SNAP 2009

Let the quantity of tea worth Rs.25 be x kg.
Now as per given
(25x+30×30)×110/100=30(30+x)
(275x+9900)=(9000+300x)
300x−275x=900=x=900/25
Therefore, x = 36 kg.

Total number of students = 480

Percentage of total students passed

= 75 % of total student = 75×480/100=360student
Now, using the condition from the question,
Let the number of boys be x
Then, 70% of x + 85% of (480 - x) = 360
=> 70×x/100+85×(480−x)/100=360
=> 70x - 85x + 40800 = 36000
=> 40800 - 36000 = 85x - 70x
=> 4800 = 15x
=> x=4800/15=320
There are 320 boys who have appeared for the examination.

300 gms of solution has 40% salt 
so the amount of salt = 300(0.4) = 120 
Now let salt added be x 
So we get 120+x/300+x = 1/2
we get 240+2x =300+x
we get x= 60

Lower number is eight times the upper number, i.e.

6 $$\times\ $$ 8 = 48

12 $$\times\ $$ 8 = 96

y $$\times\ $$ 8 = 192

y = $$\frac{192}{8}$$ = 24

Answer is option D.

Images are shifting by one position to the left side. In each step, there are two shaded and one unshaded image. Therefore, shaded triangle with index C at upper vertex will fill the blank.

Image 6 is suitable image and answer is option B.

Let us observe the digits given to us, we have 0,1,2,3,4 and 5, to make a five digit number we only need 5 digits out of the given 6 digits,
Case 1: Using digits 0,1,2,4 and 5.
The number of ways in which we can arrange these 5 digits are ⇒4×4×3×2×1
⇒96
Case 2: Using the digits 1,2,3,4 and 5
The number of ways in which we can arrange these 5 digits are ⇒5×4×3×2×1
⇒120
Therefore, the total number of cases =96+120=216

It is given that in the initial 25 days, the work done by 105 men was $$(2/5)^{th}$$ of the total work.
Thus, the total work left is $$(3/5)^{th}$$ of the total work.
MDH/W is constant 
so we get 
$$\frac{105\times\ 25\times\ 8}{\frac{2W}{5}}=\frac{M\times\ 25\times\ 9}{\frac{3W}{5}}$$
We get M =140
So additional men = 140-105 = 35

Let the total work be 24 units 
So we can say work done by A will be +3 units / day 
and work done by B will be -8 units/day
Now as per the given condition :
work done by A in 4 days = 3(4) 12 units
Now, both worked for 2  days:
so work done will be (3-8)(2) = -10 units
Now the net work will be 12-10 = 2 units
Now remaining work 24-2 =22 units
Therefore A will complete this in 22/3 = $$7\frac{1}{3}$$ days

Let the first term of the geometric progression be a and common ratio be r 
Now as per given condition :
$$a+ar= a(1+r) =12$$  (1)
$$ar^2+ar^3 = 48$$ (2)
we get $$ar^2(1+r) =48$$  (3)
Dividing (1) and (3)
we get :
$$1/r^2 = 1/4$$
we get $$1/r = 1/2  or -1/2 $$
but since terms are alternatively positive and negative
we get $$r = -2$$
Substituting
we get $$a(-1) =12 $$
we get $$a=-12$$

We are given that :
The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80
Now taking mean 
we get $$30 = a+b+8+5+10$$
so we get $$a+b = 7$$   (1)
Now variance = 6.80
so $$6.80 =( (a-6)^2 +(b-6)^2+(8-6)^2+(5-6)^2+(10-6)^2)/5$$
we get $$6.8(5) = (a-6)^2+(b-6)^2+1+16+4$$
So we get $$13 = (a-6)^2 +(b-6)^2$$
$$a=7-b$$
we get $$13 = (1-b)^2+(b-6)^2$$
we get $$b =4$$
and so $$a= 3$$