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New answer posted
9 months agoContributor-Level 10
Let A and B work for m days and C for n days to complete the work. Therefore,
. (1)
Out of the total of Rs. 18000, B gets Rs. 6000 more than C.
i.e., . (2)
On adding Eqs. (1) and (2), we get
New answer posted
9 months agoContributor-Level 10
Let f (x) = px2 + qx + k, where p, q and k are integers, and p ≠ 0
f (0) = k = 1
f (x) = px2 + qx + 1
= px2 +qx + k (Differentiate both sides with respect to x)
f' (x) = 2px + q
For maxima or minima f' (x) = 0, x =
f (x) attains maximum at x = 1
q = − 2p
f (1) = p + q + 1 = 3
1 − p = 3
p = − 2
q = 4
f (x) = −2x2 + 4x + 1
f (10) = − 200 + 40 + 1 = −159
New answer posted
9 months agoContributor-Level 10
f (x) = x – x2 + 1
g (x) = x2 + b + 3
f (2) g (1) < 0
(a – 4 + 1) (1 + 6 + 3) < 0
(a – 3) (b + 4) < 0
New answer posted
9 months agoContributor-Level 10
100! has 24 zeroes.
100! + 200! = 100! [1 + 101 * 102 * … * 200]
Which will again give 24 zeroes at the end.
New answer posted
9 months agoContributor-Level 10
Let 2P = 3Q = 4R = 5S = K
Then,
K=1200
Therefore. R = 300, S = 240
Difference=300 - 240 = 60
New answer posted
9 months agoContributor-Level 10
400 students * 80 days = 240 students * d days
d =
The food was to last 80 more days but now it is lasting 53.33 days more.
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