Exercise 9.1
Q1. If x: y=3:5, find the ratio 3x+4y:8x+5y.
x:y=3:5
\(\frac{x}{y}=\frac{3}{5}\)
5x=3y
\(x=\frac{3y}{5}\)
\(3x+4y:8x+5y=\frac{3\times3y}{5}+4y:\frac{8\times3y}{5}+5y\\ =\frac{9y+20y}{5}:\frac{24y+25y}{5}\\ =\frac{29y}{5}:\frac{49y}{5}\\ =29y:49y =29:49\)
Q2. If x: y=8:9, find the ratio (7x-4y):3x+2y.
x:y=8:9
\(\frac{x}{y}=\frac{8}{9}\)
9x=8y
\(x=\frac{8y}{9}\)
\(7x-4y:3x+2y=\frac{7\times8y}{9}-4y:\frac{3\times8y}{9}+2y\\ =\frac{56y-36y}{9}:\frac{42y}{9}\\ =20:42\\ =10:21\)
Q3. If two numbers are in the ratio 6:13 and their L.C.M is 312, find the numbers.
Let the required numbers be 6x and 13x
Then their L.C.M is 78x
78x=312
\(x=\frac{312}{78}\)
x=4
Thus,
The numbers are
\(6x=6\times4=24\)
\(13x=13\times4=52\)
Q4. Two numbers are in the ratio 3:5. If 8 is added to each number, the ratio becomes 2:3. Find the numbers
Let the required numbers be 3x and 5x
If 8 is added to each other
3x+8:5x+8=2:3
\(\frac{3x+8}{5x+8}=\frac{2}{3}\)
3(3x+8)=2(5x+8)
9x+24=10x+16
10x-9x=24-16
x=8
Thus the numbers are 3x=3(8)=24
5x=5(8)=40
Q5. What should be added to each term of the ratio 7: 13 so that the ratio becomes 2: 3
Let the number to be added be x
Then
\(\frac{7+x}{13+x}=\frac{2}{3}\)
(7+x)3=2(13+x)
3x-2x=26-21
x=5
Hence the required number is 5
Q6. Three numbers are in the ratio 2: 3: 5 and the sum of these numbers is 800. Find the numbers
Given that
Three numbers are in ratio 2:3:5
Sum of these numbers=800
Sum of the terms of the ratio=2+3+5=10
\(First\;number=\frac{2}{10}\times800\\ =160 Second\;number=\frac{3}{10}\times800\\ =240 Third\;number=\frac{5}{10}\times800\\ =400\)
Q7. The ages of two persons are in the ratio 5: 7. Eighteen years ago their ages were in the ratio 8:13. Find their present ages.
Let the required ages be 5x and 7x
18 years ago their age ratios
\(\frac{5x-18}{7x-18}=\frac{8}{13}\)
\(65x-13\times18=8\times7x-8\times18\)
65x-234=56x-144
65x-56x=234-144
9x=90
x=10
Thus the ages are 5x=\(5\times10\)
7x=\(7\times10\)
Q8. Two numbers are in the ratio 7:11. If 7 is added to each of the numbers, the ratio becomes 2 . 3. Find the numbers.
Let the required numbers be 7x and 11x
If 7 is added to each of the numbers it becomes
\(\frac{7x+7}{11x+7}=\frac{2}{3}\)
21x+21=22x+14
X=21-14=7
Thus
The numbers are 7x=\(7\times7\)
11x=\(11\times7\)
Q9. Two numbers are in the ratio 2: 7.11the sum of the numbers is 810. Find the numbers.
Two numbers are in theratio=2:7
Sum of the numbers=810
We have,
Sum of the terms in the ratio=2+7=9
First number=\(\frac{2}{9}\times810\)
=\(2\times90\)
=180
Second number=\(\frac{7}{9}\times810\)
=\(7\times90\)
=630
Q10. Divide Rs 1350 between Ravish and Shikha in the ratio 2: 3.
We have
Sum of the terms of the ratio=2+3=5
Ravish money=\(\frac{2}{5}\times1350\)
=\(2\times270\)
=Rs.540
Shikha money=\(\frac{3}{5}\times1350\)
=\(3\times270\)
=Rs.810
Q11. Divide Rs 2000 among P, Q, R in the ratio 2:3:5.
We have
Sum of the terms of the ratio=2+3+5=10
P-share= \(\frac{2}{10}\times total\;money\\ =\frac{2}{10}\times2000\\ =2\times200\\ =Rs.400\)
Q-share= \(\frac{3}{10}\times total\;money\\ =\frac{3}{10}\times2000\\ =3\times200\\ =Rs.600\)
R-share= \(\frac{5}{10}\times total\;money\\ =\frac{5}{10}\times2000\\ =5\times200\\ =Rs.1000\)
Q12. The boys and the girls in a school are in the ratio 7:4. If total strength of the school be 550, find the number of boys and girls.
We have,
The boys and girls is in the ratio 7:4
Sum of the terms in the ratio=7+4=11
Total strength=550
Boys strength= \(\frac{7}{11}\times550\\ =7\times50\\ =350\;boys\)
Girls strength= \(\frac{4}{11}\times550\\ =4\times50\\ =200\;girls\)
Q13. The ratio of monthly income to the savings of a family is 7:2. If the savings be of Rs 500, find the income and expenditure.
It is given that
The ratio of income and savings is 7:2
Savings
2x=500
So, x=250
Therefore,
\(Income=7\times250\\ =1750\)
Expenditure=Income-savings
=1750-500
=Rs.1250
Q14. The sides of a triangle are in the ratio 1 : 2 : 3. If the perimeter is 36cm, find its sides.
The sides of the triangle are in the ratio 1:2:3
Sum of the terms in the ratio=1+2+3=6
Perimeter=36cm
First side= \(\frac{1}{6}\times36\\ =6cm\)
Second side= \(\frac{2}{6}\times36\\ =12cm\)
Third side= \(\frac{3}{6}\times36\\ =18cm\)
Q15. A sum of Rs 5500 is to be divided between Raman and Amen in the rate 2: 3. How much will each get ?
We have,
Sum of the terms in the ratio=2+3=5
Raman share=\(\frac{2}{5}\times5500\\ =2\times1100\\=Rs.2200\)
Aman share=\(\frac{3}{5}\times5500\\ =3\times1100\\=Rs.3300\)
Q16. The ratio of zinc and copper in an alloy is 7 : 9. It the weight of the copper in the alloy is 11.7 kg, find the weight of the zinc in the alloy.
We have,
The ratio of zinc and copper in an alloy=7:9
Weight of copper in the alloy=11.7kg
9x=11.7kg
\(x=\frac{11.7}{9}\)
Weight of zinc in the alloy=\(1.3\times7\)
=9.10kg
Therefore weight of zinc =9.10kg
Q17. In the ratio 7: 8. if the consequent is 40, what a the antecedent
Given ratio=7:8
Consequent
8x=40
\(x=\frac{40}{8}\)
x=5
antecedent=7x=\(7\times5\)
Q18. Divide Rs 351 into two parts such that one may be to the other as 2 : 7.
Ratio=2:7
The sum of the terms in the ratio=2+7=9
First ratio=\(\frac{2}{9}\times351\\=2\times39\\=Rs.78\)
Second ratio=\(\frac{7}{9}\times351\\=7\times39\\=Rs.273\)
Q19. Find the ratio of the price of pencil to that of ball pen, if pencil cost Rs.16 per score and ball pen cost Rs.8.40 per dozen.
One score=20
It is Rs.16 per score for pencil
Pencil cost=\(\frac{16}{20}\)
=Rs.0.80
Cost of one dozen ball pen=Rs. 8.40
Cost of one ball pen=\(\frac{8.40}{12}\)
=Rs.0.70
Ratio of price of pencil to that of ball pen=\(\frac{0.80}{0.70}\)
=\(\frac{8}{7}\)
Q20. In a class, one out of every six students fails. If there are 42 students in the class, how many pass?
Given,
One out of 6 student fails
X out of 42 students
\(\frac{1}{6}=\frac{x}{42}\)
\(x=\frac{42}{6}\)
x=7
Number of students who fail=7 students
No of students who pass=Total students- number of students who fail=42-7=35 students.
