A,B,C had some sweets. The total nu,ber of sweets that A and C have are 1,000 . If B has 500 sweets, the total sweets that A and B have cannot be more than C by?
a)377
b)179
c)244
d)none
Plz tell the approach also

A hexagon of maximum area is inscribed in a equilateral triangle .What is the ratio of the area of hexagon to the ratio of equilateral triangle.
1)1:2
2)1:4
3)1:3
4)none
Plz post the approach

one mistake u made
FD + SA only = 16
BF + SA only = 4
all three = 4
so SA only = 124 - 24 = 100
look at the ques its given 'only'
but answer is correct

zeecat bhai that is 4 times ,
and all of them is four

156 is correct!

What is the approximate area of a hexagon with side lengths (in order) 5cm, 2cm, 3cm, 6cm, 1cm and 4 cm and all interior angles equal to 120 degrees ?

ABCD is a cyclic quadrilateral. Is the measure of the smallest angle less than 45 degrees ?

a) Three angles of the quadrilateral are in the ratio 1:2:3.
b) Both the diagonals of the quadrilateral are less than diameter of the circumcircle of the quadrilateral.

If a^3 + b^3 -ab <=0 for positive values of a and b, what is the maximum possible value of (a+b) ?

a) 0.8
b) 1.0
c) 1.2
d) 1.5
e) 2.0

1.) I am getting 18.5 cm.....long approach hai....

2.) I am getting using both the statements

Using option 1: I am getting angles as 45,90,135,90 or 36,72,108,144

Using option 2: 45,90,135,90 group is ruled out...Hence 36,72,108,144 are the angles

Using both the options

3.) Is it 1 ??

add 30 numb in each gro.
1000-1099,............,1300-1399,1400-1499 i.e 150
add
1099+1100,
.
.
.
1499+1500 i.e 5 more
total=155
ans.2:)

My take is 156.

Case 1: Numbers are 1xyp and 1xyq (For example: 1001 and 1002)
(p, q) => (0, 1), (1, 2), (2, 3), (3, 4), (4, 5)
x and y => 0, 1, 2, 3, 4

Total ways = 5*5*5 ways = 125 ways

Case 2: Numbers are 1xy9 and 1x(y+1)0 (For example: 1009 and 1010)

x = 0, 1, 2, 3, 4
(y, y+1) => (0, 1), (1, 2), (2, 3), (3, 4), (4, 5)

Total = 5*5 ways = 25 ways

Case 3: Numbers are 1x99 and 1(x+1)00 (For example: 1099 and 1100)
(x, x+1) => (0, 1), (1, 2), (2, 3), (3, 4), (4, 5)

Total 5 ways

Case 4: (1999, 2000)
Only 1 way here

In all 125 + 25 + 5+ 1 = 156 ways

156 is correct!

What is the approximate area of a hexagon with side lengths (in order) 5cm, 2cm, 3cm, 6cm, 1cm and 4 cm and all interior angles equal to 120 degrees ?

ABCD is a cyclic quadrilateral. Is the measure of the smallest angle less than 45 degrees ?

a) Three angles of the quadrilateral are in the ratio 1:2:3.
b) Both the diagonals of the quadrilateral are less than diameter of the circumcircle of the quadrilateral.

If a^3 + b^3 -ab <=0 for positive values of a and b, what is the maximum possible value of (a+b) ?

a) 0.8
b) 1.0
c) 1.2
d) 1.5
e) 2.0

Little Pika who is five and half years old has just
learnt addition. However, he does not know how
to carry. For example, he can add 14 and 5, but he
does not know how to add 14 and 7. How many
pairs of consecutive integers between 1000 and
2000 (both 1000 and 2000 included) can Little
Pika add? (2 marks)
(1) 150 (2) 155 (3) 156
(4) 258 (5) None of the above

My take is 156.

Case 1: Numbers are 1xyp and 1xyq (For example: 1001 and 1002)
(p, q) => (0, 1), (1, 2), (2, 3), (3, 4), (4, 5)
x and y => 0, 1, 2, 3, 4

Total ways = 5*5*5 ways = 125 ways

Case 2: Numbers are 1xy9 and 1x(y+1)0 (For example: 1009 and 1010)

x = 0, 1, 2, 3, 4
(y, y+1) => (0, 1), (1, 2), (2, 3), (3, 4), (4, 5)

Total = 5*5 ways = 25 ways

Case 3: Numbers are 1x99 and 1(x+1)00 (For example: 1099 and 1100)
(x, x+1) => (0, 1), (1, 2), (2, 3), (3, 4), (4, 5)

Total 5 ways

Case 4: (1999, 2000)
Only 1 way here

In all 125 + 25 + 5+ 1 = 156 ways

4 is rite last 1.

Little Pika who is five and half years old has just
learnt addition. However, he does not know how
to carry. For example, he can add 14 and 5, but he
does not know how to add 14 and 7. How many
pairs of consecutive integers between 1000 and
2000 (both 1000 and 2000 included) can Little
Pika add? (2 marks)

(1) 150 (2) 155 (3) 156
(4) 258 (5) None of the above

add 30 numb in each gro.
1000-1099,............,1300-1399,1400-1499 i.e 150
add
1099+1100,
.
.
.
1499+1500 i.e 5 more
total=155
ans.2:)

4 is rite last 1.

Little Pika who is five and half years old has just
learnt addition. However, he does not know how
to carry. For example, he can add 14 and 5, but he
does not know how to add 14 and 7. How many
pairs of consecutive integers between 1000 and
2000 (both 1000 and 2000 included) can Little
Pika add? (2 marks)

(1) 150 (2) 155 (3) 156
(4) 258 (5) None of the above

bhai acc to venn diagram
FD=48
FD+SA=16
FD+BF=8
and all three4
Only FD=20 (1
Total = 28
Only FD=20Similarily
SA=124
FD+SA=16
BF+SA=16
all three=4
Now only SA=124-36=88 (2
Now for BF
lets take x
now (x-(16+8+4))
X-128(3)

adding eq1+2+3 and (16+16+8+4) and putting them equal to 181 getting x now sub from 28

one mistake u made
FD + SA only = 16
BF + SA only = 4
all three = 4
so SA only = 124 - 24 = 100
look at the ques its given 'only'
but answer is correct

yah rite!

A straight line through point P of a triangle PQR
intersects the side QR at the point S and the
circumcircle of the triangle PQR at the point T. lf S
is not the centre of the circumcircle, then which
of the following is true? (3 marks)
(1) (1/PS) + (1/ST) < 2/(QS)(QR)
(2) (1/PS) + (1/ST) < 4/ QR
(3) (1/PS) + (1/ST) > 1(QS)(QR)
(4) (1/PS) + (1/ST) > 4/ QR
(5) None of the above

If S~Q then (1/ST) ~ infinity..
1 and 2 are not possible
PS*ST = QS*SR
(1/PS)+(1/ST) = (PS+ST)/(QS*SR) = D/(QS*SR)
3)D>(QS*SR) * sqrt (QS*SR)
At some points (QS*SR)>D. so.. 3 will not be true always..
4)D>4*(QS*SR)/QR
Lets make the RHS max.. by QS=SR and QR=D
RHS becomes D.. but QS=SR is not possible and hence RHS < LHS..
4 is the answer

bhai acc to venn diagram
FD=48
FD+SA=16
FD+BF=8
and all three4
Only FD=20 (1
Total = 28
Only FD=20Similarily
SA=124
FD+SA=16
BF+SA=16
all three=4
Now only SA=124-36=88 (2
Now for BF
lets take x
now (x-(16+8+4))
X-128(3)

adding eq1+2+3 and (16+16+8+4) and putting them equal to 181 getting x now sub from 28

bhai check my approach

FD+SA=16
FD+BF=8
BF+SA=16
Total students=240-59
=181
Only FD=20
Only SA=88
all three=4
Only BF= (x-28 )
Adding ( x-28 )+16+8+16+20+88+4=181
x=57
now 57-28=29

i got only SA = 100
BF and SA = 4

what is this x-28 ?

zeecat Says

29 ? easy but lengthy

bhai check my approach

FD+SA=16
FD+BF=8
BF+SA=16
Total students=240-59
=181
Only FD=20
Only SA=88
all three=4
Only BF= (x-28 )
Adding ( x-28 )+16+8+16+20+88+4=181
x=57
now 57-28=29

In a locality there are ten houses in a row. On a particular night a thief planned to steal from three houses of the locality. In how many ways can he plan such that no two of them are adjacent to each other.
120
56
73
none

Lets say we fix the first position at 1.
Now fixing the second one at 3.
So the third one can be at position = 6 ways.
Now fixing 2nd one at position 4.
3rd one will be fixed in 5 ways..
So on we get 21 ways
Now fixing the first position at 2 and
2nd one at position 4.
3rd can be at = 5 ways...
Goin like this we get 15 ways
Similarly proceeding like this we will get
21 + 15 + 10 + 6 + 3 + 1 = 56:)without using formula

yah rite!

A straight line through point P of a triangle PQR
intersects the side QR at the point S and the
circumcircle of the triangle PQR at the point T. lf S
is not the centre of the circumcircle, then which
of the following is true? (3 marks)
(1) (1/PS) + (1/ST) < 2/(QS)(QR)
(2) (1/PS) + (1/ST) < 4/ QR
(3) (1/PS) + (1/ST) > 1(QS)(QR)
(4) (1/PS) + (1/ST) > 4/ QR
(5) None of the above

Ans.4 Good question

since no two house shud be adjacent
go from reverse let us assume 3 houses were selcted
so between the remaining 7 houses there are 8 gaps and we have to select 3 from these which can be done in 8C3 ways

In a locality there are ten houses in a row. On a particular night a thief planned to steal from three houses of the locality. In how many ways can he plan such that no two of them are adjacent to each other.
120
56
73
none

X_X_X_X_X_ let thief star from house 1 then no. of ways= 5c3*3!=60
_X_X_X_X_X again starting from house no.2=> no. of ways = 5c3*3!=60 ths total =120 ?

zeecat Says

8c3 = 56 ?

bhai can you post the approach?

in a locality there are ten houses in a row. On a particular night a thief planned to steal from three houses of the locality. In how many ways can he plan such that no two of them are adjacent to each other.
120
56
73
none

8c3 = 56 ?

In a locality there are ten houses in a row. On a particular night a thief planned to steal from three houses of the locality. In how many ways can he plan such that no two of them are adjacent to each other.
120
56
73
none

is it option a? by taking a right angle triangle as a limiting case and then using the concept vthat when sum is constant product is maximum when two numbers are equal.
im weak in geometry

zeecat Says

29 ? easy but lengthy

yah rite!

A straight line through point P of a triangle PQR
intersects the side QR at the point S and the
circumcircle of the triangle PQR at the point T. lf S
is not the centre of the circumcircle, then which
of the following is true? (3 marks)
(1) (1/PS) + (1/ST) < 2/(QS)(QR)
(2) (1/PS) + (1/ST) < 4/ QR
(3) (1/PS) + (1/ST) > 1(QS)(QR)
(4) (1/PS) + (1/ST) > 4/ QR
(5) None of the above

I.


There are 240 second year students in a B-School.
The Finance area offers 3 electives in the second
year. These are Financial Derivatives,
Behavioural Finance, and Security Analysis. Four
students have taken all the three electives, and 48
students have taken Financial Derivatives. There
are twice as many students who study Financial
Derivatives and Security Analysis but not
Behavioural Finance, as those who study both
Financial Derivatives and Behavioural Finance
but not Security Analysis, and 4 times as many
who study all the three. 124 students study
Security Analysis. There are 59 students who
could not muster courage to take up any of these
subjects. The group of students who study both
Financial Derivatives and Security Analysis but
not Behavioural Finance, is exactly the same as
the group made up of students who study both
Behavioural Finance and Security Analysis. How
many students study Behavioural Finance only?
(2 marks)
(1) 29 (2) 30 (3) 32
(4) 35 (5) None of the above


Suja

29 ? easy but lengthy

bhai I believe centers of arcs should be at D and C respectively...
then according to attached figure diameter of circle(by taking options) =2*10=20 thus answer= 10 cm

The base of the triangle is 60 cm. One of the base angle is 60 degress. What is the length of shortest side?

(I)Sum of length of other 2 sides is 80
(II)Other base angle is 45 degrees

Data suficiency question .plz solve it in detail. XAT is my last hope :(

I.
let other 2 sides be a,b
cos 60 = a^2+c^2-b^2/2ac
c=60,a+b=80
So we get,
4b-a=180
nothing can be said from here.

II.
sine 75/60 = sine 45/ X
X = side opposite to smallest angle will be shorter.

There are 240 second year students in a B-School.
The Finance area offers 3 electives in the second
year. These are Financial Derivatives,
Behavioural Finance, and Security Analysis. Four
students have taken all the three electives, and 48
students have taken Financial Derivatives. There
are twice as many students who study Financial
Derivatives and Security Analysis but not
Behavioural Finance, as those who study both
Financial Derivatives and Behavioural Finance
but not Security Analysis, and 4 times as many
who study all the three. 124 students study
Security Analysis. There are 59 students who
could not muster courage to take up any of these
subjects. The group of students who study both
Financial Derivatives and Security Analysis but
not Behavioural Finance, is exactly the same as
the group made up of students who study both
Behavioural Finance and Security Analysis. How
many students study Behavioural Finance only?
(2 marks)
(1) 29 (2) 30 (3) 32
(4) 35 (5) None of the above


Suja

jackoneill Says

Using 1 how do we determine height??

you dont need height, area can be determined using hero's formula
is se we will get a equation in terms of b.
using 1/2*a*b*sin theta you will get the area of triangle
a,b- sides that form an angle of theta between them.
YOu ahve one more equation in terms of b.

Solve these two equations for b.
you now know a(given), b(found) and c = 80-b--> given in the first sentence

Either is suffficient

using 1, we can find semiperimeter = 70, we have a = 60, b = x and c = (80-x)
use hero's formula to find the area.

also using 1/2 a*b*sin theta we can find the area.
equate and find out the value of b. you can get c once you get b. can find the smallest side

using 2, we know one side and the two angles, using sine rule you can find the other two sides

Using 1 how do we determine height??

The base of the triangle is 60 cm. One of the base angle is 60 degress. What is the length of shortest side?

(I)Sum of length of other 2 sides is 80
(II)Other base angle is 45 degrees

Data suficiency question .plz solve it in detail. XAT is my last hope :(

Using the second stmt we get the 3rd angle to be 75..
Using the law of cosines here..
60^2 = a^2 + b^2 - 2ab cos75
also,
a+b = 80 from 1st stmt...
=> a^2 + b^2 + 2ab = 6400
=> 3600 = 6400 - 2ab - 2abcos75
=> 2800 = 2ab(1+cos75)
=> ab = 1120
What is the value of a or b can be determined..but which is the smaller one cannot be decided.

The base of the triangle is 60 cm. One of the base angle is 60 degress. What is the length of shortest side?

(I)Sum of length of other 2 sides is 80
(II)Other base angle is 45 degrees

Data suficiency question .plz solve it in detail. XAT is my last hope :(

Either is suffficient

using 1, we can find semiperimeter = 70, we have a = 60, b = x and c = (80-x)
use hero's formula to find the area.

also using 1/2 a*b*sin theta we can find the area.
equate and find out the value of b. you can get c once you get b. can find the smallest side

using 2, we know one side and the two angles, using sine rule you can find the other two sides

The answer is given as either.I'm still not getting the logic.

According to what you said "side opp 45 is shortest ". But the lenght of the shortest side is asked. how do we find out that??

bhai frst u find all the angles tht come to be 45,75, 60

the side 60 is opp angle 75[ which is nt ny of the base angles]
the side opp largest angle is the largest and the side opp smallest angle is smallest

so using sine rule now we cn calculate the shortest side which is opp angle 45

60/sin75 = X/sin45

i dnt knw how 1st alone is suff

using I alone, its not possible
let a = 60cm, b+c=80 and one angle given
nothing can be said about the length of the shortest
using II alone we cant find anything, even when you use hero's formula after finding s, we will not be able to find the length of the smallest side

using both together,
we have one angle and two sides,using sine formula we get one euqation for area
similarly with the angle as 45 deg, we will get another equation
using both the equations we can find the value of individual sides as
0.5*60*(80-b)sin 60 = 0.5*60*b*sin 45
solve for b, we have all the three sides now

bhai 2nd is sufficient to answer the ques coz the side 60 is opp angle 75 so its the largest and the side opp 45 will be the shortest

using sine rule 60/sin75 = X/sin45

The base of the triangle is 60 cm. One of the base angle is 60 degress. What is the length of shortest side?

(I)Sum of length of other 2 sides is 80
(II)Other base angle is 45 degrees

Data suficiency question .plz solve it in detail. XAT is my last hope :(

******Wrong post****

The base of the triangle is 60 cm. One of the base angle is 60 degress. What is the length of shortest side?

(I)Sum of length of other 2 sides is 80
(II)Other base angle is 45 degrees

Data suficiency question .plz solve it in detail. XAT is my last hope :(

can be solved by second only
from 2nd the angles are 60, 45, 75 so the side opp 45 is shortest which can be solved by sine rule

from 1st x+y = 80 > 60 x and y can assume numerous values fr which the length of ny two sides is greater than the third.. so only 2

The base of the triangle is 60 cm. One of the base angle is 60 degress. What is the length of shortest side?

(I)Sum of length of other 2 sides is 80
(II)Other base angle is 45 degrees

Data suficiency question .plz solve it in detail. XAT is my last hope

bhai I believe centers of arcs should be at D and C respectively...
then according to attached figure diameter of circle(by taking options) =2*10=20 thus answer= 10 cm

ans is rite but your figure is wrong.Q is correct we can draw circle as mentioned in Q.

Suja

bhai I believe centers of arcs should be at D and C respectively...
then according to attached figure diameter of circle(by taking options) =2*10=20 thus answer= 10 cm

bhai nothing is wrong with the question, u can make the desired figure

Consider a square ABCD of side 60 cm. lt contains
arcs BD and AC drawn with centres at A and D
respectively. A circle is drawn such that it is
tangent to side AB, and the arcs BD and AC. What
is the radius of the circle? (5 marks)
(1) 9 cm (2) 10 cm (3) 12 cm
(4) I5 cm (5) None of the above

bhai I believe centers of arcs should be at D and C respectively...
then according to attached figure diameter of circle(by taking options) =2*10=20 thus answer= 10 cm

Bhaiyo two question :
1)After ajay gave Rs500 to Raman, he had less money than Vijay. If vijay gives Rs501 to ajay. Ajay will have Rs500 more than vijay. Who had largest money and how much initially?
2)A,B,C had some sweets. The total nu,ber of sweets that A and C have are 1,000 . If B has 500 sweets, the total sweets that A and B have cannot be more than C by?
a)377
b)179
c)244
d)none
Plz tell the approach also
thankx

1. let ajay=a, vijay=v, raman=r
so according to question, r+500 vijay had more more than raman)
again v-1=a (-> vijay had more money than ajay)
so vijay had the highest amount of money intitially...
2. I dont really get the question, u sure its not incomplete?

Bhaiyo some questions from permutation and combination inme doubt hai jyada...

1)Find the number of ways in which 5 identical marbles can be kept in 10 identical boxes, if not more than one marble go into box.
2)A box has 10 mangoes,Then in how many ways 6 mangoes are selected if each mango is repeated
any number of times.
3)There are 5 different blue color, 4 different blue color and 3 different golden color. In how many ways the same color that are different can be selected.
4)There are 10 white, 8 green , 6 red balls in a box. In how many ways can one or more ball can be put in the box.

Bhaiyo two question :
1)After ajay gave Rs500 to Raman, he had less money than Vijay. If vijay gives Rs501 to ajay. Ajay will have Rs500 more than vijay. Who had largest money and how much initially?
2)A,B,C had some sweets. The total nu,ber of sweets that A and C have are 1,000 . If B has 500 sweets, the total sweets that A and B have cannot be more than C by?
a)377
b)179
c)244
d)none
Plz tell the approach also
thankx

Consider a square ABCD of side 60 cm. lt contains
arcs BD and AC drawn with centres at A and D
respectively. A circle is drawn such that it is
tangent to side AB, and the arcs BD and AC. What
is the radius of the circle? (5 marks)
(1) 9 cm (2) 10 cm (3) 12 cm
(4) I5 cm (5) None of the above