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Given - A= 5040, r₁=25%, r₂=40%, r₃=50% P=? So now - A=P[1+r₁/100] [1+r₂/100] [1+r₃/100] 5040=P [1+25/100] [1+40/100] [1+50/100] 5040=P [125/100] [140/100]]150/100] P = [5040×1000000]/125×140×150 =5040×80/14×15 =403200/210=40320/21 P=1920
3/7 of 3499.86 – 44.94% of 2000.19 + 2/7 of 2100.18 = ?
(22.03 + 89.98) ÷ 14.211 = 89.9 – 25.23% of ?
? = 54.89 × 270.08 ÷ 135.17 + 464.35 ÷ 29.03
15.232 + 19.98% of 649.99 = ? × 4.99
(7.013 – 20.04) = ? + 7.98% of 5399.98
24.96% of 380 + ? – 169.99 = 149.99% of 80
Find the approximate value of Question mark(?). No need to find the exact value.
18.07 × (47.998 ÷ 12.03) + 59.78% of 150.14 – √(255.86) = ...
447.79 ÷ √(√2400) + 30.94 × 6.07 – 5.08 × 21.96 = ?