第 8 类 RD Sharma – 第 10 章正变和逆变 – 练习 10.2 |设置 2

第 10 章正变和反变——练习 10.2 |设置 1

问题 10. 汽车以 48 公里/小时的速度可以在 10 小时内完成一段路程。它的速度应该提高多少，这样它可能只需要 8 个小时就能走完同样的距离？

解决方案：

As we know if the speed of the car increases, then it will take less time to reach the destination. So, the speed of the car and time vary inversely. Let us take a km/hr be the speed of the car

Speed of Car	48	a
Time taken	10	8

So, 48 × 10 = a × 8

=> a = 60 km/hr

So, the speed needed = 60 km/hr

And, the speed of car = 48 km/hr

So, increased speed of car = 60 – 48 = 12 km/hr

Hence, the speed of the car must be increased by 12 km/hr

问题11.一个堡垒的1200名士兵有足够28天的食物。 4天后，一些士兵被转移到另一个堡垒，因此食物现在又持续了32天。有多少士兵离开了堡垒？

解决方案：

As we know that the number of soldiers and the number of days vary inversely. Let a be the number of soldiers present in the fort

Number of soldiers	1200	a
Number of days	28	32

So, 1200 × 28 = a × 32

=> a = 900

Total soldiers that were present in the fort = 1200 

So, the number of soldiers left the fort = 1200 – 900 = 300

Hence, 900 soldiers are present in the fort and 300 soldiers left the fort

问题 12.三台喷涂机协同工作，60 分钟即可完成房屋的粉刷。 5台同样容量的机器做同样的工作需要多长时间？

解决方案：

As we know if the number of spraying machines is increased, then they will take less time to paint the house. So, the number of spraying machines and time vary inversely. Let the machines take a minute to paint the house

Number of spraying machines	3	5
Number of minutes	60	a

So, 3 × 60 = 5 × a

=> a = 36 

Hence, 5 spraying machines will take 36 minutes to paint the house.

问题 13.一群 3 位朋友住在一起，每个月消耗 54 公斤小麦。更多的朋友加入了这个群，他们发现同样数量的小麦可以持续 18 天。这个小组现在有多少新成员？

解决方案：

As we know that the number of friends and their food consumption varies inversely. Let currently there be a number of friends in the group

Number of friends in the Group	3	a
Number of days	30	18

So, 3 × 30 = a × 18

=> a = 5

Number of friends initially = 3

Number of friends now = 5

So, the number of friends that joined the group = 5 – 3 = 2

Hence, 2 new members joined the group

问题 14. 55 头奶牛可以在 16 天内放牧一块田地。 10 天内有多少头奶牛会在同一片土地上吃草？

解决方案：

As we know that the number of cows and time taken to graze the field var inversely. Let a cow will graze the field in 10 days

Number of Cows	55	a
Number of days	16	10

So, 55 × 16 = a × 10

=> a = 88

Hence, 88 cows will graze the field in 10 days 

问题 15. 18 个人可以在 35 天内收割一块地。 15天收割同一块地，需要多少人？

解决方案：

As we know that the number of men and time taken to reap the field var inversely. Let a men will reap the field in 15 days

Number of Men	18	a
Number of days	35	15

So, 18 × 35 = a × 15

=> a = 42

Hence, 42 men will reap the field in 15 days

问题 16.一个人有钱可以购买 25 个价值卢比的周期。 500个。如果每个周期花费卢比，他将能够购买多少个周期。还有125个？

解决方案：

As we know that if the prices of cycles are less, the person can buy more cycles and vice-versa. So, the number of cycles and their cost varies inversely. Let a be the number of cycles, the person can buy

Number of Cycles	25	a
Price of each Cycle	500	625

So, 25 × 500 = a × 625

=> a = 20

Hence, the person can buy 20 cycles if each cycle is costing Rs. 125 more

问题 17. Raghu 有足够的钱购买价值卢比的 75 台机器。 200个。如果他获得卢比的折扣，他可以购买多少台机器。每台机器50？

解决方案：

As we know that if the prices of machines are less, Raghu can buy more machines and vice-versa. So, the number of machines and their cost vary inversely. Let a be the number of machines, Raghu can buy

Number of Machines	75	a
Price of each Machine	200	150

So, 75 × 200 = a × 150

=> a = 100

Hence, Raghu can buy 100 machines if he gets a discount of Rs. 50 on each machine

问题 18. 如果 x 和 y 反向变化，并且

i) 当 y = 8 时 x = 3，当 x = 4 时找到 y

解决方案：

As we know x and y vary inversely

x	3	4
y	8	y1

So, 3 × 8 = 4 × y1

=> y1 = 6

ii) x = 5，当 y = 15 时，找到 x 当 y = 12

解决方案：

As we know x and y vary inversely

x	5	x1
y	15	12

So, 5 × 15 = x1 × 12

=> x1 = 25/4

iii) x = 30，当变化常数 = 900 时求 y

解决方案：

Constant of variation i.e k = 900

As we know x and y vary inversely, which means 

x × y = k

So, 30 × y = 900

Hence, y = 30

iv) y = 35，当变化常数 = 7 时求 x

解决方案：

Constant of variation i.e k = 7

As we know x and y vary inversely, which means 

x × y = k

So, x × 35 = 7

Hence, x = 1/5