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Algebraic Expressions - Evaluation 3

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Find the value of the algebraic expression by substituting the given value of the variable.


1. k + 12.6 where k = 29.8
Answer: 42.4
126 + 298 = 424
There is 1 decimal place in the numbers being added.
So the sum must have 1 decimal place.
Thus, 12.6 + 29.8 = 42.4

2. t + 42.8 where t = 26.1
Answer: 68.9
428 + 261 = 689
There is 1 decimal place in the numbers being added.
So the sum must have 1 decimal place.
Thus, 42.8 + 26.1 = 68.9

3. 6.03 + f where f = 1.15
Answer: 7.18
603 + 115 = 718
There are 2 decimal places in the numbers being added.
So the sum must have 2 decimal places.
Thus, 6.03 + 1.15 = 7.18


4. 20.97 + g where g = 14.76
Answer: 35.73
2097 + 1476 = 3573
There are 2 decimal places in the numbers being added.
So the sum must have 2 decimal places.
Thus, 20.97 + 14.76 = 35.73

5. 38.97 + d where d = 17.93
Answer: 56.90
3897 + 1793 = 5690
There are 2 decimal places in the numbers being added.
So the sum must have 2 decimal places.
Thus, 38.97 + 17.93 = 56.90

6. g − 5.1 where g = 11.0
Answer: 5.9
110 − 51 = 59
There is 1 decimal place in the numbers being subtracted.
So the difference must have 1 decimal place.
Thus, 11.0 − 5.1 = 5.9

7. k − 4.1 where k = 13.2
Answer: 9.1
132 − 41 = 91
There is 1 decimal place in the numbers being subtracted.
So the difference must have 1 decimal place.
Thus, 13.2 − 4.1 = 9.1

8. 14.41 − a where a = 9.67
Answer: 4.74
1441 − 967 = 474
There are 2 decimal places in the numbers being subtracted.
So the difference must have 2 decimal places.
Thus, 14.41 − 9.67 = 4.74

9. 17.11 − h where h = 11.65
Answer: 5.46
1711 − 1165 = 546
There are 2 decimal places in the numbers being subtracted.
So the difference must have 2 decimal places.
Thus, 17.11 − 11.65 = 5.46

10. r × 0.3 where r = 1.4
Answer: 0.42
3 × 14 = 42
There is 1 decimal place in the multiplicand and 1 in the multiplier.
So the product must have 1 + 1 = 2 decimal places.
Thus, 0.3 × 1.4 = 0.42

11. 0.8 × j where j = 9
Answer: 7.2
8 × 9 = 72
There is 1 decimal place in the multiplicand and 0 in the multiplier.
So the product must have 1 + 0 = 1 decimal place.
Thus, 0.8 × 9 = 7.2

12. g × 3.9 where g = 5.6
Answer: 21.84
56 × 39 = 2184
There is 1 decimal place in the multiplicand and 1 in the multiplier.
So the product must have 1 + 1 = 2 decimal places.
Thus, 5.6 × 3.9 = 21.84

13. m × 5.1 where m = 8.0
Answer: 40.80
80 × 51 = 4080
There is 1 decimal place in the multiplicand and 1 in the multiplier.
So the product must have 1 + 1 = 2 decimal places.
Thus, 8.0 × 5.1 = 40.80

14. 0.57 × e where e = 2
Answer: 1.14
57 × 2 = 114
There are 2 decimal places in the multiplicand and 0 in the multiplier.
So the product must have 2 + 0 = 2 decimal places.
Thus, 0.57 × 2 = 1.14

15. q ÷ 0.2 where q = 0.52
Answer: 2.6
5.2 ÷ 2 [Multiplied divisor and dividend by 10]
Now, 52 ÷ 2  
= 26 There is 1 decimal place in thedividend so the quotient must have 1 decimal place.
Thus, 0.52 ÷ 0.2 = 2.6

16. q ÷ 0.6 where q = 1.92
Answer: 3.2
19.2 ÷ 6 [Multiplied divisor and dividend by 10]
Now, 192 ÷ 6  
= 32 There is 1 decimal place in thedividend so the quotient must have 1 decimal place.
Thus, 1.92 ÷ 0.6 = 3.2

17. q ÷ 0.6 where q = 9
Answer: 15
90 ÷ 6  [Multiplied divisor and dividend by 10]
Now, 90 ÷ 6
= 15 There is 0 decimal place in thedividend so the quotient must have 0 decimal place.
Thus, 9 ÷ 0.6 = 15

18. w ÷ 0.6 where w = 23
Answer: 39
234 ÷ 6  [Multiplied divisor and dividend by 10]
Now, 234 ÷ 6
= 39 There is 0 decimal place in thedividend so the quotient must have 0 decimal place.
Thus, 23 ÷ 0.6 = 39

19. g ÷ 3.8 where g = 39.14
Answer: 10.3
391.4 ÷ 38 [Multiplied divisor and dividend by 10]
Now, 3914 ÷ 38  
= 103 There is 1 decimal place in thedividend so the quotient must have 1 decimal place.
Thus, 39.14 ÷ 3.8 = 10.3

20. y ÷ 2.2 where y = 27.06
Answer: 12.3
270.6 ÷ 22 [Multiplied divisor and dividend by 10]
Now, 2706 ÷ 22  
= 123 There is 1 decimal place in thedividend so the quotient must have 1 decimal place.
Thus, 27.06 ÷ 2.2 = 12.3

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