Ncert – Class 10 – QUADRATIC EQUATIONS- Hots Question

Question 1

A piece of cloth costs Rs200. If the piece was 5m longer and each metre of cloth costs Rs2
less the cost of the piece would have remained unchanged. How long is the piece and what is
the originalrate permetre?

Solution

Letthe length of piece = x m Rate per metre =200/x New length = ( x + 5) New rate
per metre =200/(x+5). ATQ 200/x -200/(x+5) = 2
x² + 5x − 500 = 0
x = −25, x = 20
Rate per metre =10

Question 2

Some student planned a picnic. The budget for food was Rs. 500. But 5 of these failed to go
and then cost of food for each member isincreased by Rs 5. How many students attended the
picnic?

Solution

Let no. of student= x Cost per head= 500/x When 5 failed to go
New cost per head=500/(x-5) ATQ 500/(x-5) – 500/x =5 x
2 − 5x − 500 = 0
( x − 25)( x + 20) = 0 x = 25, x = −20
No of student attended the picnic
= 25 − 5= 20

Question 3

The sum of reciprocals of Rehman’s ages (in years)3 years ago and 5 years from now is13. Find his
present age

Solution

x=7 or –3; but age cannot be negative.
Hence the present age of Rehman is 7 years

Question 4

A cottage industry produces certain number of pottery articles in a day. It was
observed on a particular day that the cost of production of each article ( in Rs.)
was found to be 3 more than twice the number of articles produced on that
day.The total cost of production on that day was
Rs. 90.
i)Frame a quadratic equation to find the number of articles.
ii)Find the number of article produced.
iii)Find the cost of each article.

Solution

i)Let the number of articles produced in a day be x.
Total cost of production that day = no. of article x cost of each article
= x(2x +3),
Given that total cost of production is Rs.90
ATQ, 2x² +3x -90 = 0
ii) No of article produced =(2x + 15 )( x – 6 ) =0 => 2x +15 = 0 or x – 6 = 0
 X = -15/2 ( impossible) or x = 6 , x = 6 (no. of articles)
Iii) Cost of each article =2x + 3 =2 x 6 +3 = Rs.15

Question 5

The longest side of a right triangle is 5 more than 4 times the shortest side. The
third side is 10 less than the sum of the other two. Let the length of the shortest
side be s cms. Find the three sides of triangle.

Solution

S, 4s+5,s+4s+5-10; using PGT (4s+5)² = s² + (5s-10)² ; s(s-9)=0; s=9 cm
,hypo=41 cm, other leg=40 cm

Question 6

Pabitra can row his boat at a speed 5 km/h in still water. If it takes him 1 hour more to row
the boat 5.25 km upstream than to return downstream, find the speed of the stream?

Solution

Stream x km/hr
Boat 5 km/hr
Upstream (5-x) km/hr
Downstream (5+x) km/hr

(2x+25)(x-2)=0
Or, x= 2 , −25/2
Hence the speed of the stream is 2 km/hr.

Question 7

A boy is going to make an open-top box by cutting equal squares from the four corners of
an20 𝑐𝑚 by 30 𝑐𝑚 sheet of cardboard and folding up the sides. If the area of the base of box
is to be 336 𝑐𝑚2 then what size square be cut from each corner?

Solution

Length = (30 − 2𝑥)
Breadth = (20 − 2𝑥)
(30 − 2𝑥)(20 − 2𝑥) = 336
𝑥² − 25𝑥 + 66 = 0
(𝑥 − 3)(𝑥 − 22) = 0
𝑥 = 3 𝑎𝑛𝑑 𝑥 = 22 𝑖𝑠 𝑛𝑜𝑡 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑖𝑡 𝑤𝑖𝑙𝑙 𝑔𝑖𝑣𝑒 𝑛𝑒𝑎𝑔𝑡𝑖𝑣𝑒 𝑣𝑎𝑙𝑢𝑒
Length = 24
Breadth = 14

Question 8

A flower bed is rectangular in shape with a length of 14 𝑚𝑒𝑡𝑒𝑟 and a width of 4 𝑚𝑒𝑡𝑒𝑟. He
wants to increase the length and width by the same amount to obtain a flower bed with an
area of 200 𝑠𝑞 𝑚𝑒𝑡𝑒𝑟. What should be the amount of increase be?

Solution

𝑙𝑒𝑛𝑔𝑡ℎ = (14 + 𝑥)
Breadth = (𝑥 + 4)
Area = 200 𝑚2
(𝑥 + 14)(𝑥 + 4) = 200
𝑥² + 18𝑥 − 144 = 0
𝑥 = 6
Length = 20 , breadth = 10

Question 9

The multiplication of the two consecutive odd numbers is 6723, then the square root the
smaller number is

Solution

Let the two consecutive odd number be x+1 and x+3
According to question
(x+1)(x+3)=6723

X²+3x+x+3=6723

X²+4x+3-6723=0

X²+4x-6720=0

X²a+84x -80x -6720
X(x+84)-80(x+84)=0
(x-80)(x+84)=0
X=80
Hence the greater number is 80+3=83

Question 10

In a cricket match Kapil took one wicket less than twice the number of wickets taken byRavi.
If the product of the no. of wickets taken by these two is 15, find the no. of wickets taken by
each.

Solution

Let no. of wicket taken by Ravi = x No. ofwickettaken by Kapil = 2x −1 ATQ (2x−1).x =
15