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Answer :  b)37
Detailed Explanation:

50 × 51 × 52 × 53 ×…× 200 => 200! - 49!

9 => 3^2
[9 is not a prime factor... So we have to find the highest power of 3... From that we can calculate power of 9...]
No. of 3's in 200! is 
200/3^1 + 200/3^2 + 200/3^3 + 200^3^4 = 66+22+7+2 = 97
No. of 3's in 49! is 
49/3^1 + 49/3^2 + 49/3^3 = 16+5+1 = 22
No. of 3's in 50 × 51 × 52 × 53 ×…× 200 is (97-22) = 75
We have to find the highest power of 9 not 3 ... So 75/2 = 37

Ans : 37


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Answer :  d)19
Detailed Explanation:

 
Let the correct time to complete the journey be x min.
Distance covered in (x + 11) minutes at 40 kmph = Distance covered in (x + 5) min. at 50 kmph
=> (x + 11)*40 = ( x + 5)* 50
 x = 19 min.




Most Importantly

For such questions, You should never waste your time in converting Speed from Kilometer per hour to kilo meter per minute or time from minutes to hours as the same gets applied in L.H.S. and R.H.S. and gets cancelled.

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Answer :  c)108710
Detailed Explanation:
Perfect squares between 4000 and 7000 lies between:
64^2,65^2,66^2.....83^2

Sum = 64^2 + 65^2 + ......82^2 + 83^2
= (1^2+2^2+3^2+....+83^2)-(1^2+2^2+....+63^2)
Apply formula for sum of squares = n*(n+1)*(2n+1)/6
= (83*(83+1)*(2*83+1)/6)-(63*(63+1)*(2*63+1)/6)
=194054 - 85344
=108710


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 Answer :  d)1/5
Detailed Explanation:
Simplest way to solve this is:
As per the divisibility rules a number is divisible by 4 if its last 2 digits are divisible by 4
1. Do not consider the first 3 digits
2. Number of ways to select last 2 digits using 1,2,3,4 & 5 numbers is => 5*4 = 20
Out of these 20 numbers, the one's divisible by 4 are : ***12, ***24, ***32, ***52
Count of such numbers = 4
Probability = 4/20 = 1/5

The detailed approach to solve this is:
The total number of possibilities is 5*4*3*2*1 = 120
for a 5 digit number to be divisible by 4 the last two digits have to be divisible by 4
this will happen when the last two digits are 12, 24, 32, 52

So we have ___,___,___,1,2
We have 3 numbers available so that would mean 3*2*1 possibilities

6 numbers with a 12 at the end
6 numbers with a 24 at the end
6 numbers with a 32 at the end
6 numbers with a 52 at the end
total 5 digit numbers divisible by 4 equals 6+6+6+6 = 24
probability = 24/120 =1/5

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 Answer :  d)25
Detailed Explanation:
Let Cost price of 1 article be Re.1.
Therefore, Cost price of 15 articles = Rs.15.
Selling price of 15 articles = (15/12)*15 = 75/4 = 18.75

Profit = Selling price - Cost price

= 18.75 - 15 = 3.75

Percentage of profit = (profit/cost price)*100.

= (3.75/15)*100 = 25% Profit

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Answer :  c)18
Detailed Explanation:
Let the number of currency 1 Miso, 10 Misos and 50 Misos be x, y and z respectively.
x+10y+50z=107
Now the possible values of z could be 0, 1 and 2.
For z=0: x+10y=107
Number of integral pairs of values of x and y that satisfy the equation:
x+10y=107 will be 11.
These values of x and y in that order are:
(7,10);(17,9);(27,8)…(107,0)
For z=1: x+10y=57
Number of integral pairs of values of x and y that satisfy the equation: 
x+10y=57 will be 6.
These values of x and y in that order are: 
(7,5);(17,4);(27,3);(37,2);(47,1) and (57,0)
For z=2: x+10y=7
There is only one integer value of x and y that satisfies the equation: 
x+10y=7 in that order is (7,0)
Therefore total number of ways in which you can pay a bill of 107 Misos:
=11+6+1= 18

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Answer :  1) 9
Detailed Explanation:
For n non parallel lines, maximum number of intersection points = n*(n-1)/2
Here n=5
max points of intersection = 5*4/2=10
Minimum number of intersection pointts = 1 (all lines intersecting at common point)
Difference = 10-1=9
Key Learning: 

Remember the formula:
Maximum number of intersection points for n non parallel lines = n*(n-1)/2

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Answer :  2) 4
Detailed Explanation:

Simple Interest Compound Interest Formula's for revision:

Assume he invested Rs.100. 
Plan A(C.I.), Interest is compunded. 
Looking at the options its obvious that we do not need to calculate interests for 1st and 2nd years.
For 3rd year C.I Total Amount is: 100 * (1+10/100)^3 = 133.1
For 4th year C.I Total Amount is: 100 * (1+10/100)^4 = 146.4

For 5th year C.I Total Amount is: 100 * (1+10/100)^5 = 161.05

Now, lets calculate plan B(S.I.) 

3years : 100+ (100*3*12)/100 => 136 => Amount = Rs.136 which is greater than the 3rd  years' value of plan A
4years : 100+ (100*4*12)/100 => 148 => Amount = Rs.148 which is greater than the 4th  years value of plan A
5years : 100 + (100*5*12)/100 =160 => Amount = Rs.160 which is less than the 5th  year's value of plan A
So, plan B is a better investment for four years.

Note: In such cases SI should always be calculated with mental calculations.
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Answer :  4) 38664

Detailed Explanation:
From these four digits '0' cannot be at first place to form four digit number.
So we have 1, 2, or 3 can be at thousand's place
Keeping one digit at thousand place(example 1XXX or 2XXX or 3XXX ) we can have 6 different four digit number. For example if 1 is fixed at thousands place following 6 four digit numbers are possible = 1023, 1032,1203,1230,1302,1320 . Similarly for  2 and 3 fixed at thousand place.
So that means 1*1000*6 + 2*1000*6 + 3*1000* 6 = 36000

Now come to hundred's place we can have all four digit at this place , '0' will not contribute any thing to sum, so in all four digits number 1,2,3 will be at 4 times on same place which means: 100*4 + 200* 4 + 300 * 4= 2400

Now come to tenth's place ...similar as hundred
So 10*4 + 20*4 + 30*4= 240

Similarly for unit place
1*4 + 2*4 + 3*4= 24
At last if you add whole the total sum of all possible four digit number formed is 38664
 

Key Learning:
1. While solving such questions, fix digits at each position and find out all combinations of numbers at other positions.
2. To understand the logic above easily, you should try to solve & master a simpler question such as sum of all the three digit numbers formed by 1,2,3.


All possible 3 digit numbers are:
123,132,213,231,312,321. Now the slowest way to solve this question is to identify all such numbers and add all these numbers to get the answers :).
Smart way to solve this would be based on the fact that 123= 1*100+2*20+3*1
Step-1 :Fix numbers at at each place (100th place, 10th place & unit place) one by one and then finding all different numbers which are possible.
Fixing a number at 100th place there are 2 different numbers which could be formed by arranging numbers at 10th & unit place. Similarly on fixing numbers at 10th place and unit place there are 2 different numbers which could be formed by arranging numbers at 100th & unit place and 100th & 10th place respectively.
Step-2 :Find sum at each place of all digits (100th place, 10th place & unit place)
So for 100th place  = (1*100+2*100+3*100)*2 = 6*100*2 = 1200
Similarly for 10th place = (1*10+2*10+3*10)*2 = 6*10*2 = 120
Similarly for unit place = (1*1+2*1+3*1)*2 = 6*1*2 = 12
Step-3 : Do sum of all numbers = 1200 + 120 + 12 = 1332

 


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Answer :  4) More than 15

Detailed Explanation:
We have 1+2+3++n=n(n+1)2.
The numbers obtained from such sums are called Triangular numbers.

As the child's answer was 575, the correct answer must be greater than 575 (as he missed a number). 
The first triangular number greater than 575 is 595 corresponding to n=34. 
If this was the sum he was trying to evaluate, the child must have missed 20=595−575. 
However, if the correct answer was the next triangular number, 630, corresponding to n=35, the only way the child could have obtained 575 is if he missed 55=630−575, but 55 is not in the list of numbers 1,…,35 so this is not possible.

Hence n=34 and number which he has missed is 20


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Answer : 1
Detailed Explanation:

We know that (ab)^n = (a^n)*(b^n)
Thus, 30^2720 = (3*10)^2720 = (3^2720)*(10^2720)
10 raised to power of any number will give all zeroes next to 1 like 10, 100, 1000, 10000, etc; so we have just to find the unit digit of 3^2720

Unit digit of 3 raised to any natural number repeats after every 4 intervals; 

Hence divide 2720 by 4 to see what is left; it is 0 off course; means we have just to find the unit digit of 3^4

Now 3^4 = 81
So the required digit = 1 (option '1')

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Answer : 0
Detailed Explanation:


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Answer : 6
Detailed Explanation:

Let 10x + y be a two digit number, where x and y are positive single digit integers
and x > 0.
Its reverse = 10y + x
Now as per question, (10y + x) – (10x + y) = 18
∴ 9(y – x) = 18
∴ y – x = 2

Thus y and x can be (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9)
∴ Other than 13, there are 6 such numbers.
Hence, option 2.

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Answer : 7/8
Detailed Explanation:


Each of the three numbers could either be even or odd.
Total number of combinations of such numbers = 2^3 = 8
There is only one combination where multiple of 3 numbers would be odd is when K1, K2 and K3 all are Odd numbers.
For rest all 7 combinations the multiple of 3 number would always be even.
Hence probability = 7/8.   


Key Learning:
When any number is multiplied by an even number, the result is always a even number




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