10:[["$","title","0",{"children":"CUET General Test - How many two-digit numbers are divisible by 3? | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined Two-digit numbers ran>= from 10 to 99.

The first two-digit number divisib<= by 3: 10 ÷ 3 is not a who<= number 11 ÷ 3 is not a who<= number 12 ÷ 3 = 4 The first two-digit number divisib<= by 3 is 12.

The last two-digit number divisib<= by 3: 99 ÷ 3 = 33 The last two-digit number divisib<= by 3 is 99.

The two-digit numbers divisib<= by 3 form the sequence: 12, 15, 18, 21, 24, ... 96, 99 Each number is 3 more than the previous one (common difference = 3).

For a sequence with constant difference: Number of terms = Last - FirstCommon difference + 1 n = 99 - 123 + 1 n = 873 + 1 n = 29 + 1 n = 30

Therefore, there are 30 two-digit numbers divisib<= by 3. The answer is Option 1: 30"}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$undefined"}],["$","meta","3",{"name":"keywords","content":"CUETGET,GET,Number System,Divisibility Rules"}],["$","meta","4",{"name":"robots","content":"index, follow"}],["$","meta","5",{"name":"category","content":"education"}],["$","meta","6",{"name":"google-site-verification","content":"google"}],["$","meta","7",{"property":"og:title","content":"CUET General Test - How many two-digit numbers are divisible by 3? | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined Two-digit numbers ran>= from 10 to 99.

The first two-digit number divisib<= by 3: 10 ÷ 3 is not a who<= number 11 ÷ 3 is not a who<= number 12 ÷ 3 = 4 The first two-digit number divisib<= by 3 is 12.

The last two-digit number divisib<= by 3: 99 ÷ 3 = 33 The last two-digit number divisib<= by 3 is 99.

The two-digit numbers divisib<= by 3 form the sequence: 12, 15, 18, 21, 24, ... 96, 99 Each number is 3 more than the previous one (common difference = 3).

For a sequence with constant difference: Number of terms = Last - FirstCommon difference + 1 n = 99 - 123 + 1 n = 873 + 1 n = 29 + 1 n = 30

Therefore, there are 30 two-digit numbers divisib<= by 3. The answer is Option 1: 30"}],["$","meta","9",{"property":"og:url","content":"https://www.afterboards.in/past-year-questions/cuet-general-test/2025/slot-10/14"}],["$","meta","10",{"property":"og:site_name","content":"AfterBoards | CUET General Test Preparation"}],["$","meta","11",{"property":"og:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","meta","12",{"property":"og:type","content":"website"}],["$","meta","13",{"name":"twitter:card","content":"summary_large_image"}],["$","meta","14",{"name":"twitter:site","content":"afterboards.in"}],["$","meta","15",{"name":"twitter:title","content":"CUET General Test - How many two-digit numbers are divisible by 3? | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined Two-digit numbers ran>= from 10 to 99.

The first two-digit number divisib<= by 3: 10 ÷ 3 is not a who<= number 11 ÷ 3 is not a who<= number 12 ÷ 3 = 4 The first two-digit number divisib<= by 3 is 12.

The last two-digit number divisib<= by 3: 99 ÷ 3 = 33 The last two-digit number divisib<= by 3 is 99.

The two-digit numbers divisib<= by 3 form the sequence: 12, 15, 18, 21, 24, ... 96, 99 Each number is 3 more than the previous one (common difference = 3).

For a sequence with constant difference: Number of terms = Last - FirstCommon difference + 1 n = 99 - 123 + 1 n = 873 + 1 n = 29 + 1 n = 30

Therefore, there are 30 two-digit numbers divisib<= by 3. The answer is Option 1: 30"}],"$L18","$L19","$L1a","$L1b","$L1c","$L1d","$L1e","$L1f","$L20","$L21"]

CUET General Test 2025 20 May Shift 2 PYQs

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