Class 8 Exam > Class 8 Notes > Mathematics (Maths) Class 8 > RS Aggarwal Solutions: Exercise 5C - Playing With Numbers
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Page 2
1 is carried over.
(1 +5 +8) = 14
1 is carried over.
? B = 4
and C = 1
? A = 6, B = 4 and C = 1
Question:23
Replace A, B, C by suitable numerals.
4 C B 6
+ 3 6 9 A
8 1 7 3
Solution:
A = 7, A +6 = 7 +6 = 13
1iscarriedover
(1 +B +9) = 17, or B = 7
1iscarriedover
A = 7, B = 7 and C = 4
1iscarriedover
? A = 7, B = 7 and C = 4
Question:24
Replace A, B, by suitable numerals.
A
+ A
+ A
B A
Solution:
A +A +A = A
with1beingcarriedover
This is satisfied if A
is equal to 5
.
When A = 5
:
A +A +A = 15
1iscarriedover
Or B = 1
? A = 5 and B = 1
Question:25
Replace A, B by suitable numerals.
6 A
- A B
3 7
Solution:
First look at the left column, which is:
6 -A = 3
This implies that the maximum value of A can be 3.
A = 3
... 1
The next column has the following:
A -B = 7
To reconcile this with equation 1
, borrowing is involved.
We know:
12 -5 = 7
? A = 2 and B = 5
Question:26
Replace A, B, C by suitable numerals.
C B 5
- 2 8 A
2 5 9
Solution:
5 -A = 9
Page 3
1 is carried over.
(1 +5 +8) = 14
1 is carried over.
? B = 4
and C = 1
? A = 6, B = 4 and C = 1
Question:23
Replace A, B, C by suitable numerals.
4 C B 6
+ 3 6 9 A
8 1 7 3
Solution:
A = 7, A +6 = 7 +6 = 13
1iscarriedover
(1 +B +9) = 17, or B = 7
1iscarriedover
A = 7, B = 7 and C = 4
1iscarriedover
? A = 7, B = 7 and C = 4
Question:24
Replace A, B, by suitable numerals.
A
+ A
+ A
B A
Solution:
A +A +A = A
with1beingcarriedover
This is satisfied if A
is equal to 5
.
When A = 5
:
A +A +A = 15
1iscarriedover
Or B = 1
? A = 5 and B = 1
Question:25
Replace A, B by suitable numerals.
6 A
- A B
3 7
Solution:
First look at the left column, which is:
6 -A = 3
This implies that the maximum value of A can be 3.
A = 3
... 1
The next column has the following:
A -B = 7
To reconcile this with equation 1
, borrowing is involved.
We know:
12 -5 = 7
? A = 2 and B = 5
Question:26
Replace A, B, C by suitable numerals.
C B 5
- 2 8 A
2 5 9
Solution:
5 -A = 9
This implies that 1 is borrowed.
We know:
15 -6 = 9
? A = 6
B -5 = 8
This implies that 1 is borrowed.
13 -5 = 8
But 1 has also been lent
?B = 4
C -2 = 2
This implies that 1 has been lent.
? C = 5
? A = 6, B = 4 and C = 5
Question:27
Replace A, B, C by suitable numerals.
A B
× 3
C A B
Solution:
(B ×3) = B Then, B can either be 0 or 5. If B is 5, then 1 will be carried. Then, A ×3 +1 = A will not be possible for any number. ? B = 0 A ×3 = A is possible for either 0 or 5. If we take A
Question:28
Replace A, B, C by suitable numerals.
A B
× B A
(B +1) C B
Solution:
A ×B = B ? A = 1
In the question:
First digit = B+1
Thus, 1 will be carried from 1+B
2
and becomes B +1
(B
2
-9) B.
? C = B
2
-1
Now, all B, B+1 and B
2
-9 are one digit number.
This condition is satisfied for B=3 or B=4.
For B< 3, B
2
-9 will be negative.
For B>3, B
2
-9 will become a two digit number.
For B=3 , C = 3
2
- 9 = 9-9 = 0
For B = 4, C = 4
2
-9 = 16-9 = 7
Required answer:
A=1, B=3, C = 0
or
A=1, B=4, C = 7
Question:29
Replace A, B, C by suitable numerals.
6) 5 A B (9 C
- 5 4
3 B
- 3 6
×
Solution:
(A -4) = 3 ? A = 7
?Also, 6 ×6 = 36 ? C = 6
36 -36 = 0 ? B = 6
? A = 7B = C = 6
Question:30
Find two numbers whose product is a 1-digit number and the sum is a 2-digit number.
Solution:
1 and 9
are two numbers, whose product is a single digit number.
? 1 ×9 = 9
Page 4
1 is carried over.
(1 +5 +8) = 14
1 is carried over.
? B = 4
and C = 1
? A = 6, B = 4 and C = 1
Question:23
Replace A, B, C by suitable numerals.
4 C B 6
+ 3 6 9 A
8 1 7 3
Solution:
A = 7, A +6 = 7 +6 = 13
1iscarriedover
(1 +B +9) = 17, or B = 7
1iscarriedover
A = 7, B = 7 and C = 4
1iscarriedover
? A = 7, B = 7 and C = 4
Question:24
Replace A, B, by suitable numerals.
A
+ A
+ A
B A
Solution:
A +A +A = A
with1beingcarriedover
This is satisfied if A
is equal to 5
.
When A = 5
:
A +A +A = 15
1iscarriedover
Or B = 1
? A = 5 and B = 1
Question:25
Replace A, B by suitable numerals.
6 A
- A B
3 7
Solution:
First look at the left column, which is:
6 -A = 3
This implies that the maximum value of A can be 3.
A = 3
... 1
The next column has the following:
A -B = 7
To reconcile this with equation 1
, borrowing is involved.
We know:
12 -5 = 7
? A = 2 and B = 5
Question:26
Replace A, B, C by suitable numerals.
C B 5
- 2 8 A
2 5 9
Solution:
5 -A = 9
This implies that 1 is borrowed.
We know:
15 -6 = 9
? A = 6
B -5 = 8
This implies that 1 is borrowed.
13 -5 = 8
But 1 has also been lent
?B = 4
C -2 = 2
This implies that 1 has been lent.
? C = 5
? A = 6, B = 4 and C = 5
Question:27
Replace A, B, C by suitable numerals.
A B
× 3
C A B
Solution:
(B ×3) = B Then, B can either be 0 or 5. If B is 5, then 1 will be carried. Then, A ×3 +1 = A will not be possible for any number. ? B = 0 A ×3 = A is possible for either 0 or 5. If we take A
Question:28
Replace A, B, C by suitable numerals.
A B
× B A
(B +1) C B
Solution:
A ×B = B ? A = 1
In the question:
First digit = B+1
Thus, 1 will be carried from 1+B
2
and becomes B +1
(B
2
-9) B.
? C = B
2
-1
Now, all B, B+1 and B
2
-9 are one digit number.
This condition is satisfied for B=3 or B=4.
For B< 3, B
2
-9 will be negative.
For B>3, B
2
-9 will become a two digit number.
For B=3 , C = 3
2
- 9 = 9-9 = 0
For B = 4, C = 4
2
-9 = 16-9 = 7
Required answer:
A=1, B=3, C = 0
or
A=1, B=4, C = 7
Question:29
Replace A, B, C by suitable numerals.
6) 5 A B (9 C
- 5 4
3 B
- 3 6
×
Solution:
(A -4) = 3 ? A = 7
?Also, 6 ×6 = 36 ? C = 6
36 -36 = 0 ? B = 6
? A = 7B = C = 6
Question:30
Find two numbers whose product is a 1-digit number and the sum is a 2-digit number.
Solution:
1 and 9
are two numbers, whose product is a single digit number.
? 1 ×9 = 9
Sum of the numbers is a two digit number.
? 1 +9 = 10
Question:31
Find three whole numbers whose product and sum are equal.
Solution:
The three whole numbers are 1, 2 and 3.
1 +2 +3 = 6 = 1 ×2 ×3
Question:32
complete the magic square given below, so that the sum of the numbers in each row or in each column or along each diagonal is 15.
6 1
5
Solution:
Taking the diagonal that starts with 6:
6 +5 +x = 15 ? x = 4
6 1
5
4
Now, taking the first row:
6 +1 +x = 15 ? x = 8
6 1 8
5
4
Taking the last column:
8 +x +4 = 15 ? x = 3
6 1 8
5 3
4
Taking the second column:
1 +5 +x = 15 ? x = 9
6 1 8
5 3
9 4
Taking the second row:
x +5 +3 = 15 ? x = 7
6 1 8
7 5 3
9 4
Taking the diagonal that begins with 8:
8 +5 +x = 15 ? x = 2
6 1 8
7 5 3
2 9 4
Question:33
Fill in the numbers from 1 to 6 without repetition, so that each side of the triangle adds up to 12.
Solution:
Page 5
1 is carried over.
(1 +5 +8) = 14
1 is carried over.
? B = 4
and C = 1
? A = 6, B = 4 and C = 1
Question:23
Replace A, B, C by suitable numerals.
4 C B 6
+ 3 6 9 A
8 1 7 3
Solution:
A = 7, A +6 = 7 +6 = 13
1iscarriedover
(1 +B +9) = 17, or B = 7
1iscarriedover
A = 7, B = 7 and C = 4
1iscarriedover
? A = 7, B = 7 and C = 4
Question:24
Replace A, B, by suitable numerals.
A
+ A
+ A
B A
Solution:
A +A +A = A
with1beingcarriedover
This is satisfied if A
is equal to 5
.
When A = 5
:
A +A +A = 15
1iscarriedover
Or B = 1
? A = 5 and B = 1
Question:25
Replace A, B by suitable numerals.
6 A
- A B
3 7
Solution:
First look at the left column, which is:
6 -A = 3
This implies that the maximum value of A can be 3.
A = 3
... 1
The next column has the following:
A -B = 7
To reconcile this with equation 1
, borrowing is involved.
We know:
12 -5 = 7
? A = 2 and B = 5
Question:26
Replace A, B, C by suitable numerals.
C B 5
- 2 8 A
2 5 9
Solution:
5 -A = 9
This implies that 1 is borrowed.
We know:
15 -6 = 9
? A = 6
B -5 = 8
This implies that 1 is borrowed.
13 -5 = 8
But 1 has also been lent
?B = 4
C -2 = 2
This implies that 1 has been lent.
? C = 5
? A = 6, B = 4 and C = 5
Question:27
Replace A, B, C by suitable numerals.
A B
× 3
C A B
Solution:
(B ×3) = B Then, B can either be 0 or 5. If B is 5, then 1 will be carried. Then, A ×3 +1 = A will not be possible for any number. ? B = 0 A ×3 = A is possible for either 0 or 5. If we take A
Question:28
Replace A, B, C by suitable numerals.
A B
× B A
(B +1) C B
Solution:
A ×B = B ? A = 1
In the question:
First digit = B+1
Thus, 1 will be carried from 1+B
2
and becomes B +1
(B
2
-9) B.
? C = B
2
-1
Now, all B, B+1 and B
2
-9 are one digit number.
This condition is satisfied for B=3 or B=4.
For B< 3, B
2
-9 will be negative.
For B>3, B
2
-9 will become a two digit number.
For B=3 , C = 3
2
- 9 = 9-9 = 0
For B = 4, C = 4
2
-9 = 16-9 = 7
Required answer:
A=1, B=3, C = 0
or
A=1, B=4, C = 7
Question:29
Replace A, B, C by suitable numerals.
6) 5 A B (9 C
- 5 4
3 B
- 3 6
×
Solution:
(A -4) = 3 ? A = 7
?Also, 6 ×6 = 36 ? C = 6
36 -36 = 0 ? B = 6
? A = 7B = C = 6
Question:30
Find two numbers whose product is a 1-digit number and the sum is a 2-digit number.
Solution:
1 and 9
are two numbers, whose product is a single digit number.
? 1 ×9 = 9
Sum of the numbers is a two digit number.
? 1 +9 = 10
Question:31
Find three whole numbers whose product and sum are equal.
Solution:
The three whole numbers are 1, 2 and 3.
1 +2 +3 = 6 = 1 ×2 ×3
Question:32
complete the magic square given below, so that the sum of the numbers in each row or in each column or along each diagonal is 15.
6 1
5
Solution:
Taking the diagonal that starts with 6:
6 +5 +x = 15 ? x = 4
6 1
5
4
Now, taking the first row:
6 +1 +x = 15 ? x = 8
6 1 8
5
4
Taking the last column:
8 +x +4 = 15 ? x = 3
6 1 8
5 3
4
Taking the second column:
1 +5 +x = 15 ? x = 9
6 1 8
5 3
9 4
Taking the second row:
x +5 +3 = 15 ? x = 7
6 1 8
7 5 3
9 4
Taking the diagonal that begins with 8:
8 +5 +x = 15 ? x = 2
6 1 8
7 5 3
2 9 4
Question:33
Fill in the numbers from 1 to 6 without repetition, so that each side of the triangle adds up to 12.
Solution:
6+2+4 = 12
4+3+5 = 12
6+1+5 = 12
Question:34
Fibonacci numbers Take 10 numbers as shown below:
a, b, (a + b), (a + 2b), (2a + 3b), (3a + 5b), (5a + 8b), (8a + 13b), (13a + 21b), and (21a + 34b). Sum of all these numbers = 11(5a + 8b) = 11 × 7th number.
Taking a = 8, b = 13; write 10 Fibonacci numbers and verify that sum of all these numbers = 11 × 7th number.
Solution:
Given:
a = 8 and b = 13
The numbers in the Fibonnaci sequence are arranged in the following manner:
1st, 2nd, (1st +2nd), (2nd +3th), (3th +4th), (4th +5th), (5th +6th), (6th +7th), (7th +8th), (8th +9th), (9th +10th)
The numbers are 8, 13, 21, 34, 55, 89, 144, 233, 377 and 610
.
Sum of the numbers = ?8 +13 +21 +34 +55 +89 +144 +233 +377 +610
= 1584
11 ×7th number = 11 ×144 = 1584
Question:35
Complete the magic square:
14 0
8 6 11
4 . 7
2 1 12
Solution:
The magic square is completed assuming that the sum of the row, columns and diagonals is 30. This is because the sum of all the number of the last column is 30.
3 14 13 0
8 5 6 11
4 9 10 7
15 2 1 12
Question:36
Tick
? the correct answer
If 5x6 is exactly divisible by 3, then the least value of x is
a
0
b
1
c
2
d
3
Solution:
b
1
If a number is exactly divisible by 3, the sum of the digits must also be divisible by 3.
5 +x +6 = 11 +x
must be divisible by 3.
The smallest value of x
is 1.
x = 1
? x +11 = 12
is divisible by 3.
Question:37
Tick
? the correct answer
If 64y8 is exactly divisible by 3, then the least value of y is
a
0
b
1
c
2
d
3
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