Working on a project, Mike and Michael together complete a piece of work in 6 days. If Mike works alone for 2 days and completes (1⁄4)th of the work, in how many days can Michael alone complete the rest of the work?
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Solution
Mike completes (1⁄4)th of work in 2 days.
∴ He completes the entire work in 8 days, working alone.
∴ He does (1⁄8)th of job in 1 day.
Mike and Michael complete the entire job together in 6 days.
∴ Their combined speed is 1⁄6th job per days.
Michael does 1⁄6 - 1⁄8 = \(\left (\frac{1}{24} \right )^{th}\) of the job in 1 day.
∴ Michael finishes the job in 24 days, working alone.
∴ Michael finishes (3⁄4)th of the job in 18 days.
Mr. and Mrs. Anderson purchased an antique jewel case for $22,000. Every month they incurred $250 maintenance over the case. At the end of 1 year, the antique piece was appraised at $80,000 and donated to a museum. However, the Andersons could claim the donation as a tax deduction and as a result, received 60% of gem’s appraised value as a tax refund. What would be the net gain of Andersons from these transactions?
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Solution
Total money spent by Andersons
= $22,000 + $(250 × 12) = $25,000
Money received by them
= 60% of $80,000 = $48,000
∴ Net gain = $48,000 - $25,000 = $23,000
Paul spent (1⁄8)th of his monthly salary for his car insurance and 2⁄5 more than the insurance for a refrigerator. After spending some amount on household expenses, he finally saved only (1⁄5)th portion of his salary. Then his monthly salary that was spent on household expenses was what fraction of the amount that was saved?
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Solution
Let Paul's monthly salary be 40 (8 × 5).
Money spent on car insurance = 5
Money spent on refrigerator = 5 + 2 = 7
∴ Money saved = 1⁄4 × 40 = 8
∴ He spent on household expenses $20 (40 - 5 -7 - 8).
∴ Required fraction is \(\frac{20}{5}=\frac{5}{2}\)
WEAR ASSOCIATES estimated that a brand new CNC machine worth $106,000 will depreciate $6000 in the first year and will thereafter depreciate $x every year. If the life of machine is 16 years and the machine will have a residual value $10000. What is the book value of machine after 10 years?
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Solution
Cost of brand new machine = 106,000.
Life is 16 years and residual value is $10,000 hence $ 96,000 will be the depreciation for 16 years.
Depreciation in first year is $6,000 and in the remaining 15 years $ 90,000. Thus annual depreciation is $6,000. Depreciation for 10 years will be $ 6,000 + (9 x $ 6,000) = $ 60,000.
Value at the end of 10 years is $ 106,000 - $ 60,000 = $ 46,000
At a certain New Year party, 1⁄3rd of the total number of men were married and the wives 5⁄7 of them were also at the party. There were 7 more women at the party than men. What is the least possible number of people attending the party?
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Solution
Let the total number of men at the party be 'm'.
Since 1⁄3 rd of them are married, m must be an integral multiple of 3. Also wives of 5⁄7 of them were at the party. So 1⁄3 of m has to be an integral multiple of 7. Hence the minimum possible number of men is 3 x 1⁄3 7 = 21 (Whenever you get fractions in the problem, multiply denominators and take that number as the
base figure.∴ 3 × 7 = 21 should be the minimum number of men who must attend the party).
∴ The minimum number of women = 21 + 7 = 28.
∴ Total number of people attending the party = 21 + 28 = 49
ABC corporation employed 1200 fresh engineering graduates this year. The training day at ABC consists of 8 sessions, each of which is conducted by one manager, and there is an average of 40 trainees per batch. If each manager conducts 5 of the 8 sessions and has 3 sessions free, but no trainee has free sessions, how many managers must the company have?
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Solution
Average number of trainees per batch = 40
∴ Total number of batches =\(\frac{1200}{40}=30\)
Each batch has 8 training sessions.
∴ Total number of-training sessions for 30 batches = 30 × 8 = 240
∴ Total number of managers required
= \(\frac{240}{5}\) = 48
Three types of tickets are available for a show. Lower class cost $10 each, middle class cost $15 each and upper class cost $25 each. There were L lower, M middle and U upper class tickets sold for the show. Which of the following gives the percentage of ticket proceeds due to sale 2⁄3 of M class tickets and 3⁄5 of U class tickets?
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Solution
Total revenue from lower class tickets = 10L
Similarly, total revenue from middle class and upper class tickets = 15M + 25U
∴ Total revenue from all tickets = 10L + 15M + 25U
Total proceeds from sale of 2⁄3 of middle class tickets and
3⁄5 of the upper class tickets= 2⁄3 × 15M + 3⁄5 × 25U = 10M + 15U
∴ Required % = \(\frac{10M+15U}{10L+15M+25U}\times 100\)
=\(\left ( \frac{2M + 3U}{2L+3M+5U} \right )\times 100\)
One alloy of metal contains 90% copper and 10% tin. Another alloy contains 95% copper and 5% tin. If they are mixed so that the mixture contains 9% tin, what percentages of copper will it contains.
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Solution
Alloy 1 contains 10% tin. Alloy 2 contains 4% tin. Mixture contains 9% tin. By allegation principle
∴ Alloy 1 and Alloy 2 must be mixed in the ratio 4 : 1.
∴ Copper content in the mixture is \(\left ( 4x\frac{90}{100} \right )+1\times \frac{95}{100}= 3.60+0.95 = 4.55\) parts
percentage 0 copper in the mixture is \(\frac{4.55}{5}\) × 100 = 91.00%
One hour after Yolanda started walking from x to y, a distance of 45 miles Bob started walking along same road from y to x. If Yolanda’s walking rate was 3 miles per hour and Bob’s was 4 miles per hour, how many miles had Bob walked when they met?
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Solution
Yolanda walked alone for 1 hour in which he covered 3 miles.
∴ Distance still to be covered by Yolanda and Bob together is 42 miles and then they meet each other.
It will be covered in \(\frac{42}{3+4}\) = 6 hours.
∴ Bob walked for 6 hours × 4 miles per hour = 24 miles
There were 36000 hardback copies of a certain novel sold before paperback version was issued. Since the launch of paperback copies 9 times of them as that of hardback were sold in the same period. If a total of 441000 copies of novel were sold in all, how many paperback copies were sold?
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Solution
Total copies sold 441000 less 36000 hard copies sold in the beginning.
∴ 405000 copies were sold after the paperback copies were released in the market. This includes.
paperback and hard copies in the ratio 9 : 1.
∴ paperback copies sold will be
\(\frac{405000}{10}\times 9=364500\)