Oops! It appears that you have disabled your Javascript. In order for you to see this page as it is meant to appear, we ask that you please re-enable your Javascript!

Linear Equations in One Variable : Exercise 2.4 (Mathematics NCERT Class 8th)


Q.1 Amina thinks of a number and subtracts from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?
Sol. Let the number be a.
Given, {\rm{8}}\left( {{\rm{a - }}{{\rm{5}} \over {\rm{2}}}} \right){\rm{ = 3a}}
8a – 20 = 3a
Transposing 3a from RHS to LHS and – 20 from LHS to RHS, we get,
8a – 3a = 20
5a = 20
Dividing by 5 on both the sides, we get,
a = 4
Therefore, the number is 4.

Q.2 A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
Sol. Let the numbers be a and 5a.
Given, 21 + 5a = 2(a + 21)
21 + 5a = 2a + 42
Transposing 2a from RHS to LHS and 21 from LHS to RHS, we get,
5a – 2a = 42 – 21
3a = 21
Dividing both the sides by 3, we get,
a = 7
Hence, 5a = 5 x 7 = 35
Thus, the numbers are 7 and 35 respectively.

Q.3 Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?
Sol. Let the digit at ten’s place and one’s place be s and (9 – s) respectively.
Hence, original number = 10s + (9 – s) = 9s + 9
Now, new number on interchanging the digits = 10(9 – s) + s = 90 – s
Given, 90 – 9s = 9s + 9 + 27
90 – 9s = 9s + 36
Transposing 9s from LHS to RHS and 36 from RHS to LHS, we get,
90 – 36 = 18s
54 = 18s
Dividing by 18 on both the sides, we get,
s = 3
Therefore, 9 – s = 9 – 3 = 6
Thus, the two digit number is 9s + 9 = 9 × 3 + 9 = 36.

Q.4 One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?
Sol. Let the digits at ten’s place and one’s place be s and 3s respectively.
Hence, original number = 10s + 3s = 13s
Now, number after interchanging = 10 x 3s + s = 30s + s = 31s
Given, 13s + 31s = 88
44s = 88
Dividing by 44 on both the sides, we get,
s = 2
Therefore, original number = 13s = 13 x 2 = 26
And number after interchanging = 31 x 2 = 62
Hence, the two-digit number might be 26 or 62.

Q.5 Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age. What are their present ages?
Sol. Let the age of Shobo be a years. Hence, Shobo’s mother’s age will be 6a years.
Given, {\rm{a + 5 = }}{{{\rm{6a}}} \over {\rm{3}}}
a + 5 = 2a
Transposing a from LHS to RHS, we get,
5 = 2a – a
5 = a
a = 5
Therefore, 6a = 6 x 5 = 30
Thus, the present ages of Shobo and his mother are 5 years and 30 years respectively.

Q.6 There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4.At the rate Rs100 per metre it will cost the village panchayat Rs 75000 to fence the plot. What are the dimensions of the plot?
Sol. Let the common ratio between the length and breadth of the plot be s.
Therefore, the length and the breadth of the rectangular plot will be 11s and 4s respectively.
Perimeter of the plot = 2(length + breadth) = 2(11s + 4s) = 30s
Given, 100 x Perimeter = 75000
100 x 30s = 75000
3000s = 75000
Dividing by 3000 on both the sides, we get,
s = 25
Therefore, length = 11s = (11 x 25) = 275
Breadth = 4s = (4 x 25) = 100
Thus, the dimensions of the plot are 275 m and 100 m respectively.

Q.7 Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him Rs 50 per metre and trouser material that costs him Rs 90 per metre. For every 2 meters of the trouser material he buys 3 metres of the shirt material. He sells the materials at 12% and 10% profit respectively. His total sale is Rs 36,660. How much trouser material did he buy?
Sol. Let trouser material and shirt material bought be 2a and 3a respectively.
Selling price of trouser material per metre = Rs \left( {90 + {{90 \times 12} \over {100}}} \right) = Rs 100.80
Selling price of shirt material per metre = Rs \left( {50 + {{50 \times 10} \over {100}}} \right) = Rs 55
Given, 100.80 x 2a + 55 x 3a = 36660
201.60a + 165a = 36660
366.60a = 36660
Dividing by 366.60 on both the sides, we get,
a = 100
Therefore, trouser material = 2a = 2 x 100 = 200
Shirt material = 3a = 3 x 100 = 300
Thus, Hasan should buy 200 m of trouser material.

Q.8 Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.
Sol. Let the number of deer be a.
So, number of deer grazing in the field = {a \over 2}
Number of deer playing nearby ={3 \over 4} x Number of remaining deer
={3 \over 4} \times \left( {a - {a \over 2}} \right) = {3 \over 4} \times \left( {{a \over 2}} \right) = {{3a} \over 8}
Given, a - \left( {{a \over 2} + {{3a} \over 8}} \right) = 9
a - \left( {{{4a + 3a} \over 8}} \right) = 9
a - \left( {{{7a} \over 8}} \right) = 9
{a \over 8} = 9
Multiplying by 8 on both the sides, we get,
a = 72
Thus, the total number of deer is 72.

Q.9 A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.
Sol. Let the age of granddaughter be s years. Hence, the age of grandfather will be 10s years.
Given, 10s = s + 54
Transposing s from RHS to LHS, we get,
10s – s = 54
9s = 54
s = 6
Therefore, age of grandfather = 10s = 10 x 6 = 60 years.
Thus, the present ages of granddaughter and grandfather are 6 and 60 respectively.

Q.10 Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.
Sol. Let the age of Aman’s son be a years. Hence, age of Aman will be 3a years.
Before ten years, age of Aman and his son will be (a – 10) and (3a – 10) respectively.
Given, (3a – 10) = 5(a – 10)
3a – 10 = 5a – 50
Transposing 3a from LHS to RHS and 50 RHS to LHS, we get,
50 – 10 = 5a – 3a
40 = 2a
Dividing by 2 on both the sides, we get,
20 = a
Therefore, age of Aman’s son = a = 20
Age of Aman = 3a = 3 x 20 = 60
Hence, the present ages of Aman and his son are 60 years and 20 years respectively.

 



2 Comments

Leave a Reply