NCERT Solutions for Class 7
Maths
Chapter 1 – Integers
Exercise 1.1
1. Following number line shows the temperature in degree Celsius
 C at
different places on a particular day:
(a) Observe this number line and write the temperature of the places
marked on it.
Ans: First find the distance between two marks on the number line. Observe
that there is total five partitions between every two consecutive numbers
written say 0,5 . Hence, one partition will represent 1 C .
Therefore, the temperature of the places marked on it is:
Place
Temperature
Banglore
22 C
Ooty
14 C
Shimla
5C
Srinagar
2 C
Lahulspiti
8 C
(b) What is the temperature difference between the hottest and the coldest
places among the above?
Class VII Maths
www.vedantu.com
1
Ans: From the table we can see that the temperature of the hottest place is
Bangalore which is 22 C and the temperature of the coldest place Lahulspiti
8 C and their difference will be
Difference = Temperature of Bangalore – Temperature of Lahulspiti
Difference  22   8   22+8=30 C .
(c) What is the temperature difference between Lahulspiti and Srinagar?
Ans: From the table we can see that the temperature of the Srinagar is 2 C
and the temperature of Lahulspiti is 8 C and their difference will be
Difference = Temperature of Srinagar – Temperature of Lahulspiti
Difference  2   8   2  8  6 C .
(d) Can we say temperature of Srinagar and Shimla taken together is less
than the temperature at Shimla? Is it also less than the temperature at
Srinagar?
Ans: From the table we can see that the temperature of the Srinagar is 2 C
and the temperature of Shimla is 5 C and their sum will be
Sum = Temperature of Shimla – Temperature of Srinagar
Sum  5   2   5  2  3 C . …..(1)
Temperature of Shimla  5 C . …..(2)
From (1) and (2), the temperature of Srinagar and Shimla taken together is less
than the temperature at Shimla.
Temperature of Srinagar  2 C . …..(3)
From (1) and (3), the temperature of Srinagar and Shimla taken together is not
less than the temperature at Srinagar.
2. In a quiz, positive marks are given for correct answers and negative marks
are given for incorrect answers. If Jack’s scores in five successive rounds
were 25, 5, 10,15 and 10 , what was his total at the end?
Ans: Jack’s scores in five successive rounds are 25, 5, 10,15 and 10 . Then the
total marks got by Jack will be the sum of all these marks i.e.,
Class VII Maths
www.vedantu.com
2
25   5   10   15  10  25  5  10  15  10
 25  15  5  10  35
Therefore, Jack scored 35 marks in the quiz.
3. At Srinagar temperature was 5 C on Monday and then it dropped by
2 C on Tuesday. What was the temperature of Srinagar on Tuesday? On
Wednesday, it rose by 4 C . What was the temperature on this day?
Ans: It is given that on Monday, temperature at Srinagar is 5 C . On Tuesday,
temperature dropped by 2 C i.e., the temperature becomes 2 less than Monday.
 Temperature on Tuesday  5 C  2 C  7 C ….. (1)
On Wednesday, temperature rose up by 4 C i.e., the temperature becomes 4 more
than Tuesday. Hence from (1),
 Temperature on Wednesday  7 C+4 C  3 C
4. A plane is flying at the height of 5000m above the sea level. At a particular
point, it is exactly above a submarine floating 1200 m below the sea level.
What is the vertical distance between them?
Ans: Given, height of a place above the sea level is 5000 m and a floating
submarine is 1200 m below the sea level. Let us denote the distance above the sea
level by  sign and the distance below the sea level by  sign.
Hence, Distance of plane from sea level  5000m
Distance of submarine from sea level  1200m
 The vertical distance between the plane and the submarine is
5000   1200   6200 m .
Class VII Maths
www.vedantu.com
3
Thus, the vertical distance between the plane and the submarine is 6200 m .
5. Mohan deposits ₹ 2,000 in his bank account and withdraws ₹ 1,642 from it,
the next day. If withdrawal of amount from the account is represented by a
negative integer, then how will you represent the amount deposited? Find
the balance in Mohan’s accounts after the withdrawal?
Ans: Deposit and Withdrawal are opposite of each other. Therefore, if we are
representing withdrawal by  , we should represent deposit by  .
Given, Deposit amount  ₹ 2,000 and Withdrawal amount  ₹1,642 . In integer
representation, deposit  2000 and withdrawal  1642 .
 Balance  2,000  1,642  ₹ 358
Therefore, the balance in Mohan’s account after withdrawal is ₹ 358 .
6. Rita goes 20 km towards east from a point A to the point B. From B, she
moves 30 km towards west along the same road. If the distance towards
east is represented by a positive integer, then, how will you represent the
distance travelled towards west? By which integer will you represent her
final position from A?
Ans: East and West are opposite of each other. Therefore, if we are representing
east by  , we should represent west by  .
Let A be the point of origin i.e., A will lie at 0 on the number line.
Given, Rita goes 20 km towards east from a point A to the point B . From B , she
moves 30 km towards west along the same road.
Plotting the given information on the number line we get,
Distance from A to B  20 km
Distance from B to final point  30 km
Distance from A to final point  20   30   10km .
Therefore, Rita is 10 km west from her initial position A and will be represented
by the integer 10 .
Class VII Maths
www.vedantu.com
4
7. In a magic square each row, column and diagonal have the same sum.
Check which of the following is a magic square.
i.
Ans: In a magic square each row, column and diagonal have the same sum. So
let us check whether the given square is a magic square or not.
Sum of row 1  5   1   4   0 …..(1)
Sum of row 2   5   2   7  7  7  0 …..(2)
Sum of row 3  0  3   3  3  3  0 …..(3)
Sum of column 1 5   5  0  5  5  0 …..(4)
Sum of column 2   1   2   3  3  3  0 …..(5)
Sum of column 3   4   7   3  7  7  0 …..(6)
Sum of diagonal 1 5   2    3  5  5  0 …..(7)
Sum of diagonal 2   4    2   0  6 …..(8)
From (1) to (8) we can conclude that this box is not a magic square because all
the sums are not equal.
ii.
Ans: In a magic square each row, column and diagonal have the same sum. So
let us check whether the given square is a magic square or not.
Sum of row 1 1   10   0  1  10  9 …..(1)
Sum of row 2   4    3   2   7  2  9 …..(2)
Sum of row 3   6   4   7   2  7  9 …..(3)
Sum of column 1 1   10   0  1  10  9 …..(4)
Class VII Maths
www.vedantu.com
5
Sum of column 2   10    3  4  13  4  9 …..(5)
Sum of column 3  0   2    7   0  9  9 …..(6)
Sum of diagonal 1 1   3   7   1  10  9 …..(7)
Sum of diagonal 2  0   3   6   9 …..(8)
From (1) to (8) we can conclude that this box is magic square because all the
sums are equal.
8. Verify a   b   a  b for the following values of aand b :
(i). Given, a  21,b  18
Ans: Given, a  21,b  18 . We have to verify a   b   a  b .
LHS  21   18  21  18  39 ….. (1)
RHS  a  b  21  18  39 ….. (2)
From (1) and (2), since LHS  RHS therefore, it is verified that
a   b   a  b .
(ii). Given, a  118,b  125
Ans: Given, a  118,b  125 . We have to verify a   b   a  b .
LHS  118   125  118  125  243 ….. (1)
RHS  118  125  243 ….. (2)
From (1) and (2), since LHS  RHS therefore, it is verified that
a   b   a  b .
(iii). Given, a  75,b  84
Ans: Given, a  75,b  84 . We have to verify a   b   a  b .
LHS  75   84   75  84  159 ….. (1)
RHS  a  b  75  84  159 ….. (2)
From (1) and (2), since LHS  RHS therefore, it is verified that
a   b   a  b .
(iv). Given, a  28,b  11
Class VII Maths
www.vedantu.com
6
Ans: Given, a  28,b  11 . We have to verify a   b   a  b .
LHS  28   11  28  11  39 ….. (1)
RHS  28  11  39 ….. (2)
From (1) and (2), since LHS  RHS therefore, it is verified that
a   b   a  b .
9. Use the sign of ,  or  in the blank to make the statements true:
(a)  8    4  __  8    4  .
Ans:  8   4   12 and  8   4   4
  12   4 .
  8   4    8   4 
(b)  3   7   19  __15  8   9 
Ans:  3  7  19   15 and 15  8   9   2
 15  2
  3  7  19   15  8   9 
(c) 23  41  11__ 23  41  11
Ans: 23  41  11  7 and 23  41  11  29
 7  29
 23  41  11  23  41  11
(d) 39   24    15  __ 36   52    36 
Ans: 39   24   15  0 and 36   52    36   20
 0  20
 39   24   15  36   52    36 
(e)  231  79  51__  399   159  81
Ans:  231  79  51  101 and  399   159  81  159
 101  159
Class VII Maths
www.vedantu.com
7
  231  79  51   399   159  81
10. A water tank has steps inside it. A monkey is sitting on the topmost step
(i.e., the first step). The water level is at the ninth step:
(i). He jumps 3 steps down and then jump back 2 steps up. In how many
jumps will he reach the water level?
Ans: Given that the monkey jumps 3 steps down and jumps back 2 steps up.
We can represent the path of the monkey as:
First jump  1  3  4 steps
Second jump  4  2  2 steps
Third jump  2  3  5 steps
Fourth jump  5  2  3 steps
Fifth jump  3  3  6 steps
Sixth jump  6  2  4 steps
Seventh jump  4  3  7 steps
Eighth jump  7  2  5 steps
Class VII Maths
www.vedantu.com
8
Ninth jump  5  3  8 steps
Tenth jump  8  2  6 steps
Eleventh jump  6  3  9 steps
Therefore, the monkey will reach ninth steps in 11 jumps.
(ii). After drinking water, he wants to go back. For this, he jumps 4 steps up
and then jumps back 2 steps down in every move. In how many jumps will
he reach back the top step?
Ans: Given that the monkey jumps four steps up and them jumps down 2 steps
down. We can represent the path of the monkey as:
First jump  9  4  5 steps
Second jump  5  2  7 steps
Third jump  7  4  3 steps
Fourth jump  3  2  5 steps
Fifth jump  5  4  1 steps
Therefore, the monkey reaches back on the first step in fifth jump.
(iii). If the number of steps moved down is represented by negative integers
and the number of steps move up by positive integers, represent his moves
in part (i) and (ii) by completing the following:
(a) 3  2  ..........  8
(b) 4  2  ..........  8
In (a) the sum (−8) represent going down by eight steps. So, what will the
sum 8 in (b) represent?
Class VII Maths
www.vedantu.com
9
Ans: (a) 3  2  3  2  3  2  3  2  3  2  3  2  3  2  3  2  8
(b) 4  2  4  2  4  2  4  2  8
Thus, sum 8 in (b) represents going up by eight steps.
Exercise 1.2
1. Write down a pair of integers whose:
a) sum is 7
Ans: The pair of integers whose sum is 7 is  4, 3 i.e., 4   3  7 .
b) difference is 10
Ans: The pair of integers whose difference is 10 is  3,7  i.e., 3  7  10
c) sum is 0
Ans: The pair of integers whose sum is 0 is  30,30  i.e., 30  30  0 .
2. Find the integers in the below questions:
a) Write a pair of negative integers whose difference gives 8 .
Ans: The pair of negative integers whose difference is 8 is  1, 9  i.e.,
1   9   1  9  8
b) Write a negative integer and a positive integer whose sum is 5 .
Ans: The pair of negative and positive integers whose sum is 5 is  9,4  i.e.,
 9   4  5 .
c) Write a negative integer and a positive integer whose difference is 3 .
Ans: The pair of negative and positive integers whose difference is 3 is
 1,2 i.e.,  1  2  1  2  3 .
3. In a quiz, team A scored 40,10,0 and team B scores 10,0, 40 in three
successive rounds. Which team scored more? Can we say that we can add
integers in any order?
Class VII Maths
www.vedantu.com
10
Ans: Given that the team A scored 40,10,0 . Therefore, total score of the
team A  40  10  0  30 .
Given that the team B scored 10,0, 40 .Therefore, total score of Team B
 10  0   40   30 .
Therefore, scores of both teams are same and we can add integers in any order due
to commutative property.
4. Fill in the blanks to make the following statements true:
(i).  5    8    8    ......
Ans: Using Commutative Property we can fill the blank as 5 .
  5   8   8   5 .
(ii). 53   ......   53
Ans: Using Zero additive property we can fill the blank as 0 .
 53  0  53
(iii). 17   ......   0
Ans: Using Zero Additive identity we can fill the blank as 17 .
 17   17   0
(iv). 13   12     ......  13   12    7  
Ans: Using Associative property we can fill the blank as 7 .
 13   12   7   13   12   7 
(v).  4   15   3     4  15   ......
Ans: Using Associative property we can fill the blank as 3 .
  4  15   3  4  15   3
Exercise 1.3
1. Find each of the following products:
Class VII Maths
www.vedantu.com
11
a) 3   1
Ans: While multiplying a negative integer and a positive integer, multiply them
as whole numbers and then put a minus sign    before the product i.e.,
3   1  3
b)  1  225
Ans: While multiplying a negative integer and a positive integer, multiply them
as whole numbers and then put a minus sign    before the product i.e.,
 1  225  225
c)  21   30 
Ans: While multiplying two negative integers, multiply them as whole numbers
and then put a plus sign    before the product i.e.,  21   30   630
d)  316    1
Ans: While multiplying two negative integers, multiply them as whole numbers
and then put a plus sign    before the product i.e.,  316    1  316
e)  15   0   30 
Ans: While multiplying two negative integers, multiply them as whole numbers
and then put a plus sign    before the product i.e.,  15  0   18  0
f)  12    11   10 
Ans: While multiplying two negative integers, multiply them as whole numbers
and then put a plus sign    before the product i.e.,
 12   11  10  132 10  1320
g) 9   3    6 
Class VII Maths
www.vedantu.com
12
Ans: While multiplying two negative integers, multiply them as whole
numbers and then put a plus sign    before the product i.e.,
9   3   6  9 18  162
h)  18    5    4 
Ans: While multiplying two negative integers, multiply them as whole numbers
and then put a plus sign    before the product i.e.,
 18   5   4  90   4  ….. (1)
While multiplying a negative integer and a positive integer, multiply them as
whole numbers and then put a minus sign    before the product i.e., from (1),
 18   5   4  360
i)  1   2    3   4
Ans: While multiplying two negative integers, multiply them as whole numbers
and then put a plus sign    before the product and while multiplying a
negative integer and a positive integer, multiply them as whole numbers and
then put a minus sign    before the product i.e.,
 1   2   3  4  2   12  24
j)  3   6    2    1
Ans: While multiplying two negative integers, multiply them as whole numbers
and then put a plus sign    before the product and while multiplying a
negative integer and a positive integer, multiply them as whole numbers and
then put a minus sign    before the product i.e.,
 3   6   2   1  18   2   36
2. Verify the following:
a) 18  7   3    18  7  18   3  
Ans: Given expression, 18  7   3  18  7  18   3 .
Class VII Maths
www.vedantu.com
13
Simplifying the given expression by first solving the square brackets.
While multiplying a negative integer and a positive integer, multiply them as
whole numbers and then put a minus sign    before the product.
 18   4  126   54
 72  72
 L.H.S.  R.H.S.
Hence verified.
b)  21   4    6     21   4     21   6  
Ans: Given expression,  21   4   6   21   4   21   6
Simplifying the given expression by first solving the square brackets.
While multiplying two negative integers, multiply them as whole numbers and
then put a plus sign    before the product
  21   10   84  126
 210  210
 L.H.S.  R.H.S.
Hence verified.
3. Solve the following:
(i). For any integer a , what is  1  a equal to?
Ans:  1  a  a, where a is an integer.
(ii). Determine the integer whose product with  1 is:
(a) 22
Ans: The integer whose product with 1 is 22 is 22 i.e.,  1   22   22 .
(b) 37
Ans: The integer whose product with 1 is 37 is 37 i.e.,  1   37   37
(c) 0
Ans: The integer whose product with 1 is 0 is 0 i.e., 1 0  0 .
Class VII Maths
www.vedantu.com
14
4. Starting from  1  5 write various products showing some patterns to
show  1   1  1 .
Ans: Consider the product,  1  5  5
Also,  1  4  4 ,  1  3  3 ,  1  2  2 ,  1  1  1, etc.
Thus, we can observe that the product of one negative integer and one positive
integer is negative integer.
Similarly,  1   1  1 i.e., the product of two negative integers is a positive
integer.
5. Find the product, using suitable properties:
(a) 26   48    48    36 
Ans: 26   48   48   36  …..(1)
Taking 48 common from (1) using distributive property we get,
  48  26   36
  48   10 
 480
(b) 8  53   125 
Ans: 8  53   125 …..(1)
Using Commutative property on (1) we get,
 53  8   125
 53   1000 
  53000
(c) 15   25    4    10 
Ans: 15   25   4    10 
 15   25   4   10 …..(1)
Using Commutative property on (1) we get,
 15   4    25  10
Class VII Maths
www.vedantu.com
15
 15   4   250
  15000
(d)  41   102 
Ans:  41  102 
  41 100  2 …..(1)
Using distributive property on (1) we get,
  41 100   41  2
  4100   82 
  4182
(e) 625   35    625   65
Ans: 625   35   625  65 …..(1)
Taking 625 common from (1) using distributive property we get,
 625   35   65
 625   100 
  62500
(f) 7   50  2 
Ans: 7   50  2  ….. (1)
Using distributive property on (1) we get,
 7  50  7  2
 350  14  336
(g)  17    29 
Ans:  17    29 
  17    30  1 …..(1)
Using distributive property on (1) we get,
  17    30    17   1
Class VII Maths
www.vedantu.com
16
 510   17 
 493
(h)  57    19   57
Ans:  57    19   57
  57    19   57  1
 57  19  57  1
…..(1)
Taking 57 common from (1) using distributive property we get,
 57  19  1
 57  20  1140
6. A certain freezing process requires that room temperature be lowered
from 40 C at the rate of 5 C every hour. What will be the room
temperature 10 hours after the process begins?
Ans: It is given that the present temperature of the room is 40 C . Also, the
temperature is decreasing the temperature every hour by 5 C .
Therefore, temperature after 1 hour is 40  5  35 C .
Temperature after 2 hours is 40  2   5   30 C .
Temperature after 3 hours is 40  3   5   25 C .
Similarly, temperature after n hours is 40  n  5 C . ….. (1)
Form (1), room temperature after 10 hours  40  10   5 C
  40  50 C  10 C
Therefore, the room temperature 10 hours after the process begins is 10 C .
7. In a class test containing 10 questions, 5 marks are awarded for every
correct answer and  2  marks are awarded for every incorrect answer
and 0 for questions not attempted.
(i). Mohan gets four correct and six incorrect answers. What is his score?
Class VII Maths
www.vedantu.com
17
Ans: Given that out of 10 questions, Mohan gets 4 correct and 6 incorrect
answers.
Marks for 1 correct answer  5
Marks for 4 correct answers  4  5  20 …..(1)
Marks for 1 wrong answer  2
Marks for 6 wrong answers  6   2   12 …..(2)
Therefore from (1) and (2), total scores of Mohan   20    12   8 .
(ii). Reshma gets five correct answers and five incorrect answers, what is
her score?
Ans: Given that out of 10 questions, Reshma gets 5 correct and 5 incorrect
answers.
Marks for 1 correct answer  5
Marks for 5 correct answers  5  5  25 …..(1)
Marks for 1 wrong answer  2
Marks for 5 wrong answers  5   2   10 …..(2)
Therefore from (1) and (2), total scores of Resham   25   10   15 .
(iii). Heena gets two correct and five incorrect answers out of seven
questions she attempts. What is her score?
Ans: Given that out of 10 questions, Heena gets 2 correct, 5 incorrect and 3
un-attempted.
Marks for 1 correct answer  5
Marks for 2 correct answers  2  5  10 …..(1)
Marks for 1 wrong answer  2
Marks for 5 wrong answers  5   2   10 …..(2)
Marks for 1 un-attempt question  0
Marks for 3 un-attempt questions  0  3  0 …..(3)
Therefore from (1), (2) and (3), total scores of Heena  10    10   0  0 .
8. 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.
Class VII Maths
www.vedantu.com
18
(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?
Ans: Given that profit of 1 bag of white cement  ₹ 8
Therefore, Profit on selling 3000 bags of white cement
 3000  8  ₹ 24,000 ….. (1)
And loss of 1 bag of grey cement  ₹ 5 .
Therefore, Loss on selling 5000 bags of grey cement
 5000  5  ₹ 25,000 …..(2)
Let us denote profit by  and loss by  . Then from (1) and (2),
Total selling  24000   25000   1000 .
Therefore, his total loss on selling cement bags is ₹ 1,000 .
(b) What is the number of white cement bags it must sell to have neither profit
nor loss. If the number of grey bags sold is 6,400 bags.
Ans: Let the number of bags of white cement bags sold be x .
Given that profit of 1 bag of white cement  ₹ 8
Therefore, Profit on selling x bags of white cement ₹ 8x ….. (1)
And loss of 1 bag of grey cement  ₹ 5 .
Therefore, Loss on selling 6400 bags of grey cement
 6400  5  ₹ 32,000 …..(2)
To have neither profit nor loss, from (1) and (2) we get,
8x  32000
32000
x
8
 x  4000
Thus, he must sell 4000 white cement bags to have neither profit nor loss.
9. Replace the blank with an integer to make it a true statement:
a)  3   _____  27
Ans: Let the number in the blank be x , then
 3  x  27
x
27
3
Class VII Maths
www.vedantu.com
19
 x  9
 3   9   27
b) 5  _____  35
Ans: Let the number in the blank be x , then
5  x  35
35
x
5
 x  7
 5   7   35
c) _____  8   56
Ans: Let the number in the blank be x , then
x   8  56
56
8
x7
 7   8  56
x
d) _____  12   132
Ans: Let the number in the blank be x , then
x   12   132
132
12
 x  11
 11   12   132
x
Exercise 1.4
1. Evaluate each of the following:
(a)  30   10
Class VII Maths
www.vedantu.com
20
Ans: While dividing a negative integer by a positive integer, divide them as
whole numbers and then put a minus sign    before the quotient. Therefore,
 30  10  
30
 3 .
10
(b) 50   5 
Ans: While dividing a positive integer by a negative integer, divide them as
whole numbers and then put a minus sign    before the quotient. Therefore,
50   5   
50
 10 .
5
(c)  36    9 
Ans: While dividing a positive integer by a positive integer, divide them as
whole numbers and then put a plus sign    before the quotient. Therefore,
 36    9  
36
 4.
9
(d)  49   49
Ans: While dividing a negative integer by a positive integer, divide them as
whole numbers and then put a minus sign    before the quotient. Therefore,
 49   49  
49
 1 .
49
(e) 13   2   1
Ans: Simplifying the given expression, 13   2  1  13   1 ….(1)
While dividing a positive integer by a negative integer, divide them as whole
numbers and then put a minus sign    before the quotient. Therefore from (1),
13   1  13 .
(f) 0   12 
Class VII Maths
www.vedantu.com
21
Ans: While dividing 0 by any integer, the quotient is 0 . Therefore
0   12   0 .
(g)  31   30    1 
Ans: While dividing a positive integer by a positive integer, divide them as
whole numbers and then put a plus sign    before the quotient. Therefore,
 30   1  30 . …..(1)
Hence from (1),  31   30   1   31   30 ….. (2)
While dividing a negative integer by a positive integer, divide them as whole
numbers and then put a minus sign    before the quotient. Therefore from (2),
 31   30    1  
31
.
30
(h)  36   12  3
Ans: While dividing a negative integer by a positive integer, divide them as
whole numbers and then put a minus sign    before the quotient. Therefore,
36
 3 …..(1)
12
While dividing a negative integer by a positive integer, divide them as whole
numbers and then put a minus sign    before the quotient. Therefore from (1),
 36  12  
 36  12  3   3  3  1.
(i)  6   5   2   1
Ans: Simplifying the given expression,  6  5   2  1   1   1
….(1)
While dividing a negative integer by a negative integer, divide them as whole
numbers and then put a plus sign    before the quotient. Therefore from (1),
 6  5   2  1  1.
Class VII Maths
www.vedantu.com
22
2. Verify that a   b  c    a  b    a  c 
1
for each of the following values
2
of a,b and c.
(a) a  12,b  4,c  2
Ans: Given, a  12,b  4,c  2 . We have to verify
a   b  c  a  b  a  c .
Substituting the given values of a,b,c in LHS we get,
L.H.S.  12   4  2   12   2  ….. (1)
While dividing a positive integer by a negative integer, divide them as whole
numbers and then put a minus sign    before the quotient. Therefore from (1),
12   4  2   6 . …..(2)
Substituting the given values of a,b,c in RHS we get,
R.H.S.  12   4 + 12  2 …..(3)
While dividing a negative integer by a positive integer, divide them as whole
numbers and then put a minus sign    before the quotient. Therefore, from
(3), 12   4 + 12  2    3 +6
 RHS  3 …..(4)
From (2) and (4), we can conclude that L.H.S.  R.H.S. Hence verified.
(b) a   10  ,b  1,c  1
Ans: Given, a  10,b  1,c  1. We have to verify
a   b  c  a  b  a  c .
Substituting the given values of a,b,c in LHS we get,
L.H.S.  10  1  1  10   2  ….. (1)
While dividing a negative integer by a positive integer, divide them as whole
numbers and then put a minus sign    before the quotient. Therefore from (1),
10  1  1  5 . …..(2)
Substituting the given values of a,b,c in RHS we get,
R.H.S.   10  1 +  10  1 …..(3)
Class VII Maths
www.vedantu.com
23
While dividing a negative integer by a positive integer, divide them as whole
numbers and then put a minus sign    before the quotient. Therefore, from
(3),  10  1 +  10  1   10  +  10 
 RHS  20 …..(4)
From (2) and (4), we can conclude that L.H.S.  R.H.S. Hence verified.
3. Fill in the blanks:
(a) 369  _____  369
Ans: While dividing any integer by 1 , the quotient is the original integer.
Therefore, 369  1  369 .
(b)  75   _____   1
Ans: While dividing any integer by itself, the quotient is 1 . Also, while
dividing a negative integer by a positive integer, divide them as whole numbers
and then put a minus sign    before the quotient. Therefore,
 75  75   1
(c)  206   _____  1
Ans: While dividing any integer by itself, the quotient is 1 . Also, while
dividing a negative integer by a negative integer, divide them as whole
numbers and then put a plus sign    before the quotient. Therefore,
 206    206   1
(d)  87   _____  87
Ans: While dividing any integer by 1 , the quotient is the original integer. Also,
while dividing a negative integer by a negative integer, divide them as whole
numbers and then put a plus sign    before the quotient. Therefore,
 87    1  87 .
(e) _____  1  87
Class VII Maths
www.vedantu.com
24
Ans: While dividing any integer by 1 , the quotient is the original integer.
Therefore,  87   1  87 .
(f) _____  48  1
Ans: While dividing any integer by 1 , the quotient is the original integer. Also,
while dividing a negative integer by a positive integer, divide them as whole
numbers and then put a minus sign    before the quotient. Therefore,
 48  48  1
(g) 20  _____  2
Ans: While dividing a positive integer by a negative integer, divide them as
whole numbers and then put a minus sign    before the quotient. Therefore,
20   10   2
(h) _____   4   3
Ans: While dividing a negative integer by a positive integer, divide them as
whole numbers and then put a minus sign    before the quotient.
 12   4  3
4. Write five pairs of integers  a,b  such that a  b  3 . One such pair is
 6, 2 
because 6   2    3  .
Ans: Five pair of integers  a,b  such that a  b  3 are:
i.
ii.
iii.
iv.
v.
The pair of integers  9,3 is such that  9   3  3 .
The pair of integers  15,5  is such that  15  5  3 .
The pair of integers 12, 4  is such that 12   4   3 .
The pair of integers  3,1 is such that  3  1  3 .
The pair of integers  21, 7  is such that 21   7   3 .
Class VII Maths
www.vedantu.com
25
5. The temperature at noon was 10 C above zero. If it decreases at the rate of
2 C per hour until mid-night, at what time would the temperature be 8 C
below zero? What would be the temperature at mid-night?
Ans: Given that the temperature at 12 noon is 10 C above the zero i.e., 10 C .
Now, the temperature decreases 2 C each hour until midnight. Therefore,
following number line represents the temperature path:
Temperature at 12 noon  10 C .
Temperature at 1PM  10  2  C  8 C .
Temperature at 2PM   8  2  C  6 C .
Temperature at 3PM   6  2  C  4 C .
Therefore, temperature at nPM  10  2n  C .
According to the question, 10  2n   8 ….(1)
Solving (1) by rearranging terms we get,
10  8  2n
n 9
Therefore, at 9pm the temperature would be 8 C below 0 C .
6. In 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?
Ans: Given that Radhika scored 20 marks and she has got 12 correct answers.
Let the number of incorrect answers be x then,
Marks given for one correct answer  3
Marks given for 12 correct answers  3  12  36 ….. (1)
Class VII Maths
www.vedantu.com
26
Marks given for one wrong answer  2
Marks given for x wrong answers  2x ….. (2)
Also, Radhika scored 20 marks. Hence from (1) and (2),
20  36   2x  ….. (3)
Solving (3) by rearranging terms we get,
2x  16
 x 8
Therefore, Radhika has attempted 8 incorrect questions.
(ii). Mohini scores  5  marks in this test, though she has got 7 correct
answers. How many questions has she attempted incorrectly?
Ans: Given that Mohini scored 5 marks and she has got 7 correct answers.
Let the number of incorrect answers be x then,
Marks given for one correct answer  3
Marks given for 7 correct answers  3  7  21 ….. (1)
Marks given for one wrong answer  2
Marks given for x wrong answers  2x ….. (2)
Also, Mohini scored 5 marks. Hence from (1) and (2),
5  21   2x  ….. (3)
Solving (3) by rearranging terms we get,
2x  26
 x  13
Therefore, Mohini has attempted 13 incorrect questions.
7. An elevator descends into a mine shaft at the rate of 6 m/min . If the
descent starts from 10 above the ground level, how long will it take to
reach -350 m ?
Ans: Given that the starting position of mine shaft is 10 m above the ground.
And its destination is 350m below the ground.
Let us denote the distance above the ground by  sign and the distance below the
ground by  sign. Therefore, starting from 10m it has to go 350m .
Total distance covered by mine shaft  10m   350  m  10  350  360m …..(1)
Now, it is given that elevator takes 1 min to cover the distance of 6m i.e.,
Class VII Maths
www.vedantu.com
27
Time taken to cover a distance of 6 m  1 minute. …..(2)
Hence from (2),
1
Time taken to cover a distance of 1m  minute. …..(3)
6
Hence from (1) and (3), time taken to cover a distance of 360m
1
  360  60 minutes  1 hour.
6
Therefore, in 1 hour the mine shaft reaches 350 below the ground.
Class VII Maths
www.vedantu.com
28