Class 7

Haryana State Board HBSE 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.7 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 2 Fractions and Decimals Exercise 2.7

Question 1.
Find :
(i) 0.4 ÷ 2
(ii) 0.35 ÷ 5
(iii) 2.48 ÷ 4
(iv) 65.4 ÷ 6
(v) 651.2 ÷ 4
(vi) 14.49 ÷ 7
(vii) 3.96 ÷ 4
(viii) 0.80 ÷ 5
Solution:
(i) 0.4 ÷ 2 = 0.2
(ii) 0.35 ÷ 5 = 0.07
(iii) 2.48 ÷ 4 = \(\frac{218}{400}=\frac{62}{100}\) = 0.62
(iv) 65.6 ÷ 6 = 10.93
(v) 651.2 ÷ 4 = 162.8
(vi) 14.49 ÷ 7 = 2.7
(vii) 3.96 ÷ 4 = 0.99
(viii) 0.80 ÷ 5 = 0.16

Question 2.
Find:
(i) 4.8 ÷ 10
(ii) 52.5 ÷ 10
(iii) 0.7 ÷ 10
(iv) 33.1 ÷ 10
(v) 272.23 ÷ 10
(vi) 0.56 ÷ 10
(vii) 3.97 ÷ 10
Solution:
(i) 4.8 ÷ 10 = 0.48
(ii) 52.5 ÷ 10 = 5.25
(iii) 0.7 ÷ 10 = 0.07
(iv) 33.1 ÷ 10 = 3.31
(v) 272.23 ÷ 10 = 27.223
(vi) 0.56 ÷ 10 = 0.056
(vii) 3.97 ÷ 10 = 0.397

Question 3.
Find :
(i) 2.7 ÷ 100
(ii) 0.3 ÷ 100
(iii) 0.78 ÷ 100
(iv) 432.6 ÷ 100
(v) 23.6 ÷ 100
(vi) 98.53 ÷ 100
Solution:
(i) 2.7 ÷ 100 = \(\frac{2.7}{100}\) = 0.027
(ii) 0.3 ÷ 100 = 0.003
(iii) 0.78 ÷ 100 = 0.0078
(iv) 432.6 ÷ 100 = 4.326
(v) 23.6 ÷ 100 = 0.236
(vi) 98.53 ÷ 100 = 0.9853

Question 4.
Find:
(i) 7.9 ÷ 1000
(ii) 26.3 ÷ 1000
(iii) 38.53 ÷ 1000
(iv) 128.9 ÷ 1000
(v) 0.5 ÷ 1000
Solution:
(i) 7.9 ÷ 1000 = \(\frac{79}{1000}\) = 0.0079
(ii) 26.3 ÷ 1000 = \(\frac{263}{1000}\) = 0.0263
(iii) 38.53 ÷ 1000 = \(\frac{38.50}{10000}\) = 0.03850
(iv) 128.9 ÷ 1000 = \(\frac{1289}{10000}\) = 0.1289
(v) 0.5 ÷ 1000 = \(\frac{05}{10000}\) = 0.0005

Question 5.
Find :
(i) 7÷ 3.5
(ii) 36 ÷ 0.2
(iii) 3.25 ÷ 0.5
(iv) 30.94 ÷ 0.7
(v) 0.5 ÷ 0.25
(vi) 7.75 ÷ 0.25
(vii) 76.5 ÷ 0.15
(viii) 37.8 ÷ 1.4
(ix) 2.73 ÷ 1.3
Solution:
(i) 7÷ 3.5 = \(\frac{70}{35}\) = 2
(ii) 36 ÷ 0.2 = \(\frac{360}{2}\) = 180
(iii) 3.25 ÷ 0.5 = \(\frac{325}{50}\) = \(\frac{65}{10}\) = 6.5
(iv) 30.94 ÷ 0.7 = \(\frac{3094}{700}=\frac{442}{100}\) = 4.42
(v) 0.5 ÷ 0.25 = \(\frac{50}{25}\) = 2
(vi) 7.75 ÷ 0.25 = \(\frac{775}{25}\) = 31
(vii) 76.5 ÷ 0.15 = \(\frac{7650}{15}\) = 510
(viii) 37.8 ÷ 1.4 = \(\frac{378}{14}\) = 27
(ix) 2.73 ÷ 1.3 = \(\frac{2.73}{130}\) = 21

Question 6.
A vehicle covers a distance of 43.2 km. in 2.4 litres of petrol. How much distance will it travel in one litre of petrol ?
Solution:
∵ 2.4 litres of petrol a vehicle covers a distance of 43.2 km.
∴ 1 litre of petrol a vehicle covers
\(=\frac{43.2}{2.4} \mathrm{~km}=\frac{432}{24}=\frac{108}{6}=18 \mathrm{~km}\)

Haryana State Board HBSE 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.6 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 2 Fractions and Decimals Exercise 2.6

Question 1.
Find:
(i) 0.2 × 6
(ii) 8 × 4.6
(iii) 2.71 × 5
(iv) 20.1 × 4
(v) 0.05 × 7
(vi) 211.02 × 4
(vii) 2 × 0.186
Solution:
(i) 0.2 × 6 = 1.2
(ii) 8 × 4.6 = 36.8
(iii) 2.71 × 5 = 13.55
(iv) 20.1 × 4 = 80.4
(v) 0.05 × 7 = 0.35
(vi) 211.02 × 4 = 844.08
(vii) 2 × 0.86 = 1.72.

Question 2.
Find the area of rectangle whose length is 5.7 cm. and breadth is 3 cm.

Solution:
Area of rectangle = l × b = 5.7× 3 = 17.1cm².

Question 3.
Find :
(i) 1.3 × 10
(ii) 36.8 × 10
(iii) 153.7 × 10
(iv) 168.07 × 10
(v) 31.1 × 100
(vi) 156.1 × 100
(vii) 3.62 × 100
(viii) 43.07 × 100
(ix) 0.5 × 10
(x) 0.08 × 10
(xi) 0.9 × 100
(xii) 0.03 × 1000
Solution:
(i) 1.3 × 10 = 13
(ii) 36.8 × 10 = 368
(iii) 153.7 × 10 = 1537
(iv) 168.07 × 10 = 1680.7
(v) 31.1 × 100 = 3110
(vi) 156.1 × 100 = 15610
(vii) 3.62 × 100 = 362
(viii) 43.07 × 100 = 4307
(ix) 0.5 × 10 = 5
(x) 0.08 × 10 = 0.8
(xi) 0.9 × 100 = 90
(xii) 0.03 × 1000 = 3

Question 4.
A two wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol ?
Solution:
Total distance cover = 55.3 × 10 = 553 km.

Question 5.
Find:
(i) 2.5 × 0.3
(ii) 0.1 × 51.7
(iii) 0.2 × 316.8
(iv) 1.3 × 3.1
(v) 0.5 × 0.05
(vi) 11.2 × 0.15
(vii) 1.07  × 0.02
(viii) 10.05 × 1.05
(ix) 101.01 × 0.01
(x) 100.01 × 1.1
(xi) 211.02 × 11.32
(xii) 13.01 × 5.01
Solution:
(i) 2.5 × 0.3 = 0.75
(ii) 0.1 × 51.7 – 5.17
(iii) 0.2 × 316.8 = 63.36
(iv) 1.3 × 3.1 = 4.03
(v) 0.5 × 0.05 = 0.025
(vi) 11.2 × 0.15 = 1.680
(vii) 1.07 × 0.02 = 0.0214
(viii) 10.05 × 1.05 = 10.5525
(ix) 101.01 × 0.01 = 1.0101
(x) 100.01 × 1.1 = 110.011.
(xi) 211.02 × 11.32 = 2388.7464
(xii) 13.01 × 5.01 = 65.1801

Haryana State Board HBSE 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.5 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 2 Fractions and Decimals Exercise 2.5

Question 1.
What is greater ?
(i) 0.5 or 0.05
(ii) 0.7 or 0.5
(iii) 7 or 0.7
(iv) 1.37 or 1.49
(v) 2.03 or 2.30
(vi) 0.8 or 0.88.
Solution:
(i) 0.5 > 0.05 [Because 5 > 0]
(ii) 0.7 > 0.5 [Because .7 > .5]
(iii) 7 > 0.7
(iv) 1.49 > 1.37 [Because 4 > 3]
(v) 2.30 > 2.03 [Because 3 > 0]
(vi) 0.88 > 0.80. [Because 8 > 0]

Question 2.
Express as rupees using decimals.
(i) 7paise
(ii) 7 rupees 7paise
(iii) 77 rupees 77paise
(iv) 50paise
(v) 235paise.
Solution:
(i) ∵ 100 paise = 1 Rupee
∴ paise = 0.07 Rupees
(ii) 7 rupees 7 paise = 7.07 Rupees.
(iii) 77 rupees 77 paise = 77.77 Rupees.
(iv) 50 paise = 0.50 Rupees
(v) 235 paise = \(\frac{235}{100}\)
Rupees = 2.35 Rupees.

Question 3.
(i) Express 5 cm in metre and kilometre.
(ii) Express 35 mm in cm, m and km.
Solution:
(i) ∵ 100 cm = 1 m and 1000 m
= 1 km
5 cm = 0.05 m and \(\frac{0.05}{1000}\) = 0.00005 km

(ii) 35 mm – 3.5 cm = \(\frac{3.5}{100}\) = 0.035 m = \(\frac{0.035}{1000}\)
= 0.000035 km

Question 4.
Express in kg.
(i) 200g
(ii) 3470g
(iii) 4 kg 8 g
(iv) 2598 mg
Solution:
(i) ∵ 1000 gm = 1 kg
200g = \(\frac{200}{1000}=\frac{2}{10}\) = 0.2kg
(ii) 3470 g = \(\frac{3470}{1000}=\frac{347}{100}\) = 3.47 kg
(iii) 4 kg 8g = 4.008 kg
(iv) 2598 mg = \(\frac{2508}{1000}\)
= \(\frac{2598}{1000 \times 1000} \mathrm{~kg}=\frac{2598}{1000000} \mathrm{~kg}\) = 0.02598 kg

Question 5.
Write the following decimal numbers in the expanded form :
(i) 20.03
(ii) 2.03
(iii) 200.03
(iv) 2.034
Solution:
(i) 20.03 = 2 x 10 + 0 x 1 + 0 x (\(\frac{1}{10}\)) + 3 x (\(\frac{1}{100}\))
(ii) 2.03 = 2 x 1 + 0 x (\(\frac{1}{10}\)) + 3 x (\(\frac{1}{100}\))
(iii) 200.03 = 2 x 100 + 0 x 10 + 0 x 1 + 0 x (\(\frac{1}{10}\)) + 3 x (\(\frac{1}{100}\)) + 4 x (\(\frac{1}{1000}\))

Question 6.
Write the place value of 2 in the following decimal numbers :
(i) 2.56
(ii) 21.37
(iii) 10.25
(iv) 9.42
(v) 63.352.
Solution:
(i) 2.56 Place value of 2 = Ones
(ii) 21.37 Place value of 2 = Tens
(iii) 10.25 Place value of 2 = Tenths
(iv) 9.42 Place value of 2 = Hundredths
(v) 63.352 Place value of 2 = Thousandths.

Question 7.
Dinesh went from place A to place B and from there to place C. A is 7.5 km from B and B is
12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. Who travelled more and by how much ?
Solution:
Total travelled distance of Dinesh = 7.5 km + 12.7 km = 20.2 km
Total travelled distance of Ayub = 9.3 km + 11.8 km = 21.1 km
or, 21.1 km – 20.2 km = 0.9 km
∴ Ayub travelled 0.9 km more.

Question 8.
Shyama bought 5 kg 300g apples and 3 kg 250g mangoes. Sarala bought 4 kg 800g apples and 4 kg 150g bananas. Who bought more fruits ?
Solution:

or, 8.950 kg – 8.550 kg = 0.400 kg
.’. Sarala bought 400 g more fruits.

Question 9.
How much less is 28 km than 42.6 km?
Solution:

Haryana State Board HBSE 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.4 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 2 Fractions and Decimals Exercise 2.4

Question 1.
Find

Solution:
0

Question 2.


Solution:

Question 3.

Solution:

Question 4.

Solution:

Haryana State Board HBSE 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.3 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 2 Fractions and Decimals Exercise 2.3

Question 1.
Find

Solution:

Question 2.
Multiply and reduce to lowest form (if possible) :

Solution:

Question 3.
Multiply the following fractions :

Solution:

Question 4.
Which is greater?

Solution:

Question 5.
Saili plants 4 saplings in a row, in her garden. The distance between two adjacent saplings is \(\frac{3}{4}\) m. Find the distance between the first and the last sapling.
Solution:
The distance between two adjacent saplings = \(\frac{3}{4}\) m
There are 4 saplings in a row.
∴ The distance between the first and the last sapling = \(\frac{3}{4}\) x 3 = \(\frac{9}{4}\) m.

Question 6.
Lipika reads a book for 1 \(\frac{3}{4}\) hours every day. She reads the entire book in 6 days. IIow many hours in all were required by her to read the book ?
Solution:
Lipika reads a book for \(\frac{7}{4}\) hours everyday.
∴ Required hours = \(\frac{7}{4} \times 6=\frac{7 \times 3}{2}=\frac{21}{2}=10 \frac{1}{2}\) hours

Question 7.
A car runs 16 km. using 1 litre of petrol. How much distance will it cover using 2\(\frac{3}{4}\) litres of petrol ?
Solution:
A car runs 16 km. using 1 litre of petrol.
∴ Total distance cover =16 x 2\(\frac{3}{4}\)
= 16 x \(\frac{11}{4}\) = 44 km.

Question 8.

Solution:

Haryana State Board HBSE 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.2 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 2 Fractions and Decimals Exercise 2.2

Question 1.
Which of the drawings (a) to (d) show :


Solution:
(i)—(d), (ii)—(b), (iii)—(a), (iv)—(c).

Question 2.
Some pictures (a) to (c) are given below. Tell which of them shows :

Solution:
(i)—(c), (ii)—(a), (iii)—(b).

Question 3.
Multiply and reduce to lowest form :

Solution:

Question 4.
Shade:
(i) \(\frac{1}{2}\) of the circle in box (a)
(ii) \(\frac{2}{3}\) of the triangles in box (b).
(iii) \(\frac{3}{5}\) of the boaxes in box (c).

Solution:

Question 5.
(i) \(\frac{1}{2}\) of (i) 24 (ii) 46
(ii) \(\frac{2}{3}\) of (i) 18 (ii) 27
(iii) \(\frac{3}{5}\) of (i) 16 (ii) 36
(iv) \(\frac{4}{5}\) of (i) 20 (ii) 35
Solution:

Question 6.
Multiply and express as a mixed fraction :

Solution:

Question 7.
Find:

Solution:

Question 8.
Vidya and Pratap went for a picnic. Their mother gave them a water bag that contained 5 litres of water. Vidya consumed \(\frac{2}{5}\) the remaining water. Pratap consumed the remaining water.
(i) How much water did Vidya drink ?
(ii) What fraction of the total quantity of water did Pratap drink ?
Solution:
(i) Vidya consumed \(\frac{2}{5}\) of the water
∴ 5 litres of \(\frac{2}{5}\) = 5 x \(\frac{2}{5}\) = 2 litres
∴ Vidya drink = 2 litres.
(ii) Now 5 litres – 2 litres = 3 litres 3
∴ Pratap drink = \(\frac{3}{1}\) litres = 3 litres.

Haryana State Board HBSE 7th Class Maths Solutions Chapter 2 Fractions and Decimals Ex 2.1 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 2 Fractions and Decimals Exercise 2.1

Question 1.
Solve :

Solution:

Question 2.
Arrange the following in descending order:
(i) \(\frac{2}{9}, \frac{2}{3}, \frac{8}{21}\)
(ii) \(\frac{1}{5}, \frac{3}{7}, \frac{7}{10}\)
Solution:

Question 3.
In a “magic square”, the sum of the numbers in each row, in each column and along the diagonal is the same. Is this a magic square ?

Solution:

Question 4.
A rectangular sheet of paper is 12 \(\frac{1}{2}\)cm long and 10\(\frac{2}{3}\)cm wide. Find its perimeter.
Solution:
Perimeter of a rectangular sheet of paper

Question 5.
Find the perimeters (i) ΔABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater ?
Solution:
(i) Perimeter of ΔABE
= AB + BC + EA


Perimeter of ΔABE is greater than the rectangle BCDE.

Question 6.
Salil wants to put a picture-in a frame. The picture is 7\(\frac{3}{5}\) cm wide. To fit in the frame the picture cannot be more than 7\(\frac{3}{10}\) cm wide. How much should the picture be trimmed ?
Solution:
The picture, should be trimmed

Question 7.
Ritu ate \(\frac{3}{5}\) part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat ? Who had the larger share 7 By how much ?
Solution:
Ritu ate \(\frac{3}{5}\) part of an apple.
Remaining apple = \(1-\frac{3}{5}=\frac{5-3}{5}=\frac{2}{5}\)
Somu ate \(\frac{2}{5}\) part of an apple \(\frac{3}{5}>\frac{2}{5}\)
Ritu had the larger share,
Now \(\frac{3}{5}-\frac{2}{5}=\frac{3-2}{5}=\frac{1}{5}\)

Question 8.
Michael finished colouring a picture \(\frac{7}{12}\) in hour. Vaibhav finished colouring the same picture in \(\frac{3}{4}\) hour. Who worked longer?
By what fraction was it longer ?
Solution:
Michael finished colouring a pictuie in \(\frac{7}{12}\) hour
Vaibhav finished colouring the same picture in \(\frac{3}{4}\) hour.

Haryana State Board HBSE 7th Class Maths Solutions Chapter 1 Integers InText Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 1 Integers InText Questions

Try These (Page 2) :

Question 1.
Number line representing integers is given below :

– 3 and – 2 are marked by E and F respectively. Which integers are marked by alphabets B, D, H, J, M, O ?
Solution:

Question 2.
Arrange 7, – 5, 4, 0, and – 4 in ascending order and then mark them on a number line to check your answer.
Solution:

Try These (Page 3) :

Question 1.
We have done various patterns with numbers in our previous class.
Can you find a pattern for each of the following ? If yes, complate them :
(a) 7,3, -1, 5, ………… , ………… , …………
(b) -2,-4,-6,-8, ………… , ………… , …………
(c) 15, 10, 5, 0, ………… , ………… , …………
id) – 11, -8,-5,-2, ………… , ………… , …………
Solution:
(a) 7, 3, -1, 5, -9 . -13 . -14
(b) -2,-4, – 6, – 8, -10 , – 12, -14
(c) 15, 10, 5, 0, -5 . -10 . – 15
(d) – 11, – 8, – 5, – 2, 1, 4, 7

Try These (Page 8) :

Question 1.
Write pair of integers whose sum gives:
(a) A negative integer.
(b) Zero.
(c) An integer smaller than both the integers.
(d) An integer smaller than only one of the integers.
(e) An integer greater than both the integers.
Solution:
(a) – 3 + (- 3) = – 3 – 3 = – 6
(b) 3 + (-3) = 3 -3 = 0
(c) 3 + (- 2) = 3 – 2 = 1
(d) – 3 + (- 1) = – 3 – 1 = – 4
(e) 4 + 5 = 9

Question 2.
Write a pair of integers whose diffeences gives:
(a) A negative integer.
(b) Zero.
(c) An integer smaller than both the integers.
(d) An integer greater than only one of the integers.
(e) An integer greater than both the integers.
Solution:
(a) – 3 – (+ 3) = – 3 – 3 = – 6
(b) – 3 – (+ 3) = – 3 + 3 = – 0
(c) – 3 – (+ 2) = – 3 – 2 = – 5
(d) – 2 – (- 1) = – 2 + 1 = – 0
(e) – 3 – (- 4) = – 3 + 4 = 1

I. Try These (Page 10) :

Question 1.
Find 4 x (- 8), 8 x (- 2), 3 x (- 7), 10 x (¬1), using number line.
Solution:

(- 8) + (— 8) + (— 8) + (— 8) = -32
4 x (-8) = -32

(- 2) + (- 2) + (- 2) + (- 2) + (- 2) + (- 2) + (- 2) + (- 2) = – 16
8 x (- 2) = – 16

(- 7) + (- 7) + (- 7) = -21
3 x (- 7) = – 21

(— 1) + (— 1) + (— 1) + (— 1) + (— 1) + (— 1) + (— 1) + (- 1) + (- 1) + (- 1) = – 10
10 x (- 1) = – 10

Try These (Page 10) :

Question 1.
Find:
(i) 6 x (- 19)
(ii) 12 x (- 32)
(iii) 1 x (-22)
Solution:
(i) 6 x (- 19) = – (6 x 19) = – 114
(ii) 12 x(- 32) = -(12×32) = -384
(iii) 7 x (- 22) = – (7 x 22) = – 154

Try These (Page 11) :

Question 1.
Find (a) 15 x (- 16)
(b) 21 x (- 32)
(c) (- 42) x 12
(d) – 55 x 15
Solution:
(a) 15 x (-16) = -240 = -(15 x 16)
(b) 21 x(- 32) = -672 = -(21 x 32)
(c) (-42) x (12) = -504 = -(42 x 12)
(d) – 55 x 15 = – 825 = – (55 x 15)

Question 2.
Check if: (a) 25 x (-21) = (-25) x 21 (6) (-23) x 20 = 23 x (-20)
Write five more such examples.
Solution:
(a) 25 x (- 21) = (- 25) x 21
-(25 x 21) = -(25 x 21)
-525 = -525
(- 23) x 20 = 23 x (- 20)
-(23 x 20) = -(23 x 20)
-460 = -460

1. 12 x (-11) = (— 11) x 12
2. 13 x (-12) = 0-13) x 12
3. 20 x (-19) = (— 20) x 19
4. (-21) x(20) = 21 x (- 20)
5. (- 24) x (23) = 24 x (— 23)

Try These (Page 12) :

Question 1.
(i) Starting from (- 5) x 4, find (- 5) x (- 6)
(ii) Starting from (- 6) x 3, find (- 6) x (- 7)
Solution:
(i) -5 x 4 = -20
-5 x 3 = -15
-5 x 2 = -10
-5 x 1 = -5
-5 x 0 = 0
– 5 x – 1 = + 5
– 5 x – 2 = + 10
– 5 x – 3 = + 15
– 5 x – 4 = + 20
– 5 x – 5 = + 25
– 5 x – 6 = + 30

(ii) -6 x 3 = -18
-6 x 2 = -12
-6 x 1 = -6
-6 x 0 = 0
-6 x – 1 = + 6
– 6 x – 2 = + 12
– 6 x – 3 = + 18
– 6 x – 4 = + 24
– 6 x – 5 – + 30
– 6 x – 6 = + 36
– 6 x – 7 = + 42

Try These (Page 12) :

Question 1.
Find (- 31) x (-100), (- 25) x (- 72), (- 83) x (- 28)
Solution:
(- 31) x (-100) = 3100
(-25) x (-72) = 1800
(-83) x (-28) = 2324

I. Try These (Page 18) :

Question 1.
(i) Is 10 x [6 + (-2)] = 10 x 6 + 10 x (-2) ?
(ii) Is (-15) x [(- 7) + (-1)] = (-15) x (- 7) + (- 5) x (- 1) ?
Solution:
(i) 10 x [6 + (- 2)] = 10 x 6 +10 x (-2) 10 x 4 = 60-20 40 = 40
(ii)(- 15) x [(- 7) + (- 1)] = (- 15) x (- 7) + (- 5) x (- 1)
— 15 x (— 8) = 105 + 5
120 = 120

I. Try These (Page 18) :

Question 1.
(i) Is 10 x [6- (-2)] = 10 x 6 – 10 x (- 2) ?
(ii) Is (- 15) x [(- 7) – (- 1)1 = (- 15) x (-7) – (- 15) x (- 1) ?
Solution:
(i) 10 x [6 + 2] = 60 + 20
10 x 8 = 80
80 = 80
(ii) – 15 x [- 7 + 1] = 105-15
15 x (- 6) = 90
=> 90 = 90

Try These (Page 18) :

Question 1.
Find (- 49) x 18; (- 25) x (- 31); 70 x (-19) + (- 1) x 70 using distributivity peroperly.
Solution:
(- 49) x 18 = (- 49) x [10 + 8]
= (- 49) x 10 + (- 49) x 8
= -490 – 392 = -882

(- 25) x (- 31) = (- 25) x [(- 30) + (- 1)]
= (- 25) x (- 30) + (- 25) x(- 1)
= 750 + 25 = 775

70 x (- 19) + (- 1) x 70
= 70 [(- 19) + (-1)]
= 70 [-19-1]
= 70 x (-20) = – 1400

Try These (Page 22) :

Question 1.
Find (a) (-100) ÷ 5, (b) (-81) ÷ 9, (c) (-75) ÷ 25, (d) (- 32) ÷ 2
Solution:
(a) (-100) ÷ 5 = -20
(b) (- 81) ÷ 9 = – 9
(c) (- 75) ÷ 25 = – 3
(d) (-32) ÷ 2 = -16

Try These (Page 23) :

Question 1.
Find (a) 125 ÷ (- 25), (b) 88 ÷ (- 5), (c) (64) ÷ (-16)
Solution:
(a) 125 ÷ (- 25) = – 5
(b) 80 ÷ (- 5) = -16
(c) 64 ÷ (- 16) = – 4

Try These (Page 23) :

Question 1.
Find (a) (- 36) ÷ (- 4), (b) (- 201) ÷ (- 3), (c) (-325) ÷ (-13)
Solution:
(a) (-36) ÷ (-4) =9
(6) (-201) ÷ (-3) = 67
(c) (-325) ÷ (-13) = 25.

Try These (Page 24) :

Question 1.
Is (i) Is 1 ÷ a = 1 ?
(ii) a ÷ (-1) = – a ? For any integer. Take different values of a and check.
Solution:
(i) 1 ÷ a = \(\frac{1}{a}=\frac{1}{a}\)
hence \(\frac{1}{a}\) ≠ 1 L.H.S ≠ R.H.S
Check, a = 1 then, 1 ÷ 1 = 1.
a = 2 then, 1 ÷ 2 = \(\frac{1}{2}\)
i.e. 1 ≠ \(\frac{1}{2}\) , hence, verified

(ii) a ÷ (- 1) = a x \(\frac{1}{-1}\) = -a
hence, L.H.S = R.H.S
i.e. – a = – a
Check, a = 1 then, 1 ÷ (- 1) = -1
– 1= – 1
a = 2
then, 2 ÷ (-1) = -2 => -2 = -2

Haryana State Board HBSE 7th Class Maths Solutions Chapter 1 Integers Ex 1.4 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 1 Integers Exercise 1.4

Question 1.
Evaluate each of the following:
(a) (-30) ÷ 10
(b) 50 ÷ ( -5)
(c) (-36) ÷ (-9)
(d) (-49) ÷ (49)
(e) 13 ÷ [(- 2) + 1]
(f) 0 ÷ (- 12)
(g) (-31) ÷ [(-30) ÷ (- 1)1
(h) [(-36) ÷ 12] ÷ 3
(i) [(-6)+ 5] ÷ [(-2) A -77
Solution:
(a) (-30) ÷ 10 = -3
(b) 50 ÷ (- 5) = -10
(c) (-36) ÷ (-9) = 4
(d) (-49) ÷ (49) = -1
(e) 13 ÷ (-1) = -13 (Meaningless)
(f) 0 ÷ (-12) = 0
(g) (- 31) ÷ [(- 30) + ( -1)]
= (-31) ÷ (-31) = 1
(h) [(- 36) ÷ 12] ÷ 3 = -3 ÷ 3 = – 1
(i) [(-6) + 6] ÷ [(-2) + 1]
= 0 ÷ (- 1) = 0 (Meaingless)

Question 2.
Verify that a ÷ (b + c) * (a ÷ b) + (a ÷ c) for each of the following values of a, b and c.
(а) a = 12, b = – 4, c = 2
(б) a = (- 10), 6 = 1, c = 1.
Solution:
(a) 12 ÷ (— 4 + 2) ≠ [12 ÷ (-4)]+ [12 ÷ 2]
12 ÷ (-2) ≠ (-3) + (6)
– 6 ≠ 3
(b) (- 10) ÷ (1 + 1) ≠ [(-10) ÷ 1] + [(- 10) ÷ 1]
(- 10) ÷ 2 ≠ (- 10) + (- 10)
– 5 ≠ – 20

Question 3.
Fill in the blanks :
(a) 369 ÷ = 369
(b) -75 ÷ = – 1
(c) (-206) + -1
(d) -87 = 87
(e) ÷ 1 = -87
(f) ÷ 48 = -l
(g) 20 ÷ = – 2
(h) ÷ (4) = – 3
Sol.
(a) 369 ÷ 1 = 369
(b) – 75 ÷ 75 = – 1
(c) ( -206) ÷ (-206) = 1
(d) (- 87) ÷ (- 1) = 87
(e) (-87) ÷ 1 = -87
(f) (-48) ÷ 48 = -1
(g) 20 ÷ (- 10) = – 2
(h) (- 12) ÷ 4 = – 3

Question 4.
Write five pairs of integers (a, 6) such that a ÷ b = – 3. One such pair is (6, – 2) because 6 ÷ (- 2) = (- 3).
Solution:
(i) (3, – 1) ⇒ .3 ÷ (- 1) = -3
(ii) (12,-4) ⇒ 12 ÷ (-4) = -3
(iii) (15,-5) ⇒ 15 ÷ (-5) = -3
(to) (18,-6) ⇒ 18 ÷ (-6) = -3
(v) (21,-7) ⇒ 21 ÷ (-7) = -3

Question 5.
The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2° per hour until mid-night, at what time would the temperature 8°C below zero ? What would be the temperature at midnight ?
Solution:

At 9 p.m. the temperature was 8 degree below zero and the temperature was – 14°C at mid night.

Question 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? (ii) Mohini scores -5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly? (iii) Rakesh scores 18 marks by attempting 16 questions. How many questions has he attempted correctly and how many has he attempted incorrectly?
Solution:
(i) Radhick scored = (3,000 x 8)
– (5,000 x 5)
Radhika has got 12 correct answers
= 12 x 3 = 36 marks
incorrect answers = 36-20 = 16 marks 16
hence no. of questions = 16/2 = 8
(ii) Mohini has got 7 correct answers
= 7 x 3 = 21 marks
incorrect answers = 21 – (- 5) = 26
hence no. of incorrect questions
= 26 ÷ 2 = 13

Question 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 – 350 m.
Solution:
Rate of Shaft = 6m / min.
Distance of descent from ground level = 10m
Total Distance = 10m + 350m = 360
Time taken to reach the mine = \(\frac{360 \mathrm{~m}}{6 \mathrm{~m}}\)
= 60m
= 1 hour

Haryana State Board HBSE 7th Class Maths Solutions Chapter 1 Integers Ex 1.3 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 1 Integers Exercise 1.3

Question 1.
Find each of the following products :
(a) 3 x (- 1)
(b) (- 1) x 225
(c) (-21) x (-30)
(d) (-316) x (- 1)
(e) (- 15) x 0 x (- 18)
(f) (-12) x(-11)x (10)
(g) 9 x(- 3) x (- 6)
(h) (-18) x (-5) x (-4)
(i) (- 1) x (- 2) x (- 3) x 4
(j) (- 3) x (- 6) x (- 2) x (- 1)
Solution:
(a) 3 x (- 1) = – 3
(b) (-1) x (225)=-225
(c) (- 21) x (- 30) = +(21×30)= + 630
(d) (-316) x (-1) = +(316 x 1) = + 316
(e) (-15) x 0 x (-18) = 0 x – 18 = 0
(f) (-12) x (-11) x 10 = + 132x 10 = 1320
(g) 9 x (- 3) x (- 6) = 9 x 18 = 162
(h) (- 18) x (- 5) x (- 4)=- 18 x 20 = – 360
(i) (- 1) x (- 2) x (- 3) x 4=2 x (- 12) = – 24
(j) (- 3) x (- 6) x (- 2) x (- 1)= 18 x 2 = 36

Question 2.
Verify the following:
(a) 18 x [7 + (- 3)] = [18 x 7] + [18 x (- 3)]
(b) (- 21) x [(- 4) + (- 6)] = [(- 21) x (- 4) + [ (-21) x (- 6)]
Solution:
(a) 18 x [7 + (- 3)] = [18 x 7] + [18 x (- 3)]
18 x [7 – 3] = 126 + (- 54)
18 x 4 = 126 – 54
72 =72
(b) (- 21) x [(- 4) + (- 6)] = [(- 21) x (- 4)] + [(- 21) x (- 6)]
– 21 x (- 10) = 84 + 126
210 =210

Question 3.
(i) For any integer what is (- 1) x a equal to ?
(ii) Determine the integer whose product with – 1 is
(a) -22 (6) 37 (c) 0
Solution:
(i) 1 x (-1) = -1
2 x (- 1) = – 2
– 2 x (- 1) = 2 etc.

(ii) (a) 22 x (- 1) = – 22
(b) (- 37) x (- 1) = 37
(c) 0 x (- 1) = 0

Question 4.
Starting from (- 1) x 5, write various products showing some pattern to show (- 1) x (- 1) = 1.
Solution:
– 1 x 5 = – 5
– 1 x 4 = – 4
-1 x 3 =-3
-1 x 2 = -2
-1 x 1 =-1
– 1 x 0 = 0
-1 x -1 = 1

Question 5.
Find the product, using suitable properties :
(a) 26 x (- 48) + (-48 x- 36)
(b) 8 x 53 x (- 125)
(c) 15 x (- 25) x (-4) x (- 10)
(d) (-41) x 102
(e) 625 x (- 35) + (- 625) x 65
(f) 7 x [50 – 2]
(g) (- 17) x (- 29)
(h) (-57) x (-19) + 57
Solution:
(a) 26 x (-48) + (-48 x – 38)
= -48[26-36] – 48 [26 – 36]
= – 48 x [-10] – 48 x [-10]= 480

(b) 8 x [53 x (- 125)]
= [8 x 53] x (- 125) 8 x (- 6625)
= 424 x(-125)-53000
= -53000

(c) 15 x [(- 25) x (- 4)] x – 10
= (15 x 100) x (- 10)
= 1500 x (- 10) = – 15000

(d) (- 41) x 102 =(- 41) x (100 + 2)
= (-41×100)+ (-41×2)
= -4100+ (-82) = -4182

(e) 625 x (- 35) + (- 625) x 65
= 625 [(- 35) + (- 1) x 65]
= 625 [-35-65]
= 625 x – 100 = – 62500

if) 7 x [50-2]
= [7 x 50] – [7 x 2]
= 350-14 = 336

(g) (- 17) x (- 29) = – 17 x [- 30 + 1]
= [(- 17) x (- 30)] + [(- 17) x 1]
= 510+ (-17) = 510-17 = 493

(h) (- 57) x (- 19) + 57
= 57 [(- 1) x (- 19) + 1]
= 57 [19 + 1] = 57 x 20 = 1140

Question 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 ?
Solution:
After 10 hours room temperature = 40°C – (5°C x 10) = 40°C – 50°C = – 10°C

Question 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 ans six incorrect answers. What is his score ?
(ii) Reshma gest five cored answers and five incorrect answer, what is her score?
(iii) Heena gets two correct and five incorrect answers out of seven questions she attempts. What is her score ?
Solution:
(i) 4 x 5 + 6 x (- 2) = 20 – 18 Mohan score = 2
(ii) 5 x 5 + 5 x (- 2) = 25-10 Reshma’s score = 15
(iii) 2 x 5 + 5 x (- 2) = 10-10 Heena’s score = 0

Question 8.
A cement company earns a profit of Rs. 8 per bag of white cement sold and a loss of Rs. 5 per hag of grey cement sold.
(a) The company sells 3,000 hags of white cement and 5,000 bags of grey cement in month, what is its profit or loss ?
(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.
Solution:
(a) = (3,000 x 8) – (5,000 x 5)
= 24,000-25,000 = -1000 loss = Rs. 1000 (b) No. of grey bags cement bags = 6,400
loss = 6,400 x 5 = Rs. 32,000 profit of white cement bags = Rs. 32,000
hence no. of white cement bags = 32000 ÷ 8 = 4000

Question 9.
Replace the blank with an integer to make it a true statement.
(a) (- 3) x …………….. = 27
(b) 5 x …………….. =(-35)
(c) …………….. x(-8) = (-56)
(d) …………….. x (- 12) = 132
Solution:
(a) (- 3) x (- 9) = 27
(6) 5 x (- 7) =-35
(c) 7 x (-8) =-56
id) (- 11) x (— 12) = 132