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# The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is

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Posts: 64111
The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is  [#permalink]

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10 Apr 2020, 01:37
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55% (hard)

Question Stats:

24% (02:06) correct 76% (01:44) wrong based on 29 sessions

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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.

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Re: The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is  [#permalink]

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10 Apr 2020, 02:42
1
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  [#permalink]

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10 Apr 2020, 03:15
1
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

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The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is  [#permalink]

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Updated on: 13 Apr 2020, 03:16
1
1
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

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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  [#permalink]

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10 Apr 2020, 09:00
1
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+5i-195) + n(5c+7b+3i-217) = 7c+11b+9i-x
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+2b-2i=22 (by subtraction)
c+b-i=11........(iv)

Again,
5c+7b+3i=217
3c+5b+5i=195
-----------------

By (v)-(iv)
8c+12b+8i=412
c+b-i=11
-----------------

(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

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Re: The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is  [#permalink]

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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  [#permalink]

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Updated on: 13 Apr 2020, 12:20
1
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  [#permalink]

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11 Apr 2020, 00:48
1
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

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Re: The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is  [#permalink]

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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 = 217

Multiplying 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 = 143

NOT SUFFICIENT

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Re: The cost of 3 chocolates, 5 biscuits, and 5 ice creams is 195. What is   [#permalink] 11 Apr 2020, 02:32