Rahul@gurome wrote:

goyalsau wrote:The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is the cost of 7 chocolates, 11 biscuits and 9 ice creams?
(A) The cost of 5 chocolates, 7 biscuits and 3 ice creams is 217.
(B) The cost of 4 chocolates, 1 biscuit and 3 ice creams is 141.

We need to find whether linear combination of two equations can result in a third equation or not. By linear combination I mean simple arithmetic operations that can done on two equations, like addition/subtraction of two equations, multiplication/division of an equation by a constant.

Please note that this the methodical solution which uses some higher level mathematical concept (namely linear independence). Solving proper GMAT question does not need such concepts.

Given: Say cost of one chocolate = a, cost of one biscuit = b and cost of one ice-cream = c. Then (3a + 5b + 5c) = 195 => (3a + 5b + 5c - 195) = 0. We have to find (7a + 11b + 9c) = ?.

Say, (7a + 11b + 9c) = x => (7a + 11b + 9c - x) = 0

Statement 1: (5a + 7b + 3c) = 217 => (5a + 7b + 3c - 217) = 0
Now the question is whether we can complete the required equation by linearly combining (3a + 5b + 5c) = 195 and (5a + 7b + 3c) = 217. If we can, then the following relation must hold for some constant m and n,

• m*(3a + 5b + 5c - 195) + n*(5a + 7b + 3c - 217) = (7a + 11b + 9c -x)

As a cannot contribute in b, b in c and so on, m and n must follow the following relations, (the relations are obtained by equating the coefficients of a, b and c)

• 1. 3m + 5n = 7
  2. 5m + 7n = 11
  3. 5m + 3n = 9

If there exists a set of value for m and n for which all the three relations are satisfied, then we can easily find x. In fact x will be equal to (195m + 217m).

For this case we can find such a set of value for m and n: m = 3/2 and n = 1/2

Sufficient.

Statement 1: (4a + b + 3c) = 141 => (4a + b + 3c - 141) = 0
Applying same procedure as above, m and n must satisfy the following relations,

• 1. 3m + 4n = 7
  2. 5m + n = 11
  3. 5m + 3n = 9

Try to solve these three relations, you'll find there is no such (m, n) for which all the three relations are satisfied. Thus we cannot complete the required equation.

Not sufficient.

Correct answer is A.

GMATGuruNY wrote:

goyalsau wrote:The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is the cost of 7 chocolates, 11 biscuits and 9 ice creams?

(A) The cost of 5 chocolates, 7 biscuits and 3 ice creams is 217.
(B) The cost of 4 chocolates, 1 biscuit and 3 ice creams is 141.

Given equation: 3c + 5b + 5i = 195.
Question: What is the value of 7c + 11b + 9i (or a multiple of this expression)?

Statement 1: 5c + 7b + 3i = 217

Try to combine this equation with the given equation so that the coefficient in front of c is a multiple of 7.

Multiplied by 3, the given equation 3c + 5b + 5i = 195 becomes 9c + 15b + 15i = 585.
Stacking this equation with the equation in statement 1, we get:

9c + 15b + 15i = 585
5c + 7b + 3i = 217

Adding the two equations, we get:
14c + 22b + 18i = 802.

Dividing the equation above by 2, we get:
7c + 11b + 9i = 401.
Sufficient.

Statement 2: 4c + 1b + 3i = 141.
Try to combine this equation with the given equation so that the coefficient in front of c is a multiple of 7.

Stacking the given equation with the equation in statement 2, we get:

3c + 5b + 3i = 195
4c + 1b + 3i = 141

Adding the two equations, we get:
7c + 6b + 6i = 336.

No way to determine the value of 7c + 11b + 9i.
Insufficient.

The correct answer is A.

GMATGuruNY wrote:

goyalsau wrote:The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is the cost of 7 chocolates, 11 biscuits and 9 ice creams?

(A) The cost of 5 chocolates, 7 biscuits and 3 ice creams is 217.
(B) The cost of 4 chocolates, 1 biscuit and 3 ice creams is 141.

Given equation: 3c + 5b + 5i = 195.
Question: What is the value of 7c + 11b + 9i (or a multiple of this expression)?

Statement 1: 5c + 7b + 3i = 217

Try to combine this equation with the given equation so that the coefficient in front of c is a multiple of 7.

Multiplied by 3, the given equation 3c + 5b + 5i = 195 becomes 9c + 15b + 15i = 585.
Stacking this equation with the equation in statement 1, we get:

9c + 15b + 15i = 585
5c + 7b + 3i = 217

Adding the two equations, we get:
14c + 22b + 18i = 802.

Dividing the equation above by 2, we get:
7c + 11b + 9i = 401.
Sufficient.

Statement 2: 4c + 1b + 3i = 141.
Try to combine this equation with the given equation so that the coefficient in front of c is a multiple of 7.

Stacking the given equation with the equation in statement 2, we get:

3c + 5b + 3i = 195
4c + 1b + 3i = 141

Adding the two equations, we get:
7c + 6b + 6i = 336.

No way to determine the value of 7c + 11b + 9i.
Insufficient.

The correct answer is A.

GMATGuruNY wrote:

goyalsau wrote:The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is the cost of 7 chocolates, 11 biscuits and 9 ice creams?

(A) The cost of 5 chocolates, 7 biscuits and 3 ice creams is 217.
(B) The cost of 4 chocolates, 1 biscuit and 3 ice creams is 141.

Given equation: 3c + 5b + 5i = 195.
Question: What is the value of 7c + 11b + 9i (or a multiple of this expression)?

Statement 1: 5c + 7b + 3i = 217

Try to combine this equation with the given equation so that the coefficient in front of c is a multiple of 7.

Multiplied by 3, the given equation 3c + 5b + 5i = 195 becomes 9c + 15b + 15i = 585.
Stacking this equation with the equation in statement 1, we get:

9c + 15b + 15i = 585
5c + 7b + 3i = 217

Adding the two equations, we get:
14c + 22b + 18i = 802.

Dividing the equation above by 2, we get:
7c + 11b + 9i = 401.
Sufficient.

Statement 2: 4c + 1b + 3i = 141.
Try to combine this equation with the given equation so that the coefficient in front of c is a multiple of 7.

Stacking the given equation with the equation in statement 2, we get:

3c + 5b + 3i = 195
4c + 1b + 3i = 141

Adding the two equations, we get:
7c + 6b + 6i = 336.

No way to determine the value of 7c + 11b + 9i.
Insufficient.

The correct answer is A.

goyalsau wrote:The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is the cost of 7 chocolates, 11 biscuits and 9 ice creams?

(A) The cost of 5 chocolates, 7 biscuits and 3 ice creams is 217.
(B) The cost of 4 chocolates, 1 biscuit and 3 ice creams is 141.

GMATGuruNY wrote:

goyalsau wrote:The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is the cost of 7 chocolates, 11 biscuits and 9 ice creams?

(A) The cost of 5 chocolates, 7 biscuits and 3 ice creams is 217.
(B) The cost of 4 chocolates, 1 biscuit and 3 ice creams is 141.

Given equation: 3c + 5b + 5i = 195.
Question: What is the value of 7c + 11b + 9i (or a multiple of this expression)?

Statement 1: 5c + 7b + 3i = 217

Try to combine this equation with the given equation so that the coefficient in front of c is a multiple of 7.

Multiplied by 3, the given equation 3c + 5b + 5i = 195 becomes 9c + 15b + 15i = 585.
Stacking this equation with the equation in statement 1, we get:

9c + 15b + 15i = 585
5c + 7b + 3i = 217

Adding the two equations, we get:
14c + 22b + 18i = 802.

Dividing the equation above by 2, we get:
7c + 11b + 9i = 401.
Sufficient.

Statement 2: 4c + 1b + 3i = 141.
Try to combine this equation with the given equation so that the coefficient in front of c is a multiple of 7.

Stacking the given equation with the equation in statement 2, we get:

3c + 5b + 3i = 195
4c + 1b + 3i = 141

Adding the two equations, we get:
7c + 6b + 6i = 336.

No way to determine the value of 7c + 11b + 9i.
Insufficient.

The correct answer is A.

The Iceman wrote:

goyalsau wrote:The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is the cost of 7 chocolates, 11 biscuits and 9 ice creams?

(A) The cost of 5 chocolates, 7 biscuits and 3 ice creams is 217.
(B) The cost of 4 chocolates, 1 biscuit and 3 ice creams is 141.

This question can be solved by two approaches:

1>hit and trial (can give quick results but may backfire if the coefficients do not strike)

2>traditional method (Appropriate way to do; basically a superset of approach 1; will take less time once you realise that you just need to establish whether a rational solution exists for a pair of variables)

GMAT should give numbers that would strike instantly. In case they don't, fall back to approach 2.

Statement A:
(3 5 5)*x
(5 7 3)*y
------
7 11 9

We need an unique solution in (x,y)such that 3x+5y=7; 5x+7y=11; 5x+3y=9 => (x,y) does have a unique solution

Statement B:
(3 5 5)*x
(4 1 3)*y
------
7 11 9

We need an unique solution in (x,y)such that 3x+4y=7; 5x+y=11; 5x+3y=9 => (x,y) does not have a unique solution

Hence A.

The Iceman wrote:

goyalsau wrote:The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is the cost of 7 chocolates, 11 biscuits and 9 ice creams?

(A) The cost of 5 chocolates, 7 biscuits and 3 ice creams is 217.
(B) The cost of 4 chocolates, 1 biscuit and 3 ice creams is 141.

This question can be solved by two approaches:

1>hit and trial (can give quick results but may backfire if the coefficients do not strike)

2>traditional method (Appropriate way to do; basically a superset of approach 1; will take less time once you realise that you just need to establish whether a rational solution exists for a pair of variables)

GMAT should give numbers that would strike instantly. In case they don't, fall back to approach 2.

Statement A:
(3 5 5)*x
(5 7 3)*y
------
7 11 9

We need an unique solution in (x,y)such that 3x+5y=7; 5x+7y=11; 5x+3y=9 => (x,y) does have a unique solution

Statement B:
(3 5 5)*x
(4 1 3)*y
------
7 11 9

We need an unique solution in (x,y)such that 3x+4y=7; 5x+y=11; 5x+3y=9 => (x,y) does not have a unique solution

Hence A.