Among 100 students, x1have birthdays in January, x2 have birthday in February, and so on. If x0 = max (x1, x2,……., x12), then the smallest possible value of x0is
We want the least value of the maximum function, which is possible when the values inside the brackets are as close as possible. As number of students must be integer, therefore values of x1, x2……………..x12 can be 8 or 9 (i.e 8,8,8,8,8,8,8,8,9,9,9,9). Therefore max value of x0 is 9.
A gentleman decided to treat a few children in the following manner. He gives half of his total stock of toffees and one extra to the first child, and then the half of the remaining stock along with one extra to the second and continues giving away in this fashion. His total stock exhausts after he takes care of 5 children. How many toffees were there in his stock initially?
Let the number of toffees be x. Toffees Given to first child = (x/2)+1 Toffees given to second child= (1/2) (x- (x/2) -1) + 1 = (x/4)+(1/2) We find the symmetry in the pattern of toffees distribution, Therefore toffees distribution done [(x/2)+1] ,[(x/4)+(1/2)] ,[(x/8)+(1/4)] ,[(x/16)+(1/8)] ,[(x/32)+(1/16)] ⇒ [(x/2)+1] + [(x/4)+(1/2)] + [(x/8)+(1/4)] + [(x/16)+(1/8)] + [(x/32)+(1/16)] = x ⇒ (62/32) = x – (31/32)x ⇒ x=62
Veeru invested Rs 10000 at 5% simple annual interest, and exactly after two years, Joy invested Rs 8000 at 10% simple annual interest. How many years after Veeru’s investment, will their balances, i.e., principal plus accumulated interest, be equal?
Let x be the number of year after which veeru amount will be equal to Joy amount. So, (principle + Interest) for Veeru after x year = 10,000 + 10,000and (principle + Interest) for Joy after (x – 2) years = 8000 + 8000 Accroding to given condtion, ⇒ 10000 + 500x = 800 + 800x – 1600 ⇒ x = 12
Leaving home at the same time, Amal reaches office at 10:15 am if he travels at 8 km/hr, and at 9:40 am if he travels at 15 km/hr. Leaving home at 9:10, at what speed, in km/hr, must be travel so as to reach office exactly at 10 am?
As, distance covered is same with both the speed, (where, t is the time taken in minutes by Amal when his speed is 8km/hr) This means, Amal started from his house at 9:00 AM and taken 75 minutes to reach office with the speed of 8 km/hr. Now, Amal starts at 9:10 Am and wanted to reach office at 10:00 AM, i.e. is 50 minutes.
A train travelled at one-thirds of its usual speed, and hence reached the destination 30 minutes after the scheduled time. On its return journey, the train initially travelled at its usual speed for 5 minutes but then stopped for 4 minutes for an emergency. The percentage by which the train must now increase its usual speed so as to reach the destination at the scheduled time, is nearest to
Let usual speed is x and time is t If speed becomes 1/3 time will become 3 times so, time taken is 3t Given that 3t –t = 30. So t is 15 min. On return journey, in 5 minutes, it will cover 1/3rd return journey, in 5 minutes, it will cover 1/3rd distance, Ti cover the remaining distance, it has 10 minutes at usual speed but as it stopped for 4 minutes, remaining time is 6 minutes. Ratio of normal time to new time is 5:3. Ratio of normal speed to new speed is 3:5. So speed increased by 2/3 or 66.77%
A solution, of volume 40 litres, has dye and water in the proportion 2 : 3. Water is added to the solution to change this proportion to 2 : 5. If one-fourths of this diluted solution is taken out, how many litres of dye must be added to the remaining solution to bring the proportion back to 2 : 3?
A straight road connects points A and B. Car 1 travels from A to B and Car 2 travels from B to A, both leaving at the same time. After meeting each other, they take 45 minutes and 20 minutes, respectively, to complete their journeys. If Car 1 travels at the speed of 60 km/hr, then the speed of Car 2, in km/hr, is
1. 100
2. 80
3. 90
4. 70
Two persons are walking beside a railway track at respective speeds of 2 and 4 km per hour in the same direction. A train came from behind them and crossed them in 90 and 100 seconds, respectively. The time, in seconds, taken by the train to cross an electric post is nearest to
If f(5 + x) = f(5- x) for every real x, and f(x) = 0 has four distinct real roots, then the sum of these roots is
f(5 + x) = f(5- x) x → x - 5 ⇒ f(5 + x - 5) = f(5- x + 5) ⇒ f (x) = f (10 - x) Given, f (x) = 0 ∴ Also, f (10 - x) = 0 Given that these are four distinct solutions ⇒ f (α) = 0, f (β) = 0 ⇒ f (10 - α) = 0, f (10 - β) = 0 Sum of these roots = α + β + 10 - α + 10 - β = 20
If a, b and c are positive integers such that ab = 432, bc = 96 and c < 9, then the smallest possible value of a + b + c is
Case I ⇒ when C = 8 ⇒ bc = 96 ⇒ b = 12 ∴ ab = 432 ⇒ a = 36 ⇒ a + b + c = 8 + 12 + 36 = 56 Case II ⇒ when C = 7 ⇒ bc = 96 ⇒ b will not be integer ∴ c = 7 not possible Case III ⇒ when C = 6 ⇒ bc = 96 ⇒ b = 16 ∴ ab = 432 ⇒ a = 432/16 = 27 ⇒ a + b + c = 49 Case IV ⇒ C = 5 (not possible) because b will not be integer Case V ⇒ C = 4 ∴ bc = 96 ⇒ b = 24 ∴ ab = 432 ⇒ a = 432/24 = 18 ∴ a + b + c = 46 No, need to check further, of 46 is the least option given
The area of the region satisfying the inequalities |x| – y ≤ 1, y ≥ 0 and y ≤ 1 is (in numerical value)
x| - y ≤ 1, y ≥ 0, y ≤ 1
If x > 0 ⇒ x – y = 1 ……… (1)
And x < 0 ⇒ x – y = 1
Or x + y = -1 ……………. (2)
Put x = 0 in (1), y = -1
Put y = 0 in (1) x = 1
Put y = 1 in (1), x = 2
Put x = 0 in (2), y = -1
Y = 0 in (2), x = -1
Y = 1 in (2), x = -2
Shaded area is trapezium = ½ [ 2 + 4] × 1 = 3.5 sq. units
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