Class 8 Exam > Class 8 Notes > Mathematics (Maths) Class 8 > RS Aggarwal Solutions: Exercise 5C - Playing With Numbers
RS Aggarwal Solutions: Exercise 5C - Playing With Numbers | Mathematics (Maths) Class 8 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
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
Read More
|
79 videos|408 docs|31 tests
|
Related Exams
About this Document
|
4.62/5 Rating |
|
Nov 21, 2024 Last updated |
Document Description: RS Aggarwal Solutions: Exercise 5C - Playing With Numbers for Class 8 2024 is part of Mathematics (Maths) Class 8 preparation.
The notes and questions for RS Aggarwal Solutions: Exercise 5C - Playing With Numbers have been prepared according to the Class 8 exam syllabus. Information about RS Aggarwal Solutions: Exercise 5C - Playing With Numbers covers topics
like and RS Aggarwal Solutions: Exercise 5C - Playing With Numbers Example, for Class 8 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RS Aggarwal Solutions: Exercise 5C - Playing With Numbers.
Introduction of RS Aggarwal Solutions: Exercise 5C - Playing With Numbers in English is available as part of our Mathematics (Maths) Class 8
for Class 8 & RS Aggarwal Solutions: Exercise 5C - Playing With Numbers in Hindi for Mathematics (Maths) Class 8 course.
Download more important topics related with notes, lectures and mock test series for Class 8
Exam by signing up for free. Class 8: RS Aggarwal Solutions: Exercise 5C - Playing With Numbers | Mathematics (Maths) Class 8
Description
Full syllabus notes, lecture & questions for RS Aggarwal Solutions: Exercise 5C - Playing With Numbers | Mathematics (Maths) Class 8 - Class 8 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 8 | Best notes, free PDF download
Information about RS Aggarwal Solutions: Exercise 5C - Playing With Numbers
In this doc you can find the meaning of RS Aggarwal Solutions: Exercise 5C - Playing With Numbers defined & explained in the simplest way possible. Besides explaining types of
RS Aggarwal Solutions: Exercise 5C - Playing With Numbers theory, EduRev gives you an ample number of questions to practice RS Aggarwal Solutions: Exercise 5C - Playing With Numbers tests, examples and also practice Class 8
tests
|
Explore Courses for Class 8 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Sample Paper
,RS Aggarwal Solutions: Exercise 5C - Playing With Numbers | Mathematics (Maths) Class 8
,RS Aggarwal Solutions: Exercise 5C - Playing With Numbers | Mathematics (Maths) Class 8
,Extra Questions
,Previous Year Questions with Solutions
,Free
,Summary
,shortcuts and tricks
,RS Aggarwal Solutions: Exercise 5C - Playing With Numbers | Mathematics (Maths) Class 8
,ppt
,MCQs
,past year papers
,Semester Notes
,video lectures
,Exam
,Objective type Questions
,mock tests for examination
,practice quizzes
,study material
,Viva Questions
,Important questions
;
Additional Information about RS Aggarwal Solutions: Exercise 5C - Playing With Numbers for Class 8 Preparation
RS Aggarwal Solutions: Exercise 5C - Playing With Numbers Free PDF Download
The RS Aggarwal Solutions: Exercise 5C - Playing With Numbers is an invaluable resource that delves deep into the core of the Class 8 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the RS Aggarwal Solutions: Exercise 5C - Playing With Numbers now and kickstart your journey towards success in the Class 8 exam.
Importance of RS Aggarwal Solutions: Exercise 5C - Playing With Numbers
The importance of RS Aggarwal Solutions: Exercise 5C - Playing With Numbers cannot be overstated, especially for Class 8 aspirants.
This document holds the key to success in the Class 8 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
RS Aggarwal Solutions: Exercise 5C - Playing With Numbers Notes
RS Aggarwal Solutions: Exercise 5C - Playing With Numbers Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to RS Aggarwal Solutions: Exercise 5C - Playing With Numbers.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, RS Aggarwal Solutions: Exercise 5C - Playing With Numbers Notes on EduRev are your ultimate resource for success.
RS Aggarwal Solutions: Exercise 5C - Playing With Numbers Class 8 Questions
The "RS Aggarwal Solutions: Exercise 5C - Playing With Numbers Class 8 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 8 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study RS Aggarwal Solutions: Exercise 5C - Playing With Numbers on the App
Students of Class 8 can study RS Aggarwal Solutions: Exercise 5C - Playing With Numbers alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the RS Aggarwal Solutions: Exercise 5C - Playing With Numbers,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of RS Aggarwal Solutions: Exercise 5C - Playing With Numbers is prepared as per the latest Class 8 syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup on EduRev and stay on top of your study goals
10M+ students crushing their study goals daily