Chocolates.

This topic has expert replies
User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

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.
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.

User avatar
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 am
Location: Bangalore, India
Thanked: 146 times
Followed by:24 members

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 :(
If the problem is Easy Respect it, if the problem is tough Attack it

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pm
Location: Milpitas, CA
Thanked: 447 times
Followed by:88 members

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.
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)

User avatar
Master | Next Rank: 500 Posts
Posts: 406
Joined: Mon Jan 25, 2010 11:36 am
Location: Syracuse, NY
Thanked: 23 times
Followed by:4 members
GMAT Score:740

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.
I'm really old, but I'll never be too old to become more educated.

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pm
Location: Milpitas, CA
Thanked: 447 times
Followed by:88 members

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.
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)

Legendary Member
Posts: 759
Joined: Mon Apr 26, 2010 10:15 am
Thanked: 85 times
Followed by:3 members

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

User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

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?
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pm
Location: Milpitas, CA
Thanked: 447 times
Followed by:88 members

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.
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)

User avatar
Legendary Member
Posts: 866
Joined: Mon Aug 02, 2010 6:46 pm
Location: Gwalior, India
Thanked: 31 times

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,,
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.

User avatar
Master | Next Rank: 500 Posts
Posts: 143
Joined: Sat Nov 06, 2010 9:06 pm
Thanked: 10 times
Followed by:1 members

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?
I don't know what to say, really. Three minutes to the biggest battle of our professional lives. You find out life's this game of inches, so is football. Because in either game - life or football - the margin for error is so small. I mean, one half a step too late or too early and you don't quite make it. One half second too slow, too fast and you don't quite catch it. I'll tell you this, in any fight it's the guy whose willing to die whose gonna win that inch. That's football guys, that's all it is. Now, what are you gonna do?

User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

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.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

User avatar
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Fri Aug 20, 2010 2:52 pm

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 ?

Legendary Member
Posts: 608
Joined: Sun Jun 19, 2011 11:16 am
Thanked: 37 times
Followed by:8 members

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.

Master | Next Rank: 500 Posts
Posts: 101
Joined: Thu Aug 25, 2011 9:39 pm

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

User avatar
Legendary Member
Posts: 626
Joined: Fri Dec 23, 2011 2:50 am
Location: Ahmedabad
Thanked: 31 times
Followed by:10 members

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)
Follow your passion, Success as perceived by others shall follow you