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1.  The circle o having a diameter 2 cm,has square inscribed in it each side of the square is then     taken as the diameter to form 4 smaller circles to' find the total area of 4 circles which is outside the circle O.

ANS:

Since the square is inscribed in the circle,
diameter of circle = diagonal of square

===> diagonal of square = 2 cm
Side of square = diagonal/sqrt(2)= 2/sqrt(2) cm

Diameter of each of the 4 circles = 2/sqrt(2)
Radius of each of the 4 circles ,r = diameter/2 = 1/sqrt(2)

Total area of 4 circles = 4 * pi * r^2
=4 * pi * 1/2
=2*pi
=6.283cm^2

2. There are two swimmers at the two end of a swimming pool..Then first meet at a distance of 50 km from the east and there second meeting will be at a distance of 20 km from the west end. What is the length of the Pool.

ANS:

50+20=70
50 frm east and 20 from west
total 70

OR

if they started swimming same time
let total distance=D,First swimmer speed which starts from west=W &second=E;
1-from first condition-
time taken by both swimmers same-> (D-50)/W =50/E (time=distance/time)
W/E=50/(D-50)..........(1)
2-Second condition--
Distance covered by first=(2D-20)
"" "" "" "" second=(D+20)
by above formula W/E=(D+20)/(2D-20) ........(2)
after solving above two equations D=130 ans-130

3. A triangle ABC has an area of 150 sq cm. The length of the largest side is 50 cm and smallest side is 10 cm. What will be the perimeter of the triangle

ANS:

ans: perimeter = 60 + 30√2 = 30(2+√2)= 102.43 cm

let height of triangle wrt largest side be h cm

then, area of triangle = 1/2 *base*height
=> 1/2 *50*h = 150
=> h = 6 cm

this height will divide traingle in two right angled triangles

base of smaller triangle = √(10^2-h^2)=√(10^2-6^2)= 8 cm
so, base of larger traingle =(50-8)= 42 cm

third side of triangle = √(42^2+h^2)= √(42^2+6^2)= 30√2 cm

perimeter of the triangle = 50 + 10 + 30√2 = 60 + 30√2 = 30(2+√2)= 102.43 cm

4. What will be the remainder if 16/12?
4 or 1.


ANS:

4

16 ==> 12 * 1 + 4 = 16

So remainder is 4...

12*1=12 ,so to get 16 we'll have to add 4, hence remainder =4.
we cannot simplify this as 16/12= 4/3. therefore, 1 is wrong.

5. Ratio of boys to girls 4:5 , when 100 girls leave the school the ratio becomes 6:7 . How many boys are there ?
a) 1600
b) 1200
c) 600
d) 800

ANS:

B/G = 4/5 => G = 5B/4 -----(1)

B/(G-100) = 6/7 ------(2)
=> 7B = 6G -600
=> 7B = 6*5B/4 -600
=> 28B = 30B -2400
=> 2B = 2400
=> B = 1200

b) 1200

6. Dhoni when hits a boundary he swings his bat. When he hits a boundary he swing his bat 4 time and when he hits a six he swings his bat 7 time .when he hit 2 boundary he swing his bat for both the boundary i.e. say he hits a four first then he swings 4 time. Next when he hits a six or four he swings his bat 4+7(for six) 4+4(for four),and so on.so for 3 boundary say 4,6,4 he swings 4,4+7,4+7+4 numbers of times. In total he swings 4+4+7+4+7+4 number of time. 1) dhoni swings his bat for 99 times, find out number of way he can hit boundary.

ANS:

total no of swings =99 which is equal to ( 4*23 + 7*1 =99)
therefore, no of ways he can hit the boundary = (4 + 4+.....22 times + (4+7)once)=23 times.

OR
There are two ways
1st way==[4,4+7,4+7+4,4+7+4+4,4+7+4+4+4,4+7+4+4+4+4]
2nd way==[4,4+4,4+4+4,4+4+4+7,4+4+4+7+7,4+4+4+7+7+4]

7. OUM
x WUL
-------
WWSO
ILOI*
UUES**
-------
USIITO
-------

ANS:

743
x 649
------
6687
2972*
4458**
------
482207
------

8. 1/2(log base 6 of (x^2-8x+16)) + 1/2(log base 6 of(x^2-18x+81)) = 1
How many real solutions of x exist?

ANS:

1/2(log base 6 of (x^2-8x+16)) + 1/2(log base 6 of(x^2-18x+81)) = 1

=> 1/2[ log base 6 of (x^2-8x+16) + log base 6 of(x^2-18x+81)]= 1

=> log base 6 of (x-4)^2 + log base 6 of (x-9)^2 = 2

=> log base 6 of [(x-4)^2*(x-9)^2] = log base 6 of 36

=> (x-4)^2 * (x-9)^2 = 36

=> (x-4)*(x-9)=6 0r (x-4)*(x-9)= -6

=> x^2-13x+30=0 0r x^2-13x+42=0

=> (x-3)(x-10)= 0 or (x-6)(x-7)= 0

=> x= 3,10,6,7

all 4 values satisfies the given eqn,

4 real solns

9. THE ARE 20 STUDENTS AND THEY NEED TO BE ALLOCATED INTO 3 CLASSROOMS.IN HOW MANY WAYS CAN THIS BE DONE
1.ASSUMING THAT THE STUDENTS ARE IDENTICAL
2.ASSUMING THAT THE STUDENTS ARE DISTINGUISHABLE

ANS:

1. 20 students have to be allocate in 3 identical classrooms
n=20,r=3
no. of ways=(n+r-1)C(r-1)= (20+3-1)C(3-1)= 22C2 = 231

2. 20 students have to be allocate in 3 DISTINGUISHABLE classroms
each student can be allocated in 3 ways so total
no. of ways = 3^20

10. GAS
*FBI
------
FTBI
SSTB
SASF
--------
SRISTI
--------

ANS:

S=1 is pretty evident as the multiplicand digits are same as the last digit of each row of multiplication.
Now we can see B+B=T
and GA1*B=11TB
using hit and trial starting from taking the value of B as 2
=>B=2 , then T=4
=>A*B=_4 , A*2=_4 : A=7 only
=>Also,2*G+1=11 : G=5
=>T+T+F=_1 : 8+F=_1 F=3
=>F+S+S=I : 3 + 1 + 1 + CARRY(1)=6=I
=>S+A=R : 1+7=8=R

Therefore,
571
*326
________
3426
1142
1713
_________
186146

11. There are m pipes from 1 to m..such that tge speed at which it fills a pool alone is equal to the time taken by other m-1 pipes(not for m=1)..suppose time taken by 9 pipe is 40min when filled alone..what would be the time taken if pipe no. 12 and 13 fill together..??

ANS:

using Efficiency consept..
1/T9=1/40, given
=>1/T9=1/T8 + 1/T7 +.......+ 1/T1 = 1/40
=>1/T10=1/T9 + (1/T8 + 1/T7 +.......+ 1/T1) = 1/40+1/40
=>1/T11=1/T10 + 1/T9 + 1/T8 + 1/T7 +.......+ 1/T1 = 1/20 + 1/20
=>1/T12=1/T11 + 1/T10 + 1/T9 + 1/T8 + 1/T7 +.......+ 1/T1 =1/10 + 1/10
=>1/T13=1/T12 + 1/T11 + 1/T10 + 1/T9 + 1/T8 + 1/T7 +.......+ 1/T1 =1/5 + 1/5
now, time taken by pipe no. 12 & 13 will be given by
1/T12 + 1/T13 = 1/5 + 2/5 =3/5
Hence,time = 5/3 =1.66min

12. Suppose you have a currency called miso in three denominations, 1 miso, 10misos and 50 misos. In how many ways can you pay a bill of 107 misos?
Option
A.17
B.16
C.18
D.15
E.19

ANS:

Let the number of currency 1 Miso, 10 Misos and 50 Misos be x, y and z respectively.
x+10y+50z=107

Now the possible values of z could be 0, 1 and 2.

For z=0: x+10y=107

Number of integral pairs of values of x and y that satisfy the equation:
x+10y=107 will be 11.

These values of x and y in that order are:
(7,10);(17,9);(27,8)…(107,0)

For z=1: x+10y=57

Number of integral pairs of values of x and y that satisfy the equation:
x+10y=57 will be 6.

These values of x and y in that order are:
(7,5);(17,4);(27,3);(37,2);(47,1) and (57,0)

For z=2: x+10y=7

There is only one integer value of x and y that satisfies the equation:
x+10y=7 in that order is (7,0)

Therefore total number of ways in which you can pay a bill of 107 Misos:
=11+6+1= 18

Ans : (C)

13.

 Multiplication Problem 1

This type of problems can be solved by back-tracking technique after finding one letter (number).

Approach (Method):

The best method which I follow is from bottom to top multiplication (You can do in vice-verse), but I found it was easy.

If you see the question, start from bottom to top like D*(ABC) and E*(ABC)

It is given that E * C = C this happens only with numbers of 5 & 6 let’s see how?

Explanation:

1 * 5 = 5

3 * 5 = _5 (Actually 15 but concentrate on last digit)

7 * 5 = _5 (35) I am not using 5 * 5 because it’s already assigned to C we should not repeat.

9 * 5 = _5 (45) so, from this don’t conclude C = 5 & E= (1, 3, 7, 9).

Because there is another number with these properties Ex: – number (6)

6 * 2 = _2 (12)

6 * 4 = _4 (24)

6 * 8 = _8 (48)

From this don’t conclude E = 6 & C = (2, 4, 8).

Let, note all the possible cases and try. For Example take C = 5 and E = odd number (1, 3, 7, 9)

	A	B	C
		D	E
	F	E	C
D	E	C
H	G	B	C

Now Replace C with 5 and redraw the following table.

And also place E as odd number (3, 7, 9) I didn’t write 1 here because it’s not going to be 1 Let’s take first E = 3 and fill the table.

	A	B	5
		D	3
	F	3	5
D	3	5
H	G	B	5

If you see the colored portion then sum 3 + 5 = B (8) lets assign 8 to B

And redraw the table, if u put B = 8, let’s check from first

3 * 5 = 5 (carry 1)

3 * 8 = 24 + 1 = 5 (carry 2) but in the table it is showing 3 it is contradictory, then we can understand that E! = 3.

So Next I will take E = 7 and again fill the table.

	A	B	5
		D	7
	F	7	5
D	7	5
H	G	B	5

	A	2	5
		D	7
	F	7	5
D	7	5
H	G	2	5

If you see the red color portion

Sum of 7 + 5 = 2 (carry 1) so we got B = 2.

Assign B = 2 and fill the table.

Now see the row which I highlighted and think what value we can keep for D after clear Observation we can get D = 3 satisfy the given condition and put D = 3.

	A	2	5
		3	7
	F	7	5
3	7	5
H	G	2	5

	1	2	5
		3	7
	8	7	5
3	7	5
4	6	2	5

To get 375 in second row the value of A is

A = 1 so put value of A in table

If you keep A = 1 it’s very simple

F = 8, G = 6, H = 4.

14. A P R
*O C T
-------------
P U R A
R O J R
R E C U
-------------
R A A J A A
What is the value of E ?

ANS:

E = 1

A P R
*O C T
-------------
P U R A
R O J R
R E C U
-------------
R A A J A A

==> R+R =A ; 2R=A
R+E+1=A ; R=e+1
U+U+J=J ; U= 0 or 5
U=0
APR
*O
----
RECU

ie
432
* 5
----
2160
====>E=1

APR
* T
----
PURA

432
* 7
----
3024

APR
* C
----
ROJR

432
* 6
----
2592

APR*OCT=RAAJAA
432*567 = 244944

15. No of ways you can fill a 3*3 grid (with 4 corners marked as a,b,c,d) if you have 3 white and 6 black marbles.
Option
a) 9C3
b) 6C3
c) 9C3+6C3
d) (9C3+6C3)/3!

ANS:

3 white marbles can be filled in (3*3=9) grids in 9C3 ways & rest 6 will be automatically filled
so, no. of ways = 9C3

if we start with black marbles then 6 black marbles can be filled in 9C6 ways & rest 3 will be automatically filled

so, no. of ways = 9C6

2nd mtd,
we have to filled 3*3=9 grid with 3 of white & 6 of black marbles
so, no. of ways = 9!/(3!*6!) = 9C3 = 9C6 = 84

16. v,w,x,y,z are non negative integers each <11, then how many distinct combinations of (v,w,x,y,z) satisfy
v(11^4)+w(11^3)+x(11^2)+y(11)+z= 151001

ANS:

v(11^4)+w(11^3)+x(11^2)+y(11)+z= 151001

=> 11*( v*11^3 + w*11^2+ x*11 + y ) + z = 11*13727 + 4

=> ( v*11^3 + w*11^2+ x*11 + y )= 13727 & z = 4

=> 11*( v*11^2 + w*11 + x ) + y = 11*1247 + 10

=> ( v*11^2 + w*11 + x )= 1247 & y = 10

=> 11*( v*11 + w ) + x = 11*113 + 4

=> ( v*11 + w )= 113 & x = 4

=> ( v*11 + w ) = 11*10 + 3

=> v = 10 & w = 3

so, v=10, w=3, x=4, y=10 ,z=4

(v,w,x,y,z)=(10,3,4,10,4)

so, only one combination

17. How many values of 'c' results in rational roots which are integer in
x^2-5x+c.
Options
a) 1
b) 3
c) 6
d) Infinite

ANS:

equation is
x^2-5x+c=0

=> x = [5 +- √(25-4c)]/2

c=0 => x = 5,0
c=4 => x = 4,1
c=6 => x = 2,3
c=-6 => x = 6, -1
.......

thus we see that, for c = 0,4,6,-6,... we get rational roots which are integers

no. of values of c = infinite

18. A square of side x is given... joining the mid point of the sides another square is made...in this way it is continued...what will be the perimeter of the the 9th square???

ANS:

area of nth square = x^2/2^(n-1)

area of 1st square = x^2
area of 2nd square = x^2/2 = x^2/2^1
area of 3rd square = x^/4 = x^2/2^2
area of 4th square = x^2/8 = x^2/2^3
......

area of 9th square = x^2/2^8

perimeter of 9th square = 4*side = 4* √(x^2/2^8) = 4*x/2^4 = 4*x/16 = x/4

19. CGD
*BQS
---------
OQSC
GOSR
QPOO
---------
DSBRSC

ANS:

846 * 753 = 637038

C G D
*B Q S
---------
O Q S C
G O S R
Q P O O
---------
D S B R S C

S+R=S ===>R=0
D=Q+1 and D*Q=R(0) ===> D=6, Q=5
Q+S+O=R(0)
ie, S+O = 5 === S=3 and O=2
by trail and error
B=7, C=8, G=4, P=9

20. 100^x (log x (base 3) - log sqrt x (base 9))... X is Always greater than 1
Option
a) 0, infinity
b) -infinity, infinity
c) infinity, -infinity
d) None

ANS:
it should be right
100^x (log x (base 3) - log sqrt x (base 9))
= 100^x (log x (base 3) - log x^(1/2) (base 3^2))
= 100^x (log x (base 3) - (1/4)*log x (base 3))
= 100^x *(3/4)*log x (base 3)
as x > 1     
log x (base 3) is always +ve , so given exp always +ve & never equal to zero or -infinity. as x inc, value of given exp tends to infinity.    so, 100^x (log3 basex - log root9 basex) lies in interval ( 0, infinity)
                                                                                                                                                                   
                                                                                                                                                                   

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