Linear Equations Made Simple : With Word Problems For Class 7th,8th,9th,10th[8]

Word Problems Related To
Linear Equations For
Class 7th,8th,9th & 10th

Q 1]. If the sides of a triangle are taken in ascending order , each side is 4cm longer than the preceding side . If the perimeter of the triangle is 66 cm , find the length of each side .

Solution :

Let the first side be = x cm
Thus , the second side = (x + 4 ) cm
Thus , the third side = (x + 4 + 4 ) = (x + 8 ) cm

Perimeter = 66cm

According to the question ,

= x + (x + 4 ) + ( x + 8 ) = 66

= x + x + 4 + x + 8 = 66

= 3x + 12 = 66

Thus , 3x = 66 -12
and , 3x = 54
so , x = 54 /3 = 18

Therefore , first side is = x = 18 cm
and , second side is = x + 4 = 18 + 4 = 22 cm
and , third side is = x + 8 = 18 + 8 = 26 cm

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Q 2]. Divide Rs. 105 into two parts such that one – half of the first exceeds one – fifth of the second part by 35 .

Solution :

Total amount : – Rs . 105
Let the 2nd part be = x

Thus , the 3rd part is = (105 – x )

According to the given question ,

= 1/2( 105 – x ) – 1/5(x) = 35

= [5(105 – x ) -2 ( 1x) ] / 10 = 35

= [ 525 -5x – 2x ] / 10 = 35

= [525 – 7x ] / 10 = 35

= 525 – 7x = 35 * 10

= 525 – 350 = 7x

= 175 = 7x

= 175 / 7 = x

= 25 = x

Thus , 1st part is = 105 – x = 105 – 25 = Rs. 80
and 2nd part is = x = Rs . 25

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Q 3 ] .A train is travelling at a speed of 65 km / hour from Mumbai to Patna . On its return journey , it travels at a speed of 60 km / hour and takes 5 hour more than the onward journey . what is the distance between Mumbai and Patna ?.

Solution:

Let’s assume the distance between Mumbai and Patna as x km .

Given : Speed from Mumbai to Patna ( onward journey ) = 65 km / hour

Step 1] – Assume time taken for onward journey = t

thus , speed = distance / time = s= d/t1 ;
t1 = d/s

thus , t1 = x / 65 ——[ 1 ] .

Step [2 ] – Speed from Mumbai to Patna ( return journey ) = 60 km / hour

Thus , time taken is = t2

thus , s = d / t2
t2 = d /s
t2 = x / 60 ————– [ 2 ]

Step [3 ] – But ,given , the return journey takes 5 hours more than time taken for onward journey .

Therefore , t2 = t1 + 5 = ——-[ 3 ]

Step [ 4 ] – Substitute equation [ 1 ] + [ 2 ] in equation [ 3 ] ;

t2 = t1 + 5
Therefore , x/ 60 = x /65 + 5

= x/ 60 – x/65 = 5

= [13x – 12x ] / 780 = 5

= [ x ] = 5 * 780

= x = 3900

Thus , the distance between Mumbai to Patna is 3900 km .

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Q 4 ] . A daily – wage worker was engaged for 30 days . She was paid Rs. 80 for each day she worked and was fined Rs. 15 for each day she was absent . At the end of the month , she received Rs . 1925 . For how many days was she absent ? .

Solution :

Given : 1] The worker was engaged for 30 days .
2] The worker was paid Rs 80 for each day she worked .
3] The worker was fined Rs . 15 for each day she was absent .
4] The worker received Rs. 1925 at the end of the month .

Solution : Total days engaged = 30 days

Step [ 1 ] – Let the number of days . the worker was absent be = x days

Therefore , the number of days the worker was present = ( 30 – x ) days


Step [ 2 ] – Paid for each day she worked = Rs 80

Therefore , paid for (30 – x ) days she worked = Rs . 80 ( 30 – x )

Fined for each day she was absent = Rs . 15
Fined for x days she was absent = 15 x


Step [ 3 ] – Total amount received at the end of the month = Rs . 1925

According , to the question ,
= 80 ( 30 – x ) – 15 x = 1925

= 2400 – 80x – 15x = 1925

= 2400 – 95x = 1925

= – 95x = 1925 – 2400

= -95x = -475

= x = -475 /-95

= x = 5

Therefore, the worker was absent for 5 days .

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Q 5 ]. A carpenter puts a beading around a wooden table whose length and breadth are int the ratio 7 : 3 . If the total cost of putting the beading is Rs. 1200 at the rate of Rs. 30 per meter , find the dimensions of the table .

Solution :

Given : 1 ] Ratio of length and breadth = 7 : 3
2] Total cost of putting the beading is = Rs . 1200
3] Rate is = Rs. 30 per meter

Find = Dimensions of the table .

Solution – Let the length = l = ( 7x ) m
breadth = b = ( 3x )m

Step [ 1 ] – Perimeter = 2 x ( l + b ) = 2 x ( 7x + 3x ) = 2 x 10x = ( 20x )m

Step [ 2 ] – The cost of beading = ?
per meter = Rs 30
(20x) m = ? = 20x * 30 = Rs 600x

But , the cost is given = Rs . 1200
therefore , 600x = 1200

And , x = 1200 / 600 = 2
x = 2

Therefore , length = 7x = 7 ( 2) = 14 m
breadth = 3x = 3 (2 ) = 6m .

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