77
B. 70%
5c2*4c1/9c3 +5c3/9c3=25/42
15/46=0.32
The numbers that lie between 100 and 1000 which are divisible by 14 are 112, 126,140 …,994
a = 112; l = 994, d = 14
n= (l−a)/d+1
= (994-112)/14+1
= 64
Sn=n/2(l+a)
= 64/2(994+112)
= 32*1106
= 35392
20 litres
cos
140 280 320 liters oil are exactly fittable in 20 litred tin
190, 166, 145, 128, 112, 100, 91
128 is incorrect. There is a difference of multiples of 3 in the number pattern, which starts at 24 and reduces by 3 each time. Therefore the correct pattern would read 190, 166, 145, 127, 112, 100, 91.
this would leave the pattern with differences in the pattern 24, 21, 18, 15, 12, 9. The next number in the pattern would be 6 lower than 91. 91 – 6 = 85. Then 85 -3 – 82, followed by 82 – 0 = 82. After this, if -3 were taken the pattern would start to rise so 82 – – 3 = 85.
Answer: 450
Explanation:
10 * 150 = 1500
3 * 100 + 5 * 150 = 1050
1500 – 1050 = 450
Hint: Assume that the speed of the stream is x and the speed of the boat in still water is x. From the statement of the question form two equations in two variables x and y. This system is reducible to linear equations in two variables. Reduce the system to a system of linear equations in two variables by proper substitutions. Solve the system of equations using any one of the methods like Substitution method, elimination method, graphical method or using matrices. Hence find the value of x satisfying both the equation. The value of x will be the speed of the stream.
Complete step-by-step answer:
Let the speed of the stream be x, and the speed of the boat in still water be y.
We have the speed of the boat upstream = y-x.
Speed of the boat downstream = y+ x.
Now since it takes 14 hours to reach a place at a distance of 48 km and come back, we have the sum of the times taken to reach the place downstream and time taken to return back upstream is equal to 14.
Now, we know that time =Distance speed
Using, we get
Time taken to reach the place =48y+x and the time taken to return back =48y−x.
Hence, we have
48y+x+48y−x=14
Dividing both sides by 2, we get
24y+x+24y−x=7 —–(i)
Also, the time taken to cover 4km downstream is equal to the time taken to cover 3km upstream.
Hence, we have 4y+x=3y−x
Transposing the term on RHS to LHS, we get
4y+x−3y−x=0 ——– (ii)
Put 1y+x=t and 1y−x=u, we have
24t+24u=7 ——-(iii)4t−3u=0 ——–(iv)
Multiplying equation (iv) by 6 and subtracting from equation (iii), we get
24t−24t+24u+18u=7⇒42u=7
Dividing both sides by 42, we get
u=742=16
Substituting the value of u in equation (iv), we get
4t−3(16)=0⇒4t−12=0
Adding 12 on both sides, we get
4t=12
Dividing both sides by 4, we get
t=18
Reverting to original variables, we have
1y+x=18 and 1y−x=16
Taking reciprocals on both sides in both equations, we have
y+ x=8 ——- (v)y−x=6 ——–(vi)
Adding equation (v) and equation (vi), we get
2y=14
Dividing both sides by 2, we get
y=7.
Substituting the value of y in equation (v), we get
7+x=8
Subtracting 7 from both sides we get
x = 8-7 =1
Hence the speed of the stream is 1 km/hr.
15km/hr
1 1 2+4
2 2 1+4
3 1+2 4
4 4 1+2
5 4+1 2
6 4+2 1
7 4+2+1 0