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Re: A certain fruit stand sold apples for $0.70 each and bananas
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17 May 2016, 18:54
Attached is a visual that should help.
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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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Updated on: 09 Jan 2017, 14:33
The algebraic explanations in this thread are valid but needlessly complicated. Plus, it is unlikely that you will be able to come up with a similar approach to such a problem on your own. There is a much simpler and more broadly applicable approach that doesn't require trial and error. Because the numbers of apples and bananas have to be integers, the easiest thing to do here is start with the total of 6.3 and subtract off .7 until you arrive at a multiple of .5. 6.3  .7 = 5.6 5.6  .7 = 4.9 4.2  .7 = 4.2 4.2  .7 = 3.5 We are now at a multiple of .5, having subtracted off 4 apples. Because the bananas are 50 cents each, this gives us 7 bananas, for a total of 11 pieces of fruit.
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Originally posted by dannythor6911 on 05 Jan 2017, 14:24.
Last edited by dannythor6911 on 09 Jan 2017, 14:33, edited 1 time in total.



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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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09 Jan 2017, 09:33
pzazz12 wrote: A certain fruit stand sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchase ?
A. 10 B. 11 C. 12 D. 13 E. 15 We are given that apples were sold for $0.70 each and that bananas were sold for $0.50 each. We can set up variables for the number of apples sold and the number of bananas sold. b = number of bananas sold a = number of apples sold With these variables, it follows that: 0.7a + 0.5b = 6.3 We can multiply this equation by 10 to get: 7a + 5b = 63 Notice that we do not have any other information to set up a second equation, as we sometimes do for problems with two variables. So, we must use what we have. Keep in mind that variables a and b MUST be whole numbers, because you can't purchase 1.4 apples, for example. Notice also that 7 and 63 have a factor of 7 in common. Thus, we can move 7a and 63 to one side of the equation and leave 5b on the other side of the equation, and scrutinize the new equation carefully: 5b = 63 – 7a 5b = 7(9 – a) b = [7(9 – a)]/5 Remember that a and b MUST be positive whole numbers here. Thus, 5 must evenly divide into 7(9 – a). Since we know that 5 DOES NOT divide evenly into 7, it MUST divide evenly into (9 – a). We can ask the question: What must a equal so that 5 divides into 9 – a? The only value a can be is 4. We can check this: (9 – a)/5 = ? (9 – 4)/5 = ? 5/5 = 1 Since we know a = 4, we can use that to determine the value of b. b = [7(9 – 4)]/5 b = [7(5)]/5 b = 35/5 b = 7 Thus a + b = 4 + 7 = 11. Answer: B
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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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07 Oct 2017, 05:43
pzazz12 wrote: A certain fruit stand sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchase ?
A. 10 B. 11 C. 12 D. 13 E. 15 Here's an approach where we test the POSSIBLE CASES. FACT #1: ( total cost of apples) + ( total cost of bananas) = 630 CENTS FACT #2: total cost of bananas is DIVISIBLE by 50, since each banana costs 50 cents. Now let's start testing POSSIBLE scenarios. Customer buys 1 apple. 1 apple costs 70 cents, which means the remaining 560 cents was spent on bananas. Since 560 is NOT divisible by 50, this scenario is IMPOSSIBLE Customer buys 2 apples. 2 apple costs 140 cents, which means the remaining 490 cents was spent on bananas. Since 490 is NOT divisible by 50, this scenario is IMPOSSIBLE Customer buys 3 apples. 3 apple costs 210 cents, which means the remaining 520 cents was spent on bananas. Since 520 is NOT divisible by 50, this scenario is IMPOSSIBLE Customer buys 4 apples. 4 apple costs 280 cents, which means the remaining 350 cents was spent on bananas. Since 350 IS divisible by 50, this scenario is POSSIBLE 350 cents buys 7 bananas. So, the customer buys 4 apples and 7 bananas for a total of [spoiler]11[/spoiler] pieces of fruit Answer: Cheers, Brent
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Re: A certain fruit stand sold apples for $0.70 each and bananas
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26 May 2018, 06:46
After I thought of an alternative method to solve this type of problems ( an equation with two variables and constraints ) I found this method : this equation 7x+5y= 63 represents a line draw the line (by choosing two points , the easiest is when x=0 ,then y= ? ; when y=0,then x=? ) after drawing the line , search for a point that has x ,y integers it's (4,7) However , this method is not practical because it requires precise drawing , but it may give you an indication ( for example , plug and try values of x , not y because x has fewer values )
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Re: A certain fruit stand sold apples for $0.70 each and bananas
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15 Jan 2019, 03:39
Consider the following equation: 2x + 3y = 30. If x and y are nonnegative integers, the following solutions are possible: x=15, y=0 x=12, y=2 x=9, y=4 x=6, y=6 x=3, y=8 x=0, y=10 Notice the following: The value of x changes in increments of 3 (the coefficient for y). The value of y changes in increments of 2 (the coefficient for x). This pattern will be exhibited by any fully reduced equation that has two variables constrained to nonnegative integers. Bunuel wrote: A certain fruit stand sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchase?
(A) 10 (B) 11 (C) 12 (D) 13 (E) 14 70x + 50y = 630 7x + 5y = 63 In accordance with the pattern illustrated above, we get the following nonnegative solutions for x and y: x=9, y=0 x=4, y=7Here  since apples and bananas are both purchased  x and y must both be positive. Thus, only the option in green is viable, with the result that x+y = 4+7 = 11.
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A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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30 Apr 2019, 20:32
You can use divisibility properties here to get to the answer. The idea is this: if you see an equation like the following, and you know that r and s are positive integers: 5r = 7s this means that 5r and 7s are exactly the same number, so they must of course have the same divisors. So, since 7 is a divisor of 7s, it must also be a divisor of 5r, and therefore of r. Similarly s must be divisible by 5. That is, when you have equations involving only integers, the primes which divide the left side of the equation must divide the right, and vice versa. So onto this question. We know: 0.7a + 0.5b = 6.3 5b = 63  7a and since 7 is a factor of the right side of this equation, it must be a factor of the left, so b must be a multiple of 7. Since b is greater than 0, and b cannot be 14 (then the total cost would be greater than $6.30), b must be 7, from which we can find that a is 4.
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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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07 Jul 2019, 06:36
Let the no of apples sold = a no of bananas sold = b Question is a+b=? Thus 70a + 50b = 630 7a + 5b = 63 Quick Tip In order to find out the value of a & b its better to find the value of 'a' as "63 7a" must leave a number which will end either with 0 or with 5. (Think about it for a second) Thus the only value which satisfies above equation is a=4 & b=7 a+b=11 (Other values of a & b will lie outside the answer choices) Answer B



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A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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07 Jul 2019, 08:45
could some one put a link about solving problem with Diophantine ??????? I think best approach to solve these problem without ANY risk !! put here a link about fastest method of Diophantus ..... thanks
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A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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07 Jul 2019, 16:32
Bunuel wrote: A certain fruit stand sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchase?
(A) 10 (B) 11 (C) 12 (D) 13 (E) 14 we know the customer bought at least one applebanana pair for $1.20 say she bought 5 pairs for $6.00, leaving her 30¢ no say she bought 4 pairs for $4.80, leaving her $1.50 yes she can buy 4 pairs for $4.80 plus 3 additional bananas for $1.50 for $6.30 total 11 B



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A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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07 Jul 2019, 23:03
First, simplify the equation 0.7A + 0.5B = 6.30 by multiplying 10 on each side of the equation and you get 7A + 5B = 63
Next, find A and B by summing the last digit through trials. The trick is to begin testing with the multiples that end with less variations.
multiples of 7A ends with digit  x7,x4,x1,x8,x5,x2,x9,x6,x0 multiples of 5B ends with digit  x0,x5
In this case, starts with multiple of 5B, which ends with 2 variations as compared to multiples of 7A, which ends with 9 variations.
Case1: last digit of 5B is 0, 5B + 7A = 63 _0 + _? = _3 ; (? = 3) The multiples of 7 that ends with 3 is 9 ; 7X9 = 63. This means A+B =9 since 5(0)+ 7(9) = 63;
Case2: last digit of 5B is 5, 5B + 7A = 63 _5 + _? = _3 ; (? = 8) The multiples of 7 that ends with 8 is 4; 7X4 = 28. This means A+B =11 since 5(7)+ 7(4) = 63
Base on the answer choices, A+B is more than 9, therefore, A+B = 11. B is the answer.



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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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12 Jul 2019, 04:48
A is the number of apples B is the number of bananas >What is being asked: A + B =????? 0.7*A + 0.5*B = 6.3 >7*A + 5*B = 63 = 7*9 >5*B = 7*9  7*A = 7*(9  A) From here, we realise that B should be an integer that has to be a multiple of 5 & 7, option can be only B = 7 and A = 4



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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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21 Jul 2019, 04:22
pzazz12 wrote: Bunuel wrote: pzazz12 wrote: A certain fruit stand sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchase ?
A. 10 B. 11 C. 12 D. 13 E. 15 Given: \(0.7b+0.5a=6.3\) Question: \(a+b=?\) \(0.7a+0.5b=6.3\) > \(7a+5b=63\). After some trial and error you'll get that only two integer pairs of (a,b) satisfy this equation: (9,0) and (4,7) as we are told that "a customer purchased both apples and bananas" then the first pair is out and we'll have: \(a=4\) and \(b=7\) > \(a+b=11\). . thank you, but can you explain me how this (9,0) and (4,7) to be solve... Let me try a different way to solve this > If an Apple is 70 cents, a Banana is 50 cents, and the total price is 6.3 dollars, so 7A plus 5B equals 63; 637A=5B => 5B= 7(9A) If (9  A) is a 5, that means that A = 4, and the total number of apples and bananas is 11. The correct answer chose is B. Answer: B
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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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21 Jul 2019, 07:26
Bunuel wrote: A certain fruit stand sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchase?
(A) 10 (B) 11 (C) 12 (D) 13 (E) 14 I have done so many DS questions in last few days that after reading the question I was like okay let's see if only (1) or only (2) is sufficient until I saw options and I am like meh I have to solve this one and there is surely an unique solution. Haha! Bunuel has provided the best solution.
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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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20 Nov 2019, 09:36
VeritasKarishma wrote: n2739178 wrote: Hi all I can't understand MGMAT's OG guide answer for this one... 1st question: why can't you exchange apples with bananas 1 for 1 until you get the right answer? 2nd question: They're banging on about 5 apples having the same value as 7 bananas, then taking 5 apples away from the 'partial solution ~(i.e. when A = 9 and B = 0) and adding 7 bananas... Why would you do that? It makes no sense to me... If you set the vlaue of the apples = bananas you get them both equalling $3.50 in value which equals $7 between both of them... which is clearly over $63 ... this is really doing my head in! thanks I do not know what exactly your book says but I am guessing this is how they have solved it: 7a + 5b = 63 Such equations have infinite solutions. We can get a single solution under particular constraints. (Will explain this later) One thing we notice right away is that one solution to this problem is a = 9 and b = 0 because 63 is divisible by 7. 7a + 5b = 63 a = 9, b = 0 a = 4, b = 7 (To get this solution, subtract 5, coefficient of b, from a above and add 7, coefficient of a, to b above) a = 1, b = 14 (Again, do the same to the solution above) a = 13, b = 7 (You will also get solutions when you add 5 to a of any other solution and subtract 7 from b of the same solution) Hence there are infinite solutions. Here the constraints are that a and b should not be negative. Also, they should not be 0 since he buys at least 1 apple and at least 1 banana. Only 1 solution satisfies these constraints so answer is a = 4 and b =7. Why this works is because when you reduce a by 5, the reduction in 7a is offset by the increase in 5b when you increase b by 7. Let this suffice for now. This is the theory of Integral solutions to equations in two variables. I will explain you the complete theory soon. Hi VeritasKarishma, Is my approach is correct ? given 0.7A+0.5B=6.30 If all apples purchased, than max 9 apples can be purchsed on other hand if all banana purchased than max 12 banana purchased as given both apples and banana purchased, hence number of bananas and apples purchsed will be 9< (A+B)<12 so only 11 satifies it.
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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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13 May 2020, 18:39
We can also eliminate answer choices D & E since if B>13, and the price of bananas = 0.50, then this will exceed $6.30



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A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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23 Jun 2020, 02:39
Bunuel wrote: A certain fruit stand sold apples for $0.70 each and bananas for $0.50 each. If a customer purchased both apples and bananas from the stand for a total of $6.30, what total number of apples and bananas did the customer purchase?
(A) 10 (B) 11 (C) 12 (D) 13 (E) 14 A > 0 and B > 0 both are positive integers. Hence (A,B) = (9,0) is ruled out. 7A + 5B = 63 So, the unit digits of 7A and 5B should yield 3. As 5B would only result in 0 and 5 as unit place, 7A should result in either 3 or 8 as units digit. 7A results in units digit as 3 only when A = 9 which is ruled out. Hence 7A must result in 8 as units place. This can be done as only when A = 4(under given conditions) i.e. 7*4 = 28. Thus, 5B = 35 => B = 7 A + B = 7 + 4 = 11 Answer B.
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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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02 Jul 2020, 12:30
This can not be solved by equations. Can only be solved by trail and error method
0.7*apples + 0.5*bananas = 6.3 This equation is satisfied by (9,0) and (4,7) Since it is said that both fruits are bought answer should be (4,7) Therefore apples + bananas = 4 + 7 = 11




Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50
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