Solve the following equation.
x + 1 = 6
⇒ x = 6 – 1
∴ x = 5
Solve the following equation.
x + 4 = 3
⇒ x = 3 – 4
∴ x = -1
Solve the following equation.
x – 3 = 5
⇒ x = 5 + 3
∴ x = 8
Solve the following equation.
2x = 8
∴ x = 4
Solve the following equation.
⇒ a = 7 × 3
∴ a = 21
Solve the following equation.
5a = -20
∴ a = -4
Solve the following equation.
2x – 5 = 1
⇒ 2x = 1 + 5
⇒ 2x = 6
∴ x = 3
Solve the following equation.
3x - 2 = -7
⇒ 3x = -7 + 2
⇒ 3x = -5
Solve the following equation.
4x + 7 = 15
⇒ 4x = 15 – 7
⇒ 4x = 8
∴ x = 2
Solve the following equation.
x + 5 = 3x
⇒ 3x – x = 5
⇒ 2x = 5
Solve the following equation.
7x = 20 – x
⇒ 7x + x =20
⇒ 8x = 20
Solve the following equation.
6(x – 1) = 5(x + 1)
⇒ 6x – 6 = 5x + 5
Bringing all the variable terms to one side and constants to other,
⇒ 6x – 5x = 5 + 6
∴ x = 11
Solve the following equation.
2x + 3 = 3(x - 1)
⇒ 2x + 3 = 3x – 3
⇒ 3x - 2x = 3 + 3
∴ x = 6
Solve the following equation.
On cross multiplying the given equation, we get,
⇒ x + 5 = 2(1 – x)
⇒ x + 5 = 2 – 2x
⇒ x + 2x = 2 – 5
⇒ 3x = -3
∴ x = -1
Solve the following equation.
On cross multiplying the given equation, we get,
⇒ 5x – 4 = 2 × 2x
⇒ 5x – 4 = 4x
⇒ 5x – 4x = 4
∴ x = 4
Solve the following equation.
On cross multiplying the given equation, we get,
⇒ 2x = -3(3x + 1)
⇒ 2x = -9x – 3
⇒ 2x + 9x = -3
⇒ 11x = -3
Solve the following equation.
On cross multiplying the given equation, we get,
⇒ 5a – 7 = 2 × 3a
⇒ 5a – 7 = 6a
⇒ 6a – 5a = -7
∴ a = -7
Solve the following equation.
On cross multiplying, we get,
⇒ 9(5m – 3) = 8 × 2m
⇒ 45m – 27 = 16m
⇒ 29m = 27
Solve the following equation.
On cross multiplying the given equation, we get,
⇒ 4x + 2 = 2(x – 4)
⇒ 4x + 2 = 2x – 8
⇒ 4x – 2x = -8 – 2
⇒ 2x = -10
∴ x = -5
Solve the following equation.
On cross multiplying the given equation, we get,
⇒ 2x + 3 = 3(x – 1)
⇒ 2x + 3 = 3x – 3
⇒ 3x – 2x = 3 + 3
∴ x = 6
Solve the following equation.
On cross multiplying, we get,
5(x + 1) = 6(x – 1)
⇒ 5x + 5 = 6x – 6
⇒ 6x – 5x = 5 + 6
∴ x = 11
Solve the following equation.
On cross multiplying the given equation, we get,
⇒ 3x + 5 = 4(2x + 7)
⇒ 3x + 5 = 8x + 28
⇒ 8x – 3x = 5 – 28
⇒ 5x = - 23
Solve the following equation.
On cross multiplying the given equation, we get,
⇒ 5x + 4 = 2(2x + 1)
⇒ 5x + 4 = 4x + 2
⇒ 5x – 4x = 2 – 4
∴ x = -2
Solve the following equation.
On cross multiplying the given equation, we get,
⇒ 11m – 1 = 2(2m + 3)
⇒ 11m – 1 = 4m + 6
⇒ 11m – 4m = 6 + 1
⇒ 7m = 7
∴ m = 1
Solve the following equation.
On cross multiplying the given equation, we get,
⇒ 2(x + 2) = 7(x – 3)
⇒ 2x + 4 = 7x – 21
⇒ 7x – 2x = 4 + 21
⇒ 5x = 25
∴ x = 5
Solve the following equation.
On cross multiplying the given equation, we get,
⇒ 5(2 – m) = -3(m + 7)
⇒ 10 – 5m = -3m – 21
⇒ 3m – 5m = -21 – 10
⇒ -2m = -31
Solve the following equation.
On cross multiplication, we get,
⇒ 4(2x – 1) = 9(x – 1)
⇒ 8x – 4 = 9x – 9
⇒ 9x – 8x = 9 – 4
∴ x = 5
Solve the following equation.
On cross multiplying the given equation, we get,
⇒ 3(3y + 5) = 5(3 – 7y)
⇒ 9y + 15 = 15 – 35y
⇒ 9y + 35y = 15 – 15
⇒ 44y = 0
∴ y = 0
Solve the following equation.
On cross multiplying the given equation, we get,
⇒ 9(2x + 1) = 5(3x – 2)
⇒ 18x + 9 = 15x – 10
⇒ 18x – 15x = -9 – 10
⇒ 3x = -19
Solve the following equation.
On cross multiplying the given equation, we get,
⇒ 17(8 – 3y) = 5y + 2
⇒ 136 – 51y = 5y + 2
⇒ 5y + 51y = 136 – 2
⇒ 56y = 134
Solve the following equation.
On cross multiplying the given equation, we get,
⇒ 3(18 – 2y) = 4(y – 4)
⇒ 54 – 6y = 4y – 16
⇒ 6y + 4y = 54 + 16
⇒ 10y = 70
∴ y = 7
Poorva is older than Durva by 6 years. The ratio of Poorva age to Durva’s age three years hence will be 4 : 3. What is Poorva’s age today?
Let the age of Poorva be x years.
Since Poorva is older than Durga by 6 years
Age of Durva = x-6
Now, age of Poorva after three years = x + 3
Also, age of Durga = (x - 6) + 3
It is given that, 3 years hence (later), their age will be in the
ratio 4:3
On cross multiplying, we get
⇒ 3(x + 3) = 4(x-3)
⇒ 3x + 9 = 4x – 12
⇒ x = 21
∴ Present age of Poorva is 21 years.
The difference of two natural numbers is 72. If we get 4 of dividing the bigger by the smaller number, find the numbers.
Let the smaller number be x
Since the difference of two numbers is 72,
⇒ The larger number is (x + 72)
According to the question, it is given that,
On cross multiplying, we get,
⇒ x + 72 = 4x
⇒ 3x = 72
⇒ x = 24
The other number is = x + 72 = 96
∴The two numbers are 24, 96
The ratio of the difference obtained by subtracting 2 from thrice a number to the sum of twice the same number and 5 is 13 : 15. What is the number?
Let the number be x
On Subtracting 2 from thrice a number = 3x - 2
Sum of twice the number and 5 = 2x + 5
It is given that, their ratio is 13:15
On cross multiplying, we get,
⇒ 45x – 30 = 26x + 65
⇒ 19x = 95
⇒ x = 5
∴The number is 5
The difference between the ages of two brothers today is 4 years. Five years ago, the ratio of their ages was 5 : 7. Find the age of the younger brother.
Let the age of younger brother be x years
Therefore, the age of elder brother is (x + 4) years
Now, Five years before, age of younger brother = x-5
Also, age of elder brother = (x + 4) -5
It is given that, 5 years ago(before), their age was in the
ratio 5:7
On cross multiplying, we get,
⇒ 7x – 35 = 5x – 5
⇒ 2x = 30
⇒ x = 15
Therefore, the age of younger brother is 15 years.
The numerator of a fraction is 3 less than its denominator. If the numerator is tripled and the denominator is increased by 2, the value of the fraction obtained is 1/2. What was the original fraction?
Let the value denominator be x
Since the numerator is 3 less than its denominator,
⇒ numerator = x-3
When numerator is tripled, new numerator = 3(x-3) = 3x-9
And, when denominator is increased by 2, new denominator = x + 2
Now it is given that,
On cross multiplying, we get,
6x – 18 = x + 2
⇒ 5x = 20
⇒ x = 4
∴Old Denominator = 4
⇒ Old Numerator = x – 3 = 1
Therefore, the original fraction was
The denominator of a fraction is bigger than its numerator by 3. If 3 is subtracted from the numerator, and 2 added to the denominator, the value of fraction we get is 1/5. Find the original fraction.
Let the numerator be x.
Since, the denominator is bigger than numerator by 3
⇒ Denominator = x + 3
If 3 is subtracted from the numerator, it becomes = x-3
And, 2 added to the denominator, it becomes = (x + 3) + 2
It is given that :-
On cross multiplying, we get,
⇒ 5x – 15 = x + 5
⇒ 4x = 20
⇒ x = 5
Sandeep has Rs 50 more than Gayatri. If, when each of them is given Rs 15, the ratio of the money they have becomes 3 : 1, how much money did Gayatri have to begin with?
Let the money that Gayatri initially has = Rs. x
Therefore, money with Sandeep = Rs. (x + 50)
When each of them is given Rs 15, the ratio of the money they have becomes 3 : 1
Now, Money with Sandeep = x + 50 + 15 = x + 65
Also, money with Gayatri = x + 15
On cross multiplying, we get,
x + 65 = 3x + 45
⇒2x = 20
⇒x = 10
Therefore, Gayatri initially had Rs. 10 with her