Quant - Number System Ex. 4

Exercise4 + Answers + Explanation

Question:
When 242 is divided by a certain divisor the remainder obtained is 8. When 698 is divided by the same divisor the remainder obtained is 9. However, when the sum of the two numbers 242 and 698 is divided by the divisor, the remainder obtained is 4. What is the value of the divisor?

(1) 11
(2) 17
(3) 13
(4) 23

Correct Answer - (3)

Explanation:
Let the divisor be d.
When 242 is divided by the divisor, let the quotient be ‘x’ and we know that the remainder is 8. Therefore, 242 = xd + 8

Similarly, let y be the quotient when 698 is divided by d.
Then, 698 = yd + 9.

242 + 698 = 940 = xd + yd + 8 + 9
940 = xd + yd + 17
As xd and yd are divisible by d, the remainder when 940 is divided by d should have been 17.
However, as the question states that the remainder is 4, it would be possible only when leaves a remainder of 4.
If the remainder obtained is 4 when 17 is divided by d, then d has to be 13.

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Question:
What is the total number of different divisors of the number 7200?

(1) 20
(2) 4
(3) 54
(4) 32

Correct Answer - (3)

Explanation:
Express 7200 as a product of prime numbers. We get 7200 = 25 * 32 * 52.
The number of different divisors of 7200 including 1 and 7200 are (5 + 1)(2 + 1)(2 + 1) = 6 * 3 * 3 = 54

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Question:
When a number is divided by 36, it leaves a remainder of 19. What will be the remainder when the number is divided by 12?

(1) 10
(2) 7
(3) 192
(4) None of these

Correct Answer - (2)

Explanation:
Let the number be ‘a’.
When ‘a’ is divided by 36, let the quotient be ‘q’ and we know the remainder is 19
=> and remainder is 19
=> a = 36q + 19

now, a is divided by 12

=> 36q is perfectly divided by 12
Therefore, remainder = 7

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Question:
A person starts multiplying consecutive positive integers from 20. How many numbers should he multiply before the will have result that will end with 3 zeroes?

(1) 11
(2) 10
(3) 6
(4) 5

Correct Answer - (3)

Explanation:
A number will end in 3 zeroes when it is multiplied by 3 10s. To get a 10, one needs a 5 and a 2.
Therefore, this person should multiply till he encounters three 5s and three 2s.
20 has one 5 (5 * 4) and 25 has two 5s (5 * 5).
20 has two 2s (5 * 2 * 2) and 22 has one 2 (11 * 2).

Therefore, he has to multiply till 25 to get three 5s and three 2s, that will make three 10s. So, he has to multiply from 20 to 25 i.e. 6 numbers.

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Question:
How many four digit numbers exist which can be formed by using the digits 2, 3, 5 and 7 once only such that they are divisible by 25?

(1) 4! - 3!
(2) 4
(3) 8
(4) 6

Correct Answer - (2)

Explanation:
The numbers that are divisible by 25 are the numbers whose last two digits are ‘00′, ‘25′, ‘50′ and ‘75′.
However as ‘0′ is not one of the four digits given in the problem, only numbers ending in ‘25′ and ‘75′ need to be considered.

Four digit number whose last two digits are ‘25′ using 3 and 7 as the other two digits are 3725 and 7325.
Similarly, four digit numbers whose last two digits are ‘75′ using 2 and 3 as the other two digits are 2375 and 3275.

Hence there are four 4-digit numbers that exist.

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