## 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?
 A 187 B 176 C 540 D 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
 A 5 B 7.5 C 4 D 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?
 A 1 B 0 C 2 D 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?
 A 10 B 12 C 9 D 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.
 A 0 B 5 C 3 D 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?
 A 955 B 980 C 797 D 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
 A 2 B 4 C 6 D 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=?
 A aone B baed C btwo D 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?
 A 1 B 2 C 3 D 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?
 A 11 B 10 C 9 D 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,  ..?
 A 179 B 178 C 456 D 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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