A man sold two cows for Rs.210 at a total profit of 5 %. He sold one cow at a loss of 10% and another at a profit of 10%. What is the price of each cow?
50,100
100,50
150,100
50,150

50 and 150
after profit, price is 210
so cost price of 2 cows = 200
x+y=200
selling price of 2 cows =210
0.9x + 1.1y =210 //1 cow sold at 10 % loss and another is at 10% profit
solving both equations x=50 and y=150

so one cow is of x = Rs. 50
and Other is of 200-x = (200-50) = Rs. 150

In a class there are less than 500 students. When it is divided by 3 it gives a whole number. Similarly when it is divided by 4, 5 or 7 gives a whole number. Find the no. of students in the class.
430
420
440
410

Just find LCM of all the numbers. Its 420.

At 6’o clock, the clock ticks 6 times. The time between first and last ticks was 30sec. How much time it takes at 12’o clock?
65 sec
60 sec
62 sec
66sec

At 6’o clock ticks 6 times means 5 interval
and also given that time b/w first and last tick is 30 sec.
so
30/5=6
i.e time of each interval is 6 sec.

Similarly at 12’o clock, clock ticks 12 times i.e 11 interval
so
1 interval time=6
so 11 interval=11*6

There is a escalator and 2 persons move down it. A takes 50 steps and B takes 75 steps while the escalator is moving down. Given that the time taken by A to take 1 step is equal to time taken by B to take 3 steps. Find the no. of steps in the escalator while it is stationary.
50
60
80
100

## 100

Let us say that the escalator moves at the rate of n steps per second. Let us also say A takes 1 step per second and B takes 3 steps per second.

Case 1: When A is coming down

A will take 50 seconds to complete 50 steps.

In 50 seconds, escalator would have moved 50n steps.

Total number of steps on the stationary escalator = 50 + 50n

Case 2: When B is coming down

B will take 25 seconds to complete 75 steps

In 25 seconds, escalator would have moved 25n steps.

Total number of steps on the stationary escalator = 75 + 25n

Total number of steps on the stationary escalator is a constant

=> 50 + 50n = 75 + 25n

=> 25n = 25

=> n = 1

Total number of steps = 50 + 50 = 75 + 25 = 100

Food grains are to be sent to city from godown. Owner wants to reach the food grains at 11 O’ clock in the city. If a truck travels at a speed of 30km/hr then he will reach the city one hour earlier. If the truck travels at a speed of 20km/h then it will reach the city one hour late. Find the distance between the godown to city. Also with which speed the truck should travel in order to reach at exactly 11 ‘O clock.
120 km, 20 kmph
100 km, 22 kmph
120 km, 24 kmph
110 km, 21.5 kmph

Let distance be x km and original time be t hr, when he travels at constant speed, say s kmph.
Clearly, x = st.
Now, x=30(t-1) and x=20(t+1). Solving these to simultaneous linear eqns, we get x=120 km and t=5 hr. So, distance between city and go-down=120km. To reach exactly at 11 o’clock, he must travel at speed s=x/t = 120/5 = 24 kmph

There are 5 burglars and once went to a bakery to rob it obviously. The first guy ate 1/2 of the total bread and 1/2 of the bread. The second guy ate 1/2 of the remaining and 1/2 of the bread. The third guy, fourth guy and fifth guy did the same. After the fifth guy there is no bread left out. How many bread are there?
31
37
23
21

First we create a general rule for amount of bread left when each burglar leaves
Lets assume we have x no of bread
At every step x/2 + 1/2 bread is eaten by the burglar
Therefore after each step compared to previous step we have [x-(x/2 + 1/2)] = x/2 – 1/2 bread left.

Now let us go to the problem
Now after 5th burglar leaves we have 0 bread left
Hence number of bread after 4th burglar leaves (or before 5th burglar enters) is calculated by
x/2 – 1/2 = 0
Therefore x = 1

Similarly after 4th burglar leaves we have 1 bread left
Hence number of bread after 3th burglar leaves (or before 4th burglar enters) is calculated by
x/2 – 1/2 = 1
Therefore x = 3

Similarly after 3rd burglar leaves we have 3 bread left
Hence number of bread after 2nd burglar leaves (or before 3rd burglar enters) is calculated by
x/2 – 1/2 = 3
Therefore x = 7

Similarly after 2nd burglar leaves we have 7 bread left
Hence number of bread after 1st burglar leaves (or before 2nd burglar enters) is calculated by
x/2 – 1/2 = 7
Therefore x = 15

Similarly after 1st burglar leaves we have 15 bread left
Hence number of bread before 1st burglar enters is calculated by
x/2 – 1/2 = 15
Therefore x = 31·····························Ans

a, d, i, p, ? what is the next term.
q
r
s
T

Y

a=1*1
d=2*2
i=3*3
p=4*4
next will be
5*5=25=Y

9, 4, 16, 6, 36, 21, 441, 421, ?
277240
277230
277241
277250

if this series is divided into series of pairs we can get
9,(4,16),(6,36),(21,441)
4*4=16
6*6=36
21*21=441
so in pair every second no is square of the first no
now in series every first no of a pais is determined by reducing a multiple of 5 from the second no of previous series such as
9-(5*1)=4
16-(5*2)=6
36-(5*3)=21
441-(5*4)=421

14,28,20,40,32,64,?
52
56
96
128