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# Problems on Numbers Questions

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A subscriber is allowed to make a certain number of telephone calls for a lumpsum charge of Rs.300. Beyond that, he is charged at a certain rate per call. Two subscribers together make 1400 calls, and were charged Rs.425 and Rs.925 respectively. If a single person had made all of the 1400 calls he would have been charged Rs.1550. How many telephone calls are allowed for the first Rs.300?

A300

B350

C400

D450

Explanation:

Here is no explanation for this answer

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If each of the three nonzero numbers a , b , and c is divisible by 3, then abc must be divisible by which one of the following the number

A8

B27

C81

D121

Explanation:

Since each one of the three numbers a, b, and c is divisible by 3, the numbers can be represented as 3p,3q, and 3r, respectively, where p, q, and r are integers.
The product of the three numbers is 3p*3q*3r =27(pqr).
Since p, q, and r are integers, pqr is an integer and therefore abc is divisible by 27

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If n = 4p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n ?

ATwo

BThree

CFour

DSix

EEight

Explanation:

since p is prime greater than 2, then p=odd, thus 4p=even, which means that it has 4 even divisors: 2, 4, 2p, and 4p.

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There are equal number of boys and girls in a class. If 12 girls entered out, twice the boys as girls remain. What was the total number of students in a class?

A48

B44

C40

D34

Explanation:

Lets assume total number of boys = y
Total number of girls = x
Given:
Equal number of girls and boys
=> x/y=1/1 => x = y
If 12 girls are entered out,
=> x-12 = x/2
=> x=2x-24 => x=24
So, girls x=24 and boys y=x=24
So, total number of students = x+y
=> 24+24 => 48

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What is the lowest possible integer that is divisible by each of the integers 1 through 7, inclusive?

A440

B420

C410

D320

Explanation:

number from 1 to 7, that number should be L.C.M of(1,2,3,4,5,6,7) = 420

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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

A110 to 120

B100 to 130

C110 to 130

D90 to 110

Explanation:

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?

A17

B15

C11

D19

Explanation:

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?

A27 sec.

B20 sec.

C22 sec.

D25 sec.

Explanation:

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?

A980

B797

C955

D618

Explanation:

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?

A9

B3

C5

D7

Explanation:

Let 'N' be the given number. N=357k+5 = 17*21k+5

If this number is divided by 17 remainders is 5 as 357k is exactly divided by 17.

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