By using Ganita Prakash Class 7 Solutions and Part 2 Chapter 2 Operations with Integers Class 7 Question Answer, students can improve their problem-solving skills.
Class 7 Maths Ganita Prakash Part 2 Chapter 2 Solutions
Class 7 Maths Operations with Integers Solutions
Class 7 Ganita Prakash Part 2 Chapter 2 Solutions Operations with Integers
Figure it Out (Page: 25)
Question 1.
Let us try to find a few more pairs of numbers from their sums and differences:
(a) Sum = 27, Difference = 9
(b) Sum = 4, Difference = 12
(c) Sum = 0, Difference = 10
(d) Sum = 0, Difference = – 10
(e) Sum = – 7, Difference = – 1
(f) Sum = – 7, Difference = – 13
Solution:
(a) Sum = 27, Difference = 9
Therefore, the correct pair is 18 and 9.
(b) Sum = 4, Difference = 12
Therefore, the correct pair is 8 and -4.
(c) Sum = 0, Difference = 10
Therefore, the correct pair is 5 and -5.
(d) Sum = 0, Difference = -10
Therefore, the correct pair is -5 and 5.
(e) Sum = -7, Difference = -1
Therefore, the correct pair is -4 and -3.
(f) Sum = -7, Difference = -13
Therefore, the correct pair is -10 and 3.
Figure it Out (Page: 31)
Question 1.
Using the token interpretation, find the values of
(a) 3 × (-2)
(b) (-5) × (-2)
(c) (-4) × (-1)
(d) (-7) × 3
Solution:
(a) 3 × (-2)
Two red tokens 3 times = (-6) So, 3 × (-2) = (-6).
(b) (-5) × (-2)
Remove 2 red tokens from the zero pairs, 5 times.
So, (-5) × (-2) = 10.
(c) (-4) × (-1)
Remove 1 red token from the zero pair, 4 times.
So, (-4) × (-1) = 4.
(d) (-7) × 3
Remove 3 green tokens from the zero pairs, 7 times.
So, (-7) × 3 = (-21).
Question 2.
If 123 × 456 = 56088, without calculating, find the value of:
(a) (-123) × 456
(b) (-123) × (-456)
(c) (123) × (-456)
Solution:
Given 123 × 456 = 56088
(a) (-123) × 456
= (-1) × 123 × 456
= (-1) × 56088
= -56088
(b) (-123) × (-456)
= (-1) × 123 × (-1) × 456
= (-1) ×(-1) × 123 × 456
= 1 × 56088 = 56088.
(c) (123) ×(-456)
= 123 × 456 ×(-1)
= 56088 ×(-1)
= -56088
Question 3.
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:
Rule for multiplying two integers:
(i) Multiply their absolute values.
(ii) If the integers have different signs, the product is negative.
(iii) If both integers have the same sign, the product is positive.
Figure it Out (Page: 33-34)
Question 1.
Find the following products.
(a) 4 × (-3)
(b) (-6) × (-3)
(c) (-5) × (-1)
(d) (-8) × 4
(e) (-9) × 10
(f) 10 × (-17)
Solution:
(a) 4×(-3)
Multiplier is positive, multiplicand is negative → product is negative.
∴ 4 × (-3) = -12.
(b) (-6) × (-3)
Both numbers are negative → product is positive.
∴ (-6) × (-3) = 18.
(c) (-5) × (-1)
Both numbers are negative → product is positive.
∴ (-5) × (-1) = 5.
(d) (-8) × 4
Multiplier is negative, multiplicand is positive → product is negative.
∴ (-8) × 4 = -32.
(e) (-9) × 10
Multiplier is negative, multiplicand is positive → product is negative.
∴ (-9) × 10 = -90.
(f) 10 × (-17)
Multiplier is positive, multiplicand is negative → product is negative.
∴ 10 × (-17) = -170.
Figure it Out (Page: 39)
Question 1.
Find the values of:
(a) 14 × (-15)
(b) -16× (-5)
(c) 36 ÷ (-18)
(d) (-46) ÷ (-23)
Solution:
(a) (As when positive number multiplied by a negative number, result is negative number).
∴ 14 × (-15) = -(14 × 15) = -210
(b) (When negative number is multiplied by negative number, result is positive number)
∴ -16 × (-5) = 80
(c) (As when positive number is divided by a negative number, result is negative number).
∴ 36 ÷(-18) = -(36 ÷ 18) = -2
(d) (When negative number is divided by a negative number; result is positive number).
∴ (-46) ÷ (-23) = 2
Question 2.
A freezing process requires that the room temperature be lowered from 32°Cat the rate of 5°Cevery hour. What will be the room temperature 10 hours after the process begins?
Solution:
As the temperature is lowered at a rate of 5 °C every hour for 10 hours.
Total drop = 5°C/ hour × 10 hours = 50°C.
⇒ To Calculate the final temperature:
Final temperature = initial temperature – Total Drop
Final temperature = 32°C-50°C = -18°C
∴ The room temperature after 10 hours will be -18°C
Question 3.
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.]
(a) 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?
(b) 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:
Profit on one white cement bag = ₹ 8
Loss on one grey cement bag = ₹5
(Profit is positive, loss is negative.)
(a) White cement sold = 3000 bags
Grey cement sold = 5000 bags
Profit from white cement = 3000 × ₹ 8 = ₹24,000.
Loss from grey cement = 5000 × 5 = ₹ 25,000.
Total profit/loss = 24,000+(-25,000) = ₹ 1,000.
∴ The company has a loss of ₹1,000.
(b) Grey cement sold = 6400 bags
Loss = 6400 × (-5) = ₹ 32,000
Let the number of white cement bags to be sold = x
Profit from white cement = x × 8
For no profit and no loss,
x × 8 + (-32,000) = 0
8 x = 32,000
x = 32,000 ÷ 8
x = 4000
∴ The company must sell 4000 bags of white cement to have neither profit nor loss.
Question 4.
Replace the blank with an integer to make a true statement.
(a) (-3) × ______ = 27
(b) 5 × ______ = (-35)
(c) ______ × (-8) = (-56)
(d) ______ × (-12) = 132
(e) ______ ÷ (-8) = 7
(f) ______ ÷ 12 = -11
Solution:
(a) (-3) × = 2 7
or 27 ÷ (-3) = -9
∴ (-3) × (-9) = 27
(b) ______ = (-35)
or -35 ÷ 5 = -7
∴ 5 ×(-7) = (-35)
(c) ______ × (-8) = (-56)
or -56 ÷ (-8) = 7
∴ 7 ×(-8) = (-56)
(d) ______ × (-12) = 132
or 132 ÷ (-12) = -11
∴ (-11) × (-12) = 132
(e) ______ ÷ (-8) = 7
or 7 ×(-8) = -56
∴ (-56) ÷ (-8) = 7
(f) ______ ÷ 12 = -11
or -11 × 12 = -132
∴ (-132) ÷ 12 = -11
Figure it Out (Page: 42-44)
Question 1.
Find the values of the following expressions:
(a) (-5) × (18+(-3))
(b) (-7) × 4 ×(-1)
(c) (-2) × (-1) × (-5) × (-3)
Solution:
(a) (-5) × (18+(-3))
= (-5) × 18 + (-5) × (-3)
= (-90) + 15
= (-75)
(b) (-7) × 4 × (-1)
= (-28) × (-1)
= 28.
(c) (-2) × (-1) × (-5) × (-3)
= 2 × (-5) × (-3)
= 2 × 15
= 30
Question 2.
Find the values of the following expressions:
(a) (-27) ÷ 9
(b) 84 ÷(-4)
(c) (-56) ÷(-2)
Solution:
(a) (-27) ÷ 9
or 9 × _____ = (-27)
(b) 84 ÷(-4)
or (-4) × _____ = 84
We know that (-4) ×(-21) = 84
∴ 84 ÷(-4) = (-21).
(c) (-56) ÷(-2)
or (-2) × _____ = (-56)
We know that (-2) × 28 = (-56)
∴ (-56) ÷ (-2) = 28.
Question 3.
Find the integer whose product with (-1) is:
(a) 27
(b) -31
(c) -1
(d) 1
(e) 0
Solution:
(a) 27 (b) -31 (c) -1 (d) 1 (e) 0
Question 4.
If 4 7-5 6+1 4-8+2-8+5= -4, then find the value of -47+56-14+8-2+8-5 without calculating the full expression.
Solution:
Given expression is
47 – 56 + 14 – 8 + 2 – 8 + 5 = -4
The expression,
-47 + 56 – 14 + 8 – 2 + 8 – 5 = -(47 – 56 + 14 – 8 + 2 – 8 + 5) = -(-4) = 4
So, the value of the expression is 4.
Question 5.
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:
(a) Following the rules for the sequence starting with -7.
The rule is if even, take half, if odd, multiply by -3, and add 1.
Start with -7 (odd)
(-7) × (-3) + 1 = 21 + 1 = 22 (even)
then 22 ÷ 2 = 11 (odd)
then 11 ×(-3) + 1 = -33 + 1 = -32 (even)
then (-32) ÷ 2 = -16 (even)
then (-16) ÷ 2 = -8 (even)
then (-8) ÷ 2 = -4 (even)
then (-4) ÷ 2 = -2 (even)
then (-2) ÷ 2 = -1 (odd)
then (-1) × (-3) + 1 = 4 (even)
then 4 ÷ 2 = 2 (even)
then 2 ÷ 2 = 1 (odd)
Hence the sequence is
(b) (i) Now for the starting number -21 (odd)
then (-21) × (-3) + 1 = 63 + 1 = 64 (even)
then 64 ÷ 2 = 32 (even)
then 32 ÷ 2 = 16 (even)
then 16 + 2 = 8 (even)
then 8 ÷ 2 = 4 (even)
then 4 ÷ 2 = 2 (even)
then 2 ÷ 2 = 1 (odd)
then 1 × (-3) +1 = -2 (even)
then (-2) ÷ 2 = -1 (odd)
then [-1 × (-3) + 1 = 4 (even)
then 4 ÷ 2 = 2 (even)
then 2 ÷ 2 = 1 (odd)
then 1 × (-3) + 1 = -2 (even)
Hence the sequence for -21 is
(ii) For the sequence, the starting number is -6
-6 is even then -6+2 = -3 (odd)
then (-3) ×(-3)+1 = 9+1 = 10 (even)
then 10 ÷ 2 = 5 (odd)
5 × (-3) +1 = -15 + 1 = -14 (even)
then -14 ÷ 2 = -7 (odd)
(-7) ×(-3)+1 = +21+1 = 22 (even)
then 22 ÷ 2 = 11 (odd)
then [11 ×(-3)+1 = -33+1 = -32 (even)
then -32 ÷ 2 = -16 (even)
then -16 ÷ 2 = -8 (even)
then -8 ÷ 2 = -4 (even)
then -4 ÷ 2 = -2 (even)
then (-2) ÷ 2 = -1 (odd)
then (-1) ×(-3)+1 = 3+1 = 4 (even)
then 4 ÷ 2 = 2 (even)
then 2 ÷ 2 = 1 (odd)
then 1 × (-3)+1 = -2
Hence, the sequence is
Observation: For numbers like -21,-6, etc., the sequences eventually reach a repeating loop of -2,-1,4,2,1,-2.
All starting numbers end up in this cycle.
Question 6.
In a test, (+4) marks are given for every correct answer and (-2) marks are given for every incorrect answer.
(a) 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?
(b) 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:
(a) Anita’s score = 40 marks
Correct answers = 15
Marks from correct answers = 15 × 4 = 60
Marks lost = 60 – 40 = 20
These 20 marks were lost because of incorrect answers:
Each incorrect answer gives -2 marks.
Number of incorrect answers
= 20 ÷ 2 = 10
Total questions in the test = correct + incorrect = 15 + 10 = 25 questions.
(b) Anil’s score = -10 marks
Correct answers = 5
Marks from correct answers = 5 × 4 = 20
Let incorrect answers be x.
Total score:
⇒ 20 – 2x = -10
⇒ -2x = -10-20
⇒ -2x = -30
⇒ x = 15
So Anil had 15 incorrect answers.
Yes, he leaves some questions unanswered.
Anil answered the question = (Total No. of Questions) – (Anil’s correct answers + Anil’s incorrect answers)
= 25 – (5+15)
= 25-20
= 5
Hence, Anil left 5 questions unanswered.
Question 7.
Pick the pattern – find the operations done by the machine shown below.
Solution:
First we find the rule
It looks like
First number – Second Number × Third Number
Solving for -16, -6, -9
Putting values
= (-16) – (-6) × (-9)
= (-16) – [-6 ×-9]
= (-16) – [6 × 9]
= (-16) – 54
= -16(16+54)
= -70 missing operation
Question 8.
Imagine you’re in a place where the temperature drops by 5°C each hour. If the temperature is currently at 8°C, write an expression which denotes the temperature after 4 hours.
Solution:
Current temperature = 8°C
The temperature drops by 5°C each hour
Total drop in temperature
= Temperature drop in 1 hour × no. of hours = 5 × 4
Temperature after 4 hours = 8-4 × 5
Hence required expression is 8-(5 × 4).
Question 9.
Find 3 consecutive numbers with a product of (a) – 6, (b) 120.
Solution:
(a) Let three consecutive numbers be n-1, n, and n+1.
Then product = (n-1)(n)(n+1) = -6
The only consecutive integers whose product is -6 are -2, -1, and 0.
But -2 × -1 × 0 = 0, not possible.
Consecutive numbers are integers.
The product is negative, so there must be an odd number of negative integers.
Integers are -2, -1, 1.
Their product = -2 × -1 × 1 = 2, not possible
Integers are-3, -2, -1.
Their product = [-3 × -2] × -1 = 6 × -1 = -6
Hence, consecutive integers are -3,-2,-1.
(b) Let three consecutive numbers be n-1, n, and n+1.
∴ (n-1)(n)(n+1) = 120
Now the cube root of 120 = 4.93
So n is likely to be 5.
∴ (5-1) × (5) × (5 + 1) = 4 × 5 × 6 = 120
Hence, consecutive integers are 4, 5, and 6.
Question 10.
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:
(a) +20
(b) + 40
(c) -50
(d) +8
(e) + 10
(f) – 2
(g) + 1
[Hint: Writing down a few multiples of 13 and 9 can help.]
(h) Is it possible to purchase an item that costs 1568 pibs?
Solution:
The currency has two denominations +13 pibs and -9 pibs. We need to determine if it is possible to make the given totals.
This is a linear equation of the form 13x – 9y = total where x and y are non-negative integers.
(a) Now total = +20
Then we need to find integers 13x – 9y = 20, x, y > 0 such that
If x = 1, then 13 – 9y = 20 ⇒ -9y = 7.
No solution.
If x = 2, then 26 – 9y = 20 ⇒ -9y = -6.
No solution.
If x = 3 then 39 – y = 20 ⇒ -9y = -19.
No solution.
If x = 4, then 52 – 9y = 20 ⇒ -9y = -32.
No solution.
If x = 5 then 65 – 9y = 20 ⇒ -9y = -45
⇒ y = 5
Hence, x = 5, y = 5 is a valid solution.
(b) + 40
Take 10 coins of +13 and 10 coins of -9:
10 × 13 – 10 × 9 = 130 – 90 = +40 pibs.
(c) -50
Take 10 coins of +13 and 20 coins of -9:
10 × 13 – 20 × 9 = 130-180 = -50 pibs.
(d) +8
Take 2 coins of +13 and 2 coins of -9:
2 × 13 – 2 × 9 = 26 – 18 = +8 pibs.
(e) + 10
Take 7 coins of +13 and 9 coins of -9 :
7 × 13-9 × 9 = 91-81 = +10 pibs.
(f) -2
Take 13 coins of +13 and 19 coins of -9:
13 × 13-19 × 9 = 169-171 = -2 pibs.
(g) + 1
Take 7 coins of +13 and 10 coins of -9 :
7 × 13-10 × 9 = 91-90 = +1 pibs.
(h) Yes, it is possible.
Take 122 coins of +13 and 2 coins of -9:
122 × 13 – 2 × 9 = 1586 – 18 = 1568 pibs.
Question 11.
Find the values of:
(a) (32 ×(-18)) ÷((-36))
(b) (32) ÷((-36) ×(-18))
(c) (25 ×(-12)) ÷((45) ×(-27))
(d) (280 ×(-7)) ÷((-8) ×(-35))
Solution:
(a) (32 ×(-18)) ÷(-36))
= 36 ×(-18) = -576
-576 ÷(-36) = 16.
(b) (32) ÷(-36) ×(-18))
= 36 ×(-18) = 648
32 ÷ 648 = \(\frac{32}{648}\) = \(\frac{4}{81}\)
(c) (25 ×(-12)) ÷ (45 × (-27))
= -300 ÷ -1215
= \(\frac{300}{1215}\) = \(\frac{20}{81}\).
(d) (280 ×(-7)) ÷((-8) ×(-35))
= -1960 ÷ 280
= -1960 / 280
= -7.
Question 12.
Arrange the expressions given below in increasing order.
(a) (-348) + (-1064)
(b) (-348) – (-1064)
(c) 348 – (-1064) (d) (-348) × (-1064)
(e) 348 × (-1064)
(f) 348 × 964
Solution:
(a) -348 + (-1064)
-348 – 1064 = -1412
(Adding two negative, the result is larger negative number)
(b) (348) – (-1064)
= -348 + 1064 = 716 (Two negative makes positive)
(c) -348-(-1064)
348+1064 = 1412 (subtracting a negative becomes adding a positive).
(d) (-348 × (-1064)
= -348 ×(-1064) = 370368.
Negative number multiply by negative number result in large positive number.
(e) 348 ×(-1064)
= 370368.
Positive number multiply with negative, results in large positive number.
(f) 348 × 964
= 335592
Positive number multiply with positive, results in positive number.
Arranging in increasing order (smallest to largest)
a-b-c-and -d
Question 13.
Given that (-548) × 972 = -532656, write the values of:
(a) (-547) × 972
(b) (-548) × 971
(c) (-547) × 971
Solution:
(a) (-547) × 972
we can rewrite using the given information
(-548) × 972 = -532656
= (-547) × 972
= (-548+1) × 972
(-548 × 972) + (1 × 972) (using distributive property)
Substituting
-532656+972 = -531684
Answer:
= -531684
(b) (-548) × 971
rewrite as:
(-548) × 971 = (-548) ×(972-1)
(-548 × 972) – (-548 × 1)
(Using Distributive Property)
Substituting
-532656 – (-548)
-532656 + 548 = -532108
Answer:
= -532108
(c) (-547) × 971
Rewrite
(-547) × 971 = (-548+1) × (972-1)
(-548 × 972) – (-548 × 1) + (1 × 972)-
(1 × 1) (Using Distributive Property)
Substituting
-532656+548+972-1
-532656+1520-1
Answer:
= -531137
Question 14.
Given that 207 × (-33+7) = -5382, write the value of -207 ×(33-7) = ______.
Solution:
207 × (-33+7) = -5382. To find the value of -207 × (33-7)
we can write the second expression in terms of the first
-207 × (33-7) = (-1 × 207) ×(-1 ×(-33+7))
= (-1) × (-1) ×(207 × (-33+7))
= 1 ×(207 ×(-33+7))
Since we are given that
207 × (-33+7) = -5382,
the value of the new expression is also -5382
Question 15.
Use the numbers 3, – 2, 5, – 6 exactly once and the operations ‘+’, ‘-‘, and ‘×’ exactly once and brackets as necessary to write an expression such that –
(a) the result is the maximum possible
(b) the result is the minimum possible
Solution:
(a) Maximum possible results:
Largest numbers are 5 and 3
Multiplying two negative (-2 and-6) gives a positive result
Few Combinations:
(5 × 3) – (-6) + (-2) = 15 + 6 – 2 = 19
– (-6 ×-2) + 5 + 3 = 12 + 5 + 3 = 20
5 ×(3-(-6)+(-2)) = 5 × (3 + 6 – 2) = 5 × 7 = 35
So, maximum possible result is 5 × (3- (-6) + (-2)) = 35
(b) Minimum Possible result:
Few Combinations:
3 × (-6) – 5 -(-2) = -18 – 5 + 2 = -21
– 5 × (-6)- 3-(-2) = -30-3 + 2 = -31
(-6-5) × 3 + (-2) = (-11) × 3 – 2 = – 33-2 = -35
(-6-5) × (-2) + 3 = (-11) × (-2) + 3 = 22 + 3 = 25
(5-3) × (-6) – ((-2) = 2 × (-6) + 2 = -12 + 2 = -10
Maximum possible results (-6-5) × 3 + (-2) = -35
Question 16.
Fill in the blanks in at least 5 different ways with integers:
Solution:
(a) 9 + (-3) × 6 = -36
0 + -6 × 6 = -36
4 + (-5 × 8) = -36
-4 + -8 × 4 = -36
12 + -8 × 6 = -36
(b) (4-1) × 4 = 12
(10-6) × 3 = 12
(5-1) × 3 = 12
(12-10) × 6 = 12
(24-10) × 6 = 12
(c) 3-(5-1) = -1
-2-(10-9) = -1
0-(5-4) = -1
1-(7-5) = -1
-1-(2-2) = -1