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This test series consist of 10 questions of quantitative aptitude from the topic number system. There is no time limit for the test and no negative marking for wrong answer. The level of the question is set at medium. Its is to test your knowledge of the topic.

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- Question 1 of 10
##### 1. Question

A number when divided by a divisor leaves a remainder of 24.

When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor?CorrectLet the original number be ‘a’

Let the divisor be ‘d’

Let the quotient of the division of aa by d be ‘x’

Therefore, we can write the relation as a/d=xa/d=x and the remainder is 24.

i.e., a=dx+24

When twice the original number is divided by d,2a is divided by d.

We know that a=dx+24. Therefore, 2a=2dx+48

The problem states that (2dx+48)/d leaves a remainder of 11.

2dx is perfectly divisible by dd and will, therefore, not leave a remainder.

The remainder of 11 was obtained by dividing 48 by d.

When 48 is divided by 37, the remainder that one will obtain is 11.

Hence, the divisor is 37.IncorrectLet the original number be ‘a’

Let the divisor be ‘d’

Let the quotient of the division of aa by d be ‘x’

Therefore, we can write the relation as a/d=xa/d=x and the remainder is 24.

i.e., a=dx+24

When twice the original number is divided by d,2a is divided by d.

We know that a=dx+24. Therefore, 2a=2dx+48

The problem states that (2dx+48)/d leaves a remainder of 11.

2dx is perfectly divisible by dd and will, therefore, not leave a remainder.

The remainder of 11 was obtained by dividing 48 by d.

When 48 is divided by 37, the remainder that one will obtain is 11.

Hence, the divisor is 37. - Question 2 of 10
##### 2. Question

In a two digit number, the unit’s digit exceeds ten’s digit by 2 and the product of the sum of the digits and the number is 144, then the number is,

Correct144 being a small number, its 2 digit factors can be enumerated directly as 12, 16, 18, 24, 36, 48, and 72. Only 24 satisfies both the criteria.

Incorrect144 being a small number, its 2 digit factors can be enumerated directly as 12, 16, 18, 24, 36, 48, and 72. Only 24 satisfies both the criteria.

- Question 3 of 10
##### 3. Question

Find the second largest prime number that will divide 5329 with 6 added to it.

CorrectDividing by 11 yields, 485. Now only we divide by 5, the second factor, resulting in, 97 which is a prime number we know by experience.

But you can test also by checking divisibility first by 7, as it is already not divisible by 3. The test fails and as already we have tested for 11 square of which 121 is larger then 97, we know for sure that 97 is a prime number.

So, 5335=5×11×97

The second largest prime factor of 5335 is then 11.IncorrectDividing by 11 yields, 485. Now only we divide by 5, the second factor, resulting in, 97 which is a prime number we know by experience.

But you can test also by checking divisibility first by 7, as it is already not divisible by 3. The test fails and as already we have tested for 11 square of which 121 is larger then 97, we know for sure that 97 is a prime number.

So, 5335=5×11×97

The second largest prime factor of 5335 is then 11. - Question 4 of 10
##### 4. Question

The product of 4 consecutive even numbers is always divisible by:

CorrectThe product of 4 consecutive numbers is always divisible by 4!.

Since, we have 4 even numbers, we have an additional 2 available with each number.

Now, using both the facts, we can say that the product of 4 consecutive even numbers is always divisible by,

=(2^4)×4!

=16×24=16×24

=384IncorrectThe product of 4 consecutive numbers is always divisible by 4!.

Since, we have 4 even numbers, we have an additional 2 available with each number.

Now, using both the facts, we can say that the product of 4 consecutive even numbers is always divisible by,

=(2^4)×4!

=16×24=16×24

=384 - Question 5 of 10
##### 5. Question

The unit’s digit of 459×459+77×77+2785×2789 is,

CorrectThe unit’s digit of 459×459+77×77+2785×2789 is the sum of unit’s digit of each of the three terms. So the resultant unit’s digit is, the unit’s digit of,

9×9+7×7+5×9=81+49+45

Again, unit’s digit of a sum of numbers is the sum of the unit’s digit. So the desired unit’s digit is,

1+9+5=15, that is, 5.IncorrectThe unit’s digit of 459×459+77×77+2785×2789 is the sum of unit’s digit of each of the three terms. So the resultant unit’s digit is, the unit’s digit of,

9×9+7×7+5×9=81+49+45

Again, unit’s digit of a sum of numbers is the sum of the unit’s digit. So the desired unit’s digit is,

1+9+5=15, that is, 5. - Question 6 of 10
##### 6. Question

How many digits altogether are required to write the numbers from 1 to 60?

CorrectLeaving aside the first 9 single digit numbers, we find that the number set 10 to 19 form one set of two digit numbers, 20 to 29 the second set and 50 to 59 the fifth and last set, altogether 5 sets of 10 two digits numbers. The number of digits of these 5 ten sets is then 50×2=100.

We have left behind the first nine natural numbers and the last of the lot 60. Together these contribute another 11 digits.

So total number of digits is 111.IncorrectLeaving aside the first 9 single digit numbers, we find that the number set 10 to 19 form one set of two digit numbers, 20 to 29 the second set and 50 to 59 the fifth and last set, altogether 5 sets of 10 two digits numbers. The number of digits of these 5 ten sets is then 50×2=100.

We have left behind the first nine natural numbers and the last of the lot 60. Together these contribute another 11 digits.

So total number of digits is 111. - Question 7 of 10
##### 7. Question

A boy writes all the numbers from 100 to 999. The number of zeroes that he uses is ‘a’, the number of 5’s that he uses is ‘b’ and the number of 8’s he uses is ‘c’.

What is the value of b+c−a?

CorrectWe can see by symmetry b=cb=c and hence all we need to calculate bb and aa

b=280 and a=180

⇒ 2b−a=2b−a= 380IncorrectWe can see by symmetry b=cb=c and hence all we need to calculate bb and aa

b=280 and a=180

⇒ 2b−a=2b−a= 380 - Question 8 of 10
##### 8. Question

In three three digit numbers each of which when divided by 5 leaves a remainder of 3. The difference between the largest and smallest is twice the difference between the first two. If the sum of the three numbers is 429, the second largest among the three is,

CorrectThe second statement means the three numbers form a series of three numbers with same difference between each two adjacent numbers. Sum of such a set of three numbers will be three times the average middle number.

Dividing 429 by 3 we get, 143.IncorrectThe second statement means the three numbers form a series of three numbers with same difference between each two adjacent numbers. Sum of such a set of three numbers will be three times the average middle number.

Dividing 429 by 3 we get, 143. - Question 9 of 10
##### 9. Question

The product of two numbers is 120 and sum of their squares is 289. The difference between the numbers is,

Correct(a−b)^2=a^2−2ab+b^2

As a^2+b^2 and ab are given we have,

(a−b)^2=289−2×120=49, and

a−b=7Incorrect(a−b)^2=a^2−2ab+b^2

As a^2+b^2 and ab are given we have,

(a−b)^2=289−2×120=49, and

a−b=7 - Question 10 of 10
##### 10. Question

Which digits should come in place of @ and # if the number 62684@# is divisible by both 8 and 5?

CorrectSince the given number is divisible by 5, so 0 or 5 must come in place of #.

But, a number ending with 5 is never divisible by 8.

So, 0 will replace #.

Now, the number formed by the last three digits is 4@0, which becomes divisible by 8, if @ is replaced by 4.

Hence, digits in place of @ and # are 4 and 0 respectively.IncorrectSince the given number is divisible by 5, so 0 or 5 must come in place of #.

But, a number ending with 5 is never divisible by 8.

So, 0 will replace #.

Now, the number formed by the last three digits is 4@0, which becomes divisible by 8, if @ is replaced by 4.

Hence, digits in place of @ and # are 4 and 0 respectively.