Speed = 44 kmph x 5/18 = 110/9 m/s
We know that, Time = distance/speed
Time = (360 + 140) / (110/9)
= 500 x 9/110 = 41 sec.
N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4.
Assume both trains meet after 'p' hours after 7 a.m.
Distance covered by train starting from A in 'p' hours = 20p km
Distance covered by train starting from B in (p-1) hours = 25(p-1)
Total distance = 200
=> 20x + 25(x-1) = 200
=> 45x = 225
=> p= 5
Me, they meet after 5 hours after 7 am, ie, they meet at 12 p.m.
Cost Price = Rs. 1950
Selling Price = Rs. 2340
Profit = S.P – C.P
Profit = Rs. 2340 – 1950 = 390
Profit % = (Profit/C.P) x 100
Profit % = (390/1950) x 100
Profit % = 20 %.
Let the value of one rupee, 50 paise and 25 paise be 11x, 9x, 5x respectively.
No. of 1 rupee coins = (11x / 1) =11x
No. of 50 paise coins = (9x / 0.5) = 18x
No. of 25 paise coins = (5x / 0.25) = 20x
11x + 18x + 9x = 342
38x = 342
x = 9
Therefore, no. of 1 rupee coins = 11 x 9 = 99 coins
No. of 50 paise coins = 18 x 9 = 162 coins
No. of 25 paise coins = 20 x 9 = 180 coins.
Let the two-digit number be 10x + y
10x + y = 7(x + y)
? x = 2y ...(i)
10(x +2 ) + (y + 2) = 6(x + y + 4) + 4
or 10x + y + 22 = 6x + 6y + 28
? 4x - 5y = 6 ...(ii)
Solving equations (i) and (ii)
We get x = 4 and y = 2.
Let 1/2 of the no. = 10x + y
and the no. = 10v + w
From the given conditions,
w= x and v = y-1
Thus the no. = 10 (y-1) + x
? 2(10x + y ) = 10 (y-1) + x
? 8y - 19x = 10 ...(i)
v + w = 7
? y-1 + x = 7
? x + y = 8
Solving equations (i) and (ii) , we get
x = 2 and y = 6.
Amount
= Rs.(25000x(1+12/100)³
= Rs.(25000x28/25x28/25x28/25)
= Rs. 35123.20.
C.I = Rs(35123.20 -25000)
= Rs.10123.20.
Total running distance in four weeks = (24 x 240) + (4 x 400)
= 5760 + 1600
= 7360 meters
= 7360/1000
=> 7.36 kms.
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.
P(E) =n(E)/n(S)=7/8.
Given that SP = Rs. 12000 - 10% = Rs. 10,800
Loss% = 4
We know that, C.P = 100/(100 - Loss%) x 100
=> 100/100-4 x 10800
=> 1080000/96
C.P = Rs. 11,250.
Suppose they meet after 'h' hours
Then
3h + 4h = 17.5
7h = 17.5
h = 2.5 hours
So they meet at => 4 + 2.5 = 6:30 pm.
Let the distance traveled be x km.
Then, x/10 - x/15 = 2
3x - 2x = 60 => x = 60 km.
Time taken to travel 60 km at 10 km/hr = 60/10 = 6 hrs.
So, Robert started 6 hours before @p.m. i.e., at 8 a.m.
Required speed = 60/5 = 12 kmph.
Let the cost price of a ball is Rs.x
Given, on selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls
The equation is :
17x - 720 = 5x
Solving the equation
we get x = 60
Therefore, cost price of a ball is Rs. 60.
Walking at 3/4th of usual rate implies that time taken would be 4/3th of the usual time. In other words, the time taken is 1/3rd more than his usual time
so 1/3rd of the usual time = 15min
or usual time = 3 x 15 = 45min = 45/60 hrs = 3/4 hrs.
Here S= {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}.
Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}.
P(E) = n(E) / n(S)
= 4/8= 1/2.
Let the numbers be a and b.
We know that product of two numbers = Product of their HCF and LCM
Then, a + b = 55 and ab = 5 x 120 = 600.
=> The required sum = (1/a) + (1/b) = (a+b)/ab
=55/600 = 11/120.
Let the numbers be 3x, 4x, 5x.
Then, their L.C.M = 60x.
So, 60x=3600 or x=60.
Therefore, The numbers are (3 x 60), (4 x 60), (5 x 60).
Hence,required H.C.F=60.
The CP of profitable cow = 9900/1.1 = 9000
and profit = Rs. 900
The CP of loss yielding cow = 9900/0.8 = 12375
and loss = Rs. 2475
so, the net loss = 2475 - 900 = 1575.
C.I.= Rs.[4000*(1+10/100)^2-4000]
=Rs.840
sum=Rs.(420 * 100)/3*8=Rs.1750.
The number of liters in each can = HCF of 80, 144 and 368 = 16 liters.
Number of c of Maaza = 368/16 = 23
Number of c of Pepsi = 80/16 = 5
Number of c of Sprite = 144/16 = 9
The total number of c required = 23 + 5 + 9 = 37 c.
S.I. = Rs.(1200*10*1)/100=rs.120
C.I. =rs[1200*(1+5/100)2-1200]=rs.123
Difference = Rs.(123-120) =Rs.3
Amount of Rs. 100 for 1 year
when compounded half-yearly = Rs.[100*(1+3/100)^2]=Rs.106.09
Effective rate=(106.09-100)%=6.09%.
Initial speed = 80km/hr
Total distance = 80 x 10 = 800km
New speed = 800/4 =200km/hr
Increase in speed = 200 - 80 = 120km/hr.
Bag consists of 25 paise, 50 paise and 1 rupee (100 paise) so the ratio becomes 25 : 50 : 100 or 1 : 2 : 4
Total value of 25 paise coins =(1 / 7 ) x 105 = 15
Total value of 50 paise coins = (2 / 7) x 105 = 30
Total value of 100 paise coins = (4 / 7) x 105 = 60
No. of 25 paise coins = 15 x 4 = 60 coins
No. of 50 paise coins = 30 x 2 = 60 coins
No. of 1 rupee coins = 60 x 1 = 60 coins.
P(first letter is not vowel) = 2/4
P(second letter is not vowel) = 1/3
So, probability that none of letters would be vowels is = 2/4×1/3=1/6.
Let S be the sample space.
Then, n(S) = 52C2=(52 x 51)/(2 x 1) = 1326.
Let E = event of getting 1 spade and 1 heart.
n(E)= number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = 13C1*13C1 = 169.
P(E) = n(E)/n(S) = 169/1326 = 13/102.
(x - 4) / 6 = 9
Multiply both sides by 6:
x - 4 = 54
Add 4 to both sides:
x = 58
(58 - 3) / 5 = 55 / 5 = 11.
Let 1kg of Rs. 100 then 840gm is of Rs. 84.
Now (label on can 1kg but contains 840kg ) so for customer it is of Rs. 100 and further gives 4% discount [he sells his article on 4% loss on cost price.]
So now S.P = Rs. 96
But actually it contains 840 gm so C.P for shopkeeper = Rs. 84
S.P = Rs. 96
C.P = Rs. 84
Profit% = {(S.P-C.P)/C.P}x100
{(96-84)/84} x 100 = 14.28571429% PROFIT.
Let the actual distance travelled be x km.
Then x/8=(x+20)/12
=> 12x = 8x + 160
=> 4x = 160
=> x = 40 km.
Divisor = 2/3 x dividend
and Divisor = 2 x remainder
or 2/3 x dividend = 2 x 5
Dividend = 2 x 5 x 3 / 2 = 15.
To find the minimum distance, we have to get the LCM of 75, 80, 85
Now, LCM of 75, 80, 85 = 5 x 15 x 16 x 17 = 20400
Hence, the minimum distance each should walk so that thay can cover the distance in complete steps = 20400 cms = 20400/100 = 204 mts.
Clearly, Rate = 5% p.a .,
Time = 3 years
S.I =Rs.1200.
So,Principal
=Rs.(100 x 1200/3x5)
=Rs.8000.
Amount
=Rs.[8000 x (1+5/100)³]
=Rs(8000x21/20x21/20x21/20)
= Rs.9261
C.I
=Rs.(9261-8000)
=Rs.1261.
Let the required no of working hours per day be x.
More pumps , Less working hours per day (Indirect Proportion)
Less days, More working hours per day (Indirect Proportion)
Pumps4 : 3Days1 : 2}?? 8:x
=> (4 * 1 * x) = (3 * 2 * 8)
=> x=12
The total number of elementary events associated to the random experiments of throwing four dice simultaneously is:
= 6*6*6*6=64
n(S) = 64
Let X be the event that all dice show the same face.
X = { (1,1,1,1,), (2,2,2,2), (3,3,3,3), (4,4,4,4), (5,5,5,5), (6,6,6,6)}
n(X) = 6
Hence required probability = n(X)n(S)=6/64=1216.
To find the LCM of 24, 36 and 40
24 = 2 x 2 x 2 x 3
36 = 2 x 2 x 3 x 3
40 = 2 x 2 x 2 x 5
Now, LCM of 24, 36 and 40 = 2 x 2 x 2 x 3 x 3 x 5
= 8 x 9 x 5
= 72 x 5
= 360.
Let C.P.= Rs. 100.
Then, Profit = Rs. 320,
S.P. = Rs. 420.
New C.P. = 125% of Rs. 100 = Rs. 125
New S.P. = Rs. 420.
Profit = Rs. (420 - 125) = Rs. 295
Required percentage = (295/420) * 100
= 70%(approx).
No. of letters in the word = 6
No. of 'E' repeated = 2
Total No. of arrangement = 6!/2! = 360.
Number of one digit pages from
1 to 9 = 9
Number of two digit pages from
10 to 99 = 90
Number of three digit pages from
100 to 200 = 101.
Income = Rs. 100
Expenditure = Rs. 80
Savings = Rs. 20
Present Expenditure 80x(15/100) = Rs. 12 = 80 + 12 = Rs. 92
Present Savings = 100 – 92 = Rs. 8
100 ------ 8
? --------- 400 => 5000
His salary = Rs. 5000.
Here n(S) = (6 x 6) = 36
Let E = event of getting a total more than 7
= {(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}
Therefore,P(E) = n(E)/n(S) = 15/36 = 5/12.
Principal = Rs.16,000;
Time=9 months = 3 quarters;
Amount
=Rs.[16000x(1+5/100)³] =[16000x21/20x21/20x21/20]
= Rs.18522.
C.I
= Rs.(18522 - 16000)
= Rs.2522.
Given,
Compound rate, R = 10% per annum
Time = 2 years
C.I = Rs. 420
Let P be the required principal.
A = (P+C.I)
Amount, A = P(1 + (r/100))n
(P+C.I) = P[1 + (10/100)]2
(P+420) = P[11/10][11/10]
P-1.21P = -420
0.21P = 420
Hence, P = 420/0.21 = Rs. 2000.
xy = 96050 ...(i)
and xz = 95625 ...(ii)
and y - z = 1 ... (iii)
Dividing (i) by (ii) we get
y/z = 96050 / 95625
= 3842 / 3825
= 226 / 225 ... (iv)
Combining (iii) and (iv) we get z = 225.
99 = 1 x 3 x 3 x 11
101= 1 x 101
176= 1 x 2x 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 1,101
divisors of 176 are 1, 2, 4, 8, 16, 22, 44, 88, 176
divisors of 182 are 1, 2, 7, 13, 14, 26, 91, 182
Hence , 176 hasthe most number of divisors.
Using pythagarous theorem,
distance travelled by first train = 36x5/18x30 = 300m
distance travelled by second train = 48x5/18x30 = 400m
so distance between them =v( 90000 + 160000) = v250000 = 500mts.
Taking the 2 investments to be 3x and 5x respectively
Total income of Raghu = (3x) x 1.25 + (5x) x 0.9 = 8.25
Therefore, Gain% = 0.25/8 x 100 = 3.125 %.