In this page we have Class 8 Maths Chapter 2 Exercise 2.5 Solutions.This exercise has questions about some more applications on simple linear equation.Here you need to form the equation from question and then solve it. The equation forms here would have linear expression on both the sides. Hope you like them and do not forget to like , social share
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Class 8 Maths Chapter 2 Exercise 2.5 Solutions
Question 1 Amina thinks of a number and subtracts 5/2 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? Answer Let the number be x.
According to the given question,
$8(x- \frac {5}{2}) = 3x$
$8x - 20 = 3x$
Transposing 3x to L.H.S and -20 to R.H.S, we obtain
$8x - 3x= 20$
5x= 20
Dividing both sides by 5, we obtain
x= 4
Hence, the number is 4.
Question 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? Answer Let the numbers bexand 5x. As per question,
$21 + 5x= 2(x+ 21)$
$21 + 5x= 2x+ 42$
Transposing 2x to L.H.S and 21 to R.H.S, we obtain
$5x - 2x= 42 - 21$
3x= 21
Dividing both sides by 3, we obtain
x= 7
5x= 5 × 7 = 35
Hence, the numbers are 7 and 35 respectively.
Question 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? Answer Let the digits at tens place and one’s place bexand 9 -xrespectively.
Therefore, original number = $10x+ (9 -x) = 9x+ 9$
On interchanging the digits, the digits at ones place and tens place will bexand 9 -xrespectively.
Therefore, new number after interchanging the digits = 10(9 -x) +x
= 90 - 10x+x
= 90 - 9x
As per question,
New number = Original number + 27
$90 - 9x= 9x+ 9 + 27$
$90 - 9x= 9x+ 36$
Transposing 9x to R.H.S and 36 to L.H.S, we obtain
$90 - 36 = 18x$
54 = 18x
Dividing both sides by 18, we obtain
3 =xand 9 -x= 6
Hence, the digits at tens place and ones place of the number are 3 and 6 respectively.
Therefore, the two-digit number is 36
Question 4 One of the two digits of a two digit number is three times the other digit. If you interchange the digit of this two-digit number and add the resulting number to the original number, you get 88. What is the original number? Answer Let the digits at tens place and one’s place bexand 3xrespectively.
Therefore, original number = 10x+ 3x= 13x
On interchanging the digits, the digits at ones place and tens place will bexand 3xrespectively.
Number after interchanging = 10 × 3x+x= 30x+x= 31x
According to the given question,
Original number + New number = 88
$13x+ 31x= 88$
44x= 88
Dividing both sides by 44, we obtain
x= 2
Therefore, original number = 13x= 13 × 2 = 26
By considering the tens place and ones place as 3xandxrespectively, the two-digit number obtained is 62.
Therefore, the two-digit number may be 26 or 62.
Question 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 this mother’s present age. What are their present ages? Answer Let Shobo’s age bexyears. Therefore, his mother’s age will be 6xyears.
According to the given question,
$(x+5)=\frac {6x}{3}$
$x+ 5 = 2x$
Transposing x to R.H.S, we obtain
$5 = 2x -x$
5 =x
6x= 6 × 5 = 30
Therefore, the present ages of Shobo and Shobo’s mother will be 5 years and 30 years respectively.
Question 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 Rs 100 per metre it will cost the village panchayat Rs 75, 000 to fence the plot. What are the dimensions of the plot? Answer Let the common ratio between the length and breadth of the rectangular plot bex. Hence, the length and breadth of the rectangular plot will be 11xm and 4xm respectively.
Perimeter of the plot = 2(Length + Breadth)
It is given that the cost of fencing the plot at the rate of Rs 100 per metre is Rs 75, 000.
So
100 × Perimeter = 75000
100 × 30x= 75000
3000x= 75000
Dividing both sides by 3000
x= 25
Length = 11xm = (11 × 25) m = 275 m
Breadth = 4xm = (4 × 25) m = 100 m
Hence, the dimensions of the plot are 275 m and 100 m respectively.
Question 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 36660. How much trouser material did he buy? Answer Let 2xm of trouser material and 3xm of shirt material be bought by him.
Per metre selling price of trouser material =Original Price + Profit
=90 + 12% Profit
=90 + 12 × 90/100
=Rs 100.80
Per metre selling price of shirt material
=Original Price + Profit
=50 + 10% Profit
=50 + 10 × 50/100
=Rs 55
We know from question that total amount of selling = Rs 36660,So
100.80 × (2x) + 55 × (3x) = 36660
201.60x+ 165x= 36660
366.60x=36660
Dividing both sides by 366.60
x= 100
Trouser material = 2xm = (2 × 100) m = 200 m
Question 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. Answer Let the number of deer bex.
Number of deer grazing in the field =x/2
So remaining would be x/2
Now Number of Deer playing near by= (3/4)(x/2)= 3x/8
Number of deer drinking water from the pond = 9
So
Multiplying both sides by 8, we obtain
8x –(4x+3x)=72
x=72
Hence, the total number of deer in the herd is 72.
Question 9 A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages Answer Let the granddaughter’s age bexyears. Therefore, grandfather’s age will be
10xyears.
According to the question,
Grandfather’s age = Granddaughter’s age + 54 years
10x=x+ 54
Transposingxto L.H.S, we obtain
10x-x= 54
9x= 54
x= 6
Granddaughter’s age= xyears = 6 years
Grandfather’s age = 10xyears = (10 × 6) years = 60 years
Question 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. Answer Let Aman’s son’s age bexyears. Therefore, Aman’s age will be 3xyears. Ten years ago, their age were (x- 10) years and (3x- 10) years respectively.
According to the question,
10 years ago, Aman’s age = 5 × Aman’s son’s age 10 years ago
3x- 10 = 5(x- 10)
3x- 10 = 5x- 50
Transposing 3xto R.H.S and 50 to L.H.S, we obtain
50 - 10 = 5x- 3x
40 = 2x
Dividing both sides by 2, we obtain
20 =x
Aman’s son’s age =xyears = 20 years
Aman’s age = 3xyears = (3 × 20) years = 60 years
Summary
Class 8 Maths Chapter 2 Exercise 2.5 Solutions has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also Download NCERT Solutions Linear equation Exercise 2.5 as pdf
This chapter 2 has total 2 Exercise 2.1,2.2 as per latest syllabus. Exercise 2.3,2.4,2.5.2.6 has been deleted as per latest syllabus. This is one of the deleted exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below
