If half of the number of white mice in a certain lab is 1⁄8 of the total number of mice and 1⁄3 of the total number of gray mice is 1⁄9 of the total number of mice, then what is the ratio of white mice to gray mice in the laboratory?
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Solution
1⁄2 no. of white mice = 1⁄8 number of total mice
∴ Total number of white mice = 1⁄4 number of total mice.
Similarly if 1⁄3 rd number of gray mice = 1⁄9 th number of total mice
⇒ total number of gray mice = 1⁄3 rd number of total mice.
∴ White mice: Gray mice = 1⁄4 : 1⁄3..... multiply both by 12,
Ratio is 3 : 4
A vendor buys milk at a certain price, adds water and sells adulterated milk at the same rate as he bought it for. He makes a 30% profit. What is the percentage of water that he adds to the milk?
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Solution
Let us assume that vendor purchased 100 liters of milk @ $1 per liter. Therefore the cost is $100. He makes 30% profit. Thus the selling price of milk is $130. The selling price per liter is same as cost price per liter i.e. $1 per liter. It means he sells 130 liters of adulterated milk. This includes 100 liters of pure milk and 30 liters of water.
(Vendor is making 30% profit by adding water, but keeping the same price. This indicates that he adds 30% water i. e. 30 ltr.)
a = 2b, 1⁄2b = c, 4c = 3d . Find the ratio of d to a.
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Solution
a = 2b, b = 2c ∴ a = 4c
Also 4c = 3d ∴ a = 3d ∴ d : a = 1 : 3
Two containers have volumes of x and y respectively and are each one third full. Contents of both are poured into the container of volume z, What part of z is full?
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Solution
Volume of contents in container 1 and 2 are x⁄3 and y⁄3
Volume of contents, when they are mixed in the third container = \(\frac{x + Y}{3}\)
∴ Part of z which is full =\(\left (\frac{x + Y}{3} \right )\)x 1⁄z = \(\frac{x + Y}{3z}\)
A man is traveling by car at the rate of 40kmph. After every 80km, he rests for 20 minutes. How long will he take to cover a distance of 240km?
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Solution
The man rests for 20 minutes after traveling 80 km. So he would take 2 halts before he travels 240km.
So time required = \(\frac{240}{40}\) hrs. + 40 min.
= 6 hrs. 40 min.
A man travels 35km partly at 4kmph and partly at 5kmph. If he covers former distance also at 5kmph he could cover 2km more in the same time. The time taken to cover whole distance at original rate is
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Solution
Time=\(\frac{Distance}{Speed}\)
In the first case the distance covered at 4 kmph should be x.
∴ Time require Will be \(\frac{x}{4}+\frac{35-x}{5}\)
In the second case the distance covered is 37 km.
∴ Time required Will be \(\frac{37}{5}\)
Time taken in both the cases is the same.
∴ \(\frac{x}{4}+\frac{35-x}{5}=\frac{37}{5}\) (LCM = 20)
∴ 5x + 140 - 4x = 148
∴ x = 8
∴ Time required with the earlier speed is
\(\frac{8kmph}{4kmph}\) = 2hrs. \(\frac{27km}{5kmph}\) = 5hrs 24 min.
Total time 7 hrs 24 min
The average of 3 prime numbers is 223 . What is the difference between the greatest and the smallest \(\frac{223}{3}\) number?
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Solution
At least two sets are possible, 71, 73, 79 and 67,73,83.
Mr. Confusion wanted to buy both pens as well as pencils. He had $190 with him. If every pen costs $15 and every pencil costs $8, what is the maximum number of items that he can buy if he has to exhaust all the money he has?
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Solution
Maximum number of items can be purchased if we buy minimum number of pens.
If we buy one pen of $15, balance amount will be $175 which is not exactly divisible by 8. If we buy two pens of $30, balance amount will be $160 and we can buy 20 pencils of $8 each. Hence answer is 22, i.e. 2 pens and 20 pencils.
If pq is a perfect square as well as a perfect cube, where p and q are natural numbers then q must be a multiple of
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Solution
From given information nothing can be said about the values of p and q as
26 = 43 = 82 = 641 = 64
∴ p can take values 2, 4, 8,64 q can take values 1,2,3,6
For any natural number n, n4 + n2 + 1 is always
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Solution
If n is even, even + even + 1 = odd
If n is odd, odd + odd + 1 = odd
So n4 + n2 + 1 is always odd.