SOLUTIONS FOR NUMBER SYSTEM PART 2

                                       

1)(112 x 5^4) = ?
A. 67000
B. 70000
C. 76500
D. 77200
Answer: Option B

Explanation:
(112 x 54) = 112 x 10 ^4 = 112 x (10/2)^4 = 1120000/16 = 70000





2) It is being given that (2^32 + 1) is completely divisible by a whole number. Which of the following numbers is completely divisible by this number?
A. (216 + 1)
B. (216 - 1)
C. (7 x 223)
D. (296 + 1)


Answer: Option D

Explanation:

Let 232 = x. Then, (232 + 1) = (x + 1).

Let (x + 1) be completely divisible by the natural number N. Then,

(296 + 1) = [(232)3 + 1] = (x3 + 1) = (x + 1)(x2 - x + 1), which is completely divisible by N, since (x + 1) is divisible by N.

3)A number when divided by 114,leaves remainder 21.If the same number is divided by 19,find the remainder
(s.s.c.,2010)
A.6
B.4
C.2
D.8
answer:2
as we know the formula dividend=divisior*quotient +remainder apply it

4. What least number must be added to 1056, so that the sum is completely divisible by 23 ?
A. 2
B. 3
C. 18
D. 21
E. None of these


Answer: Option A

Explanation:

 23) 1056 (45
      92
      ---
      136
      115
      ---
       21
      ---
    
 Required number = (23 - 21)   
                 = 2. 

5.   1397 x 1397 = ?
A. 1951609
B. 1981709
C. 18362619
D. 2031719
E. None of these
Answer: Option A

Explanation:
1397 x 1397 = (1397)2
= (1400 - 3)2
= (1400)2 + (3)2 - (2 x 1400 x 3)
= 1960000 + 9 - 8400
= 1960009 - 8400
= 1951609.

6. 5358 x 51 = ?
A. 273258
B. 273268
C. 273348
D. 273358
Answer: Option A

Explanation:
5358 x 51 = 5358 x (50 + 1)
= 5358 x 50 + 5358 x 1
= 267900 + 5358
= 273258.




7.The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder. What is the smaller number ?
A. 240
B. 270
C. 295
D. 360


Answer: Option B

Explanation:

Let the smaller number be x. Then larger number = (x + 1365).

x + 1365 = 6x + 15

5x = 1350

x = 270

Smaller number = 270.


8. (?) - 19657 - 33994 = 9999
A. 63650
B. 53760
C. 59640
D. 61560
E. None of these


Answer: Option A

Explanation:

 19657         Let x - 53651  = 9999
 33994         Then, x = 9999 + 53651 = 63650
 -----
 53651
 -----
 Let x - 53651  = 9999
       Then, x = 9999 + 53651 = 63650
9. The sum of first 45 natural numbers is:
A. 1035
B. 1280
C. 2070
D. 2140
Answer: Option A

Sn = n(n + 1)/2 = 45(45 + 1)/2 = 1035.

10.  753 x 753 + 247 x 247 - 753 x 247 (numerator)/(denominator)= ?
     753 x 753 x 753 + 247 x 247 x 247
A. 1/1000
B. 1/506
C. 253/500
D.None of these
Answer: Option A

Explanation:
Given Exp. = (a^2 + b^2 - ab)/ (a^3 + b^3)= 1 /(a+b) = 1/(753+247) =1/1000
                              
11. If the number 481 * 673 is completely divisible by 9, then the smallest whole number in place of * will be:
A. 2
B. 5
C. 6
D. 7
E. None of these


Answer: Option D

Explanation:

Sum of digits = (4 + 8 + 1 + x + 6 + 7 + 3) = (29 + x), which must be divisible by 9.

   x = 7.

12. The difference between a positive proper fraction and its reciprocal is 9/20. The fraction is:
A. 3/5
B. 3/10
C.4/5
D.4/3


Answer: Option C

Explanation:

Let the required fraction be x. Then 1 /x - x = 9/20
                                              

    1/x - x^2 /x = 9/20

     20 - 20x^2 = 9x

     20x^2 + 9x - 20 = 0

     20x^2 + 25x - 16x - 20 = 0

     5x(4x + 5) - 4(4x + 5) = 0

     (4x + 5)(5x - 4) = 0

x = 4/5

13.On dividing a number by 56, we get 29 as remainder. On dividing the same number by 8, what will be the remainder ?
A. 4
B. 5
C. 6
D. 7


Answer: Option B

Explanation:

Formula: (Divisor*Quotient) + Remainder = Dividend.

Soln:

(56*Q)+29 = D -------(1)

D%8 = R -------------(2)

From equation(2),

((56*Q)+29)%8 = R.

=> Assume Q = 1.

=> (56+29)%8 = R.

=> 85%8 = R

=> 5 = R.

14. If n is a natural number, then (6n^2 + 6n) is always divisible by:
A. 6 only
B. 6 and 12 both
C. 12 only
D. by 18 only


Answer: Option B

Explanation:

(6n2 + 6n) = 6n(n + 1), which is always divisible by 6 and 12 both, since n(n + 1) is always even.

15.What will be remainder when (67^67 + 67) is divided by 68 ?
A. 1
B. 63
C. 66
D. 67


Answer: Option C

Explanation:

(xn + 1) will be divisible by (x + 1) only when n is odd.

(6767 + 1) will be divisible by (67 + 1)

(6767 + 1) + 66, when divided by 68 will give 66 as remainder.

16. what is the place value of 5 in 3254710?(clat,2010)
A.5
B.10000
C.50000
D.54710
answer:c
expl:hence 5 is in tenthousands place so 5*10000=500000
17.The face value of 8 in the number 458926 is
A.8
B.1000
C.8000
D.8926
answer:A
expl:face value remains its same place
18.The sum of the place values of 3 in the number 503535 is ?(MBA,2005)
A.6
B.60
C.3030
D.3300
answer:c
exp:3000+30=3030
19.The sum of greatest and smallest number of five digits is
A.11,110
B.10,999
C.109,999
D.11,100
answer:c
expl:99999+100000=109999
20.what is the minimum number of four digits formed by using the digits 2,4,0,7
A.2047
B.2247
c.2407
D.2470
answer:A
(20-25) are the previous bits from RRB
21.if a and b are two numbers such that ab=0,then
A.a=0 and b=0
B.a=0 or b=0 or both
C.a=0 and b not equal to 0
$.b=0 and a not equal to 0
answer:d
exp=ab=0 a=0 or b=0 or both are zeroes
22.which of the following is a prime number
a.143
b.289
c.117
d.359
answer: d 359  which is divisible by 13
23. if you subtract -1 from +1 ,what will be the result?
a.-2
b.0
c.1
d.2
answer:d
24.217*217+183*183=?    (*=multiplication)
a.79698
b.80578
c.80698
d.81268
answer:b
25.(999)^2-(998)^2=?
a.1992
b.1995
c.1997
d.1998
answer:c apply (a+b) (a-b)=(a^2-b^2)
(26-30) are the previous bits from (BANK PO)
26.1260/14/9=?
a.9
b.10
c.81
d.810
answer:10
27.136*12*8=?(*=multiplication)
a.12066
b.13046
c.13064
d.none of the above
answer:d
28.8888+848+88-?=7337+737
a.1450
b.1670
c.1090
d.1750
answer:d
29.414*?*7=127512
a.36
b.40
c.44
d.48
answer:a replace a variable like x in the place of (?)
30.what is 786 times 964?
a.757704
b.754164
c.759726
d.749844
answer:a

31. How many of the numbers x (x being integer) with 10<= x<= 99 are 18 more than the sum of their digits
a. 9
b. 12
c. 18
d. 10
Answer: d
Explanation:
Let the number be ab. So given that
⇒10a + b = 18 + a + b
⇒9a = 18
⇒a = 2
So 20, 21, … upto 29 there are total 10 numbers possible.
32. X = 101102103104105106107……146147148149150  (From numbers 101-150).  Find out the remainder when this number is divided by 9.
a.4
b.5
c.2
d.1
Answer: c
Explaination : The divisibility rule for 9 is sum of the digits is to be divisible by 9.  So
We calculate separately, sum of the digits in hundreds place, tenths place, and units place.
Sum of the digits in hundreds place: 1 x 50 = 50
Sum of the digits in tenths place : 0 x 9 + 1 x 10 + 2 x 10 + 3 x 10 + 4 x 10 + 5 x 1 = 105
Sum of the digits in units place : (1 + 2 + 3 + …+ 9) x 5 = 225
So total = 380
So remainder = 380 / 9 = 2

33.7^1+7^2+7^3+…….+7^205. Find out how many numbers present which unit place contain 3?
a.44
b.51
c.25
d.65
Answer:b
Explaination: Units digits of first 4 terms are 7, 9, 3, 1. and this pattern repeats. So for every 4 terms we get one term with 3 in its unit digit. So there are total of 205/4 = 51 sets and each set contains one terms with 3 in its unit digit.

34. Given that 0 < a < b < c < d, which of the following the largest ?
a.(c+d) / (a+b)
b.(a+d) / (b+c)
c.(b+c) / (a+d)
d.(b+d) / (a+c)
Answer: A
Explanation: Take a = 1, b = 2, c = 3, d = 4. option A is clearly true.

35.The sum of the digits of a three digit number is 17, and the sum of the squares of its digits is 109.  If we subtract 495 from the number, we shall get a number consisting of the same digits written in the reverse order.  Find the number.
a. 773
b. 683
c. 944
d. 863
Answer: D
Explanation: Check options.  Sum of the squares should be equal to 109. Only Options B and D satisfying. When we subtract 495, only 863 becomes 368.

36.2ab5 is a four digit number divisible by 25. If a number formed from the two digits ab is a multiple of 13, then ab is
a.52
b.45
c.10
d.25
Answer:a
Explaination: For a number to be divisible by 25, last two digits of that number should be divisible by 25. So b must be either 2 or 7
it is given that ab must be divisible by 13 and in the options only 52 is divisible by 13.

37.A school has 120, 192 and 144 students enrolled for its science, arts and commerce courses. All students have to be seated in rooms for an exam such that each room has students of only the same course and also all rooms have equal number of students. What is the least number of rooms needed?
a.18
b.12
c.19
d.20
Answer:c
Explaination: We have to find the maximum number which divides all the given numbers so that number of roots get minimized. HCF of 120,192 & 144 is 24. Each room have 24 students of the same course.
Then rooms needed
120/24 + 192/24 + 144/24 = 5 +8 + 6 = 19

38.A two digit number is 18 less than the square of the sum of its digits. How many such numbers are there?
a.1
b.2
c.3
d.4
Answer: Option 2
Explaination:Take N = 10a+b.
Given that, (10a+b)+18 = K2 = (a+b)2
Given number = K2 – 18 = (10a+b)
That means, when we add 18 to the given number it should be a perfect square. So K2 takes the following values. 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, ….
1 to 16 are ruled out as if we subtract 18 from them, the resulting number is a single digit number.
Now 25 – 18 = 7
36 – 18 = 18
49 – 18 = 31
64 – 18 = 46
81 – 18 = 63
100 – 18 = 82
121 – 18 = 103
Now 63, 82 satisfies.

39.Find the greatest number that will divide 148 246 and 623 leaving remainders 4 6 and 11 respectively?
a.20
b.12
c.6
d.48
Answer:b
Explanation:
Hcf ( (148-4), (246-6), (623-11))=12

40.Find L.C.M. of 1.05 and 2.1
a. 1.3
b. 1.25
c. 2.1
d. 4.30
Answer: c
Explaination: If numbers are in decimal form, convert them without decimal places. Therefore, the numbers are 105 and 210.
41. The binary equivalent of the decimal number 125 is

a.1100100         b.1111101            c.1101100            d.1111111

 Ans: (b)
  

42. The hexa decimal equivalent of the decimal number 128 is

a.128         b.175                     c.80                        d.81
Ans: (c).



43. The decimal number 1356 expressed in octal system equals

a.2514            b.125                     c.353                      d.235

 Ans: (a)

44. The decimal conversion of the binary number (1111)2 is…..

a.31           b.15                       c.13                        d.14

 Ans: (b)

(1111)2 = 1*23 + 1*22 + 1*21 + 1*20 = 8+4+2+1 = 15

45. The sum of (101101)2 and (111011)2 is

a.1010110       b.1101000            c.1000110            d.1110010

Ans: (b).

46. The square root of (2011)5 is

a.(21)5      b.(31)5                  c.(121)5                 d.(41)5

Ans: (b).

(2011)5= 2*53 + 0*52 + 1*51 + 1*50 = 250+5+1= (256)10

The square root of (256)10 = (16)10 = (31)5

47. The sum of (6E)16 and (3B)12 is

a.(157)10       b.(137)11              c.(166)8                 d.(192)7
Ans: (a)

(6E)16 = 6*161 +14*160 = 96 + 14 = (110)10

(3B)12 = 3*121 + 11*120 = 36 + 11=(47)10

(110)10 + (47)10 = (157)10



48. The decimal equivalent of hexa-decimal number (ABC)16

a.2847                   b.2748                   c.7428                   d.1478

 Ans: (b)

(ABC)16 = 10*162 + 11*161 + 12*160  = 2560+176+12 = 2748.


49. The decimal fraction 0.75 in the binary system equals

a.0.11         b.0.00                    c.0.10                    d.0.111

 Ans: (a)


50. The octal equivalent to the binary (11010)2 is

a.26          b.32                       c.28                        d.30

ans:a
(11010)2 = 1*24 + 1*23 + 0*22 + 1*21 +0*20 = 16+8+2 = (26)10