RS Aggarwal Class 6 Math Third Chapter Whole Numbers Exercise 3E Solution

RS Aggarwal Class 6 Math Third Chapter Whole Numbers Exercise 3E Solution

EXERCISE 3E

(1) Divide and check your answer by the corresponding multiplication in each of the following:

(i) 1936 ÷16        

Solution:

Ans: 121.

(ii) 19881 ÷ 47

Ans: 423.

(iii) 257796 ÷ 341

Ans: 756

(iv) 612846 ÷ 582

Ans: 1053.

(v) 34419 ÷ 149

Ans: 231

(vi) 39039 ÷ 1001

Ans: 39

(2) Divide, and find out the quotient and remainder. Check your answer.

(i) 6971 ÷ 47

Solution:

Ans: So the quotient= 147 and remainder= 22.

(ii) 4178 ÷ 35

Ans: So, the quotient= 119 and remainder= 13.

(iii) 36195 ÷ 153

Ans: So, the quotient= 236 and remainder= 87.

(iv) 93575 ÷ 400

Ans: So, the quotient= 208 and remainder= 375.

(v) 230254 ÷ 1000

Ans: So, the quotient= 230 and remainder= 254.

(vi) 16135 ÷ 875

Ans: So, the quotient= 18 and remainder= 385 .

(3) Find the value of

(i) 65007 ÷ 1= 65007.

(ii) 0 ÷ 879= 0.

(iii) 981 + 5720 ÷ 10

= 981 + 572

= 1553.

(iv) 1507 – (625 ÷ 25)

= 1507 – 25

= 1482.

(v) 32277 ÷ (648 – 39)

= 32277 ÷ 609

= 53.

(vi) (1573 ÷ 1573) – (1573 ÷ 1573)

= 1 – 1

= 0.

(4) Find a whole number n such that n ÷ n = n.

 Ans: 

nn

= 1

So, n = 1.

(5) The product of two numbers is 504347. If one of the numbers is 317, find the other.

Ans: The other numb er is

So, the other number is 1591.

(6) On dividing 59761 by a certain number, the quotient is 189 and the remainder is 37. Find the divisor

Solution: Let the divisor is m.

Thus, Dividend= (Division x Quotient) + Remainder

Or, 59761= (m x 189) + 37

Or, 189 m= 59761-37

Or, m=

59724189

Or, m= 316.

Therefore, the division (m)= 316.

(7) On dividing 55390 by 299, the remainder is 75. Find the quotient using the division algorithm.

Solution: The quotient is m.

Thus, Dividend = ( divisor x quotient) + remainder

Or, 55390 = (299 x m) + 75

Or, 299 m = 55390 – 75

Or, m =

55313299

Or, m = 185

Therefore, the quotients (m)= 185.

(8) What least number must be subtracted from 13601 to get a number exactly divisible by 87?

Solution:

Ans: 29 subtracted from 13601 to get the number which is divisible by 87.

(9) What least number must be added to 1056 to get a number exactly divisible by 23?

Solution: Here, 23 x 45= 1035, 23 x 46= 1058

We have 1056 that means (1058 – 1056)= 2. If 2 is added to 1056 to get the number that divisible by 23.

(10) Find the largest 4-digit number divisible by 16.

Ans: 9984.

(11) Divide the largest 5 digit number by 653. Check your answer by the division algorithm.

Solution:

Ans: Quotient= 153, remainder= 90.

(12)  Find the largest 6-digit number exactly divisible by 83.

Ans: 100015.

(13) 1 dozen bananas cost RS 29. How many dozens can be purchase for RS 1392?

Solution: 1 dozen= 12

Therefore, 1392 ÷ 12= 116.

Ans: 116 bananas can be purchase for Rs 1392.

(14) 19625 trees have been equally planted in 157 rows. Find the number of trees in each row.

Solution: 19625 ÷ 157= 125.

Ans: 125 trees have been planted in each row.

(15) The population of a town is 517530. If one out every 15 is reported to be literate, find how many literate persons are there in the town.

Solution: 517530 ÷ 15= 34502.

Ans: 34502 persons are literate in the town.

(16) The cost price of 23 colour television sets is RS 570055. Determine the cost price of each TV set if each costs same.

Solution: 5700555 ÷ 23= 24785.

Ans: The cost of each TV is RS 24785.

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  1. It’s very helpful for us, the students!

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