{"id":139416,"date":"2021-02-20T08:15:35","date_gmt":"2021-02-20T08:15:35","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=139416"},"modified":"2023-09-15T01:52:41","modified_gmt":"2023-09-15T01:52:41","slug":"ncert-solutions-for-7th-class-maths-chapter-1-integers","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-7th-class-maths-chapter-1-integers\/","title":{"rendered":"NCERT Solutions for 7th Class Maths: Chapter 1-Integers"},"content":{"rendered":"\n

Class 7: Maths Chapter 1 solutions. Complete Class 7 Maths Chapter 1 Notes.<\/p>\n\n\n\n

NCERT Solutions for 7th Class Maths: Chapter 1-Integers<\/strong><\/h2>\n\n\n\n

NCERT 7th Maths Chapter 1, class 7 Maths Chapter 1 solutions<\/p>\n\n\n\n

Page: 4<\/h4>\n\n\n\n

Exercise 1.1 <\/h4>\n\n\n\n

1. Following number line shows the temperature in degree celsius (co<\/sup>) at different places on a particular day.<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

(a) Observe this number line and write the temperature of the places marked on it.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

By observing the number line, we can find the temperature of the cities as follows,<\/p>\n\n\n\n

Temperature at the Lahulspiti is -8o<\/sup>C<\/p>\n\n\n\n

Temperature at the Srinagar is -2o<\/sup>C<\/p>\n\n\n\n

Temperature at the Shimla is 5o<\/sup>C<\/p>\n\n\n\n

Temperature at the Ooty is 14o<\/sup>C<\/p>\n\n\n\n

Temperature at the Bengaluru is 22o<\/sup>C<\/p>\n\n\n\n

(b) What is the temperature difference between the hottest and the coldest places among the above?<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the number line we observe that,<\/p>\n\n\n\n

The temperature at the hottest place i.e., Bengaluru is 22o<\/sup>C<\/p>\n\n\n\n

The temperature at the coldest place i.e., Lahulspiti is -8o<\/sup>C<\/p>\n\n\n\n

Temperature difference between hottest and coldest place is = 22o<\/sup>C \u2013 (-8o<\/sup>C)<\/p>\n\n\n\n

= 22o<\/sup>C + 8o<\/sup>C<\/p>\n\n\n\n

= 30o<\/sup>C<\/p>\n\n\n\n

Hence, the temperature difference between the hottest and the coldest place is 30o<\/sup>C.<\/p>\n\n\n\n

(c) What is the temperature difference between Lahulspiti and Srinagar?<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the given number line,<\/p>\n\n\n\n

The temperature at the Lahulspiti is -8o<\/sup>C<\/p>\n\n\n\n

The temperature at the Srinagar is -2o<\/sup>C<\/p>\n\n\n\n

\u2234The temperature difference between Lahulspiti and Srinagar is = -2o<\/sup>C \u2013 (8o<\/sup>C)<\/p>\n\n\n\n

= \u2013 2O<\/sup>C + 8o<\/sup>C<\/p>\n\n\n\n

= 6o<\/sup>C<\/p>\n\n\n\n

(d) Can we say temperature of Srinagar and Shimla taken together is less than the<\/strong><\/p>\n\n\n\n

temperature at Shimla? Is it also less than the temperature at Srinagar?<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the given number line,<\/p>\n\n\n\n

The temperature at Srinagar =-2o<\/sup>C<\/p>\n\n\n\n

The temperature at Shimla = 5o<\/sup>C<\/p>\n\n\n\n

The temperature of Srinagar and Shimla taken together is = \u2013 2o<\/sup>C + 5o<\/sup>C<\/p>\n\n\n\n

= 3o<\/sup>C<\/p>\n\n\n\n

\u2234 5o<\/sup>C > 3o<\/sup>C<\/p>\n\n\n\n

So, the temperature of Srinagar and Shimla taken together is less than the temperature at Shimla.<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

3o<\/sup> > -2o<\/sup><\/p>\n\n\n\n

No, the temperature of Srinagar and Shimla taken together is not less than the temperature of Srinagar.<\/p>\n\n\n\n

NCERT 7th Maths Chapter 1, class 7 Maths Chapter 1 solutions<\/p>\n\n\n\n

2. In a quiz, positive marks are given for correct answers and negative marks are given<\/strong><\/p>\n\n\n\n

for incorrect answers. If Jack\u2019s scores in five successive rounds were 25, \u2013 5, \u2013 10,<\/strong><\/p>\n\n\n\n

15 and 10, what was his total at the end?<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question,<\/p>\n\n\n\n

Jack\u2019s score in five successive rounds are 25, -5, -10, 15 and 10<\/p>\n\n\n\n

The total score of Jack at the end will be = 25 + (-5) + (-10) + 15 + 10<\/p>\n\n\n\n

= 25 \u2013 5 \u2013 10 + 15 + 10<\/p>\n\n\n\n

= 50 \u2013 15<\/p>\n\n\n\n

= 35<\/p>\n\n\n\n

\u2234Jack\u2019s total score at the end is 35.<\/p>\n\n\n\n

3. At Srinagar temperature was \u2013 5\u00b0C on Monday and then it dropped by 2\u00b0C on Tuesday. What was the temperature of Srinagar on Tuesday? On Wednesday, it rose by 4\u00b0C. What was the temperature on this day?<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question,<\/p>\n\n\n\n

Temperature on Monday at Srinagar = -5o<\/sup>C<\/p>\n\n\n\n

Temperature on Tuesday at Srinagar is dropped by 2o<\/sup>C = Temperature on Monday \u2013 2o<\/sup>C<\/p>\n\n\n\n

= -5o<\/sup>C \u2013 2o<\/sup>C<\/p>\n\n\n\n

= -7o<\/sup>C<\/p>\n\n\n\n

Temperature on Wednesday at Srinagar is rose by 4o<\/sup>C = Temperature on Tuesday + 4o<\/sup>C<\/p>\n\n\n\n

= -7o<\/sup>C + 4o<\/sup>C<\/p>\n\n\n\n

= -3o<\/sup>C<\/p>\n\n\n\n

Thus, the temperature on Tuesday and Wednesday was -7o<\/sup>C and -3o<\/sup>C respectively.<\/p>\n\n\n\n

NCERT 7th Maths Chapter 1, class 7 Maths Chapter 1 solutions<\/p>\n\n\n\n

4. A plane is flying at the height of 5000 m 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?<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question,<\/p>\n\n\n\n

Plane is flying at the height = 5000 m<\/p>\n\n\n\n

Depth of Submarine = -1200 m<\/p>\n\n\n\n

The vertical distance between plane and submarine = 5000 m \u2013 (- 1200) m<\/p>\n\n\n\n

= 5000 m + 1200 m<\/p>\n\n\n\n

= 6200 m<\/p>\n\n\n\n

NCERT 7th Maths Chapter 1, class 7 Maths Chapter 1 solutions<\/p>\n\n\n\n

5. Mohan deposits \u20b9 2,000 in his bank account and withdraws \u20b9 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\u2019s account after the withdrawal.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Withdrawal of amount from the account is represented by a negative integer.<\/p>\n\n\n\n

Then, deposit of amount to the account is represented by a positive integer.<\/p>\n\n\n\n

From the question,<\/p>\n\n\n\n

Total amount deposited in bank account by the Mohan = \u20b9 2000<\/p>\n\n\n\n

Total amount withdrawn from the bank account by the Mohan = \u2013 \u20b9 1642<\/p>\n\n\n\n

Balance in Mohan\u2019s account after the withdrawal = amount deposited + amount withdrawn<\/p>\n\n\n\n

= \u20b9 2000 + (-\u20b9 1642)<\/p>\n\n\n\n

= \u20b9 2000 \u2013 \u20b9 1642<\/p>\n\n\n\n

= \u20b9 358<\/p>\n\n\n\n

Hence, the balance in Mohan\u2019s account after the withdrawal is \u20b9 358<\/p>\n\n\n\n

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?<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question, it is given that<\/p>\n\n\n\n

A positive integer represents the distance towards the east.<\/p>\n\n\n\n

Then, distance travelled towards the west will be represented by a negative integer.<\/p>\n\n\n\n

Rita travels a distance in east direction = 20 km<\/p>\n\n\n\n

Rita travels a distance in west direction = \u2013 30 km<\/p>\n\n\n\n

\u2234Distance travelled from A = 20 + (- 30)<\/p>\n\n\n\n

= 20 \u2013 30<\/p>\n\n\n\n

= -10 km<\/p>\n\n\n\n

Hence, we will represent the distance travelled by Rita from point A by a negative integer, i.e. \u2013 10 km<\/p>\n\n\n\n

NCERT 7th Maths Chapter 1, class 7 Maths Chapter 1 solutions<\/p>\n\n\n\n

7. In a magic square each row, column and diagonal have the same sum. Check which of the following is a magic square<\/strong>.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

First we consider the square (i)<\/p>\n\n\n\n

By adding the numbers in each rows we get,<\/p>\n\n\n\n

= 5 + (- 1) + (- 4) = 5 \u2013 1 \u2013 4 = 5 \u2013 5 = 0<\/p>\n\n\n\n

= -5 + (-2) + 7 = \u2013 5 \u2013 2 + 7 = -7 + 7 = 0<\/p>\n\n\n\n

= 0 + 3 + (-3) = 3 \u2013 3 = 0<\/p>\n\n\n\n

By adding the numbers in each columns we get,<\/p>\n\n\n\n

= 5 + (- 5) + 0 = 5 \u2013 5 = 0<\/p>\n\n\n\n

= (-1) + (-2) + 3 = -1 \u2013 2 + 3 = -3 + 3 = 0<\/p>\n\n\n\n

= -4 + 7 + (-3) = -4 + 7 \u2013 3 = -7 + 7 = 0<\/p>\n\n\n\n

By adding the numbers in diagonals we get,<\/p>\n\n\n\n

= 5 + (-2) + (-3) = 5 \u2013 2 \u2013 3 = 5 \u2013 5 = 0<\/p>\n\n\n\n

= -4 + (-2) + 0 = \u2013 4 \u2013 2 = -6<\/p>\n\n\n\n

Because sum of one diagonal is not equal to zero,<\/p>\n\n\n\n

So, (i) is not a magic square<\/p>\n\n\n\n

Now, we consider the square (ii)<\/p>\n\n\n\n

By adding the numbers in each rows we get,<\/p>\n\n\n\n

= 1 + (-10) + 0 = 1 \u2013 10 + 0 = -9<\/p>\n\n\n\n

= (-4) + (-3) + (-2) = -4 \u2013 3 \u2013 2 = -9<\/p>\n\n\n\n

= (-6) + 4 + (-7) = -6 + 4 \u2013 7 = -13 + 4 = -9<\/p>\n\n\n\n

By adding the numbers in each columns we get,<\/p>\n\n\n\n

= 1 + (-4) + (-6) = 1 \u2013 4 \u2013 6 = 1 \u2013 10 = -9<\/p>\n\n\n\n

= (-10) + (-3) + 4 = -10 \u2013 3 + 4 = -13 + 4<\/p>\n\n\n\n

= 0 + (-2) + (-7) = 0 \u2013 2 \u2013 7 = -9<\/p>\n\n\n\n

By adding the numbers in diagonals we get,<\/p>\n\n\n\n

= 1 + (-3) + (-7) = 1 \u2013 3 \u2013 7 = 1 \u2013 10 = -9<\/p>\n\n\n\n

= 0 + (-3) + (-6) = 0 \u2013 3 \u2013 6 = -9<\/p>\n\n\n\n

This (ii) square is a magic square, because sum of each row, each column and diagonal is equal to -9.<\/p>\n\n\n\n

NCERT 7th Maths Chapter 1, class 7 Maths Chapter 1 solutions<\/p>\n\n\n\n

8. Verify a \u2013 (\u2013 b) = a + b for the following values of a and b.<\/strong><\/p>\n\n\n\n

(i) a = 21, b = 18<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question,<\/p>\n\n\n\n

a = 21 and b = 18<\/p>\n\n\n\n

To verify a \u2013 (- b) = a + b<\/p>\n\n\n\n

Let us take Left Hand Side (LHS) = a \u2013 (- b)<\/p>\n\n\n\n

= 21 \u2013 (- 18)<\/p>\n\n\n\n

= 21 + 18<\/p>\n\n\n\n

= 39<\/p>\n\n\n\n

Now, Right Hand Side (RHS) = a + b<\/p>\n\n\n\n

= 21 + 18<\/p>\n\n\n\n

= 39<\/p>\n\n\n\n

By comparing LHS and RHS<\/p>\n\n\n\n

LHS = RHS<\/p>\n\n\n\n

39 = 39<\/p>\n\n\n\n

Hence, the value of a and b is verified.<\/p>\n\n\n\n

(ii) a = 118, b = 125<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question,<\/p>\n\n\n\n

a = 118 and b = 125<\/p>\n\n\n\n

To verify a \u2013 (- b) = a + b<\/p>\n\n\n\n

Let us take Left Hand Side (LHS) = a \u2013 (- b)<\/p>\n\n\n\n

= 118 \u2013 (- 125)<\/p>\n\n\n\n

= 118 + 125<\/p>\n\n\n\n

= 243<\/p>\n\n\n\n

Now, Right Hand Side (RHS) = a + b<\/p>\n\n\n\n

= 118 + 125<\/p>\n\n\n\n

= 243<\/p>\n\n\n\n

By comparing LHS and RHS<\/p>\n\n\n\n

LHS = RHS<\/p>\n\n\n\n

243 = 243<\/p>\n\n\n\n

Hence, the value of a and b is verified.<\/p>\n\n\n\n

(iii) a = 75, b = 84<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question,<\/p>\n\n\n\n

a = 75 and b = 84<\/p>\n\n\n\n

To verify a \u2013 (- b) = a + b<\/p>\n\n\n\n

Let us take Left Hand Side (LHS) = a \u2013 (- b)<\/p>\n\n\n\n

= 75 \u2013 (- 84)<\/p>\n\n\n\n

= 75 + 84<\/p>\n\n\n\n

= 159<\/p>\n\n\n\n

Now, Right Hand Side (RHS) = a + b<\/p>\n\n\n\n

= 75 + 84<\/p>\n\n\n\n

= 159<\/p>\n\n\n\n

By comparing LHS and RHS<\/p>\n\n\n\n

LHS = RHS<\/p>\n\n\n\n

159 = 159<\/p>\n\n\n\n

Hence, the value of a and b is verified.<\/p>\n\n\n\n

(iv) a = 28, b = 11<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question,<\/p>\n\n\n\n

a = 28 and b = 11<\/p>\n\n\n\n

To verify a \u2013 (- b) = a + b<\/p>\n\n\n\n

Let us take Left Hand Side (LHS) = a \u2013 (- b)<\/p>\n\n\n\n

= 28 \u2013 (- 11)<\/p>\n\n\n\n

= 28 + 11<\/p>\n\n\n\n

= 39<\/p>\n\n\n\n

Now, Right Hand Side (RHS) = a + b<\/p>\n\n\n\n

= 28 + 11<\/p>\n\n\n\n

= 39<\/p>\n\n\n\n

By comparing LHS and RHS<\/p>\n\n\n\n

LHS = RHS<\/p>\n\n\n\n

39 = 39<\/p>\n\n\n\n

Hence, the value of a and b is verified.<\/p>\n\n\n\n

NCERT 7th Sanskrit Chapter 1, class 7 Sanskrit Chapter 1 solutions<\/p>\n\n\n\n

9. Use the sign of >, < or = in the box to make the statements true.<\/strong><\/p>\n\n\n\n

(a) (-8) + (-4) [ ] (-8) \u2013 (-4)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us take Left Hand Side (LHS) = (-8) + (-4)<\/p>\n\n\n\n

= -8 \u2013 4<\/p>\n\n\n\n

= -12<\/p>\n\n\n\n

Now, Right Hand Side (RHS) = (-8) \u2013 (-4)<\/p>\n\n\n\n

= -8 + 4<\/p>\n\n\n\n

= -4<\/p>\n\n\n\n

By comparing LHS and RHS<\/p>\n\n\n\n

LHS < RHS<\/p>\n\n\n\n

-12 < -4<\/p>\n\n\n\n

\u2234 (-8) + (-4) [<] (-8) \u2013 (-4)<\/p>\n\n\n\n

(b) (-3) + 7 \u2013 (19) [ ] 15 \u2013 8 + (-9)<\/p>\n\n\n\n

Solution:-<\/p>\n\n\n\n

Let us take Left Hand Side (LHS) = (-3) + 7 \u2013 19<\/p>\n\n\n\n

= -3 + 7 \u2013 19<\/p>\n\n\n\n

= -22 + 7<\/p>\n\n\n\n

= -15<\/p>\n\n\n\n

Now, Right Hand Side (RHS) = 15 \u2013 8 + (-9)<\/p>\n\n\n\n

= 15 \u2013 8 \u2013 9<\/p>\n\n\n\n

= 15 \u2013 17<\/p>\n\n\n\n

= -2<\/p>\n\n\n\n

By comparing LHS and RHS<\/p>\n\n\n\n

LHS < RHS<\/p>\n\n\n\n

-15 < -2<\/p>\n\n\n\n

\u2234 (-3) + 7 \u2013 (19) [<] 15 \u2013 8 + (-9)<\/p>\n\n\n\n

(c) 23 \u2013 41 + 11 [ ] 23 \u2013 41 \u2013 11<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us take Left Hand Side (LHS) = 23 \u2013 41 + 11<\/p>\n\n\n\n

= 34 \u2013 41<\/p>\n\n\n\n

= \u2013 7<\/p>\n\n\n\n

Now, Right Hand Side (RHS) = 23 \u2013 41 \u2013 11<\/p>\n\n\n\n

= 23 \u2013 52<\/p>\n\n\n\n

= \u2013 29<\/p>\n\n\n\n

By comparing LHS and RHS<\/p>\n\n\n\n

LHS > RHS<\/p>\n\n\n\n

\u2013 7 > -29<\/p>\n\n\n\n

\u2234 23 \u2013 41 + 11 [>] 23 \u2013 41 \u2013 11<\/p>\n\n\n\n

(d) 39 + (-24) \u2013 (15) [ ] 36 + (-52) \u2013 (- 36)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us take Left Hand Side (LHS) = 39 + (-24) \u2013 15<\/p>\n\n\n\n

= 39 \u2013 24 \u2013 15<\/p>\n\n\n\n

= 39 \u2013 39<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Now, Right Hand Side (RHS) = 36 + (-52) \u2013 (- 36)<\/p>\n\n\n\n

= 36 \u2013 52 + 36<\/p>\n\n\n\n

= 72 \u2013 52<\/p>\n\n\n\n

= 20<\/p>\n\n\n\n

By comparing LHS and RHS<\/p>\n\n\n\n

LHS < RHS<\/p>\n\n\n\n

0 < 20<\/p>\n\n\n\n

\u2234 39 + (-24) \u2013 (15) [<] 36 + (-52) \u2013 (- 36)<\/p>\n\n\n\n

(e) \u2013 231 + 79 + 51 [ ] -399 + 159 + 81<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us take Left Hand Side (LHS) = \u2013 231 + 79 + 51<\/p>\n\n\n\n

= \u2013 231 + 130<\/p>\n\n\n\n

= -101<\/p>\n\n\n\n

Now, Right Hand Side (RHS) = \u2013 399 + 159 + 81<\/p>\n\n\n\n

= \u2013 399 + 240<\/p>\n\n\n\n

= \u2013 159<\/p>\n\n\n\n

By comparing LHS and RHS<\/p>\n\n\n\n

LHS > RHS<\/p>\n\n\n\n

-101 > -159<\/p>\n\n\n\n

\u2234 \u2013 231 + 79 + 51 [>] -399 + 159 + 81<\/p>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

(i) He jumps 3 steps down and then jumps back 2 steps up. In how many jumps will he reach the water level?<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us consider steps moved down are represented by positive integers and then, steps moved up are represented by negative integers.<\/p>\n\n\n\n

Initially monkey is sitting on the top most step i.e., first step<\/p>\n\n\n\n

In 1st<\/sup> jump monkey will be at step = 1 + 3 = 4 steps<\/p>\n\n\n\n

In 2nd<\/sup> jump monkey will be at step = 4 + (-2) = 4 \u2013 2 = 2 steps<\/p>\n\n\n\n

In 3rd<\/sup> jump monkey will be at step = 2 + 3 = 5 steps<\/p>\n\n\n\n

In 4th <\/sup>jump monkey will be at step = 5 + (-2) = 5 \u2013 2 = 3 steps<\/p>\n\n\n\n

In 5th<\/sup> jump monkey will be at step = 3 + 3 = 6 steps<\/p>\n\n\n\n

In 6th<\/sup> jump monkey will be at step = 6 + (-2) = 6 \u2013 2 = 4 steps<\/p>\n\n\n\n

In 7th<\/sup> jump monkey will be at step = 4 + 3 = 7 steps<\/p>\n\n\n\n

In 8th<\/sup> jump monkey will be at step = 7 + (-2) = 7 \u2013 2 = 5 steps<\/p>\n\n\n\n

In 9th <\/sup>jump monkey will be at step = 5 + 3 = 8 steps<\/p>\n\n\n\n

In 10th<\/sup> jump monkey will be at step = 8 + (-2) = 8 \u2013 2 = 6 steps<\/p>\n\n\n\n

In 11th<\/sup> jump monkey will be at step = 6 + 3 = 9 steps<\/p>\n\n\n\n

\u2234Monkey took 11 jumps (i.e., 9th<\/sup> step) to reach the water level<\/p>\n\n\n\n

(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?<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us consider steps moved down are represented by positive integers and then, steps moved up are represented by negative integers.<\/p>\n\n\n\n

Initially monkey is sitting on the ninth step i.e., at the water level<\/p>\n\n\n\n

In 1st<\/sup> jump monkey will be at step = 9 + (-4) = 9 \u2013 4 = 5 steps<\/p>\n\n\n\n

In 2nd<\/sup> jump monkey will be at step = 5 + 2 = 7 steps<\/p>\n\n\n\n

In 3rd<\/sup> jump monkey will be at step = 7 + (-4) = 7 \u2013 4 = 3 steps<\/p>\n\n\n\n

In 4th <\/sup>jump monkey will be at step = 3 + 2 = 5 steps<\/p>\n\n\n\n

In 5th<\/sup> jump monkey will be at step = 5 + (-4) = 5 \u2013 4 = 1 step<\/p>\n\n\n\n

\u2234Monkey took 5 jumps to reach back the top step i.e., first step.<\/p>\n\n\n\n

(iii) If the number of steps moved down is represented by negative integers and the number of steps moved up by positive integers, represent his moves in part (i) and (ii) by completing the following; (a) \u2013 3 + 2 \u2013 \u2026 = \u2013 8 (b) 4 \u2013 2 + \u2026 = 8. In (a) the sum (\u2013 8) represents going down by eight steps. So, what will the sum 8 in (b) represent?<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question, it is given that<\/p>\n\n\n\n

If the number of steps moved down is represented by negative integers and the number of steps moved up by positive integers.<\/p>\n\n\n\n

Monkey moves in part (i)<\/p>\n\n\n\n

= \u2013 3 + 2 \u2013 \u2026\u2026\u2026.. = \u2013 8<\/p>\n\n\n\n

Then LHS = \u2013 3 + 2 \u2013 3 + 2 \u2013 3 + 2 \u2013 3 + 2 \u2013 3 + 2 \u2013 3<\/p>\n\n\n\n

= \u2013 18 + 10<\/p>\n\n\n\n

= \u2013 8<\/p>\n\n\n\n

RHS = -8<\/p>\n\n\n\n

\u2234Moves in part (i) represents monkey is going down 8 steps. Because negative integer.<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Monkey moves in part (ii)<\/p>\n\n\n\n

= 4 \u2013 2 + \u2026\u2026\u2026.. = 8<\/p>\n\n\n\n

Then LHS = 4 \u2013 2 + 4 \u2013 2 + 4<\/p>\n\n\n\n

= 12 \u2013 4<\/p>\n\n\n\n

= 8<\/p>\n\n\n\n

RHS = 8<\/p>\n\n\n\n

\u2234Moves in part (ii) represents monkey is going up 8 steps. Because positive integer.<\/p>\n\n\n\n

\n\n\n\n

Exercise 1.2 Page: 9<\/h2>\n\n\n\n

1. Write down a pair of integers whose:<\/strong><\/p>\n\n\n\n

(a) sum is -7<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

= \u2013 4 + (-3)<\/p>\n\n\n\n

= \u2013 4 \u2013 3 \u2026 [\u2235 (+ \u00d7 \u2013 = -)]<\/p>\n\n\n\n

= \u2013 7<\/p>\n\n\n\n

(b) difference is \u2013 10<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

= -25 \u2013 (-15)<\/p>\n\n\n\n

= \u2013 25 + 15 \u2026 [\u2235 (- \u00d7 \u2013 = +)]<\/p>\n\n\n\n

= -10<\/p>\n\n\n\n

(c) sum is 0<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

= 4 + (-4)<\/p>\n\n\n\n

= 4 \u2013 4<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

2. (a) Write a pair of negative integers whose difference gives 8<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

= (-5) \u2013 (- 13)<\/p>\n\n\n\n

= -5 + 13 \u2026 [\u2235 (- \u00d7 \u2013 = +)]<\/p>\n\n\n\n

= 8<\/p>\n\n\n\n

(b) Write a negative integer and a positive integer whose sum is \u2013 5.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

= -25 + 20<\/p>\n\n\n\n

= -5<\/p>\n\n\n\n

(c) Write a negative integer and a positive integer whose difference is \u2013 3.<\/strong><\/p>\n\n\n\n

Solution<\/strong>:-<\/p>\n\n\n\n

= \u2013 6 \u2013 (-3)<\/p>\n\n\n\n

= \u2013 6 + 3 \u2026 [\u2235 (- \u00d7 \u2013 = +)]<\/p>\n\n\n\n

= \u2013 3<\/p>\n\n\n\n

3. In a quiz, team A scored \u2013 40, 10, 0 and team B scored 10, 0, \u2013 40 in three successive rounds. Which team scored more? Can we say that we can add integers in any order?<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question, it is given that<\/p>\n\n\n\n

Score of team A = -40, 10, 0<\/p>\n\n\n\n

Total score obtained by team A = \u2013 40 + 10 + 0<\/p>\n\n\n\n

= \u2013 30<\/p>\n\n\n\n

Score of team B = 10, 0, -40<\/p>\n\n\n\n

Total score obtained by team B = 10 + 0 + (-40)<\/p>\n\n\n\n

= 10 + 0 \u2013 40<\/p>\n\n\n\n

= \u2013 30<\/p>\n\n\n\n

Thus, the score of the both A team and B team is same.<\/p>\n\n\n\n

Yes, we can say that we can add integers in any order.<\/p>\n\n\n\n

4. Fill in the blanks to make the following statements true:<\/strong><\/p>\n\n\n\n

(i) (\u20135) + (\u2013 8) = (\u2013 8) + (\u2026\u2026\u2026\u2026)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume the missing integer be x,<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

= (\u20135) + (\u2013 8) = (\u2013 8) + (x)<\/p>\n\n\n\n

= \u2013 5 \u2013 8 = \u2013 8 + x<\/p>\n\n\n\n

= \u2013 13 = \u2013 8 + x<\/p>\n\n\n\n

By sending \u2013 8 from RHS to LHS it becomes 8,<\/p>\n\n\n\n

= \u2013 13 + 8 = x<\/p>\n\n\n\n

= x = \u2013 5<\/p>\n\n\n\n

Now substitute the x value in the blank place,<\/p>\n\n\n\n

(\u20135) + (\u2013 8) = (\u2013 8) + (- 5) \u2026 [This equation is in the form of Commutative law of Addition]<\/p>\n\n\n\n

(ii) \u201353 + \u2026\u2026\u2026\u2026 = \u201353<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume the missing integer be x,<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

= \u201353 + x = \u201353<\/p>\n\n\n\n

By sending \u2013 53 from LHS to RHS it becomes 53,<\/p>\n\n\n\n

= x = -53 + 53<\/p>\n\n\n\n

= x = 0<\/p>\n\n\n\n

Now substitute the x value in the blank place,<\/p>\n\n\n\n

= \u201353 + 0 = \u201353 \u2026 [This equation is in the form of Closure property of Addition]<\/p>\n\n\n\n

(iii) 17 + \u2026\u2026\u2026\u2026 = 0<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume the missing integer be x,<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

= 17 + x = 0<\/p>\n\n\n\n

By sending 17 from LHS to RHS it becomes -17,<\/p>\n\n\n\n

= x = 0 \u2013 17<\/p>\n\n\n\n

= x = \u2013 17<\/p>\n\n\n\n

Now substitute the x value in the blank place,<\/p>\n\n\n\n

= 17 + (-17) = 0 \u2026 [This equation is in the form of Closure property of Addition]<\/p>\n\n\n\n

= 17 \u2013 17 = 0<\/p>\n\n\n\n

(iv) [13 + (\u2013 12)] + (\u2026\u2026\u2026\u2026) = 13 + [(\u201312) + (\u20137)]<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume the missing integer be x,<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

= [13 + (\u2013 12)] + (x) = 13 + [(\u201312) + (\u20137)]<\/p>\n\n\n\n

= [13 \u2013 12] + (x) = 13 + [\u201312 \u20137]<\/p>\n\n\n\n

= [1] + (x) = 13 + [-19]<\/p>\n\n\n\n

= 1 + (x) = 13 \u2013 19<\/p>\n\n\n\n

= 1 + (x) = -6<\/p>\n\n\n\n

By sending 1 from LHS to RHS it becomes -1,<\/p>\n\n\n\n

= x = -6 \u2013 1<\/p>\n\n\n\n

= x = -7<\/p>\n\n\n\n

Now substitute the x value in the blank place,<\/p>\n\n\n\n

= [13 + (\u2013 12)] + (-7) = 13 + [(\u201312) + (\u20137)] \u2026 [This equation is in the form of Associative property of Addition]<\/p>\n\n\n\n

(v) (\u2013 4) + [15 + (\u20133)] = [\u2013 4 + 15] +\u2026\u2026\u2026\u2026<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume the missing integer be x,<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

= (\u2013 4) + [15 + (\u20133)] = [\u2013 4 + 15] + x<\/p>\n\n\n\n

= (\u2013 4) + [15 \u2013 3)] = [\u2013 4 + 15] + x<\/p>\n\n\n\n

= (-4) + [12] = [11] + x<\/p>\n\n\n\n

= 8 = 11 + x<\/p>\n\n\n\n

By sending 11 from RHS to LHS it becomes -11,<\/p>\n\n\n\n

= 8 \u2013 11 = x<\/p>\n\n\n\n

= x = -3<\/p>\n\n\n\n

Now substitute the x value in the blank place,<\/p>\n\n\n\n

= (\u2013 4) + [15 + (\u20133)] = [\u2013 4 + 15] + -3 \u2026 [This equation is in the form of Associative property of Addition]<\/p>\n\n\n\n

\n\n\n\n

Exercise 1.3 Page: 21<\/p>\n\n\n\n

1. Find each of the following products:<\/strong><\/p>\n\n\n\n

(a) 3 \u00d7 (\u20131)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

By the rule of Multiplication of integers,<\/p>\n\n\n\n

= 3 \u00d7 (-1)<\/p>\n\n\n\n

= -3 \u2026 [\u2235 (+ \u00d7 \u2013 = -)]<\/p>\n\n\n\n

(b) (\u20131) \u00d7 225<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

By the rule of Multiplication of integers,<\/p>\n\n\n\n

= (-1) \u00d7 225<\/p>\n\n\n\n

= -225 \u2026 [\u2235 (- \u00d7 + = -)]<\/p>\n\n\n\n

(c) (\u201321) \u00d7 (\u201330)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

By the rule of Multiplication of integers,<\/p>\n\n\n\n

= (-21) \u00d7 (-30)<\/p>\n\n\n\n

= 630 \u2026 [\u2235 (- \u00d7 \u2013 = +)]<\/p>\n\n\n\n

(d) (\u2013316) \u00d7 (\u20131)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

By the rule of Multiplication of integers,<\/p>\n\n\n\n

= (-316) \u00d7 (-1)<\/p>\n\n\n\n

= 316 \u2026 [\u2235 (- \u00d7 \u2013 = +)]<\/p>\n\n\n\n

(e) (\u201315) \u00d7 0 \u00d7 (\u201318)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

By the rule of Multiplication of integers,<\/p>\n\n\n\n

= (\u201315) \u00d7 0 \u00d7 (\u201318)<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

\u2235Any integer is multiplied with zero and the answer is zero itself.<\/p>\n\n\n\n

(f) (\u201312) \u00d7 (\u201311) \u00d7 (10)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

By the rule of Multiplication of integers,<\/p>\n\n\n\n

= (\u201312) \u00d7 (-11) \u00d7 (10)<\/p>\n\n\n\n

First multiply the two numbers having same sign,<\/p>\n\n\n\n

= 132 \u00d7 10 \u2026 [\u2235 (- \u00d7 \u2013 = +)]<\/p>\n\n\n\n

= 1320<\/p>\n\n\n\n

(g) 9 \u00d7 (\u20133) \u00d7 (\u2013 6)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

By the rule of Multiplication of integers,<\/p>\n\n\n\n

= 9 \u00d7 (-3) \u00d7 (-6)<\/p>\n\n\n\n

First multiply the two numbers having same sign,<\/p>\n\n\n\n

= 9 \u00d7 18 \u2026 [\u2235 (- \u00d7 \u2013 = +)]<\/p>\n\n\n\n

= 162<\/p>\n\n\n\n

(h) (\u201318) \u00d7 (\u20135) \u00d7 (\u2013 4)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

By the rule of Multiplication of integers,<\/p>\n\n\n\n

= (-18) \u00d7 (-5) \u00d7 (-4)<\/p>\n\n\n\n

First multiply the two numbers having same sign,<\/p>\n\n\n\n

= 90 \u00d7 -4 \u2026 [\u2235 (- \u00d7 \u2013 = +)]<\/p>\n\n\n\n

= \u2013 360 \u2026 [\u2235 (+ \u00d7 \u2013 = -)]<\/p>\n\n\n\n

(i) (\u20131) \u00d7 (\u20132) \u00d7 (\u20133) \u00d7 4<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

By the rule of Multiplication of integers,<\/p>\n\n\n\n

= [(\u20131) \u00d7 (\u20132)] \u00d7 [(\u20133) \u00d7 4]<\/p>\n\n\n\n

= 2 \u00d7 (-12) \u2026 [\u2235 (- \u00d7 \u2013 = +), (- \u00d7 + = -)]<\/p>\n\n\n\n

= \u2013 24<\/p>\n\n\n\n

(j) (\u20133) \u00d7 (\u20136) \u00d7 (\u20132) \u00d7 (\u20131)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

By the rule of Multiplication of integers,<\/p>\n\n\n\n

= [(\u20133) \u00d7 (\u20136)] \u00d7 [(\u20132) \u00d7 (\u20131)]<\/p>\n\n\n\n

First multiply the two numbers having same sign,<\/p>\n\n\n\n

= 18 \u00d7 2 \u2026 [\u2235 (- \u00d7 \u2013 = +)<\/p>\n\n\n\n

= 36<\/p>\n\n\n\n

2. Verify the following:<\/strong><\/p>\n\n\n\n

(a) 18 \u00d7 [7 + (\u20133)] = [18 \u00d7 7] + [18 \u00d7 (\u20133)]<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the given equation,<\/p>\n\n\n\n

Let us consider the Left Hand Side (LHS) first = 18 \u00d7 [7 + (\u20133)]<\/p>\n\n\n\n

= 18 \u00d7 [7 \u2013 3]<\/p>\n\n\n\n

= 18 \u00d7 4<\/p>\n\n\n\n

= 72<\/p>\n\n\n\n

Now, consider the Right Hand Side (RHS) = [18 \u00d7 7] + [18 \u00d7 (\u20133)]<\/p>\n\n\n\n

= [126] + [-54]<\/p>\n\n\n\n

= 126 \u2013 54<\/p>\n\n\n\n

= 72<\/p>\n\n\n\n

By comparing LHS and RHS,<\/p>\n\n\n\n

72 = 72<\/p>\n\n\n\n

LHS = RHS<\/p>\n\n\n\n

Hence, the given equation is verified.<\/p>\n\n\n\n

(b) (\u201321) \u00d7 [(\u2013 4) + (\u2013 6)] = [(\u201321) \u00d7 (\u2013 4)] + [(\u201321) \u00d7 (\u2013 6)]<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the given equation,<\/p>\n\n\n\n

Let us consider the Left Hand Side (LHS) first = (\u201321) \u00d7 [(\u2013 4) + (\u2013 6)]<\/p>\n\n\n\n

= (-21) \u00d7 [-4 \u2013 6]<\/p>\n\n\n\n

= (-21) \u00d7 [-10]<\/p>\n\n\n\n

= 210<\/p>\n\n\n\n

Now, consider the Right Hand Side (RHS) = [(\u201321) \u00d7 (\u2013 4)] + [(\u201321) \u00d7 (\u2013 6)]<\/p>\n\n\n\n

= [84] + [126]<\/p>\n\n\n\n

= 210<\/p>\n\n\n\n

By comparing LHS and RHS,<\/p>\n\n\n\n

210 = 210<\/p>\n\n\n\n

LHS = RHS<\/p>\n\n\n\n

Hence, the given equation is verified.<\/p>\n\n\n\n

3. (i) For any integer a, what is (\u20131) \u00d7 a equal to?<\/p>\n\n\n\n

Solution:-<\/p>\n\n\n\n

= (-1) \u00d7 a = -a<\/p>\n\n\n\n

Because, when we multiplied any integer a with -1, then we get additive inverse of that integer.<\/p>\n\n\n\n

(ii). Determine the integer whose product with (\u20131) is<\/strong><\/p>\n\n\n\n

(a) \u201322<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Now, multiply -22 with (-1), we get<\/p>\n\n\n\n

= -22 \u00d7 (-1)<\/p>\n\n\n\n

= 22<\/p>\n\n\n\n

Because, when we multiplied integer -22 with -1, then we get additive inverse of that integer.<\/p>\n\n\n\n

(b) 37<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Now, multiply 37 with (-1), we get<\/p>\n\n\n\n

= 37 \u00d7 (-1)<\/p>\n\n\n\n

= -37<\/p>\n\n\n\n

Because, when we multiplied integer 37 with -1, then we get additive inverse of that integer.<\/p>\n\n\n\n

(c) 0<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Now, multiply 0 with (-1), we get<\/p>\n\n\n\n

= 0 \u00d7 (-1)<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

Because, the product of negative integers and zero give zero only.<\/p>\n\n\n\n

4. Starting from (\u20131) \u00d7 5, write various products showing some pattern to show<\/strong><\/p>\n\n\n\n

(\u20131) \u00d7 (\u20131) = 1.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

The various products are,<\/p>\n\n\n\n

= -1 \u00d7 5 = -5<\/p>\n\n\n\n

= -1 \u00d7 4 = -4<\/p>\n\n\n\n

= -1 \u00d7 3 = -3<\/p>\n\n\n\n

= -1 \u00d7 2 = -2<\/p>\n\n\n\n

= -1 \u00d7 1 = -1<\/p>\n\n\n\n

= -1 \u00d7 0 = 0<\/p>\n\n\n\n

= -1 \u00d7 -1 = 1<\/p>\n\n\n\n

We concluded that the product of one negative integer and one positive integer is negative integer. Then, the product of two negative integers is a positive integer.<\/p>\n\n\n\n

NCERT 7th Maths Chapter 1, class 7 Maths Chapter 1 solutions<\/p>\n\n\n\n

5. Find the product, using suitable properties:<\/strong><\/p>\n\n\n\n

(a) 26 \u00d7 (\u2013 48) + (\u2013 48) \u00d7 (\u201336)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

The given equation is in the form of Distributive law of Multiplication over Addition.<\/p>\n\n\n\n

= a \u00d7 (b + c) = (a \u00d7 b) + (a \u00d7 c)<\/p>\n\n\n\n

Let, a = -48, b = 26, c = -36<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

= 26 \u00d7 (\u2013 48) + (\u2013 48) \u00d7 (\u201336)<\/p>\n\n\n\n

= -48 \u00d7 (26 + (-36)<\/p>\n\n\n\n

= -48 \u00d7 (26 \u2013 36)<\/p>\n\n\n\n

= -48 \u00d7 (-10)<\/p>\n\n\n\n

= 480 \u2026 [\u2235 (- \u00d7 \u2013 = +)<\/p>\n\n\n\n

(b) 8 \u00d7 53 \u00d7 (\u2013125)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

The given equation is in the form of Commutative law of Multiplication.<\/p>\n\n\n\n

= a \u00d7 b = b \u00d7 a<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

= 8 \u00d7 [53 \u00d7 (-125)]<\/p>\n\n\n\n

= 8 \u00d7 [(-125) \u00d7 53]<\/p>\n\n\n\n

= [8 \u00d7 (-125)] \u00d7 53<\/p>\n\n\n\n

= [-1000] \u00d7 53<\/p>\n\n\n\n

= \u2013 53000<\/p>\n\n\n\n

(c) 15 \u00d7 (\u201325) \u00d7 (\u2013 4) \u00d7 (\u201310)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

The given equation is in the form of Commutative law of Multiplication.<\/p>\n\n\n\n

= a \u00d7 b = b \u00d7 a<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

= 15 \u00d7 [(\u201325) \u00d7 (\u2013 4)] \u00d7 (\u201310)<\/p>\n\n\n\n

= 15 \u00d7 [100] \u00d7 (\u201310)<\/p>\n\n\n\n

= 15 \u00d7 [-1000]<\/p>\n\n\n\n

= \u2013 15000<\/p>\n\n\n\n

(d) (\u2013 41) \u00d7 102<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

The given equation is in the form of Distributive law of Multiplication over Addition.<\/p>\n\n\n\n

= a \u00d7 (b + c) = (a \u00d7 b) + (a \u00d7 c)<\/p>\n\n\n\n

= (-41) \u00d7 (100 + 2)<\/p>\n\n\n\n

= (-41) \u00d7 100 + (-41) \u00d7 2<\/p>\n\n\n\n

= \u2013 4100 \u2013 82<\/p>\n\n\n\n

= \u2013 4182<\/p>\n\n\n\n

(e) 625 \u00d7 (\u201335) + (\u2013 625) \u00d7 65<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

The given equation is in the form of Distributive law of Multiplication over Addition.<\/p>\n\n\n\n

= a \u00d7 (b + c) = (a \u00d7 b) + (a \u00d7 c)<\/p>\n\n\n\n

= 625 \u00d7 [(-35) + (-65)]<\/p>\n\n\n\n

= 625 \u00d7 [-100]<\/p>\n\n\n\n

= \u2013 62500<\/p>\n\n\n\n

(f) 7 \u00d7 (50 \u2013 2)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

The given equation is in the form of Distributive law of Multiplication over Subtraction.<\/p>\n\n\n\n

= a \u00d7 (b \u2013 c) = (a \u00d7 b) \u2013 (a \u00d7 c)<\/p>\n\n\n\n

= (7 \u00d7 50) \u2013 (7 \u00d7 2)<\/p>\n\n\n\n

= 350 \u2013 14<\/p>\n\n\n\n

= 336<\/p>\n\n\n\n

(g) (\u201317) \u00d7 (\u201329)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

The given equation is in the form of Distributive law of Multiplication over Addition.<\/p>\n\n\n\n

= a \u00d7 (b + c) = (a \u00d7 b) + (a \u00d7 c)<\/p>\n\n\n\n

= (-17) \u00d7 [-30 + 1]<\/p>\n\n\n\n

= [(-17) \u00d7 (-30)] + [(-17) \u00d7 1]<\/p>\n\n\n\n

= [510] + [-17]<\/p>\n\n\n\n

= 493<\/p>\n\n\n\n

(h) (\u201357) \u00d7 (\u201319) + 57<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

The given equation is in the form of Distributive law of Multiplication over Addition.<\/p>\n\n\n\n

= a \u00d7 (b + c) = (a \u00d7 b) + (a \u00d7 c)<\/p>\n\n\n\n

= (57 \u00d7 19) + (57 \u00d7 1)<\/p>\n\n\n\n

= 57 [19 + 1]<\/p>\n\n\n\n

= 57 \u00d7 20<\/p>\n\n\n\n

= 1140<\/p>\n\n\n\n

NCERT 7th Maths Chapter 1, class 7 Maths Chapter 1 solutions<\/p>\n\n\n\n

6. A certain freezing process requires that room temperature be lowered from 40\u00b0C at the rate of 5\u00b0C every hour. What will be the room temperature 10 hours after the process begins?<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question, it is given that<\/p>\n\n\n\n

Let us take the lowered temperature as negative,<\/p>\n\n\n\n

Initial temperature = 40o<\/sup>C<\/p>\n\n\n\n

Change in temperature per hour = -5o<\/sup>C<\/p>\n\n\n\n

Change in temperature after 10 hours = (-5) \u00d7 10 = -50o<\/sup>C<\/p>\n\n\n\n

\u2234The final room temperature after 10 hours of freezing process = 40o<\/sup>C + (-50o<\/sup>C)<\/p>\n\n\n\n

= -10o<\/sup>C<\/p>\n\n\n\n

NCERT 7th Maths Chapter 1, class 7 Maths Chapter 1 solutions<\/p>\n\n\n\n

7. In a class test containing 10 questions, 5 marks are awarded for every correct answer and (\u20132) marks are awarded for every incorrect answer and 0 for questions not attempted.<\/strong><\/p>\n\n\n\n

(i) Mohan gets four correct and six incorrect answers. What is his score?<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question,<\/p>\n\n\n\n

Marks awarded for 1 correct answer = 5<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

Total marks awarded for 4 correct answer = 4 \u00d7 5 = 20<\/p>\n\n\n\n

Marks awarded for 1 wrong answer = -2<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

Total marks awarded for 6 wrong answer = 6 \u00d7 -2 = -12<\/p>\n\n\n\n

\u2234Total score obtained by Mohan = 20 + (-12)<\/p>\n\n\n\n

= 20 \u2013 12<\/p>\n\n\n\n

= 8<\/p>\n\n\n\n

(ii) Reshma gets five correct answers and five incorrect answers, what is her score?<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question,<\/p>\n\n\n\n

Marks awarded for 1 correct answer = 5<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

Total marks awarded for 5 correct answer = 5 \u00d7 5 = 25<\/p>\n\n\n\n

Marks awarded for 1 wrong answer = -2<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

Total marks awarded for 5 wrong answer = 5 \u00d7 -2 = -10<\/p>\n\n\n\n

\u2234Total score obtained by Reshma = 25 + (-10)<\/p>\n\n\n\n

= 25 \u2013 10<\/p>\n\n\n\n

= 15<\/p>\n\n\n\n

(iii) Heena gets two correct and five incorrect answers out of seven questions she attempts. What is her score?<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question,<\/p>\n\n\n\n

Marks awarded for 1 correct answer = 5<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

Total marks awarded for 2 correct answer = 2 \u00d7 5 = 10<\/p>\n\n\n\n

Marks awarded for 1 wrong answer = -2<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

Total marks awarded for 5 wrong answer = 5 \u00d7 -2 = -10<\/p>\n\n\n\n

Marks awarded for questions not attempted is = 0<\/p>\n\n\n\n

\u2234Total score obtained by Heena = 10 + (-10)<\/p>\n\n\n\n

= 10 \u2013 10<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

8. A cement company earns a profit of \u20b9 8 per bag of white cement sold and a loss of<\/strong><\/p>\n\n\n\n

\u20b9 5 per bag of grey cement sold.<\/strong><\/p>\n\n\n\n

(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?<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

We denote profit in positive integer and loss in negative integer,<\/p>\n\n\n\n

From the question,<\/p>\n\n\n\n

Cement company earns a profit on selling 1 bag of white cement = \u20b9 8 per bag<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

Cement company earns a profit on selling 3000 bags of white cement = 3000 \u00d7 \u20b9 8<\/p>\n\n\n\n

= \u20b9 24000<\/p>\n\n\n\n

Loss on selling 1 bag of grey cement = \u2013 \u20b9 5 per bag<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

Loss on selling 5000 bags of grey cement = 5000 \u00d7 \u2013 \u20b9 5<\/p>\n\n\n\n

= \u2013 \u20b9 25000<\/p>\n\n\n\n

Total loss or profit earned by the cement company = profit + loss<\/p>\n\n\n\n

= 24000 + (-25000)<\/p>\n\n\n\n

= \u2013 \u20b91000<\/p>\n\n\n\n

Thus, a loss of \u20b9 1000 will be incurred by the company.<\/p>\n\n\n\n

NCERT 7th Sanskrit Chapter 1, class 7 Sanskrit Chapter 1 solutions<\/p>\n\n\n\n

(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.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

We denote profit in positive integer and loss in negative integer,<\/p>\n\n\n\n

From the question,<\/p>\n\n\n\n

Cement company earns a profit on selling 1 bag of white cement = \u20b9 8 per bag<\/p>\n\n\n\n

Let the number of white cement bags be x.<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

Cement company earns a profit on selling x bags of white cement = (x) \u00d7 \u20b9 8<\/p>\n\n\n\n

= \u20b9 8x<\/p>\n\n\n\n

Loss on selling 1 bag of grey cement = \u2013 \u20b9 5 per bag<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

Loss on selling 6400 bags of grey cement = 6400 \u00d7 \u2013 \u20b9 5<\/p>\n\n\n\n

= \u2013 \u20b9 32000<\/p>\n\n\n\n

According to the question,<\/p>\n\n\n\n

Company must sell to have neither profit nor loss.<\/p>\n\n\n\n

= Profit + loss = 0<\/p>\n\n\n\n

= 8x + (-32000) =0<\/p>\n\n\n\n

By sending -32000 from LHS to RHS it becomes 32000<\/p>\n\n\n\n

= 8x = 32000<\/p>\n\n\n\n

= x = 32000\/8<\/p>\n\n\n\n

= x = 4000<\/p>\n\n\n\n

Hence, the 4000 bags of white cement have neither profit nor loss.<\/p>\n\n\n\n

NCERT 7th Sanskrit Chapter 1, class 7 Sanskrit Chapter 1 solutions<\/p>\n\n\n\n

9. Replace the blank with an integer to make it a true statement.<\/strong><\/p>\n\n\n\n

(a) (\u20133) \u00d7 _____ = 27<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume the missing integer be x,<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

= (\u20133) \u00d7 (x) = 27<\/p>\n\n\n\n

= x = \u2013 (27\/3)<\/p>\n\n\n\n

= x = -9<\/p>\n\n\n\n

Let us substitute the value of x in the place of blank,<\/p>\n\n\n\n

= (\u20133) \u00d7 (-9) = 27 \u2026 [\u2235 (- \u00d7 \u2013 = +)]<\/p>\n\n\n\n

(b) 5 \u00d7 _____ = \u201335<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume the missing integer be x,<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

= (5) \u00d7 (x) = -35<\/p>\n\n\n\n

= x = \u2013 (-35\/5)<\/p>\n\n\n\n

= x = -7<\/p>\n\n\n\n

Let us substitute the value of x in the place of blank,<\/p>\n\n\n\n

= (5) \u00d7 (-7) = -35 \u2026 [\u2235 (+ \u00d7 \u2013 = -)]<\/p>\n\n\n\n

(c) _____ \u00d7 (\u2013 8) = \u201356<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume the missing integer be x,<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

= (x) \u00d7 (-8) = -56<\/p>\n\n\n\n

= x = (-56\/-8)<\/p>\n\n\n\n

= x = 7<\/p>\n\n\n\n

Let us substitute the value of x in the place of blank,<\/p>\n\n\n\n

= (7) \u00d7 (-8) = -56 \u2026 [\u2235 (+ \u00d7 \u2013 = -)]<\/p>\n\n\n\n

(d) _____ \u00d7 (\u201312) = 132<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume the missing integer be x,<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

= (x) \u00d7 (-12) = 132<\/p>\n\n\n\n

= x = \u2013 (132\/12)<\/p>\n\n\n\n

= x = \u2013 11<\/p>\n\n\n\n

Let us substitute the value of x in the place of blank,<\/p>\n\n\n\n

= (\u201311) \u00d7 (-12) = 132 \u2026 [\u2235 (- \u00d7 \u2013 = +)]<\/p>\n\n\n\n

\n\n\n\n

NCERT 7th Maths Chapter 1, class 7 Maths Chapter 1 solutions<\/p>\n\n\n\n

Exercise 1.4 Page: 26<\/p>\n\n\n\n

1. Evaluate each of the following:<\/strong><\/p>\n\n\n\n

(a) (\u201330) \u00f7 10<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

= (\u201330) \u00f7 10<\/p>\n\n\n\n

= \u2013 3<\/p>\n\n\n\n

When we divide a negative integer by a positive integer, we first divide them as whole numbers and then put minus sign (-) before the quotient.<\/p>\n\n\n\n

(b) 50 \u00f7 (\u20135)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

= (50) \u00f7 (-5)<\/p>\n\n\n\n

= \u2013 10<\/p>\n\n\n\n

When we divide a positive integer by a negative integer, we first divide them as whole numbers and then put minus sign (-) before the quotient.<\/p>\n\n\n\n

(c) (\u201336) \u00f7 (\u20139)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

= (-36) \u00f7 (-9)<\/p>\n\n\n\n

= 4<\/p>\n\n\n\n

When we divide a negative integer by a negative integer, we first divide them as whole numbers and then put positive sign (+) before the quotient.<\/p>\n\n\n\n

(d) (\u2013 49) \u00f7 (49)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

= (\u201349) \u00f7 49<\/p>\n\n\n\n

= \u2013 1<\/p>\n\n\n\n

When we divide a negative integer by a positive integer, we first divide them as whole numbers and then put minus sign (-) before the quotient.<\/p>\n\n\n\n

(e) 13 \u00f7 [(\u20132) + 1]<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

= 13 \u00f7 [(\u20132) + 1]<\/p>\n\n\n\n

= 13 \u00f7 (-1)<\/p>\n\n\n\n

= \u2013 13<\/p>\n\n\n\n

When we divide a positive integer by a negative integer, we first divide them as whole numbers and then put minus sign (-) before the quotient.<\/p>\n\n\n\n

(f) 0 \u00f7 (\u201312)<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

= 0 \u00f7 (-12)<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

When we divide zero by a negative integer gives zero.<\/p>\n\n\n\n

(g) (\u201331) \u00f7 [(\u201330) + (\u20131)]<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

= (\u201331) \u00f7 [(\u201330) + (\u20131)]<\/p>\n\n\n\n

= (-31) \u00f7 [-30 \u2013 1]<\/p>\n\n\n\n

= (-31) \u00f7 (-31)<\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

When we divide a negative integer by a negative integer, we first divide them as whole numbers and then put positive sign (+) before the quotient.<\/p>\n\n\n\n

(h) [(\u201336) \u00f7 12] \u00f7 3<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

First we have to solve the integers with in the bracket,<\/p>\n\n\n\n

= [(\u201336) \u00f7 12]<\/p>\n\n\n\n

= (\u201336) \u00f7 12<\/p>\n\n\n\n

= \u2013 3<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

= (-3) \u00f7 3<\/p>\n\n\n\n

= -1<\/p>\n\n\n\n

When we divide a negative integer by a positive integer, we first divide them as whole numbers and then put minus sign (-) before the quotient.<\/p>\n\n\n\n

(i) [(\u2013 6) + 5)] \u00f7 [(\u20132) + 1]<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

The given question can be written as,<\/p>\n\n\n\n

= [-1] \u00f7 [-1]<\/p>\n\n\n\n

= 1<\/p>\n\n\n\n

When we divide a negative integer by a negative integer, we first divide them as whole numbers and then put positive sign (+) before the quotient.<\/p>\n\n\n\n

NCERT 7th Maths Chapter 1, class 7 Maths Chapter 1 solutions<\/p>\n\n\n\n

2. Verify that a \u00f7 (b + c) \u2260 (a \u00f7 b) + (a \u00f7 c) for each of the following values of a, b and c.<\/strong><\/p>\n\n\n\n

(a) a = 12, b = \u2013 4, c = 2<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question, a \u00f7 (b + c) \u2260 (a \u00f7 b) + (a \u00f7 c)<\/p>\n\n\n\n

Given, a = 12, b = \u2013 4, c = 2<\/p>\n\n\n\n

Now, consider LHS = a \u00f7 (b + c)<\/p>\n\n\n\n

= 12 \u00f7 (-4 + 2)<\/p>\n\n\n\n

= 12 \u00f7 (-2)<\/p>\n\n\n\n

= -6<\/p>\n\n\n\n

When we divide a positive integer by a negative integer, we first divide them as whole numbers and then put minus sign (-) before the quotient.<\/p>\n\n\n\n

Then, consider RHS = (a \u00f7 b) + (a \u00f7 c)<\/p>\n\n\n\n

= (12 \u00f7 (-4)) + (12 \u00f7 2)<\/p>\n\n\n\n

= (-3) + (6)<\/p>\n\n\n\n

= 3<\/p>\n\n\n\n

By comparing LHS and RHS<\/p>\n\n\n\n

= -6 \u2260 3<\/p>\n\n\n\n

= LHS \u2260 RHS<\/p>\n\n\n\n

Hence, the given values are verified.<\/p>\n\n\n\n

(b) a = (\u201310), b = 1, c = 1<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question, a \u00f7 (b + c) \u2260 (a \u00f7 b) + (a \u00f7 c)<\/p>\n\n\n\n

Given, a = (-10), b = 1, c = 1<\/p>\n\n\n\n

Now, consider LHS = a \u00f7 (b + c)<\/p>\n\n\n\n

= (-10) \u00f7 (1 + 1)<\/p>\n\n\n\n

= (-10) \u00f7 (2)<\/p>\n\n\n\n

= -5<\/p>\n\n\n\n

When we divide a negative integer by a positive integer, we first divide them as whole numbers and then put minus sign (-) before the quotient.<\/p>\n\n\n\n

Then, consider RHS = (a \u00f7 b) + (a \u00f7 c)<\/p>\n\n\n\n

= ((-10) \u00f7 (1)) + ((-10) \u00f7 1)<\/p>\n\n\n\n

= (-10) + (-10)<\/p>\n\n\n\n

= -10 \u2013 10<\/p>\n\n\n\n

= -20<\/p>\n\n\n\n

By comparing LHS and RHS<\/p>\n\n\n\n

= -5 \u2260 -20<\/p>\n\n\n\n

= LHS \u2260 RHS<\/p>\n\n\n\n

Hence, the given values are verified.<\/p>\n\n\n\n

NCERT 7th Maths Chapter 1, class 7 Maths Chapter 1 solutions<\/p>\n\n\n\n

3. Fill in the blanks:<\/strong><\/p>\n\n\n\n

(a) 369 \u00f7 _____ = 369<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume the missing integer be x,<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

= 369 \u00f7 x = 369<\/p>\n\n\n\n

= x = (369\/369)<\/p>\n\n\n\n

= x = 1<\/p>\n\n\n\n

Now, put the valve of x in the blank.<\/p>\n\n\n\n

= 369 \u00f7 1 = 369<\/p>\n\n\n\n

(b) (\u201375) \u00f7 _____ = \u20131<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume the missing integer be x,<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

= (-75) \u00f7 x = -1<\/p>\n\n\n\n

= x = (-75\/-1)<\/p>\n\n\n\n

= x = 75<\/p>\n\n\n\n

Now, put the valve of x in the blank.<\/p>\n\n\n\n

= (-75) \u00f7 75 = -1<\/p>\n\n\n\n

(c) (\u2013206) \u00f7 _____ = 1<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume the missing integer be x,<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

= (-206) \u00f7 x = 1<\/p>\n\n\n\n

= x = (-206\/1)<\/p>\n\n\n\n

= x = -206<\/p>\n\n\n\n

Now, put the valve of x in the blank.<\/p>\n\n\n\n

= (-206) \u00f7 (-206) = 1<\/p>\n\n\n\n

(d) \u2013 87 \u00f7 _____ = 87<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume the missing integer be x,<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

= (-87) \u00f7 x = 87<\/p>\n\n\n\n

= x = (-87)\/87<\/p>\n\n\n\n

= x = -1<\/p>\n\n\n\n

Now, put the valve of x in the blank.<\/p>\n\n\n\n

= (-87) \u00f7 (-1) = 87<\/p>\n\n\n\n

(e) _____ \u00f7 1 = \u2013 87<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume the missing integer be x,<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

= (x) \u00f7 1 = -87<\/p>\n\n\n\n

= x = (-87) \u00d7 1<\/p>\n\n\n\n

= x = -87<\/p>\n\n\n\n

Now, put the valve of x in the blank.<\/p>\n\n\n\n

= (-87) \u00f7 1 = -87<\/p>\n\n\n\n

(f) _____ \u00f7 48 = \u20131<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume the missing integer be x,<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

= (x) \u00f7 48 = -1<\/p>\n\n\n\n

= x = (-1) \u00d7 48<\/p>\n\n\n\n

= x = -48<\/p>\n\n\n\n

Now, put the valve of x in the blank.<\/p>\n\n\n\n

= (-48) \u00f7 48 = -1<\/p>\n\n\n\n

(g) 20 \u00f7 _____ = \u20132<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume the missing integer be x,<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

= 20 \u00f7 x = -2<\/p>\n\n\n\n

= x = (20)\/ (-2)<\/p>\n\n\n\n

= x = -10<\/p>\n\n\n\n

Now, put the valve of x in the blank.<\/p>\n\n\n\n

= (20) \u00f7 (-10) = -2<\/p>\n\n\n\n

(h) _____ \u00f7 (4) = \u20133<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

Let us assume the missing integer be x,<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

= (x) \u00f7 4 = -3<\/p>\n\n\n\n

= x = (-3) \u00d7 4<\/p>\n\n\n\n

= x = -12<\/p>\n\n\n\n

Now, put the valve of x in the blank.<\/p>\n\n\n\n

= (-12) \u00f7 4 = -3<\/p>\n\n\n\n

4. Write five pairs of integers (a, b) such that a \u00f7 b = \u20133. One such pair is (6, \u20132) because 6 \u00f7 (\u20132) = (\u20133).<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

(i) (15, -5)<\/p>\n\n\n\n

Because, 15 \u00f7 (\u20135) = (\u20133)<\/p>\n\n\n\n

(ii) (-15, 5)<\/p>\n\n\n\n

Because, (-15) \u00f7 (5) = (\u20133)<\/p>\n\n\n\n

(iii) (18, -6)<\/p>\n\n\n\n

Because, 18 \u00f7 (\u20136) = (\u20133)<\/p>\n\n\n\n

(iv) (-18, 6)<\/p>\n\n\n\n

Because, (-18) \u00f7 6 = (\u20133)<\/p>\n\n\n\n

(v) (21, -7)<\/p>\n\n\n\n

Because, 21 \u00f7 (\u20137) = (\u20133)<\/p>\n\n\n\n

NCERT 7th Maths Chapter 1, class 7 Maths Chapter 1 solutions<\/p>\n\n\n\n

5. The temperature at 12 noon was 10o<\/sup>C above zero. If it decreases at the rate of 2o<\/sup>C per hour until midnight, at what time would the temperature be 8\u00b0C below zero? What would be the temperature at mid-night?<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question is given that,<\/p>\n\n\n\n

Temperature at the beginning i.e., at 12 noon = 10o<\/sup>C<\/p>\n\n\n\n

Rate of change of temperature = \u2013 2o<\/sup>C per hour<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

Temperature at 1 PM = 10 + (-2) = 10 \u2013 2 = 8o<\/sup>C<\/p>\n\n\n\n

Temperature at 2 PM = 8 + (-2) = 8 \u2013 2 = 6o<\/sup>C<\/p>\n\n\n\n

Temperature at 3 PM = 6 + (-2) = 6 \u2013 2 = 4o<\/sup>C<\/p>\n\n\n\n

Temperature at 4 PM = 4 + (-2) = 4 \u2013 2 = 2o<\/sup>C<\/p>\n\n\n\n

Temperature at 5 PM = 2 + (-2) = 2 \u2013 2 = 0o<\/sup>C<\/p>\n\n\n\n

Temperature at 6 PM = 0 + (-2) = 0 \u2013 2 = -2o<\/sup>C<\/p>\n\n\n\n

Temperature at 7 PM = -2 + (-2) = -2 -2 = -4o<\/sup>C<\/p>\n\n\n\n

Temperature at 8 PM = -4 + (-2) = -4 \u2013 2 = -6o<\/sup>C<\/p>\n\n\n\n

Temperature at 9 PM = -6 + (-2) = -6 \u2013 2 = -8o<\/sup>C<\/p>\n\n\n\n

\u2234At 9 PM the temperature will be 8o<\/sup>C below zero<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

The temperature at mid-night i.e., at 12 AM<\/p>\n\n\n\n

Change in temperature in 12 hours = -2o<\/sup>C \u00d7 12 = \u2013 24o<\/sup>C<\/p>\n\n\n\n

So, at midnight temperature will be = 10 + (-24)<\/p>\n\n\n\n

= \u2013 14o<\/sup>C<\/p>\n\n\n\n

So, at midnight temperature will be 14o<\/sup>C below 0.<\/p>\n\n\n\n

NCERT 7th Maths Chapter 1, class 7 Maths Chapter 1 solutions<\/p>\n\n\n\n

6. In a class test (+ 3) marks are given for every correct answer and (\u20132) 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 \u20135 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question,<\/p>\n\n\n\n

Marks awarded for 1 correct answer = + 3<\/p>\n\n\n\n

Marks awarded for 1 wrong answer = -2<\/p>\n\n\n\n

(i) Radhika scored 20 marks<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

Total marks awarded for 12 correct answers = 12 \u00d7 3 = 36<\/p>\n\n\n\n

Marks awarded for incorrect answers = Total score \u2013 Total marks awarded for 12 correct<\/p>\n\n\n\n

Answers<\/p>\n\n\n\n

= 20 \u2013 36<\/p>\n\n\n\n

= \u2013 16<\/p>\n\n\n\n

So, the number of incorrect answers made by Radhika = (-16) \u00f7 (-2)<\/p>\n\n\n\n

= 8<\/p>\n\n\n\n

(ii) Mohini scored -5 marks<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

Total marks awarded for 7 correct answers = 7 \u00d7 3 = 21<\/p>\n\n\n\n

Marks awarded for incorrect answers = Total score \u2013 Total marks awarded for 12 correct<\/p>\n\n\n\n

Answers<\/p>\n\n\n\n

= \u2013 5 \u2013 21<\/p>\n\n\n\n

= \u2013 26<\/p>\n\n\n\n

So, the number of incorrect answers made by Mohini = (-26) \u00f7 (-2)<\/p>\n\n\n\n

= 13<\/p>\n\n\n\n

NCERT 7th Maths Chapter 1, class 7 Maths Chapter 1 solutions<\/p>\n\n\n\n

7. An elevator descends into a mine shaft at the rate of 6 m\/min. If the descent starts from 10 m above the ground level, how long will it take to reach \u2013 350 m.<\/strong><\/p>\n\n\n\n

Solution:-<\/strong><\/p>\n\n\n\n

From the question,<\/p>\n\n\n\n

The initial height of the elevator = 10 m<\/p>\n\n\n\n

Final depth of elevator = \u2013 350 m \u2026 [\u2235distance descended is denoted by a negative<\/p>\n\n\n\n

integer]<\/p>\n\n\n\n

The total distance to descended by the elevator = (-350) \u2013 (10)<\/p>\n\n\n\n

= \u2013 360 m<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

Time taken by the elevator to descend -6 m = 1 min<\/p>\n\n\n\n

So, time taken by the elevator to descend \u2013 360 m = (-360) \u00f7 (-60)<\/p>\n\n\n\n

= 60 minutes<\/p>\n\n\n\n

= 1 hour<\/p>\n\n\n\n

\n\n\n\n

NCERT 7th Maths Chapter 1, class 7 Maths Chapter 1 solutions<\/p>\n\n\n\n

NCERT Solutions for 7th Class Maths: Chapter 1: Download PDF<\/strong><\/h2>\n\n\n\n

NCERT Solutions for 7th Class Maths: Chapter 1-Integers<\/p>\n\n\n\n

Download PDF<\/strong>: NCERT Solutions for 7th Class Maths: Chapter 1-Integers PDF<\/a><\/p>\n\n\n\n

Chapter-wise NCERT Solutions Class 7 Maths<\/h2>\n\n\n\n
Chapter 1 Integers<\/a><\/td><\/tr>
Chapter 2 Fractions and Decimals<\/a><\/td><\/tr>
Chapter 3 Data Handling<\/a><\/td><\/tr>
Chapter 4 Simple Equations<\/a><\/td><\/tr>
Chapter 5 Lines and Angles<\/a><\/td><\/tr>
Chapter 6 The Triangle and its Properties<\/a><\/td><\/tr>
Chapter 7 Congruence of Triangles<\/a><\/td><\/tr>
Chapter 8 Comparing Quantities<\/a><\/td><\/tr>
Chapter 9 Rational Numbers<\/a><\/td><\/tr>
Chapter 10 Practical Geometry<\/a><\/td><\/tr>
Chapter 11 Perimeter and Area<\/a><\/td><\/tr>
Chapter 12 Algebraic Expressions<\/a><\/td><\/tr>
Chapter 13 Exponents and Powers<\/a><\/td><\/tr>
Chapter 14 Symmetry<\/a><\/td><\/tr>
Chapter 15 Visualising Solid Shapes<\/a><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n


<\/strong>About NCERT<\/h2>\n\n\n\n

The National Council of Educational Research and Training is an autonomous organization of the Government of India which was established in 1961 as a literary, scientific, and charitable Society under the Societies Registration Act. Its headquarters are located at Sri Aurbindo Marg in New Delhi. Visit the Official NCERT website<\/a> to learn more. <\/p>\n\n\n\n

\n
NCERT Solutions<\/a><\/div>\n\n\n\n
NCERT Solutions for Class 7<\/a><\/div>\n\n\n\n
NCERT Solutions for Class 7 Maths<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Class 7: Maths Chapter 1 solutions. Complete Class 7 Maths Chapter 1 Notes. NCERT Solutions for 7th Class Maths: Chapter 1-Integers NCERT 7th Maths Chapter 1, class 7 Maths Chapter 1 solutions Page: 4 Exercise 1.1 1. Following number line shows the temperature in degree celsius (co) at different places on a particular day. (a) […]<\/p>\n","protected":false},"author":302,"featured_media":627307,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[1411,907],"tags":[1550],"boards":[1180],"class_list":["post-139416","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-book-solutions","category-class-7","tag-ncert-maths-class-7","boards-ncert","entry"],"yoast_head":"\nNCERT Solutions Class 7, Maths Chapter 1 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for 7th Class Maths: Chapter 1-Integers | Browse all Class 7 Maths Solutions of NCERT books- IndCareer Schools\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-7th-class-maths-chapter-1-integers\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for 7th Class Maths: Chapter 1-Integers\" \/>\n<meta property=\"og:description\" content=\"Class 7: Maths Chapter 1 solutions. Complete Class 7 Maths Chapter 1 Notes. NCERT Solutions for 7th Class Maths: Chapter 1-Integers NCERT 7th Maths\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-7th-class-maths-chapter-1-integers\/\" \/>\n<meta property=\"og:site_name\" content=\"IndCareer Schools\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/indcareer\" \/>\n<meta property=\"article:published_time\" content=\"2021-02-20T08:15:35+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-09-15T01:52:41+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/NCERT-Solutions-67-scaled.jpg\" \/>\n\t<meta property=\"og:image:width\" content=\"1600\" \/>\n\t<meta property=\"og:image:height\" content=\"900\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/jpeg\" \/>\n<meta name=\"author\" content=\"Pooja\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@indcareer\" \/>\n<meta name=\"twitter:site\" content=\"@indcareer\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Pooja\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"32 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-7th-class-maths-chapter-1-integers\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-7th-class-maths-chapter-1-integers\/\"},\"author\":{\"name\":\"Pooja\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e\"},\"headline\":\"NCERT Solutions for 7th Class Maths: Chapter 1-Integers\",\"datePublished\":\"2021-02-20T08:15:35+00:00\",\"dateModified\":\"2023-09-15T01:52:41+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-7th-class-maths-chapter-1-integers\/\"},\"wordCount\":5044,\"publisher\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#organization\"},\"image\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-7th-class-maths-chapter-1-integers\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/NCERT-Solutions-67-scaled.jpg\",\"keywords\":[\"NCERT Maths ( Class 7)\"],\"articleSection\":[\"Book Solutions\",\"Class 7\"],\"inLanguage\":\"en-US\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-7th-class-maths-chapter-1-integers\/\",\"url\":\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-7th-class-maths-chapter-1-integers\/\",\"name\":\"NCERT Solutions Class 7, Maths Chapter 1 - 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