Given numbers are 1.08 , 0.36 and 0.90
H.C.F of 108, 36 and 90 is 18 [ ? G.C.D is nothing but H.C.F]
Therefore, H.C.F of given numbers = 0.18.
The required numbers are 27, 36, 45……450.
This is an A.P. with a = 27 and d = 9
Let it has n terms.
Then Tn = 450 = 27 + (n-1) x9
450 = 27+ 9n - 9
9n = 432
n = 48.
C.I. when interest
compounded yearly=rs.[5000*(1+4/100)(1+1/2*4/100)]
= Rs. 5304.
C.I. when interest is
compounded half-yearly=rs.5000(1+2/100)^3
= Rs. 5306.04
Difference = Rs. (5306.04 - 5304) = Rs. 2.04.
sum=Rs.x
C.I=[x(1+4/100)^2-x]=(676/625x-x)=51/625
S.I=(x*4*2)/100=2x/25
x=625.
CAPITAL = 7
Vowels = 3 (A, I, A)
Consonants = (C, P, T, L)
5 letters which can be arranged in 5P5=5!
Vowels A,I = 3!/2!
No.of arrangements = 5! x 3!/2!=360
If n is divisible by 3, 5 and 12 it must a multiple of the lcm of 3, 5 and 12 which is 60.
n = 60 k
n + 60 is also divisible by 60 since
n + 60 = 60 k + 60 = 60(k + 1).
When n is divided by 8, the remainder is 3 may be written as
n = 8 k + 3
multiply all terms by 6
6 n = 6(8 k + 3) = 8(6k) + 18
Write 18 as 16 + 2 since 16 = 8 * 2.
= 8(6k) + 16 + 2
Factor 8 out.
= 8(6k + 2) + 2
The above indicates that if 6n is divided by 8, the remainder is 2.
Time taken by Akash = 4 h
Time taken by Prakash = 3.5 h
For your convenience take the product of times taken by both as a distance.
Then the distance = 14km
Since, Akash covers half of the distance in 2 hours(i.e at 8 am)
Now, the rest half (i.e 7 km) will be coverd by both prakash and akash
Time taken by them = 7/7.5 = 56 min
Thus , they will cross each other at 8 : 56am.
Let the sum be Rs. P.
Then,[p(1+10/100)2-p]=525
Sum =Rs.2500
S.I.= Rs.(2500*5*4)/100
= Rs. 500.
Difference in C.I and S.I for 2 years
= Rs(696.30-660)
=Rs. 36.30.
S.I for one years = Rs330.
S.I on Rs.330 for 1 year =Rs. 36.30
Rate
= (100x36.30/330x1)%
= 11%.
One may wer this question using a calculator and test for divisibility by 3.
However we can also test for divisibilty by adding the digits and if the result is
divisible by3 then the number is divisible by 3.
3 + 3 + 9 = 15 , divisible by 3.
3 + 4 + 2 = 9 , divisible by 3.
5 + 5 + 2 = 12 , divisible by 3.
1 + 1 + 1 + 1 = 4 , not divisible by 3.
S.I. on Rs.800 for 1 year
=Rs[840 - 800]
= Rs.40
Rate
=(100x40/800x1)%
= 5%.
Let the number be x and (15 - x)
Then, x2 + (15 - x)2= 113
x2- 15x + 56 = 0
(x-7) (x-8) = 0.
In this type of questions we need to get the relative speed between them,
The relative speed of the boys = 5.5kmph – 5kmph
= 0.5 kmph
Distance between them is 8.5 km
Time = Distance/Speed
Time= 8.5km / 0.5 kmph = 17 hrs.
Amount = Rs.(30000+4347) = Rs.34347
let the time be n years
Then,30000(1+7/100)^n = 34347
(107/100)^n = 34347/30000 = 11449/10000 = (107/100)^2
n = 2years.
By car 240 km at 60 kmph
Time taken = 240/60 = 4 hr.
By train 240 km at 60 kmph
Time taken = 400/100 = 4 hr.
By bus 240 km at 60 kmph
Time taken = 200/50 = 4 hr.
So total time = 4 + 4 + 4 = 12 hr.
and total speed = 240+400+200 = 840 km
Average speed of the whole journey = 840/12 = 70 kmph.
9 + 8 + 5 + * + 8 + 6 + 5 = 9x
41 + * = 9x
Nearest value of 9x must be 45
41 + * = 45
* = 4.
(2 x 14) men +(7 x 14) boys = (3 x 11) men + (8 x 11) boys
=>5 men= 10 boys => 1man= 2 boys
Therefore, (2 men+ 7 boys) = (2 x 2 +7) boys = 11 boys
( 8 men + 6 boys) = (8 x 2 +6) boys = 22 boys.
Let the required number of days be x.
More boys , Less days (Indirect proportion)
More work , More days (Direct proportion)
Boys22:11Work1 : 3}?? 14:x
Therefore, (22 * 1 * x) = (11 * 3 * 14)
=> x = 21
Hence, the required number of days = 21.
Originally let there be x men.
Less men, More days(Indirect Proportion)
Therefore, (x-10) : x :: 100 :110
=> (x - 10) * 110 = x * 100=> x= 110.
Speed of Man = 4.5 kmph
Speed of stream = 1.5 kmph
Speed in DownStream = 6 kmph
Speed in UpStream = 3 kmph
Average Speed = (2 x 6 x 3)/9 = 4 kmph.
Amount = Rs [7500*(1+(4/100)2] = Rs (7500 * (26/25) * (26/25) ) = Rs. 8112.
therefore, C.I. = Rs. (8112 - 7500) = Rs. 612.
We may have (1 boy and 3 girls)or(2boys and 2 girls)or(3 boys and 1 girl)or(4 boys).
Required number of ways = (6C1×4C3) + (6C2×4C2) + (6C3×4C1)+(6C4)
= (24+90+80+15)
= 209.
B runs around the track in 10 min.
i.e ,Speed of B = 10 min per round
Therefore, A beats B by 1 round
Time taken by A to complete 4 rounds
= Time taken by B to complete 3 rounds
= 30 min
Therefore, A's speed = 30/4 min per round = 7.5 min per round
Hence, if the race is only of one round A's time over the course = 7 min 30 sec.
Let distance = x km and usual rate = y kmph.
Then, x/y - x/(y+3) = 40/60 --> 2y (y+3) = 9x ----- (i)
Also, x/(y-2) - x/y = 40/60 --> y(y-2) = 3x -------- (ii)
On dividing (i) by (ii), we get:
x = 40 km.
Gold coins = 18000 , Silver coins = 9600 , Bronze coins = 3600
Find a number which exactly divide all these numbers
That is HCF of 18000, 9600& 3600
All the value has 00 at end so the factor will also have 00.
HCF for 180, 96 & 36.
Factors of
180 = 3 x 3 x 5 x 2 x 2
96 = 2 x 2 x 2 x 2 x 2 x 3
36 = 2 x 2 x 3 x 3
Common factors are 2x2×3=12
Therefore, Actual HCF is 1200
Gold Coins 18000/1200 will be in 15 rooms
Silver Coins 9600/1200 will be in 8 rooms
Bronze Coins 3600/1200 will be in 3 rooms
Total rooms will be (15+8+3) = 26 rooms.
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes,they will together (30/2)+1=16 times.
Let the sum be Rs.P.then
P(1+R/100)^3=6690…(i) and P(1+R/100)^6=10035…(ii)
On dividing,we get (1+R/100)^3=10025/6690=3/2.
Substituting this value in (i),we get:
P*(3/2)=6690 or P=(6690*2/3)=4460
Hence,the sum is rs.4460.
We are having time and speed given,
so first we will calculate the distance.
Then we can get new speed for given time and distance.
Lets solve it.
Time = 50/60 hr = 5/6 hr
Speed = 48 mph
Distance = S*T = 48 * 5/6 = 40 km
New time will be 40 minutes so,
Time = 40/60 hr = 2/3 hr
Now we know,
Speed = Distance/Time
New speed = 40*3/2 kmph = 60kmph.
Let the distance and original speed be 'd' km and 'k' kmph respectively.
d/0.8k - d/k = 20/60 => 5d/4k - d/k = 1/3
=> (5d - 4d)/4k = 1/3 => d = 4/3 k
Time taken to cover the distance at original speed
= d/k = 4/3 hours = 1 hour 20 minutes.
Let distance = x km.
Time taken at 3 kmph : dist/speed = x/3 = 20 min late.
time taken at 4 kmph : x/4 = 30 min earlier
difference between time taken : 30-(-20) = 50 mins = 50/60 hours.
x/3- x/4 = 50/60
x/12 = 5/6
x = 10 km.
Time = 2 years 4 months = 2(4/12) years = 2(1/3) years.
Amount = Rs'. [8000 X (1+(15/100))^2 X (1+((1/3)*15)/100)]
=Rs. [8000 * (23/20) * (23/20) * (21/20)]
= Rs. 11@.
:. C.I. = Rs. (11109 - 8000) = Rs. 3109.
Number of gaps between 41 poles = 40
So total distance between 41 poles = 40*50
= 2000 meter = 2 km
In 1 minute train is moving 2 km/minute.
Speed in hour = 2*60 = 120 km/hour.
Let x/5 = p and let y when divided by 5 gives q as quotient and 1 as remainder.
Then, y = 5q + 1
Now x = 5p and y = 5q + 1
x + y = 5p + 5q + 1 = 5(p + q) + 1.
Let xy be the whole number with x and y the two digits that make up the number.
The number is divisible by 3 may be written as follows
10 x + y = 3 k
The sum of x and y is equal to 9.
x + y = 9
Solve the above equation for y
y = 9 - x Substitute y = 9 - x in the equation 10 x + y = 3 k to obtain.
10 x + 9 - x = 3 k
Solve for x
x = (k - 3) / 3
x is a positive integer smaller than 10
Let k = 1, 2, 3, ... and select the first value that gives x as an integer.
k = 6 gives x = 1
Find y using the equation y = 9 - x = 8
The number we are looking for is 18.
It is divisible by 3 and the sum of its digits is equal to 9 and it is the smallest and positive whole number with such properties.
Since their HCFs are 7, numbers are divisible by 7 and are of the form 7x and 7y
Difference = 14
=> 7x - 7y = 14
=> x - y = 2
product of numbers = product of their hcf and lcm
=> 7x * 7y = 441 * 7
=> x * y = 63
Now, we have
x * y = 63 , x - y = 2
=> x = 9 , y = 7
The numbers are 7x and 7y
=> 63 and 49.
L.C.M. of 5, 6, 7, 8 = 840.
Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 x 2 + 3) = 1683.
Let x and y be the respective km's travelled by man via taxi and by his own car.
Given x + y = 90 => x = 90 - y
But according to the question,
7x + 6y = 595
7(90-y) + 6y = 595
=> 630 - 7y + 6y = 595
=> y = 630 - 595 = 35
=> x = 90 - 35 = 55
Therefore, the distance travelled by taxi is 55 kms.
99 = 1 x 3 x 3 x 11
101 = 1 x 101
176 = 1 x 2 x 2 x 2 x 2 x 11
182 = 1 x 2 x 7 x 13
So, divisors of 99 are 1, 3, 9, 11, 33, .99
Divisors of 101 are 1 and 101
Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176
Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.
Hence, 176 has the most number of divisors.
We know that
Time = Distance/speed
Required time = (10 1/4)/2 + (10 1/4)/6
= 41/8 + 41/6
= 287/24 = 11.9 hours.
Let the required length be x meters
More men, More length built (Direct proportion)
Less days, Less length built (Direct Proportion)
Men20:35
Days6 : 3}?? 56 :x
=> (20 x 6 x X)=(35 x 3 x 56)
=> x = 49
Hence, the required length is 49 m.
Distance travelled by Ramu = 45 x 4 = 180 km
Somu travelled the same distance in 6 hours.
His speed = 180/6 = 30 km/hr
Hence in the conditional case, Ramu's speed = 45 - 9 = 36 km/hr and Somu's speed = 30 + 10 = 40km/hr.
Therefore travel time of Ramu and Somu would be 5 hours and 4.5 hours respectively.
Hence difference in the time taken = 0.5 hours = 30 minutes.
LCM of 15, 20 and 36 is 180
Now 180 = 3 x 3 x 2 x 2 x 5
To make it perfect square, it must
be multiplied from 5.
So required no. = 32 x 22 x 52 = 900.
H. C. F of two prime numbers is 1.
Product of numbers = 1 x 161 = 161.
Let the numbers be a and b . Then , ab= 161.
Now, co-primes with product 161 are (1, 161) and (7, 23).
Since x and y are prime numbers and x >y ,
we have x=23 and y=7.
Therefore, 3y-x = (3 x 7)-23 = -2.
Time ( when X was 30 km ahead of Y) = (120-30)/20 =4.5h
Time ( when Y was 30 km ahead of X) = (120+30)/20 = 7.5 h
Thus, required difference in time = 3h.
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
n divided by 5 yields a remainder equal to 3 is written as follows
n = 5 k + 3 , where k is an integer.
add 2 to both sides of the above equation to obtain
n + 2 = 5 k + 5 = 5(k + 1)
The above suggests that n + 2 divided by 5 yields a remainder equal to zero.
Principal = Rs. 16000; Time = 9 months =3 quarters;
Rate = 20% per annum = 5% per quarter.
Amount = Rs. [16000 x (1+(5/100))3] = Rs. 18522.
CJ. = Rs. (18522 - 16000) = Rs. 2522.
The required number must be divisible by L.C.M. of 5,6 and 7.
L.C.M. of 5, 6 and 7 = 5 x 6 x 7 = 210
Let us divide 9999 by 210.
210) 9999 (47
840
----
1599
1470
----
129
Required number = 9999 – 129 = 9870.
Product of numbers = 11 x 385 = 4235
Let the numbers be 11a and 11b . Then , 11a x 11b = 4235 => ab = 35
Now, co-primes with product 35 are (1,35) and (5,7)
So, the numbers are ( 11 x 1, 11 x 35) and (11 x 5, 11 x 7)
Since one number lies 75 and 125, the suitable pair is (55,77)
Hence , required number = 77.
We first expand (2n + 2)2
(2n + 2)^2 = 4n^2 + 8n + 4
Factor 4 out.
= 4(n^2 + 2n + 1)
(2n + 2)2 is divisible by 4 and the remainder is equal to 0.