Let initial volume of A and B be 1 lt and 3 lt. Now 4lt of A is added. Now A = 5 lt and B = 3lt. Let % of alcohol in B is p%. So according to the question: 8×72/100 = (5 × 60/100) + (3 × p/100) On solving this we get p= 92
Two alcohol solutions, A and B, are mixed in the proportion 1:3 by volume. The volume of the mixture is then doubled by adding solution A such that the resulting mixture has 72% alcohol. If solution A has 60% alcohol, then the percentage of alcohol in solution B is
Let initial volume of A and B be 1 lt and 3 lt. Now 4lt of A is added. Now A = 5 lt and B = 3lt. Let % of alcohol in B is p%. So according to the question: 8×72/100 = (5 × 60/100) + (3 × p/100) On solving this we get p= 92
Anil, Sunil, and Ravi run along a circular path of length 3 km, starting from the same point at the same time, and going in the clockwise direction. If they run at speeds of 15 km/hr, 10 km/hr, and 8 km/hr, respectively, how much distance in km will Ravi have run when Anil and Sunil meet again for the first time at the starting point?
Time taken by Anil to complete one round = 3/15 Time taken by Sunil to complete one round = 3/10 Time taken by Anil and Sunil to meet at the starting point first time = 3/5 hrs Distance travelled by Ravi in 3/5 hrs = 8×3/5 = 4.8 kms
A and B are two railway stations 90 km apart. A train leaves A at 9:00 am, heading towards B at a speed of 40 km/hr. Another train leaves B at 10:30 am, heading towards A at a speed of 20 km/hr. The trains meet each other at
In the final examination, Bishnu scored 52% and Asha scored 64%. The marks obtained by Bishnu is 23 less, and that by Asha is 34 more than the marks obtained by Ramesh. The marks obtained by Geeta, who scored 84%, is
Bishnu scored 52% and Asha scored 64%. Difference between their actual marks = 23 + 34 =57 Difference in their percentages = 12% So 12% of Total = 57 Total = 57×100/12 Score of Geeta = (57 × 100/12) × 84/100 = 399
Let N, x and y be positive integers such that N = x + y, 2 < x < 10 and 14 < y < 23. If N > 25, then how many distinct values are possible for N? (in numerical value)
N=x + y Minimum of x + y = 3 + 15 =18 Maximum value of x + y = 9 + 22 = 31 Now as N > 25, so all values from 26 to 31 are possible. 6 values are possible
A man buys 35 kg of sugar and sets a marked price in order to make a 20% profit. He sells 5 kg at this price, and 15 kg at a 10% discount. Accidentally, 3 kg of sugar is wasted. He sells the remaining sugar by raising the marked price by p percent so as to make an overall profit of 15%. Then p is nearest to
Let cost per kg = 1 Mark Price = 1.2/kg Total cost = 35 Total selling price = 35x 1.15 = 40.25 [(5x1.2 )+ (15x 1.2 x 0.9)+ (3× 0) + (12×1.2× (1 + p/100 ))] = 40.25 P= 25
Vimla starts for office every day at 9 am and reaches exactly on time if she drives at her usual speed of 40 km/hr. She is late by 6 minutes if she drives at 35 km/hr. One day, she covers two-thirds of her distance to office in one-thirds of her usual time to reach office, and then stops for 8 minutes. The speed, in km/hr, at which she should drive the remaining distance to reach office exactly on time is
Let usual time taken is t 40 x t = 35 x (t + 6) So t = 42 mins Distance = 40 x 42/60 = 28kms So 28 x 2/3 = 56/3 kms are covered in 42/3 = 14 mins Vimla stops for 8 mins. Time left = 42 – 14 – 8 = 20 mins So 28/3 kms are to be covered in 20 mins. Speed = (28/3)/(20/60) = 28kmph
A batsman played n + 2 innings and got out on all occasions. His average score in these n + 2 innings was 29 runs and he scored 38 and 15 runs in the last two innings. The batsman scored less than 38 runs in each of the first n innings. In these n innings, his average score was 30 runs and lowest score was x runs. The smallest possible value of x is
Total score of (n+2) innings = 29x(n+2) = 29n + 58 Total score of n innings = 29n + 58 – 38 – 15 = 30n So n=5 So total score in 5 innings = 30 × 5 = 150 Maximum score in any inning = 37 So 150 – (37× 4) = 2
Dick is thrice as old as Tom and Harry is twice as old as Dick. If Dick’s age is 1 year less than the average age of all three, then Harry’s age, in years, is (in numerical value)
Let age of Tom = x So age of Dick = 3x and Harry = 6x So (x + 3x + 6x)/3 - 3x = 1 X = 3 So Harry’s age = 18
A person invested a certain amount of money at 10% annual interest, compounded half-yearly. After one and a half years, the interest and principal together became Rs 18522. The amount, in rupees, that the person had invested is (in numerical value)
P (1 +5/100)3 = 18522 P = 16000
The points (2, 1) and (-3, -4) are opposite vertices of a parallelogram. If the other two vertices lie on the line x + 9y + c = 0, then c is
The diagonals will intersect at the midpoint of the line joining (2,1) and (-3,-4). This point will be (-1/2 , -3/2). The line x+9y +c =0 will also pass through (-1/2 , -3/2) So -1/2 + 9x-3/2 + c = 0 C = 14
How many of the integers 1, 2, ….., 120, are divisible by none of 2, 5 and 7?
Euler Number of 120 is 120 x ½ x 4/5 = 48. So 48 numbers are there which are neither divisible by 2, nor by 5. But since the question is asking for the numbers which are not divisible by 7 also, so we need to find odd multiples of 7 but not to take multiples of 5 in them. So the numbers which have to be deleted are 7 x 1 = 7, 7 x 3 = 21, 7 x 7 = 49, 7 x 9 = 63, 7 x 11 = 77, 7 x 13 = 91 and 7 x 17 = 119. Hence these 7 numbers will be subtracted from the 48 numbers which we have just calculated. Hence required answer = 48 – 7 = 41.
A contractor agreed to construct a 6 km road in 200 days. He employed 140 persons for the work. After 60 days, he realized that only 1.5 km road has been completed. How many additional people would he need to employ in order to finish the work exactly on time? (in numerical value)
To complete 1.5 km, 140 persons took 60 days So to complete the remaining 4.5 km, 140 persons would have taken = 60×3 = 180 days Now to complete 180 days work in (200 – 60) = 140 days: Number of persons required = 140 × 180/140 = 180 Additional persons = 180 – 140 = 40
How many pairs (a, b) of positive integers are there such that a ≤ b and ab = 42017?
Ax B = 42017 A x B = 24034 Now A and B are factors of 24034 Total factors of the above number are 4035 So there are 4035 cases possible So there will be one case where A= B. (4035 – 1)/2 = 2017 cases will be there A>B, these cases are invalid. So 4035 – 2017 = 2018 cases
Let m and n be positive integers, If x2+ mx + 2n = 0 and x2 + 2nx + m = 0 have real roots, then the smallest possible value of m + n is
m2 – 8n > = 0 and 4n2 – 4m > = 0 Now the smallest value m can take for the first equation is m=3 and n =1, but this will not satisfy the second equation. If m= 4 then n= 2 So m + n = 6
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