College T has 1000 students. Of the 200 students majoring in one or more of the sciences,130 are majoring in Chemistry and 150 are majoring in Biology. If at least 30 of the students are not majoring in either Chemistry or Biology, then the number of students majoring in both Chemistry and Biology could be any number from
If we assume exactly 30 students are not majoring in any subject then the students who take atleast one subject = 200 - 30 = 170
We know that n(A?B) = n(A) + n(B) - n(A?B)
=> 170 = 130 + 150 - n(A?B)
Solving we get n(A?B) = 110.
i.e., Students who can take both subjects are 110
But If more than 30 students are not taking any subject, what can be the maximum number of students who can take both the subjects?
As there are 130 students are majoring in chemistry, assume these students are taking biology also. So maximum students who can take both the subjects is 130
So the number of students who can take both subjects can be any number from 110 to 130
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The present ratio of students to teachers at a certain school is 30 to 1. If the student enrollment were to increase by 50 students and the number of teachers were to increase by 5, the ratio of the teachers would then be 25 to 1 What is the present number of teachers?
Given the ratio of students and teachers = 30:1
=> students and teachers are in school 30x, x respectively.
After new recruitments of students and teachers the strength becomes (30x + 50), (x + 5) respectively.
So, the new ratio = 25 : 1
(30x + 50)/(x + 5) = 25/1
=> 30x + 50 = 25x + 125
=> 5x = 75 => x = 15
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Machine A produces bolts at a uniform rate of 120 every 40 second, and Machine B produces bolts at a uniform rate of 100 every 20 seconds. If the two machines run simultaneously, how many seconds will it take for them to produce a total of 200 bolts?
Machine A produces 120/40 = 3 bolts per sec.
Machine B produces 100/20 = 5 bolts per sec.
So, togather both will produce 8 bolts per second.
Hence, they wil take 200/8 = 25 seconds to produce 200 bolts.
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Which of the following numbers must be added to 5678 to give a reminder 35 when divided by 460?
Let x be the number to be added to 5678.
When you divide 5678 + x by 460 the remainder = 35.
Therefore, 5678 + x = 460k + 35 here k is some quotient.
=> 5643 + x should exactly divisible by 460.
Now from the given options x = 797.
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If a number is divided by 357 the remainder is 5, what will be the remainder if the number is divided by 17?
From the given information, (272738 - 13, 232342 - 17) are exactly divisible by that two digit number.
We have to find the HCF of the given numbers 272725, 232325.
HCF = 25.
So sum of the digits = 7.
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In a series of numbers , the next number is formed by adding 1 to the sum of the previous numbers, and the 10th number is 1280. Then what is the first number in the series? (series will be like this x , x+1, (x+(x+1))+1,....... )
The given series is x, x + 1, 2x + 2, 4x + 4 ......
If you observe the pattern here, the coefficient of x + 1 is in the powers of 2. So 4th term has a power of 2, 5th term has a power of 3... 10th term has a power of 8. So tenth term would be 2^8(x + 1)
= 256(x+1).
Given 256(x+1) = 1280
x = 4.
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The number of multiples of 10 which are less than 1000, which can be written as a sum of four consecutive integers is
We can write 10 = 1 + 2 + 3 + 4. So we have to find how many multiples of 10 can be written in this manner.
Let the first of the four numbers be n. So
n + (n+1) + (n+2) + (n+3) = 10k
4n + 6 = 10k
2n + 3 = 5k
n = (5k-3)/2
=2k-1+(k-1)/2
So n is integer for k = an odd number. So for k = 1, 3, 5, .... 99 we can write a number as a sum of four consecutive intezers.
So there are 50 numbers.
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Mr. Bean chooses a number and he keeps on doubling the number followed by subtracting one from it, if he chooses 3 as initial number and he repeats the operation for 30 times then what is the final result?
Let the number be N.
So N = 406x + 115.
Now divide this number by 29. As 406 is exactly divisible by 29, we have to divide 115 by 29 and find the remainder. So remainder = 28
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