Class 6 Operations on whole Numbers Exercise 4.3-2

\begin{array}{l} \text {Q7. In each of the following, fill in the blanks, so that the statement is true:} \\ (i) \quad (500+7)(300-1)=299 \times \ldots \\ (ii) \quad 888+777+555=111 \times \ldots \\ (iii) \quad 75 \times 425=(70+5)(\ldots + 85) \\ (iv) \quad 89 \times(100-2)=98 \times(100-\ldots) \\ (v) \quad (15+5)(15-5)=225 – \ldots \\ (vi) \quad 9 \times(10000+\ldots)=98766 \\ \\\text {Sol. (i) LHS } =(500+7)(300-1) \\ =507 \times 299 \\ =299 \times 507 \quad \quad [\text {Using commutativity}] \\ \text {Hence, the required answer is 507.} \\ \\\text {(ii) LHS } = 888+777+555 \\ = 111 \times 8 + 111 \times 7 + 111 \times 5 \\ =111(8+7+5) \quad \quad [\text {Using distributivity}] \\ =111 \times 20 \\ \text {Hence, the required answer is 20.} \\ \\\text {(iii) LHS } = 75 \times 425 \\ =(70+5) \times 425 =(70+5)(340+85)\quad \quad [\text {Using distributivity}] \\ \text {Hence, the required answer is 340.} \\ \\\text {(iv) LHS } = 89 \times(100-2) =89 \times 98 =98 \times 89 \quad \quad [\text {Using commutativity}] \\ =98 \times (100-11) \\ \text {Hence, the required answer is 11.} \\ \\\text {(v) LHS } =(15+5)(15-5) \\ =20 \times 10 \\ =200 =225-25 \\ \text {Hence, the required answer is 25.} \\ \\\text {(vi) RHS } =98766 = 90000 + 8766 \\ = 9 \times 10000 + 9 \times 974 \\ = 9 \times (10000 + 974) \quad \quad [\text {Using distributivity}] \\ \text {Hence, the required answer is 974.} \\ \\\text {Q8. A dealer purchased 125 colour television sets. If the cost of each } \\ \text {set is Rs. 19820, determine the cost of all sets together. } \\ \\\text {Sol. Given that cost of one television set }=\text {Rs. } 19820 \\\Rightarrow \text {Cost of 125 television sets } =19820 \times 125 \\=19820 \times (100+25) \\ =(19820 \times 100)+(19820 \times 25) \quad \quad [\text {Using distributivity}] \\ =1982000+495500 =\text {Rs. } 2477500 \\ \\\text {Q9. The annual fee charged from a student of class VI in a school is Rs 8880. } \\ \text {If there are, in all, 235 students in class VI, find the total collection.} \\ \\\text {Sol. Given that Annual fee charged from each student }= \text {Rs. } 8800 \\ \Rightarrow \text {Annual fee charged from 235 students }=8800 \times 235 \\ =\text {Rs. } 2086800 \\\therefore \text {The total collection is Rs. 2086800.} \\ \\\text {Q10. A group housing society constructed 350 flats. If the cost of construction } \\ \text {for each flat is Rs 993570, what is the total cost of construction of all the flats.} \\ \\\text {Sol. Given that cost of construction for each flat }= \text {Rs. } 993570 \\ \Rightarrow \text {Cost of construction of 350 flats }=993570 \times 350 \\ =\text {Rs. } 347749500 \\\therefore \text {The total cost of construction of all the flats is Rs. 347749500.} \\ \\\text {Q11. The product of two whole numbers is zero. What do you conclude? } \\ \\\text {Sol. The product of two whole numbers is zero means that either one } \\ \text { number or both of them are zero.} \\ \\\text {Q12. What are the whole numbers which when multiplied with itself gives the same number?} \\ \\\text {Sol. 0 and 1 are the two whole numbers when multiplied with itself gives the same number.} \\ \\\text {Q13. In a large housing complex, there are 15 small buildings and 22 large building. } \\ \text {Each of the large buildings has 10 floors with 2 apartments on each floor. } \\ \text {Each of the small buildings has 12 floors with 3 apartments on each floor. } \\ \text {How many apartments are there in all.} \\ \\\text {Sol. Given that no. of large buildings } =22 \\ \text {No. of floors in each large building } =10 \\ \text {No. of apartments on each floor of large building} =2 \\\therefore \text {Total apartment in each large building } =10 \times 2=20 \\\text {Also given that no. of small buildings } =15 \\\text {No. of floors in each small building } =12 \\ \text {No. of apartments on each floor of large building} =3 \\\therefore \text {Total apartment in each small building } =12 \times 3=36 \\\Rightarrow \text {The total apartment in entire housing complex } \\ =(22 \times 20)+(15 \times 36) \\ =440+540 = 980 \\ \therefore \text {There are 980 apartments in all.} \\ \end{array}