10类RD Sharma解–第8章二次方程式–练习8.13

问题1.一块布的价格为₹35。如果一块布的长度增加4 m，每米的成本降低₹1，则成本将保持不变。一块多长时间?

解决方案：

Let us considered the length of piece of cloth = x m

Given: The total cost = ₹ 35

So, the cost of 1 m cloth = ₹ 35/x

According to the question,

(x + 4)(35/x – 1) = 35

⇒ 35 – x + (140/x) – 4 = 35

⇒ -x + (140/x) + 31 – 35 = 0

⇒  -x + (140/x) – 4 = 0

⇒ -x² + 140 – 4x = 0

⇒ x² + 4x – 140 = 0

⇒ x² + 14x – 10x – 140 = 0

⇒ x (x + 14) -10 (x + 14) = 0

⇒ (x + 14) (x – 10) = 0

x = 10 or x = -14

Here the value of x = -14 is negative which is not possible

So, the length of piece of cloth = 10 m.

问题2。一些学生计划去野餐。伙食的预算是480卢比。但是其中有8个没有通过，因此每个成员的食费增加了10卢比。有多少学生参加了野餐?

解决方案：

Let us considered the number of students = x

Given: The total budget = ₹ 480

So, the share of each student = ₹ 480/x

According to the question,

(480/x – 8) – (480/x) = 10

⇒ $\frac {(480x - 480x + 3840)} {x(x - 8)}=10$

⇒ $\frac {3840}{(x^2 - 8x)} = 10$

⇒ 10x² – 80x – 3840 = 0

⇒ x² – 8x – 384 = 0               

⇒ x² + 16x – 24x – 384 = 0

⇒ x (x + 16) – 24 (x + 16) = 0

⇒ (x + 16) (x – 24) = 0

x = 24, or x = -16

Here the value of x = -16 is negative which is not possible

So, the number of students = 24.

The total number of students attend the picnic = 24 – 8 = 16

问题3.经销商以24卢比的价格出售一件商品，并获得该商品成本价的百分之几。查找文章的成本价格。

解决方案：

Let cost price of the article = ₹ x

Selling price = ₹ 24

Gain = x %

According to the Question,

S.P.= C.P. × (100 + Gain%)/100

24 = x(100 + x)/100

⇒ 2400 = 100x + x²

⇒ x² + 100x – 2400 = 0

⇒ x² – 20x +120x – 2400 = 0

⇒ x(x – 20) + 120(x – 20) = 0

⇒ (x – 20)(x + 120) = 0

x = 20 or x = -120

Here the value of x = -120 is negative which is not possible

Therefore, the cost price of the article = ₹20

问题4：在一组天鹅中，总数的平方根的7/2倍正在池塘中游玩。剩下的两个在水中摇摆。找到天鹅的总数。

解决方案：

Let us considered the total number of swans = x

According to the question,

7/2(√x) + 2 = x

⇒ 7√x = 2x – 4

On squaring both sides, we get

⇒ 49x = 4x + 16 – 16x

⇒ 4x² – 65x + 16 = 0

⇒ 4x² – 64x – x + 16 = 0 

⇒ 4x(x – 16) – (x – 16) = 0

⇒ (4x – 1)(x – 16) = 0

⇒ x =1/4 or x = 16

Since, number of swans is a natural number we can neglect the solution of a = 1/4

Hence, the total number of swans is 16.

问题5.如果玩具的标价降低₹2，则一个人可以花₹360购买两把玩具拖把。找到玩具的原始价格。

解决方案：

Let the original price of the toy = x

The number of toys he can buy at the original price for ₹ 360 = 360/x

According to the question,

$\frac {360} {x-2}=\frac {360} {x}+2$

⇒ 360x = (x – 2)(360 + 2x)

⇒ 360x = 360x + 2x² – 720 – 4x

⇒ x² – 2x – 360 = 0

⇒ x² – 20x + 18x – 360 = 0 

⇒ x(x – 20) + 18(x – 20) = 0

⇒ (x + 18)(x – 20) = 0

⇒ x + 18 = 0 or x – 20 = 0

⇒ x = -18 or x = 20

As, the price can’t be negative, x = -18 is neglected.

Thus, the original price of the toy is ₹ 20.

问题6：9000卢比在一定数量的人中平均分配。如果再有20个人，每人将少了₹160。查找原始人数。

解决方案：

Let’s consider the original number of people = a

Amount which each receives when a person are present = 9000/a

According to the question,

$\frac {9000} {a+20}=\frac {9000} {a}-160$

⇒ 9000a = (9000 – 160a)(a + 20)

⇒ 9000a = 9000a + 180000 – 160a² – 3200a

⇒ a² + 20a – 1125 = 0

⇒ a² + 45a – 25a – 1125 = 0

⇒ a(a + 45) – 25(a + 45) = 0

⇒ (a – 25)(a + 45) = 0

⇒ a = 25 or a = -45 (Number of people can never be negative,

                                   so we can neglect this value)

Therefore, the original number of people = 25.

问题7。一些学生计划去野餐。伙食的预算是500卢比。但是其中5人没有去，因此每个人的食费增加了5卢比。有多少学生参加了野餐?

解决方案：

Let us considered the number of students = x

Given: The total budget = ₹ 500

So, the share of each student = ₹ 500/x

The number of students failed to go = 5

According to the question,

$\frac {500} {x-5}-\frac {500} {x}=5$

$\implies \frac {500x-500x +2500} {(x)(x-5)}=5$

$\implies \frac {2500} {(x^2-5x)}=5$

⇒ 5x² – 25x – 2500 = 0

⇒ x² – 5x – 500 = 0                    

⇒ x² – 25x + 20x – 500 = 0

⇒ x(x – 25) + 20(x – 25) = 0

⇒ (x – 25)(x + 20) = 0

⇒ x – 25 = 0 or x + 20 = 0

⇒ x = 25 or x = -20

Here the value of x = -20 is negative which is not possible

So, the number of students = 25

问题8.必须在a的边界上的某个点竖立一个极点。直径13米的圆形停车场，其与边界上两个直径相对的固定门A和B的距离之差为7米。有可能这样做吗?是的话，离两扇闸门的距离应该是多少?

解决方案：

In the given circle, AB(Diameter) = 13 m

Let us considered P be the pole in the circle.

So, PB = x m and PA= (x + 7) m

In the triangle APB

AB² = AP² + PB²

⇒ 13² = (x + 7)² + (x)²

⇒ 169 = x² + 49 +14x + x²

⇒ 2x² + 14x -120 = 0

⇒ x² +7x – 60 = 0

⇒ x² +12x – 5x – 60 = 0

⇒ x(x + 12) – 5(x + 12) = 0

⇒ (x + 12)(x – 5) = 0

x = -12 or x = 5

Here the value of x = -12 is negative which is not possible

So the value valid of x = 5

Hence, PB = 5 m and PA = 5 + 7 = 12 m.

问题9：在课堂测试中，P在数学和科学中获得的分数总和为28。如果他在数学中获得3分以上，在科学方面得到4分以下。他们的分数乘积为180。在两个主题中找到他的分数。

解决方案：

Given: The sum of marks obtained by P in Mathematics and Science = 28

So let us considered the marks in Mathematics = x

and the marks in Science = 28 – x

According to the question

(x + 3) (28 – x – 4) = 180

⇒ (x + 3) (24 – x) = 180

⇒ 24x – x² + 72 – 3x = 180

⇒ 21x – x² + 72 – 180 = 0

⇒ – x² + 21x – 108 = 0

⇒ x² – 21x + 108 = 0

⇒ x² – 9x – 12x + 108 = 0

⇒ x (x – 9) – 12 (x – 9) – 0

⇒ (x – 9)(x – 12) = 0

x = 9 or x = 12

So, if we take x = 9 then the marks in 

Mathematics = 9 and marks in Science = 19

So, if we take x = 12 then the marks in 

Mathematics = 12 and marks in English = 16

问题10：在课堂测试中，Shefali在数学和英语中的分数总和是30。如果她在数学中获得2分，而在英语中少得到3分，则她的分数乘积应为210。在两个中找到她的分数科目。

解决方案：

Given: Sum of Shefali’s marks in Mathematics and English = 30

So let us considered the marks in Mathematics = x

and the marks in English = 30 – x

According to the question

(x + 2) (30 – x – 3) = 210

⇒ (x + 2) (27 – x) = 210

⇒ 27x – x² + 54 – 2x – 210 = 0

⇒ – x² + 25x – 156 = 0

⇒ x² – 25x + 156 = 0

⇒ x² – 12x – 13x +156 = 0

⇒ x (x – 12) – 13 (x – 12) = 0

⇒ (x – 12) (x – 13) = 0

x = 13 or x = 12

So, if we take x = 13 then the marks in 

Mathematics = 12 and marks in English = 18

So, if we take x = 12 then the marks in 

Mathematics = 13 and marks in English = 17

问题11：家庭手工业每天生产一定数量的陶器产品。在某一天观察到，每件商品的生产成本(以卢比计)比当日生产的商品数量多3倍。如果当天的总生产成本为₹90，则生产的商品数量和每件商品的成本。

解决方案：

Given: Total cost of production on that day = ₹90

So, Let us considered the number of articles = x

Hence the price of each article = 2x + 3

According to the question,

x (2x + 3) = 90

⇒ 2x² + 3x – 90 = 0

⇒ 2x² -12x + 15x – 90 = 0

⇒ 2x (x – 6) + 15 (x – 6) = 0

⇒ (x – 6) (2x + 15) = 0

If x – 6 = 0

So, x = 6

If 2x + 15 = 0

So x = -15/2

Here the value of x is negative which is not possible

So the value valid of x = 6

Hence the number of articles = 6

And price of each article = 2x + 3 = 2 x 6 + 3 = 12 + 3 = 15

问题12在t分钟过去下午2点由分针和时钟需要显示下午3点的时间被发现是小于T3分钟^2/4分钟。找到t。

解决方案：

As we already know that, the time between 2 pm to 3 pm = 1 h = 60 minutes

Given: At t minutes past 2 pm, the time needed by the minutes hand of a clock 

to show 3 pm was found to be 3 minutes less than t²/4 minutes.

Find: the value of t

So, $t = \frac {t^2}{4}-3 = 60$

⇒ 4t + t² – 12 = 240

⇒ t² + 4t – 252 = 0

⇒ t² + 18t – 14t – 252 = 0 

⇒ t (t + 18) – 14 (t + 18) = 0

⇒ (t + 18) (t – 14) = 0 

As we know that time cannot be negative, so t ≠ -18

Hence, t = 14 min