Question 1.
In a rectangle of perimeter one metre, one side is 5 cm longer than the other. What are the lengths of the rectangle?
Solution: 22.5 cm & 27.5 cm are the lengths of the rectangle.
A class has 4 more girls than boys. On a day when only 8 boys were absent, the number of girls was twice that of the boys. How many girls and boys are there in the class?
Question 3.
A man invested 10000 rupees, split into two schemes, at annual rates of interest 8% and 9%. After one year he got 875 rupees as interest from both. How much did he invest in each?
Answer:
If one part is x then
the remaining part is 10000 – x
x×8100+(10000−x)×9100=100
8x + 90000 – 9x = 87500
90000 – 87500 = x
2500 = x
one part = 2500 and
remaining part = 7500
Question 4.
A three and a half metre long rod is to be cut into two pieces, one piece is to be bent into a square and the other into an equilateral triangle. The length of their sides must be the same. How should it be cut?
Answer:
Total length = 3½ m
Since the sides of a square and equilateral triangle are equal, all these 7 sides are equal.
∴ Length of one side
312÷7=72÷7=12m
Length of the rod for the square
=4×12 = 2m
Length of the rod for the equilateral triangle = 3×12 = 112m
Question 5.
The distance travelled in t seconds by an object starting with a speed of u metres/second and moving along a straight line with speed increasing at the rate of a metres/second every second is given by ut + ½ at² metres. An object moving in this manner travels 10 metres in 2 seconds and 28 metres in 4 seconds. With what speed did it start? At what rate does its speed change?
Answer:
If t = 2
ut + ½ at²= 10
2u + 2a= 10
u + a = 5 — (1)
If t = 4
4u + 8a = 28
u + 2a = 7 — (2)
from (1) and (2)
Textbook Page No. 40
Question 1.
Raju bought seven notebooks of two hundred pages and five of hundred pages, for 107 rupees. Joseph bought five notebooks of two hundred pages and seven of hundred pages, for 97 rupees. What is the price of each kind of notebook?
Answer:
Cost of 200 page note book = x
Cost of 100 page note book = y
7x + 5y= 107 …………(1)
5x + 7y = 97 …………(2)
(1) × 5 ⇒ 35x + 25y = 535 …………(3)
(2) × 7 ⇒ 35x + 49y = 679 …………(4)
(4) – (3) ⇒ 24y = 144
y = 14424 = 6
Substitute y = 6 in equation (1)
7x + 30 = 107; 7x = 77
x = 777 = 11
Price of the 200 pages notebook = Rs. 11
Price of the 100 pages notebook = Rs. 6
Question 2.
Four times a number and three times number added together make 43. Two times the second number, subtracted from three times the first give 11. What are the numbers?
Answer:
Let the first number = x and
the second number = y
4x + 3y = 43 …………(1)
3x – 2y = 11 …………(2)
(1) × 3 ⇒ 12x + 9y= 129 …………(3)
(2) × 4 ⇒ 12x – 8y = 44 …………(4)
(3) -(4) ⇒ 17y = 85; y = 8517 = 5
Substitute y = 5 in equation (1)
4x + 3y = 43
4x + 15 = 43
4x = 43 – 15 = 28
∴ x = 284 = 7, y = 5
First number = 7
Second number = 5
Question 3.
The sum of the digits of two – digit number is 11. The number got by interchanging the digits is 27 more than the original number. What is the number?
Answer:
If the numbers are x and y
x + y = 11 …………(1)
10x + y + 27 = 10y + x
9x – 9y = -27
X – y = -3 …………(2)
(1) + (2) 2x = 8; x = 4
x + y = 11
4 + y = 11
y = 7
∴ Required number is 47.
Question 4.
Four years ago, Rahim’s age was three times Ramu’s age. After two years, it would just be double. What are their ages now?
Answer:
Ramu’s present age = x
Rahim’s present age = y
4 years back,
Ramu’s age = x – 4
Rahim’s age = y – 4
3(x – 4) = y – 4
3x – 12 = y – 4
3x – y = 8 ……….(1)
After 2 years,
Ramu’s age = x + 2
Rahim’s age = y + 2
2(x + 2) = y + 2
2x + 4 = y + 2
2x – y = -2 ……….(2)
(1) – (2) ⇒ x = 10
3x – y = 8; 30 – y = 8; y = 22
x = 10, y = 22
Ramu’s present age = 10
Rahim’s present age = 22
Question 5.
If the length of a rectangle is in-creased by 5 metres and breadth decreased by 3 metres, the area would decrease by 5 square metres. If the length is increased by 3 metres and breadth increased by 2 metres, the area would increase by 50 square metres. What are the length and breadth?
Answer:
length = x; breadth = y
(x + 5)(y – 3) = xy – 5
xy – 3x + 5y – 15 = xy – 5
– 3x + 5y = + 10
3x – 5y = -10 ………..(1)
(x + 3)(y + 2) = xy + 50
xy + 2x + 3y + 6 = xy + 50
2x + 3y = 44 ………..(2)
(2) × 1 ⇒ 6x-10y = -20 ……….(3)
(3) × 2 ⇒ 6x + 9y = 132 …………(4)
(3)- (4) ⇒ -19y = -152
y = −152−19
a = 2
∴ u = 3
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