# Quant Questions With Solutions- 4

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

1/x+1/y=1/z

1/6+1/y=1/z

z=6y/(6+y)

z is a non-negative integer

therefore 6y=(6+y)*k where 'k' is a non-negative integer

y=6k/(6-k)

0<(6-k)<6 data-blogger-escaped-br=""> k={1,2,3,4,5}

y={3,6,12,30}

z={2,3,4,5}

(x,y,z)={(6,3,2),(6,6,3),(6,12,4),(6,30,5)}

number of solutions=4

1/6+1/y=1/z

z=6y/(6+y)

z is a non-negative integer

therefore 6y=(6+y)*k where 'k' is a non-negative integer

y=6k/(6-k)

0<(6-k)<6 data-blogger-escaped-br=""> k={1,2,3,4,5}

y={3,6,12,30}

z={2,3,4,5}

(x,y,z)={(6,3,2),(6,6,3),(6,12,4),(6,30,5)}

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

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

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

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

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?

a)6

b)5

c)0

d)11

a)6

b)5

c)0

d)11

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:

5|70

  ----

  |14

  ----

   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

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:

5|70

  ----

  |14

  ----

   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

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

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;

d=(sqrroot(x2-x1)^2+(y2-y1)^2);x1=r,y1=r,x2=R,y2=R;

replacing values;

d=(sqroot(R-r)^2+(R-r)^2);here dis given as 8 int the ques;

8=(sqrrrot(2(R-r)^2)

8=root2(R-r)

so;R-r=4root(2);

itis because its lies on line y=x int the first quadrant.so we need to find(R-r)

accoding to distance formula;

d=(sqrroot(x2-x1)^2+(y2-y1)^2);x1=r,y1=r,x2=R,y2=R;

replacing values;

d=(sqroot(R-r)^2+(R-r)^2);here dis given as 8 int the ques;

8=(sqrrrot(2(R-r)^2)

8=root2(R-r)

so;R-r=4root(2);

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

99^n

99^1= 99

99^2= 9801

99^3= 970299

take only starting two digit, other digit will not change the solution..

=99*98*97*96*95....

=99*98*97*.....*89

so 11 will be the answer

99^1= 99

99^2= 9801

99^3= 970299

take only starting two digit, other digit will not change the solution..

=99*98*97*96*95....

=99*98*97*.....*89

so 11 will be the answer

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