Let CP = 100,
42 % increase => SP = 142
10 % discount in SP => ((142 x 10)/100) = 14.2
So 1st SP = (142 - 14.2) = 127.8, again 12 % discount in 1st SP ((127.8 x 11)/100) = 14.058
2nd SP = (127.8 - 14.058) = 113.742,
So finally CP = 100, SP = 113.742, => gain = 13.742%.

Let the Father be x years when he died

Average Age 10 years ago be A

Total Age 10 years ago = 6*A

Total Age after 10 years(Just before father's Death) = 6A + 6*10 = 6A + 60

Father Died and Baby was born => the Total number of people in the family is Same (6)
Baby born today so age of baby = 0

(6A +60 - x)/6 = 6A/6
=> A + 10 -(x/6) = A
=> x/6 = 10
=> x = 60

Therefore we can conclude that the father was 60 years old when he died.

If a number to be divisile by 88, it should be divisible by both "8" and "11"

Check for '8' :
For a number to be divisible by "8", the last 3-digit should be divisible by "8"
Here 72x23y --> last 3-digit is '23y'
So y=2 [ (i.e) 232 is absolutely divisible by "8"]

Chech for '11' :
For a number to be divisible by "11" , sum of odd digits - sum of even digits should be divisible by "11"
(7 + x + 3) - (2 + 2 + y)
(7 + x + 3) - (2 + 2 + 2)
(10 + x) - 6 should be divisible by "11"
for x = 7
=> 17 - 6 = 11 [ which is absolutely divisible by "11"]

So x = 7 , y= 2.

1st supporter recieve 440 calls
2nd supporter recieve 360 calls
3rd supporter recieve 300 calls

So total calls = 1100 calls ;
Calls this month= 1500
So remaining calls to be distributed is 400

So Now Ratio 1st:2nd:3rd ==> 440:360:300
=> 22:18:15

Now No. of More Calls 2nd supporter will get => [18/(22+18+15)] x 400
=> (18/55) x 400
=> 131 Calls
So 131 more Calls than last month.

Average speed=total distance/total time
Let distance to store be  K
then, total time =(K/20)+(K/30)=K/12
and, total time =(2K)
so average speed= 2K / (K/12) = 24kmph.

Let the average expenditure per head be Rs. p
Now, the expenditure of the mess for old students is Rs. 44p
After joining of 15 more students, the average expenditure per head is decreased by Rs. 3 => p-3
Here, given the expenditure of the mess for (44+15 = 59) students is increased by Rs. 33
Therefore, 59(p-3) = 44p + 33
59p - 177 = 44p + 33
15p = 210
=> p = 14
Thus, the expenditure of the mess for old students is Rs. 44p = 44 x 14 = Rs. 616.

Let the two no's be a and b;
Given product of the no's is p = ab;
If the each nos is increased by 2 then the new product will be
(a+2)(b+2) = ab + 2a + 2b + 4
= ab + 2(a+b) + 4
= p + 2(a+b) + 4
Hence the new product is (p+4) times greater than twice the sum of the two original numbers.

Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8.

Seats for Mathematics, Physics and Biology in a school are 5x, 7x and 8x.

There is a proposal to increase these seats by 40%, 50% and 75% respectively.

Then Seats for Mathematics, Physics and Biology in a school will be

1.4 X 5x, 1.5 X 7x and 1.75 X 8x and

ratio will be 7 : 10.5 : 14  => 2:3:4.