7th Grade > Mathematics
FRACTIONS AND DECIMALS MCQs
Fractions, Decimals
Total Questions : 227
| Page 9 of 23 pages
:
12=0.5
52=2.5
49=0.4444
520=0.25
310=0.3
7100=0.07
1002=50
10000010000000=0.01
1248=0.25
999=0.09090909
525=0.20
∴ Sum = 54.615
When rounded up to two decimal place,
Then Sum = 54.62
:
12=0.5
52=2.5
49=0.4444
520=0.25
310=0.3
7100=0.07
1002=50
10000010000000=0.01
1248=0.25
999=0.09090909
525=0.20
∴ Sum = 54.615
When rounded up to two decimal place,
Then Sum = 54.62
Answer: Option A. -> ₹ 4033
:
A
Cost of rice in one bag = 25×20.04
=₹501
Cost of dal in one bag = 20×15.28
=₹305.6
Total cost of one bag = 501 + 305.6
=₹806.6
Cost of 5 such bags = 5×806.6
=₹4033
:
A
Cost of rice in one bag = 25×20.04
=₹501
Cost of dal in one bag = 20×15.28
=₹305.6
Total cost of one bag = 501 + 305.6
=₹806.6
Cost of 5 such bags = 5×806.6
=₹4033
Answer: Option D. -> i) two ii) right iii) two iv) left
:
D
Let's take an example and visualize.
2.48×100=248.0 [Decimal point shifted two places to the right]
2480÷100=2480100=24.80 [Decimal point shifted two places to the left]
:
D
Let's take an example and visualize.
2.48×100=248.0 [Decimal point shifted two places to the right]
2480÷100=2480100=24.80 [Decimal point shifted two places to the left]
Answer: Option D. -> 10.1
:
D
0.010.001=0.01×10.001=0.01×10001
=10
0.0010.01=0.001×10.01=0.001×1001
=0.1
⟹0.010.001+0.0010.01 =10+0.1=10.1
:
D
0.010.001=0.01×10.001=0.01×10001
=10
0.0010.01=0.001×10.01=0.001×1001
=0.1
⟹0.010.001+0.0010.01 =10+0.1=10.1
Answer: Option B. -> 800 g
:
B
Total weightof fruits = 4 kg
Total number of children among whom the fruits are to be shared equally= 5
Thus, the total quantityof fruits each child would get = 45kg
= 45 ×1000grams
=40005g
= 800 g
:
B
Total weightof fruits = 4 kg
Total number of children among whom the fruits are to be shared equally= 5
Thus, the total quantityof fruits each child would get = 45kg
= 45 ×1000grams
=40005g
= 800 g
Answer: Option B. -> 8.9345
:
B
Shifting the decimal point 3 places to the right of the decimal number 0.89345, we have
0.89345×1000=893.45
∴(0.89345×1000)+8.9345−893.45
=893.45+8.9345−893.45
=0+8.9345
=8.9345
:
B
Shifting the decimal point 3 places to the right of the decimal number 0.89345, we have
0.89345×1000=893.45
∴(0.89345×1000)+8.9345−893.45
=893.45+8.9345−893.45
=0+8.9345
=8.9345
Answer: Option A. -> ₹ 4033
:
A
Cost of rice in one bag = 25×20.04
=₹501
Cost of dal in one bag = 20×15.28
=₹305.6
Total cost of one bag = 501 + 305.6
=₹806.6
Cost of 5 such bags = 5×806.6
=₹4033
:
A
Cost of rice in one bag = 25×20.04
=₹501
Cost of dal in one bag = 20×15.28
=₹305.6
Total cost of one bag = 501 + 305.6
=₹806.6
Cost of 5 such bags = 5×806.6
=₹4033
Answer: Option B. -> 8.9345
:
B
Shifting the decimal point 3 places to the right of the decimal number 0.89345, we have
0.89345×1000=893.45
∴(0.89345×1000)+8.9345−893.45
=893.45+8.9345−893.45
=0+8.9345
=8.9345
:
B
Shifting the decimal point 3 places to the right of the decimal number 0.89345, we have
0.89345×1000=893.45
∴(0.89345×1000)+8.9345−893.45
=893.45+8.9345−893.45
=0+8.9345
=8.9345
Answer: Option D. -> 10.1
:
D
0.010.001=0.01×10.001=0.01×10001
=10
0.0010.01=0.001×10.01=0.001×1001
=0.1
⟹0.010.001+0.0010.01 =10+0.1=10.1
:
D
0.010.001=0.01×10.001=0.01×10001
=10
0.0010.01=0.001×10.01=0.001×1001
=0.1
⟹0.010.001+0.0010.01 =10+0.1=10.1