Equations Quiz 1

3x+5y = 45………….(i)
x +4y = 22………(ii)

(i) - 3(ii) ⇒5y - 12y = 45 - 66
-7y = -21
y = 3

Total money = 1000
Money spent shopping = \frac{1}{4} \times 1000
= 250
Money remaining = 1000 - 250 = 750
Money spent on food = \frac{2}{3} \times 750
= 500
Time left = 750 - 500 = Rs 250

a^{3} - b^{3} - 3a^{2}b + 3ab^{2} = (a-b)^{3}
= 7^{3} = 343

x^{2} + 5x - 36 = 0
( x + 9 )( x - 4 ) = 0
X = -9, 4

a = 1, b = 4, c = 7
b^{2} - 4ac = 4^{2} - 4 \times 1 \times 7
= -12 < 0
So imaginary roots

Let the number of cows be x.
Number of workers be y.
4x+2y = [ 2 (x +y)]+ 36
4x+2y= (2x+ 2y ) + 36
2x = 36
X= 18

Let the number be N.
So N= 518x + 68.
Now divide this number by 37.
As 518 is exactly divisible by 37, we have to
divide 68 by 37 and find the remainder.
So remainder = 31

Let a, b and c be the three numbers
a^{2} + b^{2} + c^{2} = 172
( ab + bc + ca ) = 156

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 ( ab + bc + ca )
= 172 + ( 2 x 156 ) = 484
The sum of the three numbers a + b + c = \sqrt{484}
= 22

One dozen = 12
For 7 days, apples required = 133 x 12 = 1596
For one day, Apple required = \frac{1596}{7} = 228
For 19 days, apples required = 228 x 19
= 4332 Apples

Fuel for 200 cars = Fuel for 120 buses
Fuel for 1 car = Fuel for \frac{120}{200} bus
Fuel remaining = Fuel for 50 cars
Fuel for 50 cars = Fuel for \frac{120}{200} x 50 bus
= 30 buses