4 : 3

We have,10 g:40 g \(=\frac{10}{40}=1:4\)
and 25 kg: 100 kg \(=\frac{25}{100}=1:4\)
So, they are in proportion.

We know that, 1 m = 100 cm
\(\therefore\) Required ratio \(=70:100=\frac{70}{100}=7:10\)

20

3 :10

Given, ratio = 3 : 5, firstly multiply and divide the numerator and denominator with the same number.
\(\frac{3}{5}=\frac{3\times2}{5\times2}=\frac{6}{10}\ and\ \frac{3}{5}=\frac{3\times3}{5\times3}=\frac{9}{15},\) hence two equivalent rati.os of 3: 5 are 6 : 10 and 9: 15.

(a) 6 : 5
(b) 12 : 11

Length of line segment AB = 10 cm and AC: BC =1: 4
Sum of parts AC and BC = AC + BC = 1+ 4 = 5
Length of \(\bar {AC}=\frac{10\times1}{5}=2\ cm\)
Length of \(\bar{BC}=\frac{10\times4}{5}=8\ cm\)

Given, diesel required for 594 km = 108 L
\(\therefore\) Diesel required for 1 km \(=\frac{108}{594}L=\frac{108\div54}{594\div54}=\frac{2}{11}L\)
[\(\because\) HCF of 108 and 594 = 3 x3 x3 x 2 = 54]
\(\therefore\) Diesel required for 1650 km \(=\frac{2}{11}\times1650L=2\times150=300L\)
11) 1650 (150
      11
      ___
      55
      55
     ___
       X
     ___
Hence, 300 L diesel required by the truck to cover 1650 km.

We have, 33,121,9,96
\(\therefore\) Ratio of 33 to 121 \(=\frac{33}{121}=\frac{33\div11}{120\div11}\) [\(\because\) HCF of33 and 121= 11]
\(=\frac{3}{11}=3:11\)
and ratio of9 to 96 = \(\frac{9}{96}=\frac{9\div3}{96\div3}\)
[\(\because\)HCF of 9 and 96 = 3]
\(=\frac{3}{32}=3:32\)
Here, 3 : 11 \(\ne\)3 : 32 i.e. 33: 121 \(\ne\)9: 96
Therefore, 33,121,9 and 96 are not in proportion.

Given, number of holidays in one year = 73
We know that, number of days in one year = 365
\(\therefore\) Ratio of number of holidays to the number of days in one year \(=\frac {Number\ of\ holidays}{Total\ number\ of\ days}=\frac {73}{365}\)
\(=\frac{73\div73}{365\div73}=\frac{1}{5}=1:5\) [\(\because\)HCF of 73 and 365 = 73]
Hence, the required ratio is 1 : 5.

Given, Ravi walks 6 km in an hour i.e. distance covered by Ravi in one hour = 6 km.
and Roshan walks 4 km in an hour i.e. distance covered by Roshan in one hour = 4 km
\(\therefore\) Ratio of distance covered by Ravi to the distance covered by Roshan \(=\frac{Distance\ covered\ by\ Ravi\ in\ one\ hour}{Distance\ covered\ by\ Roshan\ in\ one\ hour}\)
\(=\frac{6\ km}{4\ km}=\frac{6}{4}=\frac{3}{2}=3:2\)
Hence, the required ratio is 3 : 2.

Given,the ratio of zinc and copper in alloy = 7 : 9 and total weight of alloy = 8 kg
Let the weight of zinc and copper in alloy be 7x and 9x respectively, where, x is a multiple of weight
Then,total weight = 7x + 9x = 16x
\(\therefore 16x=8\ kg \Rightarrow x=\frac{8}{16}=\frac{1}{2}\)
\(\therefore\)  Weight of copper in alloy  \(=9x=9\times\frac{1}{2}\)
\(=\frac{9}{2}\ kg=4\frac{1}{2}\ kg\)
Hence, the weight of copper is  \(4\frac{1}{2}\ kg\).

(a) Here, 25 cm : 1 m = 25 cm : 1 x 100 cm [\(\because\)1 m=100em]
= 25 cm: 100 cm
\(=25:100=\frac{25}{100}=\frac{25\div25}{100\div25}=\frac{1}{4}=1:4\)
[dividing numerator and denominator both by 25]
and Rs.40:Rs.160=40:160=\(\frac{40}{160}=\frac{4}{16}=\frac{1}{4}=1:4\)
[dividing numerator and denominator both by 10]
Here, 1: 4 = 1: 4 i.e. 25 cm : 1 m = Rs.40 : Rs.160
So, the ratios of 25 cm : 1 m and Rs.40 : Rs.160 are in proportion.
i.e. 25 cm : 1 m: : Rs.40 : Rs.160
Now, middle terms are 1 m and Rs.40 and extreme terms are 25 cm and Rs.160.
(b) Here,39L:65L=39:65=\(\frac{39}{65}\)
\(=\frac{39\div13}{65\div13}\) [\(\because\) HCF of 39 and 65 = 13]
\(=\frac{3}{5}=3:5\)
and 6 bottles: 10 bottles = 6 :10 \(=\frac{6}{10}\)
\(=\frac{6\div2}{10\div2}=\frac{3}{5}=3:5\)
[dividing numerator and denominator both by 2]
Here, 3 : 5 = 3 :5 i.e. 39 L : 65 L = 6 bottles: 10 bottles.
So, the ratio of 39 L : 65 Land 6 bottles: 10 bottles are in proportion.
i.e. 39 L : 65 L : : 6 bottles: 10 bottles
Now, middle terms of ratios are 65 Land 6 bottles and extreme terms of ratios are 39 Land 10 bottles.
(c) Here, 2kg:80kg=2:80=\(=\frac{2}{80}=\frac{2\div2}{80\div2}=\frac{1}{40}=1:40\)
[dividing numerator and denominator both by 2]
and 25 g : 625 g = 25 :625 \(=\frac{25}{625}=\frac{25\div25}{625\div25}=\frac{1}{25}=1:25\)
[\(\because\)HCF of 25 and 625 = 5 X 5 = 25]
Since, both ratios are not equal.
\(\therefore\) 2 kg : 80 kg \(\ne\) 25 g : 625 g
Hence, the given ratios are not in proportion.
(d) Here, 200mL:2.5L=200 \(\times\frac{1}{1000}L:2.5\ L\)
\([\because\ 1\ mL=\frac{1}{1000}L]\)
\(=\frac{200}{1000}L:2.5L=0.200L:2.5L\)
\(=0.200:2.5=\frac{0.200}{2.5}=\frac{2}{25}=2:25\)
[multiplying numerator and denominator both by 10]
and Rs.4:Rs.50=4:50=\(\frac{4}{50}=\frac{4\div2}{50\div2}=\frac{2}{25}=2:25\)
[dividing numerator and denominator both by 2]
Here, 2 : 25 = 2 : 25 i.e. 200 mL : 2.5 L = Rs. 4 : Rs.50
So, the ratios of 200 mL : 2.5 L and Rs. 4 : Rs. 50 are in proportion.
i.e. 200 mL : 2.5 L: : Rs.4 : Rs. 50
Now, middle terms of ratios are 2.5 L and Rs.4 and extreme terms of ratios are 200 mL and Rs.50.

Given, number of teachers = 102
and number of students = 3300
\(\therefore\)  Required ratio of teachers to the number of students \(=\frac{Number\ of\ teachers}{Number\ of\ students}=\frac{102}{3300}\)
\(\because\) 102=2x3x17and 3300=2x3x2x5x5x11
\(\therefore\) HCF of 102 and 3300=2x3=6
\(Ratio=\frac{102\div6}{3300\div6}=\frac{17}{55}=17:550\)