25
x:y=18:11
let a = 9 * 9 * 9;
let b = 12 * 12 * 12;
let c = 15 * 15 * 15;
sum = a + b + c;
edge = Math.cbrt(sum);
console.log(edge);
Ans = 8
Use simple box method
[1][2][2][2]= 1x2x2x2 =8
Logic >
once place have 2 number (2,6)
Tens place have also 2 number (7,3)
So that number is divisible by 4
Now, 2 numbers fixed at once and tens place left 2 numbers which is choice at hundredth place
Now 3 number fix left 1 number which is placed at thousands place
Total number form is 4×4 = 16
But divisible by 4 is 1x2x2x2 =8
(b) 16.66%
Earlier for ₹x we could purchase y gm of sugar.
Now we pay ₹1.2x for y gm of sugar
(As there was an increase in price so, x + 20%x = 1.2x)
At current rates for ₹x you can purchase y/1.2 gm of sugar
So the reduced consumption is y-(y/1.2)
Percentage change = (reduced consumption/ original consumption ) *100
That is (0.2/1.2) *100 = 16.66% (approx)
1 1 2+4
2 2 1+4
3 1+2 4
4 4 1+2
5 4+1 2
6 4+2 1
7 4+2+1 0
40
simple logic, Check Option Lets father present age = 40 so it full fill all condition like his son age become 10 and after five years it will become 15 and father age is 45 and it full fill second condition that fathers age is thrice as old as the son after 5 years
20&130
accountant trainee at thenational board of accountants and auditors
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13. One group of each = 2+3+5 = 10 and 1 remainder for each = 1+1+1 = 3. 10+3 = 13
120 (5×4!)
( e ) None of these
Let A, B and C be the three 6-faced dice.
Then, according to the question,
Since two dices has to be equal, that value can be any of the 6 faces, i.e., 6C1 cases.
Now for each case, 2 equal dices can be selected from 3 dices in 3C2 i.e., 3 ways.
And for each of the above, the third dice can have any of the 5 remaining faces
The possible outcomes are P(A)=61,P(B)=61,P(C)=65,P(A)=61,P(B)=65,P(C)=61 and P(A)=65,P(B)=61,P(C)=61
Hence the required probability = 61×61×65×6×3=21690=125
Barber