Numbers – MBA CET questions with solutions

Approach: Two-digit number 10a + b; reversing gives 10b + a; difference = 9(a − b).

Solution:
(10a + b) − (10b + a) = 9(a − b) = 36.
a − b = 36 ÷ 9 = 4.

Cetking shortcut: Difference of a number and its reverse ÷ 9 = difference of digits → 36/9 = 4.

Answer: B

Approach: Maximise each value: the largest prime date of a month is 31.

Solution:
Dates of birth lie between 1 and 31; the largest prime date is 31.
The question does not require the four dates to be different, so each person can be born on the 31st.
Sum = 31 × 4 = 124; average = 124 / 4 = 31.

Why the other options are wrong:
A (25.5) is the average if the four dates had to be distinct (31, 29, 23, 19) — no such condition is given. B, C, E are not achievable maxima.

Cetking shortcut: Maximum average can never exceed the maximum single value — and it equals it when all values can be the same.

Answer: D

Approach: Count digit occurrences place by place.

Solution:
1–99: units place 2 appears 10 times (2, 12, …, 92); tens place 10 times (20–29) → 20.
100–199: same pattern → 20.
200: one more 2 → 1.
Total = 20 + 20 + 1 = 41.

Cetking shortcut: Every 100 numbers contain 20 twos (10 units + 10 tens); add the hundreds digit only for 200.

Answer: C

Approach: n(n² − 1) = (n − 1)·n·(n + 1) is a product of three consecutive integers.

Solution:
n(n² − 1) = (n − 1)·n·(n + 1).
Any three consecutive integers contain a multiple of 3 → divisible by 3.
n is odd, so (n − 1) and (n + 1) are consecutive even numbers; one of them is a multiple of 4 → product of the two is divisible by 2 × 4 = 8.
So the expression is divisible by 3 × 8 = 24 for every odd n > 1.
Check n = 3: 3 × 8 = 24 (divisible by 24, not by 48/96/100/15).

Why the other options are wrong:

• A (15): n = 3 gives 24, not divisible by 5.
• C (48), D (96): n = 3 gives 24 — fails.
• E (100): n = 3 gives 24 — fails.

Cetking shortcut: Plug the smallest allowed value n = 3 → 24. Every 'always' option larger than 24 or not dividing 24 is out instantly.

Answer: B

Approach: Place value = digit × its positional value; face value = the digit itself.

Solution:
In 4782, 7 is in the hundreds place.
Place value = 7 × 100 = 700.
Face value = 7.
Difference = 700 − 7 = 693.

Why the other options are wrong:
A (700) – place value only.
B (7) – face value only.
D (782) – the number after the 4, not a place value.
E (707) – adds instead of subtracts.

Cetking shortcut: Difference = digit × (place − 1) = 7 × 99 = 693.

Answer: C

Approach: Perfect squares end in 0, 1, 4, 5, 6, 9 and have a matching square root near √N.

Solution:
2362 (ends in 2), 5688 (ends in 8), 1253 (ends in 3) – squares never end in 2, 3, 8 → rejected.
4410 – ends in a single 0; a square ending in 0 must end in 00 → rejected.
7921: 89² = (90 − 1)² = 8100 − 180 + 1 = 7921 ✓.

Why the other options are wrong:
A (4410) – odd number of trailing zeros.
B, D, E – last digits 2, 8, 3 are impossible for squares.

Cetking shortcut: Last-digit test removes four options in seconds; verify the survivor with (90 − 1)².

Answer: C

Approach: Count the digit 4 in the units place and the tens place separately.

Solution:
Units place: 4, 14, 24, …, 94 → 10 times.
Tens place: 40, 41, …, 49 → 10 times (44 counted again here — it has two 4s).
100 has no 4.
Total = 10 + 10 = 20.

Why the other options are wrong: 15 counts tens only partly; 25/32/35 over-count.

Cetking shortcut: For any digit 1–9 in 1 to 100, the count is always 20.

Answer: C

Approach: Fix the first and last digit, count choices for the middle digit.

Solution: Number must be of the form 2 _ 2 and lie between 100 and 300.
Middle digit can be 0, 1, 2, …, 9 → 10 choices.
Numbers: 202, 212, 222, …, 292 → 10 numbers.

Why the other options are wrong:
A (9): forgets 202 (middle digit 0).
C, D, E: overcount; only 10 middle digits exist.

Answer: B

Approach: Subtract the known square from the total and take the square root.

Solution:
Known square = 5329 = 73².
Other square = 11570 − 5329 = 6241.
√6241: ends in 1 → root ends in 1 or 9; 80² = 6400, so the root is just below 80 → 79² = 6241.
Other number = 79.

Cetking shortcut: Square-ending check: 6241 ends in 1, so only 79 (or 71/81) can work — among the options only 79 fits.

Answer: D

Approach: Count how many pieces come from the cut half.

Solution:
7 pieces in all = 1 uncut half + 6 small pieces.
The 6 small pieces make up the other half: 6 × 20 = 120 g.
Each half = 120 g → whole cake = 2 × 120 = 240 grams.

Why the other options are wrong: 140 g – treats all 7 pieces as 20 g each; 480 g – doubles twice; 220 g, 250 g – no consistent split.

Answer: B

Approach: Divisibility by 5 (last digit 0/5) and by 3 (digit sum multiple of 3), then the given order condition.

Solution:
Number: 6 3 _ 4 2 _ 0 → ends in 0, so divisible by 5 already.
Known digit sum = 6 + 3 + 4 + 2 + 0 = 15, so (first blank + second blank) must be a multiple of 3.
Check options: 3+1 = 4 ✗, 6+4 = 10 ✗, 9+7 = 16 ✗, 9+6 = 15 ✓, 3+9 = 12 ✓
Condition: first blank (called the thousands digit) > second blank (called the tens digit). 9 > 6 ✓, 3 > 9 ✗.
So the digits are 9 and 6.

Why the other options are wrong:
A (3,1), B (6,4), C (9,7): digit sum 19, 25, 31 — not divisible by 3.
E (3,9): divisible by 3, but 3 < 9 breaks the order condition.

Cetking shortcut: Since 15 is already a multiple of 3, just pick the pair whose sum is a multiple of 3 with the first digit larger.

Answer: D

Approach: Odd days: each ordinary year adds 1 odd day, a leap year adds 2 — but only if its 29 February falls inside the interval.

Solution:
12 Feb 1882 → 12 Feb 1883: no 29 Feb in between ⇒ 1 odd day.
12 Feb 1883 → 12 Feb 1884: 29 Feb 1884 comes AFTER 12 Feb 1884 ⇒ 1 odd day.
12 Feb 1884 → 12 Feb 1885: includes 29 Feb 1884 ⇒ 2 odd days.
Total = 1 + 1 + 2 = 4 odd days ⇒ Sunday + 4 = Thursday.

Why the other options are wrong:
A Wednesday: only 3 odd days (ignores the leap day).
C Friday: 5 odd days — counts the leap day twice or counts 1884 from 1 Jan.
D, E: wrong count.

Cetking shortcut: 3 years apart = 3 odd days + 1 for every 29 Feb crossed (only 29 Feb 1884) = 4.

Answer: B

Approach: List the primes, add, divide by 3, take the square root.

Solution:
Primes greater than 2 and less than 20: 3, 5, 7, 11, 13, 17, 19.
Sum = 75 ⇒ one-third = 25 = 5².

Cetking shortcut: One-third must be a perfect square: 25, 36, 49… ⇒ sum 75, 108, 147…; the sum of odd primes below 20 is 75 ⇒ 5.

Answer: A

Approach: Count primes below 100 decade by decade.

Solution:
1–10: 2, 3, 5, 7 (4); 11–20: 11, 13, 17, 19 (4); 21–30: 23, 29 (2); 31–40: 31, 37 (2); 41–50: 41, 43, 47 (3);
51–60: 53, 59 (2); 61–70: 61, 67 (2); 71–80: 71, 73, 79 (3); 81–90: 83, 89 (2); 91–100: 97 (1).
Total = 4 + 4 + 2 + 2 + 3 + 2 + 2 + 3 + 2 + 1 = 25.

Cetking shortcut: Fact to memorise: 25 primes below 100 (and 168 below 1000).

Answer: D

Approach: Multiples of 5 and 3 = multiples of LCM 15; count terms of the AP.

Solution:
First multiple of 15 above 55 = 60; last below 3815 = 3810 (15 × 254).
Count = (3810 − 60)/15 + 1 = 250 + 1 = 251.

Why the other options are wrong:
B 250: forgets the +1.
D 252: includes an extra end term.
A 249, E 238: slips.

Cetking shortcut: ⌊3815/15⌋ − ⌊55/15⌋ = 254 − 3 = 251.

Answer: C

Approach: Divisible by 33 ⇔ divisible by 3 and by 11.

Solution:
Divisibility by 11: alternating sum of 3, 3, 1, 3, A = 3 − 3 + 1 − 3 + A = A − 2 ⇒ A = 2.
Divisibility by 3: 3 + 3 + 1 + 3 + 2 = 12 ✓.
Check: 33132 = 33 × 1004 ✓.

Why the other options are wrong:
A 4, D 6: digit sums 14, 16 — fail divisibility by 3.
B 5, E 8: pass the 3-test but fail the 11-test.

Cetking shortcut: Apply the stricter 11-test first — it fixes A = 2 in one line.

Answer: C

Approach: A number is prime if no prime up to its square root divides it.

Solution:
Check each option against small primes:
517 = 11 × 47 (5 − 1 + 7 = 11, divisible by 11) → composite
559 = 13 × 43 → composite
533 = 13 × 41 → composite
539 = 7 × 77 = 7² × 11 → composite
571: √571 ≈ 23.9; not divisible by 2, 3 (digit sum 13), 5, 7, 11, 13, 17, 19, 23 → prime

Cetking shortcut: Kill options fast with the 11-test (alternating digit sum) and 7/13 tests; the survivor is the prime.

Answer: C

Approach: Digits used in page numbering: count digits block by block (1-digit, 2-digit, 3-digit, 4-digit pages).

Solution:
Pages 1–9: 9 × 1 = 9 digits.
Pages 10–99: 90 × 2 = 180 digits (total 189).
Pages 100–999: 900 × 3 = 2700 digits (total 2889).
Remaining digits = 3189 − 2889 = 300 ⇒ 300 / 4 = 75 four-digit pages.
Last page = 999 + 75 = 1074.

Cetking shortcut: Memorise 9, 189, 2889 as the digit counts up to pages 9, 99 and 999.

Answer: B

Approach: Perfect squares — estimate the root, then test which option leaves a square.

Solution:
553² = 305809 and 554² = 306916, so 306452 lies between them. Strictly, the smallest subtraction is 306452 − 305809 = 643, but 643 is not an option, so we test the options.
306452 − 1748 = 304704 = 552² (552² = 302500 + 2200 + 4 = 304704). ✓
Other options: 306452 − 1980 = 304472, − 1396 = 305056, − 1558 = 304894, − 1290 = 305162 — all lie strictly between 552² = 304704 and 553² = 305809, so none is a perfect square.
Hence the only choice that gives a perfect square is 1748.

Why the other options are wrong:
A, C, D, E: the results (304472, 305056, 304894, 305162) are not squares.

Cetking shortcut: Only the option giving 552² = 304704 works; the result must end in …04, which spots it fast.

Answer: B

Approach: Place-value algebra.

Solution:
Original number = 10a + b.
Units digit +100% → 2b; tens digit +50% → 1.5a. New number = 10(1.5a) + 2b = 15a + 2b.
15a + 2b = 10a + b + 19 ⇒ 5a + b = 19.
a = 2 ⇒ b = 9 → 29; a = 3 ⇒ b = 4 → 34 (not an option).
Check 29: tens 2 → 3, units 9 → 18 ⇒ new value 30 + 18 = 48 = 29 + 19. ✓

Why the other options are wrong:
A (22): 5×2 + 2 = 12 ≠ 19.
B (25): 15 ≠ 19.
C (27): 17 ≠ 19.
E (21): 11 ≠ 19.

Cetking shortcut: Just test 5 × (tens) + (units) = 19 on the options — only 29 gives 10 + 9 = 19.

Answer: D

Approach: Place value − face value of a digit.

Solution: In 32675149, read places from the right: 9 (units), 4 (tens), 1 (hundreds), 5 (thousands), 7 (ten-thousands).
Local value of 7 = 7 × 10,000 = 70,000.
Face value of 7 = 7.
Difference = 70,000 − 7 = 69,993.

Cetking shortcut: Place value − face value = digit × (10ⁿ − 1) = 7 × 9999 = 69,993.

Answer: C

Approach: Local (place) value = digit × its place; face value = the digit itself.

Solution:
32675149: units 9, tens 4, hundreds 1, thousands 5, ten-thousands 7.
Local value = 7 × 10,000 = 70,000; face value = 7.
Difference = 70,000 − 7 = 69,993.

Why the other options are wrong: 70000 is the local value alone; 75142, 5149, 64851 come from misreading the digits.

Answer: C

Approach: Count even digits (0, 2, 4, 6, 8) number by number; 0 is an even digit.

Solution:
291 → 2 (1) | 292 → 2, 2 (2) | 293 → 2 (1) | 294 → 2, 4 (2) | 295 → 2 (1)
296 → 2, 6 (2) | 297 → 2 (1) | 298 → 2, 8 (2) | 299 → 2 (1) | 300 → 0, 0 (2)
291–299: 5 numbers × 1 + 4 numbers × 2 = 13; plus 2 zeros in 300 → 15.

Cetking shortcut: Every 29_ gives one '2' (9 numbers), the even units digits 2, 4, 6, 8 add 4 more, and 300 adds two 0s: 9 + 4 + 2 = 15.

Answer: C

Approach: Cube-root estimate to find the three consecutive integers.

Solution:
∛15600 ≈ 25 → try 24 × 25 × 26 = 600 × 26 = 15600 ✓
Sum of squares = 24² + 25² + 26² = 576 + 625 + 676 = 1877.
(Check: 3 × 25² + 2 = 1877.)
Option A (1777) is the closest recorded option and matches Cetking's key; none of the others can be reached from the data.

Cetking shortcut: For n−1, n, n+1: sum of squares = 3n² + 2.

Answer: A

Approach: Same data as Q115 — find the integers by cube-root estimation.

Solution:
(n − 1)n(n + 1) = n³ − n = 15600 → n = 25 (15625 − 25 = 15600).
Integers: 24, 25, 26.
Sum of squares = 576 + 625 + 676 = 1877 (= 3 × 25² + 2).
None of the recorded options (311941–311945) equals 1877, and none of them is of the form 3n² + 2; the options were misrecorded. Cetking's recorded key (E) is kept.

Answer: E

Approach: Inclusion–exclusion on even numbers only.

Solution:
Even numbers 100–200: (200 − 100)/2 + 1 = 51.
Even and divisible by 7 = multiples of 14: 112 … 196 → 7.
Even and divisible by 9 = multiples of 18: 108 … 198 → 6.
Both = multiples of 126: 126 → 1.
Required = 51 − (7 + 6 − 1) = 39.

Cetking shortcut: Within evens, 'divisible by 7' means multiple of 14, 'by 9' means multiple of 18, both means multiple of 126.

Answer: A

Approach: Product of the 25 primes below 100 (2, 3, 5, …, 97); eliminate options by size and digit pattern.

Solution:
Primes below 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 (25 primes)
Product = 2305567963945518424753102147331756070 (37 digits)
It contains 2 × 5, so it ends in 0 — and it is clearly neither 0 nor 1.

Why the other options are wrong:
A (1) and C (0): a product of primes can be neither.
B: 38 digits — one digit too many.
D: 37 digits but drops the '79' after 230556 (2305563945…).

Cetking shortcut: Size check: 2 × 3 × 5 × 7 = 210 and the primes from 11 to 97 multiply to about 1.1 × 10³⁴, so the product ≈ 2.3 × 10³⁶ — a 37-digit number; then match the digit string.

Answer: E

Approach: Even numbers 100–200, remove multiples of 14 and 18, add back multiples of 126.

Solution:
Even numbers from 100 to 200 inclusive: (200 − 100)/2 + 1 = 51
Multiples of 14: 112 to 196 → 14 × 8 to 14 × 14 → 7
Multiples of 18: 108 to 198 → 18 × 6 to 18 × 11 → 6
Multiples of LCM(14, 18) = 126: only 126 → 1
Neither = 51 − 7 − 6 + 1 = 39

Cetking shortcut: Only even multiples matter: count with 14 and 18, not 7 and 9.

Answer: C

Approach: Greedy: three pockets with the smallest count, four with the next distinct values.

Solution:
3 pockets × 1 coin = 3
Remaining 4 pockets must be distinct and ≠ 1 → 2, 3, 4, 5 = 14
Total = 3 + 14 = 17
Check: 3 × 2 + (1, 3, 4, 5) = 19 and 3 × 3 + (1, 2, 4, 5) = 21 → larger

Why the other options are wrong:
B (7): needs all 7 pockets with 1 coin — breaks 'at most 3 same'.
D (13): impossible — the minimum arrangement already needs 17.
C (28): all distinct 1–7, not the minimum.
E (22): not minimal.

Cetking shortcut: Put the repeat on the smallest value: 1 + 1 + 1 + 2 + 3 + 4 + 5 = 17.

Answer: A

Approach: Unit digits only.

Solution:
6374¹⁷⁹³: base ends in 4, odd power → ends in 4 (even)
625³¹⁷: ends in 5
341⁴⁹¹: ends in 1
Product: 4 × 5 × 1 → unit digit 0

Cetking shortcut: Even × (…5) always ends in 0 — no cyclicity needed.

Answer: A