Question
Quantity-I: A mixture contains 50% milk and the rest
360 litres water. When (x −15) litres of water and (x+10) litres of milk are added to this mixture, the ratio of the quantity of milk to that of water in the resultant mixture becomes 5:6. Find the value of ‘x’. Quantity-II: The ratio of the quantity of sugar to that of water in a mixture (sugar + water) is 8:6, respectively. 35% of this mixture is replaced with 6 litres of sugar and 8 litres of water such that the ratio of the quantity of sugar to that of water in the resultant mixture becomes 7:5. If the measure of 25% of the initial quantity of water in the mixture is ‘y’ litres, then find the value of ‘y’. In the question, two Quantity I and II are given. You have to solve both the Quantity to establish the correct relation between Quantity-I and Quantity-II and choose the correct option.Solution
ATQ, Quantity I: Initial quantity of milk = 360×(0.50/0.50 )=360 litres Final quantity of milk = 360+x+10=(370+x) litres Final quantity of water = 360+x−15=(345+x) litres According to the question, (370+x)/(345+x) = 5/6 6(370+x)=5(345+x) 2220+6x=1725+5x x=160 So, Quantity I = 160 Quantity II: Let the initial quantity of sugar and water be 8x litres and 6x litres, respectively Final quantity of sugar = (0.65×8x)+6=(5.2x+6) litres Final quantity of water = (0.65×6x)+8=(3.9x+8) litres According to the question, (5.2x+6)/(3.9x+8) = 7/5 5(5.2x+6)=7(3.9x+8) 26x+30=27.3x+56 1.3x=26 x = 26/1.3 x = 20 Therefore, ‘y’ = 0.25 × 6x = x So, Quantity II = 20 So, Quantity I > Quantity II
If (x + 1/x) = 5, then value of x3 + 1/x3 is:
181/8 + 51/4 – 63/8 = ? + 9/2
40% are the passing marks. A student gets 250 marks yet fails by 38 marks. What is the maximum marks?
40.5 ÷ [4/5 of (32 + 18) - 29/2] = ? ÷ 102
20% of 10% of 900 + 84/12 = ?2
242 + 18 × 8 – ? = 356
84% of 800 + 70% of 640 = 14 × ?
[(120)2 ÷24 ×25] ÷ 250 =?
What will come in place of the question mark (?) in the following expression?
25% of 2000 – 50% of 500 = 100% of ?
Relevant for Exams:
