6. The number of boys and girls in a class are in the ratio 5 : 6. The number of boys is 5 less than the
number of girls. What is the total number of students in that class?
7. Tiventy years from now Shashank's age will be three times his present age. What is his present age ?
8. Sonia's father is 24 years younger than Sonia's grandfather and 25 years older than Sonia. The sum o
their ages is 104 years. What are the ages of each one of them?
9. The walk along a rectangular swimming tank is 282 m. Its length is 3 m less than twice its breadth. Wha
are the length and the breadth of the tank?
30. The sum of three consecutive multiples of 11 is 363. Find these multiples
IL Madhu has a total of 640 as currency notes in the denomination 50.20 and 10. The ratio of the
number of 50 notes and 20 notes is 3:5. If she has a total of 30 notes, how many notes of ead
denomination has she?
12. A cashier has notes of denominations 100, 750 and 10 respectively. The ratio of the numbers o
these notes is 2:4:7. The total cash with her is 47.000. How many notes of each denomination doe
she have?
Solving Linear Equations with Variables and Constants on Both
Sides
solve the equations which has expressions with variables on both sides, we transpose the expressions
ariables from R.H.S. to LH.S. and the constants from LH.S. to R.H.S.
xample 10 - Solve 4x + 3 = 6 + 2x.
. We have 4x + 3 = 6 + 2x
Answers
Answer:
Step-by-step explanation:
6 Let,
The number of boys = 7x and the number of girls =5x.
According to the condition,
7x = 5x + 8
2x = 8
x = 4
Therefore The number of boys = 7x = 7 × 4 = 28 and the number of girls =5x = 5 × 4 = 20.
7
Therefore total strength of the class = 28 + 20 = 48
let the age to be x years.
ATQ
x+20=3x
20=3x-x
20=2x
x=20/2
x=10
His present age is 10years.
8
Step-by-step explanation: sonia age = x
So fathers age = x + 25
So..... Grandfater age = x + 25 + 24
Sum of all ages = 104
So..... X + x + 25 + x + 25 + 24 = 104
3x + 74 = 104
3x = 30
X = 10
Sonia = 10 years
Father = 35 years
Grandfather = 59 years
9
2(2x-3+x)
282 /2 = 2x-3+x
141 = 3x-3
138 = 3x
x = 138/3
x = 46
2x-3 = 92-3
2x-3 = 89
10
The sum of three consecutive multiples of 11 is 363. Find these multiplesSolution:
Let us take the first number as 11 (C – 1)
First consecutive number that is a multiple of 11 is 11C
Second consecutive number that is a multiple of 11 is 11(C + 1)
Therefore, the sum of 3 multiple of 11 numbers is 11C + 11(C – 1) + 11(C + 1) = 363
Now the sum of 3 numbers are 11C + 11(C – 1) + 11(C + 1) = 363
33C = 363; C = 33,
By substituting C = 33 in numbers 11C, 11(C – 1) and 11(C + 1) we get values 110, 121, 132.
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