# word problems on simultaneous linear equations | practice set 1.5 algebra 10th

In this article **word problems on simultaneous linear equations** we give the **simultaneous equations word problems** between **practice set 1.5 algebra 10th**. also we give simultaneous equations examples.

### word problems on simultaneous linear equations:

**Ex. 1)** The perimeter of a rectangle is 80 cm. The length of the rectangle is more than double it’s breadth by 4. Find length and breadth.

**Solution :**

Let length of rectangle be l cm and breadth be b cm.

From the first condition –

2 ( l + b ) = 80

l + b = 40 —– I

From the second condition –

l = 2b + 4

l – 2b = 4 —— II

from eq. ( I ) and ( II )

Subtract eq. ll from eq. l

l + b = 40

l – 2b = 04

– + –

————–

3b = 36

b = 12

put b = 12 in eq. I

l + b = 40

l + 12 = 40

l = 40 – 12

l = 28

therefore, Length of the rectangle is 28 cm and breadth is 12 cm.

#### word problems on simultaneous linear equations:

**Ex. 2)** A certain amount is equally distributed among certain number of students. Each would get rs. 4 less if 20 students where more and each would get rs. 12 more if 30 students where less. calculate the number of student and the amount distributed.

**Solution :**

Let the number of students be a and amount given to each student by rs. b

therefore, Total amount distributed is ab

From the first condition we get,

( a + 20 ) ( b – 4 ) = ab

ab – 4a + 20b – 80 = ab

– 4a + 20b = 80

divided both side by 4, we get

– a + 5b = 20 —– l

from the second condition we get,

( a – 30 ) ( b + 12 ) = ab

ab + 12a – 30b – 360 = ab

12a – 30b = 360

divided both side by 6, we get

2a – 5b = 60 —– ll

eq. ( l ) + eq. ( ll )

– a + 5b = 20

2a – 5b = 60

—————–

a = 80

put a = 80 in equation ( l )

– a + 5b = 20

-80 + 5b = 20

5b = 100

b = 20

Total amount distributed is = ab = 80 × 20 = 1600

therefore, rs. 1600 distributed equally among 80 students.

**Ex. 3)** Two numbers differ by 3. The addition of twice the smaller number and thrice the greater number is 19. Find the numbers.

**Solution:**

Let the numbers are p and q

from the first condition , we get

p – q = 3 —– l

from the second condition , we get

3p + 2q = 19 —- ll

eq. ( l ) multiply by ( 2 ) , we get

2p – 2q = 6 —- lll

adding eq. ( ll ) and ( lll ) , we et

5p = 25

p = 5

put p = 5 in eq. ( l )

5 – q = 3

– q = 3 – 5

– q = – 2

q = 2

Therefore, The numbers are 5 and 2.

**Ex. 4)** The sum of father’s age and twice the age of his son is 70, if we double the age of the father and add it to the age of his son the sum is 95. Find their present ages.

**Answer :**

Suppose the present age of father be p and son be q

for the first condition

p + 2q = 70 —– I

for the second condition

2p + q = 95 —- II

equation ( I ) multiply by 2, we get

2p + 4q = 140 —– III

substract equation ( II ) from ( III )

2p + 4q = 140

2p + q = 95

– – = –

—————–

3q = 45

q = 15

put q = 15 in equation ( I )

P + 2 ( 15 ) = 70

p + 30 = 70

p = 70 – 30

p = 40

therefore, father’s present age 40 years and son present age is 15 years.

**Ex. 5)** The denominator of fraction is 4 more than twice it’s numerator. Denominator becomes 12 times the numerator, if both the numerators and denominator are reduced by 6. Find the fraction.

**Answer :**

Suppose the fraction be p / q

from the first condition,

Denominator = ( 2 × numerator ) + 4

that is q = 2 p + 4

– 2p + q = 4 —– I

from the second condition,

( Numerator – 6 ) / ( Denominator – 6 ) = 1 / 12

that’s ( p – 6 ) / ( q – 6 ) = 1 / 12

12 ( p – 6 ) = 1 ( q – 6 )

12 p – 72 = q – 6

12p – q = – 6 + 72

12 p – q = 66 —– II

eq. ( I ) + eq. ( II ) , we get

10p = 70

therefore, p = 70 / 10

p = 7

put p= 7 in equation ( I ) , we get

-2 ( 7 ) + q = 4

-14 + q = 4

q = 4 + 14

q = 18

therefore, the fraction = p / q = 7 / 18.

**Simultaneous linear equations:**

**Ex. 6)** Two type of boxes P and Q are to be placed in the truck having capacity of 10 tons. when 150 boxes of type P and 100 boxes of type Q are loaded in the truck, it weights 10 tons but when 260 boxes of type P loaded in the truck, it can still accommodate 40 boxes of type Q, so that it is fully loaded. Find the weight of each type of box. [ 1 ton = 1000 kg ]

**Answer :**

Let the weight of P type box is a kg. and the weight of Q type box is b kg

From the first condition,

150a + 100b = 10 tons

that is, 150a + 100b = 10000

15a + 10b = 1000

3a + 2b = 200 —— I

from the second condition,

260a + 40b = 10000

26a + 4b = 1000

13a + 2b = 500 —- II

Substract equation ( I ) from equation ( II )

13a + 2b = 500

3a + 2b = 200

– – = –

——————

10a = 300

a = 300 / 10

a = 30 kg

put a = 30 in equation I

13 ( 30 ) + 2b = 500

390 + 2b = 500

2b = 500 – 390

2b = 110

b = 55kg.

therefore, the weight of P type box is 30kg. and the weight of Q type box is 55kg.

**Ex. 7)** Out of 1900km, Vishal travelling some distance by bus and some distance by airplane. Bus travels with average speed 60km/he and the average speed of airplane is 70 km/hr. It take 5 hours to complete the journey. Find the distance Vishal travelled by bus.

**Ans. :**

Let Vishal travelled p km. by bus and q km. by airplane.

from the first condition,

p + q = 1900 —– I

Speed of bus = 60 km/hr and speed of Airplane = 700 km/hr

Time taken by bus to travel p km.

but, Speed = ( Distance / Time )

60 = p / T1

T1 = p / 60

Time taken by airplane to travel q km

Time = ( Distance / Speed )

T2 =q / 700

from the second condition,

T1 + T2 = 5

( p / 60 ) + ( q / 700 ) = 5

multiply both side by 10, we get

( p / 6 ) + ( q + 70 ) = 50

( 70 p / 420 ) + ( 6q / 420 ) = 50

70p + 6q = 21000 —— II

multiply equation ( I ) by 6, we get

6p + 6q = 11400 ——- III

Substract equation ( III ) from equation ( II )

70p + 6q = 21000

6p + 6q = 11400

– – –

———————–

64p = 9600

p= 9600 / 64

p = 1200 / 8

p = 150

put p = 150 in equation ( I ), we get

p + q = 1900

150 + q = 1900

q = 1900 – 150

q = 1750

Therefore, Vishal travelled by bus = p km. = 150 km.

and Vishal travelled by Airplane = q km. = 1750 km.

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