The minor angle between the hours hand and minutes hand of a clock was observed at 8:48 am. The minimum duration, in minutes, after 8.48 am when this angle increases by 50% is
Using the formula for angle between the hands of clock. q = M 11/2 – 30 H q = Angle between the two hands. H = Position of hour hand initially M = Position of Minute hand lastly. At 8:48 q = 48 × 11/2 – 30× 8 = 24 degree increasing 24 by 50%, we get = 36 degree. time between 8 and 9 the hands of a clock make an angle of 36 degree q = M 11/2 – 30 H 36 = M 11/2 – 30× 8 m = 552/11 = 50 (2/11) Difference in minutes = 50 (2/11) – 48 = 24/11
A mixture P is formed by removing a certain amount of coffee from a coffee jar and replacing the same amount with cocoa powder. The same amount is again removed from mixture P and replaced with same amount of cocoa powder to form a new mixture Q. If the ratio of coffee and cocoa in the mixture Q is 16 : 9, then the ratio of cocoa in mixture P to that in mixture Q is
The ratio of coffee and cocoa in the mixture Q is 16 : 9. Means ratio of final to total coffee = 16:25 By applying the formula for repeated mixture, 16/9 = 1 [1- taken Out / total]2 Taken Out/ total = 1/5 It means, in first go, 1/5 of coffee is replaced be cocoa powder and in second go, 1/5 of mixture was replaced by cocoa powder. cocoa in mixture P = 1/5 and cocoa in mixture Q is = 9/25 ratio of cocoa in mixture P to that in mixture Q is (1/5) / (9/25) = 5:9
In an examination, the average marks of 4 girls and 6 boys is 24. Each of the girls has the same marks while each of the boys has the same marks. If the marks of any girl is at most double the marks of any boy, but not less than the marks of any boy, then the number of possible distinct integer values of the total marks of 2 girls and 6 boys is
Let marks of each girl = g and marks of each boy = m (4 g + 6 b)/ 10 = 24 4 g + 6 b = 240 2 g + 3 b = 120 – ( I ) Given, marks of any girl is at most double the marks of any boy i.e. g = 2 b (max) but marks of any girl is not less than the marks of any boy i.e. g = b (min) putting max and min value of g in equation (I) , we get at g = 2 b, 4 b + 3 b = 120 ⇒ b = 17.14 at g = b, 2 b + 3 b = 120 ⇒ b = 24 we need to solve for 2 g + 6 b at g = 2 b, 4 b + 6 b ⇒ 10 b ⇒ 10 × 17.14 = 171.4 at g = b, 2 b + 6 b ⇒ 8 b ⇒ 8 × 24 = 192 So value range for 172 till 192, total 21 values
Brishti went on an 8-hour trip in a car. Before the trip, the car had travelled a total of x km till then, where x is a whole number and is palindromic, i.e., x remains unchanged when its digits are reversed. At the end of the trip, the car had traveled a total of 26862 km till then, this number again being palindromic. If Brishti never drove at more than 110 km/hr, then the greatest possible average speed at which she drove during the trip, in km/hr, was
Lets go by options, Option 1, 26862 – 8×80 = 26222 (Not a palindrome) Option 2, 26862 – 8×90 = 26142 (Not a palindrome) Option 3, 26862 – 8×110 = 25980 (Not a palindrome) Option 4, 26862 – 8×100 = 26062 (a palindrome) Correct answer to the question must be option 4.
Gita sells two objects A and B at the same price such that she makes a profit of 20% on object A and a loss of 10% on object B. If she increases the selling price such that objects A and B are still sold at an equal price and a profit of 10% is made on object B, then the profit made on object A will be nearest to
let CP of first = a and CP of second = b Profit of 20% on a. SP of first will be = 1.2 a Loss of 10% on second. SP of second will be = 0.9 b SP of both is same 1.2 a = 0.9 b or 0.9 b = 1.2 a to have a profit of 10% on b means 1.1 b when 0.9 b = 1.2 a then 1.1 b = 1.466 a that means a profit of 47% on first object A.
The salaries of three friends Sita, Gita and Mita are initially in the ratio 5 : 6 : 7, respectively. In the first year, they get salary hikes of 20%, 25% and 20%, respectively. In the second year, Sita and Mita get salary hikes of 40% and 25%, respectively, and the salary of Gita becomes equal to the mean salary of the three friends. The salary hike of Gita in the second year is
Ratio of the salaries of Sita, Gita and Mita is given as 5: 6 : 7 After respective hike of 20% , 25% and 20%, it becomes = 6: 7.5: 8.4. The second year, after Sita and Mita get salary hikes of 40% and 25%, respectively, we get = 8.4: x: 10.5. Now given, the salary of Gita becomes equal to the mean salary of the three friends which is = x (8.4 + x + 10.5) /3 = x x = 9.45 Salary of Gita increases from 7.5 to 9.45, so percentage increase will be = (9.45 – 7.5)/ 7.5 × 100 = 26%
Arvind travels from town A to town B, and Surbhi from town B to town A, both starting at the same time along the same route. After meeting each other, Arvind takes 6 hours to reach town B while Surbhi takes 24 hours to reach town A. If Arvind travelled at a speed of 54 km/h, then the distance, in km, between town A and town B is (in numerical value)
Anil invests Rs. 22000 for 6 years in a certain scheme with 4% interest per annum, compounded half-yearly. Sunil invests in the same scheme for 5 years, and then reinvests the entire amount received 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 same, then the initial investment made by Sunil, in rupees, is (in numerical value)
Let initial investment made by Sunil be ‘x’ As compounded half-yearly, time becomes 12 and rate becomes 2% for Anil and time becomes 10 and rate becomes 2% for Sunil. 22000(1.02)12 = x(1.02)10 × 1.1 x = 20808. Correct answer to the question must be 20808.
The amount of job that Amal, Sunil and Kamal can individually do in a day, are in harmonic progression. Kamal takes twice as much time as Amal to do the same amount of job. If Amal and Sunil work for 4 days and 9 days, respectively, Kamal needs to work for 16 days to finish the remaining job. Then the number of days Sunil will take to finish the job working alone, is (in numerical value)
Kamal takes twice as much time as Amal to do the same amount of job. So, if one day work of Amal is ‘2a’ then one day work of Kamal will be ‘a’. Let one day work of Sunil be ‘x’ Now, amount of job that Amal, Sunil and Kamal can individually do in a day, are inharmonic progression So 2/x = 1/2a + 1/ a x = 4/3a Means, one day work of Sunil is ‘4/3a’ Amal and Sunil work for 4 days and 9 days, respectively, Kamal needs to work for 16 days to finish the remaining job means total work will be = 8 a +12 a + 16 a = 36 a Time taken by Sunil to finish the job working alone = 36a/ (4/3a) = 27 days. Correct answer to the question must be 27 days.
Let C be the circle x2 + y2 + 4x – 6y – 3 = 0 and L be the locus of the point of intersection of a pair of tangents to C with the angle between the two tangents equal to 60°. Then, the point at which L touches the line x = 6 is
In a right-angled triangle ΔABC, the altitude AB is 5 cm, and the base BC is 12 cm. P and Q are two points on BC such that the areas of ΔABP, ΔABQ and ΔABC are in arithmetic progression. If the area of ΔABC is 1.5 times the area of ΔABP, the length of PQ, in cm, is (in numerical value)
The number of all natural numbers up to 1000 with non-repeating digits is
Considering all single digit number = 9 numbers Considering 2 digit number = _ × _ First place can be filled in 9 ways because cannot take zero. Second place can be filled in 9 ways because repetition not allowed. = 9×9 = 81 numbers Considering 3 digit number = _ × _ × _ First place can be filled in 9 ways because cannot take zero. Second place can be filled in 9 ways because repetition not allowed. Third place can be filled in 8 ways because repetition not allowed. = 9×9 × 8 = 648 numbers. Total = 9 + 81 + 648 = 738 numbers.
A lab experiment measures the number of organisms at 8 am every day. Starting with 2 organisms on the first day, the number of organisms on any day is equal to 3 more than twice the number on the previous day. If the number of organisms on nth day exceeds one million, then the lowest possible value of n is (in numerical value)
No. of organisms on first day = 2, Second = 2×2+3 = 7, Third = 7 × 2+3 = 17 Forth = 17 × 2 + 3 = 37, Fifth = 37 × 2 + 3 = 77 Sixth = 77× 2 +3 = 157 Seventh = 157× 2 + 3 = 317 When you look carefully you will realize, after 7 steps there is NO much impact of +3 so we can consider that after seventh term it only becomes double. So considering a GP with first term a =157, r =2 and Tn >10,00,000 n will be 12. So total it must be 7 + 12 =19th term. Correct answer to the question must be 19.
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