Math Olympiad Test: Linear Equations in Two Variables- 2 - Class 10 MCQ

Let unit’s digit be x and ten’s digit be y
Then number = 10y + x
x + y = 12 ...(1)
10x + y - (10y + x) = 18
9x - 9y = 18 ⇒ x - y = 2 ...(2)
Solving (1) and (2), we get x = 7, y = 5
∴ Number =  10 × 5 + 7 = 50 + 7 = 57

Let the present age of Ram be x years and the present age of Mohan be y years
x - 5 = 3 (y - 5)
⇒ x - 3y = -10 ...(1)
and x + 10 = 2(y + 10)
x - 2y = 10 ...(2)
x - 3y = -10
x - 2y = 10
Subtracting (1) and (2)
-y = - 20 ⇒ y = 20
Now x = 10 + 2y = 10 + 2 × 20 = 50

For a non-zero solution,

 Let
1/x = m, 1/y = n
Then am - bn = 0 ...(1)
ab²m + a²bn = a² + b² ...(2)
Now am - bn + 0 = 0
ab²m + a²bn - (a² + b²) = 0
By cross multiplication,

For infinitely many solutions,

On solving these equations, we get
m = 5, n = +1

Here
bx + by = ab + b² ...(1)
and ax - by = a² - b² ...(2)
Adding, x (a + b) = a (b + a) ⇒ x = a
Putting x = a in eqn (2), we have
a² - by = a² - b²
⇒ by = - b² ⇒ y = b
Hence x = a, y = b

Given
2x - ky + 3 = 0
3x + 2y - 1 = 0
For no solution, we have

Now

So,

Let the fraction be x/y
then x + y = 12 .....(1)
and

By solving eqn (1) and (2), we get x = 5, y = 7
x/y = 5/7

Let x be the larger angle and y be the smaller angle
then x + y = 180° ...(1)
and x - y = 18°
Solving eq. (1) and (2), we get
x = 99°, y = 81°

Let ₹ x be the charge of full first-class ticket and ₹ y be the reservation charge
x + y = 216 ...(1)
x + y + x/2 + y = 327
⇒ 3x + 4y = 654 ...(2)
By solving eq. (1) and (2), we get
x = 210, y = 6 

Let the number of rows be x and the number of students in each row be y.
Total number of students = xy
(x - 2) (y + 4) = xy
⇒ xy + 4x - 2y - 8 = xy
⇒ 4x - 2y = 8 ...(1)
and (x + 4) (y - 4) = xy
xy - 4x + 4y - 16 = xy
⇒ -4x + 4y = 16 ...(2)
From eqn. (1) and (2)
4x - 2y - 4x + 4y = 24
2y = 24 ⇒ y = 12
and 4x - 2 × 12 = 8
⇒ 4x = 8 + 24 = 32 ⇒ x = 8
Total no. of students = 12 × 8 = 96.

Let the larger number be x and the smaller number be y.
Now according to question, 3x = 4y + 3
3x - 4y = 3 ...(1)

and 3 × 25 - 4y = 3
⇒ 4y = 72 ⇒ y = 18
∴ Smaller number = 18

Let x litres of the 40% solution be mixed with y litres of 25% solution.
then x + y = 10 ...(1)
50% of x + 25% of y = 40% of 10

⇒ 50x + 25y = 400
⇒ 2x + y = 16 ...(2)
From (1) and (2)

∴ x =  10 - 4 = 6

Let abc be the three-digit number.
Given abc = a + b + c + 459 .......(i)
Again abc = 100a + 10b + c...........(ii)
From equation(i) and (ii), we get
100a + 10 b + c = a + b + c + 459
100a - a + 10b - b + c - c = 459
99a + 9b = 459
9(11a + b) = 459
11a + b = 51 .....(iii)
To find: ab + a = 10a + b + a
= 11a + b
∴ ab + b = 51 [From equation(iii)]

Let Swathi's first salary is Rs. x & fixed increment is Rs. y
∴ Swathi's salary after 6 years = Rs. (x + 6y)
Swathi's salary after 11 years = Rs. (x + 11y)
Swathi's salary after 8 years = Rs. (x + 8y)
∴ x + 6y = 22500 ......(1)
& x + 11y = 30000 .......(2)
From (1) & (2), we get
22500 - 6y + 11y = 30000
⇒ 5y = 30000 - 22500
= 7500
⇒ y = 7500/5 = 1500
∴∴ From (1), we get
x = 22500 - 6y
= 22500 - 6 × 1500
= 22500 - 9000
= 13500
∴∴ Swathi's salary after 8 years of service = Rs. (x + 8y)
= Rs. (13500 + 8 × 1500)
= Rs. (13500 + 12000)
= Rs. 25500