Given that Ramesh and Ganesh can together complete a work in 16 days 
After seven days of working together, Ramesh got sick and his efficiency fell by 30%. 
To the original schedule of 16 days there are 9 more days and in this 9 days Ramesh fall short by 9
× 0.3x = 2.7x where x is the work done by Ramesh in a day 
So the change = 2.7x (i.e. the gap because of drop in efficiency) 
This 2.7x is compensated by one day of Ganesh + another 0.7 of Ramesh 
2.7x = G + 0.7 x (Because the last day also he works with reduced efficiency) 
2x = G 
Hence Ganesh works twice as that of Ramesh. 

If Ganesh had worked alone after Ramesh got sick means in the 7 days Ramesh would have
completed 7/48 of the task 
Ganesh on each day can do 1/24 of the task 
Task to be completed by Ganesh is 41/48 completely 
One day he can complete 1/24 so 41/48 × 24/1 = 20.5 days 
If Ganesh had worked alone after Ramesh got sick, the no.of days needed for completing the
remaining work is 20.5 – 7 = 13.5 days

Given that a tank is emptied everyday at a fixed time point and after that either pump A or B or
both start working to fill the tank. 
On Monday, A alone completed filling the tank at 8 pm. 
On Tuesday, B alone completed filling the tank at 6 pm. 
On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the
tank. 
We have to find at what time will the tank filled on thursday if both were used simultaneously all
along 

A is doing 3 hours less work on Wednesday which B completes in 2 hours therefore B = 1.5 A 
⟹ (n)/(n-2) = 3/2 
⟹ 2n = 3n – 6 
⟹ n = 6 
A takes 6 hours and B takes 4 hours or the tank is closed at 2 pm 
When they are together open 
⟹ 1/6 + 1/4 = (2+3)/12 
⟹ Rate is 5/12 or they can fill the entire tank in 12/5 hours which is 2 hours 24 minutes 
Starting from 2 pm, they fill the entire tank by 4:24 pm 
Here the break through is figuring out the ratio or efficiency of the rate at which they fill is B : A
= 3 : 2 
The conventional way is 
⟹ (n−3)/n + 2/(n-2) = 1 

n-3 hours at the rate of 1/𝑛 per hour 
2 hours at the rate 1/(𝑛−2) per hour 
Solving this we get n = 6 and we can find solution using this n. 

Given that points A and B are 150 km apart. 
Cars 1 and 2 travel from A to B, but car 2 starts from A when car 1 is already 20 km away from A. 
Each car travels at a speed of 100 kmph for the first 50 km, at 50 kmph for the next 50 km, and at
25 kmph for the last 50 km. 
The Car 1 is 20km away from A and it travels at 100kmph and the time taken is 20/100 = 1/5hr 
So car 2 is 12 minutes behind car 1. 
We have to find the distance in km, between car 2 and B when car 1 reaches B. 

If car 1 reaches B, car 2 will take 12 minutes to reach b. 
Distance between car 2 and B is 25 × 1/5 = 5 kms 
The distance, in km, between car 2 and B when car 1 reaches B is 5 kms

Given that a jar contains a mixture of 175 ml water and 700 ml alcohol. 
It is given that 10% of the mixture is removed and it is substituted by water of the same amount
and the process is repeated once again 
Now we have to find the percentage of water in the mixture. 
Since the mixture is removed and substituted with water, we can deal with alcohol and the second
step we can find how much amount of alcohol is retained and not about how much amount of
alcohol is removed 
As 10% of alcohol is removed, 90% of alcohol is retained 
So alcohol remaining = 700 × 90% × 90% 
⟹ 700 × 0.9 × 0.9 = 567 

We totally have 875 ml overall mixture and of this 567 ml is alcohol. 
Remaining 875 – 567 = 308 is the amount of water. 
We have to find the percentage of water in the mixture i.e. 308/875 
Approximately 308 is 30% of 1000 so by this we know that 308 is more than 30% 
Hence 35.2% is the percentage of water in the given mixture. 

Given that the sum of squares of two numbers is 97 i.e. a 2 + b 2 = 97 
From the given options we have to find which one cannot be their product i.e. ab 
A. 64 ⟹ 2ab = 128 
B. −32 ⟹ 2ab = -64 
C. 16 ⟹ 2ab = 32 
D. 48 ⟹ 2ab = 96 
2ab is found because we know that 
a 2 + b 2 + 2ab ≥ 0 
a 2 + b 2 - 2ab ≥ 0 
By this we can know that 97 + 128 works but 97 - 128 doesn’t works so we can understand option
A cannot be the product and the rest can be. 
a 2 + b 2 ≥ |2ab| 
a 2 + b 2 ≥ 2ab 
a 2 + b 2 ≥ -2ab 
⟹( a 2 + b 2 ) / 2 ≥ |ab| 
So here 2ab should lie between +97 and -97 or ab should be less than 97/2 or greater than −97/2,
so except option A all the other options works so option A 64 cannot be the product 

Given that A and B are two points such that B is 350 km west of A. One car starts from A and
another from B at the same time. 
They meet after one hour or their relative velocity is (a + b)km/hr and their relative distance is
350kms 
a + b = 350 which is equal to 350 kms/hr 

If they both move towards east, then they meet in 7 hrs. 
We have to find the difference between their speeds, in km per hour. B is catching upon A, so once
again their relative distance will be 350kms. 
Relative speed = b – a , B travels faster so that it can meet a. 
⟹ 350/ 𝑏−𝑎 = 7 
⟹ b - a = 50 km/hr 
We know that b + a = 350 km/hr 
By this we can find the values of a and b but they have asked for only the difference between their
speeds which is b – a = 50km/hr

Given that a parallelogram ABCD has an area 48 sq cm. The length of CD is 8 cm and that of AD is s cm
Since the area given here is 48 sq.cm the height can be found as 6 sq.cm
We have to find which of the following options is necessarily true

Since this (shaded portion) is a right angled triangle with 6 as its height and s as its hypotenuse so we can assure that s ≥ 6 and s cannot be less than 6 so we can eliminate the other options.
By this we can come to the conclusion that s ≥ 6

Given that 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1:3 by volume

This M is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution ,then the unknown concentration of S has to be found

Equal quantities of mixture and 20% ethanol solution are mixed in equal ratio to get 31.25%
The mixture and 20% ethanol solution are mixed in equal ratio to get 31.25%. This 31.25% is 11.25% more than this 20%. This mixture must be 11.25% more than 31.25% so mixture is equal to 42.5%

This mixture 42.5% is mixed in the ratio 1:3

Using allegations we can find this 22.5% which should be in the ratio 1:3 so the other one is 7.5%
S = 42.5 + 7.5
S = 50%
Here to find the solution, do the second thing first and then go to the first thing
The second one is mixed in the ratio 1 : 1 so that the final thing should be bang in the middle

Given that a chord of length 5 cm subtends an angle of 60° at the centre of a circle. 
We have to find the length of a chord that subtends an angle of 120° at the centre of the same
circle 
sin 60° = BC/OB 
⟹ √3/2 = BC/5 
⟹ BC = 5√3/2 and AC = 5√3/2 
So AB = 5√3/2 + 5√3/2= 5√3 
The length of a chord that subtends an angle of 120° at the centre of the same circle is 5√3

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 18/25 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 (13/20) th of the overall mixture. From
here on, it is weighted averages. 
So, {(18/5)×3 + (x×4)} / 7 = 13/20 
54/25 + 4𝑥 = 91/20 
4𝑥 = 91/20 - 54/25 
4𝑥 = (455−216)/100 = 239/100 

𝑥 = 239/400 
In drum 2, A is 239/400 so B should be the remaining. Therefore, A : B = 239 : 161 

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

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

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

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 
6F – 5D = 1616 th of the tank ---(1) 
Also, Tank gets filled in 60 hours when 5 filling and 6 draining pipes are on 
5F-6D = 160160 th of the tank ---(2) 
Solving both (1) and (2) we get, 
6F - 5D = 1/6 
(-) 50 F - 60 D = 1/6 
--------------------------- 
44F = 55D 
F:D = 5:4 
Replacing values in (1), 6F – 5D = 1/6 
15D – 10D = 1/3 
D = 1/15 and F = 1/12 
When two filling pipes and one draining pipes are on, 
2(1/6) – 1(/15) = (3/30) = 1/10 th of the tank 
Therefore, they can fill the tank in 10 hours 

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, 
(x+1) 2 +2 2 = r 2 ---(1) 
x 2 +3 2 = r 2 ---(2) 
We find that there is an increase and decrease by 1 in both equations 
So, x 2 = 2 2 , x=2 
r 2 = 2 2 +3 2 = 4+9 = 13 
r = √13 cms

Given Area (ABCD) = 72 sq cm, CD = 9cms, AD = 16cms
Area of Parallelogram = Base × Height
CD × 16 = 72
CD = 72167216 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

Let a Human complete ‘H’ units of work per day and a Robot complete ‘R’ units of work per day 
Given, 
15H + 5R = (1/30) th work per day --- (1) 
5H + 15R = (1/60) th work per day --- (2) 
Dividing (1) and (2) by 5 and solving, 
3H + R = (1/150) and H + 3R = (1/300) 
9H + 3R = 1/50 
(-) H + 3R = 1/300 
--------------------------- 
8H = 1/50 - 1/300 => 8H = 1/60 
H = 1/480 
Work done by 15 Humans in a day = 15/480 = 1/32 
No of days required for 15 humans to complete the work = 32 days 

Given Max. mark = N and Pass mark = 45% of N 
A scores 36 marks and falls short of the pass mark by 68% 
This means the pass mark must be reduced by (100 - 68)% = 32% 
So, A scored 32% (45% N) = 36 
N = 250 marks 
The only option whose range falls within the obtained value is (B) 243 ≤ N ≤ 252 

Since, ABCD is a rectangle inscribed inside a circle 
ABC must be a Right triangle 
Given that radius of Circle = 13 cms 
5, 12, 13 forms a Pythagorean Triplet 
10, 24, 26 is also a Pythagorean triplet 
So, 10 and 24 are possible length and breadth of ABCD

Let the three real numbers be a, b, c. c is taken as 73 instead of 37 
So, 73ab = 37ab + 720 
36 ab = 720 
ab = 720/36 = 20 
We know, AM ≥ GM 
(a²+b²) / 2 ≥ ab 
(a 2 + b 2 ) ≥ 2ab 
(a 2 + b 2 ) ≥ 40 
(Or) 
ab = 20 
(a,b) = (√20, √20) 
(a 2 + b 2 ) = 20 + 20 = 40

As the triangle progresses infinitely and the side length decreases, it follows an infinite GP series 
As the sides decrease by half, their areas decrease by 1/4 
We know, Area of an Equilateral Triangle = √3/4 × a 2 
Area of T1 = √3/4 × 24 × 24 = 144 √3 sq cms 
Sum of an Infinite GP = a / 1−r where a = 144 √3 , r = 1/4 
Sum of areas ( T1, T2, T3,) = 144√3 / {1−(1/4)} = 4×144√3 / 3 
Therefore, Sum of areas= 192 √3 sq cms. 

Its given that B needs 25% more time than A 
So, if Time taken by A = 4x days, Time taken by B = 5x days 
A does half the work, So A works alone for 2x days --- (1) 
A and B work for 4 days --- (2) 
B works alone for 5x/20= x/4 days --- (3) 
Total working days = 13 
(1) + (2) + (3) = 13 
2x + 4 + x/4 = 13 we get, x = 4 
So, B alone can do the work in 5 × 4 = 20 days 
Hence, the answer is 20 days

Let the wholesaler bought walnuts and peanuts at Rs 3x and Rs x respectively 
He sold 8kgs of Peanuts to the shopkeeper at 10% Profit 
Cost price of the Peanuts bought by the shopkeeper = Rs. 1.1x per kg 
Similarly, he sold 16 kgs of Walnuts to the shopkeeper at 20% Profit 
Cost price of the Walnuts bought by the shopkeeper = 1.2 × 3x = Rs. 3.6x per Kg 
He lost 5 kgs of Walnuts and 3 Kgs of Peanuts in transit 
Remaining = 16 - 5 = 11 kgs of Walnuts & 8-3 = 5 kgs of Peanuts 
He mixes them together and sells them at Rs. 166 per kg, making an overall Profit of 25% 
Selling Price = 5/4 × Cost Price 
Overall Cost Price = Rs.(3.6 × 16x + 1.1 × 8x) 
Overall Selling Price = Rs. 16 × 166 
Overall Selling Price = 5/4 × Overall Cost Price 
(3.6 × 16x + 1.1 × 8x) × 5/4 = 16 × 166 
( ( 57.6 × 8.8) x ) × 5/4 = 16 × 166 
x = Rs. 32 
Cost price per Kg of Walnuts bought by the Wholesaler = Rs. 3x = 3 × 32 = >Rs. 96