The challenging part of solving word problems is translating the statements into equations. Collect as many such problems and attempt to solve them.

__Word problems
that involve linear equations__

The challenging
part of solving word problems is translating the statements into equations.

Collect as
many such problems and attempt to solve them.

** **

__Example 3.33__

The sum of
two numbers is 36 and one number exceeds another by 8. Find the numbers.

*Solution:*

Let the smaller
number be *x* and the greater number be
*x*+8

Given: the
sum of two numbers = 36

*x *+ (*x*+8) = 36

2 *x* +8 = 36

2 *x* = 36 – 8

2 *x *= 28

*x* = 28/2 = 14

Hence,

(i) The
smaller number, *x* =14

(ii) The
greater number, *x* +8=14+8 = 22

** **

__Example 3.34__

A bus is
carrying 56 passengers with some people having ₹8 tickets
and the remaining having ₹10 tickets. If the total money received
from these passengers is ₹500, find the number of passengers with
each type of tickets.

*Solution:*

Let the number
of passengers having ₹8 tickets be *y*. Then, the number of passengers with ₹10 tickets is (56−y).

Total money
received from the passengers = ₹500

That is,
*y* × 8 + (56* - y*) × 10 = 500

8y +560 −10*y *= 500

8y−10*y *= 500 – 560

− 2*y *= −60

*y *= 60/2

*y *= 30

Hence, the
number of passengers having,

(i) ₹8
tickets =30

(ii) ₹10
tickets =56−30 =26

** **

__Example 3.35__

The length
of a rectangular field exceeds its breadth by 9 metres. If the perimeter of the
field is 154*m*, find the length and breadth
of the field.

*Solution:*

Let the breadth
of the field be ‘*x*’ metres; then its length
(x+9) metres.

Perimeter
of the P = 2(length + breadth) = 2(*x* +
9 + *x*)= 2(2*x* + 9)

Given that,
2(2*x* + 9) = 154.

4*x* + 18 = 154

4*x *=154−18

4*x *= 136

*x *= 34

Hence,

(i)
Thus, breadth of the rectangular field = 34*m*

(ii) length
of the rectangular field = *x*+9 = 34+9
= 43*m*

** **

__Example 3.36__

There is
a wooden piece of length 2*m*. A carpenter
wants to cut it into two pieces such that the first piece is 40 *cm* smaller than twice the other piece. What
is the length of the smaller piece ?

*Solution:*

Let us assume
that the length of the first piece is *x cm*.

Then the
length of the second piece is (200*cm* **–** *x
cm*) i.e., (200 − *x*) *cm.*

According
to the given statement (change *m *to *cm*),

First piece
= 40 less than twice the second piece.

*x *= 2× (200 −* x*) – 40

*x *= 400 − 2*x *– 40

*x *+ 2*x *= 360

3*x* = 360

*x *= 360/3

*x *= 120

Hence,

(i) Thus
the length of the first piece is 120 *cm*

(ii) The
length of second piece is 200*cm* − 120*cm* = 80*cm*, which happens to be the smaller.

** **

**Think**

Suppose we take the second piece to be* x *and the first piece to be (200 − x), how will the steps vary? Will
the answer be different?

**Solution:**

Let 2^{nd} piece be ‘*x*’ & 1^{st}
piece is 200 − *x*

Given that 1st piece is 40 cm smaller than hence the other piece

∴ 200 − *x* = 2 × *x* − 40

200 − *x* = 2*x* – 40

∴ 200 + 40 = 2*x *+ *x*

240 = 3*x*

∴ *x* = 240 / 3 = 80

∴ 1^{st}
piece = 200 – *x *= 200 – 80 = 120 cm

2^{nd} piece = *x *= 80 cm

The answer will not change

** **

__Example 3.37__

Mother is
five times as old as her daughter. After 2 years, the mother will be four times
as old as her daughter. What are their present ages?

*Solution:*

Given condition:
After two years, Mother’s age = 4 times of Daughter's age

5 *x* + 2 = 4 ( *x* + 2)

5 *x* + 2 = 4 *x* + 8

5 *x* − 4*x*
= 8 - 2

*x *= 6

Hence daughter’s
present age = 6 years;

and mother’s
present age = 5 *x* = 5 × 6 = 30 years

** **

__Example 3.38__

The denominator
of a fraction is 3 more than its numerator. If 2 is added to the numerator and 9
is added to the denominator, the fraction becomes 5/6. Find the original fraction.

*Solution:*

Let the original
fraction be *x/y*.

Given that
*y* = *x* + 3. (Denominator = Numerator + 3).

Therefore,
the fraction can be written as *x*/(*x *+ 3).

As per the
given condition,

[(*x *+ 2) ] / [(*x *+ 3) + 9] = 5/ 6

By cross
multiplication, 6( *x* +2) = 5 ( *x* +3+9)

6 *x* +12 = 5( *x* +12)

6 *x* +12 = 5 *x* +60

6*x* - 5*x
*=60−12

*x *=60−12

*x *= 48.

*x* / *x *+ 3 = 48 / 48+3
= 48/51

Therefore,
the original fraction is

** **

__Example 3.39__

The sum of
the digits of a two-digit number is 8. If 18 is added to the value of the number,
its digits get reversed. Find the number.

*Solution:*

Let the two
digit number be *xy* (i.e., ten’s digit
is *x*, ones digit is *y*)

Its value
can be expressed as 10*x*+y.

Given, *x*+*y*
= 8 which gives *y* = 8 – *x*

Therefore
its value is 10*x*+y

= 10*x* + 8 – *x*

= 9*x* + 8.

The new number
is *yx* with value is 10*y + x*

= 10(8 −
*x*) + *x*

= 80 – 9*x*

Given, when
18 is added to the given number (*xy*) gives
new number (*yx*)

(9*x* + 8) + 18 = 80 – 9*x*

This simplifies
to 9*x* + 9*x* = 80 –8–18

18*x *= 54

*x *= 3 ⇒* y *= 8 – 3 = 5

The two digit
number is *xy *= 10*x+y* ⇒
10(3)+5 = 30+5 = 35

** **

__Example 3.40__

From home,
Rajan rides on his motorbike at 35 km/hr and reaches his office 5 minutes late.
If he had ridden at 50 km/hr, he would have reached his office 4 minutes earlier.
How far is his office from his home?

*Solution:*

Let the distance
be ‘*x*’ km. (Recall that, time = Distance / Speed)

Time taken
to cover ‘*x*’ km at 35 km/hr: T_{1}
= *x/*35 hr

Time taken
to cover ‘*x*’ km at 50 km/hr: T_{2}
= *x*/50 hr_{}

Speed 1 = 35 km/hr

Speed 2 = 50 km/hr

According
to the problem, the difference between two timings

= 4–(–5)

= 4+5 =9
minutes

= 9/60 hour
(changing minutes to hour)

Given, T_{1}
– T_{2} = 9/60

The distance
to his office* x *= 17(1/2) km.

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