@vbhvgupta said:Q 25
log 7623 = 3.8821
=> log 7623 - log10^5 = 3.8821 - 5
=> log 7623/10^5 = -0.1179
=> log 0.07623 = -0.1179
Hecne (c)
=> log 7623 - log10^5 = 3.8821 - 5
=> log 7623/10^5 = -0.1179
=> log 0.07623 = -0.1179
Hecne (c)
@simplydevesh said:@ScareCrow28@shattereddream said:How many seven-digit numbers divisible by 11 have the sum of their digits equal to 59?1)40 2)55 3)34C3 €” 23C2 4)38C3 €” 26C2 5)None of theseLet the no be, (9-a)(9-b)(9-c)...(9-g)Where the terms (9-a)...represent the digitsNow, 9-a + 9-b ....9-c = 59=> a+b+c+d+e+f+g = 4Also, 9-a+9-c+9-e+9-g-9+b-9+d-9+f = 11,22,33=> (b+d+f) - (a+c+e+g) = 2, 13, ...=> 2(b+d+f) = 6 OR (b+d+f) = 3b+d+f = 3Total = 5C2 = 10 waysAnd, a+c+e+g = 1Total = 4C3 = 4 waysTotal = 10*4 = 40 wayscan u plz explain why did u take no.s as (9-a)(9-b)(9-c)...(9-g)and "2(b+d+f) = 6 OR (b+d+f) = 3 "and" a+c+e+g = 1"
I took them as (9-a)(9-b) just to make it easier for myself...
Otherwise, equation would have been a+b+c+d+e+f+g = 59 where a>0 and all the digits
I can't solve by this method...
Now for the second query:
(b+d+f) - (a+c+e+g) = 2 AND a+b+c+d+e+f+g = 4
Replace (a+c+e+g) by 4 - (b+d+f)
And you will get the same...
(b+d+f) = 3
=> (a+c+e+g) = 1
I hope you understand it now!
Q27 and 28
@vbhvgupta said:Q27 and 28
say P=15k,D=5K and K=3k
Now,
(27) deisel amount = 510/5k
So kerosene = (510/5k) * 3k = 306
Petrol = (510/5k) * 15k =1530
Total = 2346
(28) Now if 2346 is used to take equal vol of petrol abd kerosene then each volume = 2346/18
And petrol cost = 15*2346/18 = 1955
Now,
(27) deisel amount = 510/5k
So kerosene = (510/5k) * 3k = 306
Petrol = (510/5k) * 15k =1530
Total = 2346
(28) Now if 2346 is used to take equal vol of petrol abd kerosene then each volume = 2346/18
And petrol cost = 15*2346/18 = 1955
A varies directly as the sum of the two quantities B and C. B in turn varies directly as x and C varies inveresly as x. when x=1 or 2 , A = 3. Find the value of A when x = 4.
3? 
@vbhvgupta said:A varies directly as the sum of the two quantities B and C. B in turn varies directly as x and C varies inveresly as x. when x=1 or 2 , A = 3. Find the value of A when x = 4.
@vbhvgupta said:A varies directly as the sum of the two quantities B and C. B in turn varies directly as x and C varies inveresly as x. when x=1 or 2 , A = 3. Find the value of A when x = 4.
51/10 or 51/8 ?
@vbhvgupta said:4.5
Can u share the solution?
=====
Q: I left the room at a time between 4 and 5PM and returned between 7 and 8PM and noticed that the two hands of the clock had just interchanged their positions. What is the time when I returned?
OA not available
https://www.facebook.com/pages/Quantexpert/123213681084921
=====
Q: I left the room at a time between 4 and 5PM and returned between 7 and 8PM and noticed that the two hands of the clock had just interchanged their positions. What is the time when I returned?
OA not available
https://www.facebook.com/pages/Quantexpert/123213681084921
a garrison of 900 soldiers had food stock sufficient for 30 days when the rate of consumption is 2.5 kg/day/soldier. After some days of the consumption at that rate 300 soldiers were transfered to another garrison and the balance food lasted for 25 days for the remaining soldiers. If the rate of consumption of the remaining soldiers was 3 kg/day/soldier, after how many days from the start were the soldiers transferred?
@vbhvgupta said:a garrison of 900 soldiers had food stock sufficient for 30 days when the rate of consumption is 2.5 kg/day/soldier. After some days of the consumption at that rate 300 soldiers were transfered to another garrison and the balance food lasted for 25 days for the remaining soldiers. If the rate of consumption of the remaining soldiers was 3 kg/day/soldier, after how many days from the start were the soldiers transferred?
Total food=900*30*2.5=67500
2.5*900*x+3*600*25=67500
x=10 days
@catahead said:Can u share the solution?=====Q: I left the room at a time between 4 and 5PM and returned between 7 and 8PM and noticed that the two hands of the clock had just interchanged their positions. What is the time when I returned?OA not available https://www.facebook.com/pages/Quantexpert/123213681084921
For the first one:
A = k(B + C)
B = K1*x ; C = k2*x
Now,B + C = k1*x + k2/x
=> A = k1*k*x + k*k2/x
Solve for k1*k and k*k2 by substituting both the values of x
So, A has to be 4.5
@vbhvgupta said:A varies directly as the sum of the two quantities B and C. B in turn varies directly as x and C varies inveresly as x. when x=1 or 2 , A = 3. Find the value of A when x = 4.
A = k(B + C)
B = k1x
C = k2/x
A = kk1x + kk2/x
x = 1 and 2
kk1 + kk1 = 3....i
4kk1 + kk2 = 6....ii
solving i and ii
kk1 = 1
kk2 = 2
A = 4kk1 + kk2/4 = 4 + 2/4 = 4.5 ?
@vbhvgupta said:a garrison of 900 soldiers had food stock sufficient for 30 days when the rate of consumption is 2.5 kg/day/soldier. After some days of the consumption at that rate 300 soldiers were transfered to another garrison and the balance food lasted for 25 days for the remaining soldiers. If the rate of consumption of the remaining soldiers was 3 kg/day/soldier, after how many days from the start were the soldiers transferred?
I think for this one,900 * 30 * 2.5 = (2.5 * x * 900) + (3 * 25 * 600)
So,they would be transfer after x days
@vbhvgupta said:a garrison of 900 soldiers had food stock sufficient for 30 days when the rate of consumption is 2.5 kg/day/soldier. After some days of the consumption at that rate 300 soldiers were transfered to another garrison and the balance food lasted for 25 days for the remaining soldiers. If the rate of consumption of the remaining soldiers was 3 kg/day/soldier, after how many days from the start were the soldiers transferred?
Total food=2.5 * 900 * 30=67500
Let soldiers leave after x days
Food consumption for x days=2.5 * 900 * x =2250x
After soldiers leave rest 600 soldiers will consume = 3*25*600=45000
Acc to ques,
67500=45000+2250x
x=10 days
@vbhvgupta said:a garrison of 900 soldiers had food stock sufficient for 30 days when the rate of consumption is 2.5 kg/day/soldier. After some days of the consumption at that rate 300 soldiers were transfered to another garrison and the balance food lasted for 25 days for the remaining soldiers. If the rate of consumption of the remaining soldiers was 3 kg/day/soldier, after how many days from the start were the soldiers transferred?
totl food=900x2.5x30
consumed by 600 soldiers fr 25 days=600x2.5x25
consumed by 900 soldiers=900x2.5x30-600x3x25=75x300
no f days consumed by 900 soldiers=75x300/900x2.5=75x300/75x30=10 days....
:drinking: 
:drinking:
:drinking:Bacardi+Smirnoff+Appy fizz= some start to weekend
can any one explain euler's theorem in detail manner??