Test: Estimation And Rounding- 1 - GRE MCQ

Correct Answer :- c

Explanation : q is nearest to 7.5

that means 7.450 ≤ 7.4 ≤ 7.499

P = 5.369

Q = 7.450

Adding P + Q

= 5.369 + 7.450

= 12.819

For Q = 7.499

P = 5.369

Adding P + Q

= 5.369 + 7.499

= 12.868

12.819 < P+Q < 12.868

According to options, 12.864 lies between 12.819 and 12.868

Step 1: Question statement and Inferences

We need to find the approximate value of 32045 × 643 ÷ 15987

As all the answer options have zero in unit and tens digit

We can round all the numbers to hundreds place

Step 2: Finding required values

Rounding 32045 to hundreds = 32000

Rounding 643 to hundreds = 600

Rounding 15987 to hundreds = 16000

Step 3: Calculating the final answer

32045 × 643 ÷ 15987 ≃ 32000 * 600/16000 = 1200

Answer: Option (C)

Step 1: Question statement and Inferences

We have to find approximate value of 33% of 3.683 × 42.198 ÷ 20.679

Step 2: Finding required values

As we have to find the approximate value.

Rounding to the units digit:

• 3.683 = 4
• 42.198 = 42
• 20.679 = 21

Step 3: Calculating the final answer

Let’s simplify the expression

• N’ = 4 x 42/21
• N’ = 8

The final answer

• N = 0.33 x 8 = 33*8/100 = approx 33*8/99 = approx 8/3 = 2.67

Answer: Option (C)

Steps 1 & 2: Understand Question and Draw Inferences

We are given a number p. We have to find the hundredths digit of p.

Let’s say p = a.bcde           (Here a, b, c, d and e are digits)

We are given that on rounding to the hundredths digit, p becomes 6.72

So, by comparing the form a.bcde with 6.72, we conclude that:

a = 6, b = 7 and:

If d ≥ 5, then c = 1

If d < 5, then c = 2

We need to find the value of c.

Step 3: Analyze Statement 1

Statement 1 says: The thousandths digit of p is 3. 

So, the number a.bcde = a.bc3e  

That is, d = 3

From the condition above, this clearly implies that c = 2

Thus, Statement 1 alone is sufficient to answer the question: what is the value of c?

Step 4: Analyze Statement 2

Statement 2 says:  The sum of the hundredths and the thousandths digits of p is 5    

That is, c + d = 5

We have inferred that:

If d ≥ 5, then c = 1

In this case, the sum of c + d will range from 6 to 10, inclusive

We also inferred that

If d < 5, then c = 2

This means, if d = 0, 1, 2, 3 or 4, then c = 2

Out of these possible values of d, we see that if d = 3, then c = 2 and the condition given in Statement 2 (that c + d = 5) is also fulfilled.

Therefore, using Statement 2, we can deduce that:

c = 2

Thus, Statement 2 alone is sufficient to find the value of c.

Step 5: Analyze Both Statements Together (if needed)

This step is not needed as we get a unique value for c in both steps 3 and 4.

Answer: Option (D)  

Step 1 & 2 – Understand the question and draw inferences from the question statement.

Given: d = 26.rst

To find: If t – 4 > 0

• r, s, t are digits
• This means that, r, s are t are integers such that 0 ≤ r, s, t ≤ 9

We need to find if t – 4 > 0

i.e. if t > 4

Thus, we need to find if 5 ≤ t ≤ 9

Step 3 – Analyze Statement 1 Independently

Statement 1 – If 10d were rounded to the nearest tenths, the result would be 260.3

d = 26. rst

• 10d =  26r.st

Let’s round 10d to the nearest tenths:

Marking the digit at the tenths place: 26r.st

The digit on s’s right is t

         i.            If t ≥ 5, we will add 1 to s.

       ii.            If t < 5, we will keep s as it is

Therefore, 26r. st may be rounded to 26r.s0 (if t < 5) or  26r.(s+1)0 (if t ≥ 5) 

Comparing this with the number 260.3, we get:

1. s = 2, if t ≥ 5
2. s = 3, if t < 5

Thus, information provided in Statement 1 is not sufficient to arrive at a unique answer.

Step 4 – Analyze Statement 2 Independently

Per Statement 2, s is even.

However, to answer the given question, we need to find if   5 ≤ t ≤ 9

Thus, information provided in Statement 2 is not sufficient to arrive at a unique answer.

Step 5 – Analyze Both Statements Together

Per Statement 1, we have:

s = 2, if t ≥ 5 and s = 3, if t < 5

Per Statement 2, we have:

s is even.

Combining both statements, we have s = 2

• t ≥ 5

Since t is a digit, t cannot be greater than 9.

• 5 ≤ t ≤ 9

Therefore statement 1 and statement 2 together are sufficient to arrive at a unique answer.

Answer: Option (C)

Steps 1 & 2: Understand Question and Draw Inferences

We are given that:

a = 3.78xy7                (Here, x and y are positive digits)

b = 1.37486   

We have to find the value of a + b.

Now, to find the sum of a and b, first we need to find the values of x and y so that we can determine the value of a. Since we already have the value of b, we can then add both of them and find a + b. 

So, the question here is what is the value of a?

Step 3: Analyze Statement 1

Statement 1 says: If a and b are rounded to nearest thousandths digit, then a + b = 5.162

So, we know that:

(Rounded value of a to nearest thousandths place) + (Rounded value of b to nearest thousandths place) = 5.162          ……………… (1)

Now, we can’t find the rounded value of a, but we can find the rounded value of b to nearest thousandths place.

Now, let’s round b to the nearest thousandths:

• Marking the digit at the thousandths place: 1.37486
• Since the digit on 4’s right is 8 (and therefore, greater than 5), we will add 1 to 4. 
• Therefore, 1.37486 may be rounded to 1.375  

By putting the rounded value of b in equation (1):

(Rounded value of a to nearest thousandths place) + 1.375 = 5.162

(Rounded value of a to nearest thousandths place) = 5.162 – 1.375 = 3.787

So, the rounded value of 3.78xy7 to the nearest thousandths place = 3.787

This means,

If y ≥ 5, then x = 6

If y < 5, then x = 7

Thus, we are not able to determine a unique value of x and y each.

Hence, statement 1 is not sufficient to answer the question: What is the value of a?  

Step 4: Analyze Statement 2

Statement 2 says:  If a and b are rounded to nearest ten-thousandths digit, then a + b = 5.1622

(Rounded value of a to nearest ten-thousandths place) + (Rounded value of b to nearest ten-thousandths place) = 5.1622                         ……………… (2) 

Let’s round b to the nearest ten-thousandths:

• Marking the digit at the thousandths place: 1.37486
• Since the digit on 8’s right is 6 (and therefore, greater than 5), we will add 1 to 8. 
• Therefore, 1.37486 may be rounded to 1.3749 

By putting the rounded value of b in equation (2):

(Rounded value of a to nearest ten-thousandths place) + 1.3749 = 5.1622

 (Rounded value of a to nearest ten-thousandths place) = 5.1622 – 1.3749 = 3.7873  

So, the rounded value of 3.78xy7 to the nearest ten-thousandths place = 3.7873

Now, let’s round a to the nearest ten-thousandths:

• Marking the digit at the thousandths place: 3.78xy7
• Since the digit on y’s right is 7, we will add 1 to y. 
• Now, when we add 1 to y, a becomes 3.7873
• This implies that y + 1 = 3.

 So, the value of y = 2.

Thus, a = 3.78x27

Also, by comparing 3.7873 to 3.78x27 we get to know that the value of x is 7.

Thus, we know that the value of a is 3.78727. 

So, statement 2 alone is sufficient to answer the question: What is the value of a?

Step 5: Analyze Both Statements Together (if needed)

Since statement 2 alone is sufficient to answer the question, we don’t need to perform this step.

Answer: Option (B) 

Given: j = ab.pqr

• 0 ≤ a, b, p, q, r ≤ 9

Also, p + q = 3

• 0 ≤ p ≤ 3 and 0 ≤ q ≤ 3

((p, q) may be (0,3), (1, 2), (2, 1) or (3, 0))

Also given that q + r = 9

We already concluded from the above that 0 ≤ q ≤ 3,

• 6 ≤ r ≤ 9

((q, r) may be (0, 9), (1, 8), (2, 7) or (3, 6))

Now, let’s analyze the given statements:

Statement I: If j/10 were rounded to the nearest thousandths, the result would contain the digit q

j = ab.pqr

• j/10 = a.bpqr

Let’s round j/10 to the nearest thousandths:

• Step R1 -Marking the digit at the thousandths place: a.bpqr
• Step R2 -The digit on q’s right is r.
• Step R3 – Since r > 5, we will add 1 to q.
• Step R4 – a.bpqr will be rounded to a.bp(q+1)0.

Conclusion: Statement (1) is FALSE.

Statement II: If 100j were rounded to the nearest tens, the result would contain the digit p

j = ab.pqr

• 100j = abpq.r

Let’s round 100j to the nearest tens:

• Step R1 -Marking the digit at the tens place: abpq.r
• Step R2 -The digit on p’s right is q.
• Step R3 – Since q < 5, we will keep p as it is. (Note that in fact q < 4.)
• Step R4 – abpq.r will be rounded to abp0.0

Conclusion: Statement (2) is TRUE.

Statement III: If 1000j were rounded to the nearest thousands, the result would contain the digit b.

j = ab.pqr

• 1000j = abpqr

Let’s round 1000j to the nearest thousands:

• Step R1 -Marking the digit at the thousands place: abpqr.
• Step R2 -The digit on b’s right is p.
• Step R3 – Since p < 5, we will keep b as it is. (Note that in fact p < 4.)
• Step R4 – abpqr will be rounded to ab000

Conclusion: Statement (3) is TRUE.

 Overall Conclusion: Only statements (2) and (3) are true.

Answer: Option (D)