10类RD Sharma解–第8章二次方程式–练习8.9

问题1：阿树x岁，而他的母亲维纳夫人x ²岁。 Veena太太五年后，其年龄将是Ashu的三倍。找到他们现在的年龄。

解决方案：

 Ashu’s present age is =  x years

 and Mrs. Veena’s present age is = x²  years

 Now five years hence,

 Mrs. Veena’s age will be = (x² + 5) years

 and Ashu’s age will be = (x + 5) years 

 So according to the question-

 ⇒ Mrs. Veena’s age = Three times of the Ashu’s age

 ⇒ x² + 5 = 3(x + 5)

 ⇒ x² + 5 = 3x + 15

 ⇒ x² – 3x – 10 = 0

Now for factorizing above quadratic equation-

Break coefficient of x in difference form as constant term is negative-

 ⇒ x² – (5-2)x – 10 = 0

 ⇒ x² – 5x + 2x – 10 = 0

 ⇒ x(x – 5) + 2(x – 5) = 0

⇒ (x – 5)(x + 2) = 0

⇒ Either x – 5 = 0       or     x + 2 = 0

              x = 5            or      x = -2 

So on discarding x = -2 (because age can not be negative)-

Mrs. Veena’s present age = x² = 25 years

and Ashu’s present age = x = 5 years

问题2：一个人和他儿子的年龄之和为45岁。五年前，他们年龄的乘积是当时男人年龄的四倍。找到他们现在的年龄。

解决方案：

Let present age of man = x years

And given,  

Man’s present age + son’s present  age = 45 years

⇒ Son’s present age = (45 – x) years.

Now Five years ago-

⇒ Age of man = (x – 5) years

and age of his son = (45-x) – 5     years

⇒ son’s age = (40-x) years

Now according to question (five years ago) –

Product of their ages = four times the man’s age at that time

⇒ (x – 5)(40 – x) = 4(x – 5)

⇒ (x – 5)(40 – x) – 4(x – 5) = 0

⇒ (x – 5)(40 – x – 4) = 0               [by taking common (x-5)]

⇒ (x – 5)(x – 36) = 0

⇒ either x – 5 = 0        or       x – 36 = 0

Now in first case if we take x = 5 then 

Man’s age = 5 years

His son’s age = 40 years (It is not possible)

so x = 36

⇒ Present age of Man = x = 36 years

⇒ and present age of his son = (45 – x) = 9 years

问题3：Shikha五年前的年龄与8岁以后的年龄的乘积是30，两次的年龄均以年为单位。找到她现在的年龄。

解决方案：

Let the present age of Shikha is = x years

So five years ago Shikha’s age was = (x – 5) years

And 8 years later Shikha’s age will be = (x + 8) years

Now according to question-

⇒ (five years ago Shikha’s age) × (8 years later Shikha’s age) = 30

⇒ (x – 5)(x + 8) = 30

⇒ x² + 3x – 40 = 30

⇒ x² + 3x – 70 = 0

⇒ x² + (10 – 7)x – 70 = 0 [when constant term is negative always 

                                         break the coefficient of x in the difference form]

⇒ x² + 10x – 7x – 70 = 0

⇒ x(x + 10) – 7(x + 10) = 0

⇒ (x + 10)(x – 7) = 0                             [after factorization]

⇒ Either x + 10 = 0             or          x – 7 = 0

              x = -10            or          x = 7

⇒ x = -10 is not valid as age can never be negative

So taking x = 7

Means present age of Shikha is 7 years.

问题4：拉木五年前的年龄(以年为单位)和九年后九岁的他(以年为单位)的乘积是15。确定拉木的当前年龄。

解决方案：

Let the present age of Ramu = x years

So,

Five years ago Ramu’s age was = (x-5) years

And Nine years later Ramu’s age will be = (x+9) years

Now according to question –

(Five years ago Ramu’s age)×(Nine years later Ramu’s age) = 15

⇒ (x-5)(x+9) = 15

⇒ x² + 4x – 45 = 15

⇒ x² + 4x – 60 = 0

Now performing factorization—

⇒ x² + (10 – 6)x – 60 = 0

⇒ x² + 10x – 6x – 60 = 0

⇒ x(x + 10) – 6(x + 10) = 0

⇒ (x + 10)(x – 6) = 0

⇒ Either x + 10 = 0             or          x – 6 = 0

              x = -10                  or          x = 6

So x = -10 is not valid as age can never be negative,

⇒ Taking x = 6

Means present age of Ramu is 6 years.         

问题5：是否可能出现以下情况?如果是这样，请确定他们的年龄。两个朋友的年龄之和为20岁。四年前，他们年龄的乘积是48。

解决方案：

Let A and B are two friends.

And age of A is = x years

Now According to the First condition –

(Age of A) + (Age of B) = 20

⇒ x + (Age of B) = 20

⇒ Age of B = (20 – x) years 

So four years ago,

A’s age was = (x – 4) years

and B’s age was = (20 – x – 4)= (16 – x) years

Now coming to the Second condition –

(A’s age) × (B’s age) = 48

⇒ (x-4)(16-x) = 48

⇒ 16x – x² – 64 + 4x = 48

⇒ -x² + 20x – 112 = 0

⇒ x² – 20x + 112 = 0      ———————————–(1)

Now checking for Discriminatnt (D) = b² – 4ac

Comparing  equaton(1) with ax² + bx + c = 0—

⇒ a = 1, b = -20, c = 112

⇒ D = (-20)² – 4 * 1 * 112

⇒ D = 400 – 448

⇒ D = -48

⇒ D < 0

So roots are imaginary means.

Therefore, above given situations are not possible.

问题6：一个女孩的年龄是她姐姐的两倍。因此，四年后，他们的年龄乘积(以年为单位)将为160。找到他们的当前年龄。

解决方案：

Let present age of girl is = 2x years 

So present age of her sister = x years         [because girl is twice as old as her sister

                                                                    so sister’s age will be half that of girl’s age]

Now four years hence-

Girl’s age will be = (2x + 4) years

and sister’s age will be = (x + 4) years

Now according to given condition —

⇒ (Girl’s age) × (sister’s) = 160

⇒ (2x+4)(x+4) = 160

⇒ 2x² + 8x + 4x + 16 = 160

⇒ 2x² + 12x + 16 -160 = 0

⇒ 2x² + 12x – 144 = 0

⇒ x² + 6x – 72 = 0

⇒ x² + 12x – 6x – 72 =0

⇒ x(x + 12) – 6(x + 12) = 0 

⇒ (x – 6)(x + 12) = 0

⇒ Either x – 6 = 0      or      x + 12 = 0

              x = 6       or          x = -12 

But x =-12 is not possible as age can never be negative

So taking x = 6

⇒ Present age of Girl = 2x = 12 years

and present age of her sister = x = 6 years.

问题7：3年前和5年前的Rehman年龄(以年为单位)的倒数之和是1/3。找到他现在的年龄。

解决方案：

Let Rehman’s present age = x years

So three years ago Rehman’s age was = (x-3) years

and five years after Rahman’s age will be = (x+5) years

Now according to given condition-

⇒ [1/(three years ago Rehman’s age)] + [1/(five years after Rahman’s age)] = 1/3

⇒ [1/(x – 3)] + [1/(x + 5)] = 1/3

⇒ (x + 5) + (x – 3) = (x + 5)(x – 3)/3 [after taking LCM and transferring denominator of LHS to the RHS]

⇒ 3(2x + 2) = (x² + 2x – 15)               [multiplying by 3]

⇒ 6x + 6 = x² + 2x – 15

⇒ x² – 4x – 21 = 0

⇒ x² – (7 – 3)x  – 21 = 0        [by factorization law]

⇒ x² – 7x + 3x – 21 = 0

⇒ x(x – 7) + 3(x – 7) = 0

⇒ (x – 7)(x + 3) = 0 

⇒ Either x – 7 = 0         or          x + 3 = 0

              x = 7       or          x = -3

But x = -3 is not possible as age can never be negative.

So taking x = 7 —

Means present age of Rehman = 7 years.

问题8：如果Zeba比实际年龄小5岁，那么她的年龄平方(以年为单位)将是实际年龄的5倍多11。她现在几岁?

解决方案：

Let Zeba’s present age (actual age) is = x years 

So her age if she were 5 years younger = (x – 5) years       

Now according to condition-

⇒ (x – 5)² = 5x + 11

⇒ x² – 10x + 25 = 5x + 11

⇒ x² – 15x + 14 = 0

⇒ x² – (14 + 1)x + 14 = 0

⇒ x² -14x – x + 14 = 0

⇒ x(x – 14) – (x – 14) = 0

⇒ (x – 14)(x – 1) = 0

⇒ Either x – 1 = 0           or          x – 14 = 0

But if x – 1 = 0 ⇒ x = 1, so in this case situation when she was 5 years younger is not possible.

Therefore, taking x – 14 = 0

⇒ x = 14

means Zeba’s present age is = 14 years