Students can use the Spectrum Math Grade 3 Answer Key Chapter 3 Posttest as a quick guide to resolve any of their doubts.

Question 1.
a.  Explanation:
In the addition of 3 numbers in a column form first we need to find the sum of first two numbers and add it to the third number, 23 + 13 + 27 = (23 + 13) + 27 = 36 + 27 = 63.

b.  Explanation:
In the addition of 3 numbers in a column form first we need to find the sum of first two numbers and add it to the third number, 8 + 36 + 45 = (8 + 36) + 45 = 44 + 45 = 89.

c.  Explanation:
In the addition of 3 numbers in a column form first we need to find the sum of first two numbers and add it to the third number, 72 + 38 + 43 = (72 + 38) + 43 = 110 + 43 = 153.

d.  Explanation:
In the addition of 3 numbers in a column form first we need to find the sum of first two numbers and add it to the third number, 20 + 35 + 47 = (20 + 35) + 47 = 55 + 47 = 102.

e.  Explanation:
In the addition of 3 numbers in a column form first we need to find the sum of first two numbers and add it to the third number, 86 + 93 + 10 = (86 + 93) + 10 = 179 + 10 = 189.

Question 2.
a.  Explanation:
In the addition of 3 numbers in a column form first we need to find the sum of first two numbers and add it to the third number, 123 + 427 + 192 = (123 + 427) +192 = 550 + 192 = 742.

b.  Explanation:
In the addition of 3 numbers in a column form first we need to find the sum of first two numbers and add it to the third number, 86 + 425 + 119 = (86 + 425) + 119 = 511 + 119 = 630.

c.  Explanation:
In the addition of 3 numbers in a column form first we need to find the sum of first two numbers and add it to the third number, 19 + 87 + 425 = (19 + 87) + 425 = 106 + 425 = 531.

d.  Explanation:
In the addition of 3 numbers in a column form first we need to find the sum of first two numbers and add it to the third number, 295 + 221 + 196 = (295 + 221) + 196 = 516 + 196 = 712.

e.  Explanation:
In the addition of 4 numbers in a column form first we need to find the sum of first two numbers, find the sum of last two numbers and then add the two sums, 425 + 196 + 176 + 105 = (425+196) + (176+105) = 621 + 281 = 902.

Question 3.
a.  Explanation:
There are two 4-digit numbers given, first add numbers in the ones place, then tens, then hundreds and lastly add numbers in the thousands place, 4321+1972 = 6293.

b.  Explanation:
There are two 4-digit numbers given, first add numbers in the ones place, then tens, then hundreds and lastly add numbers in the thousands place, 5916+432 = 6348.

c.  Explanation:
There are two 4-digit numbers given, first add numbers in the ones place, then tens, then hundreds and lastly add numbers in the thousands place, 8764+492 = 9256.

d.  Explanation:
There are two 4-digit numbers given, first add numbers in the ones place, then tens, then hundreds and lastly add numbers in the thousands place, 4567+1986 = 6553

e.  Explanation:
There are two 4-digit numbers given, first add numbers in the ones place, then tens, then hundreds and lastly add numbers in the thousands place, 5921+2053 = 7974.

Question 4.
a.  Explanation:
Subtract the ones.
Rename 2 hundreds and 1 tens as “1 hundreds and 11 tens.” Subtract the tens.
Rename 8 thousands and 1 hundreds as “7 thousands and 11 hundreds.” Subtract the hundreds.
Subtract the thousands.

b.  Explanation:
Subtract the ones.
Rename 8 hundreds and 7 tens as “7 hundreds and 17 tens.” Subtract the tens.
Subtract the hundreds.
Subtract the thousands.

c.  Explanation:
Subtract the ones.
Rename 6 hundreds and 5 tens as “5 hundreds and 15 tens.” Subtract the tens.
Subtract the hundreds.
As there is no thousands place digit in the second number consider it to be zero and then subtract the thousands.

d.  Explanation:
Subtract the ones.
Rename 9 hundreds and 8 tens as “8 hundreds and 18 tens.” Subtract the tens.
Subtract the hundreds.
As there is no thousands place digit in the second number consider it to be zero and then subtract the thousands.

e.  Explanation:
Subtract the ones.
Rename 1 thousands and 0 hundreds as “0 thousands 10 hundreds.”
Rename 10 hundreds and 5 tens as “9 hundreds and 15 tens.” Subtract the tens then Subtract the hundreds.
As there is no thousands place digit in the second number consider it to be zero and then subtract the thousands.

Question 5.
a.  Explanation:
Subtract the ones.
Subtract the tens.
Rename 7 thousands and 2 hundreds as “6 thousands and 12 hundreds.” Subtract the hundreds.
As there is no thousands place digit in the second number consider it to be zero and then subtract the thousands.

b.  Explanation:
Subtract the ones.
Rename 7 hundreds and 8 tens as “6 hundreds and 18 tens.” Subtract the tens.
Subtract the hundreds.
Subtract the thousands.

c.  Explanation:
Subtract the ones.
Rename 8 hundreds and 3 tens as “7 hundreds and 13 tens.” Subtract the tens.
Rename 4 thousands and 7 hundreds as “3 thousands and 17 hundreds.” Subtract the hundreds.
Subtract the thousands.

d.  Explanation:
Subtract the ones.
Rename 9 hundreds and 5 tens as “8 hundreds and 15 tens.” Subtract the tens.
Subtract the hundreds.
Subtract the thousands.

e.  Explanation:
Subtract the ones.
Rename 2 hundreds and 1 tens as “1 hundreds and 11 tens.” Subtract the tens.
Rename 3 thousands and 1 hundreds as “2 thousands and 11 hundreds.” Subtract the hundreds.
Subtract the thousands.

Question 6.
a.  Explanation:
Subtract the ones.
Rename 1 hundreds and 6 tens as “0 hundreds and 16 tens.” Subtract the tens.
Rename 8 thousands and 0 hundreds as “7 thousands and 10 hundreds.” Subtract the hundreds.
Subtract the thousands.

b.  Explanation:
Subtract the ones.
Rename 9 hundreds and 1 tens as “8 hundreds and 11 tens.” Subtract the tens.
Subtract the hundreds.
Subtract the thousands.

c.  Explanation:
Subtract the ones.
Rename 3 hundreds and 2 tens as “2 hundreds and 12 tens.” Subtract the tens.
Rename 4 thousands and 2 hundreds as “3 thousands and 12 hundreds.” Subtract the hundreds.
As there is no thousands place digit in the second number consider it to be zero and then subtract the thousands.

d.  Explanation:
Subtract the ones.
Rename 1 hundreds and 4 tens as “0 hundreds and 14 tens.” Subtract the tens.
Subtract the hundreds.
Subtract the thousands.

e.  Explanation:
Subtract the ones.
Rename 6 hundreds and 4 tens as “5 hundreds and 14 tens.” Subtract the tens.
Rename 8 thousands and 5 hundreds as “7 thousands and 15 hundreds.” Subtract the hundreds.
As there is no thousands place digit in the second number consider it to be zero and then subtract the thousands.

Round each number to the place named.

Question 7.
a.
592
hundreds
_________
600 is the rounded off number of 592,

Explanation:
To round off a number to hundreds we need to check the tens place as the tens place is more than 5 in the above question we can add 1 to the hundreds place by changing the tens and ones to zero, 592 is rounded off to 600.

b.
86 tens
_________
90 is the rounded off number of 86,

Explanation:
To round off a number to tens we need to check the ones place as the ones place is more than 5 in the above question we can add 1 to the tens place but change the ones to zero, 86 is rounded off to 90.

c.
432
hundreds
_________
400 is the rounded off number of 432,

Explanation:
To round off a number to hundreds we need to check the tens place as the tens place is less than 5 in the above question we need not change the  hundreds place but change the tens and ones to zero, 432 is rounded off to 400.

d.
981
tens
_________
980 is the rounded off number of 981,

Explanation:
To round off a number to tens we need to check the ones place as the ones place is less than 5 in the above question we need not change the  hundreds place and the tens place but change the ones to zero, 981 is rounded off to 980.

Solve each problem.

Question 8.
Jerry has 37 red marbles, ‘42 blue marbles, 13 black marbles, and 23 yellow marbles. How many marbles does Jerry have?
Jerry has __________ marbles.
Jerry has 115 marbles,

Explanation:
Number of red marbles Jerry has = 37, Number of blue marbles Jerry has = 42, Number of black marbles Jerry has = 13, Number of yellow marbles Jerry has = 23, Total number of marbles Jerry has = Number of red marbles Jerry has + Number of blue marbles Jerry has + Number of black marbles Jerry has + Number of yellow marbles Jerry has = 37 + 42 + 13 + 23 = 115.

Question 9.
In the year 1976, Mrs. Lopez was 82 years old. In what year was she born?
Mrs. Lopez was born in _____.
Mrs. Lopez was born in 1894,

Explanation:
Age of Mrs. Lopez in 1976 = 82 years old, Year in which Mrs. Lopez was born = Year – Age of Mrs. Lopez = 1976 – 82 = 1894.

Question 10.
Estella is 23 years old, Lydia is 27 years old, Toni is 42 years old, and Mai is 18 years old. What are their combined ages?
Their combined ages equal ____ years.
The combined ages are equal to 110 years,

Explanation:
Age of Estella = 23 years, Age of Lydia = 27 years, Age of Toni = 42 years, Age of Mai = 18 years, Total combined ages = Age of Estella + Age of Lydia + Age of Toni + Age of Mai = 23 + 27 + 42 + 18 = 110 years.

Question 11.
Marty earned 586 dollars one week at his job and 432 dollars the next week. Estimate about how much money Marty earned for both weeks.
Marty earned about _________ dollars for both weeks.