# Quantitative Aptitude Daily Online Free Practice Quiz for 9th Sep 2019

by Admin · 09/09/2019

## Quantitative Aptitude Quiz 9th Sep 2019

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This quiz is practice for your Quantitative Aptitude knowledge, the practice set has 10 question based on the latest syllabus of Quantitative Aptitude of competitive exams, all in multiple choice questions MCQ, there is no negative marking for wrong answer, and you can also check the answer of the question

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

1 pointsA 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?Correct

Let 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.Incorrect

Let 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

1 pointsIn 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,

Correct

144 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.

Incorrect

144 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

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

Correct

Dividing 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.Incorrect

Dividing 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

1 pointsThe product of 4 consecutive even numbers is always divisible by:

Correct

The 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

=384Incorrect

The 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

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

Correct

The 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.Incorrect

The 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

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

Correct

Leaving 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.Incorrect

Leaving 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

1 pointsA 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?

Correct

We 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= 380Incorrect

We 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

1 pointsIn 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,

Correct

The 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.Incorrect

The 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

1 pointsThe 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

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

Correct

Since 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.Incorrect

Since 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.