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The weights of the two wrestlers are in ratio 7:8. 25 Apr 2020, 19:59
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E
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GMATBusters’ Quant Quiz Question -9

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The weights of the two wrestlers are in ratio 7:8. if each of the wrestlers puts on fighting accessories weighing x pounds, what could be ratio of their new weights?
A. 11/15
B. 12/16
C. 13/17
D. 11/14
E. 12/13
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E
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The weights of the two wrestlers are in ratio 7:8. 25 Apr 2020, 20:06
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Answer E
w1/w2=7/8=7k/8k
new weight ratio-w1+x/w2+x=7k+x/8k+x=7*1+5/8*1+5=12/13
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The weights of the two wrestlers are in ratio 7:8. 25 Apr 2020, 20:12
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The weights of the two wrestlers are in ratio 7:8. if each of the wrestlers puts on fighting accessories weighing x pounds, what could be ratio of their new weights?

Total weights of two wrestlers is 7a:8a
let us assume 49:56.
49+x/56+x
49+7/56+7
56/63
8/9
i'd guess here as E 12/13.

Ans. E

A. 11/15
B. 12/16
C. 13/17
D. 11/14
E. 12/13
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The weights of the two wrestlers are in ratio 7:8. 25 Apr 2020, 20:13
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The weights of the two wrestlers are in ratio 7:8. if each of the wrestlers puts on fighting accessories weighing x pounds, what could be ratio of their new weights?

Ans: E

W1/W2 =7/8

with an additional weight. Difference between denominator - Numerator should exclude additional fighting accessories. but 8*y -7*y

For y =1 the difference is 1 and only E satisfies it.
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The weights of the two wrestlers are in ratio 7:8. 25 Apr 2020, 20:14
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The weights of the two wrestlers are in ratio 7:8. if each of the wrestlers puts on fighting accessories weighing x pounds, what could be ratio of their new weights?
A. 11/15
B. 12/16
C. 13/17
D. 11/14
E. 12/13

current ratio is 7/8
so with addition of weight we get new fraction
solve by using answer options
say option c 13/17
13-x/17-x= 7/8
we get -ve value which is wrong
now with option e
12-x/13-x = 7/8
we get value of x
96-8x=91-7x
x= 5 ( +ve)
which is sufficient additional weight is 5 kgs
IMO E
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The weights of the two wrestlers are in ratio 7:8. 25 Apr 2020, 21:01
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Answer is E


The weights of the two wrestlers are in ratio 7:8. if each of the wrestlers puts on fighting accessories weighing x pounds, what could be ratio of their new weights?
A. 11/15
B. 12/16
C. 13/17
D. 11/14
E. 12/13

Only option E allows adding "weigh" without losing the proportion:
The original 7/8 will become 7+5 / 8+5 = 12/13. The rest of the options do not allow this.
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The weights of the two wrestlers are in ratio 7:8. Updated on: 26 Apr 2020, 07:02
Originally posted by Kinshook on 25 Apr 2020, 21:11.
Last edited by Kinshook on 26 Apr 2020, 07:02, edited 4 times in total.
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Given: The weights of the two wrestlers are in ratio 7:8.

Asked: If each of the wrestlers puts on fighting accessories weighing x pounds, what could be ratio of their new weights?

Let the weight of two wrestlers be 7k & 8k respectively
After each of the wrestlers puts on fighting accessories weighing x pounds, ratio of their new weights = (7k+x)/(8k+x)

A. 11/15
(7k+x)/(8k+x) = 11/15; (7k + x)/k = 11/4; x/k = -17/4; Not feasible
B. 12/16
(7k+x)/(8k+x) = 12/16; (7k + x)/k = 12/4; x/k = -16/4 = -4; Not feasible
C. 13/17
(7k+x)/(8k+x) = 13/17; (7k + x)/k = 13/4; x/k = -15/4 ; Not feasible
D. 11/14
(7k+x)/(8k+x) = 11/14; (7k + x)/k = 11/3; x/k = -10/3 ; Not feasible
E. 12/13
(7k+x)/(8k+x) = 12/13; (7k + x)/k = 12; x/k = 5 ; feasible

IMO E

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The weights of the two wrestlers are in ratio 7:8. 25 Apr 2020, 23:48
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The answer is E.

Let the multiplier be K, then the body weights are 7K and 8K. When we added a constant to both numerator and denominator of a proper fraction, the fractions tends to shifts towards 1. Ex : 1/4 => (1+1)/(1+4) => 2/5 shift from 25% to 40%

All the options are below 87.5% (7/8) value, except E which is approx 91...something, higher than 7/8
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The weights of the two wrestlers are in ratio 7:8. 25 Apr 2020, 23:55
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Weights of two wrestlers (say W1 and W2) are in the ratio 7:8.
Let a be the proportionality constant.
So W1/W2= 7a/8a
If each wrestler puts on x pounds of fighting accessories, the new ratio becomes
W1/W2= 7a+x/8a+x
From here, the simplest way of going about the question is substituting the answer choices.
Since this is a could be true question, any set of values of a and x that work for 7a+x can be used for 8a+x to check.
Lets check for C. If 7a+x=13 then one set of possible values is a=1 and x=6. If we try using the same for 8a+x, we get 8(1)+6=14. However C says 13/17. So C is out.
Similarly if we check E, 7a+x=12. a can be 1 and x=5. Substituting in 8a+x, we get 8(1)+5= 13. So we get 12/13 which is a possible value.
Answer: E
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The weights of the two wrestlers are in ratio 7:8. 26 Apr 2020, 01:53
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Quote:
The weights of the two wrestlers are in ratio 7:8. if each of the wrestlers puts on fighting accessories weighing x pounds, what could be ratio of their new weights?
A. 11/15
B. 12/16
C. 13/17
D. 11/14
E. 12/13

Wrestlers weight = 7:8

With accessories, Wrestlers weight ratio = (7+x)/(8+x)

CONCEPT: for positive values of a, b and x
If a/b > 1 then (a+x)/(b+x) < a/b
e.g. 3/2 > 1 then (3+1)/(2+1) = 4/3 < 3/2

If a/b < 1 then (a+x)/(b+x) > a/b
e.g. 2/3 < 1 then (2+1)/(3+1) = 3/4 > 2/3

therefore
Since, 7/8<1, (7+x)/(8+x) MUST be greater than (7/8)

7/8 = 87.5% (12.5*7)

A. 11/15 ≈ 0.7 i.e. less than 0.87 so OUT
B. 12/16 = 0.75 i.e. less than 0.87 so OUT
C. 13/17 ≈ 0.7 i.e. less than 0.87 so OUT
D. 11/14 ≈ 0.78 i.e. less than 0.87 so OUT
E. 12/13 ≈ 0.9 CORRECT
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The weights of the two wrestlers are in ratio 7:8. 26 Apr 2020, 03:14
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Let’s check answer choices:
A) 11/15. —> 7+4/8+4= 11/12 (not true)
B 12/16 —> 7+5/8+5 = 12/13 ( not true)
....
Only E satisfies the ratio. —12/13

Answer (E)

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The weights of the two wrestlers are in ratio 7:8. 26 Apr 2020, 03:57
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The weights of the two wrestlers are in ratio 7:8. if each of the wrestlers puts on fighting accessories weighing x pounds, what could be ratio of their new weights?
A. 11/15
B. 12/16
C. 13/17
D. 11/14
E. 12/13

if constant X is added to proper fraction (<1 ) then the fraction value will increase.

7/8 = .87... so answer should give anything greater than this

Only 12/13 comes to 0.9 sth... you can also check by solving

7+x/8+x = 12/13
x= 5

Hence E
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The weights of the two wrestlers are in ratio 7:8. 26 Apr 2020, 04:18
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E is the answer.

Let their weight be 7kg and 8kg. Let the gear be 5kg, so
(7+5)/(8+5)=12/13

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The weights of the two wrestlers are in ratio 7:8. 26 Apr 2020, 06:51
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The weights of the two wrestlers are in ratio 7:8. if each of the wrestlers puts on fighting accessories weighing x pounds, what could be ratio of their new weights?

The ratio between x and y is 7 to 8 means x to y can be anything like 7/8or 14/16 or ....
Now, if you add 5 to each of numerator and denominator the new ratio of x to y will be 12:13.

Hence answer is E
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The weights of the two wrestlers are in ratio 7:8. 26 Apr 2020, 09:29
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7a+x=11,12,13,11,12 means only one possible value for a i.e 1 ( Now x can take=4,5,6,4,5) so we need to check just for a=1 and x=4,5,6 in second equation
8a+x=15,16,17,14,13-----------second equation
a=1, x=5 satisfies so ans is 12/13
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The weights of the two wrestlers are in ratio 7:8. 26 Mar 2022, 22:54
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