Page 311 - Start Up Mathematics_8 (Non CCE)

Solution: Let’s consider a 3-digit number 325.

3 2 5 4 2 2 5 2 9 5 7 5
¥ 1 3 ¥ 7 ¥ 1 1
9 7 5 2 9 5 7 5 2 9 5 7 5
3 2 5 X 2 9 5 7 5 X
4 2 2 5 3 2 5 3 2 5

So, 325 × 13 × 7 × 11 = 3,25,325
Explanation:
13 × 7 × 11 = 1,001
Let abc be a 3-digit number.

abc × 13 × 7 × 11 = abc × 1001 = abc × (1000 + 1) = abc × 1000 + abc

= abc000 + abc = abcabc
Example 10: Fill in the numbers from 1 to 6 (without repetition) so that each side
of the magic triangle adds upto 12.
Solution: 4 Place the largest numbers, i.e., 4, 5 and
6 at the three corners of the triangle.
Now 4 + 5 = 9, 4 + 6 = 10 and 5 + 6 = 11.
2 3

\ By placing 3 between 4 and 5, 2 between 4 and 6 and 1 between
6 1 5
5 and 6, we get the desired magic triangle.
Example 11: There are 10 hidden mines in the grid shown below. The
numbers in various squares indicate the total number of 2 0 1
mines hidden in 8 squares around the numbered square. 3 1 2
The numbered squares do not have any mines. Can you 1 0 1
detect the 10 mines? 2 1 0
Solution: 1 0 1
M 2 M × 0 × 1 × 1 1 1 1
× 3 × 1 × 2 M × 1 1 1 2 2
M 1 0 × 1 M × × 1 1
× 2 × × × × 1 0 M —mines
M 1 × × 0 × 1 × Riddle
1 1 × 1 × × 1 M I am as much more than 8, as
× 1 × 0 1 1 2 2
I am less than 52. What am I?
M × 1 × 1 × M ×

EXERCISE 19.3
1. Solve the following:
(a) X Y (b) 1 B (c) QP + QP = TQ
+ 3 7 ¥ B
9 X 9 B

(d) 2 P Q (e) R S 7 (f) A B
+ P Q 1 + 7 R S ¥ 4
Q 1 8 9 8 R C A B

303