Page 42 - SM inner class 8.cdr

E xercise 5.1

1. Express the follow ing numbers in gener al ised form:

(i) 416 (ii) 732 (iii) 809
2. Express the follow ing in usual form:
(i) 10 2 10 5 8´ + ´ + (ii) 100 9 10 0 4´ + ´ +
(iii) 100 6 10 6 6´ + ´ +
3. Fill in the blanks.
(i) 100 3 10´ + ´........... + 7 357 (ii) 100 4 10 5 1´ + ´ + =.............
=
=
+
´
(iii) 100 ´............. + 10 3 7 737 (iv) 100 ´....... + 10 + = pqr
q
.
r
(v) 100.x + 10.y + = .............
z
4. A two digit number exceeds the sum of the digits of that number by 18 . If the digit at the units place is
double the digit in the tens place, what is the number?
5. The differ ence between a two-digit number and the number obtained by inter chang ing the position of
its digits is 63. What is the differ ence between the two digits of that number?
6. In a two digit number, the digit in the units place is four times the digit at the tens place and sum of the
digit is equal to 10 . What is the number?

Math Puzzles or Cryptarithms Solution: To answer the above problem, we
know that 2Y = X, thus X must be even number.
Cryptarithms are a type of mathematical
We also know that 2X = Z, thus Z is also even
puzzle in which the digits are replaced by
number. Since the digits are limited from 0 to 9,
symbols (typically letters of the alphabet).
then we can simply try Y = 2, this will give us
The solution involves finding the original X = 4 and Z = 8. Thus, {X = 4 ,Y = 2 and Z = 8} is
digits to form a number. Creating and solving one of the solutions to the puzzle above.
such puzzles is known as cryptarithmetic. 42

General Rules: + 42
84
1. Each alphabet takes only one number
from 0 to 9 uniquely. Example 7. Find M and Y in the addition
2. Two single digit numbers sum can be Y + Y + Y = MY
maximum 19 with carryover. So carry Solution: Y + Y + Y = MY
over in problems of two number addition 3Y =10M + Y
is always 1. 2Y =10M

3. Try to solve left most digit in the given Y = 5 M (i.e. Y is divisible by 5. Hence Y = 0 or 5)
problem. From above, if Y = 0, Y + Y + Y = 0 + 0 + 0 = 0,
4. If a ´ b = kb, then the following are the M = 0
=
possibilities if Y = 5, Y + Y + Y = 5 + 5 + 5 15,MY =15
(3 ´ 5 = 15; 7 ´ 5 = 35; 9 ´ 5 = 45) or (2 ´ 6 = Hence M =1, Y = 5
12; 4 ´ 6 = 24; 8 ´ 6 = 48) Example 8. In 5A1 - 23 A = 325, 5A1 and 23A
Example 6. Find the value of X,Y and Z in are three digit numbers, then find A.
given problem. Solution: First we expand these number as
XY
-
+
(500 10+ A + 1 ) (200 30 + A ) = 325
+ XY
+
-
-
500 200 10 - A + 1 30 = 325
A
ZX
300 9 - 29 = 325 (By solving)
+
A
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Mathematics In Focus - 8