Two mechanics A and B work on a project for 10 days. Their daily wages are given below :
- Daily wage of \[A = Rs.100\]
- Daily wage of \[B = Rs.150\]
What will be sum of their combined wages?
The combined wage for A and B
For \[1\] day \[=Rs.(100+150)=Rs.250\]
\[\therefore\] Total wage for \[10\] days
\[= Rs.(250\times10)=Rs.2500/-\]
The average wage of \[500\] workers was found to be \[Rs.200\]. Later on, it was discovered that the wages of two workers were misread as \[180\] and \[20\] instead of \[80\] and \[220\]. The correct average wage is :
Total wages of \[500\] workers
\[= 500 × 200 = 100000\]
Now, according to the question, Correct Average
\[=\frac{\left(100000-180-20+80+220\right)}{500}\]
\[=\frac{100100}{500}\]
\[=Rs.200.20\]
A, B and C together earn Rs.150 per day while A and C together earn Rs.94 and B and C together earn Rs.76. The daily earning of ‘C’ is
The daily earning of ‘\[C\]’
\[=\] Daily earning of \[(A + C)\] and \[(B + C) –\] Daily earning of \[(A + B + C)\]
\[= 94 + 76 – 150 = Rs.20\]
A certain number of men can complete a job in \[30\] days. If there were \[5\] men more, it could be completed in \[10\] days less. How many men were in the beginning?
Let initially the number of men be \[x\].
According to question,
\[M_1W_1D_1=M_2W_2D_2\]
\[x × 30 = (x + 5) × (30 – 10)\]
\[x × 30 = 20x + 100\]
\[30x – 20x = 100\]
\[10x = 100\]
\[x = 10\]
A work could be completed in \[100\] days by some workers. However, due to the absence of \[10\] workers, it was completed in \[110\] days. The original number of workers was :
Let the original number of workers \[= x\].
Then, \[x × 100 = ( x – 10) × 110\]
\[\implies 10 x = 11 x – 110 \]
\[\implies x = 110\]
A labour was appointed by a contractor on the condition that he would be paid \[Rs.75\] for each day of his work but would be fined at the rate of \[Rs.15\] per day for his absence, apart from losing his wages, After \[Rs.20\] days, the contractor paid the labour \[Rs.1140\]. The number of days the labourer abstained from work was
Total salary for \[20\] days
\[=Rs.(75 × 20) = Rs.1500\]
Actual salary received \[= Rs.1140\]
Difference \[= Rs.(1500 – 1140) = Rs.360\]
Money deducted for \[1\] day’s absence from work
\[= Rs.(15 + 75) = Rs.90\]
\[\therefore\] Number of days he was absent
\[= \frac{360}{90}= 4\] days
It is required to find the highest common factor of \[5750\] and \[5000\], because his daily wage is their common factor.
Hence, the daily wage is \[Rs.250\].
A, B and C completed a work costing \[Rs.1,800\]. A worked for \[6\] days, B for \[4\] days and C for \[9\] days. If their daily wages are in the ratio of \[5 : 6 : 4\], how much amount will be received by A?
The ratio of wages of A, B, and C respectively
\[= 5 × 6 : 6 × 4 : 4 × 9\]
\[= 30 : 24 : 36\]
\[= 5 : 4 : 6\]
\[\therefore\] Amount received by A
\[\frac{5}{5+4+6}\times 1800\]
\[=\frac{5}{15}\times 1800\]
\[=Rs.600\]
Required ratio
\[= 15 × 22 : 11 × 25\]
\[= 6 : 5\]
Let daily wages of a man be \[Rs. x \].
\[= Rs. (2.75 – 0.5)\]
\[= Rs. 2.25\]