business mathematics

Simple Problems on Numbers

Simple Problems on Numbers

Simple problems:

1.What least number must be added to 3000 to obtain a number
exactly divisible by 19?

Solution:
On dividing 3000 by 19 we get 17 as remainder
Therefore number to be added = 19-17=2.

2.Find the unit's digit n the product 2467 153 * 34172?

Solution:
Unit's digit in the given product=Unit's digit in 7 153 * 172
Now 7 4 gives unit digit 1
7 152 gives unit digit 1
7 153 gives 1*7=7.Also 172 gives 1
Hence unit's digit in the product =7*1=7.

3.Find the total number of prime factors in 411 *7 5 *112 ?

Solution:
411 7 5 112= (2*2) 11 *7 5 *112
= 222 *7 5 *112
Total number of prime factors=22+5+2=29

4.The least umber of five digits which is exactly
divisible by 12,15 and 18 is?
a.10010 b.10015 c.10020 d.10080

Solution:
Least number of five digits is 10000
L.C.Mof 12,15,18 s 180.
On dividing 10000 by 180,the remainder is 100.
Therefore required number=10000+(180-100)
=10080.
Ans (d).

5.The least number which is perfect square and is divisible
by each of the numbers 16,20 and 24 is?
a.1600 b.3600 c.6400 d.14400

Solution:
The least number divisible by 16,20,24 = L.C.M of 16,20,24=240
=2*2*2*2*3*5
To make it a perfect square it must be multiplied by 3*5.
Therefore required number =240*3*5=3600.
Ans (b).

6.A positive number which when added to 1000 gives a sum ,
which is greater than when it is multiplied by 1000.
The positive integer is?
a.1 b.3 c.5 d.7

Solution:
1000+N>1000N
clearly N=1.

7.How many numbers between 11 and 90 are divisible by 7?

Solution:
The required numbers are 14,21,28,...........,84.
This is an A.P with a=14,d=7.
Let it contain n terms
then T =84=a+(n-1)d
=14+(n-1)7
=7+7n
7n=77 =>n=11.

8.Find the sum of all odd numbers up to 100?

Solution:
The given numbers are 1,3,5.........99.
This is an A.P with a=1,d=2.
Let it contain n terms 1+(n-1)2=99
=>n=50
Then required sum =n/2(first term +last term)
=50/2(1+99)=2500.

9.How many terms are there in 2,4,6,8..........,1024?

Solution:
Clearly 2,4,6........1024 form a G.P with a=2,r=2
Let the number of terms be n
then 2*2 n-1=1024
2n-1 =512=29
n-1=9
n=10.

10.2+22+23+24+25..........+28=?

Solution:
Given series is a G.P with a=2,r=2 and n=8.
Sum Sn=a(1-r n)/1-r=Sn=2(1-28)/1-2.
=2*255=510.

11.Find the number of zeros in 27!?

Solution:
Short cut method :
number of zeros in 27!=27/5 + 27/25
=5+1=6zeros.

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:

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 d67.

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).

Complex Problems:

24.Six bells commence tolling together and toll at intervals
of 2,4,6,8,10,12 seconds respectively. In 30 minutes how many
times do they toll together?

Solution:
To find the time that the bells will toll together we have
to take L.C.M of 2,4,6,8,10,12 is 120.
So,the bells will toll together after every 120 seconds
i e, 2 minutes
In 30 minutes they will toll together [30/2 +1]=16 times

25.The sum of two numbers is 15 and their geometric mean is
20% lower than their arithmetic mean. Find the numbers?
a.11,4 b.12,3 c.13,2 d.10,5

Solution:
Sum of the two numbers is a+b=15.
their A.M = a+b / 2 and G.M = (ab)1/2
Given G.M = 20% lower than A.M
=80/100 A.M
(ab)1/2=4/5 a+b/2 = 2*15/5= 6
(ab)1/2=6
ab=36 =>b=36/a
a+b=15
a+36/a=15
a2+36=15a
a2-15a+36=0
a2-3a-12a+36=0
a(a-3)-12(a-3)=0
a=12 or 3.
If a=3 and a+b=15 then b=12.
If a=12 and a+b=15 then b=3.
Ans (b).

26.When we multiply a certain two digit number by the
sum of its digits 405 is achieved. If we multiply the
number written in reverse order of the same digits
by the sum of the digits,we get 486.Find the number?
a.81 b.45 c.36 d. none

Solution:
Let the number be x y.
When we multiply the number by the sum of its digit
405 is achieved.
(10x+y)(x+y)=405....................1
If we multiply the number written in reverse order by its
sum of digits we get 486.
(10y+x)(x+y)=486......................2
dividing 1 and 2
(10x+y)(x+y)/(10y+x)(x+y) = 405/486.
10x+y / 10y+x = 5/6.
60x+6y = 50y+5x
55x=44y
5x = 4y.
From the above condition we conclude that the above
condition is satisfied by the second option i e b. 45.
Ans (b).

27.Find the HCF and LCM of the polynomials x2-5x+6 and x2-7x+10?
a.(x-2),(x-2)(x-3)(x-5)
b.(x-2),(x-2)(x-3)
c.(x-3),(x-2)(x-3)(x-5)
d. none

Solution:
The given polynomials are
x2-5x+6=0................1
x2-7x+10=0...............2
we have to find the factors of the polynomials
x2-5x+6 and x2-7x+10
x2-2x-3x+6 x2-5x-2x+10
x(x-2)-3(x-2) x(x-5)-2(x-5)
(x-3)(x-2) (x-2)(x-5)
From the above factors of the polynomials we can easily
find the HCF as (x-3)and LCM as (x-2)(x-3)(x-5).
Ans (c)

28.The sum of all possible two digit numbers formed from
three different one digit natural numbers when divided by
the sum of the original three numbers is equal to?
a.18 b.22 c.36 d. none

Solution:
Let the one digit numbers x,y,z
Sum of all possible two digit numbers
=(10x+y)+(10x+z)+(10y+x)+(10y+z)+(10z+x)+(10z+y) = 22(x+y+z)
Therefore sum of all possible two digit numbers when
divided by sum of one digit numbers gives 22.

29.A number being successively divided by 3,5,8 leaves
remainders 1,4,7 respectively. Find the respective
remainders if the order of divisors are reversed?

Solution:
Let the number be x.
3 - x
5 y - 1
8 z - 4
1 - 7

z=8*1+7=15
y=5z+4 = 5*15+4 = 79
x=3y+1 = 3*79+1=238
Now 8 238
5 29 - 6
3 5 - 4
1 - 2
Respective remainders are 6,4,2.

30.The arithmetic mean of two numbers is smaller by 24
than the larger of the two numbers and the GM of
the same numbers exceeds by 12 the smaller of the numbers.
Find the numbers?
a.6,54 b.8,56 c.12,60 d.7,55

Solution:
Let the numbers be a,b where a is smaller and b
is larger number.
The AM of two numbers is smaller by 24 than the
larger of the two numbers.
AM=b-24
AM of two numbers is a+b/2.
a+b/2 = b-24
a+b = 2b-48
a = b-48...................1
The GM of the two numbers exceeds by 12 the smaller
of the numbers
GM = a+12
GM of two numbers is (ab)1/2
(ab) 1/2= a+12
ab = a2+144+24a
from 1 b=a+48
a(a+48)= a2+144+24a
a2+48a = a2+144+24a
24a=144=>a=6
Therefore b=a+48=54.
Ans (a).

31.The sum of squares of the digits constituting a positive
two digit number is 13,If we subtract 9 from that number
we shall get a number written by the same digits in the
reverse order. Find the number?
a.12 b.32 c.42 d.52.

Solution:
Let the number be x y.
the sum of the squares of the digits of the number is 13
x2+y2=13
If we subtract 9 from the number we get the number
in reverse order
x y-9=y x.
10x+y-9=10y+x.
9x-9y=9
x-y=1
(x-y)2 =x2+y2-2x y
1 =13-2x y
2x y = 12
x y = 6 =>y=6/x
x-y=1
x-6/x=1
x2-6=x
x2-x-6=0
x+2x-3x-6=0
x(x+2)-3(x+2)=0
x=3,-2.
If x=3 and x-y=1 then y=2.
If x=-2 and x-y=1 then y=-3.
Therefore the number is 32.
Ans (b).

32.If we add the square of the digit in the tens place
of the positive two digit number to the product of the
digits of that number we get 52,and if we add the square
of the digit in the unit's place to the same product
of the digits we get 117.Find the two digit number?
a.18 b.39 c.49 d.28

Solution:
Let the digit number be x y
Given that if we add square of the digit in the tens place
of a number to the product of the digits we get 52.
x2+x y=52.
x(x+y)=52....................1
Given that if we add the square of the digit in the unit's plac
e to the product is 117.
y2+x y= 117
y(x+y)=117.........................2
dividing 1 and 2 x(x+y)/y(x+y) = 52/117=4/9
x/y=4/9
from the options we conclude that the two digit number is 49
because the condition is satisfied by the third option.
Ans (c)

33.The denominators of an irreducible fraction is greater
than the numerator by 2.If we reduce the numerator of the
reciprocal fraction by 3 and subtract the given fraction
from the resulting one,we get 1/15.Find the given fraction?

Solution:
Let the given fraction be x / (x+2) because given that
denominator of the fraction is greater than the numerator by 2
1 – [(x – 1/(x+2))/3] = 1/15.
1 – (x2+2x-1) /3(x+2) = 1/15
(3x+6-x2-2x+1)/3(x+2) = 1/15
(7-x2+2x)/(x+2) = 1/5
-5x2+5x+35 = x+2
5x2-4x-33 = 0
5x2-15x+11x-33 = 0
5x(x-3)+11(x-3) = 0
(5x+11)(x-3) = 0
Therefore x=-11/5 or 3
Therefore the fraction is x/(x+2) = 3/5.

34.Three numbers are such that the second is as much
lesser than the third as the first is lesser than
the second. If the product of the two smaller numbers
is 85 and the product of two larger numbers is 115.
Find the middle number?

Solution:
Let the three numbers be x,y,z
Given that z – y = y – x
2y = x+z.....................1
Given that the product of two smaller numbers is 85
x y = 85................2
Given that the product of two larger numbers is 115
y z = 115...............3
Dividing 2 and 3 x y /y z = 85/115
x / z = 17 / 23
From 1
2y = x+z
2y = 85/y + 115/y
2y2 = 200
y2 = 100
y = 10

35.If we divide a two digit number by the sum of its digits
we get 4 as a quotient and 3 as a remainder. Now if we
divide that two digit number by the product of its digits
we get 3 as a quotient and 5 as a remainder .
Find the two digit number?

Solution:
Let the two digit number is x y.
Given that x y / (x+y)
quotient=4 and remainder = 3
we can write the number as
x y = 4(x+y) +3...........1
Given that x y /(x*y) quotient = 3 and remainder = 5
we can write the number as
x y = 3 x*y +5...............2
By trail and error method
For example take x=1,y=2
1............12=4(2+3)+3
=4*3+3
! =15
let us take x=2 y=3
1..............23=4(2+3)+3
=20+3
=23
2.............23=3*2*3+5
=18+5
=23
the above two equations are satisfied by x=2 and y=3
Therefore the required number is 23.

36.First we increased the denominator of a positive
fraction by 3 and then it by 5.The sum of the
resulting fractions proves to be equal to 2/3.
Find the denominator of the fraction if its numerator is 2.

Solution:
Let us assume the fraction is x/y
First we increasing the denominator by 3 we get x/(y-3)
Then decrease it by 5 we get the fraction as x/(y-5)
Given that the sum of the resulting fraction is 2/3
x/(y+3) + x/(y-5) = 2/3
Given numerator equal to 2
2*[ 1/y+3 + 1/y-5] =2/3
(y-5+y+3) / (y-3)(y+5) =1/3
6y – 6 = y2-5y+3y-15
y2-8y-9 = 0
y2-9y+y-9 = 0
y(y-9)+1(y-9) = 0
Therefore y =-1 or 9.

37.If we divide a two digit number by a number consisting
of the same digits written in the reverse order,we get 4
as quotient and 15 as a remainder. If we subtract 1 from
the given number we get the sum of the squares of the
digits constituting that number. Find the number?
a.71 b.83 c.99 d. none

Solution:
Let the number be x y.
If we divide 10x+y by a number in reverse order
i e,10y+x we get 4 as quotient and 15 as remainder.
We can write as
10x+y = 4(10y+x)+15......................1
If we subtract 1 from the given number we get square of the digits
10x+y = x2+y2.....................................2
By using above two equations and trail and error method
we get the required number. From the options also we can
solve the problem. In this no option is satisfied so answer is d.
Ans (d)