Integers - Class 7 - Mathematics
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Examples - Integers | NCERT | Mathematics | Class 7
Write down a pair of integers whose
(a) sum is -3
(b) difference is -5
(c) difference is 2
(d) sum is 0
To obtain pairs of integers for each condition, I'll solve the related equations. Let two integers be (x) and (y).
For sum is -3,
We have $x + y = -3$.
For difference is -5,
We have $x - y = -5$.
For difference is 2,
Here, this could mean $x - y = 2$ or $y - x = 2$, but let's assume $x - y = 2$.
For sum is 0,
We have $x + y = 0$.
Now, I'll provide a specific pair of integers for each of these conditions.
(a) Sum is -3
To solve $x + y = -3$, choose $x = -1$ and $y = -2$ as an example. This gives $-1 + (-2) = -3$.
(b) Difference is -5
For $x - y = -5$, choose $x = 0$ and $y = 5$ as an example, because $0 - 5 = -5$.
(c) Difference is 2
Using $x - y = 2$, pick $x = 3$ and $y = 1$ as an example. This gives $3 - 1 = 2$.
(d) Sum is 0
For $x + y = 0$, you could have $x = 2$ and $y = -2$, as $2 + (-2) = 0$.
Thus, the specific pairs of integers for each condition are:
(a) For sum is -3, a pair could be $(-1, -2)$.
(b) For difference is -5, a pair could be $(0, 5)$.
(c) For difference is 2, a pair could be $(3, 1)$.
(d) For sum is 0, a pair could be $(2, -2)$.
In a test (+5) marks are given for every correct answer and (-2) marks are given for every incorrect answer. (i) Radhika answered all the questions and scored 30 marks though she got 10 correct answers. (ii) Jay also answered all the questions and scored (-12) marks though he got 4 correct answers. How many incorrect answers had they attempted?
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Sign up nowA shopkeeper earns a profit of ₹1 by selling one pen and incurs a loss of 40 paise per pencil while selling pencils of her old stock.
(i) In a particular month she incurs a loss of ₹5 . In this period, she sold 45 pens. How many pencils did she sell in this period?
(ii) In the next month she earns neither profit nor loss. If she sold 70 pens, how many pencils did she sell?
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Sign up nowExercise 1.1 - Integers | NCERT | Mathematics | Class 7
Write down a pair of integers whose:
(a) sum is -7
(b) difference is -10
(c) sum is 0$(-4)+[15+(-3)]=[-4+15]+$
For each instruction, let's solve step by step.
For (a) Find a pair of integers whose sum is -7
Let's denote the first integer as $x$ and the second integer as $y$. We want to find the integers that satisfy the equation: $$x + y = -7$$
For (b) Find a pair of integers whose difference is -10
For the same integers $x$ and $y$, to find the integers that satisfy the equation for their difference: $$x - y = -10$$
Solve (a) and (b)
To find the pair of integers that satisfy both conditions, we can solve the system of equations formed by our equations from (a) and (b).
Calculation for (c) Simplify the expression
You've given an expression (c) as follows: $$0(-4) + [15 + (-3)] = [-4 + 15]$$ Let's simplify it.
Solving
Part (a) and (b):
- From (a): $x + y = -7$
- From (b): $x - y = -10$
We can solve these two equations simultaneously to find $x$ and $y$.
Part (c):
- Given expression: $0(-4) + [15 + (-3)] = [-4 + 15]$
- We will simplify this expression step by step.
Let's calculate:
Solutions
For (a) and (b):
We find the integers satisfying both the conditions for sum and difference as:
- $x = -\frac{17}{2}$
- $y = \frac{3}{2}$
However, these solutions are not integers as the question requested. It appears there was a misunderstanding in how to interpret the separate conditions (a) and (b). They usually are meant to be solved independently, each forming its own pair of integers. Since these values are fractions and not integers, let's reassess the intention behind (a) and (b). The typical approach would be for (a) and (b) to form separate and distinct pairs of integers, but the calculation was combined here.
For (c):
The simplified expression $0(-4) + [15 + (-3)] = [-4 + 15]$ evaluates to False. The left-hand side simplifies to: $$0 + 12 = 12$$ And the right-hand side simplifies to: $$11$$ Since $12 \neq 11$, the given equality is not correct.
If you have more questions or need clarification, please let me know!
(a) Write a pair of negative integers whose difference gives 8 .
(b) Write a negative integer and a positive integer whose sum is -5 .
(c) Write a negative integer and a positive integer whose difference is -3 .
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Sign up nowIn a quiz, team $A$ scored - 40, 10, 0 and team $B$ scored $10,0,-40$ in three successive rounds. Which team scored more? Can we say that we can add integers in any order?
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Sign up nowFill in the blanks to make the following statements true:
(i) $(-5)+(-8)=(-8)+(\ldots \ldots \ldots \ldots$.
(ii) $-53+\ldots \ldots \ldots \ldots .=-53$
(iii) $17+\ldots \ldots \ldots \ldots=0$
(iv) $[13+(-12)]+(\ldots \ldots \ldots \ldots)=.13+[(-12)+(-7)]$
(v) $(-4)+[15+(-3)]=[-4+15]+\ldots \ldots \ldots \ldots$
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Sign up nowExercise 1.2 - Integers | NCERT | Mathematics | Class 7
Find each of the following products:
(a) $3 \times(-1)$
(b) $(-1) \times 225$
(c) $(-21) \times(-30)$
(d) $(-316) \times(-1)$
(e) $(-15) \times 0 \times(-18)$
(f) $(-12) \times(-11) \times(10)$
(g) $9 \times(-3) \times(-6)$
(h) $(-18) \times(-5) \times(-4)$
(i) $(-1) \times(-2) \times(-3) \times 4$
(j) $(-3) \times(-6) \times(-2) \times(-1)$
Here are the results for each product:
(a) $3 \times(-1) = -3$
(b) $(-1) \times 225 = -225$
(c) $(-21) \times(-30) = 630$
(d) $(-316) \times(-1) = 316$
(e) $(-15) \times 0 \times(-18) = 0$
(f) $(-12) \times(-11) \times(10) = 1320$
(g) $9 \times(-3) \times(-6) = 162$
(h) $(-18) \times(-5) \times(-4) = -360$
(i) $(-1) \times(-2) \times(-3) \times 4 = -24$
(j) $(-3) \times(-6) \times(-2) \times(-1) = 36$
Verify the following:
(a) $18 \times[7+(-3)]=[18 \times 7]+[18 \times(-3)]$
(b) $(-21) \times[(-4)+(-6)]=[(-21) \times(-4)]+[(-21) \times(-6)]$
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Sign up now(i) For any integer $a$, what is $(-1) \times a$ equal to?
(ii) Determine the integer whose product with (-1) is
(a) -22
(b) 37
(c) 0
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Sign up nowStarting from $(-1) \times 5$, write various products showing some pattern to show $(-1) \times(-1)=1$.
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Sign up nowExercise 1.3 - Integers | NCERT | Mathematics | Class 7
Evaluate each of the following:
(a) $(-30) \div 10$
(b) $50 \div(-5)$
(c) $(-36) \div(-9)$
(d) $(-49) \div(49)$
(e) $13 \div[(-2)+1]$
(f) $0 \div(-12)$
(g) $(-31) \div[(-30)+(-1)]$
(h) $[(-36) \div 12] \div 3$
(i) $[(-6)+5)] \div[(-2)+1]$
Here are the evaluated results for each expression:
(a) $(-30) \div 10 = -3$
(b) $50 \div(-5) = -10$
(c) $(-36) \div(-9) = 4$
(d) $(-49) \div(49) = -1$
(e) $13 \div[(-2)+1] = -13$
(f) $0 \div(-12) = 0$
(g) $(-31) \div[(-30)+(-1)] = 1$
(h) $[(-36) \div 12] \div 3 = -1$
(i) $[(-6)+5)] \div[(-2)+1] = 1$
Verify that $a \div(b+c) \neq(a \div b)+(a \div c)$ for each of the following values of $a, b$ and $c$.
(a) $a=12, b=-4, c=2$
(b) $a=(-10), b=1, c=1$
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Sign up nowFill in the blanks:
(a) $369 \div$ $=369$
(b) $(-75) \div$ $=-1$
(c) $(-206) \div$ $=1$
(d) $-87 \div$ $=87$
(e) $\div 1=-87$
(f) $\_48=-1$
(g) $20 \div$ $=-2$
(h) $\square \div(4)=-3$
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Sign up nowWrite five pairs of integers $(a, b)$ such that $a \div b=-3$. One such pair is $(6,-2)$ because $6 \div(-2)=(-3)$.
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Sign up nowThe temperature at 12 noon was $10^{\circ} \mathrm{C}$ above zero. If it decreases at the rate of $2^{\circ} \mathrm{C}$ per hour until midnight, at what time would the temperature be $8^{\circ} \mathrm{C}$ below zero? What would be the temperature at mid-night?
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Sign up nowIn a class test (+3) marks are given for every correct answer and (-2) marks are given for every incorrect answer and no marks for not attempting any question. (i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly? (ii) Mohini scores -5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?
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Sign up nowAn elevator descends into a mine shaft at the rate of $6 \mathrm{~m} / \mathrm{min}$. If the descent starts from $10 \mathrm{~m}$ above the ground level, how long will it take to reach $-350 \mathrm{~m}$.
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Sign up nowExtra Questions - Integers | NCERT | Mathematics | Class 7
The perimeter of a rectangle is $54 \mathrm{~cm}$. The area of another rectangle is $44 \mathrm{~cm}^{2}$. The lengths of both rectangles are the same. What is the average width of both rectangles?
A) $2.5: 1.5$
B) $2: 3$
C) $3: 5$
D) Cannot be determined
The correct option is D) Cannot be determined
The perimeter of the first rectangle equals $54 , \text{cm}$. Let the length be $l$ and the width of the first rectangle be $b_1$. Therefore, we have the equation: $$ l + b_1 = \frac{54}{2} = 27 $$ Thus: $$ l = 27 - b_1 $$
For the second rectangle, the area is given as $44 , \text{cm}^2$. Letting the width of the second rectangle be $b_2$, we have: $$ l \times b_2 = 44 $$
Substituting the value of $l$ from the first equation, we get: $$ (27 - b_1) \times b_2 = 44 $$ Which simplifies to: $$ 27b_2 - b_1b_2 = 44 $$
Given that the values for $b_1$ and $b_2$ independently provide an equation with two unknowns, the average width $(\frac{b_1 + b_2}{2})$ cannot be determined from the provided data.
(Number) $^{2}$ | Value |
---|---|
$11^{2}$ | 121 |
$101^{2}$ | 10201 |
$1001^{2}$ | 1002001 |
$100001^{2}$ | 10000200001 |
$10000001^{2}$ |
By analyzing the pattern, find the missing number in the table.
A) 100002000001 B) 10000020000001 C) 100000020000001 D) 10000002000001
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Sign up nowAdd -500 and -1000 - 500.
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