Exams

Let 'K' be the total number of sweets.
Given total number of students = 112
If sweets are distributed among 112 children,
Let number of sweets each student gets = 'L'

=> K/112 = L ....(1)
But on that day students absent = 32 => remaining = 112 - 32 = 80
Then, each student gets '6' sweets extra.

=> K/80 = L + 6 ....(2)
from (1) K = 112L substitute in (2), we get
112L = 80L + 480
32L = 480
L = 15

Therefore, 15 sweets were each student originally supposed to get.

Distance = 70 x 1 ½ = 105 km

Relative Speed = 85 – 70 = 15

Time = 105/15 = 7 hrs

4:30 + 7 hrs = 11.30 p.m.

The main memory in computers is where the programs and data are kept when the processor is actively using them.

The main storage region of a computer is of two kinds. They are ::

RAM -- Random Access Memory

ROM -- Read Only Memory.

The main memory is defined as the central storage in a computer.

First is the type of houses most commonly found in the second.

Let varun's age = x yrs.

From the given conditions,

x+1  =  2(x-12)

x = 25 years.

Length of the first carpet = (1.44)(6) = 8.64 cm

Area of the second carpet = 8.64(1 + 40/100) 6 (1 + 25/100)

= 51.84(1.4)(5/4) sq m = (12.96)(7) sq m

Cost of the second carpet = (45)(12.96 x 7) = 315 (13 - 0.04) = 4095 - 12.6 = Rs. 4082.40