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655-705 Level|   Algebra|      
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Bunuel
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An apple cost $a, a banana costs $b, and a clementine costs $c, where 15 Jun 2020, 10:45
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B
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An apple cost $a, a banana costs $b, and a clementine costs $c, where a, b and c are positive integers. The cost of b apples, b bananas, and a + b clementines is $77. What is the cost of one apple, two bananas, and one clementine?

A. 17
B. 18
C. 19
D. 21
E. 23

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B
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An apple cost $a, a banana costs $b, and a clementine costs $c, where 15 Jun 2020, 11:02
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According to question:

ab + b^2 + (a+b)c = 77

=> ab + bc + ac + b^2 = 77
=> a(b+c) + b(b+c) = 77

=> (a+b)(b+c) = 77 = 7*11

=> Either: a+b = 7, and b+c = 11
Or: a+b = 11, b+c = 7

Now,

Cost of one apple, two bananas, and one clementine = a+2b+c = (a+b) + (b+c)
=> Cost = 7 + 11 = 18

=> Answer: B
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An apple cost $a, a banana costs $b, and a clementine costs $c, where 15 Jun 2020, 11:05
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Equation:
a*b +b*b + (a+b)*c = 77
b(a+b) + (a+b)*c = 77
(b+c)*(a+b) = 77
Prime factorize 77 i.e. 11*7

Either (a+b) or (b+c) = 11 and other equation will be 7
Lets assume (a+b) = 7 (eq -1) and (b+c) = 11 (eq -2)
Assume value of a = 2, then b = 5. Substitute value of b in second equation; c = 6

Cost of a +2b + c = 2 + 5*2 + 6 = 18

If we substitute any value of a in the equation 1, we will end up getting 18 as value

Answer = 18 (B)
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An apple cost $a, a banana costs $b, and a clementine costs $c, where 15 Jun 2020, 12:03
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One apple costs a
Thus b apples cost a x b;

One banana costs b
Thus b bananas cost b\(^2\);

One clementine costs c
Thus a + b clementines cost (a + b) x c;

Now,

(a x b) + b\(^2\) + [(a + b) x c] = 77

Solving we get (a + b) x (b + c) = 77;

Which is nothing but 7 x 11;

Now a + 2b + c = 7 + 11 = 18;

IMO ans is B.
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An apple cost $a, a banana costs $b, and a clementine costs $c, where 15 Jun 2020, 21:07
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An apple cost $a, a banana costs $b, and a clementine costs $c, where a, b and c are positive integers. The cost of b apples, b bananas, and a + b clementines is $77. What is the cost of one apple, two bananas, and one clementine?

A. 17
B. 18
C. 19
D. 21
E. 23

Solution:
Given:
• Cost of 1 apple = $a
• Cost of 1 banana = $b
• Cost of 1 clementnes = $c
Required: the cost of one apple, two bananas, and one Clementine ?

That s, a + 2b + c =?
Here, ab +b^2 +c (a +b) = 77
Or, b (a+ b) + c(a +b) = 77
Or, (a +b)(b +c) = 77
Therefore, (a +b) = 7 or (b + c)= 11
And
(a +b) = 11 or (b+c)= 7
Now, a + 2b + c
= (a + b) +(b +c)
= 11 +7
= 18.
Answer: B
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An apple cost $a, a banana costs $b, and a clementine costs $c, where 03 Aug 2024, 07:58
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