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The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is
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10 Apr 2020, 01:37
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24% (02:06) correct 76% (01:44) wrong based on 29 sessions
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Competition Mode Question 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? (1) The cost of 5 chocolates, 7 biscuits and 3 ice creams is 217. (2) The cost of 4 chocolates, 1 biscuit and 3 ice creams is 141. Are You Up For the Challenge: 700 Level Questions
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Re: The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is
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10 Apr 2020, 02:42
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?
(1) The cost of 5 chocolates, 7 biscuits and 3 ice creams is 217. (2) The cost of 4 chocolates, 1 biscuit and 3 ice creams is 141.
Given: \(3C + 5B + 5IC = 195\) \(7C + 11B + 9IC = ?\) if we subtract the two equations, we'll get \(4C + 6B + 4C \);\(2(2C + 3B + 2C)\). so if we know this value, we can find the total cost.
statement 1: \(5C + 7B + 3IC = 217\). add this to given equation \((5C+3C) + (7B+5B) + (3IC+5IC) = (217+195)\) \(8C + 12B + 8IC = 522\); \(4(2C + 3B + 2C) = 522\) knowing the value of \((2C + 3B + 2C) \) is sufficient to calculate \(7C + 11B + 9IC\).
statement 2: \(4C + 1B + 3IC = 141\) this is not sufficient. even if we add this equation with the given equation, we get \(7C + 6B + 8IC\), so we still the value of \(5B + IC = ?\). not sufficient Ans: A



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Re: The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is
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10 Apr 2020, 03:15
3C +5B + 5I =195  (1) From first statement 5C + 11B + 3 I =217
Multiplying eq (1) by 3 and eq (2) by 1 and adding both of them we get 14C + 22B + 18 I = 412 7C + 11B + 9I = 206 Sufficient From statement 2 4C + 11B + 9I = 141 This statement is not sufficient Option A is the answer
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The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is
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Updated on: 13 Apr 2020, 03:16
Let chocolate, biscuits, and ice cream be c, b, r, respectively.
Known: 3c + 5b + 5r = 195.
Q. 7c + 11b + 9r = ?
(1) 5c + 7b + 3r = 217. > 3*(3c + 5b + 5r) + (5c + 7b + 3r) = 3*195 + 217 14c + 22b + 18r = 802 7c + 11b + 9r = 401 SUFFICIENT
(2) 4c + b + 3r = 141. > (3c + 5b + 5r) + (4c + b + 3r) 7c + 6b + 8r, but we are no closer to 7c + 11b + 9r. Thus, we need one more equation to solve c, b, and r. NOT SUFFICIENT
FINAL ANSWER IS (A)
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Originally posted by chondro48 on 10 Apr 2020, 04:09.
Last edited by chondro48 on 13 Apr 2020, 03:16, edited 3 times in total.



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Re: The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is
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10 Apr 2020, 09:00
The question has 3 unknowns and potentially 2 equations (A,B) or we have a choice of 3 unknowns and 3 equations (C). So it is tempting to choose C. But the question does not ask to evaluate all 3 unknowns. It is merely asking whether we can get to another equation. Question stem says: 3c+5b+5i=195
To find out: 7c+11b+9i=?=x(say)
(A) 3c+5b+5i=195 5c+7b+3i=217
So there must be numbers m and n (say), such that m(3c+5b+5i195) + n(5c+7b+3i217) = 7c+11b+9ix If we equate LHS and RHS we get 3 equations  3m+5n=7 .......(i), 5m+7n=11......(ii), 5m+3n=9......(iii) From (ii) & (iii) m =3/2, n=1/2 Putting the value of m and n in (i) to check if it satisfies. Yes it does. So m and n are unique solutions. So we can get to 7c+11b+9i from the given information.
For those who want to know how, here it is  5c+7b+3i=217 3c+5b+5i=195  2c+2b2i=22 (by subtraction) c+bi=11........(iv)
Again, 5c+7b+3i=217 3c+5b+5i=195  8c+12b+8i=412 (by addition)....(v)
By (v)(iv) 8c+12b+8i=412 c+bi=11  7c+11b+9i=401 (answer)
(B) Exactly employing the same method as we did in A, we get 3 equations 4m+3n=7..........(vi) 11m+5n=11..........(vii) 3m+5n=9..............(viii)
From (vii) & (viii) we get m=1/4 and n=33/20 We put these values in (vi) and it does not satisfy. Therefore we do not have a unique solution. Hence we would not be able to get to 7c+11b+9i from the given information
Answer: A



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Re: The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is
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10 Apr 2020, 11:34
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?
(1) The cost of 5 chocolates, 7 biscuits and 3 ice creams is 217. (2) The cost of 4 chocolates, 1 biscuit and 3 ice creams is 141.
stem: 3c+5b+5i = 195....7c+11b+9i=? 1) 5c+7b+3i=217, left with 2 variables to solve not suffiicient 2) 4c+1b+3i=141, left with 2 variables to solve not sufficient 1+2, can be solvable. Ans C



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The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is
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Updated on: 13 Apr 2020, 12:20
Quote: 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?
(1) The cost of 5 chocolates, 7 biscuits and 3 ice creams is 217. (2) The cost of 4 chocolates, 1 biscuit and 3 ice creams is 141 If \(3c+5b+5i=195\), what is the value of \(7c+11b+9i\)? Statement 1: \(5c+7b+3i=217\)Multiplying \(3c+5b+5i=195\) by 3, we get: \(9c+15b+15i=585\) Adding together \(9c+15b+15i=585\) and \(5c+7b+3i=217\), we get: \(14c+22b+18i=802\) \(7c+11b+9i=401\) SUFFICIENT. Statement 2: \(4c+b+3i=141\)Adding together \(3c+5b+5i=195\) and \(4c+b+3i=141\), we get: \(7c+6b+8i=336\) No way to determine the value of \(7c+11b+9i\). INSUFFICIENT. One way to prove that S2 is INSUFFICIENT: Case 1: c=15 Statement 2 equation > \(4*15+b+3i=141\) > \(b+3i=81\) Prompt equation > \(3*15+5b+5i=195\) > \(5b+5i=150\) > \(b+i=30\) Subtracting the prompt equation from the Statement 2 equation, we get: \(2i=51\) \(i=25.5\) Since \(b+i=30\), \(b=4.5\) In this case, \(7c+11b+9i=(7*15)+(11*4.5)+(9*25.5)=105+49.5+229.5=384\) Case 1: c=25 Statement 2 equation > \(4*25+b+3i=141\) > \(b+3i=41\) Prompt equation > \(3*25+5b+5i=195\) > \(5b+5i=120\) > \(b+i=24\) Subtracting the prompt equation from the Statement 2 equation, we get: \(2i=17\) \(i=8.5\) Since \(b+i=24\), \(b=15.5\) In this case, \(7c+11b+9i=(7*25)+(11*15.5)+(9*8.5)=175+170.5+76.5=422\) Since \(7c+11b+9i\) can be different values, INSUFFICIENT.
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Originally posted by GMATGuruNY on 10 Apr 2020, 12:44.
Last edited by GMATGuruNY on 13 Apr 2020, 12:20, edited 1 time in total.



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Re: The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is
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11 Apr 2020, 00:48
3 chocolates+5 biscuits+5 ice creams =195. 7 chocolates+11 biscuits+9 ice creams=?
(1) The cost of 5 chocolates, 7 biscuits and 3 ice creams is 217.
3 chocolates+5 biscuits+5 ice creams =195 5 chocolates+7 biscuits+3 ice creams=217
adding both we get, 8 chocolates+12 biscuits+8 ice creams=412 subtracting both , 1 chocolates 1 biscuits+1 ice creams=11
adding above we get the required value.....................SUFFICIENT
(2) The cost of 4 chocolates, 1 biscuit and 3 ice creams is 141.
even after adding and subtracting we wont be getting the required value...........INSUFFICIENT
OA:A



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Re: The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is
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11 Apr 2020, 02:32
Quote: 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?
(1) The cost of 5 chocolates, 7 biscuits and 3 ice creams is 217. (2) The cost of 4 chocolates, 1 biscuit and 3 ice creams is 141. Given: 3c + 5b + 5i = 195 Question: 7c + 11b + 9i = ?STatement 1: 5c + 7b + 3i = 217Multiplying by 3 both sides in 3c + 5b + 5i = 195 3c + 15b + 15i = 585 Adding 5c + 7b + 3i = 217 in above equation (9c + 15b + 15i) + (5c + 7b + 3i) = 585+217 14c + 22b + 18i = 802 Dividing by 2 both sides 7c + 11b + 9i = 401 SUFFICIENT STatement 2: 4c + 1b + 3i = 143NOT SUFFICIENT Answer: Option A
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Re: The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is
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