Operations with Integers - NCERT Solutions Class 7 Maths (Ganita Prakash)

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Operations with Integers – NCERT Solutions Class 7 Maths (Ganita Prakash) includes all the questions with solutions given in the NCERT Class 7 Maths (Ganita Prakash).

NCERT Solutions Class 7

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Operations with Integers – NCERT Solutions

Q.1: Using the token interpretation, find the values of: {tex}3 \times(-2){/tex}

Solution:

Three groups of {tex}-2:-2+(-2)+(-2)=-6{/tex}

Q.2: Using the token interpretation, find the values of : {tex}(-5) \times(-2){/tex}

Solution:

Multiply two negative numbers = positive result: {tex}5 \times-2=+10{/tex}

Q.3: Using the token interpretation, find the values of: {tex}(-4) \times(-1){/tex}

Solution:

Multiply two negative numbers = positive result: {tex}-4 \times-1=+4{/tex}

Q.4: Using the token interpretation, find the values of: {tex}(-7) \times 3{/tex}

Solution:

Three groups of {tex}-7:-7+(-7)+(-7)=-21{/tex}

Q.5: If {tex}123 \times 456=56088{/tex}, without calculating, find the value of : {tex}(-123) \times 456{/tex}

Solution:

One number is negative, so the answer is also negative.
{tex} (-123) \times 456=-56088 {/tex}

Q.6: If {tex}123 \times 456=56088{/tex}, without calculating, find the value of: {tex}(-123) \times(-456){/tex}

Solution:

Both numbers are negative, so the answer is positive.
{tex} (-123) \times(-456)=56088 {/tex}

Q.7: If {tex}123 \times 456=56088{/tex}, without calculating, find the value of : {tex}123 \times(-456){/tex}

Solution:

One number is negative, so the answer is also negative.
{tex} 123 \times(-456)=-56088 {/tex}

Q.8: Try to frame a simple rule to multiply two integers.
Consider the numbers represented by the following tokens:

We can see that all of them represent the number (- 2). Now, take 4 times each of these token sets. That is, place each set into the empty bag 4 times.

Solution:

Two negatives (- and -): multiply to give a positive result.
Negative and positive (- and +): multiply to give a negative result.
Negative and positive (- and +) or vice versa, again gives a negative result.

Q.9: Find: {tex}4 \times(-3){/tex}

Solution:

Different signs, product is negative.
{tex} 4 \times(-3)=-12 {/tex}

Q.10: Find: {tex}(-6) \times(-3){/tex}

Solution:

Same sign, product is positive.
{tex} (-6) \times(-3)=18 {/tex}

Q.11: Find: {tex}(-5) \times(-1){/tex}

Solution:

Same sign, product is positive.
{tex} (-5) \times(-1)=5 {/tex}

Q.12: Find: {tex}(-8) \times 4{/tex}

Solution:

Different signs, product is negative.
{tex} (-8) \times 4=-32 {/tex}

Q.13: Find: {tex}(-9) \times 10{/tex}

Solution:

Different signs, product is negative.
{tex} (-9) \times 10=-90 {/tex}

Q.14: Find: {tex}10 \times(-17){/tex}

Solution:

Different signs, product is negative.
{tex} 10 \times(-17)=-170 {/tex}

Q.15: Find the value of: {tex}14 \times(-15){/tex}

Solution:

Different signs, result is negative.
{tex} 14 \times(-15)=210 {/tex}

Q.16: Find the value of: {tex}-16 \times(-5){/tex}

Solution:

Same signs (both negative), result is positive.
{tex} 16 \times(-5)=80 {/tex}

Q.17: Find the value of: {tex}(-46) \div(-23){/tex}

Solution:

Same signs (both negative), result is positive.
{tex} 46 \div(-23)=2 {/tex}

Q.18: A freezing process requires that the room temperature be lowered from {tex}32^{\circ} \mathrm{C}{/tex} at the rate of {tex}5^{\circ} \mathrm{C}{/tex} every hour. What will be the room temperature 10 hours after the process begins?

Solution:

The room temperature drops {tex}5^{\circ} \mathrm{C}{/tex} every hour for 10 hours.
Total drop:
{tex} 5 \times 10=50^{\circ} {C} {/tex}
Starting temperature:
{tex} 32^{\circ} {C} {/tex}
Room temperature after 10 hours:
{tex} 32^{\circ} C-50^{\circ} C=-18^{\circ} C {/tex}
Final Answer:
The room temperature will be {tex}-18^{\circ} \mathrm{C}{/tex} after 10 hours.

Q.19: A cement company earns a profit of ₹ 8 per bag of white cement sold and a loss of ₹ 5 per bag of grey cement sold. [Represent the profit/ loss as integers.]

The company sells 3,000 bags of white cement and 5,000 bags of grey cement in a month. What is its profit or loss?
If the number of bags of grey cement sold is 6,400 bags, what is the number of bags of white cement the company must sell to have neither profit nor loss.

Solution:

Representing profit/loss as integers:
Profit per white cement bag: +8 (₹ 8 profit)
Loss per grey cement bag: -5 (₹ 5 loss)

Profit/Loss Calculation
White cement: {tex}3,000 \times 8=24,000{/tex}
Grey cement: {tex}5,000 \times(-5)=-25,000{/tex}
Total profit/loss: {tex}24,000+(-25,000)=-1,000{/tex}
Answer: The company has a loss of ₹ 1,000.
No profit/no loss situation
Let {tex}x{/tex} be the number of white cement bags required.
Profit: {tex}8 x{/tex}
Loss: {tex}6,400 \times(-5)=-32,000{/tex}
For no profit or loss:
{tex} 8 x-32,000=0 {/tex}
{tex} 8 x=32,000 {/tex}
{tex} x=4,000 {/tex}
Answer: The company must sell 4,000 bags of white cement to have no profit or loss.

Q.20: Replace the blank with an integer to make a true statement.

{tex}(-3) \times \quad=27{/tex}
{tex}5 \times {/tex} ________ = (-35)

Solution:

{tex}(-3) \times_{-}=27{/tex}
To find the blank, divide 27 by -3:
{tex} -=\frac{27}{-3}=-9 {/tex}
{tex}5 \times_{-}=(-35){/tex}
To find the blank, divide -35 by 5:
{tex} -=\frac{-35}{5}=-7 {/tex}

Q.21: Replace the blank with an integer to make a true statement.

________ {tex}\times(-8)=(-56){/tex}
________ {tex}\times(-12)=132{/tex}

Solution:

Divide both sides by -8:
{tex} -=\frac{-56}{-8}=7 {/tex}
Divide both sides by -12:
{tex} -=\frac{132}{-12}=-11 {/tex}

Q.22: Replace the blank with an integer to make a true statement.

________ {tex}\div(-8)=7{/tex}
________ {tex}\div 12=-11{/tex}

Solution:

________ {tex}\times (-8)=7{/tex}
Divide both sides by -8:
{tex}-=\frac{7}{-8}=-\frac{7}{8}{/tex}
But the instruction says “integer”, so there is no integer that makes this statement true. Normally, for multiplication with integers, there wouldn’t be a solution unless 7 is a multiple of 8.
{tex}-\div-12=-11{/tex}
Multiply both sides by -12:
{tex}-=-11 \times-12=132{/tex}

Q.23: Find the value of the expression: {tex}(-5) \times(18+(-3)){/tex}

Solution:

{tex} 18+(-3)=15 {/tex}
Then multiply:
{tex} (-5) \times 15=-75 {/tex}

Q.24: Find the value of the expression: {tex}(-7) \times 4 \times(-1){/tex}

Solution:

Multiply in steps:
{tex} (-7) \times 4=-28 {/tex}
{tex} -28 \times(-1)=28 {/tex}

Q.25: Find the value of the expression: {tex}(-2) \times(-1) \times(-5) \times(-3){/tex}

Solution:

Multiply in steps:
{tex} (-2) \times(-1)=2 {/tex}
{tex} 2 \times(-5)=-10 {/tex}
{tex} -10 \times(-3)=30 {/tex}

Q.26: Find the value of the expression: {tex}84 \div(-4){/tex}

Solution:

{tex}84 \div(-4)=-21{/tex}

Q.27: Find the value of the expression: {tex}(-27) \div 9{/tex}

Solution:

{tex}-27 \div 9=-3{/tex}

Q.28: Find the value of the following expression: {tex}(-56) \div(-2){/tex}

Solution:

Divide -56 by -2:
{tex} -56 \div-2=28 {/tex}

Q.29: Find the integer whose product with (-1) is: 27

Solution:

{tex} x \times(-1)=27 {/tex}
{tex} x=-27 {/tex}

Q.30: Find the integer whose product with (-1) is: -31

Solution:

{tex} x \times(-1)=-31 {/tex}
{tex} x=31 {/tex}

Q.31: Find the integer whose product with (-1) is: -1

Solution:

{tex} x \times(-1)=-1 {/tex}
{tex} x=1 {/tex}

Q.32: Find the integer whose product with (-1) is: 0

Solution:

{tex} x \times(-1)=0 {/tex}
{tex} x=0 {/tex}

Q.33: Find the integer whose product with (-1) is: 1

Solution:

{tex} x \times(1)=0 {/tex}
{tex} x=0 {/tex}

Q.34: If {tex}47-56+14-8+2-8+5=-4{/tex}, then find the value of {tex}-47+56-14 +8-2+8-5{/tex} without calculating the full expression.

Solution:

Observe that the second expression
{tex} -47+56-14+8-2+8-5 {/tex}
is the negative of the first expression:
{tex} -(47-56+14-8+2-8+5) {/tex}
Therefore,
{tex} \text { Value of second expression }=-(-4)=+4 {/tex}

Q.35: Do you remember the Collatz Conjecture from last year? Try a modified version with integers. The rule is – start with any number; if the number is even, take half of it; if the number is odd, multiply it by -3 and add 1; repeat. An example sequence is shown below.

Try this with different starting numbers: (– 21), (– 6), and so on. Describe the patterns you observe.

Solution:

Starting with -21
{tex} -21 \rightarrow 64 \rightarrow 32 \rightarrow 16 \rightarrow 8 \rightarrow 4 \rightarrow 2 \rightarrow 1 {/tex}
Starting with {tex} -6 \rightarrow-3 \rightarrow 10 \rightarrow 5 \rightarrow-14 \rightarrow-7 \rightarrow 22 {/tex}{tex}\rightarrow 11 \rightarrow 32 \rightarrow-16 \rightarrow-8 \rightarrow-4 \rightarrow-2 \rightarrow1{/tex}
Every starting number eventually reaches 1.
Negative starting values may first become positive or oscillate, but the sequence stabilizes.
The pattern always ends in a cycle 4 {tex}\rightarrow{/tex} 2 {tex}\rightarrow{/tex} 1, just like the standard Collatz sequence.

Q.36: In a test, {tex}(+4){/tex} marks are given for every correct answer and {tex}(-2){/tex} marks are given for every incorrect answer.

Anita answered all the questions in the test. She scored 40 marks even though 15 of her answers were correct. How many of her answers were incorrect? How many questions are in the test?
Anil scored (-10) marks even though he had 5 correct answers. How many of his answers were incorrect? Did he leave any questions unanswered?

Solution:

Given:
– Marks for correct answer = +3
– Marks for incorrect answer {tex}=-2{/tex}
Anita:
{tex} 40=(15 \times 3)+(x \times-2) {/tex}
{tex} 40=45-2 x \Rightarrow 2 x=5 \Rightarrow x=2.5 {/tex}
Since answers must be whole numbers, marks per question can vary in problems, but assuming integer values:
If {tex}x=2{/tex} wrong answers (rounded),
Total questions {tex}=15+2=17{/tex}
Anil’s Situation
{tex} -10=(5 \times 3)+(y \times-2) {/tex}
{tex} -10=15-2 y \Rightarrow 2 y=25 \Rightarrow y=12.5 {/tex}
Anil made 12 or 13 wrong answers.
If test questions are 17 (as Anita’s total),
Anil attempted {tex}5+12=17{/tex} or {tex}5+13=18{/tex}.
So, Anil probably did not leave any question unanswered.

Q.37: Imagine you’re in a place where the temperature drops by {tex}5^{\circ} \mathrm{C}{/tex} each hour. If the temperature is currently at {tex}8^{\circ} \mathrm{C}{/tex}, write an expression which denotes the temperature after 4 hours.

Solution:

The temperature is {tex}8^{\circ} \mathrm{C}{/tex}.
Every hour, it drops by {tex}5^{\circ} \mathrm{C}{/tex}.
temperature after 4 hours:
Expression: {tex}8-(5 \times 4){/tex}
Answer: {tex}-12^{\circ} \mathrm{C}{/tex}.

Q.38: Find 3 consecutive numbers with a product of -6.

Solution:

The three consecutive numbers with a product of -6 are {tex}-3,-2,-1{/tex}. Because {tex}(-3) \times(-2) \times(-1)=-6{/tex}.

Q.39: Find 3 consecutive numbers with a product of 120.

Solution:

The three consecutive numbers with a product of 120 are {tex}4,5,6{/tex}.
Because {tex}4 \times 5 \times 6=120{/tex}.

Q.40: An alien society uses a peculiar currency called ‘pibs’ with just two denominations of coins- a+13 pibs coin and a-9 pibs coin. You have several of these coins. Is it possible to purchase an item that costs + 85 pibs?
Yes, we can use 10 coins of +13 pibs and 5 coins of -9 pibs to make a total of + 85. Using the two denominations, try to get the following totals:

+20
+40
-50
+8
+10
-2
+1
Is it possible to purchase an item that costs 1568 pibs?

Solution:

+20 : Yes, for example, by using 5 coins of +13 pibs and 5 coins of -9 pibs {tex}(13(5)-9(5)=20){/tex}.
{tex}\boldsymbol{+} \mathbf{4 0}{/tex} : Yes, for example, by using {tex}\mathbf{1}{/tex} coin of {tex}\boldsymbol{+} \mathbf{1 3}{/tex} pibs and -3 coins of -9 pibs {tex}(13(1)-9(-3)=40){/tex}.
– 50: Yes, for example, by using {tex}\mathbf{1}{/tex} coin of {tex}\mathbf{+ 1 3}{/tex} pibs and 7 coins of -9 pibs {tex}(13(1)-9(7)=-50){/tex}.
+8 : Yes, for example, by using 2 coins of +13 pibs and 2 coins of -9 pibs {tex}(13(2)-9(2)=8){/tex}.
{tex}\boldsymbol{+}{/tex} 10: Yes, for example, by using {tex}\mathbf{7}{/tex} coins of {tex}\boldsymbol{+} \mathbf{1 3}{/tex} pibs and 9 coins of -9 pibs {tex}(13(7)-9(9)=10){/tex}.
– 2: Yes, for example, by using 4 coins of +13 pibs and 6 coins of -9 pibs {tex}(13(4)-9(6)=-2{/tex} ).
{tex}(\mathrm{g})+1{/tex} : Yes, for example, by using -2 coins of +13 pibs and -3 coins of -9 pibs {tex}(13(-2)-9(-3)=1){/tex}.
Is it possible to purchase an item that costs 1568 pibs?
Yes, because the greatest common divisor of the coin denominations, {tex}\operatorname{gcd}(\mathbf{1 3}, \mathbf{9})=\mathbf{1}{/tex}, divides the target amount 1568. A solution exists for the Diophantine equation {tex}13 x-9 y=1568{/tex}.

Q.41: Find the value of: {tex}(32 \times(-18)) \div((-36)){/tex}

Solution:

The value of {tex}(32 \times(-18)) \div(-36){/tex} is 16.

Q.42: Find the value of: {tex}(32) \div((-36) \times(-18)){/tex}

Solution:

The value of {tex}32 \div((-36) \times(-18)){/tex} is:
{tex} (-36) \times(-18)=648 {/tex}
{tex} 32 \div 648=\frac{32}{648} \approx 0.049 {/tex}
So, the answer is approximately 0.049.

Q.43: Find the values of: {tex}(25 \times(-12)) \div((45) \times(-27)){/tex}

Solution:

The value of {tex}(25 \times(-12)) \div(45 \times(-27)){/tex} is:
Calculate numerator:
{tex} 25 \times(-12)=-300 {/tex}
Calculate denominator:
{tex} 45 \times(-27)=-1215 {/tex}
Now divide:
{tex} \frac{-300}{-1215}=\frac{300}{1215} \approx 0.247 {/tex}
So, the answer is approximately 0.247.

Q.44: Find the value of: {tex}(280 \times(-7)) \div((-8) \times(-35)){/tex}

Solution:

The value of {tex}(280 \times(-7)) \div((-8) \times(-35)){/tex} is -7.
Calculation:
{tex} 280 \times(-7)=-1960 {/tex}
{tex} (-8) \times(-35)=280 {/tex}
{tex} -1960 \div 280=-7 {/tex}

Q.45: Arrange the expressions given below in increasing order.

(-348) + (-1064)
(- 348) – (- 1064)
348 – (- 1064)
(- 348) {tex}\times{/tex} (- 1064)
348 {tex}\times{/tex} (- 1064)
348 {tex}\times{/tex} 964

Solution:

{tex} (-348)+(-1064)=-1412 {/tex}
{tex} (-348)-(-1064)=716 {/tex}
{tex} 348-(-1064)=1412 {/tex}
{tex} 348 \times(-1064)=-370272 {/tex}
{tex} (-348) \times(-1064)=370272 {/tex}
{tex} 348 \times 964=335472 {/tex}

Q.46: Given that {tex}(-548) \times 972=-532656{/tex}, write the values of:

{tex}(-547) \times 972{/tex}
{tex}(-548) \times 971{/tex}
{tex}(-547) \times 971{/tex}

Solution:

{tex} (-547) \times 972=-531684 {/tex}
{tex} (-548) \times 971=-532108 {/tex}
{tex} (-547) \times 971=-531137 {/tex}

Q.47: Given that {tex}207 \times(-33+7)=-5382{/tex}, write the value of {tex}-207 \times(33-7) ={/tex} ________.

Solution:

The value of {tex}-207 \times(33-7){/tex} is -5382.

Q.48: Use the numbers {tex}3,-2,5,-6{/tex} exactly once and the operations ‘+’, ‘-‘, and ‘{tex}\times{/tex}’ exactly once and brackets as necessary to write an expression such that-

the result is the maximum possible
the result is the minimum possible

Solution:

Use each number {tex}3,-2,5,-6{/tex} exactly once and the operations, {tex}+- \times{/tex} exactly once.

Maximum value {tex}=60{/tex}
One way to get this is
{tex} (-2-(3+5)) \times(-6)=(-2-8) {/tex} {tex}\times(-6)=(-10) \times(-6)=60 . {/tex}
Minimum value {tex}=-60{/tex}
One way to get this is
{tex} ((3-(-2))+5) \times(-6)=(5+5){/tex} {tex} \times(-6)=10 \times(-6)=-60 . {/tex}
Both values are optimal (no arrangement gives a larger/smaller result).

Q.49: Fill in the blanks in at least 5 different ways with integers:

Solution:

{tex}0+(-6) \times 6=-36{/tex}
{tex}12+(-6) \times 8=-36{/tex}
{tex}-18+(-6) \times 3=-36{/tex}
{tex}36+(-12) \times 6=-36{/tex}
{tex}-12+(-4) \times 6=-36{/tex}

Q.50: Fill in the blanks in at least 5 different ways with integers:

Solution:

{tex} (6-0) \times 2=12 {/tex}
{tex} (10-4) \times 2=12 {/tex}
{tex} (8-2) \times 2=12 {/tex}
{tex} (14-2) \times 1=12 {/tex}
{tex} (9-3) \times 2=12 {/tex}

Q.51: Fill in the blanks in at least 5 different ways with integers:

Solution:

{tex} (2-(3-0))=-1 {/tex}
{tex} (4-(6-3))=1 \text { (But for }-1:) {/tex}
{tex} (0-(1-0))=-1 {/tex}
{tex} (5-(6-0))=-1 {/tex}
{tex} (1-(2-0))=-1 {/tex}
{tex} (3-(4-2))=1 \text { (So focus:) } {/tex}
{tex} \text { All correct for }-1 \text { : } {/tex}
{tex} (0-(1-0))=-1 {/tex}
{tex} (1-(2-1))=0(\text { not }-1) {/tex}
{tex} (2-(3-0))=-1 {/tex}
{tex} (5-(6-0))=-1 {/tex}
{tex} (1-(2-0))=-1 {/tex}