9 \times (-13-7)$$
                                         $$=9 \times (-20)$$
                                         $$= -180$$


Simplify:
$$(-16) \times (-8) + (-4) \times (-8)$$

Using the Distributive Law of Multiplication $$[(a \times b) + (a \times c) = a (b + c)]$$
Let,
a = -16, b = -4, c = -8
Then, 
=$$(-16) \times (-8) + (-4) \times (-8)$$
The above equation can be written as,
=$$\{((-16) +(-4)) \times (-8)\}$$ .....$$[\because (a \times b) + (a \times c) = a \times (b + c)]$$
=$$\{(-16-4) \times (-8)\}$$ ..... $$[\because (+ \times - = -)]$$
= $$\{(-20 \times (-8)\}$$ ......$$[\because (- \times - = +)]$$
=160


Simplify:
$$(-11) \times (-15) + (-11) \times (-25)$$

Using the Distributive Law of Multiplication $$[(a \times b) + (a \times c) = a (b + c)]$$
Let,
a = -11 b = -15, c = -25
Then, 
=$$(-11) \times (-15) + (-11) \times (-25)$$
The above equation can be written as,
=$$(-11) \times \{(-15) +(-25)\}$$ .....$$[\because (a \times b) + (a \times c) = a \times (b + c)]$$
=$$(-11) \times \{(-15-25)\}$$ ..... $$[\because (+ \times - = +)]$$
= $$(-11) \times \{-40\}$$ ......$$[\because (- \times - = +)]$$
=440


Simplify:
$$(-26) \times 72 + (-26) \times 28$$

Using the Distributive Law of Multiplication $$[(a \times b) + (a \times c) = a (b + c)]$$
Let,
a = -26, b = 72, c = 28
Then, 
=$$(-26) \times (72) + (-26) \times (28)$$
The above equation can be written as,
=$$\{(-26) \times (72 + 28)\}$$ .....$$[\because (a \times b) + (a \times c) = a \times (b + c)]$$
=$$\{(-26) \times (100)\}$$ ..... $$[\because (+ \times - = +)]$$
=-2600


Simplify:
$$10 \times (-12)+ 5 \times (-12)$$

Using the Distributive Law of Multiplication $$[(a \times b) + (a \times c) = a (b + c)]$$
Let,
a = 10, b = 5, c = -12
Then, 
=$$10 \times (-12) + 5 \times (-12)$$
The above equation can be written as,
=$$\{(10+5) \times (-12)\}$$ .....$$[\because (a \times b) + (a \times c) = a \times (b + c)]$$
=$$\{(15) \times (-12)\}$$ ..... $$[\because (+ \times - = -]$$
=-180


 Fill in the blanks to make the statements true
$$(-80) + 0 +(-90) =$$ _______.

$$\textbf{Addition of three numbers}$$
$$ (-80) + 0 + (–90) = -80 + 0 - 90$$                          
                                   $$= - 80 - 90$$
                                   $$= -170$$

$$\therefore$$  $$\bf{(-80) + 0 + (-90) =}\ \underline{\bf{-170}}$$


If x, y and z are integers then $$(x+ $$ _____ $$)+z=$$ _______ $$+(y+$$ ______ $$)$$

$$(X+Y)+Z=X+(Y+Z)$$

By the associate property of integers.Y, X, Z will filled the blanks.


11 ( 5) = ( _____ _____ ) = _____

11 ( 5) = ( 11 x 5 ) =  _- 55_


( 23) (42) = ( 42) _____

( 23) (42) = ( 42) _23_


( 9) 20 = _____

( 9) 20 = _$$-180$$_


[12 ( 7)] 5 = _____ [( 7) _____ ]

[12 ( 7)] 5 = _12_   [( 7) _5_ ]


Write a pair of integers whose product is 12 and there lies seven integers between them (excluding the given integers).

Let -2 and 6 be the required integer.
Therefore, A pair of the integer (-2,6)
There are total seven integers i.e. -1,0,1,2,3,4,5 that lie between -2 and 6


(a)Write a positive integer and a negative integer whose sum is a negative integer.
(b) Write a positive integer and a negative integer whose sum is a positive integer.
(c) Write a positive integer and a negative integer whose difference is a negative integer.
(d) Write a positive integer and a negative integer whose difference is a positive integer.
(e) Write two integers which are smaller than – 5 but their difference is – 5.
(f) Write two integers which are greater than – 10 but their sum is smaller than – 10.
(g) Write two integers which are greater than – 4 but their difference is smaller than – 4.
(h) Write two integers which are smaller than – 6 but their difference is greater than – 6.
(i) Write two negative integers whose difference is 7.
(j) Write two integers such that one is smaller than –11 and other is greater than –11 but their difference is–
(k) Write two integers whose product is smaller than both the integers.
(l) Write two integers whose product is greater than both the integers.

(a) $$2,-3$$
$$\because 2+(-3)=-1$$

(b) $$3,-2$$
$$\because 3+(-2)=1$$

(c) $$4,-1$$
$$\because-1-(4)=-5$$

(d) $$4,-1$$
$$\because 4-(-1)=5$$
Because, with two numbers $$a,b$$, difference can be $$a-b$$ or $$b-a$$
 
(e) $$-7,-12$$
$$\because-7<-5,-12<-5$$ 
And $$-12-(-7)=-5$$

(f) $$-5,-6$$
$$\because -5>-10,-6>-10$$
And $$-5+(-6)=-11<-10$$

(g) $$2,-3$$
$$\because 2>-4,-3>-4$$
And$$-3-2=-5<-4$$

(h) $$-7,-8$$
$$\because-7<-6,-8<-6$$ 
And $$-7-(-8)=-7+8=1>-6$$

(i) $$-2,-9$$
$$\because -2-(-9)=-2+9=7$$

(j)$$-18,-7$$
$$\because-18<-11 ;-7>-11 $$
And$$-18-(-7)=-18+7=-11$$

(k) $$-1,2$$
$$\because(-1) \times(2)=-2,$$ 
Also $$-2<-1$$ and $$-2<2$$

(l)$$2,3$$
$$\because 2 \times 3=6$$ 
Also $$2<6$$ and $$3<6$$


Below $$u, v, w$$ and $$x$$ represent different integers, where $$u=-4$$ and $$x \neq 1 .$$ By using following equations, find each of the values:
$$u \times v=u$$
$$x \times w=w$$
$$u+x=w$$
(a) $$v$$
(b) $$w$$
(c) $$x$$
Explain your reason using the properties of integers.

(a) Given, $$u \times v=u$$ and $$u=-4$$
$$\therefore-4 \times v=-4$$
$$\Rightarrow v=1$$ is the multiplicative identity

(b) Given $$x \times w=w \quad \because x \neq 1$$
Therefore, $$x \times w=w$$ only possible when $$w=0$$
that is $$x\times0=0,\ x\neq1$$
So, $$w=0$$

(c) Given, $$u+x=w$$
putting $$\mathfrak{u}=-4$$ and $${w}=0$$
$$\therefore-4+x=0$$
$$\Rightarrow x=4(\text { transposing }-4 \text { to Right hand side })$$

So, $$v=1,x=4,w=0$$


Fill in the blanks to make the statement true:
For rational numbers $$\dfrac{a}{b}, \dfrac{c}{d}$$ and $$\dfrac{e}{f}$$ we have $$\dfrac{a}{b} \times \left ( \dfrac{c}{d} + \dfrac{e}{f} \right ) = ....... + .........$$

By using distributive property over addition
$$\dfrac{a}{b} \times \left ( \dfrac{c}{d} + \dfrac{e}{f} \right ) \\=\dfrac{a}{b} \times \dfrac{c}{d} + \dfrac{a}{b} \times \dfrac{e}{f}\\=\dfrac{ac}{bd} + \dfrac{ae}{bf}$$


Fill in the blanks to make the statement true:
$$\dfrac{1}{5} \times \left [ \dfrac{2}{7} + \dfrac{3}{8} \right ] = \left [ \dfrac{1}{5} \times \dfrac{2}{7} \right ] + ........$$

Applying didtributive law over addition
$$a\times(b+c)=a\times b+b\times c$$

We get $$\dfrac{1}{5} \times \left [ \dfrac{2}{7} + \dfrac{3}{8} \right ] = \left [ \dfrac{1}{5} \times \dfrac{2}{7} \right ] +\left [  \dfrac{1}{5} \times \dfrac{3}{8} \right ]$$


Using distributive law, find the squares of 
a) 101
b) 72

(a) 101
$$(101)^2 = 101 \times 101 = 101(100+1)$$
$$ = 10100+ 101 = 10201$$
b) 72
$$(72)^2 = 72 \times 72 = 72 (70+2)$$
$$ = 5040 + 144 = 15184$$


Supply the missing information for the diagram.
1793399_228132c255f24528b4b9938dbf3ed3ca.png

A machine $$(\times x^y)$$ with input $$a$$ will give an output $$ax^y$$
So, in diagram both $$x$$ and $$y$$ are missing
Here, output is  given to be $$36$$ cm 
And, input, so let's assume it be $$x$$
Power of both machines are $$1$$, as they are not written.

When a piece of length $$x\ cm$$ is inserted in $$( \times 4)$$ and $$( \times 3)$$ hooked machine, it will produce a piece that is $$36\ cm$$ in length.
Hence, $$x\times 4\times 3=36$$
 $$\Rightarrow x\times 12=36$$
$$\Rightarrow x=\cfrac{36}{12}=3\ cm$$.
So, missing information is left length is $$3\ cm$$


Stretching Machine Suppose you have a stretching machine which could stretch almost anything. For example, if you put a $$5$$ metre stick into a $$(\times 4)$$ stretching machine ( as shown below), you get a $$20$$ meter stick. Now it you put $$10\ cm$$ carrot into a $$(\times 4)$$ machine, how long will it be when it comes out ?
1793317_d2c9c09f139d44d8933ed95fb015a5a7.png

From the equation,
If Input $$=5\ m$$, then output $$=20\ m$$
i.e $$5\times4=20$$

Similarly, 
If input $$=10\ cm$$
Then, output $$=10\times 4=40\ cm$$


Simplify the following by using suitable property. Also name the property.
$$\left[ \dfrac{1}{5} \times \dfrac{2}{15} \right] - \left[\dfrac{1}{5} \times \dfrac{2}{5} \right]$$


1927329_1792110_ans_16fe0a1d56234e47a138b7dc1b619c0a.jpg


Class 7 Maths Extra Questions