Search

NCERT Exemplar Solution for CBSE Class 10 Mathematics: Quadratic Equations (Part-IVB)

In this article you will get CBSE Class 10 Mathematics chapter 4, Quadratic Equations: NCERT Exemplar Problems and Solutions (Part-IVB). Every question has been provided with a detailed solution. All the questions given in this article are very important to prepare for CBSE Class 10 Board Exam 2017-2018.

Jun 30, 2017 17:40 IST
facebook IconTwitter IconWhatsapp Icon

Class 10 Maths NCERT Exemplar, Quadratic Equations NCERT Exemplar Problems, NCERT Exemplar Problems, Class 10 NCERT ExemplarHere you get the CBSE Class 10 Mathematics chapter 4, Quadratic Equations: NCERT Exemplar Problems and Solutions (Part-IVB). This part of the chapter includes solutions of Question Number 4 to 8 from Exercise 4.4 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Quadratic Equations. This exercise comprises only of the Long Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.

NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.

Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Quadratic Equations:

Exercise 4.4

Long Answer Type Questions (Q. No. 4-8)

Question4. A train, travelling at a uniform speed for 360 km, would have taken 48 min less to travel the same distance, if its speed were 5 km/h more. Find the original speed of the train.

Solution:

Let the original speed of the train = x km/h

Distance covered = 360 km

Time taken by train with original speed,

Ignoring x = – 50, as speed cannot be negative, we have x = 45.

Hence, the original speed of the train = 45 km/h.

Question5. If Zeba were younger by 5 year than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age, what is her age now?

Solution:

Let Zeba’s present age be = x years

∴ Zeba’s age when she was 5 year younger = (x – 5) years

According to the question,

⟹ (x – 5)2 = 5x + 11

⟹  x2 + 25 – 10x = 5x + 11

⟹  x2 – 15x + 14 = 0

⟹  x2 – 14xx + 14 = 0 [By splitting middle term]

⟹  x (x – 14) –1 (x – 14) = 0

⟹  (x – 1) (x – 14) = 0

⟹  (x – 1) = 0 or (x – 14) = 0

⟹  x = 14 or 1

Ignore x = 1 because then her age when she was 5 year younger would be, x – 5 = – 4, which is unacceptable.

Hence, Zeba's present age is 14 year.

Question6. At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.

Solution:

Let, Nisha’s present age be = x year

Therefore, according to the first condition, Asha's present age = x2 + 2

Nisha grows to her mother's present age after [(x2 + 2) – x] years.

Then, Asha's age will become (x2 + 2) +  [(x2 + 2) –x] year.

According to the question,

    (x2 + 2) + [(x2 + 2) – x] = 10x – 1

⟹  2x2x + 4 = 10x – 1

⟹  2x2 – 11x + 5 = 0

⟹  2x2 – 10xx + 5= 0

⟹  2x (x – 5) –1(x – 5) = 0

⟹  (x – 5) (2x – 1) = 0

⟹  (x – 5) = 0 or (2x – 1) = 0

⟹  x = 5 or 1/2

Hence, present age of Nisha = 5year

And present age of Asha = x2 + 2 = (5)2 + 2 = 25 + 2 = 27 year.

Question7. In the centre of a rectangular lawn of dimensions 50 m × 40 m, a rectangular pond has to be constructed, so that the area of the grass surrounding the pond would be 1184 m2 [see figure]. Find the length and breadth of the pond.

 

Solution:

Given,  length of the rectangular lawn = L = 50 m

And breadth of the rectangular lawn = B = 40 m

Suppose, the width of area of the grass surrounding the pond be x meter.

 

Then, length of rectangular pond = l = 50 – (x + x) = 50 – 2x

And breadth of rectangular pond = b = 40 – (x + x) = 40 – 2x

Now, Area of grass surrounding the pond = L × B – l × b = 1184 m2

⟹ 50 × 40 – (50 – 2x) (40 – 2x) = 1184

⟹ 2000 – (2000 – 80x – 100x + 4x2) = 1184

⟹ 80x + 100x – 4x2 = 1184  

x2 – 45x + 296 = 0

x2 – 37x – 8x + 296 = 0  [By splitting middle term]

x (x – 37) – 8 (x – 37) = 0

⟹ (x – 8) (x – 37) = 0

x = 8 or 37

If x = 37

Then length of pond =50 – 2x =  –24

And breadth of pond = 40 – 2x = –34

But length and breadth cannot be negative, so ignore x = 37.

Hence, with x = 8

Length of pond = 50 – 2x = 50 – 16 = 34 m

And breadth of pond = 40 – 2x = 40 – 16 = 24 m.

Question8. At t minutes past 2 pm, the time needed by the minute hand of a clock to show 3 pm was found to be 3 min less than t2/4 min. Find t.

Solution:

Total time taken by minute hand to run from 2 pm to 3 pm = 60 minutes.

According to question,

As time can not have negative value,

Therefore, required value of t is14 minutes..

You may also like to read:

CBSE Class 10 Mathematics Syllabus 2017-2018

CBSE Class 10 NCERT Textbooks & NCERT Solutions

NCERT Solutions for CBSE Class 10 Maths

NCERT Exemplar Problems and Solutions Class 10 Science: All Chapters

Related Stories