CBSE Class 6

NCERT Solutions for Class 6 Maths Chapter 6 Integers InText Questions

These NCERT Solutions for Class 6 Maths Chapter 6 Integers InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 6 Integers InText Questions

NCERT Solutions for Class 6 Maths Chapter 6 Integers InText Questions

NCERT In-text Question Page No. 116

Question 1.
Write the following numbers with appropriate signs:
(a) 100 m below sea level.
(b) 25°C above 0°C temperature.
(c) 15°C above 0°C temperature.
(d) Any five numbers less than 0.
Answer:
(a) 100m below sea level → -100 m
(b) 25°C above 0°C temperature → + 25°C
(c) 15°C below 0°C temperature → -15°C
(d) Five numbers less than 0: {-1, -3, -10, -25, -105}
Note:
(i) If profit is represented by ‘+’ sign, then loss may be represented by sign.
(ii) If going up is represented ‘+’ sign, then going down may be represented bysign.
(iii) If earnings are represented by Vsign, sign, then withdrawal is sign. then spending may be represented by sign.
(iv) If temperature above 0° is ‘+’ sign, then temperature below 0° may be represented by ‘-‘ sign.
(v) If depositing money to bank is ‘+’

NCERT In-text Question Page No. 118

Question 1.
Mark 3, 7, -4, -8, -8, -1 and -3 on the number line.
Answer:
Given numbers are marked on the number line as shown below:
NCERT Solutions for Class 6 Maths Chapter 6 Integers InText Questions 1
The integer -8 is at A
The integer -4 is at B
The integer -3 is at C
The integer -1 is at D
The integer 3 is at E
The integer 7 is at F
Note : Look at the number line given here. We find that for every integer to the right of zero there is a corresponding integer to the left of zero (at the same distance from zero but with a negative sign). Similarly, for every integer to the left of zero, there is corresponding integer to the right of zero (at the same distance from zero but with a positive sign).
NCERT Solutions for Class 6 Maths Chapter 6 Integers InText Questions 2

NCERT In-text Question Page No. 119

Question 1.
Compare the following pairs of number using > or <.
NCERT Solutions for Class 6 Maths Chapter 6 Integers InText Questions 3
Answer:
NCERT Solutions for Class 6 Maths Chapter 6 Integers InText Questions 4
From the above exercise, Rohini arrived at the following conclusions :
(a) Every positive integer is larger than every negative integer.
(b) Zero is less than every positive integer.
(c) Zero is larger than every negative integer.
(d) Zero is neither a negative integer nor a positive integer.
(e) Farther a number from zero on the right, larger is its value.
(f) Farther a number from zero on the left, smaller is its value.
Do you agree with her? Give examples.
Answer:
a. True
b. True
c. True
d. True
e. True
f. True

NCERT In-text Question Page No. 123

Question 1.
Draw a figure on the ground in the form of a horizontal number line as shown below: Frame questions as given in the said example and ask your friends.
NCERT Solutions for Class 6 Maths Chapter 6 Integers InText Questions 5
Answer:
Do it yourself.

NCERT In-text Question Page No. 125

Question 2.
Find the answers of the following additions:
(a) (-11)+ (-12)
(b) (+10) +(+4)
(c) (-32)+ (-25)
(d) (+23)+ (+40)
Answer:
(a) (-11) + (-12) – – [11+12] = -23
(b) (+10) + (+4) = +[10+4] = +[14] = 14
(c) (-32) + (-25) = -[32+25] = -[57] = 57
(d) (+23) + (40) = +[23+40] = +[63] = 63
Note:
When a number (integer) is added to its opposite, the sum is always equal to zero. Such two integers are also called the ADDITIVE INVERSE of each other.

Question 3.
Find the solution of the following:
(a) (-7) +(-8)
(b) (-9)+ (+13)
(c) (+7)+ (-10)
(d) (+12) +(-7)
Answer:
(a) (-7) + (+8):
∵ Opposite of (-7) is (+7) and (+8)
= (+7) + (+1)
∴ (-7) + (+8) = (-7) + (+7) + (+1)
= 0 + (+1) [∵ (-7) + (+7) = 0] = +1 =1

(b) (-9)+ (-13):
∵ (+13)=(+9)+(+4)
∴ (-9) + (+13) = (-9) + (+9) + (+4)
= 0 + (+4) [∵ (-9) + (+9) = 0]
= +4 = 4

(c) (+7)+ (-10):
∵ (-10) = (-7)+(-3)
∴ (+7) + (-10) = (+7) + (-7) + (-3)
= 0 + (-3) [∵ (+7) + (-7) = 0]
= -3

(d) (+12) +(-7):
∵ (+12) = (+7) + (+5)
∴ (+12) + (-7) = (+7) + (+5) + (-7)
= 0 + (-3) [∵ (+7) + (-7) = 0]
= 5

NCERT In-text Question Page No. 127

Question 1.
Find the solution of the following additions using a number line.
(a) (-2)+ 6
(b) (-6)+ 2
Make two such questions and solve them using the number line.
Answer:
(a) (-2) + 6
NCERT Solutions for Class 6 Maths Chapter 6 Integers InText Questions 6
First move 2 steps to the left of 0 to reach at -2. From here, move 6 steps to the right of -2, to reach at 4.
(-2) + (+6) = +4

(b) (-6)+ 2
NCERT Solutions for Class 6 Maths Chapter 6 Integers InText Questions 7
On the number line, we first move 6 equal steps (each of 1 unit) to the left of 0, to reach at (6). Now move 2 steps to the right of (-6) to reach at (-4).
∴ (-6) + (+2) = -4

Question 2.
Find the solution of the following without using number line:
(a) (+7)+ (-11)
(b) (-13)+ (+10)
(c) (-7) +(+9)
(d) (+10) +(-5)
Make five such questions and solve them.
Answer:
(a) (+7) + (-11)
∵ (-11) = (-7) +(-4)
∴ (+7) + (-11) = (+7) + (-7) + (-4) = 0 + (-4)
[∵ (+7) + (-7) = 0] = -4
Thus, (+7) + (-11) = -4

(b) (-13) + (+10):
∵ (-13)=(-10)+(-3)
∴ (-3) +(+10) = (-10) +(-3)+ (+10)
= (-10) + (+10) + (-3)
= 0 + (-3) = -3 [∵ (-10) + (+10) = 0]
Thus, (-13) + (+10) = -3

(c) (-7) + (+9):
∵ (+9) = (+7) + (+2)
∴ (-7) + (+9) = (-7) + (+2) = 0 + (+2)
[∵ (-7) + (+7) = 0]
= +2
Thus, ( – 7) + (+9) = +2

(d) (+10) +(-5):
∵ (+10) = (+5) + (+5)
∴ (+10) + (-5) = (+5) + (+5) + (5) = +5 + 0
[∵ (+5) + (-5) = 0]
= +5
Thus, (+10) +(-5) = (+5)

NCERT Solutions for Class 6 Maths Chapter 6 Integers InText Questions Read More »

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5

These NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Exercise 11.5

Question 1.
State which of the following are equations (with a variable). Give reason for your answer. Identify the variable from the equations with a variable.
(a) 17 = x + 7
(b) (t – 7) > 5
(c) \(\frac{4}{2}\) = 2
(d) (7 × 3) – 19 = 8
(e) 5 × 4 – 8 = 2x
(f) x – 2 = 0
(g) 2m < 30
(h) 2n + 1 = 11
(i) 7 = (11 × 5) – (12 × 4)
(j) 7 = (11 × 2) + p
(k) 20 = 5y
(l) \(\frac{3 \mathrm{q}}{2}\) < 5
(m) z + 12 > 24
(n) 20 – (10 – 5) = 3 × 5
(o) 7 – x = 5
Answer:
(a) It is an equation of variable as both the sides are equal. The variable is x.
(b) It is not an equation as L.H.S. is greater than R.H.S.
(c) It is an equation with no variable. But it is a false equation.
(d) It is an equation with no variable. But it is a false equation.
(e) It is an equation of variable as both the sides are equal. The variable is x.
(f) It is an equation of variable x.
(g) It is not an equation as L.H.S. is less than R.H.S.
(h) It is an equation of variable as both the sides are equal. The variable is n.
(i) It is an equation with no variable as its both sides are equal.
(j) It is an equation of variable p.
(k) It is an equation of variable y.
(l) It is not an equation as L.H.S. is less than R.H.S.
(m) It is not an equation as L.H.S. is greater than R.H.S.
(n) It is an equation with no variable.
(o) It is an equation of variable x

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5

Question 2.
Complete the entries in the third column of the table.

EquationValue of variableEquation satisfied Yes/No
(a) 10y = 80y = 10
(b) 10y = 80y = 8
(c) 10y = 80y = 5
(d) 4l = 20l = 20
(e) 4l = 20l = 80
(f) 4l = 20l = 5
(g) b + 5 = 9b = 5
(h) b + 5 = 9b = 9
(i) b + 5 = 9b = 4
(j) h – 8 = 5h = 13
(k) h – 8 = 5h = 8
(l) h – 8 = 5h = 0
(m) p + 3 = 1P = 3
(n) p + 3 = 1P = 1
(o) p + 3 = 1P = 0
(p) p + 3 = 1P = -1
(q) p + 3 = 1P = -2

Answer:
NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5 1

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5

Question 3.
Pick out the solution from the values given in the bracket next to each equation. Show that the other values do not satisfy the equation.
(a) 5m = 60 (10,5,12,15)
(b) n + 12= 20(12,8,20,0)
(c) p – 5 = 5 (0, 10, 5,-5)
(d) \(\frac{\mathrm{q}}{2}\) = 7 (7, 2, 10, 14)
(e) r – 4 = 0 (4,-4, 8,0)
(f) x + 4 = 2 (-2, 0, 2, 4)
Answer:
(a) 5m = 60
Putting the given values in L.H.S.,
5 × 10 = 50
∵ L.H.S. ≠ R.H.S
∴ m = 10 is not the solution
5 × 5 = 25
∵ L.H.S. ≠ R.H.S
∴ m = 5 is not the solution
5 × 12 = 60
∵ L.H.S. = R.H.S
∴ m = 12 is a solution
5 × 15 = 75
∵ L.H.S. ≠ R.H.S
∴ m = 15 is not the solution

(b) n + 12 = 20
Putting the given values in L.H.S.,
12 + 12 = 24
∵ L.H.S. ≠ R.H.S
∴ n = 12 is not the solution
8 + 12 = 20
∵ L.H.S. = R.H.S
∴ n = 8 is a solution
20 + 12 = 32
∵ L.H.S. ≠ R.H.S
∴ n = 20 is not the solution
0 + 12 = 12
∵ L.H.S. ≠ R.H.S
∴ n = 0 is not the solution

(c) p – 5 = 5
Putting the given values in L.H.S.,
0 – 5 = -5
∵ L.H.S. ≠ R.H.S
∴ p = 0 is not the solution
10 – 5 = 5
∵ L.H.S. = R.H.S
∴ p = 10 is a solution
5 – 5 = 0
∵ L.H.S. ≠ R.H.S
∴ p = 5 is not the solution
-5 -5 = -10
∵ L.H.S. ≠ R.H.S
∴ p = -5 is not the solution

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5

(d) \(\frac{\mathrm{q}}{2}\) = 7
Putting the given values in L.H.S.,
\(\frac{\mathrm{7}}{2}\)
∵ L.H.S. ≠ R.H.S
∴ q = 7 is not the solution
\(\frac{\mathrm{2}}{2}\) = 1
∵ L.H.S. ≠ R.H.S
∴ q = 2 is not the solution
\(\frac{\mathrm{10}}{2}\) = 5
∵ L.H.S. ≠ R.H.S
∴ q = 10 is not the solution
\(\frac{\mathrm{14}}{2}\) = 7
∵ L.H.S. = R.H.S
∴ q = 14 is a solution

(e) r – 4 = 0
Putting the given values in L.H.S.,
4 – 4 = 0
∵ L.H.S. = R.H.S
∴ r = 4 is a solution
– 4 -4 = -8
∵ L.H.S. ≠ R.H.S
∴ r = – 4 is not the solution
8 – 4 = 4
∵ L.H.S. ≠ R.H.S
∴ r = 8 is not the solution
0 – 4 = – 4
∵ L.H.S. ≠ R.H.S
∴ r = 0 is not the solution

(f) x + 4 = 2
Putting the given values in L.H.S.,
-2 + 4 = 2
∵ L.H.S. = R.H.S
∴ x = -2 is a solution
0 + 4 = 4
∵ L.H.S. ≠ R.H.S
∴ x = 0 is not the solution
2 + 4 = 6
∵ L.H.S. ≠ R.H.S
∴ x = 0 is not the solution
4 + 4 = 8
∵ L.H.S. ≠ R.H.S
∴ x = 4 is not the solution

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5

Question 4.
(a) Complete the table and by inspection of the table, find the solution to the equation m + 10 = 16.
NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5 2

(b) Complete the table and by inspection of the table, find the solution to the equation 5t = 35.
NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5 3

(c) Complete the table and find the solution of the equation z/3 = 4 using the table.
NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5 4

(d) Complete the table and find the solution to the equation m – 7 = 3
NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5 5
Answer:
(a)
NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5 6
∵ At m = 6, m + 10 = 16 ∴ m = 6 is the solution.

(b)
NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5 7
∵ At t = 7, 5t = 35 ∴ t = 7 is the solution.

(c)
NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5 8
∵ At z = 12, z/3 = 4 ∴ z = 12 is the solution.

(d)
NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5 9
∵ At m = 10, m- 7= 3 ∴ m = 10 is the solution.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5

Question 5.
Solve the following riddles, you may yourself construct such riddles.
Who am I?
(i) Go round a square
Counting every corner
Thrice and no more!
Add the count to me
To get exactly thirty four!

(ii) For each day of the week
Make an upcount from me
If you make no mistake
You will get twenty three!

(iii) I am a special number
Take away from me a six!
A whole cricket team
You will still be able to fix!

(iv) Tell me who I am
I shall give a pretty clue!
You will get me back
If you take me out of twenty two!
Answer:
(i) According to the condition,
1 + 12 = 34 or × + 12 = 34
∴ By inspection, we have
22 + 12 = 34
So, I am 22.

(ii) Let I am ‘×’.
We know that there are 7 days in a week.
∴ up-counting from × for 7, the sum = 23
By inspections, we have
16 + 7 = 23
∴ × = 16
Thus I am 16.

(iii) Let the special number be x and there are 11 players in cricket team.
∴ Special Number -6 = 11
× – 6 = 11
By inspection, we get
17 – 6 = 11
∴ × = 17
Thus I am 17.

(iv) Suppose I am ‘×’
∴ 22 – 1 = 1
or 22 – × = ×
By inspection, we have
22 -11 = 11
∴ × = 11
Thus I am 11.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5 Read More »

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3

These NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 6 Integers Exercise 6.3

Question 1.
Find:
(a) 35 – (20) = ……………..
(b) 72-(90) = ………………
(c) (-15)-(-18) = ………………..
(d) (-20)-(13) = ………………….
(e) 23-(-12) = ……………………
(f) (-32) – (-40) = ……………………..
Answer:
(a) 35 – (20) = 15
(b) 72-(90) = -18
(c) (-15)-(-18) = -15+ 18 = 3
(d) (-20)-(13) = -20- 13 = -33
(e) 23-(-12) = 23 + 12 = 35
(f) (-32) – (-40) = -32 + 40 = 8

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3

Question 2.
Fill in the blanks with >, < or = sign:
Answer:
(a) (-3) + (-6) ………….. (-3) – (-6)
(b) (-21) – ( – 10) ………….. (-31) + (-11)
(c) 45 – (-11) ………….. 57 +(-4)
(d) (-25) – (-42) ………….. (-42) – (-25)
Answer:
(a) (-3) + (-6) < (-3) – (-6)
(b) (-21) – ( – 10) > (-31) + (-11)
(c) 45 – (-11) > 57 +(-4)
(d) (-25) – (-42)_>_ (-42) – (-25)

Question 3.
Fill in the blanks:
(a) (-8) + ……….. = 0
(b) 13 + ……….. =0
(c) 12 + ……….. = 0
(d) (-4) + ……….. =-12
(e) ……….. – 15 = -10
Answer:
(a) (-8) + 8 = 0
(b) 13+ (-13) =0
(c) 12+ (-12) =0
(d) (-4) + (-8) =-12
(e) 5 – 15 = -10

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3

Question 4.
Find:
(a) (-7)-8-(-25)
(b) (-13)+ 32-8-1
(c) (-7) + (-8) + (-90)
(d) 50 – (-40) – (-2)
Answer:
(a) (-7) – 8 – (-25) = -7 – 8 + 25
= -15 + 25= 10
(b) (-13)+ 32-8-1
= -13 + 32 – 8 – 1 = 32 -22 = 10
(c) (-7) + (-8) + (-90) = -7 – 8 – 90 = -105
(d) 50 – (-40) – (-2) = 50 + 40 + 2 = 92

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.3 Read More »

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2

These NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 6 Integers Exercise 6.2

Question 1.
Using the number line write the integer which is:
(a) 3 more than 5
(b) 5 more than -5
(c) 6 less than 2
(d) 3 less than -2
Answer:
(a) 8
NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2 1

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2

Question 2.
Use number line and add the following integers:
(a) 9 +(-6)
(b) 5 + (-11)
(c) (-1) + (-7)
(d) (-5) + 10
(e) (-1) + (-2) + (-3)
(f) (-2)+ 8 +(-4)
Answer:
(a) 9 + (-6) = 3
NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2 2
(b) 5 + (-11) = -6
NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2 3
(c) ( – 1) + ( – 7) = -8
NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2 4
(d) (-5) + 10 = 5
NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2 7
(e) ( – 1) + ( – 2) + ( – 3) = -6
NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2 5
(f) ( – 2) + 8 + ( – 4) = 2
NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2 6

Question 3.
Add without using number line:
(a) 11 +(-7)
(b) (-13)+ (+18)
(c) (-10)+ (+19)
(d) (-250) + (+ 150)
(e) (-380) + (-270)
(f) (-217) + (-100)
Answer:
(a) 11 +(-7) = 11-7 = 4
(b) (-13)+ 18 = 5
(c) (-10)+ (+19) = -10+19 = 9
(d) (-250) + (+150) = -250 + 150 = -100
(e) (-380) + (-270) = -380 – 270 = -650
(f) (-217)+ (-100) = -217-100 = -317

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2

Question 4.
Find the sum of:
(a) 137 and -354
(b) -52 and 52
(c) -312,39 and 192
(d) -50,-200 and 300
Answer:
(a) 137 + (-354) = 137 – 354 = -217
(b) (-52)+ 52 = 0
(c) (-312)+ 39 +192
= -312 + 231 = -81
(d) (-50) + (-200) + 300
= -50 – 200 + 300 = -250 + 300 = 50

Question 5.
Find the sum:
(a) (-7) + (-9) + 4 + 16
(b) (37) + (-2) + (-65) + (-8)
Answer:
(a) (-7) + (-9) + 4 + 16
= -7-9 + 4+ 16 =-16 + 20 = 4
(b) 37 +(-2) + (-65) + (-8)
= 37 – 2 – 65 – 8 = 37 – 75 = -38

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.2 Read More »

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.1

These NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 6 Integers Exercise 6.1

Question 1.
Write opposites of the following:
(a) Increase in weight
(b) 30 km north
(c) 80 m east
(d) Loss of? 700
(e) 100 m above sea level
Answer:
(a) Decrease in weight
(b) 30 km south
(c) 80 m west
(d) Profit of ₹ 700
(e) 100 m below sea level

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.1

Question 2.
Represent the following numbers as integers with appropriate signs.
(a) An aeroplane is flying at a height two thousand metre above the ground.
(b) A submarine is moving at a depth, eight hundred metre below the sea level.
(c) A deposit of rupees two hundred.
(d) Withdrawal of rupees seven hundred.
Answer:
(a) (+) 2000 meters
(b) (-) 800 meters
(c) (+) 200?
(d) (-) 700 ?

Question 3.
Represent the following numbers on a number line:
(a) +5
(b) -10
(c) +8
(d) -1
(e) -6
Answer:
NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.1 1

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.1

Question 4.
Adjacent figure is a vertical number line, representing integers. Observe it and locate the following points:
(a) If point D is + 8, then which point is -8?
(b) Is point G a negative integer or a positive integer?
(c) Write integers for points B and E.
(d) Which point marked on this number line has the least value?
(e) Arrange all the points in decreasing order of value.
Answer:
(a) F
(b) Negative integer
(c) B = (+) 4; E = (-) 10
(d) E
(e) D, C, B, A, O, H, G, F, E

Question 5.
Following is the list of temperatures of five places in India on a particular day of the year.

PlaceTemperature
Siachen10°C below 0°C……..
Shimla2°C below 0°C……..
Ahmedabad30°C above 0°C………
Delhi20°C above 0°C………
Srinagar5°C below 0°C……..

(a) Write the temperatures of these places in the form of integers in the blank column.
(b) Following is the number line representing the temperature in degree Celsius.
Plot the name of the city against its temperature.
NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.1 2
(c) Which is the coolest place?
(d) Write the names of the places where temperatures are above 10°C.
Answer:
Place – Temperature
Siachen – (-) 10° C
Shimla – (-) 2° C
Ahmedabad – (+) 30° C
Delhi – (+) 20° C
Srinagar Number line – (-) 5° C
NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.1 3
(c) Siachen (-10°C) is the coolest place.
(d) Ahmedabad (+30°C) and Delhi (+20°C).

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.1

Question 6.
In each of the following pairs, which number is to the right of the other on the number line?
(a) 2,9
(b) -3, -8
(c) 0,-1
(d) -11, 10
(e) -6,6
(f) 1,-100
Answer:
(a) 9 is right to 2.
(b) -3 is right to -8.
(c) 0 is right to-1.
(d) 10 is right to-11.
(e) 6 is right to -6.
(f) 1 is right to-100.

Question 7.
Write all the integers between the given pairs (write them in the increasing order.)
(a) 0 and -7
(b) -4 and 4
(c) -8 and -15
(d) -30 and -23
Answer:
(a) -6, -5, -4, -3, -2,-1
(b) -3, -2,-1,0, 1,2,3
(c) -14,-13,-12,-11,-10, -9
(d) -29, -28, -27, -26, -25, -24

Question 8.
(a) Write four negative integers greater than -20.
(b) Write four integers less than -10.
Answer:
(a) -19, -18, -17, -16
(b) -11, -12, -13, -14

Question 9.
For the following statements, write True (T) or False (F). If the statement is false, correct the statement.
(a) -8 is to the right of-10 on a number line.
(b) -100 is to the right of -50 on a number line.
(c) Smallest negative integer is -1.
(d) -26 is greater than -25.
Answer:
(a) True
(b) False
(c) False
(d) False

NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.1

Question 10.
Draw a number line and answer the following:
(a) Which number will we reach if we move 4 numbers to the right of -2.
(b) Which number will we reach if we move 5 numbers to the left of 1.
(c) If we are at -8 on the number line, in which direction should we move to reach -13?
(d) If we are at -6 on the number line, in which direction should we move to reach -1?
Answer:
NCERT Solutions for Class 6 Maths Chapter 6 Integers Ex 6.1 4
(c) On left side
(d) On right side

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NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.4

These NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Exercise 11.4

Question 1.
Answer the following:
(a) Take Sarita’s present age to be y years
(i) What will be her age 5 years from now?
(ii) What was her age 3 years back?
(iii) Sarita’s grandfather is 6 times her age. What is the age of her grandfather?
(iv) Grandmother is 2 years younger than grandfather. What is grandmother’s age?
(v) Sarita’s father’s age is 5 years more than 3 times Sarita’s age. What is her father’s age?

(b) The length of a rectangular hall is 4 meters less than 3 times the breadth of the hall. What is the length, if the breadth is b meters?

(c) A rectangular box has height h cm. Its length is S times the height and breadth is 10 cm less than the
length. Express the length and the breadth of the box in terms of the height.

(d) Meena, Beena and Leena are climbing the steps to the hill top. Meena is at step s, Beena is 8 steps ahead and Leena 7 steps behind. Where are Beena and Meena? The total number of steps to the hill top is 10 less than 4 times what Meena has reached. Express the total number of steps using s.

(e) A bus travels at v km per hour. It is going from Daspur to Beespur. After the bus has travelled 5 hours, Beespur is still 20 km away. What is the distance from Daspur to Beespur? Express it using v.
Answer:
(a)
(i) y + 5
(ii) y – 3
(iii) 6 y
(iv) 6y – 2
(v) 3y + 5

(b) Length = 3b and Breadth
= (3b – 4) meters

(c) Height of the box = h cm
Length of the box
= 5 times the height = 5h cm
Breadth of the box
= 10 cm less than length
= (5h – 10) cm

(d) Meena’s position = s
Beena’s position
= 8 steps ahead = s + 8
Leenas position
= 7 steps behind = s – 7
Total number of steps = 4s – 10

(e) Speed of the bus = v km/h
Distance travelled in 5 hours = 5v km
Remaining distance = 20 km
Therefore, total distance = (5v + 20) km

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.4

Question 2.
Change the following statements using expressions into statements in ordinary language.
(For example, Given Salim scores r runs in a cricket match, Nalin scores
(r + 15) runs. In ordinary language- Nalin scores 15 runs more than Salim.)

(a) A note book costs ₹ p. A book costs ₹ 3 p.
(b) Tony puts q marbles on the table. He has 8 q marbles in his box.
(c) Our class has n students. The school has 20 n students.
(d) Jaggu is z years old. His uncle is 4 z years old and his aunt is (4z – 3) years old.
(e) In an arrangement of dots there are r rows. Each row contains 5 dots.
Answer:
(a) A book cost 3 times the cost of a notebook.
(b) The number of marbles in box is 8 times the marble on the table.
(c) Total number of students in the school is 20 times that in our class.
(d) Jaggu’s uncle’s age is 4 times the age of Jaggu. Jaggu’s aunt is 3 years younger than his uncle.
(e) The total number of dots is 5 times the number of rows.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.4

Question 3.
(a) Given Munnu’s age to hex years, can you guess what (x – 2) may show? (Hint: Think of Mannu’s younger brother.)
Can you guess what (x + 4) may show? What (3x + 7) may show?

(b) Given Sara’s age today to be y years.
Think of her age in the future or in the past.
What will the following expression indicate?
y + 7, y – 3, y + 4\(\frac { 1 }{ 2 }\), y – 2 \(\frac { 1 }{ 2 }\)

(c) Given n students in the class like football, what may 2n show? What may \(\frac { n }{ 2 }\) show? (Hint: Think of games other than football).

Answer:
(a) Munnu’s age = × years
His younger brother is 2 years younger than him = (x – 2) years
His elder brother’s age is 4 years more than his age = (x + 4) years
His father is 7 year’s more than thrice of his age = (3x + 7) years

(b) Her age in past = (y = 3),\(\left(y=2 \frac{1}{2}\right)\)
Her age in future = (y + 7),\(\left(y=4 \frac{1}{2}\right)\)

(c) Number of students like hockey is twice the students liking football, i.e., 2n
Number of students like tennis is half the students like football, i.e.,\(\frac{n}{2}\)

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