A box contains some red and some yellow bal

Question

GMATBusters’ Quant Quiz Question -7 
A box contains some red and some yellow balls. If one red ball is removed, one seventh of the remaining balls would be red; if one yellow ball is removed, one-sixth of the remaining balls would be red. If n denotes the total number of balls in the box, then the sum of the digits of n is 
(a) 6 
(b) 7 
(c) 8 
(d) 9
(e) 10

Raxit85 · Answer

Let total red & yellow ball be x & y
total balls = x + y.
 one red ball is removed ⇒ red ball becomes = x – 1
 Total remaining balls. = x + y – 1
 As per statement,
1/7 (x + y – 1) = x – 1
= x + y – 1 = 7 x – 7
 =6x – y – 7 + 1 = 0
 = 6x – y – 6 = 0

1/6 (x + y –1) = x
 = x + y – 1 = 6x
 = y – 1 = 5x
 = y – 5x = 1
 = y = 6x – 6
 = 6x – 6 – 5x = 1
So, x = 7
 6 × 7 – y – 6 = 0
 = 42 – y – 6 = 0
 So, y = 36
 x + y = 7 + 36 = 43
Sum of digits = 4 + 3 = 7
Ans. B

ashwini2k6jha · Answer

let r be the nos of red balls and y be the numbers of y balls. 
Hence n=r+y 
given , r+y-1/7=r-1
solving 6r-y=6--------1
also given, r+y-1/6=r
solving y-5r=1 ----------------2
1+2 Add
we get r=7
from 1, y=36
Hence n=r+y=43
Sum of digit=7 Answer B

techwave · Answer

Answer: B

Red balls = X

Yellow balls = Y

X= {(n–1)/7} +1
X= (n–1)/6
{(n–1)/7} +1=(n–1)/6
6n+36=7n–7 
n=43
Sum of digits = 7

Kinshook · Answer

Given: 
1. A box contains some red and some yellow balls. 
2. If one red ball is removed, one seventh of the remaining balls would be red;
3. if one yellow ball is removed, one-sixth of the remaining balls would be red.

Asked: If n denotes the total number of balls in the box, then the sum of the digits of n is 
(a) 6 
(b) 7 
(c) 8 
(d) 9
(e) 10

1. A box contains some red and some yellow balls. 
Let the number of red balls and yellow balls be r & y respectively

2. If one red ball is removed, one seventh of the remaining balls would be red;
(r-1)/(r+y-1) = 1/7
7r - 7 = r + y - 1
6r -y = 6 (1)

3. if one yellow ball is removed, one-sixth of the remaining balls would be red. 
r/(r+y-1) = 1/6
6r = r + y - 1
5r - y + 1 = 0 (2)

y = 6r - 6 = 5r + 1
r = 7
y = 36

If n denotes the total number of balls in the box, then the sum of the digits of n is 
n = r + y = 7 + 36 = 43
Sum of digits of n = 43 = 4+3= 7

IMO B

firas92 · Answer

Number of yellow balls  = y 
Number of red balls  = r

\frac{r-1}{r+y-1} = \frac{1}{7} 
 7(r-1) = r+y-1

Also,
 \frac{r}{r+y-1} = \frac{1}{6} 
 6r = r+y-1

So we have,  7(r-1)=6r  or  r=7

Put  r=7  in any of the above equatios, we get  y=36

So total number of balls  = 7+36 = 43

Sum of digits of  43 = 4+3 = 7

Answer is  (B)

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prathameshrathi · Answer

The answer is B) 7
Explanation:
n is total no. of balls.
1 is removed from it means n-1.
n-1 is 1/7th of ..... meaning n-1/7 is the whole number
similarly, n-1 is 1/6th of ..... meaning n-1/6 is a whole number
so n-1 should be HCF of 7 and 6 i.e. 4
n-1=42
n=43
4+3=7

Archit3110 · Answer

let red balls be ; x and yellow balls be y
so 
 If one red ball is removed, one seventh of the remaining balls would be red; 
we have (x+y-1) = 1/7 * x
simplify we get 6x-y=6--(1)
and if one yellow ball is removed, one-sixth of the remaining balls would be red
(x+y-1) * 1/6= x
simplify we get 
y-1 = 5x--(2)
put y = 5x+1 in ( 1)
we get x=7 and y = 36
x+y= 43 and sum of 4+3 ; 7
IMO B

A box contains some red and some yellow balls. If one red ball is removed, one seventh of the remaining balls would be red; if one yellow ball is removed, one-sixth of the remaining balls would be red. If n denotes the total number of balls in the box, then the sum of the digits of n is
(a) 6
(b) 7
(c) 8
(d) 9
(e) 10

shameekv1989 · Answer

A box contains some red and some yellow balls. If one red ball is removed, one seventh of the remaining balls would be red; if one yellow ball is removed, one-sixth of the remaining balls would be red. If n denotes the total number of balls in the box, then the sum of the digits of n is
(a) 6
(b) 7
(c) 8
(d) 9
(e) 10

Let R be the total number of red balls

i) When 1 red is taken out =>  \frac{n-1}{7}+1 = R

ii) When 1 yellow is taken out =>  \frac{n-1}{6} = R

\frac{n-1}{7}+1 = \frac{n-1}{6}

\frac{n+6}{7} = \frac{n-1}{6}

6n+36 = 7n-7 => n = 43

Sum of digits of n = 4+3 = 7

Answer - B

bM22 · Answer

Let the number of red balls be x and no. of Yellow balls be y.

So x+y = n

Now as per given question if 1 red ball is removed, 1/7th of remaining balls are red.
=> 1/7(n-1) = x-1 ------ eq1
=> n-1 = 7x-7

& as per the second part of the question, if one yellow ball is removed then 1/6th of remaining balls are red.
=> 1/6(n-1)= x ------------- eq2
=> n-1 = 6x

comparing the 2 equations we see that 6x = 7x-7 => x = 7.
Substituting the value of x in eq 2, we get n-1 = 42, => n= 43.

therefore sum of the digits of n = 7.

Answer B

AnirudhaS · Answer

\frac{1}{7} {(r-1)+y}=r-1
or, 6r-y=6.......(i)

\frac{1}{6} {(y-1)+r}=r
or, 5r-y=-1............(ii)

Solving (i) and (ii)
r=7 and y = 36
therefore, n=43.
Sum of digits = 4+3=7
Answer: B

edlc313 · Answer

My method was to divide it into two formulas and use substitution. I'm sure that there is probably a more clever way to do it.

(r - 1) / (r - 1 + y) = 1 / 7

and

(y - 1) / (y - 1 + r) = 5 / 6

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A box contains some red and some yellow bal

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