A triangle ABC has area 32 sq units and its side BC, of length 8 units, lies on the line x = 4. Then the shortest possible distance between A and the point (0,0) is
Given Area (△ABC) = 32 sq units and one of the length BC = 8 units on the line x = 4
Let us draw a graph and plot the given values.
Since the base lies on x = 4 and has a vertical height is of length = 8 units, A can either lie on the line x = 12 or on x = - 4
However, since we need to find the shortest possible distance between A and the origin, A should lie on the line x = - 4
So, shortest possible distance to A from the point (0,0) = 4 units
How many two-digit numbers, with a non-zero digit in the units place, are there which are more than thrice the number formed by interchanging the positions of its digits?
Let the two-digit number be xy, which can be expressed as 10x + y
Given that the two-digit number is more than thrice the number obtained by interchanging the digits
So, 10x + y > 3 × (10y + x)
=> 10x + y > 30y + 3x
=> 7x > 29y
Approximately, x > 4y
Let us fix values for y and check for conditions,
For y = 1, x can take the values of 5 , 6 , 7 , 8 , 9 (As, 51 > (3 × 15), 61 > (3 ×16), 71> (3 × 17), 81 > (3 × 18), 91 > (3 × 19))
Similarly, for y = 2, the condition satisfies only for x = 9 (As, 92 > (3 × 29))
The remaining values of y does not satisfy the given conditions.
So, Total = 5 + 1 = 6 numbers
A water tank has inlets of two types A and B. All inlets of type A when open, bring in water at the same rate. All inlets of type B, when open, bring in water at the same rate. The empty tank is completely filled in 30 minutes if 10 inlets of type A and 45 inlets of type B are open, and in 1 hour if 8 inlets of type A and 18 inlets of type B are open. In how many minutes will the empty tank get completely filled if 7 inlets of type A and 27 inlets of type B are open? (TITA)
Let A be the rate at which pipe A fills the tank in one minute and B be the rate at which pipe B fills in one minute
From the question, we can infer that 10A + 45B = (Since it takes 30 minutes to fill, th of the tank gets filled in 1 minute)
Gopal borrows Rs. X from Ankit at 8% annual interest. He then adds Rs. Y of his own money and lends Rs. X+Y to Ishan at 10% annual interest. At the end of the year, after returning Ankit’s dues, the net interest retained by Gopal is the same as that accrued to Ankit. On the other hand, had Gopal lent Rs. X+2Y to Ishan at 10%, then the net interest retained by him would have increased by Rs. 150. If all interests are compounded annually, then find the value of X + Y. (TITA)
Let us draw a flow diagram to understand the transaction
Gopal receives 10% of X + Y from Ishan as interest, from which he pays the interest of 8% of X to Ankit
So, Gopal would gain 2% of X + 10% of Y as interest and Ankit would gain 8% Interest
Its given that both the interest values are same, so 2% X + 10% Y = 8% X
3% X = 5% Y — (1)
Its also given that, if Gopal lend Rs. X + 2Y to Ishan at 10% interest, the net interest retained would increase by Rs. 150
So, the increase in 10% Y accounts for Rs. 150
10% Y = 150 => Y = Rs. 1500
Substituting the value in (1), we get 5 × 1500 = 3 × X => X = 2500
So, X + Y = 2500 + 1500 = Rs 4000
The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u = and v = . If x ≥ z, then the minimum possible value of x is (TITA)
Given that AM (x, y, z) = 80
So, x + y + z = 80 × 3 = 240 ----- (1)
Also, AM (x, y, z, u, v) = 75
So, x + y + z + u + v = 75 × 5 = 375 ----- (2)
Equation (2) – (1), we get
u + v = 375 - 240 = 135
Substituting this value in (1), we get x + z = 210
It’s given that x ≥ z
x would take the minimum value when x = z
=> 2x = 210
=> x (min) = 105
Given that f(x) = max where x is any positive real number.
The minimum possible value of f(x) has to found.
The minimum possible value is obtained if the two curves intersect or
One of these would be the minimum possible value
It is given that x is a positive real number. So, we have to consider only when x = 4
When x = 4
5x = 5 × 4 = 20
The minimum possible value of f(x) is 20
For two sets A and B, let AΔB denote the set of elements which belong to A or B but not both. If P = {1,2,3,4}, Q = {2,3,5,6,}, R = {1,3,7,8,9}, S = {2,4,9,10}, then the number of elements in (PΔQ)Δ(RΔS) is
Given that for two sets A and B, A △ B denote the set of elements which belong to A or B but not both.
This is the diagram which represents A △ B ,the set of elements which belongs to A or B i.e. A intersection B subtracted from A union B
We have to find the number of elements in (P △ Q) △ (R △ S) if
P = {1 , 2 , 3 , 4} Q = {2 , 3 , 5 , 6 ,} R = {1 , 3 , 7 , 8 , 9} S = {2 , 4 , 9 , 10}
For P △ Q, we have to find the elements which do not to belong to both P and Q such that
(P △ Q) = {1, 4 , 5 , 6}
(R △ S) = {1 , 2 , 3 , 4 , 7 , 8 , 10}
(P △ Q) △ (R △ S) = {2 , 3 , 5 , 7 , 6 , 8 , 10}
The number of elements (P △ Q) △ (R △ S) is 7
If A = { - 35n - 1: n = 1,2,3,…} and B = {35(n-1) : n = 1,2,3,…} then which of the following is true?
A. Neither every member of A is in B nor every member of B is in A
B. Every member of A is in B and at least one member of B is not in A
C. Every member of B is in A.
D. At least one member of A is not in B
B = {35(n-1) : n = 1 , 2 , 3 , …}
When n = 1
{35(0)} = 0
when n = 2
{35(1)} = 35
when n = 3
{35(2)} = 70
So B is more predictable here since it has every multiple of 35 ie. {0 , 35 , 70 , 10 , 140 , …}
From this we can see that A is a subset of B ie. Every element of A is in B and not the vice versa
Therefore every member of A is in B and at least one member of B is not in A
Hence, the answer is Every member of A is in B and at least one member of B is not in A
The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. If three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%. If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%. A fourth solution, D, is produced by mixing B and C in the ratio 2 : 7. The ratio of the strength of D to that of A is
Given that the strength of the salt solution is p% if 100 ml of the solution contains p grams of salt
It is also given that three salt solutions A , B , C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%.
We have to find the ratio of the strength of D : A
Subtracting equations 1 and 2, we get
2A – 2C = 0.6 or A – C = 0.3
Since we could not find anything from the above methods, we can eliminate the number part and get the ratio going
A + 2B + 3C = 1.2
3A + 2B + C = 1.8
So let us multiply eqn 1 and 2 with 3 and 2,
3A + 6B + 9C = 6A + 4B + 2C
2B + 7C = 3A
Therefore the ratio D : A = 1 : 3