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New answer posted
a year agoContributor-Level 10
For every day's work, P1 can afford to miss 3 days.
Hence, to break even it has to be 7 days' work in 28 days.
New answer posted
a year agoContributor-Level 10
Let Ab = X km and speed of bus be Y km/hr and time be T hrs original equation becomes
By increasing speed by 7 km/hr equation becomes
By decreasing speed by 5 km/hr equation becomes
On solving equation 1, 2 & 3 we get the value of y as and
and
AB = X =
≈ 31 km
New answer posted
a year agoContributor-Level 10
A + 4 = B + 5 = C + 6 = D + 7 = P + B + C + D + 8
A + 4 = B + 5 A – 1 = B
A + 4 = C + 6 A – 2 = C
A + 4 = D + 7 A – 3 = D
Now, A + 4 = A + A – 1 + A – 2 + A – 3 + 8
A
So, A + B + C + D =
New answer posted
a year agoContributor-Level 10
Let the ages of Akshat, Rishab, Gaurav be X, Y, Z years respectively.
X = 2 Z . (1)
X + Z = 2 Y . (2)
(X + 6 + Y + 6) = 3 (Z + 6) &nb
New answer posted
a year agoContributor-Level 10
Total number of passenger is the Rajdhani Express = 10 * 20 = 200
So, in the 9 boggies the minimum number of total passengers = 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 = 144
Hence, the minimum number of passenger in one boggie can be (200? 144) = 56
New answer posted
a year 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
a year 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
a year 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
a year agoContributor-Level 10
100! has 24 zeroes.
100! + 200! = 100! [1 + 101 * 102 * … * 200]
Which will again give 24 zeroes at the end.
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