1> the milk and water in vessels a and b are in ratio 4:3 and 2:3.in what
ratio the liquids in both vessels are mixed to obtain a new mixture in vessel c
contain half milk and half water ?
let the total liquid in container = x
quantity of A in container 1 = 4x/7
quantity of B in container 1 = 3x/7
quantity of A in container 2 = 2x/5
quantity of B in container 2 = 3x/5
new ratio of A and B = 1/2:1/2
let quantity container 1 mix = n
let quantity container 2 mix = m
=>now Quantity of A = 4xn/7+2xm/5 = x(4n/7+2m/5)
=>now Quantity of B = 3xn/7 + 3xm/5 = x(3n/7 + 3m/5)
As new quantity of A and B are equal
=> 4n/7+2m/5 = 3n/7 + 3m/5
=> n/7= m/5
=> n/m =7/5
2> numbers are formed by using 1,2,3,4,5(repetetion not allowed)sum of all
nos. formed by
these nos.
ans is 3999960I think this formula useful for this type problems..
(sum of digits)*(n-1)!(11111......n times)
here n means no. of numbers in the problem
for example in given problem n=5;
3> Ram leaves to his office 30 min late as compared to other days.He travel
with speed 25% less than other days speed and finally he reach office 50 min
late.what is the time taken by ram in normal days.
Ans-60 min
4> if
a,b,c,d,e are integer in increasing order n a+b+c+d+e=FG,FG can be written as
10F+G then what is the max. value of G.
a)4
b)5
c)3
ans is b.) 5
5> if we take five numbers as...18,19,20,21,22...their sum is 100...bt FG is
two digit...so the no. are 17,18,19, 20, 21...which gives sum as 95..
95= 10*9 + 5..so max values of G is 5
6> how
many 6 digit n0. can b formed frm digits 1 2 3 4 5 6 7 so that d digit
shouldn't repeat and the second last digit is even
a) 6480
b) 320
c) 2160
d) 720
The second last digit can be filled by only three numbers i.e 2,4and6 in
3p1 ways.Remaining 6 nos can be placed in 5 places in 6p5 ways,hence the ans
=6p5 * 3p1=2160
7> a girl sees mirror reflection of analog clock while going for school at
home n reaches school in 20 min .when she reaches school she finds tht she ws
2hrs 20 min late. wat as the time when she rchd school.?
She reached school at 7:20am.
Left home at 7:00am but she saw time in mirror reflection as 5:00am.
The school time was 5:00 am.
more questions on time work
regarding profit distribution accrdng to work done.
8> A committee of 6 is chosen from 8 men and 5 women so as to contain at least
2 men and 3 women. How many different committees could be formed if two of the
men refuse to serve together?
possible combinations can be either 2M+4W or 3M+3W
Total possible combinations = 8c2*5c4 + 8c3*5c3 = 700
Combinations in which both men are together = 5c4 + 6c1*5c2 = 65
Combination in which they are not together = 700-65 = 635
9> There are 3 buckets of 8,5 and 3 liter...out of whicg only 8 liter bucket
is fullyfill.... u have to exact 4-4 lit liquid in 8 and 5 liter bucket by
using only these buckets in minimum no of steps ?the correct answer is seven
steps
explain:-
8(lit bucket) 5(lit bucket) 3(lit bucket)
initialy-8 0 0
1) 3 5 0
2) 3 2 3
3) 6 2 0
4) 6 0 2
5) 1 5 2
6) 1 4 3
7) 4 4 0
9> If a
number 272758 and 273437 is divided by N then they give remainder as 13 and 17
respectively. find the sum of digits of number N.
since 13 n 17 are remainders
thus subtracting gives 272758-13=272745 n 273437-17=273420
then GCF of 272745,273420 gives 45
ANS:4+5=9 (easy to find gcf,N may be less than 45 also,GO ACCORDING TO THE
OPTIONS)
10> Rank of the word GOOGLE in the dictionary of words formed by letters of the
word GOOGLE taking all letters at a time?
Given word = GOOGLE
Arrange in Alphabetical order:-
=> EGGLOO
We will follow the sequence of given word characters…like
First take G then…O…then….O……G..…L…..E…
So, For G:
Coefficient is 1 bcz number of characters before G in ‘EGGLOO’ is only 1
(that is E) and,
And remaining letters can be arranged in 5!/(2!*2!) ways
For O:
Coefficient is 3 bcz number of characters before O in EGGLOO is 4 but we
will not count the above already traversed character (ie E) so Coeffiecint of O
=3 and,
And remaining letters can be arranged in 4!/2! ways
Now we can calculate so on.... for remaining characters also....
So,
G => 1*5!/(2!*2!) = 30
O => 3*4!/(2!) = 36
O => 3*3! = 18
G => 1*2! = 2
L => 1*1! = 1
E => 0*0! = 0
So total words before GOOGLE are:-
30+36+18+2+1 = 87
Hence, GOOGLE IS 88th Word.
That is, Rank = 88
11> A merchant buys 20 kg wheat at Rs.30/kg and 40 kg wheat at Rs.25/kg and
mixed them. He sold one-third of mixture at Rs.26, what should be the cost/kg
of the remaining mixture, so as to make a profit of 25% on the whole?
Total investment CP= 20*30 + 40*25= 1600
Total Mixture= 20+40=60
Sold 1/3 of 60= 20kg at Rs26/kg
Remaining mixture= 60-20=40
Earned 20*26= Rs520
To get a profit of 25% on whole total SP= 1.25*1600= 2000
remaining amount= 2000-520=1480
Hence, cost/kg of remaining mixture 1480/40=37
i think it shld be 37rs/kg
12> 1-2+3-4+5......-98+99 =?
sol>ans is 50.
as 1-2+3-4+5-6+7-....-98+99
can be written as
=(1+3+5+7+....+99)-(2+4+6+8+...98)
=(1+3+5+7+....+99)-2*(1+2+3+..+49)
=(1+3+5+7+....+99)-2*49*50/2
(1+3+5+7+....+99)-2450
=2500-2450 (as it is in AP as 1 1st term 99 last term)
50 ans
13> Prime factorisaton of int N is A*A*B*C. a,b,c all are distinct prime
integers.
how many factors does N have ??
sol>A^p*B^q*C^r=(p+1)*(q+1)*(r+1)
hence the ans is 12
Three friends Mohit,Manohar and Prashant decided to share a lot of bananas.
Each of them had half of the total plus half a banana in order. After each of
them took their share twice, no bananas were left.
1.How many bananas were there?
2.How many bananas did Mohit get?
3.How many bananas did Manohar get?
(2^n)-1 is the formula
here n is roatation
according to question its twice means 2*3=6
(2^6)-1=63
14> If the sum of n terms of two series of A.P are in the ratio 5n+4:9n+6 .find
the ratio of their 13th terms
Ratio of sum of n terms 2 A.P = 5n+4 : 9n+6
So, Ratio Sum of 13 terms = 5*13+4 : 9*13+4 (put n=13)= 69 : 123
Now, Ratio Sum of 12 terms = 5*12+4 : 9*12+4 (put n=12)= 64 : 114
So to claculate 13th term we will simple subtract sum of first 12 terms
from sum of first 13 terms i.e, Sum(13) - Sum(12) = = 69 : 123 - 64 : 114
(The important thing is tht we will not subtract it by taking LCM as these
are ratios, We will subtract Numerator to Numerator & Denomenator to
Denomenator) so, Ratio of 13th term of 2 APs will be = 5 : 9
15> A grand father has 3 grand children. Age difference of two children among
them is 3. Eldest child age is 3 times the youngest child’s age and the
eldest child age is two year more than the sum of age of other two children.
What is the age of the eldest child?
take ages x,y,z;
y-x=z-y=3;=>y=x+3;
z=3x;=>z=x+y+2 =>z=2x+5;
.ie 3x=2x+5=>x=5 then z=15
16> A boy buys 18 sharpeners (brown or white) for Rs. 100. For every white
sharpener, he pays one rupee more than the brown sharpener. What is the cost of
white sharpener & how much did he buy?
If he bought x white sharpeners @ Rs (y+1) and (18-x) brown sharpeneres @
Rs y per sharpener, then
x*(y+1)+x*(18-x)=100
x= 100-18y
Only integral value of x less than 18 will be 10.
then x=10, y=5 so he bought 10 white sharpeners @ Rs 6 per sharpeners and 8
brown sharpeners @ Rs 5 per sharpener.
17> when m+n is divided by 12 the reminder is 8 and m-n is divided by 12 the
reminder is 6.what is the reminder when mn is divided by 6??
Remainder is always 1
This answer can be proved with multiple values... 2 of them are :-
1. m=7 n=1
2. m=19 n=1
For m=7,n=1
m+n = 8 divide by 12 gives remainder 8
m-n = 6 divide by 12 gives remainder 6
so, m*n = 7 divide by 6 gives remainder 1
For m=19,n=1
m+n = 20 divide by 12 gives remainder 8
m-n = 18 divide by 12 gives remainder 6
so, m*n = 19 divide by 6 gives remainder 1
18> If p(x)=a x^4+b x^3+c x^2+d x+e.....has roots at
x=1,2,3,4......p(0)=48...then p(5)=?????
P(x)=Q(x-1)(x-2)(x-3)(x-4)
as x=1,2,3,4 are roots of equation...
P(0)=48=Q(-1)(-2)(-3)(-4)
by solving Q=2
so, P(x)=2(x-1)(x-2)(x-3)(x-4)
put x=5
P(5)=2(5-1)(5-2)(5-3)(5-4)
P(5)=48 is the ans.
19> A man asks 5 people to make a guess about the amount of money in his pocket
which is less than 50.
A guess that the amt is a multiple of 10..
B guess that the amt is a multiple of 12..
C guess that the amt is a multiple of 15..
D guess that the amt is a multiple of 18...
E guess that the amt is a multiple of 30
Which of the following guesses are correct?
a.AE
b. AB
c. BC
d. DE
AE:30 is multiple of both 10 and 30 which is less than 50. ... possible
answer.
AB: there is no multiple of 10 and 12 less than 50. ..... not possible
answer.
BC: Same logic as that of AB. ..... not possible answer.
DE: Same logic as that of AB. ..... not possible answer.
Hence answer is AE.
20> Addition of 641+852+913=2456 is incorrect. Whats the largest digit that can
be changed to make the decision corret?
4 in 641 must be replaced with 9.
so 4 is the ans.
21> If X^Y denotes X raised to the power
of Y, find out last two
digits of (2957^3661)+(3081^3643)
Options
o 42
o 38
o 98
o 22
sol. 98 is the ans
7^1 is 7
and
5*1=5
so last to digit of (2957^3661) is 57
same for
1^3 is 1
8*3=24 unit digit of 24 is 4
last 2 digit of(3081^3643) is 41
41+57=98....
22> A clock showed 5 min past 3'0 clock on Sunday evening when the correct time
was 3'0 clock it looses uniformly and was observed to be 10 min slow on the
subsequent Tuesday at 9pm . when did the clock show the correct time
A clock showed 5 min past 3'0 clock on Sunday evening when the correct time
was 3'0 clock it looses uniformly and was observed to be 10 min slow on the
subsequent Tuesday at 9pm .
In 54 hrs , it lost 15 mins.
so 5 mins are lost in 5*54/15 = 18 hrs
so watch showed correct time at 1500+1800-2400= 0900 hrs on Monday i.e 9AM
Monday
23> there are 5 sweets - jamun, kulfi, peda.
laddu and jilebi that i wis t eat on 5 consecutive days, monday throufh friday,
one sweet a day, based on following self imposed constraints:
1) laddu is not eaten on monday
2) if jamun is eaten on monady , then laddu must be eaten on friday
3) if jamun is eaten on tuesday, kulfi should be eaten on monday
4) peda is eaten the day following the day eating jelabi
based on the above , peda can be eaten on any day expect??
monday bcoz the day before jalebi is required to be taken bt its sunday and
no sweet is eaten on sunday
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