Quantitative Aptitude - Arithmetic Ability Questions

we can treat every two consecutive terms as one.
So, we will have a total of 100 terms of the nature:
(2 + 5) + (7 + 6) + (12 + 7).... => 7, 13, 19,....

We know the sum of n terms  nn+12

Now, a= 7, d=6 and n=100
Hence the sum of the given series is
S= 100/2 x[2 x 7 + 99 x 6]
=> 50[608]
=> 30,400.

K+2 M+1 N-2 L,  P+2 R+1 S-2 Q,   V+1 W+2 Y+1 Z,   J+2 L+1 M-2 K  and  W+2 Y+1 Z-2 X.

Except VWYZ, all the other groups follow similar pattern.

Let A's age 10 years ago be x years . Then B's age 10 years ago be 2x years

 Total of their present ages = x + 10 + 2x + 10

= 3x+20

= 3 * 5 + 20 = 35 years

The given 3b - 7 < 32 can be solved as

3b - 7 < 32

3b < 39

b < 13.

4 persons can be selected from 9 in 9C4 ways =126

Fvaourable events =4C2*5C2 =60

So,required probability = 60/126 = 10/21

$$\begin{gathered} 0.01+0.0064=0.01+6410000 \\ =0.01+8100 =0.01+0.08=0.3 \end{gathered}$$

Let the required weight be x kg.

Less weight, Less cost (Direct Proportion)

 250 : 200 :: 60 : x   =>    250 x x = (200 x 60)

=>

=> x = 48.