# Quant Questions With Solutions- 1

1.

__Find the number of ways in which you can fill a 3x3 grid(with four courners defined as a,b,c,d) if u have 3 white marbles and 6 black marbles.__

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 beplaced in those 6 spaces. So, that can be done in 1 way.

Hence the answer would be 9C3.

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 (banglestouching 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

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.

a)100(QA/PD)

b)100(PD/QA)

c)100-100(QA/PD)

d)100PD-100(QA/PD)

a)100(QA/PD)

b)100(PD/QA)

c)100-100(QA/PD)

d)100PD-100(QA/PD)

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)

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)?

a)0

b)1/3

c)1/2

d)1

b)1/3

c)1/2

d)1

Let x=(e(e(e.........)^1/2)^1/2)^1/2)

now squaring both side we get

x^2=e(e(e(e.........)^1/2)^1/2)^1/2)

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

now squaring both side we get

x^2=e(e(e(e.........)^1/2)^1/2)^1/2)

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

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

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

(a+b)(b+c)(c+a)=(ab+ac+b2+bc)(c+a)

=abc+ac2+b2c+bc2+a2c+a2b+b2a+abc

=0

=> abc=a2b+ab2+ac2+a2c+b2c+bc2+3abc

=> a2b+ab2+ac2+a2c+b2c+bc2+2abc=0

(a+b)(b+c)(c+a)=(ab+ac+b2+bc)(c+a)

=abc+ac2+b2c+bc2+a2c+a2b+b2a+abc

=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

a)11/20

b)3/5

c)13/20

d)18/25

a)11/20

b)3/5

c)13/20

d)18/25

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.

PQR~PMR

hence PR/RM=QR/PR.

=>PR^2=RM*QR

putting the values & from(1)

=>RM=18/5....(2)

Also

PQR~PQM

=>PQ/QM=QR/PQ

=>PQ^2=QM*QR

putting the values & from(1)

=>QM=32/5.....(3)

Also,

PQM~PMR

=>PM/QM=RM/PM

=>PM^2=QM*RM

from(2) & (3)...

PM=24/5......(4)

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

=>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.

PQR~PMR

hence PR/RM=QR/PR.

=>PR^2=RM*QR

putting the values & from(1)

=>RM=18/5....(2)

Also

PQR~PQM

=>PQ/QM=QR/PQ

=>PQ^2=QM*QR

putting the values & from(1)

=>QM=32/5.....(3)

Also,

PQM~PMR

=>PM/QM=RM/PM

=>PM^2=QM*RM

from(2) & (3)...

PM=24/5......(4)

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.

a)2

b)pi-2

c)2-pi/4

d)2-pi/2

a)2

b)pi-2

c)2-pi/4

d)2-pi/2

area of cirle 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 hav drwan its fig u'll find that to obtain the required ans u hav 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 substracted 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))

=pi-(pi-2)

=2.

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 hav drwan its fig u'll find that to obtain the required ans u hav 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 substracted 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))

=pi-(pi-2)

=2.

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