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GMAT Club World Cup 2026 (Day 4): A textile workshop must produce

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GreenPill

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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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option D. instead of solving quadratic equation we can use options. substitute i from options and s = 1. w = 14(7s+5i) ---- 1. w/6i = w/21s - 14 ---- 2 subs 1 in 2 then use options

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09 Jul 2026, 10:29

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My answer is D.) 7

Equations based on question info-

Let rates be s and i
1) [work][/7s+5i]= 14
2) [work][/21s] - [work][/6i]= 14

Equating both equations, we get
[1][/7s+5i] = [1][/21s] - [1][/6i]

Now, I substituted the answers because of the complex equation. Only 7 satisfies the above equation. (substituting [i][/s] = 7 i.e. i=7s). Therefore, answer is D.

Bunuel

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work per hour = s,i
7s+5i = W/14
now, time taken in hrs for s and i respectively = x,y
y+14 = x
6i = 1/y; 21s = 1/x
very very lengthy solving, putting i/s = a,
quadratic equation: 10a^2-63a-49 = 0. putting options and checking unit digit gives only 7.

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Bunuel

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Industrial Loom = I, Standard Loom = S
We know that together 7S and 5I complete work in => 14 hours

Let 21S complete work in => x hours
Hence 1S will complete work in => 21x hours

Also, 6I will complete work in => (x-14) hours
Hence 1I completes work in => 6(x-14 hours)

We have to compare the rates of I and S
Rate = Qty / hours
Hence Rate of S = 1/21x
and Rate of I = 1/ 6(x-14)

taking I/S, we get
Ratio = 21x / 6(x-14)
Simplifies to 7x/2x-28

Here i substitued each of the values to see where x was coming to be a real number

A. 7x = 3(2x-28)
7x = 6x -84, X is negative hence Exclude A

similarly, was getting fractional values for options B,C and E hence exclude them
for D, 7x = 7(2x-28)
x = 2x-28
hence x = 28 hours
As we are getting a whole value of X, Answer is D: 7 hours

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09 Jul 2026, 11:37

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There will be 2 equations for this problem. Let's assume Production rate (Hourly) for Standard loom = X & Inustrial Loom =Y ; Total Production ( From Job) = T, Ratio = R = Y/X
First equation will be: (7X+5Y)*14= T
Second equation will be : (T/6Y)+14 = (T/21X)
Solving above equation by replaing T in equation 2 by 1, we will get , Y/X= 7.

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09 Jul 2026, 11:45

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Assume work rate of standard loom= s
work rate of industrial loom =i
7 standard loom+ 5 industrial loom job in 14 hours
so 14/7s+14/5i=1--- equation (1)

6 industrial looms takes 14 hours less than 21 standard looms

(1/6i)=(1/21s)-14 equation(2)

solve these two equations to get 7 as answer

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09 Jul 2026, 12:03

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S: rate of stnd loom
I : rate of industrial loom

work=rate*time
W = 14(7S+5I)
W/6I = W/21S - 14
14 = W/21S - W/6I
14 = 14(7S+5I)/21S - 14(7S+5I)/6I
1 = (7S+5I)/21S - (7S+5I)/6I

Question is asking for I/S=?

Let s=1, then 7(1)+5I is a multiple of 21 since the result of this fraction has to be an integer.

A) I = 3, 7+5(3) = 22 not multiple of 21
B) I = 5, 7+5(5) = 32 not multiple of 21
C) I=6, 7+5(6) = 37 not multiple of 21
D) I=7, 7+5(7) = 42 multiple of 21
E) I=9, 7+5(9) = 52 not multiple of 21

Answer D

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09 Jul 2026, 12:05

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Bunuel

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Standard Loom = s units / hour
Industrial Loom = i units / hours

Total Work = 14(7s + 5i)

To find : i/s

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

Cancelling 14

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

Take LCM

42si + 30i^2 - 147s^2- 105si = 126si

30i^2 - 147 s^2 = 189si

30(i/s)^2 - 147 = 189(i/s)

Let i/s = c

10c^2 - 63c - 49 = 0

10c(c-7)+7(c-7) = 0

c = 7 or c = -7/10

Can't be negative, so c = i/s = 7

Option D

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I used the ratios formula directly and reached to 10r^2-63r-49=0. Try the answers, i tried 7 in the 2nd attempt and got the answer as 7:1. Which is D as our option.

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Let the rates for standard loom be x units per hour and that for the industrial loom be y units per hour

The equation would be (7x+ 5y)*14 = total work

Testing 7 ---> i.e x = 1 and y = 7

Total work done is 588 units

Time taken by 6 industial units = 588/ 6 * 7 = 14 hours

Time taken by 21 standard looms = 588/21 = 28 hours

Since this is 14 hours more it exactly matches

Answer D

Bunuel

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Let the time taken by one Standard Loom to complete the job be T and by one Industrial Loom be t.
Also, let their rates be s and i respectively.
Thus, s = 1/T and i = 1/t

What we have to find is i/s = (1/t)/(1/T) = T/t = r (let it be denoted by r for ease of calculation)
Thus, i = sr and T = tr______ (1)

Now what we are given is that 7 standard looms and 5 industrial looms combined take 14 hours.
Since we can add rates,
7s + 5i = 1/14
From (1),
7s + 5sr = 1/14
s*(7+5r) = 1/14 _____ (2)

Since 1 standard loom takes T hours to complete the job, 21 standard looms woll take T/21 hours.
Similarly, 6 industrial looms will take t/6 hours.

We are given that,
T/21 - t/6 = 14
i.e. (1/i)*(1/21) - (1/s)*(1/6) = 14
1/21i - 1/6s = 14

Again from (1), we have
1/21sr - 1/6s = 14
(1/s)(1/21r - 1/6) = 14________ (3)

From (2) we have,
1/s = 14*(7+5r)

Hence, (3) can be expressed as,
14*(7+5r) (1/21r - 1/6) = 14

Simplifying this, we finally get a quadratic equation,
10r^2 - 63r - 49 = 0

At this point, I could not factorize it and hence I just started plugging in values from answer choices. 7 satisfied the eqn.

Hence, answer is (D)

This is a very lengthy method and took me almost 5 minutes time to get here honestly. Is there a faster method to do this?

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For the first condition, I am using the algebraic formula where time multiplied by the combined work rate equals 1 completed job. This is expressed as: $(7x+5y)14=1$. For the second condition, I am using the same formula but rearranged for time, which gives $\frac{1}{21x}-\frac{1}{6y}=14$. This means the difference in time to complete 1 job between two groups is 14 hours (standard: $x$ and industrial: $y$). Next, I used the second equation and substituted the "1" with the the expression from the first formula, so it looks like this $\frac{(7x+5y)14}{21x}-\frac{(7x+5y)14}{6y}=14$. To simplify, I divided the entire equation by 14, so it looks like this:$ \frac{(7x+5y)}{21x}-\frac{(7x+5y)}{6y}=1$. But I need to find the ratio, so solving for x and y is no longer useful. Therefore, I defined the ratio as $r=\frac{y}{x}$. To substitute this value, I rearranged I to $y=rx$. After replacing the values we got the equation $\frac{(7+5r)}{21}-\frac{(7+5r)}{6r}=1$. By testing the options, letter D yields $1=1$, giving us the correct answer.

Bunuel

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09 Jul 2026, 13:11

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Let the Standard Loom's solo time to complete the task be S and for Industrial be I.
If the total work is W:
Working together combined rate is : 7S+5I.

W=14(7S+5I)....(i)

Also given:
(W/6I) = (W/21S) - 14.. (ii)

I replaced W in (ii) and got a quadratic equation of I/S

10(I/S)^2 - 63 ( I/S) - 49 = 0.
Now using the options if I use I/S = 7 I get the answer = 0 for the LHS.

Ans is D IMO.

Bunuel

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W = 14(7+5r)
difference equation: W/21 - W/6r = 14
Sub in W and divide everything by 14:

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

DONT TRY SOLVE ALGEBRAICALLY, THIS IS A TIME SINK.

INSTEAD:

Reason which is what the gmat is about: industrial looks are much faster, that points towards a large ratio, so starting from big end (7 or 9) is smart.

7 would land it.

r=7 job W = 14(7+35) = 588 units.

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09 Jul 2026, 15:56

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This question was absolutely brutal for me, I got very lost in the algebra so I decided to just test answer choices.

Let s be the rate for the standard looms and i be the rate for the industrial looms. We are looking for i / s.

Start with b. If i / s = 5 , then i = 5 and s = 1.

From first equation 14(7+25)= Total job. 32*14= 384.

Now use second equation to check the difference between 6 industrial and 21 standard, which should be 14. Remember that time = W / rate,

So 384/21 - 384/30 = 14? Can't be.

Move on to check D.

If i / s = 7, then i = 7 and s = 1.

14(7+5(7)) = 42*14=420+160+8=588.

588/21 - 588/42 = 14?

Yes. Therefore, we found the answer - D. If anyone has a faster more intuitive way to solve please help because this took such a long time. Thanks.

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09 Jul 2026, 16:31

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Standard looms=x and Industrial looms= y
Question we need to find is y/x
So its given, 7x + 5y = total work /14 so total work= 14(7x + 5y)
Also given total work/6y = total work/21x - 14
=> TW/21x - TW/6y = 14 Now substitute value of total work from above
So 14(7x+5y)/21x - 14(7x+5y)/6y = 14
=>7x/21x + 5y/21x - 7x/6y - 5y/6y = 1
=> 1/3 +(5/21)(y/x) -(7/6)(x/y) - 5/6 = 1
=> put y/x = a, now we need to find a.
(5/21)a - (7/6)/a = 1-1/3+5/6= 3/2
x by 42
10a - 49/a = 3 x 21
so 10a^2 - 63a - 49=0
Solving quadratic we will get (10a + 7) (a-7) =0
lets take a-7 = 0 so a=7
Thats our answer. D

Bunuel

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09 Jul 2026, 17:13

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work rate for standard loom be S and industrial loom be I, let ratio of I to S be R.
(7S+5I)14= W (Work done)
(7+5R)14=W/S (R=I/S)

also we have, 1/6I=1/21S - 14
W/6I - W/21S =-14

substituting for I/S
W/6RS -W/21S= -14
W/S (1/6R - 1/21)= -14

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

10R^2 -63R -49=0

R=7 = I/S

Bunuel

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09 Jul 2026, 17:39

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Let speed of standard loom- s and industrial loom is i
(7s+5i)14=w
Let time taken by 6 industrial loom to finish the work be t and time taken by 21 standard loom is t+14
w=6*i*t=21*s*(t+14)
using the information we have
(7s+5i)*14=6it
t=((7s+5i)/6i)*14
Now substituting the value of t in the second equation
(7s+5i)*14=21s(14+((7s+5i)/6i)*14)
on simplifying this
7/6*(s/i)+11/6=1/3+5/21*(i/s)

Let i/s=x= answer we are looking for
3/2=5x/21-7/6x: cross multiply to form an equation
10x^2-49=63x
x=7,-7
Hence answer is D

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09 Jul 2026, 18:22

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I feel that the hardest part of this question is managing the algebra, at first it seems that I am in the wrong path

There are 2 looms,

1. Standard -> assume it can complete 1 task in t_s with rate s -> s = 1/t_s
2. Industrial -> complete 1 task in t_i with rate i -> i = 1/t_i

We're given 2 relationships
1. 7 standard looms working with 5 industrial looms took 14 hours to complete 1 task.

This suggests:
7*(1/t_s) + 5*(1/t_i) = 1/14
or
7s + 5i = 1/14

2. 6 industrial looms working ALONE can complete 1 task faster than 21 standard looms for 14 hours.

This suggests:
t_i/6 = t_s/21 - 14
or
1/6i = 1/21s -14

Now, solving for i/s

From the first equation: 14 = 1/(7s+5i)
Substitute in the second equation: 1/6i = 1/21s - 1/(7s+5i)

Manipulating to get the quadratic equation:
1/6i = (7s+5i-21s)/(21s*(7s+5i))
21s*7s+21s*5i = 6i*5i - 6i*14s

Divide them by 3:
7s*7s + 7s*5i = 2i*5i - 2i*14s
49s^2 + 35si = 10i^2 -28si
49s^2 + 63si -10i^2 = 0

Solving the quadratic equation:
(7s + 10i)(7s - i) = 0

There are 7s = -10i and 7s = i
or i/s = -7/10 or 7.

Obviously, the answer is 7 or choice D.

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09 Jul 2026, 20:22

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s be rate of production of standard loom & i for industrial loom.

7 standard looms and 5 industrial looms working together can complete the job in 14 hours.
So, Total work done, W = (7s + 5i)×14

6 industrial looms working alone take 14 hours less than 21 standard looms working alone.
So, (W/6i) = (W/21s) - 14
Or, (14(7s+5i)/6i) = (14(7s+5i)/21s) - 14
Or, (7s+5i)/6i = (7s+5i)/21s - 1

Say, i/s = r
Then (7+5r)/6r = (7+5r)/21 - 1
Or, 7(7+5r) = 2r(7+5r) - 42r
Or, 10r^2 - 63r - 49 = 0
Or, (r-7)(10r+7)=0, r>0
Hence, r=7 ...Ans...