Today's Calibrated GRE Daily Quiz (20 Questions)

Question 1

Question 1 Quantitative arithmetic

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When the positive integer

n

is divided by

7

, the remainder is

3

. What is the remainder when

2n + 5

is divided by

7

?

Answer

Correct Answer:

4

Step-by-Step Proof:

By the division algorithm, since

n

leaves a remainder of

3

when divided by

7

,

n

can be written in the form:

n = 7k + 3

for some non-negative integer

k

. Substitute this representation into

2n + 5

:

2n + 5 = 2(7k + 3) + 5 = 14k + 6 + 5 = 14k + 11

Now express

14k + 11

in terms of multiples of

7

:

14k + 11 = 14k + 7 + 4 = 7(2k + 1) + 4

Since

4

is an integer such that

0 \le 4 < 7

, the remainder when

2n + 5

is divided by

7

is

4

. Alternatively, pick a concrete value for

n

: let

n = 3

. Then

2(3) + 5 = 11

. Dividing

11

by

7

yields a quotient of

1

and a remainder of

4

.

Common ETS Trap Warning:

Test-takers often forget to reduce the combined constant modulo

7

. After computing

2(3) + 5 = 11

, forgetting to divide

11

by

7

leads to confusion, or simply adding

3 + 5 = 8 \equiv 1

by omitting the factor of

2

multiplying

n

.

🎯 Key Rule:

Remainder arithmetic is linear: substitute the remainder directly into the algebraic expression and take the resulting value modulo the divisor.

Practice More arithmetic Drills →

Question 2

Question 2 Quantitative arithmetic

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If

n

is an integer such that the units digit of

n^2

is 6, what is the sum of all possible units digits of

(n+1)^2

?

Answer

Correct Answer: 14

Step-by-Step Proof:

Since the units digit of

n^2

is 6,

n

must end in 4 or 6 because only

4^2=16

and

6^2=36

give a units digit of 6. If

n

ends in 4, then

n+1

ends in 5 and

5^2=25

, giving a units digit of 5. If

n

ends in 6, then

n+1

ends in 7 and

7^2=49

, giving a units digit of 9. The sum of the possible units digits is

5+9=14

.

Common ETS Trap Warning:

Distractors 5 and 9 correspond to considering only one of the two possible endings for

n

; 10 arises from mistakenly adding the digits 4 and 6; 18 results from incorrectly multiplying the two possible units digits.

🎯 Key Rule:

When a square ends in 6, the original number must end in 4 or 6; always consider both possibilities before computing related expressions.

Practice More arithmetic Drills →

Question 3

Question 3 Quantitative algebra

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If

3x - 7 = 2(x + 4) - 5

, what is the value of

x

?

Answer Box: (Enter decimal e.g. 2.4 or fraction e.g. 12/5)

Answer

Correct Answer: 10

Step-by-Step Proof:

Expand the right-hand side of the linear equation:

3x - 7 = 2(x + 4) - 5

3x - 7 = 2x + 8 - 5

3x - 7 = 2x + 3

Subtract

2x

from both sides:

x - 7 = 3

Add

7

to both sides:

x = 10

The solution is

10

.

Common ETS Trap Warning:

A frequent sign error occurs when distributing the

2

into the parentheses or incorrectly adding

7

to

3

(e.g., subtracting

7

to get

-4

or forgetting to distribute the

2

to the

4

).

🎯 Key Rule:

Always expand and combine like terms on each side of a linear equation before transposing variables across the equals sign.

Practice More algebra Drills →

Question 4

Question 4 Quantitative data analysis

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A university department surveyed

120

undergraduate students regarding their enrollment in three foundational courses: Mathematics, Physics, and Chemistry. The survey revealed the following: -

75

students are enrolled in Mathematics -

52

students are enrolled in Physics -

40

students are enrolled in Chemistry -

28

students are enrolled in both Mathematics and Physics -

22

students are enrolled in both Mathematics and Chemistry -

18

students are enrolled in both Physics and Chemistry -

12

students are enrolled in all three courses How many of the students surveyed are enrolled in none of these three courses?

Answer

Correct Answer: 9

Step-by-Step Proof:

By the Principle of Inclusion-Exclusion for three sets

M

,

P

, and

C

:

|M \cup P \cup C| = (|M| + |P| + |C|) - (|M \cap P| + |M \cap C| + |P \cap C|) + |M \cap P \cap C|

Substitute the values provided in the problem:

|M| + |P| + |C| = 75 + 52 + 40 = 167

|M \cap P| + |M \cap C| + |P \cap C| = 28 + 22 + 18 = 68

|M \cap P \cap C| = 12

Calculate the total number of students enrolled in at least one of the three courses:

|M \cup P \cup C| = 167 - 68 + 12 = 111

The number of students enrolled in none of the three courses is the total surveyed minus the union:

\text{None} = 120 - 111 = 9

Common ETS Trap Warning:

Test takers often subtract the three-way intersection (

12

) instead of adding it back in, calculating

167 - 68 - 12 = 87

, leading to an incorrect result of

120 - 87 = 33

. Alternatively, examinees mistakenly assume the two-way intersections are 'mutually exclusive of the triple intersection' and subtract 12 from each pair first before using the standard formula.

🎯 Key Rule:

Always apply the alternating signs of Inclusion-Exclusion: add singletons, subtract pairs, and add back the triple intersection.

Practice More data analysis Drills →

Today's Calibrated GRE Daily Quiz (20 Questions)

Daily 20-Question Calibration Blueprint

Easy Calibration

Medium Calibration

Hard Calibration

Sample Problems from Today's Challenge (Thursday, October 29, 2026)

Why 20 Questions a Day Outperforms Weekend Cramming

🧠 The Testing Effect & Active Recall

🎯 Reflex Trap Resistance

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