Is it possible to get ten whether you remove one from nine or eleven? How?
It is possible if you write numbers in their roman forms.
11 = XI
If I (1) is removed, it becomes X = 10
9 = IX
If I (1) is removed, it becomes X = 10
How can you make the number one disappear by adding something to it?
By adding letter "G" to it. One becomes "Gone".
Can you solve the below algebraic mathematical equation?
(J+O+I+N+T)3 = JOINT
(1+9+6+8+3)*(1+9+6+8+3)*(1+9+6+8+3) = 19683
J = 1, O = 9, I = 6, N = 8, T = 3
What is the value of √16% ?
40%
Explanation: br>√16%
=√14/√100
=4/10
=40/100=40%
Can you complete below number series by replacing "?" with the correct number.
10 # 10 # 20 # ? # 110 # 300 # 930
45
Pattern goes like
10
10 * 0.5 + 5 => 10
10 * 1.0
+ 10 => 20
20 * 1.5 + 15 => 45(Ans)
45 * 2.0 + 20 => 110
110 * 2.5 + 25 => 300
300 * 3.0 + 30 => 930
A mother bought three dress for her triplets daughters(one for each) and put the dresses in the dark. One by one the girls come and pick a dress.
What is the probability that no girl will choose her own dress?
1/3
Explanation
Assuming D1 is the dress for Sister1, D2 is the dress for Sister2 and D3 is the dress for Sister3.
Therefore the total number of cases are illustrated below.
Sister1 Sister2 Sister3
D1 D2 D3
D1 D3 D2
D2 D1 D3
D2 D3 D1 ..... (1)
D3 D1 D2 .... (2)
D3 D2 D1
In both steps (1) & (2), no one gets the correct Dress.
Therefore probability that no sister gets the correct dress is 2/6 = 1/3
By using all numbers, i.e. 123456789 and subtraction/addition, operators number 100 can be formed in many ways.
Example 98 + 7 + 6 - 5 - 4 - 3 + 2 - 1 = 100
But if we add a condition that use of the number 32 is the must. Then there are limited solutions.
One of such solution is: 9 - 8 + 76 + 54 - 32 + 1 = 100
Can you tell any other solution?
9 - 8 + 7 + 65 - 4 + 32 - 1 = 100
9 - 8 + 76 + 54 - 32 + 1 = 100
Other ways of making number 100 (withiout using number 32)
98 + 7 + 6 - 5 - 4 - 3 + 2 - 1 = 100
98 - 7 - 6 - 5 - 4 + 3 + 21 = 100
9 - 8 + 76 - 5 + 4 + 3 + 21 = 100
98 - 7 + 6 + 5 + 4 - 3 - 2 - 1 = 100
98 + 7 - 6 + 5 - 4 + 3 - 2 - 1 = 100
98 + 7 - 6 + 5 - 4 - 3 + 2 + 1 = 100
98 - 7 + 6 + 5 - 4 + 3 - 2 + 1 = 100
98 - 7 + 6 - 5 + 4 + 3 + 2 - 1 = 100
98 + 7 - 6 - 5 + 4 + 3 - 2 + 1 = 100
98 - 7 - 6 + 5 + 4 + 3 + 2 + 1 = 100
9 + 8 + 76 + 5 + 4 - 3 + 2 - 1 = 100
9 + 8 + 76 + 5 - 4 + 3 + 2 + 1 = 100
You need to fill number in the bricks in the image below such that the top brick is sum of two brick below it.
It can be solved as shown below.
Can you solve the equation by finding the value ofA) HorseB) Cowboy bootC) Horseshoe
Horse = 10, Cowboy-Boot = 1 ,HorseShoe= 2
= Step1 =
3Horse = 30
=> Horse = 10 .....(A)
= Step2 =
1Horse + 2HorseShoe + 2HorseShoe = 18
10 + 4HorseShoe =18
HorseShoe = 2 .....(C)
= Step3 =
2HorseShoe - 2Cowboy-Boot = 2
4 - 2Cowboy-Boot = 2
=> Cowboy-Boot = 1 ... (B)
= Step4 =
Cowboy-Boot+Horse*HorseShoe = 1+10*2 =>21
What will be the best approach to finding all the prime numbers less than 75 that leave an odd reminder when we divide them with 5?
Except 2 and 3, all the prime numbers are in a form of (6k +1) or (6k -1)
Therefore, you can use the formula to write down the prime numbers as:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 61, 67, 71, 73
Now, you can separate numbers as per the requirement that are:
3, 11, 13, 23, 31, 41, 43, 53, 61, 71 and 73