In a certain set of weights for each positive integer n less than 10, every weight weighs 2n grams. What is the least number of such weights with a combined weight of 108 grams?
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Solution
For the nine positive integers less than 10, weights are
21,22,23,24.........29
∴ To add to 108 grams, we need 4 weights, 26, 25, 23, 22
i.e. 64 + 32 + 8 + 4 = 108 grams.
If one of the following is product of two 3 digit integers 8K7 and 6L9, then the product must be
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Solution
Consider multiplication of 8K7 and 6L9.
Multiplication of digit in units place = 7 × 9 = 63.
∴ The number must end in 3.
So (C), (D) and (E) are ruled out.
Now consider product of digits in 100's place
= 8 ∴ 6 = 48
∴ The number must begin with 48, 49, 50, depending upon the carry.
Therefore (A)cannot be the answer as it begins with 47
On a number line, what no. is halfway between \(\frac{1}{75}\) and \(\frac{1}{50}\)
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Solution
Number that is half-way \(\frac{\frac{1}{50}+\frac{1}{75}}{2}=\frac{1}{60}\)
A dress shop is having a sale in which for every dress purchased at the full price, a second dress that has the same price or lower price may be purchased for $1. If during the sale, a customer buys 6 dresses priced at $40, $42, $48, $50, $52, $60, what is the least amount the customer can expect to pay for dresses?
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Solution
If the customer buys 3 dresses worth $42, $50 and $60, he can get remaining 3 dresses worth $40, $48, and $60 at $1 each.
∴ 60 + 50 + 42 + (1 × 3) = $155 is the minimum amount he pay.
Find maximum area of a rectangle with perimeter 32.
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Solution
Perimeter = 32.
∴ 2(1+W) = 32 ∴ 1 + w = 16
A rectangle has the maximum area when it transforms into a square.
∴ 1 = w = 8
∴ Area = 64
x is a positive odd integer and y is a negative even integer, which of the following could be a negative odd integer?
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Solution
(A) ⇒ yx = (negative even) positive odd
∴ yx will be negative even integer.
(B) ⇒ xy = negative even integer.
(C) ⇒ x - y would be definitely positive.
(D) X2 + Y ⇒ (+ve odd no.) + (-ve even number)
The sum is defmitely odd but it can be +ve or -ve depending upon magnitudes of X2 and y. So it could be a possibility.
(E) x + y2 is depending positive.
0 is the centre of the circle below. What fraction of the circular region is shaded?
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Solution
Since vertically opposite angles are equal sum of central angles of two shaded regions
= 360 - 2(150) = 60
∴ \(\frac{60}{360}\) = 1⁄6 is required fraction.
\(\frac{0.0015\times 10^{m}}{0.03\times 10^{k}}=5\times 10^{7}\) ∴ m – k = ?
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Solution
\(\frac{15\times 10^{-4}}{3\times 10^{-2}}\times 10^{m-k}= 5\times 10^{-2} \times 10^{m-k}\)
∴ 5 × 10m-k-2 = 5 × 107
∴ m - k - 2 = 7
∴ m - k = 9
Water is poured into empty cylindrical tank at a constant rate for 6 minutes. After the water has been poured into the tank, depth of water is 10 ft. Radius of tank is 7 ft. Which of the following is the best approximation for rate at which water was poured into the tank?
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Solution
Volume of cylindrical tank = πr2h
= \(\frac{27}{7}\times 7\times 7\times 10= 1540\)
∴ 1540 cubic ft. of water is poured in 6 minutes.
∴ Water poured per minute
= \(\frac{1540}{6}\) = 256.67
p and q are real numbers.
Let f(p,q) =\(\frac{Pq}{P^{2}+q^{2}},\)Let g(p) =\(\frac{p^{2}+2}{3}\) Then \(\frac{f[3, g(4)]}{f(2,4)}\)=
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Solution
g(4)=\(\frac{(4)^{2}+2}{3}=\frac{18}{3}=6\)
∴ f(3,6) = \(\frac{3\times 6}{(3)^{2}+(6)^{2}}=\frac{18}{45}=\frac{2}{5}\)
∴ Numerator = 2⁄5
Denominator = [(2, 4)
∴ f(2 4)= \(\frac{2\times 4}{4+16}=\frac{8}{20}=\frac{2}{5}\)
∴ \(\frac{Numerator}{Denominator}=\frac{\frac{2}{5}}{\frac{2}{5}}=1\)