Given Area (ABCD) = 72 sq cm, CD = 9cms, AD = 16cms
Area of Parallelogram = Base × Height
CD × 16 = 72
CD = cms
We extend CD till P such that ∠APD = 90°
So, again by applying the area formula considering CD as base
CD × AP = Area of Parallelogram ABCD
9 × AP = 72 AP = 8 cms
We have AP = 8 cms, AD = 16 cms We observe that the ratio of the sides form a 1 : √3 : 2 triangle Therefore DP = 8√3 cms Area (△APD) = 1/2 × AD × DP = 1/2 × 8 × 8√3 sq cms Area (△APD) = 32 √3 sq cms
In a circle, two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm. If the distance between these chords is 1 cm, then the radius of the circle, in cm, is
Let set A be a set having first 50 perfect squares and set B be a set having Square of first 20 positive prime numbers which have three factors. Then sum of all elements in set (B - A) is
Given that Chords lie on the same side of diameter with lengths 4 cms and 6 cms
Draw a perpendicular from the origin to both the chords and mark the points of intersection as P and Q respectively
Consider radius of circle as ‘r’ and distance OP as ‘x’
Draw lines from origin to the end of the chord and mark the points as D and B respectively
Thus, △OQB and △OPD form a right Triangle
Applying Pythagoras Theorem on both triangles,
We find that there is an increase and decrease by 1 in both equations r = √13 cms
A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate, and all draining pipes drain at the same rate. The empty tank gets completely filled in 6 hours when 6 filling and 5 draining pipes are on, but this time becomes 60 hours when 5 filling and 6 draining pipes are on. In how many hours will the empty tank get completely filled when one draining and two filling pipes are on? [TITA]
Given, Tank gets filled in 6 hours when 6 filling and 5 draining pipes are on
Let, F be the rate at which a single filling pipe fills the tanks and D be the rate at which a single draining pipe drains the pipe
Also, Tank gets filled in 60 hours when 5 filling and 6 draining pipes are on
Solving both (1) and (2) we get,
44F = 55D
F:D = 5:4
15D – 10D = 1/3
D = 1/15 and F = 1/12
When two filling pipes and one draining pipes are on,
Therefore, they can fill the tank in 10 hours
Given that out of 200 students, 105 like Pizza and 134 like Burger
In order to find the possible number of students who like only Burger, we need to find the range between the minimum and maximum possible number of students
Minimum possible number of Students who like only Burger:
For this, we must try to optimize the Venn diagram in such a way that it has maximum intersection
No of Students who like only Burger = 134 – 105= 29
Maximum possible number of Students who like only Burger:
For this, we must optimize the Venn diagram in such a way that it has minimum intersection
We know that total number of Students = 200
So, Minimum number of students who like both = 105 + 134 – 200 = 39 students
Students who like Burger = 134
Max. no of Students who like only Burger = 134 – 39 = 95
So range => 29 ≤ n ≤ 95
From the options, only (D) 93 lies within range
In an apartment complex, the number of people aged 51 years and above is 30 and there are at most 39 people whose ages are below 51 years. The average age of all the people in the apartment complex is 38 years. What is the largest possible average age, in years, of the people whose ages are below 51 years?
Given that the Average age of all the residents in the complex = 38 years
Number of people having ages 51 and above = 30
Number of people having ages below 51 ≤ 39
Since we need to maximize the average age of people below 51 and to maintain the combined average as 38 years, we need to consider the ages of people above 51 as 51 (Any more than that would lower the other value)
Also, We need to take the Number of people having ages below as 39 (Again, for maximizing the value)
So, 13 : 38-x = 39 : 30 = 13 : 10
38 - x = 10
x = 28
Average age of people whose ages are below 51 = 28 years
Points A, P, Q and B lie on the same line such that P, Q and B are, respectively, 100 km, 200 km and 300 km away from A. Cars 1 and 2 leave A at the same time and move towards B. Simultaneously, car 3 leaves B and moves towards A. Car 3 meets Car 1 at Q, and Car 2 at P. If each car is moving in uniform speed then the ratio of the speed of Car 2 to that of Car 1 is
It is given that points A, P, Q and B lie on the same line such that P, Q and B are 100 km, 200 km and 300 km away from A. Let us draw the diagram first. All of them are on the same straight line and P, Q lie between A and B.
Cars 1 and Cars 2 leave A at the same time and move towards B. Simultaneously, Car 3 leaves B and moves towards A. Let us add more details to the diagram. C1 and C2 move towards B and C3 moves towards A. C3 meets C1 at Q and C3 meets C2 at P.
Let us assume the speeds of Car 1, 2, and 3 to be C1, C2 and C3 respectively. Car 1 is travelling quicker than Car 2. When Car 1 travels 200 km to reach Q, Car 3 has travelled 100 km or ratio of their speeds, C1 : C3 is 2 : 1.C1 and C2 move towards B and C3 moves towards A. C3 meets C1 at Q and C3 meets C2 at P.
Let us assume the speeds of Car 1, 2, and 3 to be C1, C2 and C3 respectively.When Car 3 travels 200 km to reach P, Car 2 has travelled 100 km or ratio of their speeds, C3 : C2 is 2 : 1. Now we have all the ratios. We know that C1 : C3 is 2 : 1 and C3 : C2 is 2 : 1. We can see that C3 is the common link
So, taking LCM of C3, we get,
We can see that ratio of speeds of Car 2 to Car 1 is 1 : 4. Hence, C2 : C1 = 1 : 4
There are two drums, each containing a mixture of paints A and B. In drum 1, A and B are in the ratio 18 : 7. The mixtures from drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7. In drum 2, then A and B were in the ratio
A. 251 : 163
B. 239 : 161
C. 220 : 149
D. 229 : 141
Given that two drums each containing a mixture of paints A and B. In drum 1, A and B are in the ratio 18 : 7. Drums 1 and 2 are mixed in the ratio 3 : 4 and in the final mixture A and B are in the ratio 13 : 7. For these kinds of questions do not consider separately as A and B. Deal with either A or B as a share of overall
Here A is of overall
In D2, let us assume the proportion of A with respect to the overall mixture is x. 3 parts of D1 is mixed with parts of D2 to give proportion of A of which is of the overall mixture. From here on, it is weighted averages.
In drum 2, A is so B should be the remaining. Therefore, A : B = 239 : 161
On a triangle ABC, a circle with diameter BC is drawn, intersecting AB and AC at points P and Q, respectively. If the lengths of AB, AC, and CP are 30 cm, 25 cm, and 20 cm respectively, then the length of BQ, in cm, is (TITA)
Let ABC be the triangle on which a circle of diameter BC is drawn, intersecting AB and AC at points P and Q respectively. The lengths of AB ,AC and CP are 30cm , 25cm and 20 cm respectively we have to find the length of BQ in cm.
Key thing here is ,this is semicircle so this angle P and Q should be 90° and now we are looking to do Pythagoras theorem
Now we have to find BQ .So we can take the area of the triangle formula which is equal to × base × height
area of the triangle formula = × base × height × AB × PC = × AC × BQ × 30 × 20 = × 25 × BQ
BQ = 24 cm
cms
Then the maximum possible value of f(x) is [TITA]
be positive integers such that 
Suppose, their arithmetic mean is one less than the arithmetic mean of 
then the largest possible value of 
of overall
of the overall mixture. From here on, it is weighted averages.
so B should be the remaining. Therefore, A : B = 239 : 161
× base × height