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06 Nov 2014, 20:56
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64% (03:01) correct
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In a joint family there are 3 couples. The ages (all in years) of the husbands are 4X, 3X, and 2X. The age of each women is 4 years less than her husband's age. The product of ages of first couple is more than the sum of the products of the ages of the other two couples by 832. Find the value of X.
A) 12
B) 14
C) 16
D) 18
E) 20
A) 12
B) 14
C) 16
D) 18
E) 20
07 Nov 2014, 04:12
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Ashishmathew01081987
In a joint family there are 3 couples. The ages (all in years) of the husbands are 4X, 3X, and 2X. The age of each women is 4 years less than her husband's age. The product of ages of first couple is more than the sum of the products of the ages of the other two couples by 832. Find the value of X.
A) 12
B) 14
C) 16
D) 18
E) 20
A) 12
B) 14
C) 16
D) 18
E) 20
4x(4x - 4) - 832 = 3x(3x - 4) + 2x(2x - 4);
x(3x + 4) = 832.
Plug-in options. C fits.
Answer: C.
08 Nov 2014, 02:51
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Ashishmathew01081987
In a joint family there are 3 couples. The ages (all in years) of the husbands are 4X, 3X, and 2X. The age of each women is 4 years less than her husband's age. The product of ages of first couple is more than the sum of the products of the ages of the other two couples by 832. Find the value of X.
A) 12
B) 14
C) 16
D) 18
E) 20
A) 12
B) 14
C) 16
D) 18
E) 20
My short cut approach: find the last digit instead of plug in.
Three couples:
------------- 1st--------2nd--------3rd
Husband------4x---------3x---------2x
Wife---------(4x - 4)----(3x - 4)----(2x - 4)
We have equation:
4x(4x - 4) = 3x(3x - 4) + 2x(2x -4) + 832
3x^2 + 4x - 832 = 0
Let group: (3^2 + 4x) - 832 = 0
The last digit of 832 is 2, so in order to have "0", logically, last digit of (3x^2 + 4x) should be 2
A) 3*12^2 + 4*12 --> last digit = 2 + 8 = 0
B) 3*14^2 + 4*14 --> last digit = 8 + 6 = 4
C) 3*16^2 + 4*16 --> last digit = 8 + 4 = 2 BINGO
D) 3*18^2 + 4*18 --> last digit = 2 + 2 = 4
E) 3*20^2 + 4*20 --> last digit = 0 + 0 = 0
Only C stands and is correct.
Hope it helps.
