Question

Chakradhar and Dhruva have two spools of 'manjha' to fly their kites. Each of the
manjhas has tiny knots at regular intervals, which helps in keeping track of the length
of the manjha that is used. Each of the persons has the same length of manjha. While
Chakradhar's manjha has knots at intervals of 10 feet, Dhruva's has knots at intervals
of 12 feet and it has 10 knots less than that of Chakradhar. Each of the manjhas has
knots at only one of the extreme ends. Then the length of the manjha with either of
those (in feet) is:
A) 610
B) 590
C) 600
D) 1190​

Answer 1

Answer:

e length of manjha. While

Chakradhar's manjha has knots at intervals of 10 feet, Dhruva's has knots at intervals

of 12 feet and it has 10 knots less than that of Chakradhar. Each of the manjhas has

knots at only one of the extreme ends. Then the length of the manjha with either of

those (in feet) is:

A) 610

B) 590

C) 600

D) 1190

Step-by-step explanation:

e length of manjha. While

Chakradhar's manjha has knots at intervals of 10 feet, Dhruva's has knots at intervals

of 12 feet and it has 10 knots less than that of Chakradhar. Each of the manjhas has

knots at only one of the extreme ends. Then the length of the manjha with either of

those (in feet) is:

A) 610

B) 590

C) 600

D) 1190

Answer 2

Given:

Each of the manjhas has tiny knots at regular intervals, which helps in keeping track of the length of the manjha that is used. Each of the persons has the same length of manjha. While Chakradhar's manjha has knots at intervals of 10 feet, Dhruva's has knots at intervals of 12 feet and it has 10 knots less than that of Chakradhar.

To find:

Length of the manjha

Solution:

Let x be the number of knots in Chakradhar's Spool

The number of notes in Dhruva's spool = x- 10 [ As given in the question]

Total length of manjha of Chakradhar = Notes at distance * Total number of knots = 10 * x = 10x

Total length of manjha of Dhruva = Notes at distance * Total number of knots = 12 * x-10 = 12(x-10)

Now, it is given that length of both manjhas are equal

So,

10x = 12(x-10)

10x = 12x - 120

12x-10x = 120

2x = 120

x = 120/2 = 60

Total knots in Chakradhar's Spool = 60

Length of manjha = 10x = 10*60 = 600 feet

Hence, answer is C. 600

#SPJ2