7770-7077=693
30%
Total 55
Manisha is a girl name so 54 boys
1 girl
1. 1g
2. 3g with 1g counter
3. 3g
4. 3g plus 1g
5. 3g plus 1 g plus 1g weighed medicine
6. 9g with 3g counter
7. 9g with 1g of counter and 1g weighed medicine
8. 9g with 1g counter
9. 9g
10. 9g plus 1g
11. 9g plus 3g with 1g on counter
12. 9g plus 3g
13. All 3 weights on one side.
Answer:
11 days.
Step-by-step explanation:
In the question,
Time taken by Ramesh to finish a piece of work = 20 days
Time taken by Sushil to finish a work = 25 days
Time for which they worked together = 5 days
Sushil left after = 5 days
So,
One day work of Ramesh is,
One day work of Sushil is,
So,
Work done in 5 days is given by,
Therefore, Remaining work is given by,
Now, as the Sushil left the remaining work was done by Ramesh,
Time taken by Ramesh for the remaining work is,
Therefore, the remaining work will be completed in 11 days by Ramesh.
thanks
brother
The last person covered 120.71 meters.
It is given that the platoon and the last person moved with
uniform speed. Also, they both moved for the identical
amount of time. Hence, the ratio of the distance they
covered – while person moving forward and backword – are
equal.
Let’s assume that when the last person reached the first
person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X)
meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-
X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total
distance covered by the last person is 100 meters, as he
ran the total lenght of the platoon (50 meters) twice.
TRUE, but that’s the relative distance covered by the last
person i.e. assuming that the platoon is stationary.
8*1024+3*64+3
8192+192+3
8387
1000/0011/1000/0111
So ie 7. 1s here
Total number of pairs is NC2^{N}C_2NC2. Number of pairs standing next to each other = N. Therefore, number of pairs in question = NC2^{N}C_2NC2 – N = 28/2 = 14. If N = 7,
7C2 – 7 = 21 – 7 = 14….
N =7
Initially (2*33 + 1*24)/(2+1) = 30p a gm
So, cost for 100g = 100*30 = 3000p
Later, (1*33 + 2*24)/(1+2) = 71/3 p a gm.
So, cost for 100g = 100*(71/3) = 2366.67p
He saves 3000 – 2366.67 = 633.33p
Concept of weighted average.
3
225
8:20
c
A