Start learning 50% faster. Sign in now
Let the time be = t hrs , speed be= 48 km/hr , distance = 48× t Distance travelled by B forms an AP = 40, 40½, 41, 41½ , ………………………. t terms Where First term is 40 and Common difference is 1/2 Now, Sum of n terms = Sn = n/2 [2a + (n-1)d] = distance covered by A Sn = t/2 [2 × 40 + (t-1) 1/2] 48t= t/2 [2 × 40 + (t-1) 1/2] 96 = 80 + (t – 1) 1/2 16 = (t – 1) 1/2 33 = t Therefore A and B will meet after 33 hours
(2/?) x (3/16) x (2/15) x 60 = 1/3
84% of 8400 + 42% of 6120 =?
[192 ÷ 6 × 5] ÷ (? + 3) = 20
{(80% of 650 + 25 × 12) – 20 × ?} = 760
10 × 100 ÷ 5 + 9 = ?
72 × 2 = ? + 104 – 14
5/13 × 104 + 1(2/9) × 198 = 133 + ?
55.55% of 30000 – 1111 = ? × 1111
((67)32 × (67)-18/ ? = (67)⁸
Find the simplified value of the given expression:
8 of 7 ÷ 2 × 5² + √36 – 10