In an office,3⁄4 of staff can neither speak German nor French. However 1⁄5 can speak German and 1⁄6 can speak French. What part of whole staff can speak both?
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Solution
Let total no. of staff be 120. (Multiply all the denominators.)
∴ Staff who can speak German or French is 30 (1/4 of 120).
1/5 of 120 = 24 can speak German.
1/6 of 120 = 20 can speak French.
∴ T = G + F - Common
30 = 24 + 20 - Common
∴ Common = 14 who can speak both.
65% of 30% of p = q then p is approximately
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Solution
\(\frac{65}{100}\times \frac{30}{100}\times p=q\) ∴ P=\(\frac{10,000q}{65\times 30}\)
∴ \(P=\frac{200q}{39}\)
∴ p = 5q
A college has raised 75% of amount it needs for a new building by receiving an average donation of $60 from the people already solicited. The people already solicited represent 60% of people the college will ask for donations. If college is to raise exactly the amount needed for new building, how much must the remaining people donate per person?
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Solution
Let total no. of donors = 100
∴ 60 donors donating $60 each contributed to a total of $3600.
$3600 = 75% of actual total.
∴ Actual total = $4800 ∴ $1200 have been donated by 40 people.
∴ Donation per person =\(\frac{\$ 1200}{40}=\$ 30\)
A cloth manufacturer has determined that he can sell 100 blazers a week at a selling price of $200 each. For each rise of $4 in the selling price he will sell 2 blazers less every week. If in a certain week he sells blazers for $x each, how many dollars will he receive from sales of blazers in that week?
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Solution
First we need to work out the no. of $4 price raises. Since x - 200 is the increase in price,\(\left ( \frac{x-200}{4} \right )\) is the no. of $4 increases.
Thus the no. of blazers sold will decrease by
\(2\left ( \frac{x-200}{4} \right )\)
∴ The no. of blazers sold will be
100 - 2\(\frac{(x-200)}{4}\)= 100 - x⁄2 + 100 = 200 - x⁄2
So the amount receive is x\(\left (\frac{200-x}{2} \right )\)
=\(\frac{200x-x^{2}}{2}\)
BD ⊥ AC, AB = BC = a
Find area of Δ ABC
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Solution
By Pythagoras theorem, AD = CD = \(\sqrt{a^{2}-x^{2}}\)
∴ AD+ CD = 2\(\sqrt{a^{2}-x^{2}}\)
∴ Area (ΔABC) = \(\frac{x\times 2\sqrt{a^{2}-x^{2}}}{2}\)
=\(x\sqrt{a^{2}-x^{2}}\)
x > 2, Y > -1, then
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Solution
x > 2 ∴ -x < -2
Also y > -1 ⇒ 2y > -2
∴ - x < -2 and ∴ - 2 < 2y
∴ - x < -2 < 2y ⇒ - x < 2y
A wall with no windows is 11 ft. high and 20 ft. long. A large roll of wall paper costs $25 and will cover 60 square ft. of wall. A small roll of wallpaper costs $6 and will cover 10 square ft. of wall. What is the least cost for wallpaper enough to cover the wall?
(Only complete wallpaper is allowed to be pasted. Cut pieces are not allowed.)
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Solution
Area of wall = 11 x 20 = 220 sq. ft. 180 sq.It. can be covered by using 3 wallpapers of $25 each i.e. $75.
The remaining 40 sq.ft. will be covered by using 4 wallpapers of $6 each i.e. $24.
∴ Minimum cost = 75 + 24 = $99
In a group of people, solicited by charity, 30% contributed $40 each, 45% contributed $20 each and rest contributed $2 each. If charity received a total of $300 from people who contributed $2, how much was contributed by entire group?
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Solution
100 - (30 + 45) = 25% contributed $2
∴ (25% of x) x 2 = $300
∴ 25% of x = 150 ∴ x = 600 people
Total contribution = (30% of 600 × $40) + (45% of 600 × $20) + (25% of 600 × $2)
= 7200 + 5400 + 300
= $12,900
Mary, John and Karen ate lunch together. Karen’s meal cost 50% more than John’s meal and Mary’s meal cost 5/6 as much as Karen’s meal. If Mary paid $2 more than John, how much was the total that three of them paid?
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Solution
Let John's meal cost $x
∴ Karen's costs $1.5x
∴ Mary's costs $5⁄6 × 1.5 = $1.25x
∴ 125x - x = $2 ∴ 0.25x = 2 ∴ x = 8
∴ Total bill = 8(x + 1.5x + 1.25x)
= 8(3.75x) = $30
Jim’s weight is 140% of Marcia’s weight. Bob’s weight is 90% of Lee’s weight. Lee weighs twice as much as Marcia. What % of Jim’s weight is Bob’s weight?
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Solution
Let Marcia's weight =100 pounds.
∴ Jim's weight = 140 pounds (1.4 × 100)
∴ Lee's weight = 200 pounds (2 × 100)
∴ Bob's weight = 180 pounds (0.9 × 200)
The required % = \(\frac{180}{140}\) × 100 = 9⁄7 × 100
\(\frac{900}{7}=128\frac{4}{7}\)