Test: Equations- 1 - CA Foundation MCQ

Take -7x to R.H.S. and 5 to L.H.S.,

Now on R.H.S., we have -3x+7x =4x and on L.H.S., we have -5+1= -4

So the equation becomes

-4 = 4x

=> x = -1

LCM of 4 and 3 = 12 , then 3(x+4) +4(x-5)=11 × 12 then 140/7 = 20 Answer

If we cross multiply; 

45x = 30*2
45x  =  60

x = 60/ 45 

x  = 4/3 

Taking the L.C.M. on R.H.S. as 4, and then on cross-multiplication, we get

4(x+24) = 5(16+x)

=> x = 16

In option 2,

Took LCM .. SO u get 9x+36+2x+20=144.. so 11x= 88.. x=8

(p+2)(p-3)+(p+3)(p-4)=p(2p-5)

p^2-3p+2p-6+p^2-4p+3p-12=2p^2-5p

2p^2-2p-18=2p^2-5p

-2p+5p=18

3p=18

p=6

Put x=10
10/0.5-1/0.05+10/0.005-1/0.0005
= 20-20+2000-2000
=0

Hence, x=10 is the solution of the given equation.

Let the 2 nos. be x and x+2

Their sum is 52, so: 

52 = x + (x+2) 

52 = x + x + 2 

50 = x + x 

50 = 2x 

25 = x 

So your unknown numbers x and (x+2) are 25 and 27.

Given :-

one side = a = 4cm

nd diagonal = h = 5cm

Let, the other side be 'b' 

Now,

By applying pythgoras theorem ;

find base,

b^2 = h^2 - p^2

b^2 = (5)^2 - (4)^2

b^2 = 25 - 16 

b^2 = 9

b = 3

So, the two sides are 4cm and 3cm 

Then ;

Area of rect. = length * breadth

ar. = 4 * 3

ar. = 12 cm^2

Let the first part be x
other part is 56 - x
three times the first part exceeds one third of the second party by 48
3x = 1/3(56-x) + 48
9x = 56-x + 144
9x + x = 200
10x = 200
x = 20

First part, x =  20
Second part, 56 - x = 36

x = 10's digit of a two digit number(Let)

 y = units digit(Let)

n=Resulting equal digits(Let)

Condition 1-the sum of a two digit number is 10

x + y = 10

further simplifying

y = (10-x)

 Condition 2-if 18 be subtracted from it the digits in the resulting number will be equal

10x + y - 18 = 10n + n

10x + y = 11n+ 18

Substitutituting values found above

10x + (10-x) = 11n + 18

10x - x = 11n + 18 - 10

9x = 11n+ 8

If we solve it we will found the values as - x = 7, y = 3, z = 5 and finally we can say the no. is 73.

Let the number be 'x'.
By the problem,
(x/4)=(x/6)+4
⇒(x/4)-(x/6)=4
⇒(3x-2x)/12=4
⇒x/12=4
⇒x=4×12
⇒x=48.
Therefore, the number is 48.

Verification:
48/4=48/6+4
12=8+4
12=12
∴LHS=RHS.
So,the number is 48.
 

F-10 = 4(S-10) on solving this one gets F - 4S = -30....1

F = 10 = 2(S+10) on solving this one gets F - 2S = 10....2

solve both the equation and you get F= 50 S = 20

Let two no is x and y X›y

X*y=3200

x/y=2

x=2y

2y*y=3200

2y^2=3200

y^2=1600

y=✓1600

y=40

then x=2*40

x=80

Let numerator be: x

Let denominator be: x+2

fraction = x/(x+2)

(x+5)/(x+2) = x/(x+2) + 1

(x+5)/(x+2) - x/(x+2) = 1

(x+5 -x)/(x+2)= 1

5/(x+2) = 1

5 = x+2

5-2 = x

x = 3

The fraction is 3/5.

Mr. Roy has = x

Mr. Pual has = x-4

Mr. Singh has = x-5 

Together they have = 51

Mr. Roy + Mr. Pual + Mr. Singh = 51

x+(x-4)+(x-5)=51

3x-9=51

3x=51+9

3x=60

divide by 3

3x/3=60/3

x=20 

Mr. Roy has = x = 20

Mr. Pual has = x-4 = 20-4 = 16

Mr. Singh has = x-5 = 20-5= 15 

20+16+15=51

51=51

Let the digit at Tens place be x

The digit at unit place be y

The Number is 10x + y

ACCORDING TO THE QUESTION :-

x = 3y ... eq. 01

According to the second condition :

10x + y - 54 = 10y + x

=> 9x - 9y = 54

=> 9(x - y) = 9 × 6

=> x - y = 6 ... eq.02

Now we can solve the Equation 01 and 02 by substitution Method :-

x - y = 6

=> 3y - y = 6 [ from eq.01 , x = 3y ]

=> 2y = 6

=> y = 3

Now we can substitute the value of y in eq. 01 to get value of x

x = 3y

=> x = 3 × 3

=> x = 9

Hence the number is

Hence the number is 10x + y = 90 + 3 = 93 (ANS)

Let it be x

x/2-x/5=15

3x/10=15

x/10=5

x=50

Here 3x+4y=7.........(1)

4x-y=3........(2)

eq(1) * 1 --- 3x +4y =7
.
eq(2) * 4--- 16x -4y =12

adding eq(1) with eq(2)

................................................................
19x=19

=>x=19/19=1 ........

from eq(1) 3x + 4y=7

=>3*1 + 4y=7

=>4y = 7-3=4

=> y= 4/4=1 .

According to the given question,

i ) x + y = p + q

x = p + q - y ---------( 1 )

ii ) x / p + y / q = 2

Multiply each term with pq we get

qx + py = 2pq ---------( 2 )

Put x value , from ( 1 ) in equation ( 2)

q( p + q - y) + py = 2pq

pq + q^2 - qy + py = 2pq

- qy + py = 2pq - pq - q^2

y ( - q + p) = pq - q^2

y ( p - q ) = q ( p - q )

y = q ( p - q )/ ( p - q )

y = q -----( 3 )

Put y = q in ( 1 )

x = p + q - q

x = p

Therefore ,

x = p and y = q

Put x=1/4 and y=1/3 in equation only option a is satisfied the equation LHS=RHS

By option checking,

3xy = 10(y- x) so 3.2.5 =10(5-2)30=30

given here

x-3y = 0...(1)

x+2y = 20...(2)

subtract equations (1) and (2)

we get ,

-5y = -20

=> y = (-20)/(-5)

=> y = 4

by (1).

x-3×4= 0

=> x= 12

Thus:-

Value of X = 12

Value of Y = 4

I want to show you a good method to solve these equations.  Take time to read it and follow it.  This is the method in the International standards.  It is easy when you understand it.

  1.5  x  + 2.4 y  -  1.8  =  0 

  2.5  x  -  7 y   + 2.5   =  0

For solving :  a x + b y + c  =  0     and    d x + e y + f  = 0 ,  we do as mentioned in the diagram enclosed.  This is the easier way.

   -1.8      1.5      2.4      -1.8

   2.5      2.5        -7       2.5

        y                    x

x =   (-7*-1.8  -  2.4*2.5 ) / (2.5 * 2.4  - 1.5*-7)

   = (12.6  -  6) / (6 + 10.5)  =  6.6/16.5 =  0.4  = 2/5

  y =  (2.5*1.5 -  -1.8*2.5) / (2.5*2.4 - 1.5*-7) 

   = (3.75 + 4.5) / 16.5    =  8.25 / 16.5  = 0.5 = 1/2