Total number of ways = ( select 3 boys from group of 4 boys ) * ( 2 girls from group of 4 girls )
= ( ⁴C₃ ) * ( ⁴C₂ )
= 24

• Total number of letters = 6 and R is repeated twice.
• Required arrangements = (6! / 2!) = 360

1800 = 2³ * 3² * 5² then
⁵C₂ * ⁴C₂ * ⁴C₂ = 360

Here $ represent the – operator so X – X = 0
First condition x $ (y $ z) = x – (y – z) = x – y + z.
It can be written as x $ y + z
2012 – 0 + 2012 – 1912 = 2112

The LCM for 2088 is (2 * 2) * 2 * (3 * 3) * 29.
In order to make it perfect square multiply 2 * 29 the answer is 58

(m+n)/12 remainder is 8 so m + n = 8 + 12 = 20
(m-n)/12 remainder is 6 so m – n = 6 + 12 = 18
Solving this m = 19,n = 1
( m * n ) / 6 = ( 19 * 1 ) / 6 remainder is 1.

The digital sum of number is (1+2+3+4+5+6+7+8+9+0+1+2+3+4+5+6+7+8+9) = 90 = (9+0) = 9 So it is divisible by 3 or 9
The given divisor 6561 = 9⁴
So 9²⁴/9⁴ Reminder is 0.

• If we remove the remainder from number, 5678-35 = 5643
• 5643 should be divisible by 460, means the unit digit must be 5 if we add any number from options.
• The only number 797 can give the unit digit 5, So answer is 797.

Let the number = N.
Now N = 5K
K = 3L + 2
L = 2M + 1
K = 3 (2M + 1) + 2
= 6M + 5
N = 5 (6M + 5)
= 30 M + 25.

For M = 0 we get the least number as 25. Now when 25 is divided by 2, we get 12 as quotient and 1 as remainder.
When 12 is divided by 3 we get 4 as quotient, and 0 as remainder. When 4 is divided by 5 we get 4 as remainder.

• A leap year there are 366 days means 52 weeks and 2 days. so already we have 52 fridays.
• Now the rest two days can be (sun, mon), (mon,tues), (tues, wed), (wed, thrus), (thrus, fri), (fri, sat), (sat, sun) so, the probability of 53 fridays = 2/7.

• If a month has 31 days, and it starts with sunday, Then Sundays, Mondays, tuesdays are 5 for that month. If this month starts with monday, then mondays, tuesdays, and wednesdays are 5 and remaining days are 4 each. so this month start with Monday

Given A = x³y² , B = xy³
A = x*x*x*y*y
B = x*y*y*y
The highest common factor between A and B is xy².

Y = 4Z — (i)
2X = 3Y => Y = 2X/3 — (ii)

To get the ratio of X, Y, Z we can represent Y and Z in terms of X.

From (i) & (ii) ,

2X/3 = 4Z
Z = X/6
So X : Y : Z = X : 2X/3 : X/6
= 6 : 4 : 1
So Y’s share in Rs. 16500 = (4/11) * 16500 =Rs. 6000

Let number of pens sold on tuesday be x given that the percentage increase in the number of pens sold by a trader from Tuesday to Wednesday is same as the percentage decrease in the number of pens sold from Monday to Tuesday.

Percentage increase in sales of pen from tuesday to wednesday = Percentage decrease in sales from monday to tuesday = 20% then on wednesday trader sold 16 more pens than those sold on tuesday.

So number of pens sold on wednesday = x + 16 and this is 20% more than number of pens sold on tuesday
[1 + (20/100) ] * x = x + 16
6x/5 = x + 16
x = 80
So number of pens sold on tuesday x = 80.

Party 2 and 3 Got 12 and 43% Respectively = 55% as a whole.
55% of 16000 = 8800 and Party 1 got 45% Vote = 45% of 16000 = 7200.
Margin of Votes = 8800 – 7200 = 1600

The fruit content in both the fresh fruit and dry fruit is the same.
Given, fresh fruit has 68% water. So remaining 32% is fruit content.
weight of fresh fruits is 100kg Dry fruit has 20% water. So remaining 80% is fruit content.
Let weight if dry fruit be y kg.
fruit % in freshfruit = fruit % in dryfruit
(32/100) * 100 = (80/100 ) * y
we get y = 40 kg.

Let the total vote is x. then 30% of x = (30/100) * x=1200, or x = (1200 * 100) / 30 or x = 4000
Now, 30% of votes already given…so other 70% of x votes = (70/100) * 4000 = 2800 votes.
Now other needed votes = (70 – 30)% = 40% votes.
So 40% votes of x = (40/100) * 4000 = 1600;
Now he have needed only % of votes = (100 * 1600) / 2800 = 160000 / 2800 = 57.14285714%

40,000 * 3/4 * 3/4 * 3/4 = 16875

Paper A : (18/70) * 100 ~= 25.71%
Paper B : (14/30) * 100 ~= 46.66%.
Therefore the solution is Paper B