Let the distance covered by dog in 1 leap is x and hare
covered 1 leap is y.
then, 3x = 4y
=> x =(4/3) y
=> 4x =(16/3) y
Then, The ratio of speeds of dog and hare = Ratio of distances covered by them
in the same time
= 4x : 5y
= (16/3)y : 5y
=(16/3) : 5
= 16:15 Ans...=16:15
there are two boxes,one containing 39 red
balls & the other containing 26 green balls.you are allowed to move the
balls b/w the boxes so that when you choose a box random & a ball at random
from the chosen box,the probability of getting a red ball is maximized.this
maximum probability is
a)60 b)50 c)80 d)30
the probability of getting a red ball should be maximized so
we will keep 1 red ball in the first box and transfer remaining 38 red balls to
other box which has green. The reason for doing this only this combination will
make the probability of getting red ball is maximum (i.e) 1.
In the question its mentioned that first we have to choose a box at random.
So 2 boxes are there, probability of selecting a bag is 1/2
Probability of getting red ball from first box is 1
Probability of getting green ball from second box is 38/64
So (1/2)*1 for first box and (1/2)*(38/64) for box 2
answer is (1/2)*1 + ((1/2)*(38/64) which is approximately 0.8
Next number in the given series 1, 7, 8, 49,
50, 56, 57, 343
344 because
1*7=7
7+1-8
7*7=49
49+1=50
8*7=56
56+1=57
49*7=343
343+1=344
ans= 344
q.1 in how many ways can 3 postcards can be
posted in 5 postboxes?
q.2.in how many ways can 5 postboxes hold 3 post cards?
q.1 in how many ways can 3 postcards can be posted in 5
postboxes?
first post card can be posted in any of the 5 post boxes =5c1 ways
second post card can be posted in any of the 5 post boxes =5c1 ways
third post card can be posted in any of the 5 post boxes =5c1 ways
number of ways=5c1*5c1*5c1=125=5^3
q.2.in how many ways can 5 postboxes hold 3 post cards?
first post box can hold 3 post cards in 3c1 ways
second post box can hold 3 post cards in 3c1 ways
third post box can hold 3 post cards in 3c1 ways
fourth post box can hold 3 post cards in 3c1 ways
fifth post box can hold 3 post cards in 3c1 ways
so number of ways =3c1*3c1*3c1*3c1*3c1=3^5
A number has exactly 3 prime factors, 125
factors of this number are perfect squares and 27 factors of this number are
perfect cubes. overall how many factors does the number have???
Answer for this question is 729
the explanation goes like this
Let a,b,c be three prime no.
since the no. of perfect square as a factor is 125
which can only be if every prime no. has power in the multiple of 2 including
zeros(bcoz any no. to the power of zero is a perfect square)
so power must be 0,2,4,6,8
thereby 5*5*5=125
Since no. of cube as a factor is 27
therefore power in the prime no. will be in the multiple of 3
so power are 0,3,6
hence the no. N=a^8*b^8*c^8
hence total no. of factor is (8+1)(8+1)(8+1)=729
there r three buckets..of 8,5 n 3 litres...out
of which only 8 ltr bucket is fully filled...u hv to fill exact 4-4 ltr liquid
in 8 and 5 litre bucket by using only these buckets in minimm num of
steps......
8ltr 5ltr 3ltr
initially 8 0 0
step1 3 5 0
step2 3 2 3
step3 6 2 0
step4 6 0 2
step5 1 5 2
step 6 1 4 3
step 7 4 4 0
total 7 steps
The climb from foot to top of a hill 800
meters, Jack can climb at 16 meters per minute and rests for two minutes or
20meters per 2 minutes and rest for one minute. Paul can climb at 10 meters per
one minute and rest for one minute or16 meters per minute and rest for 2
minutes. If take has to reach the top in exactly two hours. What is the maximum
number of rests that he can take?
a) 41 b) 42 c) 40 d) 43
total=800m
16m per min means for 1m=800/16 =50min to climb
10m per min means for 1m=800/10=80min to climb for paul .....the total time is
2hours==120min....so,time taken for rest by paul is 120-80=40min
length of minute hand is 5.4 cm, area covered
by this in 10 min is ?
a)50.97 b)57.23 c)55.45 d)59.14
t will be
(3.14*5.4*5.4)*10/60 = 15.26 cm2.
Raj tossed 3 dices and there results are noted
down then what is the probability that raj gets 10?
a) 1/72 b) 1/9 c) 25/216 d)1/8
ossible
event.......(1,3,6)(1,4,5)(1,5,4)(1,6,3)(2,2,6)(2,3,5)(2,4,4)(2,5,3)(2,6,2)(3,1,6)(3,2,5)(3,3,4)(3,4,3)(3,5,2)(3,6,1)(4,1,5)(4,2,4)(4,3,3)(4,4,2)(4,5,1)(5,1,4)(5,2,3)(5,3,2)(5,4,1)(6,1,3)(6,2,2)(6,3,1)
total 27 possible event
so probability = 27/216 i.e 1/8
answer-------d(1/8)
- Probability of a sum of 3: 1/216 = 0.5%
- Probability of a sum of 4: 3/216 = 1.4%
- Probability of a sum of 5: 6/216 = 2.8%
- Probability of a sum of 6: 10/216 = 4.6%
- Probability of a sum of 7: 15/216 = 7.0%
- Probability of a sum of 8: 21/216 = 9.7%
- Probability of a sum of 9: 25/216 = 11.6%
- Probability of a sum of 10: 27/216 = 12.5%
- Probability of a sum of 11: 27/216 = 12.5%
- Probability of a sum of 12: 25/216 = 11.6%
- Probability of a sum of 13: 21/216 = 9.7%
- Probability of a sum of 14: 15/216 = 7.0%
- Probability of a sum of 15: 10/216 = 4.6%
- Probability of a sum of 16: 6/216 = 2.8%
- Probability of a sum of 17: 3/216 = 1.4%
- Probability of a sum of 18: 1/216 = 0.5%
Apple costs L rupees per kilogram for first
30kgs and Q rupees per kilogram for each additional kilogram. If the price of
33 kilograms is 11.67and for 36kgs of Apples is 12.48 then the cost of first 10
kegs of Apples is
a) 3.50 b) 10.53 c) 1.17 d)2.8
30L+3Q=11.67
30L+6Q=12.48
------------
3Q=.81 Q= .27
from that L=0.362 cost of 10 kg apple is 10*.362=3.6
The letters in the word ABUSER are permuted in
all possible ways and arranged in alphabetical order then find the word at
position 49 in the permuted alphabetical order?
a) ARBSEU
b) ARBESU
c) ARBSUE
d) ARBEUS
AB**** =4!= 24 ways
AE****=4!=24 ways next word is 49th so
AR**** in alphabetical order **** will be BESU
ans is B
A is twice efficient than B. A and B can both
work together to complete a work in 7 days. Then find in how many days A alone
can complete the work?
ans is 10.5
gn:
A+B=(1/7)
A=2B
=>B=1/21
A=2/21
=>A take (21/2)=10.5 days to complete the work.
In 8*8 chess board what is the total number of
squares.
1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2=204....
formula=1^2+2^2+3^2+....+n^2
X,Y,W,Z ARE INTEGER THE EXPRESSION X-Y-Z IS
EVEN&Y-W-Z ODD IF X IS EVEN THEN WHICH OF FOLLOWING IS TRUE(a)Y MUST BE
ODD(b)Y-Z MUST BE ODD(c)W MUST BE ODD(d)Z MUST BE ODD
Ans.(c)W must be ODD
X-Y-Z=even and X even no.
hence (y-z) must be even because (even -even = even)
and y-w-z=y-z-w=odd and (y-z) is even so W must be odd because (even - odd=odd)
hence option (c) is correct
two cyclist begin training on oval racecourse at
same timethe professional cyclist complete each lap in 4 sec noves take 6
mintue how many mintue after start will both cyclist pass at exactly same spot
where they begin to cycle(a)10(b)8(C)14(D)12
ans is d.
simply LCM of 4 and 6.
which of following true:A occur only B or C
occur B occur D&E occurf occur if Cnot occur g occur if A and F occur
option A occur whenever F occur2)F never occur3) G not occur if D not
occur4)none of these
3) g nt occure i d not occure
a=>b or c
b=>d & e
f=>c not
g =>a & f
1) a occure whenevr f :- depen on c where as 'a' can occure if 'c' not ..so
this is not true
2) f nevr occure => depnd on c only n c is indipendent ...if c not occure f
will occure ....so this is also not true
3) g not occure if d not:-g depnd on a & f ...if d not ocuure b will not
occure (as b=>d&e) a will occure if c occure...if c occure f will not
occure ...hence g will not occure...this is true
House1 is older than H2, H3 is taller than
H4,H4 is older than H1...(i)H1 is older than H3 (ii)H2 is taller than H4
(a)only 1 (b)both (c)neither (d)only2
c)neither, is the answer.
We can't relate heights of 2 & 4 coz relation b/w heights of 4 and 2 isn't
given(4 is 'OLDER' than 2, we can't say anything about their height);also no
relation is stated b/w ages of 1 & 3, so neither of i and ii is correct.
cost of 3 mango,4 pears cost 55 and cost of 1
apple, 5 pears cost 70 ,and cost of 2 mango, 5 apple cost 60...what the total
cost of 1 app,1 mango and 1 pea?
answer is 35
3 mango,4 pear 3x+4y=55
5 pear ,1 apple 5y+z=70
5 apple, 2 mango 2x+5z=60
add all 5x+9y+6z=185 equation 1
equation 2 is 3x+4y = 55
by using z=70-5y put in to equation 2 and we get 5x-26y=-235 equation 3
subtract eq.2 - eq 3
we get y=10 , z= 20, x= 5
so x+y+z=35
(1) From 52 cards 3 cards drawn randomly prob
of getting 1 spade , 1 red queen and 1 black king??
13C1*2C1*2C1/52C3
= ANS
Or
13C1*2C1*1C1/52C3
1 black king involve in the spade series,so just one has left
(2^1096
+ 2^2248 + 2^2n) find the value of n so that the value will be perfect square
(a)2011
(b)2012 (c)2010 (d)2008
GOOOOOOOOD like a number e.g(54321) was there
then what should be added to that number so the remainder will be 35 when
divided by 424.
if v divide 54321 wid 424 we get 49 as remainder
so we must add 424-49=375 to get 0 as remainder...but we want remainder 35 so
375+35=410 should b added to 54321 to get remainder 35....
Q- two points r there two person from A
running to same direction with speed 20 km/hr & 15/hr respectively and from
other end another person running to opposite direction with 30 km/hr? distance
bw them 100 km??at what time they will meet?
avg speet of 20 and 15 is =120/7 km/h (2ab/a+b)
let they will meet after time t.
t=100*7/120 -100/30
solvin t=2.5 hour
In how many ways can 20 identical pencils be
distributed among three girls so
that each gets at least 1 pencil?
now A'+B'+C''=17,
number of non negative integers of A',B',C' are (17+3-1)C(3-1)
so total Combinations are 19c2=19*9 = 171,
simple formula for this kind of problems is
(n-1)c(r-1)
so 19c2= 171
if p(x)=ax^4+bx^3+cx^2+dx+e has roots at x=1,2,3,4 &
p(0)=48.what is p(5)
48
if 1 is a root then x-1 is
a factor of p(x)
similarly x-2 is a factor,x-3 ,x-4 are factors
but p(x) is 4th degree polynomial therefore it can be in the form
p(x) = Q(x-1)(x-2)(x-3)(x-4) , where Q is a constant
but given p(0) =48
therefore 24 Q =48
Q=2
p(x) =2(x-1)(x-2)(x-3)(x-4)
p(5)= 2*4*3*2*1 = 48
1!+2!+3!...+50! divided by 5!
remainder will be ?
1!+2!+3!...+50! divided by 5!
remainder will be whatever is obtained by dividing 1!+2!+3!+4! with 5!.
so remainder is obtained by dividing (1+2+6+24)= 33 with 5! ( 120).
so remainder is 33.
if there are Six periods in each working day
of a school, In how many ways can one arrange 5 subjects such that each subject
is allowed at least one period?
we have 5 sub and 6 periods so their arrangement is 6P5 and
now we have 1 period which we can fill with any of the 5 subjects so 5C1
6P5*5C1=3600
An article manufactured by a company consists
of two parts X and Y. In the process of manufacturing of part X, 9 out 100
parts many be defective. Similarly , 5 out of 100 are likely to be defective in
the manufacturer of Y. Calculate the probability that the assembled product
will not be defective?
a)0.6485
b)0.6565
c)0.8645
d)none of these
ans=c (0.8645)
probablity of nondefective of x= 91/100=.91
probablity of non defective of y= 95/100=.95
so, probablity of nondefective product=.91*.95=0.8645
There are six multiple choice questions in the
examination. How many sequences of anwers are possible, if the first two
question have 3 choices each, the next two have 4 choice each and last two have
5 choices each?
3c1*3c1*4c1*4c1*5c1*5c1=3600
for any one of first three option of Q.1 we have three options in Q.2 4 option
in Q.3.....this goes on...
ind the area (in square units) of the triangle
formed by 2x+3y=5, y=x and X-Axis.
the points of intersection for lines:
2x+3y=5 and x=y => 5x=5 => x=1 and y=1 =>(1,1)
x=y and y=0(x-axis) => x=0 and y=0 =>(0,0)
2x+3y=5 and y=0(x-axis) => 2x=5 => x=5/2 and y=0 =>(5/2,0)
area of triangle with vertices (x1,y1),(x2,y2),(x3,y3) is
|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|/2
=|1(0-0)+0(0-1)+5/2(1-0)|/2
=|1+5/2|/2
=7/4 sq units...
Arun was all bent on building a new house. He
carefully got the blue print of his house designed buy his friend Ashwin, a
civil engineer. He wanted to build a room of dimension 27 by 48 ft and lay
tiles in this room. Each tile was of dimension 2 by 3 ft. How many tiles should
Arun buy?
27*48/2*3=216
Roy is now 4 year older than Erik and half of
that amount than Lewis.If in two years Roy will be twice as old as Erick,then
in two year what would be Roy age multiplied by Lewis age??
ans is 48
firstly two equation from 1st sentence ..
suppose roy=x, erik=y, lewis=z
so x=y+4....(1)
2nd is x=z+2...(2) (Roy is now 4 year older than Erik and half of that amount
than Lewis so 4/2=2)
also after 2 years erik and roy relation wil b
x+2=2(y+2).. (3)
so put this into 1st equ. u will get y=2..
and then x=6,z=4(by substituting other)
nw asking multiplication after two years of roy n lewis is (6+2)*(4+2)=48..
if there are maximum acute angles then how
many convex hectagons are there
there can be max 3 acute angles in a convex polygon which is
a triangle.
The number of bacteria was growing in a city
exponentially.at 4 pm yesterday , the number of bacteria was 400 and at 6 pm
yesterday it was 3600.How many bacteria were there at 7pm yesterday?
The number of bacteria was growing in a city
exponentially.at 4 pm yesterday , the number of bacteria was 400 and at 6 pm
yesterday it was 3600.How many bacteria were there at 7pm yesterday?
.there are 12 children in a family .the
youngest 1 is a boy.then find the probabality
of finding boy in the family.
a)4096
b)2048
c)2
d)2!
Probability cannot be more than 1..it ranges from 0 to 1..
there can be two options..either a boy/girl in a family
asked is to find a boy=1/2
given 12 children but it is already mentioned that the youngest is a boy so no
need to count that..
hence,1/2*1/2*1/2......(11 times)=1/2048 option b..but the option is wrong..it
should be 1/2048
A volume of 10936 l is in a conatiner of
sphere . how many semispheres of volume 4 l each will be required to transfer
all the water into small semispheres?
v10936/4=2734.....is the ans
a volume of A are having in a container of
sphere. how many semi hemispheres of B volume each will be required to transfer
all the A in to semi hemispheres?
ans is A/B.
My name is PREET.But my son accidentally types
the by interchanging a pair of letters in my name.What is the probability that
despite this interchange, the name remains unchanged? a.5% b.20% c.25% d.12.5%
10%
Selecting any two - 5C2 = 10[ treating E AND E as distinct)
Selecting EE = 1
1/10 * 100 = 10%
