Let’s assume total work to be 420 units

Anil = 420/60 = 7 units/day

Bimal = 420/84 = 5 units /day

After 10 days, total work left = 420 -70 = 350

Now to do 350 units of work in 14 days, has to work 25 unit per day

C’s work= 25-7-5= 13 units a day

In term of work    A           B          C
                            168       70       182

Profit of C = 182/420 * 21000 = 9100

Let the initial quantity (Capacity of Container) = x 

Milk / Total = 16/25 = [(x -9)/x] 2=> 4/5 = (x - 9)/x=> x = 45 

Height of Cone = 27 cm 
The Larger cone after passing a plane has a smaller cone with height 9 cm (27-18) 
The heights are in the ratio 9 : 27 or 1 : 3 
Volumes will be in ratio of (Height) 3 
Volume of smaller cone : Volume of whole cone à 1 : 27 
Volume of remaining cone = 27x – 1x = 26x 
26x – x = 225cc 
x = 9cc 
Volume of original cone = 27*9 = 243 cc

We need the sum of a+b+c to be minimum hence the values of a,b ,c must be as close as
possible 
ab = 432 = 18*24 
bc = 96 = 24*4 
a + b + c = 18 + 24 + 4 = 46

Original speed = S & Original time = T 
New Speed = S/3, speed and time are inversely proportional hence New time = 3T 
The difference between times = 3T-T = 30 mins 
Original time taken = 15 mins 
Assume Distance is 15 Kms and Speed = 1 km/hr

In 5 mins the train travelled 5 kms and 4 mins it stopped. 
In 6 mins it has to cover 10 kms 
The original speed will be increased by a factor of 1.66 or 67%

Area = 0.5*4*2 – 0.5*2*1 = 4-1 = 3

As f(x) = 0 we must try to convert f(5+x) = f(5-x) in terms of f(x) 
We substitute x as x-5 
f(x) = f(10-x) 
Lets assume x = A as a root, then we simultaneously get 10-A as a root 
Similarly we assume x = B as a root, then we get 10-B as a root 
the 4 roots are A, B, 10-A, 10-B 
The sum of roots is 20 

2 years simple interest accrued by Veeru is Rs. 1000 
10000 + 1000 + 10000*0.05*x = 8000 + 8000*0.1*x 
3000 = 300x 
x = 10 years 
2 years of Veeru’s initial investment hence answer = 10+2 = 12 years 

A + (B+C)/2 = 5 à 2A + B + C = 10 - Eq 1 
B + (A+C)/2 = 7 à A + 2B + C = 14 - Eq 2 
Subtracting, Eq2 – Eq1 
B = 4 + A 
Adding Eq 1 and Eq 2 
3(A+B) + 2C = 24 – Eq 3 
Now since A B C are positive and B = 4 + A hence A+B cannot be equal to 4 
In Eq 3 all components have to be even thereby eliminating 5 and 7. 
Hence A+B = 6. 

Assume length of Train as L and Original speed as S 
To cross the first person (2 kmph) train needs 90 seconds at S-2 km/hr and to cross the second
person (2 kmph) train needs 100 seconds at S-4 km/hr 
Speeds are inversely proportional to time 
(S-2)/(S-4) = (100/90) 
S = 22 
(S-2) = L * 3600 / 90 
L = 0.5 km 

T = (0.5 * 3600) / 22 = 82 seconds

Cost of Desktop = D 
Cost of Laptop = L 
D + L = 50000 à Eq 1 
1.2*D + 0.9*L = 51000 à Eq 2 
Multiplying Eq 1 by 0.9 and subtracting with Eq 2 
0.3*D = 6000 
D = Rs. 20000 

AOC is a right-angle triangle and AC = 10 
Perpendicular of a right-angle triangle = P*B/H = 8*6 / 10 = 4.8 
Area of circle = P *4.8*4.8 / (0.5*12*16) = 6* P / 25 

These questions work with a backward approach. 
The 5 child gets half of what was remaining after 4 child + 1, this essentially means the 5 child
got 2 toffees. 
Now the gentleman had 2 toffees after giving half of what was remaining after child 3 + 1, this
means he had 2*(2+1) = 6 left after giving the third child. Hence the 4-child got 
0.5*6 + 1 = 4 and 2 were left for 5-child. 
Similarly, after giving second child he had 2*(6+1) = 14 toffees 
After giving first child he had 2*(14+1) = 30 toffees 
Initially he had 2*(30+1) = 62 toffees 

The even numbers will be of the form 
1100, 1122, 1144, 1166, 1188 
2200, 2222, 2244, 2266, 2288 and so on 
There will be 45 such numbers from 1100 to 9988, the mean will be the 23 number which is 5544

We have been given the ratio of the volume and weights of the volume (1-unit cube of A will
have a weight of 5x). 
Hence to find the actual weights we must multiply the volume and weights of volume. 
A : B : C = (3*5 : 4*2 : 7*6) = 15 : 8 : 42 
Total weight = 15x + 8x + 42x = 65x = 130 
x = 2 
Therefore weight of C = 42*2 = 84 Kgs

Solution = 40 litres 
Dye = 16 litres 
Water = 24 litres 
New amount of water in the solution will be 5*8 = 40 litres, i.e 16 litres of water has been added 
Solution = 56 litres 
We remove ¼ or 14 litres, which leaves 42 litres with dye = 12 litres and water = 30 litres 
We need 8 litres of dye to get the proportion to 2:3 

Let’s Assume there are 400 people 
Young – 112 
Old – 288 
Illiterates = 140 
Literates = 260 
Young literate = 65 
Old Literates = 195 
Old illiterates = 93 
Old illiterates / Total illiterates = 93/140 = 66%

Let’s assume (x + (1/x)) = A 
A – 3A + 2 = 0 
A = 2,1 
(x + (1/x)) = 2 
X = 1, 
(x + (1/x)) = 1 à This case is not possible as (x + (1/x)) >= 2 if x is positive and (x + (1/x)) <= -2 if
x is negative. 
Therefore, we have 1 distinct real root.

Let’s start with numbers with products 3 4 5 6 
113, 131, 311 
122, 212, 221 
114, 141, 411 
115, 151, 511 
123, 132, 213, 312, 231, 321 
116, 161, 611 
21 Solutions 

At 8 km/hr Amal reaches at 10:15 am and at 15km/hr Amal reaches at 9:40 am. 
Lets assume the original time takes as T mins and at 15km/hr the time taken would be T-35 mins 
Ratio of speeds is inverse of ratio of time. 
(8/15) = (T-35/T)

T = 75 mins 
Distance = 8 * (75/60) = 10 kms 
Time = (50/60) hr 
Speed = 10 * (50/60) = 12 km/hr

The natural numbers are divided into groups as (1), (2, 3, 4), (5, 6, 7, 8, 9), ….. and so on.
If you observe that the first group ends with 1², the second group ends with 2²and the third group ends with 3² 
Hence, the 15th group ends with 15² = 225.
The 14th group ends with 14² = 196.
Therefore, the 15th group contains (197, 198, . . . . . . . . . . ,225)

Given that the numbers 3 and 5 should be present in every subset and contain at least 3 numbers in it.
First, we need to find the subsets possible
{3, 5, 1, 2, 4, 6, 7, 8} except 3 and 5, remaining all numbers have two possible outcomes that either it is in the set or out of the set.
So, the number of possible subsets containing at least 3 numbers is 2⁶-1 = 63
(In this 2⁶, we have a possibility that nothing is present from {1, 2, 4, 6, 7, 8}. So, we should remove that possibility.)
From these 63, we need to remove the subsets which have {3, 5, 7, 8}
{3, 5, 7, 8, 1, 2, 4, 6} except 3, 5, 7, 8 remaining all numbers {1, 2, 4, 6} have two possible outcomes that either it is in the set or out of the set.
So, the number of sets possible is 2⁴ = 16
Therefore, the answer would be 63 - 16 = 47.

Let's consider a three-digit number to be ‘abc’.
Given that, three-digit numbers increase by 198 when the three digits are arranged in the reverse order.
100c + 10b + a - 100a - 10b - c = 198
99c - 99a = 198
c - a = 2
So, the difference between the hundreds place digit and the units place digit is 2.
The possible combinations are:
1 _ 3, 2 _ 4, 3 _ 5, 4 _ 6, 5 _ 7, 6 _ 8, 7 _ 9.
We have 10 numbers for each combination.
Hence, the total numbers are 70.

Given that onion is sold for 5 consecutive months at the rate of Rs 10, 20, 25, 25, and 50 per kg respectively.
Let’s consider the amount spend on onions in these five months to be 100, 100, 100, 50, 50.
Hence, the number of kgs in these five months should be 10, 5, 4, 2, 1 respectively.
The average expense on onions per kg over these five months is given by

Therefore, the closest average would be 18.

Speed of faster train = 160/12 m/s 
Let's convert it to km/hr = (160/12) x (18/5) = 48 km/hr 
Speed of slower train = 48-6= 42 kmph 
Trains are moving in opposite directions we must add the speeds and convert to m/s 48+42 =
90x5/18 = 25m/s 
(160+x)/14 =25 160 + x = 350 
x = 190m

Let’s consider the total number of pens to be
(100 +x)
Let's consider the fixed-wage of the labour to be ‘w’
case (i): the net profit is 300
CP = 8(100 +x) + w
 SP = 12 x 100 + 11 x = 1200 + 11x 
Profit = SP - CP
300 = 1200 + 11x - 8(100 +x) - w
w - 3x = 100 → eq(1)
Case (ii): the net loss is 300
CP = 8(100 +x) + w
SP = 12 x 100 + 9 x = 1200 + 9x
Loss = CP - SP
300 = 8(100 +x) + w - 1200 - 9x
w - x = 700 → eq(2)

By solving equation (1) & (2) we get w = 1000.
Therefore, the wage of the employee is 1000.
Alternate solution:
We can also do this with a little intuition. The Rs 2 decrease per pen results in 300 loss from the case of 300 profit.
The net value of 600 resulted by selling the remaining pens at Rs 2 lesser.
If the number of remaining pens is x, then 2x = 600

So, x = 300.
Total pens = 100 + 300 = 400.
We can get the wage of an employee(w) by considering profit/loss
Profit = 300
100(12) + 300(11) - 400(8) -w = 300
w = 1000

In a regular hexagon, each internal angle is equal to 120°.
From isosceles triangle ABC, we know the length of two sides and including angle.
We will be able to find the third side (AC) using the Pythagoras theorem or the sine rule.
Hence, AC = 2√3cm
Given that, T is the midpoint.
So, CT = 1cm.
From the right-angled △ ACT,
AC² + CT² = AT²
AT² = (2 √3)² + (1)² = 13
AT = √(13)   

Let's consider the number of patients admitted in hospital A and hospital B to be 'x' and ‘x + 21' 
Given that, the sum of recovery days for patients in hospitals A and B were 200 and 152, respectively.
The average recovery days for patients admitted in hospital A was 3 more than the average in-hospital B.

By solving the above equation we get x = 35.
Hence, the number of patients admitted to hospital A is 35.