Additional Questions - Chapter 1 - Number Systems - Term 2 - 7th Maths Guide Samacheer Kalvi Solutions

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Additional Questions and Answers
Exercise 1.1
Question $1 .$
Express as rupees using decimals.
(i) 4 paise
(ii) 4 rupees 4 paise
(iii) 44 rupees 44 paise
(iv) 50 paise
(v) 625 paise
Solution:
We know that 100 paise $=₹ 1$
1 paise $=₹ \frac{1}{100}$
(i) 4 paise $=₹ 4 \times \frac{1}{100}=₹ \frac{4}{100}=₹ 0.04$
(ii) 4 rupees 4 paise $=₹ 4+₹ 0.04=₹ 4.04$
(iii) 44 rupees 4 paise $=₹ 44+44$ paise $=₹ 44+₹ \frac{44}{100}=₹ 44+₹ 0.44=₹ 44.44$
(iv) 625 paise $=600$ paise $+25$ paise $=₹ 6+₹ \frac{25}{100}=₹ 6+₹ 0.25=₹ 6.25$

Question $2 .$
Express $7 \mathrm{~cm}$ in metre and kilometer.
Solution:
$7 \mathrm{~cm}=\frac{7}{100} \mathrm{~m}=0.07 \mathrm{~m}$
$7 \mathrm{~cm}=\frac{7}{10000} \mathrm{~km}=0.00007 \mathrm{~km}$

Question $3 .$
Write the following decimal numbers in the expanded form.
(i) $30.04$
(ii) $3.04$
(iii) $300.04$
Solution:
(i) $30.04=3 \times 10+0 \times 1+0 \times \frac{1}{10}+4 \times \frac{1}{100}=3 \times 10+\frac{4}{100}$
(ii) $3.04=3 \times 1+0 \times \frac{1}{10}+4 \times \frac{1}{100}=3 \times 1+\frac{4}{100}$
(iii) $300.04=3 \times 100+0 \times 10+0 \times 1+0 \times \frac{1}{10}+4 \times \frac{1}{100}=3 \times 100+\frac{4}{100}=3 \times 100+\frac{4}{100}$
 

Question $4 .$
Write the place value of 2 in the following decimal numbers.
(i) $2.47$
(ii) $26.89$
(iii) $36.28$
Solution:
(i) $2.47$ Place value of 2 in $2.47$ is ones.
(ii) $26.89$ Place value of 2 in $26.89$ is Tens.
(iii) $36.28$ Place value of 2 in $36.28$ is tenths

Exercise $1.2$
Question $1 .$
Explain the following as fractions.
(i) Ajar containing $3.6$ litres of milk.
(ii) A cup containing $9.63 \mathrm{mg}$ of medicine.
Solution:
(i) $3.6=3+\frac{6}{10}=3+\frac{3}{5}=3 \frac{3}{5}$ litre of milk
(ii) $9.63=9+\frac{6}{10}+\frac{3}{100}=\frac{900+60+3}{100}=\frac{963}{100} \mathrm{mg}$ of medicine
 

Question $2 .$
Convert into decimal.
(i) Three hundred three and nine hundredths.
(ii) Six and fifty five thousands
Solution:
(i) Three hundred three and nine hundredths
$=303+\frac{9}{100}=303+0 \times \frac{1}{10}+9 \times \frac{1}{100}=303.09$
(ii) Six and fifty five thousands
$6+\frac{55}{100}=6+\frac{5}{100}+\frac{5}{1000}=6+\frac{0}{10}+\frac{5}{100}+\frac{5}{1000}=6.055$

Question $3 .$
Find the decimal form of (i) $194+20+3+\frac{7}{10}+\frac{2}{100}$
(ii) $111+11+1+\frac{1}{10}+\frac{1}{1000}$
Solution:
(i) $194+20+3+\frac{7}{10}+\frac{2}{100}=217+7 \times \frac{1}{10}+2 \times \frac{1}{100}=217.72$
(ii) $111+11+1+\frac{1}{10}+\frac{1}{1000}=123+1 \times \frac{1}{10}+0 \times \frac{1}{100}+1 \times \frac{1}{1000}=123.101$
 

Exercise $1.3$

Question $1 .$
Maya bought $5 \mathrm{~kg} 300 \mathrm{~g}$ bananas and $3 \mathrm{~kg} 250 \mathrm{~kg}$ oranges. Diya bought $4 \mathrm{~kg} 800 \mathrm{~g}$ apples and $4 \mathrm{~kg}$ $150 \mathrm{~g}$ of mangoes. Who bought more fruits.
Solution:
Total fruits bought by Maya $=5 \mathrm{~kg} 300 \mathrm{~g}+3 \mathrm{~kg} 250 \mathrm{~g}=8 \mathrm{~kg} 550 \mathrm{~g}=8.550 \mathrm{~kg}$
Total fruits Diya bought $=4 \mathrm{~kg} 800 \mathrm{~g}+4 \mathrm{~kg} 150 \mathrm{~g}=8 \mathrm{~kg} 950 \mathrm{~g}=8.950 \mathrm{~kg}$
Comparing the whole number parts, they are equal.
Comparing thet tenths place we get $9>5$.
$\therefore 8.950 \mathrm{~kg}>8.550 \mathrm{~kg}$
$\therefore$ Diya bought more fruits.
 

Question 2 .
Which is greater $28 \mathrm{~km}$ or $42.6 \mathrm{~km}$.
Solution:
Comparing the whole number part $42>28$.
$42.6 \mathrm{~km}$ is greater than $28 \mathrm{~km}$.

Exercise $1.4$
Question $1 .$
Show that the following numbers in a number line.
(i) $0.2$
(ii) $1.8$
(iii) $1.1$
(iv) $2.6$
Solution:

Question 2 .
Write the decimal numbers represented by the points A, B, C, D, E and F.

Solution:

$\mathrm{A}(0.5) ; \mathrm{B}(1.2) ; \mathrm{C}(2.3) ; \mathrm{D}(2.8) ; \mathrm{E}(3.4) ; \mathrm{F}(3.9)$