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Chocolates.

by goyalsau » Tue Dec 07, 2010 3:28 am
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.
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by shovan85 » Tue Dec 07, 2010 9:11 am
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.
IMO A

Let c be chocolate, b be biscuits, i be ice cream.
So, 3c+5b+5i = 195 ...(1)
7c+11b+9i = ?

(A) 5c+7b+3i = 217 ...(2)

Subtract (2) from (1) => (1) - (2)
=> -2c -2b +2i = -22 => - c - b + i = -11 .....(3)

Add (2) with (1) => (1) + (2)
=> 8c + 12b + 8i = 412 ......(4)

Add (3) and (4) gives 7c+11b+9i = ___ SUFFICIENT.

(B) 4c+1b+3i = 141 ....(5)

Add (1) and (5) => 7c + 6b + 8i = 336 ...(6)

(1) - (5) => -c + 4b + 2i = 54 ...(7)

I cannot go further here :(
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by Rahul@gurome » Tue Dec 07, 2010 11:14 am
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.
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by tomada » Tue Dec 07, 2010 11:36 am
I wonder if a good number of people will see 2 equations and 3 variables and assume that it can't be solved.
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by Rahul@gurome » Tue Dec 07, 2010 12:04 pm
tomada wrote:I wonder if a good number of people will see 2 equations and 3 variables and assume that it can't be solved.
Then they will make a fundamental mistake! :)
This question doesn't require to solve the equations. In fact it is more about constructing a new equation from two given equation.
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by clock60 » Tue Dec 07, 2010 12:39 pm
i see that we all came to the same result but slightly different ways. my thinking
3c+5b+5a=195, we need to find
7c+11b+9a=x?, here i notice that if we multiply the first equation with 2 and then substract the coefficient will be 1
2*(3c+5b+5a)=195*2

6c+10b+10a=195*2
7c+11b+9a=x, now substract
c+b-a=x-195*2, so we need to find what is c+b-a=? to find x
(1)5c+7b+3a=217
3c+5b+5a=195, again substract. and left with 2c+2b-2a=22. c+b-a=11-we need this figure
so suff
(2) we can`t derive useful info insuff
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by goyalsau » Tue Dec 07, 2010 8:33 pm
Rahul@gurome wrote:
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.
Thanks, Rahul
Your Method is Awesome, Solution's that was given by shovan and clock60 is also Brilliant , But i must say your solution was like a Sure shot way,

I trying to calculate the individual cost of chocolate , biscuit & one ice-cream With the Help of the derived values of M and N But just not able to figure out How to do it :(

I know the problem is not asking to determine the individual values of items, But i was just wondering, Can it be Done?
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by Rahul@gurome » Wed Dec 08, 2010 2:39 am
goyalsau wrote:I trying to calculate the individual cost of chocolate , biscuit & one ice-cream With the Help of the derived values of M and N But just not able to figure out How to do it :(

I know the problem is not asking to determine the individual values of items, But i was just wondering, Can it be Done?
No, we can't.
To be able to do so we must have at least three equations as there are three unknowns. But we have only two equations available if we take one statement at a time.
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by goyalsau » Wed Dec 08, 2010 2:59 am
Rahul@gurome wrote:
goyalsau wrote:I trying to calculate the individual cost of chocolate , biscuit & one ice-cream With the Help of the derived values of M and N But just not able to figure out How to do it :(

I know the problem is not asking to determine the individual values of items, But i was just wondering, Can it be Done?
No, we can't.
To be able to do so we must have at least three equations as there are three unknowns. But we have only two equations available if we take one statement at a time.
Thanks Rahul,,
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by edvhou812 » Thu Mar 03, 2011 1:40 am
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.
I think I understand the explanation, but it seems like I would go over 2 minutes while checking if the values of m and n make each statement true. Is there a faster way?
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chocolates, biscuits and ice cream

by GMATGuruNY » Thu Mar 03, 2011 5:08 am
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 + 5i = 195
4c + 1b + 3i = 141

Adding the two equations, we get:
7c + 6b + 8i = 336.

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

The correct answer is A.
Last edited by GMATGuruNY on Fri Apr 10, 2020 12:43 pm, edited 1 time in total.
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by rabbulnawaz » Fri May 27, 2011 8:51 am
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.
Hi, By following your method, we can solve this question in 2 mins. But can you please explain me, how did you come to a conclusion that adding/subtracting the given equation should give a multiple of 7 for the constant of "C".
Is it because, in the question the constant of "C" is 7 ?
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by saketk » Thu Oct 20, 2011 1:00 am
3C+5B+5I = 195 -- equation (1)

to find -- 7C+11B+9I = ?

Stmt A--
5C+7B+3I= 217 equation (2)

Add the equations 1 and 2--

8C+12B+8I= 412 equation (3)

subtract the equation (1) from (2)

2C+2B-2I= 22

C+B-I = 11

multiply by (-1) both sides

I-C-B = -11 equation (4)

Add equation (3) & (4)

we will get
7C+11B+9I= 401

STATEMENT A is sufficient.


stmt 2 --
4C+1B+3I= 141 --equation (5)

we perform the similar operation as done in case of stmt 1

add equation (1)_ & (5)

7C+6B+6i = 336 equation (6)

We have to find the value of 7C+11B+9I .

We cannot arrive at this combination because any operation on the equation (6) will mean that we will not be able to maintain the value of C.

hence this stmt is insufficient.
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by prashant misra » Wed Nov 02, 2011 5:11 am
thanks shovan and gmatguruY i understood the way you provided the solutions to the questions.earlier i was trying them as two individual linear equations an ended up choosing wrong answer
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by ronnie1985 » Thu Mar 29, 2012 10:35 am
Going for a combination of factors to arrive at answer helps, but is lengthy and depends on luck sometimes. But the coefficient method is definitely very lengthy, I would not attempt to solve the question in the exam, but this question is solvable and the answer is (A)
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