Posted on - 10-05-2017

Bank PO

We need to frame equations of one or two unknown in invariably in every problem. Sometimes we get three equations in three unknowns. In general, we need as many equations as the variables we will have to solve for. So, for solving for the values fo three unknowns, we need three equations (and hence the problem should give three conditions from which we can frame three equations). Solving the equations by itself is not a difficult task. The important part of the problem is framing the equation/equations. Once the equations are framed, solving them is very easy.

An equation like 2x + 4 = 26 is an equation in one variable. We have only one variable X whose value we have to find out.

Step I: Keep variables on L.H.S and constants on R.H.S i.e 2x=26-4=22.

Step II: Dividing both sides of equation with coefficient of x i.e 2.

i.e = = x = 11

**Two
equations in two unknowns**

A set of equations like

2x + 3y = 8 → (1)

5x + 4y = 13 → (2)

is called simultaneous equations in two unknown. Here, we have two variables (or unknowns) x and y whose values we have to find out.

**Step I: **Using
both the equations we first eliminate one variable (so that we can then have
one equation in one unknown).

For this purpose, we multiply equation (1) with 5 (the co-efficient of x in the second equation) and multiply equation (2) with 2 (the co-efficient of x in the first equation) to eliminate x. Thus we have.

(1) x 5 = 10x + 15y = 40 → (3)

(2) x 2 = 10x + 8y = 26 →

Now, subtracting equation (4) from eqn (3) we have, 7y = 14 → (4)

This is one equation in one unknown.

**Step II:** Solve
for the value of one variable from the equation (In one unknown) obtained from
step I above.

Therefore, y = 2.

**Step II:** Substitute
this value in one of two equations to get the value of the second variable.

Substituting the value of y in eqn (1) or equation (2), we get x = 1

Therefore the values of x and y that satisfy the given set of equations are x = 1 and y = 2

**Worked
out examples**

The cost of 4 chairs and 3 tables is Rs.1800. The cost of 4 chair and 4 tables is Rs.2300. Find the cost of each chair.

Let the costs of one there and one table be Rs.C and Rs.T respectively:.

4C + 3T = 1800 → (1)

5C + 4T = 2300 → (2)

{(2) x 3} - {(1) x 4} we get C = 300

So Cost each chair = Rs.300.

The sum of the digits of a two-digit number is 12. If 36 is subtracted from it, the reversed number is obtained. Find the number.

The sum of the digits=12

x + y = 12 → (1)

&⇒ 10x + y – 36 = 10y +
x

&⇒ 10x + y - (10y + x) =
36

&⇒ 9(x - y) = 36

&⇒ x – y = 4

Adding (1) and (2) , we get

2x = 16

x = 8

From (1) y =12 – x = 4

The number is 84

The cost of 3 pens, 4 erasers and 5 sharpeners is Rs.40. The cost of 5 pens, 7 erasers and 9 sharpeners is Rs.70. Find the total cost of one of each.

Let the cost of each pen, eraser and sharper be p, e and s respectively.

3p + 4e + 5s = 40 → (1)

5p + 7e + 9s = 70 → (2)

Multiplying the equations (1) b 2 and subtracting the equating (2) from it. We get. p + e + s = 10.

Hence the total cost of one of each is Rs.10.

**Note:** In
this case, there are three unknowns and two equations containing these
variables. As the number of equations is less than the number of variables,
most what is asked for is not a complete solution. It is just a certain
combination of one or more variables which sometimes can be found out.

Manoj and Anuj have some chocolates with each of them, if Manoj gives 5 chocolates to Anuj gives the same number of chocolates to Manoj, Manoj would have twice as many chocolates as Anuj. Find the number of chocolates Anuj has.

Let the number of chocolates with Manoj and Anuj be S and R respectively. If Manoj gives 5 chocolates to Anuj, He would have S-5 chocolates and Anuj would have R+5 chocolates.

So S-5=R+5

S-R=10

If Anuj Gives 5 chocolates to Manoj, he would have R-5 chocolates and Manoj would have S+5 chocolates.

S+5=2(R-5)

2R-S=15 → (2)

Adding (1) and (2) we get R = 25.

A,B,C,D have a total Rs.200 with them. The amount with B is Rs.10 more than that with A. The amount with C is Rs.20 more than that with B. The amount with D equals the sum of the amounts with B and C together. Find the amount with A.

Let the amounts with A,B,C,D be a,b,c and d respectively.

a+b+c+d=200

b=a+10

c=b+20=a+30

d=b+c=2a+40

a+b+c+d=5a+80=200

a=24

The Present age of father is two years more than thrice his son’s age 10 years later. The age of the father ten years later will be 2 years more than 7 times his son’s present age. Find the present age of the father.

Let the present age of the father and the son be f and s respectively.

f=3(s+10) +2=3s+32

f+10=7s+2

Substituting f=3s+32 in the above equation,

3s+32+10=7s+2

40=4s

10=s

f=3(10)+32=62 years