Medium Problems:
12.The difference between two numbers 1365.When the larger
number is divided by the smaller one the quotient is 6 and
the remainder is 15.The smaller number is?
a.240 b.270 c.295 d.360
Solution:
Let the smaller number be x, then larger number =1365+x
Therefore 1365+x=6x+15
5x=1350 => x=270
Required number is 270.
13.Find the remainder when 231 is divided by 5?
Solution:
210 =1024.
unit digit of 210 * 210 * 210 is 4
as 4*4*4 gives unit digit 4
unit digit of 231 is 8.
Now 8 when divided by 5 gives 3 as remainder.
231 when divided by 5 gives 3 as remainder.
14.The largest four digit number which when divided by 4,7 or 13 leaves a remainder of 3 in each case is?
a.8739 b.9831 c.9834 d.9893.
Solution:
Greatest number of four digits is 9999
L.C.M of 4,7, and 13=364.
On dividing 9999 by 364 remainder obtained is 171.
Therefore greatest number of four digits divisible by 4,7,13
=9999-171=9828.
Hence required number=9828+3=9831.
Ans (b).
15.What least value must be assigned to * so that th number 197*5462 is divisible by 9?
Solution:
Let the missing digit be x
Sum of digits = (1+9+7+x+5+4+6+2)=34+x
For 34+x to be divisible by 9 , x must be replaced by 2
The digit in place of x must be 2.
16.Find the smallest number of 6 digits which is exactly divisible by 111?
Solution:
Smallest number of 6 digits is 100000
On dividing 10000 by 111 we get 100 as remainder
Number to be added =111-100=11.
Hence,required number =10011.
17.A number when divided by 342 gives a remainder 47.When the same
number is divided by 19 what would be the remainder?
Solution:
Number=342 K + 47 = 19 * 18 K + 19 * 2 + 9=19 ( 18K + 2) + 9.
The given number when divided by 19 gives 18 K + 2 as quotient
and 9 as remainder.
18.In doing a division of a question with zero remainder,a candidate
took 12 as divisor instead of 21.The quotient obtained by him was 35.
The correct quotient is?
a.0 b.12 c.13 d.20
Solution:
Dividend=12*35=420.
Now dividend =420 and divisor =21.
Therefore correct quotient =420/21=20.
19.If a number is multiplied by 22 and the same number is
added to it then we get a number that is half the square
of that number. Find the number.
a.45 b.46 c.47 d. none
Solution:
Let the required number be x.
Given that x*22+x = 1/2 x2
23x = 1/2 x2
x = 2*23=46
Ans (b)
20.Find the number of zeros in the factorial of the number 18?
Solution:
18! contains 15 and 5,which combined with one even number
gives zeros. Also 10 is also contained in 18! which will
give additional zero .Hence 18! contains 3 zeros and the
last digit will always be zero.
21.The sum of three prime numbers is 100.If one of them
exceeds another by 36 then one of the numbers is?
a.7 b.29 c.41 d.67
Solution:
x+(x+36)+y=100
2x+y=64
Therefore y must be even prime which is 2
2x+2=64=>x=31.
Third prime number =x+36=31+36=67.
22.A number when divided by the sum of 555 and 445 gives
two times their difference as quotient and 30 as remainder .
The number is?
a.1220 b.1250 c.22030 d.220030.
Solution:
Number=(555+445)*(555-445)*2+30
=(555+445)*2*110+30
=220000+30=220030.
23.The difference of 1025-7 and 1024+x is divisible by 3 for x=?
a.3 b.2 c.4 d.6
Solution:
The difference of 1025-7 and 1024+x is
=(1025-7)-(1024-x)
=1025-7-1024-x
=10.1024-7 -1024-x
=1024(10-1)-(7-x)
=1024*9-(7+x)
The above expression is divisible by 3 so we have to
replace x with 2.
Ans (b).
12.The difference between two numbers 1365.When the larger
number is divided by the smaller one the quotient is 6 and
the remainder is 15.The smaller number is?
a.240 b.270 c.295 d.360
Solution:
Let the smaller number be x, then larger number =1365+x
Therefore 1365+x=6x+15
5x=1350 => x=270
Required number is 270.
13.Find the remainder when 231 is divided by 5?
Solution:
210 =1024.
unit digit of 210 * 210 * 210 is 4
as 4*4*4 gives unit digit 4
unit digit of 231 is 8.
Now 8 when divided by 5 gives 3 as remainder.
231 when divided by 5 gives 3 as remainder.
14.The largest four digit number which when divided by 4,7 or 13 leaves a remainder of 3 in each case is?
a.8739 b.9831 c.9834 d.9893.
Solution:
Greatest number of four digits is 9999
L.C.M of 4,7, and 13=364.
On dividing 9999 by 364 remainder obtained is 171.
Therefore greatest number of four digits divisible by 4,7,13
=9999-171=9828.
Hence required number=9828+3=9831.
Ans (b).
15.What least value must be assigned to * so that th number 197*5462 is divisible by 9?
Solution:
Let the missing digit be x
Sum of digits = (1+9+7+x+5+4+6+2)=34+x
For 34+x to be divisible by 9 , x must be replaced by 2
The digit in place of x must be 2.
16.Find the smallest number of 6 digits which is exactly divisible by 111?
Solution:
Smallest number of 6 digits is 100000
On dividing 10000 by 111 we get 100 as remainder
Number to be added =111-100=11.
Hence,required number =10011.
17.A number when divided by 342 gives a remainder 47.When the same
number is divided by 19 what would be the remainder?
Solution:
Number=342 K + 47 = 19 * 18 K + 19 * 2 + 9=19 ( 18K + 2) + 9.
The given number when divided by 19 gives 18 K + 2 as quotient
and 9 as remainder.
18.In doing a division of a question with zero remainder,a candidate
took 12 as divisor instead of 21.The quotient obtained by him was 35.
The correct quotient is?
a.0 b.12 c.13 d.20
Solution:
Dividend=12*35=420.
Now dividend =420 and divisor =21.
Therefore correct quotient =420/21=20.
19.If a number is multiplied by 22 and the same number is
added to it then we get a number that is half the square
of that number. Find the number.
a.45 b.46 c.47 d. none
Solution:
Let the required number be x.
Given that x*22+x = 1/2 x2
23x = 1/2 x2
x = 2*23=46
Ans (b)
20.Find the number of zeros in the factorial of the number 18?
Solution:
18! contains 15 and 5,which combined with one even number
gives zeros. Also 10 is also contained in 18! which will
give additional zero .Hence 18! contains 3 zeros and the
last digit will always be zero.
21.The sum of three prime numbers is 100.If one of them
exceeds another by 36 then one of the numbers is?
a.7 b.29 c.41 d.67
Solution:
x+(x+36)+y=100
2x+y=64
Therefore y must be even prime which is 2
2x+2=64=>x=31.
Third prime number =x+36=31+36=67.
22.A number when divided by the sum of 555 and 445 gives
two times their difference as quotient and 30 as remainder .
The number is?
a.1220 b.1250 c.22030 d.220030.
Solution:
Number=(555+445)*(555-445)*2+30
=(555+445)*2*110+30
=220000+30=220030.
23.The difference of 1025-7 and 1024+x is divisible by 3 for x=?
a.3 b.2 c.4 d.6
Solution:
The difference of 1025-7 and 1024+x is
=(1025-7)-(1024-x)
=1025-7-1024-x
=10.1024-7 -1024-x
=1024(10-1)-(7-x)
=1024*9-(7+x)
The above expression is divisible by 3 so we have to
replace x with 2.
Ans (b).
