OG 12 - Q65 (Algebra)

[rahul.s]

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

OA: B

[thephoenix]

let a +b=n---->n=b-a
0.7a+0.5b=6.3

0.7a+0.5n-0.5a=6.3
0.5n+0.2a=6.3
now plug for n ; keep in mind that a has to be an int(bcoz no one purchase half apple)
for n=11 a=4
hence B

[ajith]

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

OA: B

to get a total of 6.30 there are only a few combination possible

1. 9 apples and 0 bananas

(not possible since the question gives an indication that both apples and bananas are purchased)

2. 4 apples and 7 bananas
(possible)


( There is only 1 mutiple of 7 which ends in 3 that is 9*7 =63
There is only 1 multiple of 7 under 63 which ends in 8 that is 4*7 = 28)

Hence B

[sars72]

no of bananas = b
no of apples = a

0.7a+ 0.5b = 6.30

multiplying by 10--> 7a + 5b = 63
--> b is multiplied by 5 and so its units digit is 0 or 5
units digit of 62 is 3
which means the units digit of b added to 0 or 5 should give 3 as the units digit

so its either 3 or 8
no multiple of 7 ends in 3, so it must be 28 i.e. 7*4

7*4 + 5*7 = 63

--> Total = 4+ 7 =11 = Choice B

[sanju09]

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

OA: B

Well, the number of apples and bananas bought got to be positive integers only. One may wonder of how to solve (answer uniquely) a single linear equation in two variables here, but the first statement that I have made is nearly equivalent to a second dissimilar equation/information in the same two variables, and the question may now be answered uniquely. See, how we could go about it.

If the customer purchases, a apples and b bananas, then


$0.70 a + $0.50 b = $6.30 such that b = (63 - 7 a)/5. There must be a unique pair of positive integers for a and b, if this question really holds. For 7 a, a number < 63, ending with a '3' or an '8' could give us b, a positive integer. We've 7*4 = 28 and 7*9 = 63 to work on, but, since the customer purchased both apples and bananas, therefore both a and b are positive integers. Can't take 63 for 7 a. Take 7 a = 28 and hence a = 4 and b = 7 to answer the total number of apples and bananas that the customer has purchased as a + b = 4 + 7 = [spoiler]11[/spoiler].

[spoiler]B[/spoiler]

[sanju09]

no of bananas = b
no of apples = a

0.7a+ 0.5b = 6.30

multiplying by 10--> 7a + 5b = 63
--> b is multiplied by 5 and so its units digit is 0 or 5
units digit of 62 is 3
which means the units digit of b added to 0 or 5 should give 3 as the units digit

so its either 3 or 8
no multiple of 7 ends in 3, so it must be 28 i.e. 7*4

7*4 + 5*7 = 63

--> Total = 4+ 7 =11 = Choice B

Everything is fine E X C E P T