Exercise 7A
Question 1
Express each of the following ratios in its simplest form.
(a) A length of 6 cm to a length of 8 cm.
Sol :
$\frac{6}{8}=\frac{3}{4}$
Ans : 3 : 4
(b) An area of 20 cm square to an area of 45 cm square.
Sol :
$\frac{20}{45}=\frac{4}{9}$
Ans 4 : 9
(c) A volume of 26 litres to a volume of 65 litres.
Sol :
$\frac{26}{65}$
Ans 26 : 65
(d) A population of 50 thousand to a population of 1.5 lakh.
Sol :
$\frac{0.5}{1.5}=\frac{1}{3}$
Ans 1 : 3
(e) $\frac{1}{4} : 2$
Sol :
$=\dfrac{\frac{1}{4}}{2}$
$=\frac{1}{8}$
Ans 1 : 8
(f) $\frac{1}{2}: \frac{1}{8}$
Sol :
$=\frac{1}{2} \times \frac{8}{1}$
$=\frac{8}{2}$
=4
Ans 4 : 1
(g) 0.75 : 1
Sol :
$=\frac{0.75}{100}$
Ans 3 : 4
(h) 14 cm to 1 m
Sol :
$=\frac{14}{100}=\frac{7}{50}$
Ans 7 : 50
(i) 25 cm to 50 mm
Sol :
$=\frac{250}{50}=\frac{5}{1}$
Ans 5 : 1
(j) 15 m to 1 km
Sol :
$=\frac{15}{1000}=\frac{3}{200}$
Ans 3 : 200
Question 2
Ravi earns Rs. 84,000 a year and spends Rs. 63,000 a year. Find the simplest form the ratio of
(a) Ravi’s income to his savings
Sol :
Ravi's savings
$=(84000-63000)$
=21000
ATQ,
84000 : 21000
$\frac{84}{21}=\frac{28}{7}$
or 28 : 7
(b) Money that Ravi saves to the money he spends.
Sol :
21000 : 63000
$\frac{21}{63}=\frac{7}{21}=\frac{1}{3}$
or
1 : 3
Question 3
There are 12,000 students in a university, out of which 5,600 are girls. Find the simplest form, the ratio of
(a) number of girls to the number of students
Sol :
5600 : 12000
$\frac{56}{120}=\frac{28}{60}=\frac{7}{15}$
(b) number of boys to the number of girls
Sol :
Number of boys=12000-5600
=6400
ATQ,
6400 : 5600
$\frac{64}{56}=\frac{8}{7}$
(c) number of boys to the number of students
Sol :
6400 : 12000
$\frac{64}{120}=\frac{16}{30}=\frac{8}{15}$
Question 4
Which is a greater ratio of the following pairs?
(a) 3:2 or 4:3
Sol :
$\frac{3}{2}=\frac{3\times 3}{2\times 3}=\frac{9}{6}$
$\frac{4}{3}=\frac{4\times 2}{3\times 2}=\frac{8}{6}$
Ans 3 : 2
(b) 7:10 or 8:11
Sol :
$\frac{7}{10}=\frac{7\times 11}{10\times 11}=\frac{77}{110}$
$\frac{8}{11}=\frac{8\times 10}{11\times 10}=\frac{80}{110}$
Ans 8 : 11
(c) 32 : 22 or 3 : 2
Sol :Ans 32 : 22
(d) 3 : 5 or 0.66 : 1
Sol :
$\frac{3}{5}=\frac{3\times 10}{5\times 10}=\frac{30}{50}$
$\frac{0.66}{100}=\frac{33}{50}$
Ans 0.66 : 1
(e) $\frac{1}{5} : \frac{1}{7}$ or $\frac{1}{4} : \frac{1}{9}$
Sol :
$\frac{1}{5} : \frac{1}{7}=\dfrac{\frac{1}{5}}{\frac{7}{1}}$
$=\frac{7\times 4}{5\times 4}=\frac{28}{20}$
$\frac{1}{4} : \frac{1}{9}=\dfrac{\frac{1}{4}}{\frac{1}{9}}$
$=\frac{9\times 5}{4\times 5}=\frac{45}{20}$
Ans $\frac{1}{4} : \frac{1}{9}$
Question 5
Write the ratios – 4:9, 5:8 and 3:7 in order, from the least to the greatest.
Sol :
4 : 9
$=\frac{4}{9}$
$=\frac{4\times 56}{9\times 56}$
$=\frac{224}{504}$
5 : 8
$=\frac{5}{8}$
$=\frac{5\times 63}{8\times 63}$
$=\frac{315}{504}$
3 : 7
$=\frac{3}{7}$
$=\frac{3\times 72}{7\times 72}$
$=\frac{216}{504}$
Ans 3:7 , 4:9, 5:8
Question 6
If a:b = 4:7 and b:c = 5:3, then find a:b:c
Sol :
a:b=4:7
=(4×5):(7×5)
=20:35
b:c=5:3
=(5×7):(3×7)
=35 : 21
∴a : b : c= 20 : 35 : 21
Question 7
If a:b = 5:6 and b:c = 9:4, then find a:c
Sol :
a : b=5 : 6
=(5×3):(6×3)
=15 : 18
b : c=9 : 4
=(9×2):(4×2)
∴a:c=15:8
=15 : 8
Question 8
The sides of a triangle are in the ratios, 1:1.5:2 and its perimeter is 18 cm. Find the length of each side.
Sol :
Let, the sides of triangle =x, 15x, 2x
ATQ,
x+1.5x+2x=18
4.5x=18
$x=\frac{180}{4.5}$
∴x=4
∴The sides of triangle=4,(1.5×4),(2×4.5)
=4 cm,6 cm ,8 cm
Question 9
In an examination, 100 marks are distributed among 3 questions so that they are very proportional to 8,5,7. Find the marks of each question.
Sol :
100 marks distributed in 8 : 5 : 7
ATQ,
8x+5x+7x=100
20x=100
$x=\frac{100}{20}$
∴x=5
Marks for 1st question=(8×5)=40
Marks for 2nd question=(5×5)=25
Marks for 3rd question=(5×7)=35
Question 10
The length of three ribbons are in the ratio 7:5:9. If the sum of the lengths of the three ribbons is 42 cm, find the length of the smallest ribbon.
Sol :Question 11
Shyam lost his weight in the ratio 9:5. His original weight was 90 kg. What is his new weight?
Sol :Shyam lost his weight in the ratio 9 : 5
∴9x=90
x=10
∴Shyam's new weight=(5×10)=50kg
Question 12
A, B and C contributed Rs. 25,000, Rs, 50,000, Rs. 75,000 respectively in a business and their share of his profits are proportional to the capital they contributed. If the profits are Rs. 48,000, what is the share of each of them?
Sol :
A,B,C's capital in a ratio=25000 : 50000 : 75000
=25 : 50 : 75
=5 : 10 : 15
=1 : 2 : 3
Total profit 48000
A's share $=48000 \times \frac{1}{6}$
=8000
B's share $=48000 \times \frac{2}{6}$
=16000
C's share $=48000 \times \frac{3}{6}$
=24000
Question 13
A legacy of Rs. 1,08,000 is to be divided among 3 sons in the ratio $\frac{3}{2} : \frac{9}{4} : 3$. How much does each of them receive?
Sol :
Ratio of the share of 3 son $=1\frac{1}{2} : 2\frac{1}{4} : 3$
$=\frac{3}{2} : \frac{9}{4} : 3$
=6 : 9 : 12
=2 : 3 : 4
Share of 1st son$=108000 \times \frac{2}{9}$
=24000
Share of 2nd son$=108000 \times \frac{3}{9}$
Share of 3rd son$=108000 \times \frac{4}{9}$
=48000
Question 14
A map drawn to a scale of 1 cm to 10 km, measures 24 cm by 18 cm. If the areas of land and water represented are in the ratio of 7:2, find in square km, the land area represented.
Sol :
Given : 1 cm to 10 km
∴24 cm=24×10=240 km
∴18 cm=18×10=180 km
Area of map=(240×180)sq km
=43200 sq km
Sum of terms in ratio=7+2=9
Area of land$=43200 \times \frac{7}{9}$
=33600 sq km
Question 15
Solve:
(a) Increase 180 in the ratio 6:9.
Sol :
6x=180
x=30
Ans (9×30)=270
(b) Decrease Rs. 900 in the ratio 5:3
Sol :
5x=900
$x=\frac{900}{5}$
x=180
=(3×180)
=540
(c) Find the multiplying factor which decreases 120 kg to 84 kg.
Sol :
120 : 84
40 : 28
10 : 7
(d) Two distances are in the ratio 15:8, the larger is 60 km, what is the smaller?
Sol :
15x=60
∴$x=\frac{60}{15}=4$
∴Smaller distance=(8×4)=32km
(e) A boy worked 8 hours a day. In what ratio , did his earnings change when, the pay was raised from Rs. 50 per hours Rs. 500 a day.
Sol :
Number of hours the boy worked in a day=8
His wage=50 per hour
So, per day wage becomes =(50×8)
=400
Now his wage per day=500
∴Ratio 400 : 500
=4 : 5
Question 16
A photo measuring 7.5 cm by 5 cm is enlarged, so that the larger side becomes 18 cm. What does the shorter side become? In what ratio is the area increased?
Sol :
Area of photo=(7.5×5)=37.5sq cm
∴7.5x=18
x=2.4
∴5x=5×5.4=12
∴New area=(18×12)sq cm
=216 sq cm
∴Ratio=216 : 37.5
=5.76 : 1
The shorter side becomes 12 cm
The ratio is 5.76 : 1
Question 17
At the beginning of a war, the numbers of war planes possessed by two powers of 1.6 :1. the weaker power having 400. In a general engagement, each power loses the same number of planes but, the ratio is changed to 2:1. How many planes does each lose?
Sol :
Let, 1st power have planes=1.6x
2nd power have planes=x
ATQ,
Less power have planes=400
∴x=400
1.6x=(1.6×400)
1.6x=640
New ratio= 2 : 1
∴640-y : 400-y= 2 : 1
$\frac{640-y}{400-y}=\frac{2}{1}$
640y=800-2y
2y-y=800-640
y=160
∴160 planes less by both to get new ratio 2 : 1
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