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6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Solutions: Let us say, the shorter side of the rectangle be x m.
Then, larger side of the rectangle = (x + 30) m
Ncert solutions class 10 chapter 4-6
As given, the length of the diagonal is = x + 30 m
Therefore,
Ncert solutions class 10 chapter 4-7
⇒ x2 + (x + 30)2 = (x + 60)2
⇒ x2 + x2 + 900 + 60x = x2 + 3600 + 120x
⇒ x2 – 60x – 2700 = 0
⇒ x2 – 90x + 30x – 2700 = 0
⇒ x(x – 90) + 30(x -90) = 0
⇒ (x – 90)(x + 30) = 0
⇒ x = 90, -30
However, side of the field cannot be negative. Therefore, the length of the shorter side will be 90 m.
and the length of the larger side will be (90 + 30) m = 120 m.
7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution: Let us say, the larger and smaller number be x and y respectively.
As per the question given,
x2 – y2 = 180 and y2 = 8x
⇒ x2 – 8x = 180
⇒ x2 – 8x – 180 = 0
⇒ x2 – 18x + 10x – 180 = 0
⇒ x(x – 18) +10(x – 18) = 0
⇒ (x – 18)(x + 10) = 0
⇒ x = 18, -10
However, the larger number cannot considered as negative number, as 8 times of the larger number will be negative and hence, the square of the smaller number will be negative which is not possible.
Therefore, the larger number will be 18 only.
x = 18
∴ y2 = 8x = 8 × 18 = 144
⇒ y = ±√144 = ±12
∴ Smaller number = ±12
Therefore, the numbers are 18 and 12 or 18 and -12.
8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution: Let us say, the speed of the train be x km/hr.
Time taken to cover 360 km = 360/x hr.
As per the question given,
⇒ (x + 5)(360-1/x) = 360
⇒ 360 – x + 1800-5/x = 360
⇒ x2 + 5x + 10x – 1800 = 0
⇒ x(x + 45) -40(x + 45) = 0
⇒ (x + 45)(x – 40) = 0
⇒ x = 40, -45
As we know, the value of speed cannot be negative.
Therefore, the speed of train is 40 km/h.
9. Two water taps together can fill a tank in 9 3/8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution: Let the time taken by the smaller pipe to fill the tank = x hr.
Time taken by the larger pipe = (x – 10) hr
Part of tank filled by smaller pipe in 1 hour = 1/x
Part of tank filled by larger pipe in 1 hour = 1/(x – 10)
As given, the tank can be filled in 9 3/8 = 75/8 hours by both the pipes together.
Therefore,
1/x + 1/x-10 = 8/75
x-10+x/x(x-10) = 8/75
⇒ 2x-10/x(x-10) = 8/75
⇒ 75(2x – 10) = 8x2 – 80x
⇒ 150x – 750 = 8x2 – 80x
⇒ 8x2 – 230x +750 = 0
⇒ 8x2 – 200x – 30x + 750 = 0
⇒ 8x(x – 25) -30(x – 25) = 0
⇒ (x – 25)(8x -30) = 0
⇒ x = 25, 30/8
Time taken by the smaller pipe cannot be 30/8 = 3.75 hours, as the time taken by the larger pipe will become negative, which is logically not possible.
Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 – 10 =15 hours respectively.
10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speeds of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
Solution: Let us say, the average speed of passenger train = x km/h.
Average speed of express train = (x + 11) km/h
Given, time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance. Therefore,
(132/x) – (132/(x+11)) = 1
132(x+11-x)/(x(x+11)) = 1
132 × 11 /(x(x+11)) = 1
⇒ 132 × 11 = x(x + 11)
⇒ x2 + 11x – 1452 = 0
⇒ x2 + 44x -33x -1452 = 0
⇒ x(x + 44) -33(x + 44) = 0
⇒ (x + 44)(x – 33) = 0
⇒ x = – 44, 33
As we know, Speed cannot be negative.
Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train will be 33 + 11 = 44 km/h.
11. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
Solutions: Let the sides of the two squares be x m and y m.
Therefore, their perimeter will be 4x and 4y respectively
And area of the squares will be x2 and y2 respectively.
Given,
4x – 4y = 24
x – y = 6
x = y + 6
Also, x2 + y2 = 468
⇒ (6 + y2) + y2 = 468
⇒ 36 + y2 + 12y + y2 = 468
⇒ 2y2 + 12y + 432 = 0
⇒ y2 + 6y – 216 = 0
⇒ y2 + 18y – 12y – 216 = 0
⇒ y(y +18) -12(y + 18) = 0
⇒ (y + 18)(y – 12) = 0
⇒ y = -18, 12
As we know, the side of a square cannot be negative.
Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.
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