\(9548 + 7314 = 8362 + x\). What is \(x\) ?

**Answer: **C

\(9548 + 7314 = 8362 + x\)

\(\therefore x = 9548 + 7314 - 8362\)

\(= 8500\)

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If the decimal representation of a number is non-terminating, non-repeating then the number is

**Answer: **D

No answer description available for this question.

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The largest 4 digit number exactly divisible by 88 is:

**Answer: **A

Largest 4-digit number = 9999

\(\frac{9999}{88} = 113\tfrac{55}{88}\)

Required number \(= (9999 - 55) = 9944\)

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What is the remainder when $3^7$ is divided by $8$?

**Answer: **C

No answer description available for this question.

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How many three digit numbers can be formed using the digits 1,2,3,4,5,6,7 and 8 without repeating the digits and such that the tens digit is greater than the hundreds digit and less than the units digit?

**Answer: **B

Ten's digit = 7 ⇒ units digit =8 ⇒ Hundred's digit =1,2,3,4,5,6.

⇒ Number of ways =1×6

Ten's digit = 6 ⇒ units digit =7,8 ⇒ Hundred's digit =1,2,3,4,5.

⇒ Number of ways =2×5

Ten's digit = 5 ⇒ units digit =6,7,8 ⇒ Hundred's digit =1,2,3,4

⇒ Number of ways =3×4

Ten's digit = 4 ⇒ units digit =5,6,7,8 ⇒ Hundred's digit =1,2,3.

⇒ Number of ways =4×3

Ten's digit = 3 ⇒ units digit =4,5,6,78 ⇒ Hundred's digit =1,2

⇒ Number of ways =5×2

Ten's digit = 2 ⇒ units digit =3,4,5,6,7,8 ⇒ Hundred's digit =1

⇒ Number of ways =6×1

Total number of ways =6+10+12+12+10+6= **56**.

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

**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=2b=2 or 77.

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

We also have that abab is divisible by 13.

The multiples of 13 are 13,26,39,52,65,7813,26,39,52,65,78 and 491491.

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

Hence, ab=52

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Three gold coins of weight 780gm, 840gm and 960gm are cut into small pieces, all of which have the equal weight. Each piece must be heavy as possible. If one such piece is shared by two persons, then how many persons are needed to give all the pieces of gold coins?

**Answer: **A

No answer description available for this question.

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Two prime numbers A, B

If A and B are twin primes with B>23, then which of the following numbers would always divide A+B?

**Answer: **A

Any prime number greater than 3 will be in the form of 6x+1 or 6x−1.

Thus, both prime number are twins:

Let first be 6x−1

and 2nd be 6x+1

Sum=12x

Thus it is always divisible by 12.

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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 divided by the divisor, the remainder obtained is 4.

The value of the divisor?

**Answer: **C

Let the divisor be dd.

When 242 is divided by the divisor, let the quotient be 'xx' and we know that the remainder is 8.

Therefore, 242=xd+8242=xd+8

Similarly, let yy be the quotient when 698 is divided by dd.

Then, 698=yd+9698=yd+9.

242+698=940=xd+yd+8+9242+698=940=xd+yd+8+9

940=xd+yd+17940=xd+yd+17

As xd and ydxd and yd are divisible by dd, the remainder when 940 is divided by dd should have been 17.

However, as the question states that the remainder is 4, it would be possible only when 17/d leaves a remainder of 4.

If the remainder obtained is 4 when 17 is divided by dd, then dd has to be **13**

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A girl wrote all the numbers from 100 to 200. Then she started counting the number of one's that has been used while writing all these numbers. What is the number that she got?

**Answer: **C

From 100 to 200 there are 101 numbers.

There are 100, 1's in the hundred place.

10, 1's in tens place and10, 1's in unit place.

Thus, the answer is 100+10+10=100+10+10= **120.**

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