Answer is 45
First we need to subtract those reminders from the respective numbers, then we have to find the hcf of two numbers(numbers got from the subtraction) then you will get the answer.
So,
After subtraction you will get
3026-11 = 3015
5053-13 = 5040
HCF of these two numbers
5 | 3015 5040
3 | 603 1008
3 | 201 336
| 67 112
We can’t find a common diviser since 67 is a prime number
So the HCF = 5 * 3 * 3
= 45
5*3*3 = 45
Speed = (45×518)m/sec=252m/sec
Total distance covered = (360+140) m = 500 m.
∴Required time= 500×225sec = 40 sec.
Let n be the number of days it takes A and B, working together, to finish. And we know B=A+10 and B=3A, so:
3A=A+10
2A=10
A=5
Then B=15
So:
1/A + 1/B=1/n where n is the total amount of days. So:
1/5 + 1/15=1/n
3n+n=15
n=15/4 days
Profit one with 3 & 2 nd 2.4
5
Summation of 301 – summation of 99
=41550
Summation of n=((n*n+1)/2)
Let the journey distance be : x miles
Now x/40+x/30=8 (time=distance/speed)
Solving this for x we get. x=960/7
therefore, x=137(approx)
(137.14 exact)
Question is not completed
The letters A, B, C, D, E, F and G, not necessarily in that order, stand for seven consecutive integers from 1 to 10
D is 3 less than A
B is the middle term
F is as much less than B as C is greater than D
G is greater than F
1. The fifth integer is
(a) A
(b) C
(c) D
(d) E
(e) F
ans:a
2.A is as much greater than F as which integer is
less than G
(a) A
(b) B
(c) C
(d) D
(e) E
ans:a
3. If A = 7, the sum of E and G is
(a) 8
(b) 10
(c) 12
(d) 14
(e) 16
4. An integer T is as much greater than C as C is
greater than E. T
can be written as A + E. What is D?
(a) 2
(b) 3
(c) 4
(d) 5
(e) Cannot be determined
ans:a
( e ) Apple
when know that 13 is factor of 39 so it will also not be able to divide it differently. so the remainder will be same 20 but as we know that 20 is greater then 13 so we can divide it once again and in the casE of 13 the remainder will be 20-13=7
Answer: 3121 gold coins
Let total no of coins be M
Let the disbursement D to each son:
D1 = 1 + (M – 1)/5 = (M + 4)/5
D2 = 1 + ( M – D1 -1)/5 = (D1) * 4/5
D3= (D2) * 4/5
D4= (D3) * 4/5
D5= (D4) * 4/5
Total disbursements to sons=
= ∑D= (M+4)*1/5[ 1+4/5+(4/5)(4/5)+ (4/5)(4/5)(4/5)+(4/5)(4/5)(4/5)(4/5) ]
= (2101/3125)*(M+4)
Thus balance left for daughters =M-{(2101/3125)*(M+4)}
=(1024M-8404)/3125
This balance should be a positive integer ( assuming M and all disbursements are full coins )
Thus 1024M-8404 should be a multiple of 3125….so….
1024M – 8404 = N*3125 where N is an integer
Using Python code:
n=int(input(“Enter num n: “))
X=int()
a=int()
a=0
X=’ ‘
for a in range(0,n+1):
a=a+1
X= (3125*a + 8404)/1024
if (3125*a + 8404)% 1024== 0:
print(X,a)
Enter num n: 10000
3121.0 1020
6246.0 2044
9371.0 3068
12496.0 4092
15621.0 5116
18746.0 6140
21871.0 7164
24996.0 8188
28121.0 9212
We get minimum value of N = 1021 and M = 3121 gold coins
Is it not 24?
1^1,2^2,3^3,4:^4,5^5,6^6
1,4,27,256,3125,46656