290862
Rahul purchased 4kg 500g rice, 5kg 650g sugar and 12kg 600g atta. The total weight of his purchases was __________.
1 10.150kg
2 22.750kg
3 25.350kg
4 18.250kg
Explanation:
10.150kg B 22.750kg Weight of rice = 4kg 500g = 4.500kg Weight of sugar = 5kg 650g = 5.650kg Weight of atta = 12kg 600g = 12.600kg \(\therefore\) Total weight of all items = (4.500 + 5.650 + 12.600)kg = 22.750kg.
04. DECIMALS
290863
Sum of 0.3, 0.03 and 0.003 is _____
1 0.999
2 0.393
3 0.636
4 0.333
Explanation:
0.999 D 0.333 To add 3 decimal numbers of different decimal values we pick the number with the highest decimal value and expand all the other numbers to the same value by adding 0 after the decimal point. 0.300 + 0.030 + 0.003 = 0.333
04. DECIMALS
290864
What is the value of \(2.\overline{6}-1.\overline{(9)}\)
1 \(0.\overline{6}\)
2 \(0.\overline{9}\)
3 \(0.\overline{7}\)
4 \(0.{7}\)
Explanation:
\(0.\overline{6}\) The value of \(2.\overline{6}-1.\overline{9}\) is \(=\Big(2+\frac{6}{9}\Big)-\Big(1+\frac{9}{9}\Big)\) \(=\frac{6}{9}=0.\overline{6}\)
290862
Rahul purchased 4kg 500g rice, 5kg 650g sugar and 12kg 600g atta. The total weight of his purchases was __________.
1 10.150kg
2 22.750kg
3 25.350kg
4 18.250kg
Explanation:
10.150kg B 22.750kg Weight of rice = 4kg 500g = 4.500kg Weight of sugar = 5kg 650g = 5.650kg Weight of atta = 12kg 600g = 12.600kg \(\therefore\) Total weight of all items = (4.500 + 5.650 + 12.600)kg = 22.750kg.
04. DECIMALS
290863
Sum of 0.3, 0.03 and 0.003 is _____
1 0.999
2 0.393
3 0.636
4 0.333
Explanation:
0.999 D 0.333 To add 3 decimal numbers of different decimal values we pick the number with the highest decimal value and expand all the other numbers to the same value by adding 0 after the decimal point. 0.300 + 0.030 + 0.003 = 0.333
04. DECIMALS
290864
What is the value of \(2.\overline{6}-1.\overline{(9)}\)
1 \(0.\overline{6}\)
2 \(0.\overline{9}\)
3 \(0.\overline{7}\)
4 \(0.{7}\)
Explanation:
\(0.\overline{6}\) The value of \(2.\overline{6}-1.\overline{9}\) is \(=\Big(2+\frac{6}{9}\Big)-\Big(1+\frac{9}{9}\Big)\) \(=\frac{6}{9}=0.\overline{6}\)
290862
Rahul purchased 4kg 500g rice, 5kg 650g sugar and 12kg 600g atta. The total weight of his purchases was __________.
1 10.150kg
2 22.750kg
3 25.350kg
4 18.250kg
Explanation:
10.150kg B 22.750kg Weight of rice = 4kg 500g = 4.500kg Weight of sugar = 5kg 650g = 5.650kg Weight of atta = 12kg 600g = 12.600kg \(\therefore\) Total weight of all items = (4.500 + 5.650 + 12.600)kg = 22.750kg.
04. DECIMALS
290863
Sum of 0.3, 0.03 and 0.003 is _____
1 0.999
2 0.393
3 0.636
4 0.333
Explanation:
0.999 D 0.333 To add 3 decimal numbers of different decimal values we pick the number with the highest decimal value and expand all the other numbers to the same value by adding 0 after the decimal point. 0.300 + 0.030 + 0.003 = 0.333
04. DECIMALS
290864
What is the value of \(2.\overline{6}-1.\overline{(9)}\)
1 \(0.\overline{6}\)
2 \(0.\overline{9}\)
3 \(0.\overline{7}\)
4 \(0.{7}\)
Explanation:
\(0.\overline{6}\) The value of \(2.\overline{6}-1.\overline{9}\) is \(=\Big(2+\frac{6}{9}\Big)-\Big(1+\frac{9}{9}\Big)\) \(=\frac{6}{9}=0.\overline{6}\)
290862
Rahul purchased 4kg 500g rice, 5kg 650g sugar and 12kg 600g atta. The total weight of his purchases was __________.
1 10.150kg
2 22.750kg
3 25.350kg
4 18.250kg
Explanation:
10.150kg B 22.750kg Weight of rice = 4kg 500g = 4.500kg Weight of sugar = 5kg 650g = 5.650kg Weight of atta = 12kg 600g = 12.600kg \(\therefore\) Total weight of all items = (4.500 + 5.650 + 12.600)kg = 22.750kg.
04. DECIMALS
290863
Sum of 0.3, 0.03 and 0.003 is _____
1 0.999
2 0.393
3 0.636
4 0.333
Explanation:
0.999 D 0.333 To add 3 decimal numbers of different decimal values we pick the number with the highest decimal value and expand all the other numbers to the same value by adding 0 after the decimal point. 0.300 + 0.030 + 0.003 = 0.333
04. DECIMALS
290864
What is the value of \(2.\overline{6}-1.\overline{(9)}\)
1 \(0.\overline{6}\)
2 \(0.\overline{9}\)
3 \(0.\overline{7}\)
4 \(0.{7}\)
Explanation:
\(0.\overline{6}\) The value of \(2.\overline{6}-1.\overline{9}\) is \(=\Big(2+\frac{6}{9}\Big)-\Big(1+\frac{9}{9}\Big)\) \(=\frac{6}{9}=0.\overline{6}\)
