A bus starts at 9 am and follows a fixed route every day. One day, it traveled at a constant speed of 60 km per hour and reached its destination 3.5 hours later than its scheduled arrival time. Next day, it traveled two-thirds of its route in one-third of its total scheduled travel time, and the remaining part of the route at 40 km per hour to reach just on time. The scheduled arrival time of the bus is
Let the usual time taken = t hours Given, distance, d = 60 × (t + 3.5) Next day, 2d/3 of the distance covered in t/3 of the time Remaining distance, d/3 = 40 × 2t/3 => d = 80t => 80t = 60t + 210 => t = 10.5 hours The usual scheduled arrival time = 9 AM + 10.5 hrs = 7:30 PM
When 3333 is divided by 11, the remainder is
Three circles of equal radii touch (but not cross) each other externally. Two other circles, X and Y, are drawn such that both touch (but not cross) each of the three previous circles. If the radius of X is more than that of Y, the ratio of the radii of X and Y is
The given shape can be drawn as follows.
Due to symmetry, the center of both the circles X and Y coincides at O (let)
Let the radius of e circles of equal radii = R
Let the radius of smaller circle Y = OA
And radius of larger circle X = OC = OA + AC = OA + 2R
Also, joining the center of 3 circles having radius R forms an equilateral triangle with each side = 2R
Now, OB is circum-radius of equilateral triangle thus formed
OB = OA + AB = OA + R = (2/√3) × R
=> OA = 2R/√3 – R = (2 - √3)R/√3
Also, OC = OA + 2R = (2 - √3)R/√3 + 2R = (2 + √3)R/√3
Required ratio = OC : OA = (2 + √3)R/√3 : (2 - √3)R/√3
= (2 + √3)2 : (2 - √3)(2 + √3)
= 7 + 4√3 : 1
ABCD is a trapezium in which AB is parallel to CD. The sides AD and BC when extended, intersect at point E. If AB = 2 cm, CD = 1 cm, and perimeter of ABCD is 6 cm, then the perimeter, in cm, of AEB is
Given perimeter of trapezium ABCD = 6
=> AB + BC + CD + DA = 6
Let BC = y and DA = x
=> 2 + y + 1 + x = 6
=> x + y = 3
Since, AB//CD and CD = 1/2 AB
=> D is mid-points of AE
and C is mid-point of BE
So, DE = x = AD
And CE = y = BC
Perimeter of AEB
= AB + BE + AE
= 2 + 2y + 2x
= 2 + 2(x + y)
= 8 units
A company has 40 employees whose names are listed in a certain order. In the year 2022, the average bonus of the first 30 employees was Rs. 40000, of the last 30 employees was Rs. 60000, and of the first 10 and last 10 employees together was Rs. 50000. Next year, the average bonus of the first 10 employees increased by 100%, of the last 10 employees increased by 200% and of the remaining employees was unchanged. Then, the average bonus, in rupees, of all the 40 employees together in the year 2023 was
In 2022, Let the sum of employees (1-10) = a Let the sum of employees (11-30) = b Let the sum of employees (31-40) = c Given a + b = 30 × 40000 = 12 lakhs and b + c = 30 × 60000 = 18 lakhs Also given, a + c = 20 × 50000 = 10 lakhs Adding all 3 equations, 2 (a + b + c) = 40 lakhs => a + b + c = 20 lakhs Solving, a = 2 lakhs, b = 10 lakhs and c = 8 lakhs Now, if average is increased by certain percentage, sum will also increased by the same percentage In 2023, Sum of employees (1-10) = 2 lakhs × 2 = 4 lakhs and sum of employees (31-40) = 8 lakhs × 3 = 24 lakhs Sum of employees (11-30) = 10 lakhs (remains unchanged) Total sum = 4 + 10 + 24 = 38 lakhs The new average of all employees in 2023 = 3800000/40 = 95000
Anil invests Rs 22000 for 6 years in a scheme with 4% interest per annum, compounded half-yearly. Separately, Sunil invests a certain amount in the same scheme for 5 years, and then reinvests the entire amount he receives at the end of 5 years, for one year at 10% simple interest. If the amounts received by both at the end of 6 years are equal, then the initial investment, in rupees, made by Sunil is
The coordinates of the three vertices of a triangle are: (1, 2), (7, 2), and (1, 10). Then the radius of the in circle of the triangle is
The given vertices are (1, 2), (7, 2) and (1, 10) The distance between (1, 2) and (7, 2) = 6 units The distance between (1, 2) and (1, 10) = 8 units The distance between (7, 2) and (1, 10) = 10 units So, the sides of the given triangle satisfies the Pythagoras Theorem, therefore forms a right angled triangle Area of the right angled triangle = 1/2 × 6 × 8 = 24 units Also, the area of the triangle = r × s where r is the in-radius and s is the semi-perimeter, s = (6+8+10)/2 = 12 Hence, r × 12 = 24 => r = 2 units
A fruit seller has a stock of mangoes, bananas and apples with at least one fruit of each type. At the beginning of a day, the number of mangoes make up 40%of his stock. That day, he sells half of the mangoes, 96 bananas and 40% of the apples. At the end of the day, he ends up selling 50% of the fruits. The smallest possible total number of fruits in the stock at the beginning of the day is
At the beginning, let the total number of fruits = 5x => Mangoes at the beginning = 40% of 5x = 2x Let the Apples at the beginning = 5a => Bananas at the beginning = 3x – 5a At the end of the day, Mangoes sold = 2x/2 = x, Bananas sold = 96 and Apples sold = 40% of 5a = 2a Given, x + 96 + 2a = 50% of 5x => 1.5x = 2a + 96 => 3x = 4a + 192 The smallest possible value of a = 3 (at least 1 fruit of each type) Solving, x = 68 (smallest) Hence, the smallest possible total number of fruits at the beginning = 5x = 340
Amal and Vimal together can complete a task in 150 days, while Vimal and Sunil together can complete the same task in 100 days. Amal starts working on the task and works for 75 days, then Vimal takes over and works for 135 days. Finally, Sunil takes over and completes the remaining task in 45 days. If Amal had started the task alone and worked on all days, Vimal had worked on every second day, and Sunil had worked on every third day, then the number of days required to complete the task would have been
Given, A + V = W/150 and V + S = W/100 Let work, W = 300 units A + V = 2 units/day and V + S = 3 units/day Also, 75A + 135V + 45S = 300 => 75A + 75V + 15V + 45V + 45S = 300 => 75 × 2 + 15V + 45 × 3 = 300 => V = 1 unit/day => A = 1 unit/day and S = 2 units/day Now A works every day, while V works on every 2nd day and S words on every 3rd day, that makes the cycle of 6 days Work done by A on 1st day = 1 unit Work done by A and V on 2nd day = 1 + 1 = 2 units Work done by A and S on 3rd day = 1 + 2 = 3 units Work done by A and V on 4th day = 1 + 1 = 2 units Work done by A on 5th day = 1 unit Work done by A, V and S on 6th day = 1 + 1 + 2 = 4 units Work completed in 6 days = 1 + 2 + 3 + 2 + 1 + 4 = 13 units After that the cycle will repeat Work done in (6 × 23 = 138 days) = 299 Next day, work done by A = 1 unit Hence, total number of days = 139
P, Q, R and S are four towns. One can travel between P and Q along 3 direct paths, between Q and S along 4 direct paths, and between P and R along 4 direct paths. There is no direct path between P and S, while there are few direct paths between Q and R, and between R and S. One can travel from P to S either via Q, or via R, or via Q followed by R, respectively, in exactly 62 possible ways. One can also travel from Q to R either directly, or via P, or via S, in exactly 27 possible ways. Then, the number of direct paths between Q and R is
Given, paths between P and Q = 3, Q and S = 4 and P and R = 4
Let the number of paths between Q and R = p and R and S = q
Given, paths between P to Q to S + paths between P to R to S + paths between P to Q to R to S
= 3 × 4 + 4 × q + 3 × p × q = 62
=> 4q + 3pq = 50
=> q (4 + 3p) = 50
Possible values, q = 2 and p = 7 or q = 5 and p = 2
Also, paths between Q to R + paths between Q to P to R + paths between Q to S to R
= p + 3 × 4 + 4 × q = 27
=> p + 4q = 15
Now, q = 2 and p = 7 satisfies
Hence, the number of direct paths between Q and R = p = 7
A function f maps the set of natural numbers to whole numbers, such that f(xy) = f(x) f(y) + f(x) + f(y) for all x, y and f(p) =1 for every prime number p. Then, the value of f(160000) is
f(xy) = f(x) f(y) + f(x) + f(y) Given, f(p) = 1 where p is a prime number => f(2) = 1 and f(5) = 1 => f(10) = f(2 × 5) = f(2) f(5) + f(2) + f(5) = 1 × 1 + 1 + 1 = 3 => f(100) = f(10 × 10) = f(10) f(10) + f(10) + f(10) = 3 × 3 + 3 + 3 = 15 => f(10000) = f(100 × 100) = f(100) f(100) + f(100) + f(100) = 15 × 15 + 15 + 15 = 255 Now, f(4) = f(2 × 2) = f(2) f(2) + f(2) + f(2) = 1 × 1 + 1 + 1 = 3 => f(16) = f(4 × 4) = f(4) f(4) + f(4) + f(4) = 3 × 3 + 3 + 3 = 15 Now, f(160000) = f(10000 × 16) = f(10000) f(16) + f(10000) + f(16) = 255 × 15 + 255 + 15 = 4095
A vessel contained a certain amount of a solution of acid and water. When 2 litres of water was added to it, the new solution had 50% acid concentration. When 15 litres of acid was further added to this new solution, the final solution had 80% acid concentration. The ratio of water and acid in the original solution was
Let the initial water = w and initial acid = a Given, a = 50% (a + w + 2) => a = w + 2 Also, a + 15 = 80% (a + 15 + w + 2) => 5a + 75 = 8a + 60 => a = 5 and w = 3 Hence, initially water : acid = 3 : 5
Bina incurs 19% loss when she sells a product at Rs. 4860 to Shyam, who in turn sells this product to Hari. If Bina would have sold this product to Shyam at the purchase price of Hari, she would have obtained 17% profit. Then, the profit, in rupees, made by Shyam is
Let the cost price of Bina = 100x Given, 100x – 19% of 100x = Rs 4860 => x = 60 So, the cost price of Bina = 100x = Rs 6000 Certain Price = 6000 + 17% of 6000 = Rs 7020 Profit of Shyam = 7020 – 4860 = Rs 2160
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