4, 5, 15, 49, 201, 1011, 6073
Correct option is A)
2nd term =(1stterm×1+2)=(4×1+2)=6
3rd term =(2ndterm×2+3)=(6×2+3)=15
4th term =(3rdterm×3+4)=(15×3+4)=49
5th term =(4thterm×4+5)=(49×4+5)=201 and so on
∴5 is wrong.
20
x (A to B by car) + y (B to A by cycle) is 7 hours
x (A to B by car) + x (B to A by car) is 7 minus 3 is 4 hours
2x is 4 and hence x is 2
2 + y is 7 and hence y is 5
y (A to B by cycle) + y (B to A by cycle) is 5 + 5 is 10 hours
these r all factors of 30
1*30=30
2*15=30
3*10=30
5*6=30
like that
sorry the above ans was typed by mistake the correct ans is…..
1027.05 ? 314.005 + 112.25 = ?
ans:
1027.05 – 314.005 + 112.25 = 825.295
Let cost for apple be a Cost for banana be b and Orange be c
So by first value expression becomes 17a + 13b + 9c =130 ———-1 therefore if you further solve a = (130 – 13b – 9c)/17 ———- 2 the second expression becomes: 13c + 7a + 10b = 100 ———- 3 If you put value of a in second expression it becomes: 13c + 7[(130 – 3b – 9c)/17] +10b = 100
Further if you solve you get value of b:
b = 10 – 2c ———-4
put value of b from 4 in 1
17a + 13 [10 – 2c] + 9c = 130
Further if you solve you find value of a
a = c ———-5
Put 5 in 3
13c + 7c + 10b = 100
further solve you get: c = 1 ———-6
from 5 and 6
a = c = 1 ———-7
Substitute value of c in expression 4
b = 10 – 2c b = 10 -2 * 1 b = 8 ———-8
therefore a + b + c = 10
One day’s work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10.
C leaves 4 days before completion of the work, which means only A and B work during the last 4 days.
Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10.
Remaining Work = 7/10, which was done by A,B and C in the initial number of days.
Number of days required for this initial work = 7 days.
Thus, the total numbers of days required = 4 + 7 = 11 days.
Answer: 8:10, 7:10
Explanation:
The bus b1, which started at P, reached S at 10:40, passing through the intermediary cities Q and R.
The time taken to travel from P to S
= 3 * 40 + 2 * 15 = 150minute
(journey)+(stoppage) = 2 hrs 30 minutes.
Hence, b1 started at 10:40 – 2:30 = 8:10 at P.
b2 reached Q, starting at U, through the city T, S and R.
The time taken by it to reach S = 4 * 40 + 3 * 15 = 205 minutes = 3 hr 25 minutes.
Hence, b2 started at, 10:35 – 3:25 = 7:10, at U.
1)Port
2)Trop
A die is thrown twice n(s)=36
The probability of getting same number is (1,1)(2,2)(3,3)
(4,4)(5,5)(6,6) i.e n(A)=6
Ans. P(A)=n(A)/n(s)
= 6/36
=1/6.
So, The Answer is 1/6
15