## Quants Number System H

Question 1 |

What is the minimum number of square marbles required to tile a floor of length 5 metres 78 cm and width 3 metres 74 cm?

187 | |

176 | |

540 | |

748 |

Question 1 Explanation:

As we have to use a whole number of marbles, the side of the square should a factor of both 5 m 78 cm and 3m 74. And it should be the highest factor of 5 m 78 cm and 3m 74.
5 m 78 cm = 578 cm and 3 m 74 cm = 374 cm.
The HCF of 578 and 374 = 34.
Hence, the side of the square is 34.
The number of such square marbles required,
=578×37434×34=578×37434×34
=17×11==17×11= 187 marbles

Question 2 |

Find the number, when 15 is subtracted from 7 times the number, the result is 10 more than twice of the number

5 | |

7.5 | |

4 | |

15 |

Question 2 Explanation:

Let the number be x.
7x -15 = 2x + 10 => 5x = 25 => x = 5

Question 3 |

What is the least value of x, so that 23x57 is divisible by 3?

1 | |

0 | |

2 | |

3 |

Question 3 Explanation:

Hint: SUM of all digits should be divisible by 3.

Question 4 |

A set M contains element all even number between 1 and 23 and all odd numbers 24 and 100. if all the elements of the set multiplied than how many trailing 0, resulting number will contain?

10 | |

12 | |

9 | |

11 |

Question 4 Explanation:

the result trailing 0 occurs when one 2 and one 5 multiplied.
here we see the set M:
( 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 25, 35, 45, 55, 65, 75, 85, 95 )
after 22 I have taken only the term which contains factor of 5
now we find prime factor
2 = 2
4 = 2 * 2
6 = 2 * 3
8 = 2 * 2 * 2
10 = 2 * 5
12 = 2 * 2 * 3
14 = 2 * 7
16 = 2 * 2 * 2 * 2
18 = 2 * 3 * 3
20 = 2 * 2 * 5
22 = 2 * 11
25 = 5 * 5
35 = 5 * 7
45 = 5 * 9
55 = 5 * 11
65 = 5 * 13
75 = 5 * 5 * 3
85 = 5 * 17
95 = 5 * 19
here we find
total number of 5 is 12
and
total number of 2 is 19
and
we get total number of 5 * 2 pair is 12
so number of trailing zero will be 12

Question 5 |

Find remainder of (9^1+9^2+.........+9^n)/6
n is multiple of 11.

0 | |

5 | |

3 | |

can't be determined |

Question 5 Explanation:

9/6 remainder is 3
9^2/6 remainder is 3
9^3/6 remainder is 3
9 to the power of any number when divided by 6 ,the remainder will always be 3.
Now, ( 3+3+3 ......11 times)/6 =(3*11)/6; therefore the remainder will be 3.
If we take the even multiple of 11, then remainder will be zero.
Therefore answer is cannot be determine.

Question 6 |

Which of the following numbers must be added to 5678 to give a remainder of 35 when divided by 460?

955 | |

980 | |

797 | |

618 |

Question 6 Explanation:

(5678+x)=(460*y)+35
5678-35+x=460*y
5643+x=460*y
so in x last digit must be 7 so ans should be c

Question 7 |

What is the unit digit of(2^3)^123456

2 | |

4 | |

6 | |

9 |

Question 7 Explanation:

(2^3^)^123456=2^3*123456=2^370368
2 cycle size=4
so,370368/4=remainder(0)
2 unit values=2,4,8,6=so,ans is 6

Question 8 |

IF a=0,b=1,c=2.....................z=25
Then one+one=?

aone | |

baed | |

btwo
| |

none |

Question 8 Explanation:

This problem is based on Base 26 rather than regular base 10 (decimal system) that we normally use. In base 10 there are 10 digits 0 to 9 exist. In base 26 there are 26 digits 0 to 25 exist. To convert any number into base 26, we have to divide the number with 26 and find the remainder. (Study this Base system chapter).
Here, ONE + ONE =
E has value of 4. So E + E = 8 which is equal to I.
Now N + N = 13 + 13 = 26. But in base 26, there is no 26. So (26)10=(10)26(26)10=(10)26
So we put 0 and 1 carry over. But 0 in this system is A.
Now O + O + 1 = 14 + 14 + 1 = 29
Therefore, (29)10=(13)26(29)10=(13)26
But 1 = B and 3 = D in that system. So ONE + ONE = BDAI

Question 9 |

how many squares are there in series 2006,2013,2020,2027....2300?

1 | |

2 | |

3 | |

4 |

Question 9 Explanation:

as there r only 3 squares b/w 2006-2300 that are 45^2=2205, 46^2=2116, 47^2=2209
now this series is an ap with 7 differnce.
so the difference of these squares & 2006 must be a multiple of 7.
only 2209-2006=203 which is multiple of 7.
so correct ans is 1.

Question 10 |

99^n is such a number begin with 8, least value of n?

11 | |

10 | |

9 | |

n does not exist |

Question 10 Explanation:

99(100 - 1) = 9900-99= 9801
9801(100 - 1) = 980100-9801= 971299
971299(100 - 1) = 97129900 - 971299 = 96157601
.....
.....
observe the pattern, 98, 97, 96, .... for power of 2, 3, 4, ...
So for 90 the power could be 10.
For 11,we get a number starts with 8.
or his is not rigorous, but this is what I will do if I were asked such a question in an exam.
We want (100−1)n=100n−(n1)100n−1+…(100−1)n=100n−(n1)100n−1+….
These are really the terms that contribute to the answer, so definitely n<10n<10 is not possible. Even for n=10n=10, we get a number starting with '9' here and the subsequent term is +ve and everything together is going to have a positive effect.
This leaves us with n=11n=11 which I have verified as the correct value.
More Intuition
We know that 99×99=980199×99=9801. Now when we further multiply, the essential digits come from 98×9998×99 which is 97029702 (ignore the 99 at the end). One more multiplication with 9999 and the product will start with 96…96…. This goes on and at n=11n=11, we get the value 89…89…. (This was essentially explained using the binomial expansion above)

Question 11 |

What is the next number of the following sequence 7, 14, 55, 110, ..?

179 | |

178 | |

456 | |

148 |

Question 11 Explanation:

7 + 7 = 14
14 + 41 = 55
55 + 55 = 110
110 + 011 = 121

There are 11 questions to complete.

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