M4maths Previous Puzzles

29 Jul 2015 Four digit number

A four digit number is such that the product of all of its digits is 126. The sum of all the digits is equal to the 2 digit number formed by using thousands digit and tens digit (Thousand digit in tens place & ten digit in units place) which in turn is equal to 19. Then difference of units and thousand place of the number is, given that this difference is positive.

OPtion

1) 0
2) 1
3) 2
4) 3
5) 4
6) 5
7) 6
8) 7
9) 8
10)9

Solution

Let the number is 1000a+100b+10c+d
According to given conditions
a*b*c*d = 126
a+b+c+d = 10a+c = 19
so a = 1 and c = 9
b*d = 126/9 = 14
b+d = 9
on solving b = 2, d = 7 and b = 7, d = 2
so difference = 7-1 = 6 and 2-1 = 1

Option 6)

Correct Option: 6 Best Solution (1)

Diksha tiwari   9 years AGO

Let 4 digit number be abcd
a*b*c*d=126=1*2*7*9
sum of all digits=ac=19
therefore, a=1 c=9
hence diff can be 2-1=1 or 7-1=6

Like? Yes (2) | No   

28 Jul 2015 Time and work

A cistern which has leak in the bottom is filled in 15 hours. Had there been no leak, it could have been filled in 12 hours. If the cistern is full, the leak can empty it is

OPtion

1) 3 hours
2) 6 hours
3) 10 hours
4) 12 hours
5) 14 hours
6) 15 hours
7) 28 hours
8) 30 hours
9) 24 hours
10)60 hours

Solution

Work done by the leak in 1 hours = (1/12 - 1/15) = 1/60
Therefore lean can empty the full cistern in 60 hours.

Option 10)

Correct Option: 10 Best Solution (29)

Kethineni Vinodh   9 years AGO

1/12+1/leak=1/15
1/leak=1/15-1/12
1/leak=-1/60
minus indicates leak i.,e 60 hours

Like? Yes (11) | No   

neha gupta   9 years AGO

Let the pipe which fills the cistern fills it in x hours , so its one hour work is 1/x
and the leak which empties it empties it in y hours , so its one hour work is 1/y
according to ques
1/x - 1/y = 1/15
and
1/x = 1/12
so x=12
1/12 - 1/15 = 1/y
on solving it we get
y = 60
thus its answer is 60 hours

Like? Yes (10) | No (1)  

Abhishek Mohanty   9 years AGO

let B be the reason to fill the cistern without any leak then work done by B in hour =(1/12).

similarly let A can empty the cistern in x hours without B filling simultaneously.then work done by B in 1 hour=(1/x).

Net work work done in 1 hour simultaneously by A & B =(1/12)-(1/x)=(x-12)/12x.
so total time to fill the cistern=[12x/(x-12)]=15 hours (given).

Solving we get x=60.

Like? Yes (5) | No   

Devendra Marghade   9 years AGO

If leakage can empty the full tank in 'x' hours, then
(1/12)-(1/x)=1/15
On solving, x=60
Option (10)

Like? Yes (4) | No   

UJJWAL KANT   9 years AGO

cistern with leak will fill 1/15 of cistern in an hour
cistern without leak will fill 1/12of cistern in an hour
hence to completely empty the tank 1/12-1/15=1/60
i.e 60 hours

Like? Yes (5) | No (1)  

Piyush Telang   9 years AGO

Let 'F' be the hours in which the cistern can be completely filled without any hole in it & 'E' be the hours in which the hole can empty the cistern when it is completely filled.
Such that : In 1 hour, one-Fth part of the cistern is filled and one-Eth part of the cistern is emptied respectively.
Then 1/F - 1/E = 1/15;
which means in 1 hour one-fifteenth of the cistern is filled having a hole.
And 1/F = 1/12;
which means in 1 hour one-twelfth of the cistern is filled without a hole.
1/12 - 1/E = 1/15
=> 1/E = 1/12-1/15 = 1/60
=> E = 60 hours.
Thus the leak can empty the full cistern in 60 hours.

Like? Yes (5) | No (2)  

Rahul kumar   9 years AGO

let x be time taken by leak to empty full tank
1/12-1/15=1/x
x=60hrs

Like? Yes (4) | No (1)  

pankaj chaurasia   9 years AGO

tank can be fill full in (without leak)=12 hour
tank can be fill full in(with leak)=15 hour
now,
tank filled in 1 hour=1/12
tank fill in 1 hour with leak=1/15
therefore,,,,3 hour is taken extra when there is been a leak in tak,,
so tank can be emptied in
1/12 - 1/15=1/60
now , do 1/60 time will be used to empty the tank in 1 hour therefore
total time is reciprocal of 1/60= 60 hour

Like? Yes (3) | No   

Kshitiz Khandelwal   9 years AGO

filling rate x l/hr
leak rate y l/hr
ATQ
15(x-y)=12x
on solving
x=5y
which gives
leak will empty it in 60 hours

Like? Yes (3) | No (1)  

praveendas   9 years AGO

leaked water from the tank in one hour is (1/12-1/15)=1/60
in order to make it full empty time required is 60 hours

Like? Yes (2) | No (1)  

shivani gupta   9 years AGO

1/x-1/y=1/15
1/12-1/y=1/15
1/12-1/15=1/y
1/60=1/y
y=60 hrs

Like? Yes (1) | No   

Ria Chandra   9 years AGO

consider the time taken by the pipe to fill the tank as p and time with which tank gets empty as e.
so 1/p-1/e=1/15(given)
since, 1/p=1/12
so 1/e=1/60
hence, answer is 60 hours

Like? Yes (1) | No   

SUNIDHI   9 years AGO

in one hour the leak empties 1/60th part of the tank in 60 hrs it will empty the tank

Like? Yes (1) | No   

Dinesh kumar   9 years AGO

(1/12)-(1/x)=1/15
(1/12)-(1/15)=1/x
1/60=1/x
X=60

Like? Yes (2) | No (1)  

Narendra Gopi   9 years AGO

1/x-1/y=1/15
-1/y=1/15-1/12
1/y=1/60
y=60hrs

Like? Yes (1) | No   

Vivek   9 years AGO

The time it takes to fill the cistern =12 hrs
The total time taken by the cistern and the leak =15 hrs
Therefore the time taken by cistern alone x
1/x = 1/12 -1/15
=1/60
x = 60 hrs
Hence the option is 10)

Like? Yes (1) | No   

anand vijay   9 years AGO

1 hour work if there is no leak in the tank= 1/12
1 hour work if there is a leak in the tank=1/15
so,time taken by the leak to empty the tank in 1 hour=1/12-1/15
1/60
total time to empty the tank=60 hour
option 10)

Like? Yes (1) | No   

Manish Dwivedi   9 years AGO

let the volume of cistern be v
so v ltrs can be filled in 12 hrs
so in 1 hr v/12 ltrs can be filled
so in 15 hrs total volume filled=15*v/12=1.25v
so in 15 hrs 0.25v volume is leaked
so total time to empty the cistern is 60hrs

Like? Yes (1) | No   

Subhash Gupta   9 years AGO

let required hours=h
then
1/12-1/h=1/15
h=60

Like? Yes (1) | No   

Haripriya   9 years AGO

1/12-1/15
(4-5)/60
1/60
so 60

Like? Yes (1) | No   

Soma Naresh   9 years AGO

let us assume that the work done by leaked pipe and other be x and y.
given work done by both is 15 hours
x+y=1/15
cistern filled by other pipe without having leak at 12 hours
y=1/152
then
x=(1/15)-(1/12)
x=-1/60
means x takes 60 hours to empty tank

Like? Yes (1) | No   

Altaf Hussain   9 years AGO

Let the capacity be 60. Rate at which it is filled with leak is 60/15=4 and rate at which it is filled Without leak is 60/12=5. Rate of leak is 1. It takes 60 hours.

Like? Yes (1) | No   

sonu kumar suman   9 years AGO

because in one hour it fill 1/12 if there had no leak.and if leak then it fill 1/15 in one hour.so,leak in one hour is 1/12 - 1/15=1/60.so for total time to empty full cistern is 60.

Like? Yes (1) | No   

Sushmita Ghanty   9 years AGO

part of the cistern emptied by the leak in 1 hr = 1/12 - 1/15 = 1/60
hence the leak will empty the cistern in 60 hrs

Like? Yes (1) | No   

VIKASH RAY   9 years AGO

when there is no leak the tank filled in 12 hrs.
so,in 1 hrs. the part of the tank filled is=1/12
let, due to the leak the tank can be empty in x hrs.
so, the part of the tank empty in 1 hrs. is=1/x
so,
(1/12)-(1/x)=(1/15)
on solving this we get x=60
so the tank can be empty in 60 hrs.

Like? Yes (1) | No   

GAUTAM CHAURASIA   9 years AGO

tank filling time with leak and without leak is 12 hour and 15 hour respectively.
now, talk about 1 hour,
tank filling time with leak and without leak is 1/12 hour and 1/15 hour respectively.
so,,,due to leak total extra time consumed in tank filling process in one hour is[ (1/12) - (1/15)= (1/60) ]

hence total time will be sufficient for empty the leak is 60 hours(reserve of 1/60)

Like? Yes (1) | No   

Subham Mahanta   9 years AGO

work done by the leak in 1 hour = (1/12 - 1/15) = 1/60
so, if the cistern is full, the leak can empty it in 60 hours
ans. is option (10)

Like? Yes (1) | No   

ANUSHKA   9 years AGO

Let The time taken to empty be x hrs.

so, 1/12-1/x = 1/15
1/x = 1/12 - 1/15
1/x = 1/60

Therefore, time taken to empty is 60 hours.

Like? Yes (1) | No   

aman aditya   9 years AGO

let cistern capacity is 12 liter
so if it took 15 hr with leakage which means it had 15 liter if water.
so it leaked 3 liter of water in 15hr.
so total water is 12 liter so it will take 60 hr

Like? Yes (1) | No (1)  

27 Jul 2015 Sequence

What is the next number of the following sequence
67, 871, 13936, 306592, ......

OPtion

1) 9197760
2) 7971392
3) 7972546
4) 8891168
5) 7664800
6) 8277984
7) 8584576
8) 7664888
9) 7684364
10)None of these

Solution

67
67*(6+7) = 67*13 = 871
871*(8+7+1) = 871*16 = 13936
13936*(1+3+9+3+6) = 13936*22 = 306592
Similarly
306592*(3+0+6+5+9+2) = 306592*25 = 7664800

Option 5)

Correct Option: 5 Best Solution (40)

anand kumar   9 years AGO

multiply number with sum of digits to get next number
67x(7+6)=871
871x(8+7+1)=13936
306592x(3+0+6+5+9+2)=7664800

Like? Yes (1) | No   

anand vijay   9 years AGO

sum of the digit of each number in the following series....
6+7, 8+7+1, 1+3+9+3+6, 3+0+6+5+9+2
13,16,22,25,?
difference between the no 1st and second is 3
2nd and 3rd is 6
3rd and 4th is 3
so it follows the pattern 3,6,3,6
hence fifth no in the series would be 7664800
7+6+6+4+8+0+0=31
option 5)

Like? Yes (1) | No   

Devendra Marghade   9 years AGO

Multiply the number with its sum to get next number
67*(6+7)=871
871*(8+7+1)=13936
13936*(1+3+9+3+6)=306592
306592*(3+0+6+5+9+2)=7664800
Option (5)

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praveendas   9 years AGO

some of the digits in a series is 13, 16 ,22, 25 ,the next sum in the series is31...
check out in the given ans by adding its digits

Like? Yes (1) | No   

~Sandeep~   9 years AGO

Next Number =
Sum of digits of previous number * Previous number

Like? Yes (1) | No   

Vivek   9 years AGO

Adding the tens and units place we get
67 -> 6+7=13
then multiplying the number with the above sum we get the next number i.e
next number = 67 * 13
= 871
Similarly 871 -> 8+7+1 =16
next number = 871 * 16 = 13936
Similarly if we the last number
306592 -> 3+0+6+5+9+2 =25
And so the next number = 306592 * 25 = 7664800
So the answer is option 5)

Like? Yes (1) | No   

neha gupta   9 years AGO

67 * (6+7) = 871
871 * (8+7+1) = 13936
13936 * (1+3+9+3+6) = 306592
306592 * (3+0+6+5+9+2) = 306592 * 25 = 7664800
so next term is 7664800 .

Like? Yes (1) | No   

UJJWAL KANT   9 years AGO

if we do sum of digits then we get
13 16 22 25 31
i.e pattern is 3 6 3 6
hence next numbers digits summation must be 31
i.e 7664800

Like? Yes (1) | No   

Ayush Singhal   9 years AGO

871/67=13
here 6+7=13
13936/871=16
here 8+7+1=16
..
..
and so on
so answer is
306592*25=7664800

Like? Yes (1) | No   

Ajay Kumar Gupta   9 years AGO

the sum of all digits are like this sequence...
13,16,22,25
so in this series the next no will be- 31
a/c to options the sum of all digits of 7664800 is 31.

Like? Yes (1) | No   

Altaf Hussain   9 years AGO

67*13=871
871*16=13936
13936*22=306592
306592*25=7664800

13, 13+3=16, 16+6=22, 22+3=25

Like? Yes (1) | No   

abhishek kumar gupta   9 years AGO

Digital sum of every number is a is a difference of 3 , 6 ,3 , 6

that is
6+7=13
8+7+1=16
1+3+9+3+6=22
3+0+6+5+9+2=25
each coming up number is a difference of 3 and 6 alternately so next number must be 25+6=3 and it is 76648000 ,that is 7+6+6+4+8+0+0+0=31.

Like? Yes (1) | No   

Padal Shivani   9 years AGO

67*(6+7)=871
871*(8+7+1)=13936
-------
-----
306592*(3+0+6+5+9+2)-=7664800

Like? Yes (1) | No   

shivani gupta   9 years AGO

Given seq is next term =(next -1)term*(digit of (next-1)term)

6+7=13*67=871
8+7+1=16*871=13936
1+3+9+3+6=22×13936=306592
3+6+5+9+2=25*306592=7664800
So ans is option 5

Like? Yes (1) | No   

GITANJALI GOSWAMI   9 years AGO

6+7=13
8+7+1=16
1+3+9+3+6=22
3+0+6+5+9+2=25
7+6+6+4+8+0+0=31
option5 is correct.

Like? Yes (1) | No   

Arju   9 years AGO

ans-5-7664800
Explanation=871/67=13
13936/871=16
306592/13936=22
therefore 303692*25=7664800

Like? Yes (1) | No   

sudiptaa   9 years AGO

871/67= 13
13936/871=16
306592/13936=22
13,16,22,25
so d = 3 ,6,3,6,
306952*25=7664800

Like? Yes (1) | No   

Pinku Burnwal   9 years AGO

sequence is 67,871,13936,306592
1st number is 67,
2nd number=67*(6+7)=871
3rd number=871*(8+7+1)=13936
4th number=13936*(1+3+9+3+6)=306592
5th number=306592*(3+0+6+5+9+2)=7664800

Like? Yes (1) | No   

Parth Thakkar   9 years AGO

67*13=871
871*16=13936
13936*22=306592
306592*25=7664800

Like? Yes (1) | No   

ROHIT KUMAR   9 years AGO

Series is progress as
67*(6+7)=871
871*(8+7+1)=13939
13936*(1+3+9+3+6)=306592
306592*(3+0+6+5+9+2)=7664800

Like? Yes (1) | No   

S R Y Farhan   9 years AGO

6+7 =13
67*13 = 871
8+7+1 = 16
871*16 = 13936
and so on :)

Like? Yes (1) | No   

Dinesh thatti   9 years AGO

67*13 = 871
871*16 = 13936
13936 * 22 = 306592
306592 * 25 = 7664800

Like? Yes (1) | No   

Dhanraj Wanjare   9 years AGO

67, 871, 13936, 306592, .............
6+7=13;
67*13=871;

8+7+1=16;
871*16=13936;

1+3+9+3+6=22;
13936*22=306592;

3+0+6+5+9+2=25;
306592*25=7664800;

so option 5 is satisfied.

Like? Yes (1) | No   

GAUTAM CHAURASIA   9 years AGO

Now
67,871,13936,306592,....
First Number is 67 i.e Now 871/67 = 13 = 6+7,
Second Number is 871 i.e 13936/871 = 16 = 8+7+1,
Third Number is 13936 i.e 306592/13936 = 22 = 1+3+9+3+6,
Fourth Number 306592 i.e 3+0+6+5+9+2=25, 306592*25=7664800,

Hence Sixth Number would be " 7664800 "

Like? Yes (1) | No   

roshni.s   9 years AGO

1st term 67
6+7=13
2nd term= (67*13)=871
8+7+1=16
3rd term= (871*16)=13936
1+3+9+3+6=22
4th term= (13936*22)=306592
3+0+6+5+9+2=25
5th term=( 306592*25)=7664800

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keerthi choudhry   9 years AGO

6+7 , 8+7+1 , 1+3+9+3+6 , 3+0+6+5+9+2 , 7+6+6+4+8+0+0
13 16 22 25 31
/ / / /
3 6 3 6
/ / /
3 3 3
therefore, the answer is option: 5
7+6+6+4+8+0+0=31

Like? Yes (1) | No   

vimal tiwari   9 years AGO

sum of digits are in series

67 = > 6+7 =13 => 1+3 =4
871 => 8+7+1 = 16 => 1+6=> 7

hence 4 ,7,4,7...
next will be of sum 4..

5th option is correct

Like? Yes (1) | No   

Sushmita Ghanty   9 years AGO

67
6+7 = 13
67*13= 871
8+7+1= 16
871*16= 13936
1+3+9+3+6= 22
13936*22= 306592
3+0+6+5+9+2=25
306592*25 = 7664800

Like? Yes (1) | No   

RAVAL   9 years AGO

67=6+7=13
so, 67*13=871
871=8+7+1=16
so, 871*16=13936
same as,
306592=3+6+5+9+2=25
so, 306592*25=7664800

Like? Yes (1) | No   

prashanth   9 years AGO

67*13=871
871*16=13936
13936*22=306592
306592*25=7664800

Like? Yes (1) | No   

shiva   9 years AGO

67*(6+7) = 871
871*(8+7+1) = 13936
13936*(1+3+9+3+6)= 306592
306592*(3+0+6+5+9+2)=7664800

Like? Yes (1) | No   

anubhav kumar gupta   9 years AGO

67*(13)=871
871*(8+7+1)=13936
similarly,
,
,
306592*(3+0+6+5+9+2)=7664800

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Nazia Noorani   9 years AGO

67, 67*13 , 67*13*(13+3) , 67*13*16*13+9 , 67*13*16*22*(13+12)

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PRASHANT KUMAR   9 years AGO

871=67*(6+7)
13936=871*(8+7+1)
so REQD ANS=306592*25=7664800

Like? Yes (1) | No   

pankaj chaurasia   9 years AGO

871/67=13(6+7=13)
13936/871=16(8+7+1=16)
??therefore 3+0+6+5+9+2=25
now next no in the sequence is 25�306592=7664800

Like? Yes (1) | No   

SWARNIKA   9 years AGO

6+7=13,67*13=871
8+7+1=16,871*16=13936
1+3+9+3+6=22,13936*22=306592
3+0+6+5+9+2=25,306592*25=7664800

Like? Yes (1) | No   

Kirty   9 years AGO

sum of digits of the no. is multiplied with itself gives the result
So the addition of the digits of 306592 is 25 and 25 is multiplied with the no. gives 7664800

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PUSHPENDRA KUMAR   9 years AGO

67, 871, 13936, 306592, .....
Simple puzzle nothing in there :
In the given series every no's digits added and then multiply by itself .
Explanation:
1. 67*(6+7)=67*13=871 (2nd no)
2. 871*(8+7+1)=871*16=13936 (3rd no)
3. 13936*(1+3+9+3+6)=13936*22=306592 (4th no)
4. 306592*(3+0+6+5+9+2)=306592*25=7664800 (5th no)(Ans.)

Like? Yes (1) | No   

Ashu Kumar   9 years AGO

871%67=13
13936%871=16
306592%13936=22
SINCE WE SEE THAT 16-13=3,22-16=6..
AND PATTERN ARE 3,6,3,6..
SO NEXT TERM=25*306592=7664800

Like? Yes (1) | No   

Soma Naresh   9 years AGO

871=67�(6+7)=67�13
13936=871�(8+7+1)=871�16
306592=13936�(1+3+9+3+6)
=13936�22
306592�(3+0+6+5+9+2)=306592�25
=7664800

Like? Yes (1) | No   

24 Jul 2015 Time and work

John and Nick together can complete a work in 10 days. They plan to do the job in a different way. In first day John do it whereas Nick do it for the next 2 days and this pattern is repeated again and again. If they have to finish the work in 18 days, then in how many days the man who finish the work do this work alone.

OPtion

1) 12
2) 10
3) 15
4) 16
5) 18
6) 24
7) 21
8) 27
9) 30
10)20

Solution

Let John do it alone in x days and Nich alone in y days,
then according to first given condition
1/x + 1/y = 1/10 (1)
According to second condition
18/3 (1/x + 2/y) = 1
1/x + 2/y = 1/6 (2)
On solving (1) & (2) y = 15 & x = 30
So Nick will finish the work in 18 days & he complete it in 15 days
Option 3)

Correct Option: 3 Best Solution (3)

Devendra Marghade   9 years AGO

If John and Nick working alone completes the work in 'J' and 'N' days respectively, then
(1/J)+(1/N)=1/10 ---(i)
As their 3 days work will be (1/J)+(2/N) which is equal to 1/6 of the 18 days ,in which they have to complete the work.
So (1/J) +(2/N)=1/6 ---(ii)
From (i) & (ii), N=15, J=30
So John finishes who alone can complete the work in 15 days.
Answer option(3)

Like? Yes (1) | No   

Piyush Telang   9 years AGO

John takes 'J' days to complete the work alone and Nick takes 'N' days to complete the work alone.
Then 1 day's worth of work for John and Nick are 1/J and 1/N respectively.
--> When both work together simultaneously then the work is completed in 10 days.
=> 1/J + 1/N = 1/10
which means that one-tenth of work is completed in 1 day when both work together simultaneously.
--> When they work alternatively such that on 1st day John works and on 2nd and 3rd day Nick works, then the works is completed in 18 days and at last day Nick finishes the work.
=> 1/J + 2/N = 3 x ( 1/18 )
which means that when they work alternatively total of 3 days are covered and out of 18 days worth of work they complete one-sixth of work in 3 days.
--> So solving the 2 equations, we get N =15 days.
Thus, Nick takes 15 days alone to complete the work and John takes 30 days to complete the work alone.
As Nick finishes the work at the 18th day the answer is 15 days.

Like? Yes (1) | No   

ADITYA SARIPALLI   9 years AGO

let the rate of work done by John = r1
let the rate of work done by Nick = r2
let the total work needs to be done = w

=> 10(r1+r2) = w -------> (1)

At the same time if they do the work in batches of 3 days (John 1 day & Nick 2 days)
then to complete the work (w) it takes 18 days.
Moreover it would be 'Nick' who will be completing the last batch.

Thus for 1st batch (i.e w/6) it takes ...
=> (r1+2r2) = (w/6) -------> (2)

Thus from (1) & (2), we have

10(r1 + r2) = 6(r1 + 2r2)
=> 2r1 = r2 -------> (3)

from (1) & (3), we have

10(3r1) = w OR 10((3/2) * r2) = w
=> (w/r1) = 30 OR (w/r2) = 15

Therefore,
No Of days to complete the job if John works alone = (w/r1) = 30
No Of days to complete the job if Nick works alone = (w/r2) = 15


Thus ans = 15 days.

Like? Yes (1) | No   

23 Jul 2015 Compound Interest

The compound interest on a sum for 2 years is Rs. 832 and the simple interest on the same sum for the same period is Rs. 800. The difference between the compound interest and the simple interest for 3 years will be

OPtion

1) Rs. 48.35
2) Rs. 46.25
3) Rs. 86.50
4) Rs. 92.25
5) Rs. 97.65
6) Rs. 66.56
7) Rs. 68.25
8) Rs. 98.56
9) Rs. 96.48
10)None of these

Solution

SI for 1 years = Rs. 400
SI on Rs. 400 for 1 year = Rs.(832-800)
Rate = 32/400 * 100 = 8%
Difference for 3rd year = SI on Rs. 832
Rs. 832* 8/100 * 1 = 66.56
Total difference = Rs.(32+66.56) = Rs. 98.56
Option 8)

Correct Option: 8 Best Solution (15)

swarupa nuggu   9 years AGO

Given that simple interest for 2 years is Rs.800

i.e., Simple interest for 1st year is Rs.400
and simple interest for 2nd year is also Rs.400

Compound interest for 1st year will be 400
and Compound interest for 2nd year will be 832 - 400 = 432

you can see that compound interest for 2nd year is more than simple interest for 2nd year by 432 - 400 = Rs.32

i.e, Rs. 32 is the interest obtained for Rs.400 for 1 year

Rate, R = 100×SIPT =100×32400×1 =8%

Difference between compound and simple interest for the 3rd year
= Simple Interest obtained for Rs.832

=PRT100 =832×8×1100 =Rs. 66.56

Total difference between the compound and simple interest for 3 years
= 32 + 66.56 = Rs.98.56

Like? Yes (23) | No (5)  

vini   9 years AGO

Given SI = 800 for two years
The SI and CI for first year is same = Rs . 400
CI for second year is (832-400= 432) so, extra Rs 32 is interest for previous year SO, R% of 400 = 32
R= 8%
Principal = Rs 5000
Now, SI for three years is Rs 1200
CI for three years is Rs 1298.56
The difference is 98.56

Like? Yes (17) | No (5)  

Dhanraj Wanjare   9 years AGO

CI=832 n=2 years SI=800, T=2 years
CI = P(1+R/100)^n - P; AND SI = P*R*T/100;
832=P(1+R/100)^2 - P; 800=P*R*2/100;
832=P{[(100+R)^2/10000]-1} P=800*100/2R;
832*10000=P[(10000+R^2+2*100*R)-10000];
8320000=P[R^2 + 200R];
8320000/P=R^2 + 200R
8320000/(800*100/2R)=R^2+200R;
208R=R^2+200R;
8R=R^2;
R=8;

P=800*100/2R
P=800*100/2*8;
P=5000;

NOW n=T=3 YEARS R=8% P=5000

CI = P(1+R/100)^3 - P; AND SI=P*R*T/100;
CI = 5000(1+8/100)^3 -5000; SI=5000*8*3/100; CI = 500*108^3/100^3 - 5000; SI=1200;
CI = 1298.56;

RESULT = CI - SI;
RESULT = 1298.56 - 1200;
RESULT = 98.56;

OPTION 8 IS SATISFIED ;

Like? Yes (6) | No (1)  

praveendas   9 years AGO

let find the amount which gives 800 on smple intrest for 2 years and 832 on cmpnd intrest for 2 years is 5000...
therfore for three years on simple intrest it is 1200
and for compound it is 1298.56..
1298.56-1200=98.56

Like? Yes (7) | No (3)  

KUMAR HARSH   9 years AGO

difference in C.I and S.I in 2years =Rs.32
S.I for 1year =Rs.400
S.I for Rs.400 for one year =Rs.32
rate=[100*32)/(400*1)%=8%
difference between in C.I and S.I for 3rd year
=S.I on Rs.832= Rs.(832*8*1)/100=Rs.66.56
Total difference between the compound and simple interest for 3 years
= 32 + 66.56 = Rs.98.56

Like? Yes (5) | No (1)  

Anuj Rai   9 years AGO

Compound interest for 1st year will be 400
and Compound interest for 2nd year will be 832 - 400 = 432

you can see that compound interest for 2nd year is more than simple interest for 2nd year by 432 - 400 = Rs.32

i.e, Rs. 32 is the interest obtained for Rs.400 for 1 year

Rate, R = 100×SIPT=100×32400×1=8%

Difference between compound and simple interest for the 3rd year
= Simple Interest obtained for Rs.832

=PRT/100=832×8×1100=Rs. 66.56

Total difference between the compound and simple interest for 3 years
= 32 + 66.56 = Rs.98.56

Like? Yes (4) | No (1)  

Veerraj Chauhan   9 years AGO

Given that simple interest for 2 years is Rs.800

i.e., Simple interest for 1st year is Rs.400
and simple interest for 2nd year is also Rs.400

Compound interest for 1st year will be 400
and Compound interest for 2nd year will be 832 - 400 = 432

you can see that compound interest for 2nd year is more than simple interest for 2nd year by 432 - 400 = Rs.32

i.e, Rs. 32 is the interest obtained for Rs.400 for 1 year

Rate, R = 100×SIPT=100×32400×1=8%

Difference between compound and simple interest for the 3rd year
= Simple Interest obtained for Rs.832

=PRT100=832×8×1100=Rs. 66.56

Total difference between the compound and simple interest for 3 years
= 32 + 66.56 = Rs.98.56

Like? Yes (5) | No (2)  

sudiptaa   9 years AGO

SI for 1st year =400
SI for 2nd year=400

CI for 1st year = 400
CI for 2nd year = 832-400= 432

compound interest for 2nd year is more than simple interest for 2nd year by 432 - 400 = Rs.32

i.e, Rs. 32 is the interest obtained for Rs.400 for 1 year

Rate, R = 100*SI*P*T=100×32400×1=8%

Difference between compound and simple interest for the 3rd year
= Simple Interest obtained for Rs.832

=PRT/100=832×8×1100=Rs. 66.56

Total difference between the compound and simple interest for 3 years
= 32 + 66.56 = Rs.98.56

Like? Yes (5) | No (2)  

Chitra.S   9 years AGO

Given that simple interest for 2 years is Rs.800

i.e., Simple interest for 1st year is Rs.400
and simple interest for 2nd year is also Rs.400

Compound interest for 1st year will be 400
and Compound interest for 2nd year will be 832 - 400 = 432

you can see that compound interest for 2nd year is more than simple interest for 2nd year by 432 - 400 = Rs.32
i.e, Rs. 32 is the interest obtained for Rs.400 for 1 year



Difference between compound and simple interest for the 3rd year
= Simple Interest obtained for Rs.832



Total difference between the compound and simple interest for 3 years
= 32 + 66.56 = Rs.98.56

Like? Yes (3) | No (1)  

SHAKE MAHAMAD GOUSE   9 years AGO

difference in C.I and S.I in 2years =Rs.32
S.I for 1year =Rs.400
S.I for Rs.400 for one year =Rs.32
rate=[100*32)/(400*1)%=8%
difference between in C.I and S.I for 3rd year
=S.I on Rs.832= Rs.(832*8*1)/100=Rs.66.56
S.I for 3 years is Rs.1232=Rs.(1232*8*1)/100=Rs.98.56

Like? Yes (2) | No   

SURAJ NIRALA   9 years AGO

On applying the firmula of si and ci ,
two equation form where pr=40000.
putting tge value of pr in ci formula we fet r=8%
now on finding the intersts the difference is 98.56

Like? Yes (2) | No (1)  

Dinni   9 years AGO

Difference in C.I. and S.I. for 2 years = Rs. 32.

S.I. for one year = Rs. 400.

∴ S.I. on Rs. 400 for one year = Rs. 32.

So, Rate = 100∗32400∗1%=8%.

Hence, difference in C.I. and S.I. for 3rd year

= S.I. on Rs. 832 = Rs. (832∗8∗1100) = Rs. 66.56.

Total difference = Rs. (32 + 66.56) = Rs. 98.56.

Like? Yes (3) | No (2)  

anuj gupta   9 years AGO

Given that simple interest for 2 years is Rs.800

i.e., Simple interest for 1st year is Rs.400
and simple interest for 2nd year is also Rs.400

Compound interest for 1st year will be 400
and Compound interest for 2nd year will be 832 - 400 = 432

you can see that compound interest for 2nd year is more than simple interest for 2nd year by 432 - 400 = Rs.32

i.e, Rs. 32 is the interest obtained for Rs.400 for 1 year

Rate, R = 100×SIPT=100×32400×1=8%

Difference between compound and simple interest for the 3rd year
= Simple Interest obtained for Rs.832

=PRT100=832×8×1100=Rs. 66.56

Total difference between the compound and simple interest for 3 years
= 32 + 66.56 = Rs.98.56

Like? Yes (1) | No   

Riyashree Mukherjee   9 years AGO

diff of ci &si for 2 years =32
find for another one yr and then add

Like? Yes (1) | No (1)  

GITANJALI GOSWAMI   9 years AGO

diff of ci and si=32
si of 1 year=400
rate=8%
diff of ci and si on third year=si on rs.832=66.56
total diff=(32+66.56)=98.56

Like? Yes (2) | No (2)  

22 Jul 2015 Sequence

What is the next number of the following sequence
1/2, 6/1, 35/6, 7134/35,.......

OPtion

1) 1781245585 / 7134
2) 1780425685 / 7134
3) 1780246585 / 7136
4) 1782425485 / 7144
5) 1781425855 / 7144
6) 1730425585 / 7136
7) 1781425686 / 7134
8) 1782145685 / 7145
9) 1782245585 / 7136
10)1780242485 / 7134

Solution

First term = 1/2
1+2 = 3; |2-1| = 1
Next term = 3*1 / (1/2) = 6/1

6+1 = 7; |1-6| = 5
Next term = 7*5 / (6/1) = 35/6

35+6 = 41; |6-35| = 29
Next term = 41+29 / (35/6) = 7134/35

Similarly
Next
Option 1)

Correct Option: 1 Best Solution (4)

Deepak kushwaha   9 years AGO

FOR DENOMINATOR
numerator of previous fraction is equal to denominator of next fraction
FOR NUMERATOR

1(6-1)(6+1)=35
6(35-6)(35+6)=7134
35(7134-35)(7134+35)=1781245585
SO NEXT FRACTION MUST BE
1781245585/7134(option 1)

Like? Yes (1) | No   

praveendas   9 years AGO

only numerator of denominators 7134 which divided by 7169....
in the above given series farward numerator is divided by the sum of previous numerator and denominator....

Like? Yes (1) | No   

pankaj kumar   9 years AGO

solu:- addition of numerator & denominator * subtraction of numerator & denominator * denominator = numerator of next sequence
:- (1+2 ) *(1-2) * 2 = 6, (6+1)*(6-1)* 1= 35 , repeating the next term then ans is option (1)

Like? Yes (1) | No   

Pranati Kundu   9 years AGO

previous numarator =next denominator
1st term numerator*(6+1)*(6-1)= the third term35
6*(35+6)*(35-6)=7134
35*(7134+35)*(7134-35)=1781245585

Like? Yes (1) | No   

21 Jul 2015 Time and Distance

A man can row three-quarters of a km against the stream in 45/4 minutes and return 15/2 minutes. The speed of the main in still water is

OPtion

1) 2 km/hr
2) 6 km/hr
3) 7 km/hr
4) 8 km/hr
5) 1 km/hr
6) 3 km/hr
7) 9 km/hr
8) 4 km/hr
9) 5 km/hr
10)None of these

Solution

Speed upstream = 3/4 * 2/15 * 60 km/hr = 4 km/hr
Speed downstream = 3/4 * 2/15 * 60 km/hr = 6 km/hr
Therefore
Speed in still water = 1/2 (4+6) km/hr = 5 km/hr

Option 9)

Correct Option: 9 Best Solution (37)

Sushmita Ghanty   9 years AGO

three quarters of a km= 750 m
45/4 mins= 675 secs, 15/2 mins= 450 secs
rate of upstream = (750/675)m/sec= 10/9 m/sec
rate of downstream = (750/450)m/sec = 5/3 m/sec
The speed of the main in still water= 1/2 * (10/9 + 5/3) = 25/18 m/sec = (25/18 * 18/5)= 5 km/hr

Like? Yes (26) | No (3)  

Priya   9 years AGO

distance=3/4 km
speed upstream= (3/4)/(45/4)km/min=1/15
speed downstream=(3/4)/7.5km/min=1/10
speed in still water= 1/2*(1/15+1/10)=1/12km/min= 5 km/kr

Like? Yes (18) | No (3)  

Veerraj Chauhan   9 years AGO

Rate Downstream = (3/4)/(15/2)= 1/10 km/min
= (1/10)*60= 6 km/hr

Rate Upstream = (3/4)/(45/4)= 1/15 km/min
= (1/15)*60= 4 km/hr

Rate in still water = 1/2 (6+4) = 5km/hr

Like? Yes (10) | No (1)  

Narendra Gopi   9 years AGO

We can write three-quarters of a kilometre as 750 metres,

and 11 minutes as 675 seconds.

Rate upstream =750/675m/sec =10/9m/sec.

Rate downstream =750/450 m/sec=5/3m/sec.

Rate in still water = 1/2(10/9+5/3)m/sec = 25/18m/sec
= 25/18x18/5km/hr
= 5 km/hr.

Like? Yes (7) | No (1)  

ABHINAV PUSHPAM   9 years AGO

Distance = 3/4 km
Time taken to travel upstream = 11 1/4 minutes = 45/4 minutes
= 45/(4×60)hours = 3/16 hours
Speed upstream = Distance/Time=(3/4)/(3/16) = 4 km/hr
Time taken to travel downstream = 7 1/2 minutes = 15/2 minutes
= 15/(2×60) hours = 1/8 hours
Speed downstream = Distance/Time=(3/4)/(1/8)= 6 km/hr
Rate in still water = (6+4)/2=10/2=5 kmph

Like? Yes (2) | No   

aman aditya   9 years AGO

speed of man against the stream=4km/hr
speed of man with the stream= 6km/hr
speed in still water=5km/hr

Like? Yes (2) | No   

Archie   9 years AGO

Dist covered =3/4 km
Speed upstream ={(3/4)/(3/16)}km/hr=4km/hr
Speed downstream={(3/4)/(1/8)}km/hr=6kmph
Speed of man in still water ={4+6}/2 =5km/hr

Like? Yes (2) | No   

Pulkit   9 years AGO

Basically for against the stream take velocity as (velocity in still water - velocity against the stream) and for downstream add both velocities and apply v=km/hr

Like? Yes (1) | No   

Soma Naresh   9 years AGO

upstream speed(x)=0.750 km/(45/4) min
x=4 kmph
downstream speed (y) =0.750 km/(15/2) min
y=6 kmph
speed in still water=0.5 (x+y)
0.5 (4+6)
0.5�10
5kmph

Like? Yes (1) | No   

sudiptaa   9 years AGO

Rate upstream = 750/675 m/sec = 10/9 m/sec.

Rate downstream = 750/450 m/sec = 5/3 m/sec.
Rate in still water = 1/2( 10/9 +5/3) m/sec
= 25/18 m/sec
= 25/18 x 18/5 km/hr
= 5 km/hr.

Like? Yes (1) | No   

Abhisheak Kumar Missu   9 years AGO

speed of man in upstream=3/4*4/45=1/15km/m
speed of man in downstream=3/4*2/15=1/10km/m
speed of man in still water=1/15+1/10/2=5/60km/m=5km/hr

Like? Yes (4) | No (3)  

praveendas   9 years AGO

if he covers 3/4 distance in 45/4 min..in order to cover the full distance required time is5min.from this the speed is4km/hr
towards downstream the time taken to cover 3/4 km is 15/2..therfore to cover 1 km the required time is10 min..the speed is 6km/hr..
the avg of these two gives..speed of the man in still water which is (6+4)/2=5km/hr

Like? Yes (1) | No   

shivani gupta   9 years AGO

rat upstream=750/675m/sec=10/9 m/sec
rate downstream=750/450=5/3m/sec
rate instill water=1/2(a+b)
1/2(10/9+5/3)25/18m/sec=5km/hr

Like? Yes (1) | No   

Dinesh kumar   9 years AGO

3/4 km against the stream = 45/4 min
3/4 km parallel to stream = 15/2 min
3 km against the stream = 45 min
3 km parallel to stream = 30 min
Therefore ,against stream the speed is 4km/hr
for the stream the speed is 6km/hr

stream speed is (6-4)/2=1km/hr

Like? Yes (3) | No (2)  

PRITI MISHRA   9 years AGO

s=vt
velocity of man against the stream is (3/4) / ((45/4)/60) = 4 KM/H
velocity of man towards the stream is (3/4) / ((15/2)/60) =6
speed of the man in still water= (4+6)/2 = 5 km/hr

Like? Yes (2) | No (1)  

anand vijay   9 years AGO

let the speed of the man in still water be x and the speed of the stream be y
hence we have the two equation
going against the stream 3/4(x-y)=45/60*4
x-y = 4.......................(i)
going with the stream 3/4(x+y)=15/60*2
x+y = 6.......................(ii)
on solving (i) & (ii) we get
x=5, and y=1
hence , the speed of the man in still water= 5km/hr
option 9)

Like? Yes (1) | No   

pankaj kumar   9 years AGO

a man speed is 4 kmph against stream and return speed is 6 kmph, so the speed of man in still water is as average speed 5 kmph

Like? Yes (1) | No   

Piyush Telang   9 years AGO

Distance Traveled Upstream and Downstream (Dist) = 3/4 km
Time Taken while going upstream (Tu)= 45/4 min = 3/16 hours
Time Taken while going downstream (Td) = 15/2 min = 1/8 hours
Speed while going upstream (U) = (Dist)/(Tu) = (3/4) / (3/16) = 4 km/hr
Speed while going downstream (D) = (Dist)/(Td) = (3/4) / (1/8) = 6 km/hr
Speed of Still water is given as = ( U + D ) / 2 = (4+6)/2 = 5 km/hr

Like? Yes (1) | No   

SOMA NARESH   9 years AGO

upstream speed(x)=0.750 km/(45/4) min
x=4 kmph
downstream speed (y) =0.750 km/(15/2) min
y=6 kmph
speed in still water=0.5 (x+y)
0.5 (4+6)
0.5×10
5kmph

Like? Yes (1) | No   

Ashu Kumar   9 years AGO

LET SPEED OF STREAM IS U & MAN IN STILL WATER IS V
V-U=3*4*60/(4*45)=4
V+U=3*2*60/(4*15)=6
SO V=5

Like? Yes (1) | No   

Riyashree Mukherjee   9 years AGO

Rate at still water=1/2(rate of upstream+rate of downstream)=5km/hr

Like? Yes (1) | No   

Ravi shankar   9 years AGO

(b-w)*(1/4)=1 and (b+w)*(1/8)=3/4. solving this we get b=5kmph

Like? Yes (1) | No   

krishna chandrika   9 years AGO

speed of upstream = (3*4*60)/(4*45) = 4km/hr
speed of down stream = (3*2*60)/(4*15) = 6km/hr

speed of man in still water = 1/2(speed of upstream+speed of down stream)
=1/2(4+6)
=5km/hr

Like? Yes (1) | No   

Ajay Kumar Gupta   9 years AGO

We can write three-quarters of a kilometre as 750 metres,

and 45/4 minutes as 675 seconds.

Rate upstream = (750/675) m/sec = 10/9 m/sec.
Rate downstream =(750/450) m/sec = 5/3 m/sec.
Rate in still water =1/2(10/9+5/3)m/sec = 25/18 m/sec
=(25/18)*(18/5)=5 km/hr

Like? Yes (1) | No   

Kowshik Biswas   9 years AGO

man + water speed =6 km/h
(in 30/4 min it can row 3/4 km so speed of man + water =3*4*60/4*30=6 km/h )
man - water speed =4 km/h
(in 45/4 min it can row 3/4 km so speed of man - water =3*2*60/4*15=4 km/h )
so, man + water = 6 km/h (speed)
man - water = 4 km/h (speed)
so,man speed in steel water is 5 km per hour

Like? Yes (1) | No   

Deepak kushwaha   9 years AGO

Speed of man in upstream is 4kmph
Speed of man in downstream is 6kmph
Let the speed of man in still water be x & speed od of water be y
then,
x+y=6..........(1)
x-y=4...........(2)
from eqn(1)&(2)
x=5(option 9)

Like? Yes (1) | No   

prem kishor   9 years AGO

against the stream
750/v-s=45/4

750/v+s=15/2

after solving eqn we will get v=5km/hr

Like? Yes (2) | No (1)  

vimal tiwari   9 years AGO

speed of man = (going speed - return speed )/2;
going speed = (3/4) / (45/(4*60));
return speed = (3/4) / (15/(2*60));

speed of man = 5 km/hr

Like? Yes (1) | No   

siddharth prakash singh   9 years AGO

Assuume speed in still water =x and speed of stream=y. Down stream speed will be x+y and upstream speed will be x-y.
Solving given conditions. We get two equations giving x=5 and y=1.

Like? Yes (1) | No   

Arju   9 years AGO

Answer: Option 9

Explanation:

We can write three-quarters of a kilometre as 750 metres,

and 45/4 minutes as 675 seconds and return 15/2minutes as 450seconds.

Rate upstream = 750/675 m/sec = 10/9 m/sec.

Rate downstream = 750/450 m/sec = 5/3 m/sec.

Rate in still water = 1/2( 10/9 + 5/3) m/sec

= 25/18 m/sec

= 25/18 x 18/5 km/hr
= 5 km/hr.

Like? Yes (1) | No   

anubhav kumar gupta   9 years AGO

total distance =3 /4km
let still water speed be x
and during upstream & downstream be = y
therefore,
speed during upstream i.e (x-y)=3/4 km*((4*60)/(45*1))
=4 -----(i)
speed during downstream i.e (x+y)=3/4*((2*60)/(15*1))
=6 -------(ii)
adding equation (i) and (ii)
we get
x+y+x-y=4+6
2x=10
x=5km/hr

Like? Yes (1) | No   

UJJWAL KANT   9 years AGO

Given distance =3/4km
for upstream
time=45/(4*60)=3/16 hr
speed=4kmph
for downstream
time=1/8hr
speed=6kmph
now
spd upstrm=spd of boat-spd of strm
4=SB-SS.............(1)

spd dwmstrm=spd of boat+spd of strm
6=SB+SS.............(2)
on solving both eqn 1 and 2 we get
spd of boat =5kmph

Like? Yes (1) | No   

sourav pal   9 years AGO

We can write three-quarters of a kilometre as 750 metres,

and 11 minutes as 675 seconds.

Rate upstream = 750 m/sec = 10 m/sec.
675 9

Rate downstream = 750 m/sec = 5 m/sec.
450 3

Rate in still water = 1 10 + 5 m/sec
2 9 3

= 25 m/sec
18

= 25 x 18 km/hr
18 5

= 5 km/hr.

Like? Yes (1) | No (1)  

Anuj Rai   9 years AGO

3/4th of a kilometer is 750

Time 45/4*60=675

upstream = 750/675=10/9
downstream = 750/450=5/3
rate in still water 1/2(10/9+5/3)
25/18
he speed(in km/hr) of the man in still water is
25/18*18/5=5kmph

Like? Yes (2) | No (2)  

SURAJ NIRALA   9 years AGO

let the speed of man in still water is x and the speed of the speed of stream is y km/h
now spead of man in the direction of stream will be (x+y) km/h
speed of man in the opposite direction of stream will be (x-y) km/h
Now as per question
(3/4)/(x-y) = (45/4)/60
i.e.
x-y =4 …………(1)
(3/4)/(x+y) =(15/2)/60
i.e.
x+y = 6 ………..(2)
On solving equation 1 and 2 we get
X=5 & y= 1
i.e. speed of man = x= 5km/h

Like? Yes (3) | No (3)  

Dinni   9 years AGO

We can write three-quarters of a kilometre as 750 metres,

and 11 minutes as 675 seconds.

Rate upstream = 750 m/sec = 10 m/sec.
675 9
Rate downstream = 750 m/sec = 5 m/sec.
450 3
Rate in still water = 1 10 + 5 m/sec
2 9 3
= 25 m/sec
18
= 25 x 18 km/hr
18 5
= 5 km/hr.

Like? Yes (1) | No (1)  

RAKESH ARYA   9 years AGO

(x-y)*(45/4) = (x+y)*(15/2)
=> x = 5y
=> y = 1/5

then , (x-x/5) * 45/( 4 * 60) = 3/4
=> x =5 km/hr

Like? Yes (1) | No (1)  

20 Jul 2015 Decimal Fractions

The L.C.M. of 3, 2.7, .09 is

OPtion

1) .9
2) .018
3) 1.8
4) .18
5) 2.7
6) 18
7) .27
8) .027
9) 27
10)9

Solution

With equal number of decimal places, the given number 3.00, 2.70, .09
L.C.M. of 300, 270, 9 is 2700
Therefore L.C.M. of given number is 27.00

Option 9)

Correct Option: 9 Best Solution (55)

RAKESH   9 years AGO

L.C.M. of 3, 2.7, .09 = lcm(3,27,9)/hcf(1,10.100)= 27/1 = 27

Like? Yes (27) | No (9)  

anand vijay   9 years AGO

3,2.7,0.09 can be written in fraction form as
3/1, 27/10, 9/100
since, L.C.M of two or more fractions is given by L.C.M of numerators/ H.C.F of denominators ( A/c to how to prepare for Quantitative Aptitude for CAT by ARUN SHARMA page no.:-26 fifth edition)
LCM of numerators 3,27,9 is 27
HCF of denominators 1,10,100 is 1
therefore , 27/1
=27
option 9)

Like? Yes (10) | No (1)  

UJJWAL KANT   9 years AGO

3, 2.7, 0.09
these can be written as
3/1, 27/10, 9/100
Lcm = Lcm of numerator/Hcf of denominator
Lcm= 27/1=27

Like? Yes (6) | No   

praveendas   9 years AGO

lcm 0f numerator/hcf of denominators=27 is the ans
lcm of 3,27,9 is 27
hcf of 1,10,100 is 1
therfore 27/1=27

Like? Yes (5) | No   

Piyush Telang   9 years AGO

LCM of 3, 2.7, 0.09 is calculated as fractions
=> 300/100 , 270/100, 9/100
=> LCM of numerators / HCF of Denominators
=> 2700 / 100 => 27
Thus LCM of 3, 2.7, 0.09 is 27

Like? Yes (7) | No (3)  

Abhishek   9 years AGO

3, 2.7, .09

Now, expressing each of the numbers without the decimals as the product of primes we get

300 = 3 × 2 x 2 x 5 x 5

270 = 3 x 3 x 3 x 5 x 2

9 = 3 x 3

L.C.M. of 300, 270, 9 = (2^2) x (3^3) x (5 ^2) = 4 x 27 x 25 = 2700

Therefore, L.C.M. of 3, 2.7, .09 = 27.00 (taking 2 decimal places)

Like? Yes (7) | No (3)  

ABHINAV PUSHPAM   9 years AGO

LCM of 300, 270, 9 is 2700
So, Answer is 27
option 9)

Like? Yes (4) | No (1)  

Amit Roy   9 years AGO

the given numbers with equal numbers of decimal places are 3.00,2.70 and 0.09
L.C.M of 300,270,9 is 2700.
therefore L.C.M of given number=27.00
therefore option 9 is correct!!

Like? Yes (2) | No   

Sanjoy   9 years AGO

L.C.M. = L.C.M. of Numerators / H.C.F. of Denominators
The L.C.M. of 3/1,27/10 and 9/100 is
= 27

Like? Yes (2) | No   

Abhisheak Kumar Missu   9 years AGO

L.C.M of 3/1,27/10,9/100
=L.C.M of 3,27,9/H.C.F of 1,10,100
=27

Like? Yes (2) | No   

Kirty   9 years AGO

without decimal places the nos. are 300,270,9
so l.c.m of 300,270,9 is 2700
thus L.c.m of 3,2.7,0.09 is 27

Like? Yes (2) | No   

Dhanraj Wanjare   9 years AGO

=>3,2.7,0.09 LCM
=>31, 27/10, 9/100
=>(L.C.M. Of 3, 27, 9)/(H.C.F. of 1,10,100)
=>LCM 27/ HCF 1
=> so 27/1
=> 27
option 9 satisfied

Like? Yes (2) | No   

bhavya dubey   9 years AGO

27/3=9
27/2.7=10
27/.09=300
and it is the lowest number which is divisible by all the numbers

Like? Yes (2) | No   

lavanya   9 years AGO

first i have muliplied with 100 and then the final lcm is divided by 100

Like? Yes (2) | No   

vivek kashyap   9 years AGO

GIVEN NOS=3.00=300
2.70=270
.09=9
LCM(300,270,9)=2700
SO LCM OF NO IS 27.00

Like? Yes (1) | No   

shashank   9 years AGO

rest all options dnt satisfy the minimum conditions fr lcm....while 27 does it. it is completly divisble by 3, .09 and .27

Like? Yes (1) | No   

shivani gupta   9 years AGO

the l.c.m of 3.00 ,2.70,.09 is the LCM of(300,270 and 9) 2^2 *3^3*5^2=900 then it is wriiten as 27.00
so ans is 27

Like? Yes (2) | No (1)  

Santosh Kumar   9 years AGO

L.C.M of rational number=L.C.M of numntr/H.C.F of denmntr
so,L.C.M=L.C.M of 3 ,27,9/H.C.F of 1,10,100
hence,27/1=27ans

Like? Yes (1) | No   

Rahul kumar   9 years AGO

lcm of fraction =lcm of numrator/ hcf of denominator
lcm of 3,27,9=27
hcf of 1,10,100=1
so
lcm =27/1=27

Like? Yes (1) | No   

k sureshbabu   9 years AGO

among the options 27 is the least common multiple
3,9,12,15,18,21,24,27
2.7,5.4,8.1,...........27
0.09,0.18,0.27,.........,27
or
Option should be divisible by each number with remainder zero

Like? Yes (1) | No   

Priya   9 years AGO

for decimal numbers we multiply all by maximum number to make all numbers integers.
here we multiply all by 100.
so numbers become 300,27,9. LCM (300,27,9)=2700.
finalyy 2700/100=27
so ans -(9)27

Like? Yes (1) | No   

Ashu Kumar   9 years AGO

LCM OF 3,27/10 AND 9/100=LCM OF (3/1,27/10,9/100) =LCM OF NOMINATIOR /HCF OF DOMINATOR=27/1=27

Like? Yes (1) | No   

rajkumar   9 years AGO

given 3,2.7,0.09

multiply with 100 we get, 300,270,9
factors are 3*3*10*3*10
now divide with 100
we get 3*3*3=27
so option 9) is correct

Like? Yes (1) | No   

m.lokeswari   9 years AGO

Multiply all the numbers with 100 then find the lcm of 300,270,9
We get 2700
Finally divide it by 100
The answer is 27

Like? Yes (1) | No   

Balaselvaraj M   9 years AGO

first decimal no's are change into integer.
thus we get 300,27,9.
then take lcm for these no's and finallu resultant no is divide by 100.
we get 27

Like? Yes (1) | No   

Ankita Sharma   9 years AGO

no.'s are 3,2.7,.09
converting them to whole no's by multiplying by 100 we get,
300,270,9
calculating the LCM we get the LCM to be
2700
since we multiplied it by 100 in the starting to get reqd LCM we will divide it by 100 and get LCM=27

Like? Yes (1) | No   

Soma Naresh   9 years AGO

3 , 27� 10 ^ - 1, 9� 10 ^ - 2
( 3 ^ 1 � 10 ^ 0 ), ( 3 ^ 3 � 10 ^ - 1), ( 3 ^ 2 � 10 ^ - 2)
LCM of numbers is the product of highest power terms in the each number
so 3^3�10^0=27

Like? Yes (1) | No   

SOMA NARESH   9 years AGO

3 , 27× 10 ^ - 1, 9× 10 ^ - 2
( 3 ^ 1 × 10 ^ 0 ), ( 3 ^ 3 × 10 ^ - 1), ( 3 ^ 2 × 10 ^ - 2)
LCM of numbers is the product of highest power terms in the each number
so 3^3×10^0=27

Like? Yes (1) | No   

samar abbas   9 years AGO

Step 1. First taking all no in same format i.e: 3.00 , 2.70 . 0.09
Step 2. Multiply by 100 ,
Step 3. Now taking lcm of 300 , 270,9
Step 4. Divide by 100.

Like? Yes (3) | No (2)  

RaviKiran   9 years AGO

300/100,270/100,9/100
LCM of numerators and HCF of denominaors

Like? Yes (1) | No   

keerthi choudhry   9 years AGO

The given nos are 3,2.7,.09
the lcm ,
since the largest decimal point is 2 we can write the given numbers as,
300,270,9
thus,
300=2^2*3*5^2
270=2*3^3*5
9=3^2
=> 2^2*3^3*5^2=2700
on taking 2decimal points,we get the answer as 27
option:9

Like? Yes (1) | No   

sai bhargav   9 years AGO

LCM( 3/1 , 27/10 , 9/100 )=LCM(3,27,9)/HCF(1,10,100)
=>27/1
=1

Like? Yes (1) | No   

Sushmita Ghanty   9 years AGO

lcm of 3,2.7,.09
= (lcm of 300,270,9)/100
=2700/100
=27

Like? Yes (1) | No   

Govind raj   9 years AGO

Simple method first by 0.03 then 3 then 10 and solve we get 27

Like? Yes (1) | No   

Charulata Lodha   9 years AGO

Given numbers can be written as
3/1, 27/10 , 9/100

So lcm=lcm of num/lcm of denom

Lcm=27/1

Like? Yes (1) | No   

naveen   9 years AGO

because we r doing 3/1,27/10,9/100
in this we r doing lcm so
lcm of numerator/hcf of denominator
by this we get lcm as 27 and hcf as 1
27/1=27

Like? Yes (1) | No   

Karna   9 years AGO

lcm of numerator is 27 and hcf of denominator is 1
so the ans is 27

Like? Yes (1) | No   

Manish Dwivedi   9 years AGO

l.c.m of a/b+c/d is equal to l.c.m of numerator divided by h.c.f of denominator

Like? Yes (1) | No   

G Mounika   9 years AGO

Multiply each number with 100
Then the numbers become 300,270,9
The L.C.M of 300, 270, 9 is 2700
So the L.C.M of 3.00, 2.70, 0.09 is 27.

Like? Yes (1) | No   

Murali Krishna.P   9 years AGO

to remove the decimal value the three numbers should be multiplied by 100.
Now the new numbers are 300,270,9.
and the lcm of the above numbers is 2700 and the ans should be divided 100.

and therefore 2700/100=27.

Like? Yes (1) | No   

Nikita   9 years AGO

it has to b greater than 3, and a multiple of 2.7, satisfying 1st condition out of 9, 18 and 27 is option (9) 27

Like? Yes (1) | No   

Kshitiz Khandelwal   9 years AGO

Lcm of decimals = (LCM of numerators)/(HCF of deno minators)
In this ques
Nos. are 3/1, 27/10, 9/100
LCM of Numerators = 27(3, 27,9)
HCF of denominators = 1(1,10,100)

Hence LCM is 27/1 = 27

Like? Yes (1) | No   

madhu   9 years AGO

lcm of 3,2.7,.09 is 27
convert given numbers to whole numbers 300,270,9
lcm of 300,270,9 is 3*3*5*2*10*3=2700
hence 2700/100=27

Like? Yes (1) | No   

Navin prasad K   9 years AGO

L.C.M is the product of highest powers of all the factors so 2^2, 3^3,5^2=2700 after simplification 27 will get

Like? Yes (1) | No   

kavita singh   9 years AGO

LCM of fraction = lcm (3,9,27)/HCF(1,10,100)
LCM of numerator is 27 and
HCF is 1
so answer is option number 9

Like? Yes (1) | No   

KUMAR HARSH   9 years AGO

lcm of 3,27/10,9/100
lcm of 3,27,9=27
hcf of 1,10,100=1
lcm of 3,27/10,9/100=27

Like? Yes (1) | No   

yashas   9 years AGO

300/100, 270/100, 9/100 are given nos.
LCM of these no's = LCM of numerator ÷ LCM of denominator = 2700/100 = 27
Option 9)

Like? Yes (1) | No   

Soumen Dofadar   9 years AGO

multiply every number by 100 then find the lcm then divide it by 100

Like? Yes (1) | No   

shivjee yadav   9 years AGO

i can write as 3/1,2.7/10,09/100...then lcm of this is lcm of nominator by hcf of denominator =27/1=27

Like? Yes (1) | No (1)  

Dinesh kumar   9 years AGO

First make three numbers into whole numbers.
multiply by 100.
300,270,9
by taking l.c.m we get 2700
divide by 100
The answer will be 27

Like? Yes (1) | No (1)  

SAYAN CHATTERJEE   9 years AGO

the LCM of 30/10, 27/10 & 9/10 is 270/10
=27 (9)

Like? Yes (1) | No (1)  

PRITI MISHRA   9 years AGO

3, 27/10, 9/100
L.C.M will be= (L.C.M of 3,27&9) / (H.C.F of 1,10,100)
L.C.M will be= 27 / 1
=27

Like? Yes (1) | No (1)  

DEEPAK RAJ   9 years AGO

Numers are 3/1, 27/10 and 9/100.
LCM of this = (LCM of numerators)/(HCF of denomonators)

So, LCM = 27/1 = 27

Like? Yes (1) | No (1)  

Dinni   9 years AGO

LCM of 300, 270, 9 is 2700
So LCM of 3.00, 2.70, 0.09 is 27.00

Like? Yes (1) | No (1)  

roshni.s   9 years AGO

multiply those numbers with 100 for easy solving... then take LCM of those ... ans will be 2700... atlast remove two zeros... ans is 27...

Like? Yes (1) | No (1)  

17 Jul 2015 Problem on ages

The age of a father 10 years ago was thrice the age of his son. Ten years hence, the father's age will be twice that of his son. The ratio of their present ages is

OPtion

1) 5:2
2) 11:2
3) 8:3
4) 5:2
5) 13:4
6) 3:1
7) 9:2
8) 7:2
9) 7:3
10)None of these

Solution

Let son's age 10 years ago be x.
Father's age 10 years ago = 3x
2(x+20) = (3x+20)
x = 20
10 years ago, father's age = 60 years and son's age = 20 years
Ration of their present age = 70:30 = 7:3

Option 9)

Correct Option: 9 Best Solution (73)

G Mounika   9 years AGO

Present age of the son = x years.
10 years ago age of the son = x-10
10 years ago age of a father = 3(x-10)
Ten years hence age of son = x+10
Ten years hence age of father = 2(x+10)

from the above data
2(x+10) = 3(x-10)+20
2x+20 = 3x - 30 +20
x=30.
Ten years ago father's age = 3(x-10) = 3 * 20 =60.
Present father's age = 60+10 = 70.
Hence ration of their ages = 70 :30 = 7:3.

Like? Yes (31) | No (8)  

sudiptaa   9 years AGO

Present age of father = x years
present age of son = y years

x-10= 3(y-10) .............(1)
x+10= 2(y+10).............(2)

solving 1 and 2
x= 70
y= 30
ratio = 7:3

Like? Yes (14) | No (5)  

Devendra Marghade   9 years AGO

Let the present age of father=f and son=s
Given, f-10=3(s-10) or f-3x=-20 ---(i) and
f+10=2(x+10) or f-2x=10 ---(ii)
On solving (i) & (ii), f=70, s=30
So ratio of their present ages=70/30= 7/3

Answer option(9)

Like? Yes (9) | No (1)  

Piyush Telang   9 years AGO

Let X be the present age of father and Y be the present age of son.
Then X - 10 = 3 ( Y - 10 ) => X = 3Y - 20 &
X + 10 = 2 ( Y + 10 ) => X = 2Y + 10
Then Y = 30 and X = 70
Ratio Between their present ages = 7:3

Like? Yes (7) | No (1)  

sourav pal   9 years AGO

Let the ages of father and son 10 years ago be 3x and x years respectively.

Then, (3x + 10) + 10 = 2[(x + 10) + 10]

3x + 20 = 2x + 40

x = 20.

Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.

Like? Yes (13) | No (11)  

Abhimanyu Kumar   9 years AGO

Assume son's present age to be x

Son's 10 yrs ago=x-10
so father's 10 yrs ago = 3(x-10)

father's present=3x-30+10=3x-20
father's 10 yrs hence=3x-10
Son's 10 yrs hence=x+10

now,
3x-10=2(x+10)
So x=30
So the required ratio is
3x-20/x = 90-20/30
= 7/3

Like? Yes (4) | No (2)  

Kethineni Vinodh   9 years AGO

Let the ages of father and son 10 years ago be 3x and x years respectively.

Then, (3x + 10) + 10 = 2[(x + 10) + 10]

3x + 20 = 2x + 40

x = 20.

Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.

Like? Yes (2) | No   

Ashu Kumar   9 years AGO

LET PRESIDENT AGE OF FATHER IS F,SON IS S
10 YEAR AGO ,F-10=3(S-10)
AFTER 10 YEAR ,F+10=2(S+10)
S=30
F=70
SO RATIO=7/3

Like? Yes (3) | No (1)  

prem kishor   9 years AGO

let father's age=x
and son's age=y
hence, x-3y=-20
x-2y =10
after solving these two equestion we can get x=70 and y=30 so the ration of x/y=7/3

Like? Yes (2) | No (1)  

NIRBHAY KUMAR   9 years AGO

Let the present age of father =x
and the present age of son=y
A/C to question, 10 yrs ago
x-10=3(y-10) ...eq1
and 10 yrs hence
x+10=2(y+10) ......eq2
solving these two eqs..... we get
x=70 and y=30

Like? Yes (2) | No (1)  

anand kumar   9 years AGO

let 10 years back sons age was x, and fathers was 3x. 3x+20=2(x+20)
x=20
so, 10 year back father was 60, son was 20. now father 70, son 30. 70/30=7:3

Like? Yes (1) | No   

ANKUR ANAND   9 years AGO

x-10=3(y-10)
and x+10=2(y+10)
Solving these two equations we get
x=70 and y=30
so the required ration is 7:3

Like? Yes (1) | No   

Narendra Gopi   9 years AGO

son's present age=x years
son's age 10 years ago=x-10
father's age 10 years ago=3*(x-10)
after 10 years son's age=x+10
after 10 years father's age=2*(x+10)
so 3*(x-10)+20=2(x+10)
3x-30+20=2x+20
3x-10=2x+20
son's present age x=30,
and father's present age is= 3(x-10)+10
=3(30-10)+10
=3*20+10
=70
Ratio of theirs age is 70/30=7/3

Like? Yes (1) | No   

vivek kashyap   9 years AGO

Let the ages of father and son 10 years ago be 3x and x years respectively.

Then, (3x + 10) + 10 = 2[(x + 10) + 10]

3x + 20 = 2x + 40

x = 20.

Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.

Like? Yes (1) | No   

praveendas   9 years AGO

let assume the ages of father and son be x and y
10 years ago it was x-10=3*(y-10)
10 years hence it wil be x+10=2*(y+10)
solve above two equations for x and y.
we get x=70 and y=30...
the ratio is7:3

Like? Yes (1) | No   

Shiva Sharma   9 years AGO

Let x and y is current age of father and son
x-10 = 3*(y-10)
x+10 = 2*(y+10)
after solving two equation
x=70
y=30
so x:y = 7:3

Like? Yes (4) | No (3)  

charul kardam   9 years AGO

let present age of father be x
and present age of son be y
then 10 years ago
x-10=3(y-10).... eq (1)
after 10 years
x+10=2(y+10)... eq (2)
solving 1&2 x=70
y=30
hence ratio is 7:3

Like? Yes (1) | No   

BINOD BIHARI BEHERA   9 years AGO

10 yrs ago : f=3x, s=x
present: f=3x+10, s=x+10
after 10 yrs : f=3x+20, s=x+20
now 3x+20= 2(x+20)
so x =20
present: f=70 and s=30
so f:s= 7:3

Like? Yes (1) | No   

Soumen Maity   9 years AGO

let say, present age of father is 'f' and son is 's'
So, f-10=3*(s-10)
or, 3s-f=20.......................(i)

and, f+10=2*(s+10)
or, f-2s=30......................(ii)

solving these two we get, f=70, s=30
so present ratio of their ages is 7:3

Like? Yes (1) | No   

Dheeraj Kumar   9 years AGO

f age : x
s age :y

x-10=3(y-10)
x+10=2(y+10)

after solving this eq. we will get y=30; x=70
hence ratio=7:3

Like? Yes (1) | No   

Soma Naresh   9 years AGO

If F and S be present ages of father and son respectively
10 years ago
F-10=3 (S-10)
10 years hence
F+ 10=2 (S+10)
solving F=70
S=30
ratio=7:3

Like? Yes (1) | No   

rahul shaw   9 years AGO

father present age: f
son present age: s
two eq are:
(f-10)=3(s-10) and (f+10)=2(s+10)
after solving two eqn...
we get f=70 and s= 30
ratio=7:3

Like? Yes (1) | No   

krishna chandrika   9 years AGO

F-10 = 3(S-10)
F+10 = 2(S+10)

Solving the 2 eq, F = 70, S=30

Ratio of their ages F:S = 7:3

Like? Yes (1) | No   

SOMA NARESH   9 years AGO

F and S be present ages of father and son respectively
10 years ago
F-10=3 (S-10)
10 years hence
F+ 10=2 (S+10)
solving F=70
S=30
ratio=7:3

Like? Yes (1) | No   

manisha   9 years AGO

let 10 yrs before father's age=3x
son's age=x
then 20 yrs after father's age=3x+20
son's age=x+20
so given in the question 3x+20=2(x+20)
3x+20=2x+40
x=20
so
present age of father 3x+10=70
son's age x+10=30
so ration is 7;3

Like? Yes (1) | No   

Anil Gautam   9 years AGO

present age of father be x and that of son be y
so,10 years ago (x-10)=3(y-10)
10 years hence (x+10)=2(y+10)
solving above equations
y=30 and x=70
hence the ratio is 7:3

Like? Yes (1) | No   

bhavya dubey   9 years AGO

suppose father's age=x
and son's age=y
as per question before 10 years (x-10)=3(y-10)......equation 1
as per question after 10 years (x+10)=2(y+10)......equation 2
solving eqution 1 and eqution 2 we get x=70 and y=30
ratio=70/30=7/3

Like? Yes (1) | No   

Riyashree Mukherjee   9 years AGO

Let the present age of the son=x.
ten yrs ago his age=(x-10)
fathers age ten yrs ago=3(x-10)
ten yrs hence=3(x-10)+10+10=2(x+10)
x=30
Father:son=70:30
7:3

Like? Yes (3) | No (2)  

Pinku Burnwal   9 years AGO

let present age of father be x,
& of son be y
10yrs ago, x-10=3(y-10)
10yrs hence, x+10=2(y+10)
we get x=70 & y=30
option 9

Like? Yes (2) | No (1)  

UJJWAL KANT   9 years AGO

present age of father and son be f an s
ten years ago
f-10=3(s-10)......(1)
ten years hence
f+10=2(s+10).....(2)
solving eqn1 and 2 we get
s=30 and by putting it in eqn we get f=70
hence their ratio is 70/30=7:3

Like? Yes (1) | No   

Subham Mahanta   9 years AGO

let, father's age = x years
son's age = y years
so, 10 years ago, (x-10)=3(y-10)........equation (i)
10 years hence, (x+10)=2(y+10)..........equation (ii)
from [ (ii) - (i) ] , we get,
y = 30
so, from equation (ii) we plot the value in equation (ii) , we get,
x= 70
so, the ratio of their present ages are = x : y = 70 : 30 = 7 : 3
ans is option (9)

Like? Yes (1) | No   

Shanti Kumari   9 years AGO

x-3y=-20
x-2y=10
y=30
x=70

Like? Yes (1) | No   

Rahul kumar   9 years AGO

let fathers present age=x
sons present age=y
a/c to ques
x-10=3(y-10)........eqn 1
also
x+10=2(y+10).....eqn2
solving both eqn we get x=70,y=30
so,x:y=7:3

Like? Yes (1) | No   

Nitish Varshney   9 years AGO

let x be father age
y be son age
x-10=3(y-10)
x+10=2(y+10)
x=70
y=30
7:3

Like? Yes (1) | No   

Anubhav jain   9 years AGO

Fathers age's=X
Son age=Y
Ist cond.
X-10=3(y-10)
X-3Y+20=0...........(1)
IIst Cond.
X+10=2(Y+10)
X-2Y-10=0..............(2)
Solve
then X=70,Y=30
Ratio is =7/3

Like? Yes (1) | No   

santu ghosh   9 years AGO

10 yrs ago ,
son age =x
so father age =3x
10 yrs after,
son=x+20
father=3x+20
so,2(x+20)=3x+20
0r,x=20
hence ratio of thier present ages is 70:30=7:3

Like? Yes (1) | No   

aashish kumar   9 years AGO

let present ages of father and son be x nd y respectively.
10 years ago
(x-10)=3 (y-10).....eq 1
after 10 years
(x+10)=2 (y+10)......eq 2
solving both equations we get value of x=70, y=30
ratio=x/y=70/30=7/3

Like? Yes (1) | No   

Murali Krishna.P   9 years AGO

Let the ages of father and son 10 years ago be 3x and x years respectively.

Then, (3x + 10) + 10 = 2[(x + 10) + 10]

3x + 20 = 2x + 40

x = 20.

Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.

Like? Yes (1) | No   

santhanalakshmi   9 years AGO

let take father's age X and son's age 3Y before ten years
so X=3Y
Y=X/3
so X=X/3________________(1)
present age will be X+10=X/3+10______________(2)
after 10 years their ages will be
X+10+10=X/3+10+10___________________(3)
solving this equation we get
x=60
by substituting X=60 in equation 2
we get the ages as 70:30
so the answer is 7:3

Like? Yes (1) | No   

karthiga   9 years AGO

F - 10 = 3(s - 10)
F - 10 = 3S +20
F+ 10 = 2(s + 10)
F - 10 = 3S - 30;
2S +10 -10 = 3S - 30;
S = 30;
F = 70;
Ratio is 7:3.

Like? Yes (1) | No   

Lavanya Rajendran   9 years AGO

(y-10)=3(x-10)
y=3x-20
(y+10)=2(x+10)
y=2x+10
3x-20=2x+10
x=30
(y-10)=60
y=70
x=40
y=100
100/40
5:2

Like? Yes (1) | No   

sai   9 years AGO

10 years ago F-10=3(S-10)
10 years hence F+10=2(S+10)
by solving we get F=70, S=30
hence F/S=7/3

Like? Yes (1) | No   

sainath somisetty   9 years AGO

consider father: F
son : S
10 years ago: (F-10)=3(S-10)
10 years later: (F+10)=2(S+10)
solving this two equations we get age of father as 70 and son as 30
ratio of their ages is 7:3

Like? Yes (2) | No (1)  

k sureshbabu   9 years AGO

ten years back their age are 60:20
after 10 year 80:40
present 70:30
so finally7:3
is the answer

Like? Yes (2) | No (1)  

champavat jaydipsinh padamsinh   9 years AGO

f-3s=-20
f-2s=10
so s=30 and f=70
so f/s=70/30=7/3

Like? Yes (1) | No   

vertika agrawal   9 years AGO

f-10=3(s-10)
f+10=2(s+10)
solve this.f=70 s=30
ratio is 7:3

Like? Yes (1) | No   

Chirag Mathur   9 years AGO

let father's and son's present age is F and S respectively.
10 years ago=
F-10 = 3(S-10)............(1)
after 10 year from present=
F+10 = 2(S+10).............(2)

after solving both equation (1) & (2)
F=70 & S=30
that is the present age ,so the ratio would be == 7:3

Like? Yes (1) | No   

kumari jyoti   9 years AGO

Let father's present age=x
& son's present age=y.
Then 10years ago,
X-10=3(y-10).....(1)
And 10 years hence,
X+10=2(y+10)......(2)
Solving the eqns. We get,
X:y=7:3
So, option 9 is the answer.

Like? Yes (1) | No   

sahil arora   9 years AGO

let father present age be x and son's present age be y
then 10 years ago
x-10=3(y-10)-------------->(1)
10 years hence
x+10=2(y+10)---------------->(2)

solving both equation (1) & (2)
we get y=30
and we'll calculate x i.e 70

Like? Yes (2) | No (1)  

KUMAR HARSH   9 years AGO

let son age 10 year ago was x
so father age is 3x
10 year hance,
2(x+20)=3x+20
x=20
present age ratio=(3x+10):x+10=3:7

Like? Yes (2) | No (1)  

DEEPAK RAJ   9 years AGO

Let 10 years ago, Age of the son = x years
then, Fathers's son = 3x years

10 years from hence, 3x + 20 = 2(x+20)
So, x = 20 years.
So. current age of the son = 30 years
Current age of father = 70 years

Ratio = 7:3

Like? Yes (3) | No (3)  

SAYAN CHATTERJEE   9 years AGO

let present age of father be x and son be y
so, (x-10)= 3*(y-10)
& (x+10)= 2*(y+10)
solving, we get, x=70, y=30
so ratio of the ages is 7:3 (9)

Like? Yes (1) | No (1)  

shivjee yadav   9 years AGO

let son age = x and , father age =y
according to question
10 years ago,y-10=3(x-10).....eq(1)
now,10 years hence,y+10=2(x+10)...(2)
solving both eq we gate x=30,y=70
hence ratio is 7:3

Like? Yes (2) | No (2)  

Dinni   9 years AGO

Assume son's present age to be x

Son's 10 yrs ago=x-10
so father's 10 yrs ago = 3(x-10)

father's present=3x-30+10=3x-20
father's 10 yrs hence=3x-10
Son's 10 yrs hence=x+10

now,
3x-10=2(x+10)
So x=30
So the required ratio is
3x-20/x = 90-20/30
= 7/3

Like? Yes (3) | No (3)  

Deepak kushwaha   9 years AGO

10 yeasrs ago
F-10=3(S-10)---------------(1)
10 years from present
F+10=2(S+10)---------(2)
on solving eqn(1) and eqn(2)
we get F=70 and S=30
F:S=7:3(option 9)

Like? Yes (2) | No (2)  

anand vijay   9 years AGO

let x be the present age of father and y be the present age of son
then we have two linear equation
x-10 = 3(y-10)
x-3y = -20...........(i)
x+10 = 2(y+10)
x-2y=10..............(ii)
on solving (i) & (ii) we get
x=70, y=30
ratio of present ages is=x:y
=70:30
=7:3
option 9)

Like? Yes (1) | No (1)  

Amit Roy   9 years AGO

Let,
at present son's age=x and father's=y
therefore 10 years ago ,
y-10/x-10=3/1
or,3x-y=20-------(i)
again after 10 years,
y+10/x+10=2/1
or,2x-y=-10------(ii)

solving equ (i) and (ii)
we got,x=30
therefore y=70
hence y/x=7:3
option 9 is correct!!

Like? Yes (1) | No (1)  

Abhisheak Kumar Missu   9 years AGO

let the age of father be x and son be y.
ten years ago,
(x-10)=3(y-10)...............(1)
ten years hence
(x+10)=2(y+10)...................(2)
solving these two equation
x=70
y=30
ratio= 7:3

Like? Yes (2) | No (2)  

Inoka   9 years AGO

Assume Fathers current age is X and Son age is Y.
So The age of a father 10 years ago was thrice the age of his son = X - 10 = 3 (Y-10) --> Statement 1
Ten years hence, the father's age will be twice that of his son = X +10 = 2(Y+10) --> Statement 2

after resolve above 2 statements X = 70, Y = 30 so the ratio is 7: 3
Answer is 9)

Like? Yes (1) | No (1)  

sonali gaur   9 years AGO

Let the present age of father be 'x' & that of son be 'y'
10 years ago father's age = x-10 & son's age= y-10
(x-10)= 3(y-10)
x-3y= -20........(1)
10 years later father's age = x+10 & son's age will be y+10
(x+10)=2(y+10)
x-2y = 10.............(2)
Solving (1) & (2)
x=70 and y=30
Ratio (x:y)= 7:3

Like? Yes (1) | No (1)  

Anuj Rai   9 years AGO

Let the ages of father and son 10 years ago be 3x and x years respectively.

Then, (3x + 10) + 10 = 2[(x + 10) + 10]

3x + 20 = 2x + 40

x = 20.

Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.

Like? Yes (1) | No (1)  

pankaj kumar   9 years AGO

solu:- let father current age x & son age y . then 10 years ago age will be
equ 1 : x-10= 3(y-10)
and 10 years after age is :- x+10= 2(y+10)
after solving equ x= 70 & y = 30

Like? Yes (2) | No (2)  

Santosh Kumar   9 years AGO

let present age of father =y and present age of son=x.
then 10 years ago,
(x-10)3=y-10
i.e3x-y=20
10 yrs later
2(x+10)=(y+10)
i.e2x-y=-10
solving these two eqn we get,
x=30 nd y=70
hence,ratio of father age ans son age =70/30=7/3

Like? Yes (1) | No (1)  

ROHIT KUMAR   9 years AGO

Let current age of father= x and son= y
then according to question 10 year back
x-10 = 3(y-10)------------ 1
after 10 years
x+10 =2(y+10)-------------2
solving equation 1 and 2
x=70
y=30
so ratio is 7:3

Like? Yes (1) | No (1)  

shivani gupta   9 years AGO

present age of father=x years
present age of son=y years
ten years before father age =(x-10)
10 years before son age=(y-10)
so acc to que
(x-10)=3(y-10)
simillarly 10 years after,
(x+10)=2(y+10)
solving both eq,
we get
x=70
y=30
ie 7:3
so ans is 7:3 option 9

Like? Yes (1) | No (1)  

GITANJALI GOSWAMI   9 years AGO

present age of father=x
son=y
x+10=2(y+10)
x-10=3(y-10)

x=70
y=30
x:y=7:3

Like? Yes (1) | No (1)  

Rajkumar Swansi   9 years AGO

Let son's present age to be x

Son's 10 years ago= x-10
so father's 10 years ago = 3(x-10)

Father's present= 3x-30+10 = 3x-20
Father's 10 yrs hence= 3x-10
Son's 10 yrs hence=x+10

Then
3x-10=2(x+10)
So x=30
So the required ratio is
3x-20/x = 90-20/30
= 7/3

Like? Yes (1) | No (1)  

Biren Patel   9 years AGO

10 years befoe:
x-10 =3(y-10)
10 years hence:
x+10=2(y+10)
by solving this
y=30
x=70
x:y = 7:3

Like? Yes (1) | No (1)  

KAVITHA   9 years AGO

Let the ages of father and son 10 years ago be 3x and x years respectively.

Then, (3x + 10) + 10 = 2[(x + 10) + 10]

3x + 20 = 2x + 40

x = 20.

Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.

Like? Yes (1) | No (1)  

Thupakula vasu   9 years AGO

Let father age 3x ,sons age x

Then, (3x + 10) + 10 = 2[(x + 10) + 10]

3x + 20 = 2x + 40

x = 20.

Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.

Like? Yes (1) | No (1)  

shyamala   9 years AGO

x-10=3(y-10)
x+10=2(y+10)
on solving,we get
x=70,y=30
ratio=7:3

Like? Yes (1) | No (1)  

poojasaisri   9 years AGO

f-10=3s-30
f+10=2s+20
f=70
s=30

Like? Yes (1) | No (2)  

pooja jha   9 years AGO

let ten year ago son's age be x
then father age was 3x.
10 year hence son's age will be x+10+10=x+20.
and father age will be 3x+20.
therefore 3x+20=2(x+20).
so x=20.
therefore sons present age=20+10=30
and fathers present age =3*20+10=70.
therefore required ratio=70:30=7:3.

Like? Yes (3) | No (4)  

16 Jul 2015 Arithmetic Progression

If x, (x^3 + 1) and x^4 are in arithmetic progression, then possible sum of the 3 terms of arithmetic progression will be (Given taht x is real number)

OPtion

1) 27, 0
2) 25, 1
3) 20, -1
4) 27, -1
5) 25, 0
6) 20, 1
7) 27, 1
8) 25, -1
9) 20, 0
10)18, 2

Solution

According to given conditions
x^3 + 1 = (x^4 + x)/2
2x^3 + 2 = x^4 + x
2x^3 - x^4 + 2 - x = 0
x^3 ( 2-x) + (2-x) = 0
x^3 = -1 so x = -1
x = 2
So AP will be (1) 2, 2^3+1, 2^4, i.e 2, 9 16 (Sum = 27)
and (2) -1, (-1)^3+1, (-1)^4, i.e. -1, 0, 1 (Sum = 0)

Option 1)

Correct Option: 1 Best Solution (45)

SAYAN CHATTERJEE   9 years AGO

x, x^3+1 & x^4 are in AP
hence, 2x^3+2= x^4+x
=> x^4-2x^3+x-2=0
this has two factors x=-1 & x=2
so, the summation of the terms are 0 & 27 (1)

Like? Yes (1) | No   

ramakrishnateja   9 years AGO

given that x,x^3+1,x^4 are in a.p.hence x^3+1-x=x^4-(X^3+1)
x^4-2x^3+x=2. so x=2 and x=-1. sum=27,0

Like? Yes (1) | No   

sudiptaa   9 years AGO

x=2
series : 2, 9,16
sum=27
x= -1
series : -1,0,1
sum=0

Like? Yes (1) | No   

RAMPALA JYOTHI   9 years AGO

x^3+1-x=x^4-x^3-1;
x^4-2x^3+x-2=0;
there fore x= -1 & 2 (x is real);
sum of three terms:
if x=-1
x+x^3+1+x^4=0;
if x=2
x+x^3+1+x^4=27;
so ans: 27,0...................

Like? Yes (1) | No   

himanshu    9 years AGO

A.P=a,a+d,a+2d
given x,x^3+1,x^4
x=a
a+d=x^3+1
a+2d=x^4
put the value of d
we get x^4 -2x^3+x-2=0
put x=2 and x= -1 we get 0
therefore add x+x^3+1+x^4 we get 27,0

Like? Yes (1) | No   

UJJWAL KANT   9 years AGO

since x, x^3+1 and x^4 are in AP
x^3+1-x=x^4-(x^3+1)
from here we get two values of x i.e
x=2 and x=-1
putting these value we get
for x=2 AP will be 2, 9, 16 and sum=27
lly for x=-1 AP will be -1, 0, 1 and sum will be 0

Like? Yes (1) | No   

sudhanshu ranjan   9 years AGO

on solving the equation x^4 - 2x^3 + x - 2 =0 we get x=2 and x= -1

Like? Yes (1) | No   

Balaselvaraj M   9 years AGO

they tell us given series is a arithmatic progression and x is an integer.so we take 2 it satisfies the given data.

Like? Yes (1) | No   

Ananya Singh   9 years AGO

(x+x^4)/2=x^3 +1
x=(2,-1)
sum=n/2(2a+(n-1)d)
sum=27,0

Like? Yes (1) | No   

Deeksha Joshi   9 years AGO

since 3 terms are in ap so they hv same diffence .
therefore
x3+1-x=x4-x3-1
x4-2x3+x-2=0
(x+1)(x-2)(x2-x+1)=0
since x is real so x=-1,2

so
two sequences will b= -1,0,1
and 2,9,16
sum of the ap is= n*(a+l)/2
so sum of 2 aps is= 27 and 0.

Like? Yes (1) | No   

aashish kumar   9 years AGO

assuming x=2.
x+(x^3+1)+x^4
=2+(2^3+1)+2^4
=2+9+16
=27

Like? Yes (1) | No   

Soma Naresh   9 years AGO

difference
x^3+1-x=x^4-(x^3+1)
2x^3+2-x^4-x=0
2 (x^3+1)-x (x^3+1)=0
(2-x)(x^3+1)=0
x= 2,-1
x=2
sum=27
x=-1
sum=0

Like? Yes (1) | No (1)  

Deepak kushwaha   9 years AGO

here we have AP
x,x^3+1,x^4
we know that in AP
mean of first and third digit is equal to second digit
(x+x^4)/2=(x^3+1)
x^4+x=2x^3+2
x^4-2x^3+x-2=0
x^3(x-2)+1(x-2)=0
(x^3+1)(x-2)=0
on solving we get
x=2,-1
on putting these value we get
AP1=2,9,16
AP2=-1,0,1
sum of AP1=27
sum of AP2=0
possible sum is 27,0 (option 1)

Like? Yes (3) | No (3)  

NIRBHAY KUMAR   9 years AGO

The progression satisfies on x=2
now progression will be 2,9,16
And the sum of this progression is 27.

Like? Yes (1) | No (1)  

SOMA NARESH   9 years AGO

difference between terms is
x^3 +1 - x = x^4 -(x^3+1)
x^3 +1 - x = x^4 -x^3-1
2x^3 +2 = x^4+x
2(x^3+1)-x(x^3+1)=0
x = 2 x = -1
Sum when x=2 is 2+ 2^3+1+ 2^4 = 2+9+16 = 27
Sum when x=2 is -1+(-1+1)+1 = 0

Like? Yes (1) | No (1)  

anand vijay   9 years AGO

x,x^3+1,x^4 are in a.p.
x^3+1-x=x^4-x^3+1.....(i)
x(x-1)(x^2-x-1)=2
x=-1,2,3
on putting the value of x in equation (i) only x=-1,2
satisfy the equation
hence,
sum of a.p=x+x^3+1+x^4
=27 (x=2),0(x=-1)
option (i) is the right answer

Like? Yes (1) | No (1)  

ROHIT KUMAR   9 years AGO

For arithmetic progression x satisfies two value
I.e x= -1 and 2
then sum of 3 terms will be 0 and 27

Like? Yes (2) | No (2)  

pooja jha   9 years AGO

x,x^3+1 and x^4 are in ap.
so x^3+1-x=x^4-x^3-1.
x^4-2x^3+x-2=0
therefore x=2 and x=-1 are the two solution.
therefore sum of three numbers are 27 and 0 on putting x=2 and x=-1 respectively in the equation

Like? Yes (1) | No (1)  

Dinesh kumar   9 years AGO

x, (x^3 + 1) and x^4 are in arithmetic progression
Therefore (X^3+1)-X=X^4-(X^3+1)
X^4-2X^3+X-2=0
X^3(X-2)+1(X-2)=0
(X^3+1)(X-2)=0
X^3=-1 X=2
X=-1 X=2
sum of the 3 terms of arithmetic progression will be
X+(X^3+1)+X^4
sub X=-1
-1+(-1+1)+(-1)^4=0
sub X=2
2+(8+1)+2^4=11+16=27
Therefore ans is 27,0

Like? Yes (1) | No (1)  

karthiga   9 years AGO

when x=-1;
A.P = -1 , 0 , 1;
Sum is 0;
when x= 2;
A.P = 2 , 9 , 16;
Sum is 27;

Like? Yes (1) | No (1)  

Kethineni Vinodh   9 years AGO

2b=a+c
then (x^3+1)(x-2)=0
=>x = 2 or -1
sum= 27 or 0

Like? Yes (1) | No (1)  

SURAJ NIRALA   9 years AGO

2nd term - 1st=3rd- 2nd term
equivalent to x^4-2x^3+x-2=0

x=2. require number is 2, 9, 16
sum 27

Like? Yes (1) | No (1)  

Ashu Kumar   9 years AGO

X,X^3+1AND X^4 ARE IN AP
SO 2(X^3+1)=X+X^4
SO X=-1,2
FOR X=-1 TERMS ARE -1,0AND 1
SO SUM=0
FOR X=2 TERMS ARE 2,9AND 16
SO SUM=27
SUM=27,0

Like? Yes (2) | No (2)  

Amit Roy   9 years AGO

as it is an AP series so x^3+1-x=x^4-(x^3+1)
there after solving this equality we found x=2 and -1.
putting the value of x(=2) we found series is 2,9,16 i.e sum=27 & x(=-1) we found -1,0,1 i.e sum=0.

Like? Yes (2) | No (2)  

Vibhu Banka   9 years AGO

since the given terms are in A.P.,
so, 2(x^3+1)=x + x^4
this gives, x=2,-1
therefore possible sum= 27 or 0

Like? Yes (1) | No (1)  

Dinni   9 years AGO

X^4-2X^3+X-2=0.
X=2,-1.
X=2, then sum=2+9+16=27.
X=-1, then sum=-1+0+1=0.
So, 27 &0 are the answers.

Like? Yes (1) | No (1)  

shivjee yadav   9 years AGO

A/C to AP (X^3+1)-X=X^4-(X^3+1)

This progression satisfied on two values of x=-1, 2
SO,Answer will be 27,0

Like? Yes (1) | No (1)  

Abhishek   9 years AGO

Since terms are in A.P. (x^4 + x)/2 = x^3+1
on simplifying it, we get x^4 - 2 x^3+x-2 = 0
two of its roots are 2,-1 other two roots are imaginary.

now sum of terms using 2 as a root 2+2^3+1+2^4=27.
using -1 as root (-1)+(-1)^3+1+(-1)^4 = -1+(-1)+1+(1) = 0
so sum of 3 terms are 27 , 0

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Rahul kumar   9 years AGO

since given series is in ap we can take common diff as
2nd -1st =3rd -2nd
so we get
x^3+1-x=x^4-x^3-1
solving it we get x=2,-1
put once x=2&x=-1
also Sn=n/2(a+l)
we will get 2 series
1) 2,9,4 whose sum is 27
2)-1,0,1 whose sum is 0

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Deepak   9 years AGO

Real roots of the equation are: x = 2 and x = -1,
Sum of three terms: x^4 + x^3 + x +1; 27, 0

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sai   9 years AGO

given that the 3 are in a.p and x is a real number..the number that satisfies the condition is 2.. then the sum is 2+9+16=27

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GITANJALI GOSWAMI   9 years AGO

as it is in Ap,differences between successive terms are equal.So
(x^3+1)-x=x^4-(x^3+1)
=>x^4-2x^3+x-2=0
=>x^3(x-2)+1(x-2)=0
=>x=2
or
x=-1

so summation will be
for x=2,2+(8+1)+16=27
for x=-1,-1+(-1+1)+(-1)^4=0

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vivek   9 years AGO

if x= -1 thn,
-1,0,1..are in ap.
so sum of three term of series will be..27,0.

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Pinku Burnwal   9 years AGO

in A.P
2(2nd term)=1st term+3rd term
2(x^3+1)=x(x^3+1)
which give x=3,-1
which give sum of three term=27, 0
option 1

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PRITI MISHRA   9 years AGO

since, x, (x^3 + 1) and x^4 are in arithmetic progression
2(x^3+1)=x+x^4
2(x^3+1)=x(x^3+1)
(x^3+1) (2-x)=0
x=2,1,1,-1
hence,
for x=2, we have
x+(x^3 + 1) +x^4=27
and for x=-1
sum will be 0

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Ajay Kumar Gupta   9 years AGO

the no. is x=2 and -1
the terms are 2,9,16
and -1,0,1
so the sum of three terms are 27,0

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Kirty   9 years AGO

by equating common difference
x^3+1-x=x^4-(x^3+1)
X^4+2*x^3+x-2=0
solving this we get as x=-1 or x=2
by putting x=2 the sum comes to be 27
and by putting x=-1 the sum comes to be 0

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novel_liker   9 years AGO

values of x are calculated as -1,-1,-1,2
arithmatic series : -1,0,-1 so sum is 0
arithmatic series : 2,9,16 so sum is 27
hence option 1 satisfies (27,0)

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shiva   9 years AGO

Given that X, X^3+1, X^4 are in A.P and also given that X is a real number
Lets assume X=2 , -1
If X=2 then the terms are 2 ,9 ,16
Here the common difference d=7
If X= -1 then the terms in A.P are -1 , 0, 1
Here the common difference d= -1
Hence the possible sum of three terms is 27 , 0

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Manish kumar   9 years AGO

(x^3+1)-x=x^4 -(x^3+1)
=>x^4-2x^3+x-2=0
aftr solving we get
x=2,-1(real no.)
thus sum=27,0

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SAURABH SINGH   9 years AGO

when we put value like x=-1 their sum becomes zero.
and if we put values like x=2 their sum becomes 27.

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rahul shaw   9 years AGO

as numbers are in Arithmetic Possession
so
diff between nos are same...
(x^3+1)-x=(x^4)-(x^3-1)
after solving it...we get x=2 ,-�?1
given: x is a real no. so x= 2 and 0
adding all terms we get 27,0

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saikumar   9 years AGO

sum of ap=b=(a+c)/2
=x^3+1=(x+X^4)
x=2,-1
sum of terms=n/2(a+l)
if x=2 by sub above eqn ans=27
if x=-1
ans=0
(27,0)

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Shiva Sharma   9 years AGO

In AP 2*b=a+c
so 2*(x^3+1) = x+x^4
x^4+x-2*x^3-2=0
x^3+1=0 and x-2=0
x=2 and x^3 = -1
so sum = 27 and 0

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Piyush Telang   9 years AGO

Since x, x^3+1 and x^4 are in AP
Then 2 * ( x^3 + 1 ) = x + x^4 => x^4 - 2x^3 + x = 2
=> x ( x^3 -2x^2 + 1 ) = 2
=> x = 2 and x^3 -2x^2 + 1 = 2
=> x^3 -2x^2 = 1 => x^2 ( x - 2 ) = 1
=> x^2 = 1 and x- 2 = 1
=> x = +1 , -1 and 3
Thus possible values of x are -1, +1, +2 and +3.
But for values +1 and +3 the x, x^3+1 and x^4 does not follow to be in AP i.e. ( 1, 2, 1 ) and ( 3, 28, 81 )
Thus Sum is either 27 or 0 for the series ( 2, 9, 16 ) and ( -1, 0, 1 ) respectively.

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