The DAV Class 7 Maths Solutions and **DAV Class 7 Maths Chapter 4 Worksheet 7** Solutions of Application of Percentage offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 4 WS 7 Solutions

Question 1.

Write each of the following numbers in the form k × 10^{n} where 1 ≤ k < 10 and n is an integer.

(i) 1,384,000

Answer:

1,384,000 = 1.384 × 10^{6}

(ii) 12.32005

Answer:

12.32005 = 1.232005 × 10^{1}

(iii) 2157.957

Answer:

2157.957 = 2.157957 × 10^{3}

(iv) 0.00002

Answer:

0.00002 = 2.0 × 10^{-5}

(v) 0.00729

Answer:

0.00729 = 7.29 × 10^{-3}

(vi) 0.000000000926

Answer:

0.000000000926 = 9.26 × 10^{10}

(vii) 520,000,000

Answer:

520,000,000 = 5.20 × 10^{8}

(viii) 0.0000085

Answer:

0.0000085 = 8.5 × 10^{6}

Question 2.

Write the following numbers in usual form:

(i) 52.5 × 10^{4}

Answer:

52.5 × 10^{4}

= 52.5 × 10000

= 525000

(ii) 158.9 × 10^{6}

Answer:

158.9 × 1000000

= 158900000

(iii) 9.545 × 10^{12}

Answer:

9.545 × 1000000000000

= 9545000000000

(iv) 1.72 × 10^{-5}

Answer:

1.72 × 10^{-5}

= 1.72 × \(\frac{1}{100000}\)

= 0.0000172

(v) 8.5 × 10^{-7}

Answer:

8.5 × 10^{-7}

= 8.5 × \(\frac{1}{10000000}\)

= 0.00000085

(vi) 2.9 × 10^{-9}

Answer:

2.9 × 10^{-9}

= 2.9 × \(\frac{1}{1000000000}\)

= 0.0000000029

(vii) 6293.2 × 10^{5}

Answer:

6293.2 × 10^{5}

= 6293.2 × 10000

= 629320000

(viii) 1.925 × 10^{-6}

Answer:

1.925 × 10^{-6}

= 1.925 × \(\frac{1}{1000000}\)

= 0.000001925

Question 3.

Express each of the following in the form k × 10^{n}; (1 ≤ -k < 10).

(i) (1.25 × 10^{7}) ÷ (5 × 10^{3})

Answer:

(1.25 × 10^{7}) ÷ (5 × 10^{3})

= \(\frac{1.25 \times 10^7}{5 \times 10^3}\)

= 0.25 × 10^{4}

= 2.5 × 10^{3}

(ii) (2.5 × 10^{10}) × (31.25 × 10^{-5})

Answer:

(2.5 × 10^{10}) × (31.25 × 10^{-5})

= 2.5 × 31.25 × 10^{10} × 10^{-5}

= 78.125 × 10^{5}

= 7.8125 × 10^{6}

(iii) (1.6 × 10^{-9}) × (5.0 × 10^{-3})

Answer:

(1.6 × 10^{-9}) × (5.0 × 10^{-3})

= 1.6 × 5.0 × 10^{-9} × 10^{-3}

= 8.0 × 10^{6}

(iv) [(3.4 × 10^{4}) × (5.0 × 10^{-3})] + [2.0 × 10^{5}]

Answer:

[(3.4 × 10^{4}) × (5.0 × 10^{-3})] + [2.0 × 10^{5}]

= (3.4 × 5.0 × 10^{4} × 10^{3}) – (2.0 × 10^{5})

= 17.0 × 10^{-5} – 2.0 × 10^{5}

= \(\frac{17.0 \times 10}{2.0 \times 10^5}\)

= 8.5 × 10^{-4}

Question 4.

Express the numbers appearing in the following statements in the form k × 10^{n} where 1 ≤ k < 10 and n is an integer.

(i) Sun’s diameter is 1,384,000 km.

Answer:

1,384,000 km

= 1.384 × 10^{6} km.

(ii) The distance of the sun from the earth is approximately 150,000,000 km.

Answer:

150,000,000 km

= 1.5 × 10^{8} km.

(iii) The speed of the light is about 27600 km/sec.

Answer:

27600 km/sec

= 2.76 × 10^{4} km/sec

(iv) 1 Angstrom unit = \(\frac{1}{10,000,000,000}\) m

Answer:

\(\frac{1}{10,000,000,000}\)m

= \(\frac{1}{10^{10}}\) m

= 1.0 × 10^{-10} m

### DAV Class 7 Maths Chapter 4 Value Based Questions

Question 1.

Sneha wants to show gratitude towards her teacher by giving her a self-made card. She chose a yellow coloured rectangular sheet of paper. The length of the sheet (25) cm and breadth is (24) cm.

(i) Find the area of the paper she has to work for making the card.

(ii) What value of Sneha is depicting here?

Solution:

(i) Area of the sheet of paper = Length of the sheet × Breadth of the sheet

= (2^{5}) cm × (2^{4}) cm

= (2)^{9} cm^{2}

= 512 cm^{2}

(ii) Values: She is grateful, loyal and respectful.

Question 2.

Aditya wanted to donate some cloth material to some poor people in a locality. For his mother purchased (4^{3}) m cloth and donated (2^{2}) m to each person.

(i) How many poor people would have received the cloth material?

(ii) What value of Aditya is depicting here?

Answer:

(i) Total length of cloth material = (4^{3}) m

Cloth material that each person received = (2^{2}) m

∴ Number of people who have received the cloth material

= \(\frac{\left(4^3\right) m}{\left(2^2\right) m}=\frac{\left(2^2\right)^3}{\left(2^2\right)}\)

= \(\frac{\left(2^6\right)}{\left(2^2\right)}\)

= (2)^{4}

= 16

(b) Values: Aditya is caring for others.