Question
48 grams of an alloy 'A' containing only gold, silver and
copper in ratio 1:1:4 respectively is mixed with 30 grams of alloy 'B' which contains only silver and copper in ratio 1:2 respectively. Find the total ratio of silver to copper in the final mixture.Solution
Quantity of silver in 48 gram of alloy 'A' = 48 × (1/6) = 8 gram
Quantity of copper in 48 gram of alloy 'A' = 48 × (4/6) = 32 gram
Quantity of silver in 30 gram of alloy 'B' = 30 × (1/3) = 10 gram
Quantity of copper in 30 gram of alloy 'B' = 30 × (2/3) = 20 gram
Required ratio = (8 + 10) : (32 + 20) = 18 : 52 = 9 : 26
(0.125)³÷ (0.25)² x (0.5)² = (0.5)?-3
54 - 78 + 1518 + 1653 – 1252 = ?
In the question, two Quantities I and II are given. You have to solve both the Quantity to establish the correct relation between Quantity-I and Quantit...
22% of 280 + 34% of 1080 × 5/12 =? + 16% of 460
800 + 900 X (3)-2 - ? = 25 X 60 ÷ 2Â
What value should come in the place of (?) in the following questions?
√289 * 2 +3√729 * 3 + 24 ÷ 8 = ?2
Evaluate:
7.5 + 12 ÷ (4 × 1.5) − 2.4
√? = 120 - 102 + ∛125
Simplify: 48 ÷ 4 × 3 + 5 × (6 − 2)
√4761 ÷ 23 + √12769 = ? × 58
Relevant for Exams:
