Quant Capsules by Shashank Prabhu  Set 12

A jar contains a mixture of two liquids A and B in the ratio 4:1. When 10 litre of the mixture is replaced with liquid B,the ratio becomes 2:3. The volume of liquid A present in the earlier was?
Let total solution be 10x liters
A>8x, B>2x
(8x8)/(2x2+10)=2/3
24x24=4x+16
x=2 and so, A > 16 liters.A number x is chosen randomly from first 50 natural numbers. What is the probability that (x + 336/x) is less than or equal to 50.
x^2 + 336 < = 50x
x^2  50x + 336 < = 0
(x42)(x8) < = 0
8 < = x < = 42
Total of 35 values satisfy this inequality
Total values possible are 50
Probability is 35/50 = 7/10Distance b/w x& y is 220km. trains P& Q leave station x at 8 am & 9:51 am resptvly at speeds 25 & 20kmph towards station y. train R leaves station y at 11:30am at 30kmph towards x. when will P be at equal distance from Q and R?
12:48
12:30
12:45
11:48 pmWhen R starts moving, the diagram would look like this:
x...(33 km)...Q...(54.5 km)...P...(132.5 km)...R(y)
The midpoint of Q and R will keep on moving towards x at a speed of 25 kmph. The midpoint of Q and R at 1130 is 93.7554.5 = 39.25 km from P. So, relative speed is 50 kmph and time taken to meet is 39.25/50 = 78.5 minutes ~12:48 pmIf x, y, z are positive numbers and ax + by + cz = bx + cy + az = cx + ay + bz = 0, then which of the following is/are definitely true?
i. a + b + c = 0
ii. a = b = c
iii. a^2 + b^2 + c^2 = 1
a. Only i
b. Only ii
c. Only i and iii
d. Only i and iiAdd all three equations
ax+by+cz+ay+bz+cx+az+bx+cy=0
(a+b+c)(x+y+z)=0
As x, y, z are positive integers, a+b+c=0P1, P2 and P3 are three consecutive prime numbers and P1 × P2 × P3 = 190747. What is the value of P1 + P2 + P3?
a. 169
b. 179
c. 163
d. 173190747 lies between 50^3 and 60^3. So, we need the three numbers to be somewhere around this range. So, we have to choose among 47, 53, 59, 61 (as 43, 47, 53 will be closer towards 125000 than 216000 and 59, 61, 67 will comfortably breach 216000). Out of the two triplets possible, only one ends in a 7.
In how many ways can 18 identical balls be distributed among 3 identical boxes?
a+b+c=18
20c2=190 ordered solutions in total
Out of these, cases where all 3 are equal = 1 ordered solution
Cases where exactly 2 are equal = 2a+b=18 = 101 (accounting for the case already considered) = 9*3 = 27 ordered solutions
Rest of the ordered solutions are made of distinct values = 190127 = 162 ordered solutions
So, 162/3! = 27 cases
Total of 1+9+27 = 37 solutions
.
What is the probability that the product of two integers chosen at random has the same unit digit as the two integers?Same digit in the units place is possible in case of only 4 digits namely, 0,1,5 or 6. Total digits eligible for units digit place are 10*10 ways = 100
Probability =4/100 or 1/25A sum of money is distributed among certain people. The second person receives Re.1 more than the first, the third Rs. 2 more than the second, the fourth Rs. 3 more than the third, and so on. The first person gets Re.1 and the last person Rs. 67. What is the total number of people?
General term is n(n1)/2+1
first term is n(n1)/2+1=1.... n=1
nth term is n(n1)/2+1=67... n=12
Total 12 terms. Although, it would be far easier to write down the terms and then get the answer.Consider ab, a two digit number and it’s square cde, a three digit number. For how many values of ab is (a+b)^2 = c + d + e ? a, b, c, d and e are all positive integers
c+d+e can be one or more among 1, 4, 9, 16, 25 only.
c+d+e=1, a+b=1, (10)^2=100
c+d+e=4, a+b=2, (11)^2=121, (20)^2=400
c+d+e=9, a+b=3, (30)^2=900, (12)^2=144, (21)^2=441
c+d+e=16, a+b=4, (31)^2=961, (13)^2=169, (22)^2=484
c+d+e=25, a+b=5, nothing is possible
Total 9 casesab is a two digit positive number such that ab is divisble by a as well as b. Find sum of all possible values of ab (in numerical value)
10a+b=ka
10a+b=mb
b=(k10)a
10a+ka10a=mak10ma
kmk+10m=0
k(1m)10(1m)=10
(k10)(m1)=10
(15,3) 1 case (15)
(12,6) 4 cases (12,24,36,48)
(11,11) 9 cases (11,22,33...99)
Total 14 cases