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

Answer=66 sec

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

Answer 421 * 421

14,28,20,40,32,64,?

52

56

96

128

Answer

56

it have two sequence.

1st sequence is 14,20,32,

2nd sequence is 28,40,64,

in this, 1st sequence increase like +6,+12,+24,+48….

2nd sequence increase like +12,+24,+48……..