eLitmus Sample Quants Questions from Previous Papers

1. Find the number of ways in which you can fill a 3×3 grid(with four corners defined  as a,b,c,d) if u have 3 white marbles and 6 black marbles?

soft tab viagra It is a 3*3 grid, so there are a total of 9 spaces where you can place the marbles . Let us first place  the 3 white marbles, that can be done in 9C3 ways. There are 6 spaces available and 6 black marbles are to be placed in those 6 spaces. So, that can be done in 1 way.
Hence the answer  would be 9C3.

2. If X=123456 and Z=X-Y then for how many values of Y, Z is divisible by 48, 98, 105?

X=123456 LCM of 48,98,105 is 11760. So if a number is divisible by 11760, then it will be divisble by 48,98,105. On dividing X by 11760 we get dividend = 10. So for 10 values of y, z is divisible by 11760. Thus for 10 values of y ,z value found is divide by 48,,98,105. These values of Y are 111696, 99936, 88176, 76416, 64656, 52896, 41136, 29376, 17616, 5856

3. There are 6 Bangles each of 4cms in diameter. These are to be placed in a salver(plate), what should be the minimum radius of the salver, so that each bangles are kept without overlapping (bangles touching each other)?
One bangle at the center and the remaining 5 surrounding them, so radius of the center +the diameter of the outer bangle = 4+2=6cms

4. Let Sn denote the sum of first n terms of an A.P. If S2n = 3Sn, then the ratio S3n/Sn is equal to?

We know that Sn = n(n+1)/2
now it is given that S2n = 3Sn
=> 2n(2n+1)/2 = 3n(n+1)/2
=> 2(2n+1) = 3(n+1)
=> 4n+2 = 3n+3
=> n = 1
then S3n/Sn = [3n(3n+1)/2]/[n(n+1)/2] = 3*4/1*2 = 6

5. In a strange twist of hearts, P politicians of a country agreed to an average donation of Rs. D each. Q of these politicians, who had pledged an average of Rs. A never donated the pledged money. Which of the following expressions represents the percent of the pledged money that was actually donated.


We know that Sn = n(n+1)/2
now it is given that S2n = 3Sn
=> 2n(2n+1)/2 = 3n(n+1)/2
=> 2(2n+1) = 3(n+1)
=> 4n+2 = 3n+3
=> n = 1
then S3n/Sn = [3n(3n+1)/2]/[n(n+1)/2] = 3*4/1*2 = 6 pledged money = P*D
Actually donated money = PD-QA
percent of the pledged money thas ws actually donated = 100*(PD-QA)/PD = 100-100(QA/PD)

6. What is the value of log(e(e(e…..)^1/2)^1/2)^1/2)?

Let x=(e(e(e………)^1/2)^1/2)^1/2)
now squaring both side we get
i.e., x^2=e*x [since x=(e(e(e………)^1/2)^1/2)^1/2)]
therefore x=e [since x can not be zero]
Finally, log(x)=log(e)=1

7. How many values of  c in the equation x^2-5x+c result in rational  roots which are integers ?

c=4 => x=1,4
c=-6 => x=-1,6
c=-14 => x=-2,7
c=-24 => x=-3,8
c=-36 => x=9,-4
c=-50 => x=-5,10
c=-55 => x=-6,11 and so on…
Hence, Infinite is the answer

8. If 1/a + 1/b + 1/c=1/(a+b+c) where a+b+c != 0, a*b*c != 0 what is the value of (a+b)(b+c)(c+a)?
a)equals 0
b)greater than 0
c)less than 0
d)cannot be determined

0 (ab+bc+ca)/abc=1/(a+b+c)
=> abc=a2b+ab2+ac2+a2c+b2c+bc2+3abc
=> a2b+ab2+ac2+a2c+b2c+bc2+2abc=0

9. PR is a tangent to a circle at point P.Q is another point on the circle such that PQ is the diameter and RQ cuts the circle at point M. If the radius of the circle is 4 units and PR=6 units then find the ratio of the perimeter of triangle PMR to the triangle PQR

Since PR is the tangent,Hence angle QPR is 90.
=>Triangle PQR is rit-angled triangle.
=>hyp QR=10. (Pythagoras Theorem)……..(1)

Also PM is perpendicular to QR (Since triangle PQR is made inside circle with diameter as hyp
hence PMQ is rit-angled…itz a property of circle)
=>three triangles PQR,PMQ & PMR are similar to each other.
(For similarity of such type of triangles u can refer any 10th std book).

NOw in that pic consider the triangles PQR & PMR.
hence PR/RM=QR/PR.
putting the values & from(1)

putting the values & from(1)

from(2) & (3)…

Hence peimeter of triangle PMR=PM+RM+PR
Putting the obtained values perimeter=72/5

Also perimeter of triangle PQR=PQ+QR+PR=24.

Hence th ratio PMR/PQR=72/(5*24)=3/5

10. The circle O having a diameter of 2cm, has a square inscribed in it.each side of the square is then taken as a diameter to form 4 smaller circles O’.find the total area of all four O’ circles which is outside the cirlce O.
area of circle O=pi*(1)^2 = pi

Nw the diameter of circle is also the diagonal of the square.
Hence each side of square will be sqrt(2).

=>Area of square=2

since each side of square is also the diameter of other 4 circles.
Hence summation of area of 4 circles=2*pi………..(1)

If u have drawn its fig you’ll find that to obtain the required ans u have to subtract the area of 4 semi-circles formed on the side of the square from the each of the small portion outside the square.
To get that area of small portion =area of circle O-area of square =pi-2…….(2)

this small portion has to be subtracted from the four semi-circles.
Hence, area of 4 semi-circles=2*pi/2= pi……[from (1)]
required ans=total area of 4 semi-circles – area of small portion(from (2))

11. If v,w,x,y,z are non negative integers each less than 11, then how many distinct combinations are possible of(v,w,x,y,z) which satisfy v(11^4) +w(11^3)+ x(11^2)+ y(11) +z = 151001 ?

Changing 151001 to base 11 number we get,
it will be
A34A4 i.e.
10 3 4 10 4 v(11^4) +w(11^3)+ x(11^2)+ y(11) +z=151001 where v=10, w=3,x=4, y=10, z=4

12. In a certain examination paper there are n questions. For j=1,2,3,…..n, there are 2^(n-1) students whoanswered j or more question wrongly. If the total number of wrong answers is 4096 then the value of n is

Given that,
2^(n-1) = 4096 = 2^12
==> n=13

13. How many six digit number can be formed using the digits 1 to 6, without repetition, such that the number is divisble by the digit at it’s unit’s place?

As each number will contain all the six digits and the sum of digits is = 1+2+3+4+5+6 = 21 which is divisible by 3. So each number is divisible by 3.
The numbers ending with digit 1 will be divisible 1.
The numbers ending with digit 2 will be divisible 2.
The numbers ending with digit 3 are divisible 3.
The numbers ending with digit 5 will be divisible 5.
The numbers ending with digit 6 will be divisible 6.
Except the numbers ending with last two digits as 14, 34 and 54 all other numbers ending with 4 are divisible by 4.

The no. of numbers ending with last two digits 14,34 and 54 are = 3*4*3*2 = 72. (As for a number to be divisible by 4 last two digits must be divisible by 4)
so six digit number that can be formed using the digits 1 to 6, without repetition, such that the number is divisble by the digit at unit’s place = 720-72= 648

14. A natural number has exactly 10 divisors including 1 and itself. how many distinct prime factors can this natural number can have?

Consider the no is 512, the divisors of this no. are
so only one prime factors are there i.e. 2;
if the no is 48 the divisors are
so the prime factors are 2,3;

15. If m and n are two positive integers, then what is the value of mn? Given:

(1)7m + 5n= 29
(2) m + n= 5

7m + 5n = 29
m + n = 5(multiply both sides by 7)
subtract equation second from first,we get
7m + 5n =29
7m + 7n = 35
n = 3
substitute value of n in any equation.
m = 2
mn = 6.

16.  A natural number has exactly 10 divisors  including 1 and itself. How many distinct prime factors can this natural number have?
A. Either 1 or 2
B. Either 1 or 3
C. Either 2 or 3
D. Either 1,2 or 3
Ans) Either 1 or 2
Check on 29 i.e 512 , 39, 59 which have only 1 prime factor and 80 , 48 which are having 2
prime factors and total of 10 divisors.
Let us also consider the case of 3 prime factors. Let x, y ,z be the three prime factors of a
number. Therefore 1 , x ,y ,z ,xy ,yx, zx ,xyz must be the factors of that nos . We have
minimum 8 such factors with xyz as the nos or the factor of the nos.
When xyz is the nos then we will have exactly 8 divisors but if the nos is greater thn xyz that
is a multiple of xyz , either the nos is multiplied by any of these prime factors x , y , z then
we will get at least 12 divisors. So we don’t get 3 prime factors with 10 divisors.
17. What is the remainder when 128^1000 is divided by 153?

128^1000 = (153-25)^1000 = (25^1000)mod153 = (625^500)mod153 = [(4*153+13)^500]mod153 = (13^500)mod153 = (169^250)mod153 = [(153+16)^250]mod153 = (16^250)mod153 = (256^125)mod153 = [(153+103)^125]mod153 = (103^125)mod153 = [(2*3*17+1)^125]mod153
At this point,observe that 153=17*(3^2);
Now,therefore clearly (2*3*17+1)^125 = [(125C124)*{(2*3*17)^1}*(1^124)+1]mod153.
Actually,the above line can be written since only except the last two
terms,every term of the expansion of (2*3*17+1)^125 has [(2*3*17)^2],i.e,
[153*68] as one of its factors.
Now, [(125C124)*{(2*3*17)^1}*(1^124)+1]= 125*2*3*17 + 1;
[125*2*3*17 + 1]/(153) = [125*2*3*17 + 1]/(3^2*17) = (125*2*3*17)/(3^2*17) + 1/(3^2*17) = 250/3 + 1/(3^2*17) =83 + (1/3)+ 1/(3^2*17) = 83 + (52/153);
which means [125*2*3*17 + 1] = 83*153 + 52;
which again implies 128^1000 = [125*2*3*17 + 1]mod153 = {83*153 + 52}mod153 = 52mod153 ;
So, remainder is 52.

18. Given a,b,c are in GP and a < b < c. How many different different values of a, b, c satisfy (log(a) + log(b) + log(c) ) = 6?

abc=10^6 because of b^2=ac=100,
and the combinations are
(2,100,5000), (4,100,2500), (5,100,2000), (10,100,1000), (50,100,200),(25,100,400),(20,100,500).

19. What will be the remainder when expression 2^2+22^2+222^2+2222^2+….+22222…48times^2 is divided by 9?

First, let us consider a general case :
(222222222222…..{2 is repeated n times})^2
=[2(111111111111……..{1 is repeated n times})]^2
=4(111111111111……..{1 is repeated n times})^2
=4{(10)^(n-1) + (10)^(n-2) + (10)^(n-3) + (10)^(n-4) + …. +1}^2…….[(eqn1)]
Now, we know 9 divides (10^k -1) , since (10^k -1)=(999999….k times);
Where k is ofcourse a +ve integer. So,(10^k -1) is of the form (9A + 1);[A being a +ve integer].
Now, looking at the [(eqn1)] ,
we can find that here k =(n-1), (n-2), (n-3),….,1.
Now,for 10^(n-1) = 9*(a1) + 1;
10^(n-2) = 9*(a2) + 1;
10^(n-3) = 9*(a3) + 1;
10^(n-4) = 9*(a4) + 1;
10^1 =9 +1;
1 = 1
So,now in [(eqn1)] , we can write
(222222222222…..{2 is repeated n times})^2
=4(9*A +{1 + 1 + 1 + 1 + ………[1 is added n times]})^2
=4*(9*A + n)^2; here, A =(a1+a2+a3+a4+….+1),of course an integer.
=4*81*(A^2) + 4*2*(9*A)*n + 4*(n^2);……………[(eqn 2)]
Now this a general expression for (222222222222…..{2 is repeated n times})^2.
For the given problem,we can find that n=1,2,3,4,5,6,….48.
In the [(eqn 2)],since 1st two terms are already divisible by 9,so we only have
to consider the last term,i.e, 4*(n^2);
Summing up [4*(n^2)] for n=1,2,…48. we get
=4*[48*(48+1)*(2*48 +1)/6] ; using the formula[1^2 +2^2 +..+n^2=n(n+1)(2n+1)/6]
=4(9- 1)(9*5 +4)(9*11 -2)
=9*B + 4(-1)*(4)*(-2); B an integer, to know its value is not important
=9*B + 32
=9*B + 9*3 + 5;
So, clearly the remainder will be 5.

20. Given that a number Q < 200, calculate sum of all Q such that when Q divided by 5 or 7 gives remainder 2?

for 5
7 12 17………197….39 terms sum=3978
for 7
9 16 23 30……. 198…28 terms sum =2848
but some terms are counted more than 1 ..ie which are common in 5 and 7
eg 37 72 107 142 177..sum=535
so 6876-535=6341

21. Given a number N, 2*N has 28 factors and 3*N has 30 factors. Calculate how many factors will be there in 6*N?

It is told in question that 2*N has 28 factors.
28 = 7 * 4 or 14*2 or 7 * 2 * 2
Manav is considering the case of (7 * 4) only.
For this type of questions always start like
N = 2^p * 3^q * 5^r * ……so on(beacuse 2,3,5,7,11,13,17.. are prime factors)
since the maximum limit is not given in question so it is not possible to give a single answer…
For the question how 7*4 = 28 is coming in MANAV’s Solution—
Lets for an example u need to represent 48 in prime factorization–
u will write – 48 = 2^4 * 3^1
Now if u want to know how many factors does 48 has.. we need to do [(4+1) * (1+1)]= 10 factors total
IN General N = 2^p * 3^q * 5^r * 7^s…..so on
Total no of factors are = (p+1) * (q+1) * (r+1) * (s+1)……

22. Heinz produces tomato puree by boiling tomato juice. The tomato puree has only 20% water while tomato juice has 90% water. How many liters of tomato puree will be obtained from 20 liters of tomato juice?
a) 2 liters
b) 2.4 liters
c) 2.5 liters
d) 6 liters

20 liters of juice contains 18 liter water and 2 liter tomato pure.
now let x liters of tomato pure be produced ,then
x- 20x/100 = 2
x= 2*100/80 =2.5

23. P. T.  Usha and Shelly John decide to run a marathan between ramnagar and jamnagar. Both start from ramnagar at 1 pm. On the way are two towns: ramgarh and rampur seprated by a distance of 15 km .P. T. Usha reaches ramgarh in 90 minutes running at a constant speed of 40 km/hr, she takes additional 30 minutes to reach rampur. Between rampur and jamnagar she maintains at averagespeed of V km/hr. Shelly John being a professional marathon runner, maintains a constant speed of 18 km/hr.they both reach jamnagar togethar after n hours. What could be the total time taken by P. T.  Usha?

a) 5 hours
b) 15 hours
c) 41 hours
d) all of the above


total distance covered by shelly =18n
total distance covered by usha =60+15+(n-2)v=75+nv-2v
18n =75+nv-2v
18n-nv= 75-2v
n(18-v) =75-2v
n =75-2v/18-v
substituting the values given in the options it is found that all the first 3 values satisfy the equation obtained.
Hence “All of the above” is the answer

24. A fresher recruitment event of Hire All Smart People ltd(HASPL) at elitmus happens in 2 cities Bangalore and Delhi. The interview call is sent to everyone who is above 70th percentile(top 30% of the pool)70% and 80% of called people accept the interview calls in Bangalore and Delhi respectively 90% and 80% of the people who accepted the call,turned out on the day of the interview respectively in Bangalore and Delhi.The ‘offer’ ratio is 2 out of 3 and 3 out of 4 of the people who turned up in Bangalore and Delhi respectively. If amongst people who apply there are 1000 people above 70th percentile in each of the locations. What percentage of 70th percentile from elitmus got offered in HASPL when results of both location are taken together?


fraction of people selected from bangalore = 0.7*0.9*2/3 = 0.42 or 42%
fraction of people selected from Delhi = 0.8*0.8*3/4 =0.48 or 48%
percentage of 70th percentile from elitmus got offered in HASPL when results of both location are taken together = (42+48)/2= 45 %

25. Eden park is a cricket field while jubilee park is a football field in Mastnagar, all cricket fields are circular and football fields are rectangular of square. Along the boundary of all fields there are advertisement displays.in mastnagar, the length of advertisement displays has to be same across allplaying fields. Area of jubliee park is 468 sqm, the farthest distance between any two points in jubliee park football field is 10*(1.43)m. Find the approximate area of eden park?

If Area of jubliee park is 468 sqm.the farthest distance between any two points in jubliee park football field is 10sqrt(10),
and l and b are length and breadth of rectangular park, then
l^2+b^2 = (10sqrt(10))^2= 1000
l*b = 468 sq mtrs
(l+b)^2= 1000+2*468=1936
perimeter of rectangle= perimeter of circular field of Eden park = 2*44=88 mtrs
radius of eden park = 88/2*pi= 44*7/22= 14 mtrs
Area of eden park = Pi*r^2 = 14*14*22/7 = 44*14 = 616 sq mtrs

26. From a pack of 52 playing cards, three cards are drawn random. Find the probability of drawing a king, a queen and a jack.

A king can be drawn in 4C1 ways
Similarly, a queen and a Jack can be drawn in 4C1, 4C1.

27. Find the total numbers in the range 100 to 1000, where in the product of individual digits of the number gives 24 (For instance: 234 gives 2*3*4 = 24)

A product of 24 can be achieved by (1,3,8), (1,4,6), (2,2,6), (2,3,4)
Now these numbers can be arranged in 3! ways respectively
However 2,2,6 can be arranged in 3!/2! ways
Hence, the answer is 3*3! + 1*3!/2! = 18+3 = 21

28. There is cask full of milk. E litres are drawn from the cask, it is then filled with water. This process is repeated. Now the ratio of milk to water is 16:9.What is the capacity of the cask in litres?

if x is capacity of cask, then
x ltr milk .. 0 ltr water
after first draw and water filling by E ltrs,
x-E ltr milk and E ltr water
after 2nd draw and water filling by E ltrs,
x- [2E-(E^2/x)] and 2E -( E^2)/x ltr water
As per condition
x- [2E-(E^2/x)] /[ 2E -( E^2)/x] = 16/9

solving, we get

29. What should come in the place of (?) in the given series?


A +5==>F +5==>K +5==>P
C +4==>G +4==>K +4==>O
E +3==>H +3==>K +3==>N

30. If (9+9^2+9^3+…….9^n) is divided by 6, remainder will be? (n is a multiples of 11)

now since n is a multiple of 11 then
3(11)=33 i.e. rem =3
again 3(22)==66 rem =0
3(33)=99 rem=3
if n is odd then remainder is 3
if n is even the remainder is 0
ans remainder is 3,0

31. If x,y,z are non negative integers and x=6. Find the number of solutions of 1/x +1/y = 1/z?

z is a non-negative integer
therefore 6y=(6+y)*k where ‘k’ is a non-negative integer
0<(6-k)<6 data-blogger-escaped-br=””> k={1,2,3,4,5}
number of solutions=4

32. How many 8 digit numbers divisible by 25 can be formed with 0,1,2,3,4,5,6,7 ,if repetition of digits is not allowed?

last 2 digit number can be 25,50,75
25 – rest 6 digits will b considered bt 0 can not be at first place so 5* 5!
75 – Same as 25 so 5*5!
50 – 0 is already used so we have all left 6 digit no. to be used so 6!
5*5!+5*5!+6! = 1920

33. How many numbers are there between 0 and 1000 which on division by 2, 4, 6, 8 leave remainders
1, 3, 5, 7 respectively?

To find the smallest number satisfying the given condition, find the LCM of 2, 4, 6, 8, which is 24
Now, the given factors and remainders differ by 1, so subtract 24 by 1
Hence, 23 is the number.
To find the next number add 24 to 23 and so on..
Hence the series would be 23, 47, 71 … 983, which is in AP.
Using l=a+(n-1)*d, we get n=21.
Hence the answer is 21

34. How many six digit numbers are possible using 1,2,3,4,5,6 such that the number formed is divisible by the digit at its units place?

If digt place has 1 then possible numbers = 120(5!)
if 2 then possible numbers = 120
if 3 then possible numbers =120
if 4 then possible numbers = 48 (4!*2)
if 5 then possible numbers = 120
if 6 then possible numbers = 20
so totl no=648

35. Amit can complete a piece of work in 2.25 days. Badri takes double the time taken by amit.chetan takes double dat of badro,and das takes double dat of chetan to complete the same task. They are split into two groups(of one or more persons)such that the differnce b/w the times taken by the two groups to complete the same work is minimum.what could be the compostion of the faster group?
a)amit and das
b)badri nd chetan
c)badri,chetan nd das
d)amit alone

A = 2.25 days => 1 day = 1/2.25 = 0.44 work/day
B = 4.5 days => 0.22 work / day
C = 9 days => 0.11 work per day
D = 18 days => 0.055 work / day

now add all 1 day work of b,c,d then we ll get 0.345 which is less than 1 days amit work…
so if we all add in amit’s work the it will definetly increase the differnce
so group form like, amit in a single group and another group consist of b,c,d

36. How many numbers are there whose factorial ends with 17 zeros?
To solve this first we find a number which ends with 17 zeros and further we will find the total numbers with that condition
Let X! be the number which ends with 17 zeros
Now, X!/5 + X!/25 = 17 —-[1]

(For those who don’t know how we obtained this,
The number of trailing zeros of a number can be found out using the following method
For instance, consider we have to find the total number of trailing zeros of 70!
We would do that as follows:
&nbsp —-
&nbsp |14
&nbsp —-
&nbsp&nbsp 2
Now, total number of zeros trailing is 14+2=16 )
In equation [1], we find that no number satisfy the given condition i.e. 70!, 70!..74! have 16 zeros and 75! has 18 zeros
Hence the answer is 0

37. Two persons A and B do a work in 30 and 40 days respectively. If both do together, A start the first day and on other day work done by exactly one of them. Finally they divide the earning in ratio 1:1. How many days the work be completed?

Here earning ratio is 1:1 so half(1/2) of work done by both(A and B).
now work done by A in one day = 1/30
work done by B in one day = 1/40
total time taken by A for 1/2 of work = (1/2)/(1/30) = 15 days
total time taken by B for 1/2 of work = (1/2)/(1/40) = 20 days
so total time required to complete the work = 15+20 = 35 days

38. Two circles lying in the first quadrant, touch each other externally. Both the axes makes tangents with both the circles. If the distance between the two centre of the circles is 8 cm, find the difference in their radii?

First draw a rough figure assigning Centers o1 and o2 for smaller and larger circles respectively
Suppose Radius of smaller and larger circle are r1 and r2 respectively
By radius of smaller circle we can find out the length of PQ
PQ=r1/sin 45=root(2)
r2=PQ=root(2) (Since angle of arc is 60)
Now find the Area covered by arcs=pi/4+pi/3
Area which not covered in common region=ar(o1PQ)+ar(o2PQ)=root(3)/2 +1/2
So Area of common region=7pi/12-(1+root(3))/2

39. Two circles intersect each other @ two points P and Q …smaller circle has radius 1cm…if arc extended by smaller circle is 90 degree and that of the larger circle in 60 degree at their corresponding centres .Find out the common area of the circles?

let the radius of first circle be (R,R)

let the radius of second cirlce br(r,r)
itis because its lies on line y=x int the first quadrant.so we need to find(R-r)
accoding to distance formula;
replacing values;
d=(sqroot(R-r)^2+(R-r)^2);here dis given as 8 int the ques;

40. The value of 99^n is a number which starts with digit 8. What can be the minimum value of n?

99^1= 99
99^2= 9801
99^3= 970299
take only starting two digit, other digit will not change the solution..
so 11 will be the answer

41. How many positive integers are there that are not larger than 1000 and are neither perfect squaresnor perfect cubes?

We use the Inclusion-Exclusion Principle. There are 1000 integers from 1 to 1000;
among these numbers, 31 are perfect squares (indeed, 312 = 961 ≤ 1000, but 322 > 1000), 10 are perfect cubes (this is because 103 = 1000), and 3 are both squares and cubes (these three numbers are 16 = 1, 26 = 64, and 36 = 729). Thus, by the Inclusion-exclusion principle, there are
1000 − (31 + 10) + 3 = 962
numbers that are neither perfect squares nor perfect cubes.

42. There are 9 players including Mic and Jordan standing in a row. What is the probability of being 2 or less players between Mic and Jordan?

total no. of ways arranging 9 people is = 9factorial. case-1–
both sit together
hence no. of ways of arranging them is 8fact*2fact.
1person sit between them
hence no. of way of arranging them is 7(2fact*7fact).
2 person sit between them
hence no. of ways arranging them is 7(2fact*6fact)
hence total no. of ways ==8!2!+7(2!7!)+6(2!7!)
divide this by 9!..
ans.. 7/12

43. If the decimal number 120 when expressed to the base a,b and c equals 60,80,100 respectively, then which of the following statement is true?
a) a,b,c are in geometric progression
b) a,b,c are in arithmetic progression
c) a,b,c are in harmonic progression
d) a-b-c=1

120 base 10= 60 base a i.e. a=(120/60)*10 =>a=20
120 base 10=80 base b i.e. b=(120/80)*10 =>b=15
120 base 10= 100 base c i.e. c=(120/100)*10 =>c=12
We can see clearly it is neither in A.P. nor in G.P. also not satisfying last option
Now for H.P. b=2ac/(a+c)
condition satisfied
Ans. c) a,b,c are in harmonic progression

44. If a=b*c then  for any value of n, the equation  (a-b)^n-(c-b)^n+c^n is always divisible by
a) bc
b) b but not c always
c)c but not b always
d)non of above

put n=1 then the given (a-b)^n-(c-b)^n+c^n is divisible by -> bc
put n=2 then the given (a-b)^n-(c-b)^n+c^n is divisible by -> bc
put n=3 then the given (a-b)^n-(c-b)^n+c^n is divisible by -> bc
for for any value of n the given expression is divisible by bc
option A.

45. A and B pick up a ball at random from a bag containing M red, N yellow and O green balls one after the other, replacing the ball every time till one of them gets a red ball.The first one to get the red ball isdeclared as the winner.If A begins the game and the odds of his winning the game are 3 to 2, then findthe ration M:N.

odds in favour of winning the event = (prob that an event will occur):(prob. that an event will not occur)
if odds in favour of a winning the event is 3:2
it means 3 if for red and 2 is for yelow and green.
since the bag contains all three types so none can be zero.
that means yellow:green =1:1.
therefore M:N=3:1

46. There is a circular table and 60 people can sit in that. If there are N number of people sitting and amonster comes and want to occupy a seat such that he has someone on his side. What is the minimum value of N?
a) 15
b) 20
c) 29
d) 30

Say, one person is sitting in seat no. 2 then d monster sits in seat no. 1 or 3 (monster will find a person beside it)
Again there is a person sitting in seat no. 5, then d monster can sit in seat no 4 and 6…
again there is a person sitting in seat no. 8
Thus d pattern of ppl sitting is-
sit no. 2,5,8,11,14,17,20,23,26,29,32,35,38,41,44,47,50,53,56,59
thus min no. of n is 20

47. How many 3 digits numbers less than 1000 are there in which, if there is a 3 it is followed by 7? Given that no two digit of the number should be same.

suppose d 3 digit no. starts with 1
10__(This blank space can b filled in 7 ways 2,4,5,6,7,8,9)
in d above case we cannot take 3 since it should be followed by 7 which is not possible here.
now we take-
11__ this space can also be filled in 7 ways
12__ 7 ways
13__ this space can be filled only by 1 way and dat is 7.
14__ again 7 ways
15__ 7 ways
16__ 7 ways
17__ 7 ways
18__ 7 ways
19__ 7 ways
that gives us 57 ways in total

We will follow d same procedure for 3 digits starting with 2,4,5,6,8,9(but not in case of 3 and 7)

thus we get 57*7= 399 ways

now lets check condition for 3. 3 is always followed by 7 and nothing else
37__ this space can be filled in 8 ways(0,1,2,4,5,6,8,9)
no other condition is possible in case of 3.
that makes our total to be 399+8=407

now lets check condition for 7
70__ this can be filled in 7 ways
71__ 7 ways
72__ 7 ways
73__ this condition is not possible since 3 should be followed by 7 but we cannot repeat 7.
74__ 7 ways
75__ 7 ways
76__ 7 ways
78__ 7 ways
79__ 7 ways
this gives us 56 ways
therefore our total becomes 407+56=463 and this is the answer.

48. How many numbers are there between 1100 – 1300 which are divisible by and that all 4-digits of number should be odd (ex:1331)

1300-1100=200 now divide it by 2 we have 100 even number hence 100 odd no’s
now we see 50 are in each 1100-1200 and 1200-1300 now we see in 1200 to 1299 2 will always be there
hence no number frm here in 1100-1199 we have 1113,1119,1131,1137,1155,1173,1179,1191,1197
hence 9 no’s

49. Three dice are rolled simultaneously. Find the probability of getting at least one six.
a) 1/6*5/6*5/6
b) 5/6*5/6*5/6
c) 3*1/6*5/6*5/6
d) none of the above
condition for no sixes is-
i.e. 125/216
therefore for atleast 1 six-
therefore option d)

50. If there are n numbers of square cubes and you are given two colors, black and white. How many unique square cubes can u color?

Let B= Black color W=white color
If all 6 face colured with B and none with W -> 1 unique cube
5 B and 1 W->1
4 B and 2 W->2
3 B and 3 W->2
2 B and 4 W->2
1 B and 5 W->1
0 B and 6 W->1
———- Answer=10