Operations with IntegersClass 7 Mathematics NCERT Solutions

To Find: Pairs of numbers given their sum and difference.

Let: The two numbers be $x$ and $y$.

Formulas: Given the sum ($S$) and difference ($D$): $x + y = S$ $x - y = D$ Adding the two equations gives $2x = S + D$, so $x = \frac{S+D}{2}$. Subtracting the second equation from the first gives $2y = S - D$, so $y = \frac{S-D}{2}$.

(a) Sum = 27, Difference = 9 $x = \frac{27 + 9}{2} = \frac{36}{2} = 18$ $y = \frac{27 - 9}{2} = \frac{18}{2} = 9$ Final Answer: The numbers are 18 and 9.

(b) Sum = 4, Difference = 12 $x = \frac{4 + 12}{2} = \frac{16}{2} = 8$ $y = \frac{4 - 12}{2} = \frac{-8}{2} = -4$ Final Answer: The numbers are 8 and -4.

(c) Sum = 0, Difference = 10 $x = \frac{0 + 10}{2} = \frac{10}{2} = 5$ $y = \frac{0 - 10}{2} = \frac{-10}{2} = -5$ Final Answer: The numbers are 5 and -5.

(d) Sum = 0, Difference = -10 $x = \frac{0 + (-10)}{2} = \frac{-10}{2} = -5$ $y = \frac{0 - (-10)}{2} = \frac{10}{2} = 5$ Final Answer: The numbers are -5 and 5.

(e) Sum = -7, Difference = -1 $x = \frac{-7 + (-1)}{2} = \frac{-8}{2} = -4$ $y = \frac{-7 - (-1)}{2} = \frac{-6}{2} = -3$ Final Answer: The numbers are -4 and -3.

(f) Sum = -7, Difference = -13 $x = \frac{-7 + (-13)}{2} = \frac{-20}{2} = -10$ $y = \frac{-7 - (-13)}{2} = \frac{6}{2} = 3$ Final Answer: The numbers are -10 and 3.

To Find: The values of the given expressions.

(a) $14 \times (-15)$ $14 \times (-15) = -(14 \times 15) = -210$ Final Answer: $-210$

(b) $-16 \times (-5)$ $(-16) \times (-5) = +(16 \times 5) = 80$ Final Answer: $80$

(c) $36 \div (-18)$ $36 \div (-18) = -(36 \div 18) = -2$ Final Answer: $-2$

(d) $(-46) \div (-23)$ $(-46) \div (-23) = +(46 \div 23) = 2$ Final Answer: $2$

Given: Initial Temperature = $32^{\circ} \mathrm{C}$ Rate of temperature change = $-5^{\circ} \mathrm{C}$ per hour (since it is lowered) Time = 10 hours

To Find: The room temperature after 10 hours.

Solution: Total change in temperature = Rate of change $\times$ Time Total change = $(-5^{\circ} \mathrm{C} / \text{hour}) \times 10 \text{ hours} = -50^{\circ} \mathrm{C}$

Final Temperature = Initial Temperature + Total change in temperature Final Temperature = $32^{\circ} \mathrm{C} + (-50^{\circ} \mathrm{C})$ Final Temperature = $32 - 50 = -18^{\circ} \mathrm{C}$

Final Answer: The room temperature will be $-18^{\circ} \mathrm{C}$ after 10 hours.

Given: Profit on white cement = +₹8 per bag Loss on grey cement = -₹5 per bag

Solution: Profit from 3,000 bags of white cement = $3000 \times 8 = \text{₹}24,000$ Loss from 5,000 bags of grey cement = $5000 \times (-5) = -\text{₹}25,000$ Total Profit/Loss = Profit from white cement + Loss from grey cement Total = $24000 + (-25000) = -\text{₹}1,000$

Final Answer: The company has a loss of ₹1,000.

Solution: Loss from 6,400 bags of grey cement = $6400 \times (-5) = -\text{₹}32,000$ To have neither profit nor loss, the total must be ₹0. Profit from white cement + Loss from grey cement = 0 Profit from white cement + $(-32000) = 0$ Profit from white cement = ₹32,000

Let $w$ be the number of white cement bags sold. $w \times 8 = 32000$ $w = 32000 \div 8$ $w = 4000$

Final Answer: The company must sell 4,000 bags of white cement.

To Find: The integer that completes each statement.

(a) $(-3) \times \_\_ = 27$ Let the integer be $x$. $(-3) \times x = 27 \implies x = 27 \div (-3) = -9$. Answer: -9

(b) $5 \times \_\_ = (-35)$ Let the integer be $x$. $5 \times x = -35 \implies x = -35 \div 5 = -7$. Answer: -7

(c) $\_\_ \times (-8) = (-56)$ Let the integer be $x$. $x \times (-8) = -56 \implies x = -56 \div (-8) = 7$. Answer: 7

(d) $\_\_ \div (-12) = 132$ Let the integer be $x$. $x \div (-12) = 132 \implies x = 132 \times (-12) = -1584$. Answer: -1584

(e) $\_\_ \div (-8) = 7$ Let the integer be $x$. $x \div (-8) = 7 \implies x = 7 \times (-8) = -56$. Answer: -56

(f) $[\_\_ \div 12 = -11]$ Let the integer be $x$. $x \div 12 = -11 \implies x = -11 \times 12 = -132$. Answer: -132

To Find: The values of the given expressions.

(a) $(-5) \times (18+(-3))$ First, solve the expression inside the parentheses: $18 + (-3) = 15$. Then multiply: $(-5) \times 15 = -75$. Final Answer: -75

(b) $(-7) \times 4 \times (-1)$ Multiply from left to right: $(-7) \times 4 = -28$. Then: $(-28) \times (-1) = 28$. Final Answer: 28

(c) $(-2) \times (-1) \times (-5) \times (-3)$ Multiply from left to right: $(-2) \times (-1) = 2$ $2 \times (-5) = -10$ $(-10) \times (-3) = 30$ Alternatively, since there are an even number (4) of negative integers, the result will be positive. $2 \times 1 \times 5 \times 3 = 30$. Final Answer: 30

To Find: The values of the given expressions.

(a) $(-27) \div 9$ A negative divided by a positive is a negative. $27 \div 9 = 3$. Final Answer: -3

(b) $84 \div (-4)$ A positive divided by a negative is a negative. $84 \div 4 = 21$. Final Answer: -21

(c) $(-56) \div (-2)$ A negative divided by a negative is a positive. $56 \div 2 = 28$. Final Answer: 28

To Find: The values of the given multiplications.

(a) $3 \times (-2)$ This means placing 2 negative tokens into a bag 3 times. This results in $2+2+2=6$ negative tokens. Final Answer: $3 \times (-2) = -6$

(b) $(-5) \times (-2)$ This means removing 2 negative tokens from a bag 5 times. To do this, we first add 10 zero pairs (10 positive and 10 negative tokens). Then we remove the 10 negative tokens. This leaves 10 positive tokens in the bag. Final Answer: $(-5) \times (-2) = 10$

(c) $(-4) \times (-1)$ This means removing 1 negative token from a bag 4 times. We add 4 zero pairs and remove the 4 negative tokens. This leaves 4 positive tokens. Final Answer: $(-4) \times (-1) = 4$

(d) $(-7) \times 3$ This means removing 3 positive tokens from a bag 7 times. We add 21 zero pairs and remove the 21 positive tokens. This leaves 21 negative tokens. Final Answer: $(-7) \times 3 = -21$

Given: $123 \times 456 = 56088$

To Find: The values of the given expressions using the rules of integer multiplication.

(a) $(-123) \times 456$ Since one integer is negative and the other is positive, the product will be negative. The magnitude is the same. Final Answer: $-56088$

(b) $(-123) \times (-456)$ Since both integers are negative, the product will be positive. The magnitude is the same. Final Answer: $56088$

(c) $(123) \times (-456)$ Since one integer is positive and the other is negative, the product will be negative. The magnitude is the same. Final Answer: $-56088$

To Find: The products of the given integers.

(a) $4 \times (-3)$ $4 \times 3 = 12$. A positive times a negative is a negative. Final Answer: $-12$

(b) $(-6) \times (-3)$ $6 \times 3 = 18$. A negative times a negative is a positive. Final Answer: $18$

(c) $(-5) \times (-1)$ $5 \times 1 = 5$. A negative times a negative is a positive. Final Answer: $5$

(d) $(-8) \times 4$ $8 \times 4 = 32$. A negative times a positive is a negative. Final Answer: $-32$

(e) $(-9) \times 10$ $9 \times 10 = 90$. A negative times a positive is a negative. Final Answer: $-90$

(f) $10 \times (-17)$ $10 \times 17 = 170$. A positive times a negative is a negative. Final Answer: $-170$

To Argue: The given statements are true based on the concept of additive inverse.

Argument: Subtracting a number is mathematically equivalent to adding its additive inverse. The additive inverse of an integer 'a' is '-a'. In this case, the number being subtracted is 18. Its additive inverse is -18. Therefore, subtracting 18 from 7 is the same as adding -18 to 7.

Verification: LHS: $7 - 18 = -11$ RHS: $7 + (-18) = -11$ Since LHS = RHS, the statement is true.

Argument: Subtracting a number is equivalent to adding its additive inverse. The number being subtracted is -12. Its additive inverse is $-(-12)$, which is 12. Therefore, subtracting -12 from 4 is the same as adding 12 to 4.

Verification: LHS: $4 - (-12) = 4 + 12 = 16$ RHS: $4 + 12 = 16$ Since LHS = RHS, the statement is true.

Given: First Movement = -4 Final Position = 5

To Find: The second movement.

Formula: Final Position = First Movement + Second Movement

Solution: Let the second movement be $x$. $5 = -4 + x$ $x = 5 - (-4)$ $x = 5 + 4$ $x = 9$

Final Answer: The second movement is 9.

Given: A series of movements: $1, -2, 3, -4, \ldots, 9, -10$.

To Find: The final position of the coin, which is the sum of all movements.

Solution: The total movement is the sum of the series: $S = 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + 9 - 10$. We can group the terms in pairs: $S = (1 - 2) + (3 - 4) + (5 - 6) + (7 - 8) + (9 - 10)$ Each pair sums to -1: $S = (-1) + (-1) + (-1) + (-1) + (-1)$ There are 5 such pairs. $S = 5 \times (-1)$ $S = -5$

Final Answer: The final position of the coin is -5.