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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