The price of sugar increases by 15%. By what percentage should the consumption of sugar be decreased so that the expenditure on the purchase of sugar remains the same? [Give your answer correct to 2 decimal places.]
596 0643d140332185cce373eec85To solve this problem, let's denote:
After the price increases by 15%, the new price becomes 1.15P.
To keep the expenditure constant, the new quantity consumed (let's call it Q') can be calculated using the formula:
New expenditure = New price × New quantity
Setting the new expenditure equal to the initial expenditure:
PQ = (1.15P) * Q'
Now, solve for Q':
Q' = PQ / (1.15P)
Simplify:
Q' = Q / 1.15
Now, let's find the percentage decrease in consumption:
Percentage decrease = [(Q - Q') / Q] * 100
Substituting the value of Q':
Percentage decrease = [(Q - (Q / 1.15)) / Q] * 100
Percentage decrease = [(Q * (1 - 1/1.15)) / Q] * 100
Percentage decrease ≈ [(1 - 0.8696) * 100] ≈ 13.04%
Therefore, the consumption of sugar should be decreased by approximately 13.04% to keep the expenditure on the purchase of sugar the same after a 15% increase in price.
In an examination, 92% of the students passed and 480 students failed. If so, how many students appeared in the examination?
540 064cce0cd8f85ca71558f105fLet's denote the total number of students who appeared in the examination as 𝑥x.
Given that 92% of the students passed, it means 8% failed because the total percentage is 100%.
We can set up the equation:
8% of 𝑥=4808% of x=480To find 8% of 𝑥x, we multiply 𝑥x by 81001008 (which is the same as multiplying by 0.08):
0.08𝑥=4800.08x=480Now, we can solve for 𝑥x:
𝑥=4800.08x=0.08480𝑥=6000x=6000So, 6000 students appeared in the examination.
The sum of weights of A and B is 80 kg. 50% of A's weight is $${5\over6}$$ times the weight of B. Find the difference between their weights.
608 064ccef6ba919c8488e304799Let's denote the weight of A as 𝑥x kg and the weight of B as 𝑦y kg.
Given:
We can solve these two equations to find the values of 𝑥x and 𝑦y, and then calculate the difference between their weights.
From equation 2: 0.5𝑥=56𝑦0.5x=65y Multiply both sides by 2 to get rid of the fraction: 𝑥=56𝑦×2x=65y×2 𝑥=106𝑦x=610y 𝑥=53𝑦x=35y
Now substitute this expression for 𝑥x into equation 1: 53𝑦+𝑦=8035y+y=80 83𝑦=8038y=80 Multiply both sides by 3883: 𝑦=80×38y=80×83 𝑦=30y=30
Now that we have found the weight of B, we can find the weight of A using equation 1: 𝑥+30=80x+30=80 𝑥=80−30x=80−30 𝑥=50x=50
So, the weight of A is 50 kg and the weight of B is 30 kg.
Now, let's find the difference between their weights: Difference=Weight of A−Weight of BDifference=Weight of A−Weight of B Difference=50−30Difference=50−30 Difference=20Difference=20
Therefore, the difference between their weights is 20 kg.
If 40% of (A+B) = 60% of (A-B) then $${2A-3B}\over {A+B}$$ is
463 065424ed827398ecb26d95af50.15% of $$33{1\over 3}\%$$ of ₹ 10000 is :
334 0653f6eb2dfda58cb3c9373bfIf x% of $${25\over 2}\%$$ is 150, then the value of x is :
398 0653f7cb899dc28caf022bf23if 50% of (x-y) =30% of (x+y), then what percentage of x is y?
337 0653f7d3bc298bbcb027eedf6