Insta–DART (Daily Aptitude and Reasoning Test) 7 December 2020

Insta–DART (Daily Aptitude and Reasoning Test) 2020 - 21

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Considering the alarming importance of CSAT in UPSC CSE Prelims exam and with enormous requests we received recently, InsightsIAS has started Daily CSAT Test to ensure students practice CSAT Questions on a daily basis. Regular Practice would help one overcome the fear of CSAT too. We are naming this initiative as Insta– DART – Daily Aptitude and Reasoning Test. We hope you will be able to use DART to hit bull’s eye in CSAT paper and comfortably score 100+ even in the most difficult question paper that UPSC can give you in CSP-2021. Your peace of mind after every step of this exam is very important for us.

Looking forward to your enthusiastic participation (both in sending us questions and solving them on daily basis on this portal).

Wish you all the best ! 🙂

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Question 1 of 5

1. Question

When writing numbers from 1 to 10,000, how many times is the digit 9 written?

Correct

Answer : c

The digits 9 occurs in the thousands place in 1000 numbers.

It occurs in the hundreds place in 1000 numbers and so on

The digit occurs 4000 times.

Incorrect

Answer : c

The digits 9 occurs in the thousands place in 1000 numbers.

It occurs in the hundreds place in 1000 numbers and so on

The digit occurs 4000 times.

Question 2 of 5

2. Question

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

Correct

Answer : a

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

Answer : a

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.

Question 3 of 5

3. Question

How many natural numbers below 660 are divisible by 5 and 11 but not by 3?

Correct

Answer : a

If the number is divisible by 5 and 11 it must be divisible by 55.

The numbers are less than 660.

Hence, dividing 659 by 55 gives the number of multiples of 55 = 11 (ignoring fraction part).

The 11 multiples of 55 which are less than 560, but of these 11 multiples some can be multiples of 3.

The numbers of such, multiples is the quotient of 11 by 3.

Quotient of 113=3

Out of 11 multiples of 55, 3 are multiples of 3.

Hence, numbers less than 660 and divisible by 5 and 11 but not by 3=11−3=8

Incorrect

Answer : a

If the number is divisible by 5 and 11 it must be divisible by 55.

The numbers are less than 660.

Hence, dividing 659 by 55 gives the number of multiples of 55 = 11 (ignoring fraction part).

The 11 multiples of 55 which are less than 560, but of these 11 multiples some can be multiples of 3.

The numbers of such, multiples is the quotient of 11 by 3.

Quotient of 113=3

Out of 11 multiples of 55, 3 are multiples of 3.

Hence, numbers less than 660 and divisible by 5 and 11 but not by 3=11−3=8

Question 4 of 5

4. Question

2ab5 is a four-digit number divisible by 25. If the number formed from the two digits ab is a multiple of 13, then ab=?

Correct

Answer : c

We have given that the number 2ab52ab5 is divisible by 25.

Any number divisible by 25 ends with the last two digits 00, 25, 50, or 75.

So, b5b5 should equal 25 or 75.

Hence, b=2 or 7.

Since a is now free to take any digit from 0 through 9, ab can have multiple values.

We also have that ab is divisible by 13.

The multiples of 13 are 13,26,39,52,65,,26,39,52,65,78 ….

Among these, the only number ending with 2 or 7 is 52.

Hence, ab=52

Incorrect

Answer : c

We have given that the number 2ab52ab5 is divisible by 25.

Any number divisible by 25 ends with the last two digits 00, 25, 50, or 75.

So, b5b5 should equal 25 or 75.

Hence, b=2 or 7.

Since a is now free to take any digit from 0 through 9, ab can have multiple values.

We also have that ab is divisible by 13.

The multiples of 13 are 13,26,39,52,65,,26,39,52,65,78 ….

Among these, the only number ending with 2 or 7 is 52.

Hence, ab=52

Question 5 of 5

5. Question

The positive integers m and n leave remainders of 2 and 3, respectively, when divided by 6. m>n.

What is the remainder when m–n is divided by 6?

Correct

Answer : c

We are given that the numbers M and N ,when divided by 6, leave remainders of 2 and 3, respectively.

Hence, we can represent the numbers m and n as 6p+2 and 6q+3, respectively, where p and q are suitable integers.

Now,

m−n=(6p+2)−(6q+3)

=6p−6q−1

=6(p−q)−1

A remainder must be positive, so let’s add 6 to this expression and compensate by subtracting 6:

6(p−q)−1=6(p−q)−6+6−1

=6(p−q)−6+5

=6(p−q−1)+5

=6(p−q−1)+5

Thus, the remainder is 5

Incorrect

Answer : c

We are given that the numbers M and N ,when divided by 6, leave remainders of 2 and 3, respectively.

Hence, we can represent the numbers m and n as 6p+2 and 6q+3, respectively, where p and q are suitable integers.

Now,

m−n=(6p+2)−(6q+3)

=6p−6q−1

=6(p−q)−1

A remainder must be positive, so let’s add 6 to this expression and compensate by subtracting 6:

6(p−q)−1=6(p−q)−6+6−1

=6(p−q)−6+5

=6(p−q−1)+5

=6(p−q−1)+5

Thus, the remainder is 5

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