ncert-solutions for class 8 mathematics chapter 2 exercise 2 - 4
Answers
Answer:
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?
Answer :
Let the number be x.
According to the given question,
8(x-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.
Q2 :
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 be x and 5x. According to the 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 x 7 = 35
Hence, the numbers are 7 and 35 respectively.
Q3 :
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 ones place be x and 9 - x respectively.
Therefore, original number = 10x + (9 - x) = 9x + 9
On interchanging the digits, the digits at ones place and tens place will be x and 9 - x
respectively.
Therefore, new number after interchanging the digits = 10(9 - x) + x
= 90 - 10x + x
= 90 - 9x
According to the given 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 = x and 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 9x + 9 = 9 x 3 + 9 = 36
Q4 :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 ones place be x and 3x respectively.
Therefore, original number = 10x + 3x = 13x
On interchanging the digits, the digits at ones place and tens place will be x and 3x respectively.
Number after interchanging = 10 x 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 x 2 = 26
By considering the tens place and ones place as 3x and x respectively, the two-digit number
obtained is 62.
Therefore, the two-digit number may be 26 or 62.
Q5.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 be x years. Therefore, his mother's age will be 6x years.
According to the given question,
After 5 years Shobo's mother preasent age/3
x+5=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.
Q6 :
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 be x. Hence, the
length and breadth of the rectangular plot will be 11x m and 4x m respectively.
Perimeter of the plot = 2(Length + Breadth)=[2(11x+4x)]m=30xm
It is given that the cost of fencing the plot at the rate of Rs 100 per metre is Rs 75, 000.
100 × Perimeter = 75000
100 × 30x = 75000
3000x = 75000
Dividing both sides by 3000, we obtain
x = 25
Length = 11x m = (11 × 25) m = 275 m
Breadth = 4x m = (4 × 25) m = 100 m
Hence, the dimensions of the plot are 275 m and 100 m respectively.