甲、乙两个工程队修路,最终工作量分配8400元工资,两队按原计划的工作效率,乙队应获得5040元的工资,实

2025-03-15 15:43:52
推荐回答(2个)
回答1:

原来甲乙的工作效率比为:
(8400-5040):5040,
=3360:5040,
=2:3,
甲提高工效后,甲乙的工作效率比为:
(3360+960):(5040-960)
=4320:4080,
=18:17;
设甲开始的工效为“2”,那么乙的工效为“3”,设甲在提高工效后还需a天完成任务,得:
(2×4+4a):(3×4+3a)=18:17,
解得a=
40
7
(天),
于是共有工程量为:
(2×4+4×
40
7
)+(3×4+3×
40
7
),
=(8+
160
7
)+(12+
120
7
),
=20+40,
=60;
所以原计划修好这条路的天数为:
60÷(2+3),
=60÷5,
=12(天).
答:两队原计划12天能修好这条路.

回答2:

  【分析与解题】开始时甲队拿到8400—5040=3360元,甲乙的工资比等于甲乙的工效比,即为 3360:5040=2:3 ;

  甲提高工效后,甲乙的工资及工效比为

  (3360+960):(5040—960)=18:17 ;

  设甲开始的工效为“2”,那么乙的工效为“3” ,设甲在提高工效后还需 x 天完成任务 ;

  即:(2 × 4+4x ):(3 × 4+3x)=18:17,

  化简为:216+54 x =136+68 x

  解得:a=40 / 7(天),

  共有工程量为:4 × 5+7 × 40 / 7 = 60 ,

  所以原计划 60÷(2+3)=12(天)

  答:两队原计划12天能修好这条路.

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