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I set s to be the rate of a standard loom and m to be the rate of an industrial loom

(7s + 5m) = 1/14 because the can complete the job in 14 hours.

to compare individual times, let T = time for the 6 industrial looms to finish the job

6mT = 1 -> T = 1/6m

Since the 21 standard looms take 14 more hours than 6 industrial looms,

21s(T+14) = 1, sub T from the earlier equation

21s (1/6m +14) = 1 -> 21s/6m + 294s = 1 -> 7s/2m +294s =1


We can refer back to our first equation 7s + 5m = 1/14, solve for 7s and s, then sub into 7s/2m +294s =1. So

(1/14-5m)/2m +294(((1/14)-5m)/7) = 1

Simplify

1/28m - 210m = .5 multiply both sides by 28m to get rid of the fraction

1 - 5880m^2 =14m - > 5880m^2 +14m -1 =0.

Use the quadratic equation and get 1/84 as the only positive rate ( the other root -1/70 makes no sense because we are producing not taking away). so m = 1/84.

Plug into original equation to get s

7s + 5(1/84) = 1/14, s = 1/588.

The ratio of m/s = (1/84)/(1/588) which equals 7
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Bunuel
A textile workshop must produce a certain number of identical fabric rolls. The workshop has standard looms and industrial looms. Each standard loom produces fabric at the same constant rate, and each industrial loom produces fabric at the same constant rate. If 7 standard looms and 5 industrial looms, working together, can complete the job in 14 hours, and 6 industrial looms working alone can complete the job in 14 hours less time than 21 standard looms working alone, what is the ratio of the production rate of one industrial loom to the production rate of one standard loom?

A. 3
B. 5
C. 6
D. 7
E. 9


 


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Let Total Job = T
production rate of one standard loom - s
production rate of one industrial loom - i
given, 7 standard and 5 industrial can complete in 14 hours
T = 14(7s + 5i)
Also given,
Time for 6 industrial looms = \(\frac {T} {6i}\)
Time for 21 standard looms = \(\frac {T} {21s}\)
given, \(\frac {T} {6i} = \frac {T} {21s} -14\)
so,
\(\frac {14(7s+5i)} {6i} = \frac {14(7s+5i)} {21s} -14\)
\(\frac {7s+5i} {6i} = \frac {7s+5i} {21s} -1\)
\(7s(7s+5i) = 2i(7s+5i) - 42is\)
\(49s^2 + 35is = 14is + 10i^2 - 42 is\)
\(10i^2-63is-49s^2 =0\)
\(10i^2-70is + 7is -49s^2 = 0 \)
\((10i+7s)(i-7s)=0\)
\(\frac {i}{s} = \frac {-7}{10}, \frac{i}{s} = 7\)
production rates cannot be negative, so i/s = 7
D
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If the Work is 'W', production rate of standard loom is 'S' and production rate of Industrial loom is 'L', time taken by S to complete the job 'W' is t,

then, as per question stem,
Combined Rate 7S+5I = W/14

6I(t-14) = 21S(t) = W

i.e. 7S+5I = W/14 = (6I(t-14)) /(t-14) = 6I

7S + 5I = 6I
Then 7S = I
Then I/S = 7

Hence Answer is 7


Bunuel
A textile workshop must produce a certain number of identical fabric rolls. The workshop has standard looms and industrial looms. Each standard loom produces fabric at the same constant rate, and each industrial loom produces fabric at the same constant rate. If 7 standard looms and 5 industrial looms, working together, can complete the job in 14 hours, and 6 industrial looms working alone can complete the job in 14 hours less time than 21 standard looms working alone, what is the ratio of the production rate of one industrial loom to the production rate of one standard loom?

A. 3
B. 5
C. 6
D. 7
E. 9


 


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let std machine rate = x
industrial rate =y

together work done in 14 hours ==> 7x+5y=1/14 ... eq1

1/(21x) - 1/(6y) = 14 ... eq2 :as the difference in time was 14 hours working the individual machines alone

let m=y/x and substitute in eq1
7x +5mx = 1/14 ==> x=1/(14*(7+5m)

now substitue x& y in terms of m in eqn 2 :
10m^2 - 63m - 49 =0
==> (10m+7)(m-7) =0
m has to be positive hence m=7 (Ans.)
Bunuel
A textile workshop must produce a certain number of identical fabric rolls. The workshop has standard looms and industrial looms. Each standard loom produces fabric at the same constant rate, and each industrial loom produces fabric at the same constant rate. If 7 standard looms and 5 industrial looms, working together, can complete the job in 14 hours, and 6 industrial looms working alone can complete the job in 14 hours less time than 21 standard looms working alone, what is the ratio of the production rate of one industrial loom to the production rate of one standard loom?

A. 3
B. 5
C. 6
D. 7
E. 9


 


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S = production rate standard loom
I = production rate industrial loom
W = total work

From the first condition

W = 14(7S + 5I)

From the second condition

W/6I = W/21S - 14

Substituting W = 14(7S + 5I)

14(7S + 5I)/6I = 14(7S + 5I)/21S - 14

Divide by 14

(7S + 5I)/6I = (7S + 5I)/21S - 1

Ratio: r = I/S. That means I=rs, Substituting that and cancelling would give:

(7 + 5r)/6r = (7 + 5r)/21 - 1

(7 + 5r)/6r = (5r - 14)/21

21(7 + 5r) = 6r(5r - 14)

147 + 105r = 30r^2 - 84r

30r^2 - 189r - 147 = 0

10r^2 - 63r - 49 = 0

r = (63 +/- 77)/20

r = 140/20 = 7

Therefore,

I/S = 7
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S- rate for std loom and i - rate for industrial loom. We know W=R*T Therefore (7s+5i)*14 =total work W we also know T=W/R so W/6i=W/21s-14
14= W/21s- W/6i simplifying we get 21*6si= (7s+5i)(6i-21s)
126si=-63si-147s^2+30i^2
30i^2-147s^2-189si=0
10i^2-49s^2-63si=0
10i(i-7s)+7s(i-7s)=0
(10i+7s)(i-7s)=0
i/s=-7/10
or
i/s =7
Ans is 7

Bunuel
A textile workshop must produce a certain number of identical fabric rolls. The workshop has standard looms and industrial looms. Each standard loom produces fabric at the same constant rate, and each industrial loom produces fabric at the same constant rate. If 7 standard looms and 5 industrial looms, working together, can complete the job in 14 hours, and 6 industrial looms working alone can complete the job in 14 hours less time than 21 standard looms working alone, what is the ratio of the production rate of one industrial loom to the production rate of one standard loom?

A. 3
B. 5
C. 6
D. 7
E. 9


 


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Let s be rate of the standard loom & i be teh rate of industrial loom.
W be the total job.

To find: r = i/s

Given,


1. W = 14 (7s + 5i)

2. W/6i = W/21s - 14

Solving both, we get,


14 (7s + 5i)/6i = 14 (7s + 5i)/21s - 14

(7s + 5i)/6i = (7s+5i)/21s - 1

Putting i/s = r,


(7 + 5r)/6r = (7+5r)/21 - 1
10r^2 - 63r - 49 = 0

Solving for positive root,
r = 140/20 = 7

Answer : D (7)
Bunuel
A textile workshop must produce a certain number of identical fabric rolls. The workshop has standard looms and industrial looms. Each standard loom produces fabric at the same constant rate, and each industrial loom produces fabric at the same constant rate. If 7 standard looms and 5 industrial looms, working together, can complete the job in 14 hours, and 6 industrial looms working alone can complete the job in 14 hours less time than 21 standard looms working alone, what is the ratio of the production rate of one industrial loom to the production rate of one standard loom?

A. 3
B. 5
C. 6
D. 7
E. 9


 


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A textile workshop must produce a certain number of identical fabric rolls. The workshop has standard looms and industrial looms. Each standard loom produces fabric at the same constant rate, and each industrial loom produces fabric at the same constant rate. If 7 standard looms and 5 industrial looms, working together, can complete the job in 14 hours, and 6 industrial looms working alone can complete the job in 14 hours less time than 21 standard looms working alone, what is the ratio of the production rate of one industrial loom to the production rate of one standard loom?

A. 3
B. 5
C. 6
D. 7
E. 9


let rate of standard loom be x and that of industrial be y
work = rate * time
given
7x+5y= work/ 14 ;
7x+5y = w/14 ;
w = 14* ( 7x+5y)

and w/21x - w/6y = 14
target find y/x
we now know
(7x+5y) * 14/ 21x - ( 14*(7x+5y) ) / 6y = 14
solve we get
7x + 5y / 21x - 7x + 5y / 6y = 1
solve for ratio y/x = rate
in terms of ratio we get ratio as
7:1

OPTION D ; 7 is correct
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there are two looms. standard and industrial.

say rate of one standard loom= x
rate of one industrial loom= y

find = y/x

assume job= w

7 standard and 5 industrial work together for 14 hrs to finish the job.

so work = (7x+5y) 14

now 21 standard looms work together to finish the job in certain hours.

time = w/21x

now 6 industrial looms work together to finish the job.
time = w/ 6y

but this time is 14 hrs less than 21 standards took.

w/21x - w/6y = 14
14(7x+5y)( 2y - 7x/ 42xy) = 14
(7x+5y)(2y-7x/ 42xy )= 1

solving this we get,
10y^2 - 49x^2 = 63xy
10y^2 - 63xy - 49x^2 = 0

we need to find y/x. divide the eqn by x^2

10y^2 / x^2 - 63y /x - 49= 0

now y/x = p

10p^2- 63p -49 = 0

using quadratic eqn, we can find value of p.
upon solving this , we get p = 7 or -7/10

-7/10 is not possible.

hence p = y/x = 7

choice D
Bunuel
A textile workshop must produce a certain number of identical fabric rolls. The workshop has standard looms and industrial looms. Each standard loom produces fabric at the same constant rate, and each industrial loom produces fabric at the same constant rate. If 7 standard looms and 5 industrial looms, working together, can complete the job in 14 hours, and 6 industrial looms working alone can complete the job in 14 hours less time than 21 standard looms working alone, what is the ratio of the production rate of one industrial loom to the production rate of one standard loom?

A. 3
B. 5
C. 6
D. 7
E. 9


 


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Okay so this one is a little tricky in solving, and those who know the concepts of Work rate and time will come to equations easily and then comes the solving part.

So basically as the question requires rate that is I/S ( I = rate of industrial looms ;S = rate of standard looms).
First lets write down the equations here

7S + 5I = 1/14 (assuming that work is one unit)

1/21S = 1/6I +14
now for eq 1 divide LHS and RHS by S; and for eq 2 multiply LHS and RHS by S
what u will get is

7 +5K = 1/14S
1/21 = 1/6K + 14 S

Here K = I/S ; this is the ratio we need to get.
Now in these equation the problem is the "S" so to eliminate it we can take the ratio of these equations (first we will rearrange it so that S gets cancelled)


1/(7+5K) = 14 S
1/21 -1/6K = 14 S

divide them and "S" will get cancelled. then rearrange it to get
{7+5K / 21 } - {7+5K / 6K} = 1

now for a shortcut just use the value from the options to see which one fits :)
the ans will be 7.
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Let s be the production rate of standard loom in 1 hr and i be the production rate of industrial loom
--> 7s + 5i = 1/14 with 1 is the total work (for easier calculation)
We need to find the ratio i/s. Call it r -> r = i/s
Plug in we have s(7+5r)=1/14 -> s = 1/[14x(7+5r)]

We also know that 1/6i=1/21s - 14. Plug in s value we have:
1/6i = 1/6rs = 14(7+5r)/6r
1/21s = 14(7+5r)/21= 2(7+5r)/3
--> Subtitute we have: 7(7+5r)/3r = 2(7+5r)/3-14. Simplify we have 10r^2-63r-49=0
49 is 7^2 so by guessing r = 7 -> Choose D...
Bunuel
A textile workshop must produce a certain number of identical fabric rolls. The workshop has standard looms and industrial looms. Each standard loom produces fabric at the same constant rate, and each industrial loom produces fabric at the same constant rate. If 7 standard looms and 5 industrial looms, working together, can complete the job in 14 hours, and 6 industrial looms working alone can complete the job in 14 hours less time than 21 standard looms working alone, what is the ratio of the production rate of one industrial loom to the production rate of one standard loom?

A. 3
B. 5
C. 6
D. 7
E. 9


 


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Let the production rates of one standard and one industrial loom be S & I.

Total work : (7S + 5I)*14
Also, Time for 6 industrial looms = Time for 21 standard looms - 14
So, 14(7S + 5I)/(6I) = 14 (7S + 5I)/(21S) - 14

Let I/S = x

Then,
(7+5x)/ (6x) = (7+5x)/21 - 1
-> 21(7+5x) = 6x(5x-14)
-> \(10x^2 - 63x - 49 = 0\)
-> \(10x^2 - 70x + 7x - 49 = 0\)
-> 10x(x-7) + 7(x-7) = 0
-> (10x+7)(x-7) = 0
-> x = 7 or -7/10

Since x is a ratio of production rates, x > 0

Therefore, x=7

Ans : D
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Bunuel
A textile workshop must produce a certain number of identical fabric rolls. The workshop has standard looms and industrial looms. Each standard loom produces fabric at the same constant rate, and each industrial loom produces fabric at the same constant rate. If 7 standard looms and 5 industrial looms, working together, can complete the job in 14 hours, and 6 industrial looms working alone can complete the job in 14 hours less time than 21 standard looms working alone, what is the ratio of the production rate of one industrial loom to the production rate of one standard loom?

A. 3
B. 5
C. 6
D. 7
E. 9


 


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rate of one standard loom be 1
rate of one industrial loom be r.

Since 7 standard looms and 5 industrial looms finish the job in 14 hours, the total work can be written as: 14(7 + 5r)

The time for 21 standard looms is: 14(7 + 5r) / 21 t

The time for 6 industrial looms is: 14(7 + 5r) / 6r

14(7 + 5r) / 21 - 14(7 + 5r) / (6r) = 14

Divide everything by 14:

(7 + 5r)/21 - (7 + 5r)/(6r) = 1

Using option choices let's take r = 7

(7 + 35)/21 - (7 + 35)/42 = 42/21 - 42/42 = 2 - 1

= 1

This matches.

Option C
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After translating this question, I understood it would form some kind of quadratic equation that can be a bit tricky to solve because numbers are not as small. So, I may have to use the answer choices.

Here is the equation I formed:
Time taken for 21 standard looms to complete the work is x, so for 6 industrial looms, this time becomes x-14

This gives us individual rates:

Rate of 1 standard loom= 1/21x ; Rate of 1 industrial loom = 1/6x-84

We fit this into the first equation given to us for 7 standard looms and 5 industrial looms:

7/21x + 5/6x-84 = 1/14- (1)

Now, we have to compute ratio of rates of 1 industrial loom to 1 standard loom: 21x/6x-84 - (2)

If we use options:
A: 3 - The ratio can not be 3, if we put this into x, we will get negative x and time taken can not be negative

B: 5- This gives an integer value of x, which could be true but still becomes complex

C- 6- Again integer value- eliminate for now

D- 7- This gives us a positive integer value of x, if we put into 1, the equation satisfies. I will go with this option

This is a shorter way to do it. We can solve the quadratic equation to get the value of x and then put it into (2) to find the ratio
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The answer is D.7
Setup(job=1, standard rates s, industrial rate i, ratio r=i/s)
7s+5i=1/14
1/(21s)-1/(6i)=14
Substituting gives 10r^2-63r-49=0, which factors to (r-7)(10r+7)=0, so r=7
One industrial loom produces fabric 7 times as fast as one standard loom

Bunuel
A textile workshop must produce a certain number of identical fabric rolls. The workshop has standard looms and industrial looms. Each standard loom produces fabric at the same constant rate, and each industrial loom produces fabric at the same constant rate. If 7 standard looms and 5 industrial looms, working together, can complete the job in 14 hours, and 6 industrial looms working alone can complete the job in 14 hours less time than 21 standard looms working alone, what is the ratio of the production rate of one industrial loom to the production rate of one standard loom?

A. 3
B. 5
C. 6
D. 7
E. 9


 


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Let S = Rate of 1 Standard loom and I = Rate of 1 Industrial loom.

From the 1st condition: 14(7S + 5I) = 1

Let X = I/S (the required ratio)

Then --> S = 1 / (14(7+5X))

From the 2nd condition:
Let T = the time taken by 6 Industrial Looms.
Then, 21 Standard looms take T + 14 hours.

So, T = 1/(6I) --> 1/(6xs)

Substitute into the Standard Loom equation:
21(S) (T+14)=1
21S ((1/(6XS)+14)) = 1
7/2X +294S = 1

Substituting value of S --> 7/(2X) +21/(7+5X)= 1

Simplifying - 10X^2 - 63X - 49 =0
X = 7 or -0.7

Since the productive rate cannot be -ve, x =7

Answer (D) - 7
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Standard Loom Rate - S
Industrial LR - i

Work - 1

What we have

1 = 14(7s +5i)
Also

1/21s - 1/6i = 14

Ratio i/s =r

i = rs

since 1 = 14s(7 +5r)
s = [1][/14 (7 +5r)]

1/21s - 1/6rs = 14

Subsitituing s we get
(7+5r)/21 - (7 +5r)/6r =1

Checking for ans

r=7

we get it right
so Ans is D


Bunuel
A textile workshop must produce a certain number of identical fabric rolls. The workshop has standard looms and industrial looms. Each standard loom produces fabric at the same constant rate, and each industrial loom produces fabric at the same constant rate. If 7 standard looms and 5 industrial looms, working together, can complete the job in 14 hours, and 6 industrial looms working alone can complete the job in 14 hours less time than 21 standard looms working alone, what is the ratio of the production rate of one industrial loom to the production rate of one standard loom?

A. 3
B. 5
C. 6
D. 7
E. 9


 


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