The sum of weights of A and B is 80 kg. 50% of A's weight is
(A) 20 kg
(B) 10 kg
(C) 25 kg
(D) 15 kg
Let'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.
In an examination, 92% of the students passed and 480 students failed. If so, how many students appeared in the examination?
(A) 5800
(B) 6200
(C) 6000
(D) 5000
Let'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.
In an examination, 34% of the students failed in mathematics and 42% failed in English. If 20% of the students failed in both subjects, then find the percentage of students who passed in both subjects.
(A) 40%
(B) 41%
(C) 43%
(D) 44%
To find the percentage of students who failed in at least one subject (A or B or both), we can use the principle of inclusion-exclusion:
Percentage of students who failed in at least one subject (A or B or both) = Percentage of students who failed in M + Percentage of students who failed in E - Percentage of students who failed in both subjects
= M + E - B
= 34% + 42% - 20% = 76% - 20% = 56%
So, 56% of the students failed in at least one subject.
Now, to find the percentage of students who passed in both subjects, we subtract the percentage of students who failed in at least one subject from 100%:
Percentage of students who passed in both subjects = 100% - Percentage of students who failed in at least one subject = 100% - 56% = 44%
Therefore, the percentage of students who passed in both subjects is 44%.
A is 80% more than B and C is
(A) 30
(B) 15
(C) 25
(D) 20
If (x+20)% of 250 is 25% more than x% of 220, then 10% of (x + 50) is what per cent less than 15% of x?
(A)
(B)
(C)
(D)
Akhil takes 30 minutes extra to cover a distance of 150 km if he drives 10 km/h slower than his usual speed. How much time will he take to drive 90 km if he drives 15 km per hours slower than his usual speed?
(A) 2h 45m
(B) 2h 30m
(C) 2h
(D) 2h 15m
Let's use the information given to calculate Akhil's usual speed first.
We know that Akhil takes 30 minutes (0.5 hours) extra to cover a distance of 150 km when he drives 10 km/h slower than his usual speed.
Let "S" be Akhil's usual speed in km/h. So, his slower speed would be (S - 10) km/h.
The time taken to cover a distance is equal to the distance divided by the speed: Time = Distance / Speed
At his usual speed, it takes him: Time at usual speed = 150 km / S hours
At the slower speed, it takes him: Time at slower speed = 150 km / (S - 10) hours
The difference in time between these two scenarios is 0.5 hours (30 minutes): Time at slower speed - Time at usual speed = 0.5 hours
Now, we can set up the equation and solve for S:
(150 km / (S - 10)) - (150 km / S) = 0.5
To solve this equation, we'll first get a common denominator: [150S - 150(S - 10)] / [S(S - 10)] = 0.5
Now, simplify and solve for S: [150S - 150S + 1500] / [S(S - 10)] = 0.5
1500 / [S(S - 10)] = 0.5
Now, cross-multiply: 2 * 1500 = S(S - 10)
3000 = S^2 - 10S
S^2 - 10S - 3000 = 0
Now, we can solve this quadratic equation for S using the quadratic formula:
S = [-(-10) ± √((-10)^2 - 4(1)(-3000))] / (2(1))
S = [10 ± √(100 + 12000)] / 2
S = [10 ± √12100] / 2
S = [10 ± 110] / 2
Now, we have two possible values for S, but we'll take the positive one because speed cannot be negative:
S = (10 + 110) / 2 = 120/2 = 60 km/h
So, Akhil's usual speed is 60 km/h.
Now, we want to find out how much time he will take to drive 90 km when he drives 15 km/h slower than his usual speed, which would be (60 - 15) = 45 km/h.
Time = Distance / Speed Time = 90 km / 45 km/h = 2 hours
Akhil will take 2 hours to drive 90 km when he drives 15 km/h slower than his usual speed.
Manjula is observing her image in a plane mirror. The distance between herself and the mirror is 5 m. She moves 1 m towards the mirror. The distance between herself and her image now is
(A) 1 m
(B) 4 m
(C) 8 m
(D) 10 m
A takes 2 hours more than B to cover a distance of 40 km. If A doubles his speed, he takes
(A)
(B)
(C)
(D)
A man rows at 5 km/h in still water. If the river is running at 1 km/h, it takes him 75 minutes to row to a place and back. How for is the place?
(A) 5 km
(B) 3 km
(C) 4 km
(D) 5 km
4 men can complete a piece of work in 2 days. 4 women can complete the same piece of work in 4 days whereas 5 children can complete the same piece of work in 4 days. If 2 men, 4 women and 10 children work together, in how many days can the work be completed?
(A) 1 day
(B) 3 days
(C) 2 days
(D) 4 days
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