RBSE Solutions for Class 5 Maths Chapter 2 Addition and Subtraction

Rajasthan Board RBSE Solutions for Class 5 Maths Chapter 2 Addition and Subtraction Textbook Exercise Questions and Answers.

RBSE Class 5 Maths Solutions Chapter 2 Addition and Subtraction

Exercise:

Solve : 

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6. 
580 + 27 + 306 = ...................
Solution:

 
Question 7.
473 - 296 = ..................
Solution:

Question 8. 
364 + 67 - 199 = ....................
Solution:
4 + 7 - 9 = 2
6 + 6 - 9 = 3
3 - 1 = 2
⇒ 364 + 67 - 199 = 232

Question 9.
37 - 15 + 10 = ..................
Solution:
7 - 5 + 0 = 2
3 - 1 + 1 = 3
⇒ 37 - 15 + 10 = 32

Question 10.
185 boys and 162 girls are studying in Government Higher Secondary School. Udlivas. What is the total number of students studying in this school ? 
Solution: 
No. of boys = 185
No. of girls = 162
Total no. of Students = 185 + 162

Thus, total number. of students = 347 

Question 11. 
In a garden 255 roses, 156 marigold, 178 jasmine plants are there. What is the total number of flowering plants ?
Solution: 
No. of Rose plants = 225 
No. of Marigold plants = 156 
No. of Jasmine plants = 178
Total No. of plants = 225 + 156 + 178

Total No. of flowering plants = 559

Question 12. 
Gita's father is at the mart to sell pots. This time her father built 523 pots, of which 484 were sold out. Then how many pots are left unsold. 
Solution: 
Number of total pots built = 523
No. of pots sold = 484 
Remaining pots = 523 - 484

Thus, 39 pots are left.

Question 13.
Add and subtract with Indian numbers.

Solution:

 

Practice:

Question 1.
Solve: 

Solution:

Exercise:

Question 1. 

Solve:

Solution:

Question 2. 
Add: 
(i) 8725 and 907 
(ii) 7865 and 2107 
(iii) 5113, 1999 and 638 
(iv) 8999 and 1001 
Solution: 

Question 3. 
Subtract:
(i) 840 from 3944 
(ii) 2407 from 4817 
(iii) 4999 from 6000
(iv) 7986 from 8344 
Solution: 

Question 4. 
Find out the sum of greatest four digit number and greatest three digit number. 
Solution: 
Greatest number of four digits = 9999 
Greatest number of three digits = 999 
Sum = 9999 + 999 = 10998 

Thus, total sum is 10998.

Question 5. 
Ramesh has deposited ₹ 2850 in January and ₹ 3650 in February to his bank account. What is the total deposited amount during these two months by Ramesh ? 
Solution: 
Rupees deposit in January = ₹ 2850 
Rupees deposit in February = ₹ 3650 
Total deposit = 2850 + 3650 = 6500

Thus, total deposit is ₹ 6500.

 

Question 6. 
Anshu bought a bicycle in ₹ 2999 and Ruchi bought another bicycle in ₹ 2650. Find out how much extra money Anshu paid as compared to Ruchi. 
Solution: 
Cost of Anshu's cycle = ₹ 2999
Cost of Ruchi's cycle = ₹ 2650 
Difference between cost of two cycles = 2999 - 2650 = 349

Thus Anshu paid ₹ 34 g more as compared to Ruchi.

Question 7. 
Sum of two numbers is 7678. If one number is 4613, then find out the second number. 
Solution: 
Sum of two numbers = 7678
One number = 4613 
Let other number = a
Then, a + 4613 = 7678

⇒ a = 7678 - 4613 = 3065 
(For side change we take -ve sign inplace of +ve) 
Thus, other number is 3065. 

Question 8. 
Mohan has ₹ 10000. He bought wheat for ₹ 4500 and rice for ₹ 1600. How much money was left with him ? 
Solution: 
Total money Mohan has = ₹ 10000
Cost of wheat = ₹ 4500
Cost of rice = ₹ 1600 
Total cost of both = 4500 + 1600 = 6100

Left money = ₹ 10000 - ₹ 36100 = ₹ 3900
Thus, ₹ 3900 was left with him 

Question 9. 
During three phases of pulse polio campaign in bagwas panchayat, total number of children who were administered polio drop are 8975. In the first phase 2780 children were administered and in the second phase 2925 children were administered. Find out how many children were administered polio drop in the third phase ? 
Solution: 
Total Number of children who administered polio drop = 8976
Number of children administered polio drop in phase I = 2780 
Number of children administred polio drop in phase II = 2925 
Total Number of children administered polio drop in both phases = 2780 + 2925 = 5705

No. of children administered polio drop in phase IIIrd = 8976 - 5705 = 3271 
Thus, 3721 children administered
polio drop in IIIrd phase. 

 

Question 10. 
Add and Subtract with indian numbers: 

Solution: 

Question 11. 
Shyam bought an almirah for ₹ ६५९० and a bed for ₹२६२४. How much money has Shyam spent in total ? 
Solution: 
Cost of almirah = ₹ ६५९० 
Cost of bed = ₹ २६२४ 
Total expenses = ₹ ६५९० + ₹ २६२४

Thus, Shyam spent ₹ ६२०४ in total. 

Question 12. 
Rama bought a table for ₹ ३५४० and Radha bought another table for ₹ २६६०, What is the price difference between both the tables ? 
Solution: 
Cost of Rama's table = ₹ ३५४० 
Cost of Radha's table = ₹ २६६०
Difference between two values = ₹ ३५४० - ₹ २६६० = ४६०
Thus, difference between two values is ४६०. 

Question 13. 
Add the greatest and the smallest four digit numbers. 
Solution: 
Greatest number of ४ digits = ६६६६
and smallest number = 9000
Sum of two numbers = ६६६६ + 9000 = 90६६६ 

Thus, sum of two numbers is 90६६६. 

Question 14. 
२६४ students appeared in an exam. 9६४ of them passed. Find out the total number of failed students. 
Solution: 
Total students appeared = २६४ 
Passed students = 9६४
Failed students = २६४ - 9६४ = ७०

Thus, ७० students failed.

 

Question 15. 
Addition of two numbers is ६७३२. If one number is ३९४६, then find out the second number. 
Solution: 
Addition of two numbers = ६७३२
One number = ३९४६ 
Let a be second number. 
then, a + ३९४६ = ६७३२
⇒ a = ६७३२ - ३९४६ = ५९९६

Thus, second number be ५९९६. 

Question 16. 
For sewing three curtains we need 9२५ meter of cloth and for five tents we need २९६ meter of cloth. How much cloth is required for curtains and tents both ? 
Solution: 
For three curtains, cloth required = 9२५ m 
For five tents, cloth required = २९६ m 
Total cloth required for tent and curtain = 9२५ + २९६ = ४99
Thus, required cloth = ४99 m

Important Questions:

Multiple Choice Questions:

Question 1. 
Value of 9९५६ + 9५४७ will be : 
(a) ३४०३
(b) ३५०३ 
(c) ३६०३ 
(d) ३३०३ 
Solution:
(a) ३४०३

Question 2. 
In 3528 tens digit is :
(a) 5 
(b) 2 
(c) 8
(d) 3 
Solution:
(b) 2 

Question 3. 
Value of 9345 - 6234 will be : 
(a) 3191 
(b) 3111 
(c) 3110 
(d) 3991 
Solution:
 (b) 3111 

 
Question 4. 
Value of 72860 - 56280 will be : 
(a) 16580 
(b) 16480 
(c) 16380 
(d) 16481 
Solution:
(a) 16580 

Question 5. 
How more than 10000 from 9999 ? 
(a) 11 
(b) 101 
(c) 1 
(d) 10
Solution:
(c) 1 

Question 6. 
Value of 2000 + 3009 will be : 
(a) 5000 
(b) 5009 
(c) 5019 
(d) 5090 
Solution:
(b) 5009

Question 7. 
Value of ६५३०० - ७४९५० will be: 
(a) २०६५०
(b) २०४५० 
(c) २०४३६
(d) २०४४० 
Solution:
(b) २०४५०

Question 8. 
Value of 3045 + 3189 will be : 
(a) 6234 
(b) 6245 
(c) 6244 
(d) 6274 
Solution:
(a) 6234

Fill in the blanks:

Question 1. 
Nine is written as .............. in Indian numeral system. 
Solution:
६

Question 2. 
9९५६ + 9४७४ = ...................
Solution:
३३३०

Question 3. 
२५४९ - 9६५४ = .....................
Solution:
९६४

Question 4. 
4654 - 2314 = ........... 
Solution:
2340

Question 5. 
3235 + 3121 = ..........
Solution:
6356

Short Answer Type Questions:

Question 1. 
Solve the following: 

Solution:

Question 2. 
Match the following: 

Solution: 
१. (f), 
२. (e), 
३. (a), 
४. (g), 
५. (d), 
६. (b), 
७. (c).

Question 3. 
A farmer needs 4510 bricks to an make a well and 3485 bricks to make · a drain. Calculate, how many bricks he needs in total. 
Solution: 
Bricks to make a well = 4510
Bricks to make a drain = 3485 
So, total bricks = 4510 + 3485 = 7995

Question 4. 
Subtract 3 digit largest number from 4-digit smallest number. 
Solution: 
4-digit smallest number = 1000 
3-digit largest number = 999

 

Question 5. 
There are 4300 sheep-goat in a herd. Out of them 2075 are goats, then find number of sheep in the herd. 
Solution: 
Total sheep-goat = 4300 and
goats = 2075 g 
Number of sheep = 4300 - 2075 = 2225

Thus, number of sheep is 2225. 

Question 6. 
In pulso polio programme, 3842 children administered in step 1 and 2975 children administered in step II. Find how many children administered in both steps. 
Solution: 
Number of children who administered in step I = 3842 
Number of children who administered in step II = 2975 
Total number of children who administered in both steps = 3842 + 2975 = 6817

Thus, 6817 children administered in two steps.