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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 + 20.3 where k = 10.3
Answer: 30.6
203 + 103 = 306
There is 1 decimal place in the numbers being added.
So the sum must have 1 decimal place.
Thus, 20.3 + 10.3 = 30.6


2. t + 36.8 where t = 22.5
Answer: 59.3
368 + 225 = 593
There is 1 decimal place in the numbers being added.
So the sum must have 1 decimal place.
Thus, 36.8 + 22.5 = 59.3


3. 6.65 + f where f = 1.58
Answer: 8.23
665 + 158 = 823
There are 2 decimal places in the numbers being added.
So the sum must have 2 decimal places.
Thus, 6.65 + 1.58 = 8.23



4. 24.18 + g where g = 14.37
Answer: 38.55
2418 + 1437 = 3855
There are 2 decimal places in the numbers being added.
So the sum must have 2 decimal places.
Thus, 24.18 + 14.37 = 38.55


5. 34.20 + d where d = 13.32
Answer: 47.52
3420 + 1332 = 4752
There are 2 decimal places in the numbers being added.
So the sum must have 2 decimal places.
Thus, 34.20 + 13.32 = 47.52


6. g − 6.5 where g = 11.8
Answer: 5.3
118 − 65 = 53
There is 1 decimal place in the numbers being subtracted.
So the difference must have 1 decimal place.
Thus, 11.8 − 6.5 = 5.3


7. k − 3.5 where k = 14.6
Answer: 11.1
146 − 35 = 111
There is 1 decimal place in the numbers being subtracted.
So the difference must have 1 decimal place.
Thus, 14.6 − 3.5 = 11.1


8. 12.68 − a where a = 7.31
Answer: 5.37
1268 − 731 = 537
There are 2 decimal places in the numbers being subtracted.
So the difference must have 2 decimal places.
Thus, 12.68 − 7.31 = 5.37


9. 16.11 − h where h = 12.26
Answer: 3.85
1611 − 1226 = 385
There are 2 decimal places in the numbers being subtracted.
So the difference must have 2 decimal places.
Thus, 16.11 − 12.26 = 3.85


10. r × 0.3 where r = 1.5
Answer: 0.45
3 × 15 = 45
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 × 1.5 = 0.45


11. 0.6 × j where j = 7
Answer: 4.2
6 × 7 = 42
There is 1 decimal place in the multiplicand and 0 in the multiplier.
So the product must have 1 + 0 = 1 decimal place.
Thus, 0.6 × 7 = 4.2


12. g × 3.3 where g = 5.7
Answer: 18.81
57 × 33 = 1881
There is 1 decimal place in the multiplicand and 1 in the multiplier.
So the product must have 1 + 1 = 2 decimal places.
Thus, 5.7 × 3.3 = 18.81


13. m × 5.2 where m = 9.2
Answer: 47.84
92 × 52 = 4784
There is 1 decimal place in the multiplicand and 1 in the multiplier.
So the product must have 1 + 1 = 2 decimal places.
Thus, 9.2 × 5.2 = 47.84


14. 0.50 × e where e = 5
Answer: 2.50
50 × 5 = 250
There are 2 decimal places in the multiplicand and 0 in the multiplier.
So the product must have 2 + 0 = 2 decimal places.
Thus, 0.50 × 5 = 2.50


15. q ÷ 0.2 where q = 0.42
Answer: 2.1
4.2 ÷ 2 [Multiplied divisor and dividend by 10]
Now, 42 ÷ 2  
= 21 There is 1 decimal place in thedividend so the quotient must have 1 decimal place.
Thus, 0.42 ÷ 0.2 = 2.1


16. q ÷ 0.6 where q = 2.04
Answer: 3.4
20.4 ÷ 6 [Multiplied divisor and dividend by 10]
Now, 204 ÷ 6  
= 34 There is 1 decimal place in thedividend so the quotient must have 1 decimal place.
Thus, 2.04 ÷ 0.6 = 3.4


17. q ÷ 0.3 where q = 3
Answer: 12
36 ÷ 3  [Multiplied divisor and dividend by 10]
Now, 36 ÷ 3
= 12 There is 0 decimal place in thedividend so the quotient must have 0 decimal place.
Thus, 3 ÷ 0.3 = 12


18. w ÷ 0.3 where w = 9
Answer: 30
90 ÷ 3  [Multiplied divisor and dividend by 10]
Now, 90 ÷ 3
= 30 There is 0 decimal place in thedividend so the quotient must have 0 decimal place.
Thus, 9 ÷ 0.3 = 30


19. g ÷ 3.5 where g = 38.85
Answer: 11.1
388.5 ÷ 35 [Multiplied divisor and dividend by 10]
Now, 3885 ÷ 35  
= 111 There is 1 decimal place in thedividend so the quotient must have 1 decimal place.
Thus, 38.85 ÷ 3.5 = 11.1


20. y ÷ 2.5 where y = 29.75
Answer: 11.9
297.5 ÷ 25 [Multiplied divisor and dividend by 10]
Now, 2975 ÷ 25  
= 119 There is 1 decimal place in thedividend so the quotient must have 1 decimal place.
Thus, 29.75 ÷ 2.5 = 11.9



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