Number System: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 2)

Natural numbers are 1, 2, 3, and so on. Adding 0 to this set gives 0, 1, 2, 3, …. This exact set — all natural numbers plus zero — is by definition called the whole numbers. So the answer is whole numbers.

Since PQ = 90, P and Q must be a factor pair of 90. Listing the factor pairs of 90 and their sums: for each divisor d of 90, P=d and Q=90/d give a sum of d+90/d. This produces the possible sums 19, 21, 23, 33, 47, 91. The value 98 is not among these possible sums, so it can NOT be P+Q.

Since PQ = 48, P and Q must be a factor pair of 48. Listing the factor pairs of 48 and their sums: for each divisor d of 48, P=d and Q=48/d give a sum of d+48/d. This produces the possible sums 14, 16, 19, 26, 49. The value 52 is not among these possible sums, so it can NOT be P+Q.

Since PQ = 96, P and Q must be a factor pair of 96. Listing the factor pairs of 96 and their sums: for each divisor d of 96, P=d and Q=96/d give a sum of d+96/d. This produces the possible sums 20, 22, 28, 35, 50, 97. The value 104 is not among these possible sums, so it can NOT be P+Q.

Since PQ = 120, P and Q must be a factor pair of 120. Listing the factor pairs of 120 and their sums: for each divisor d of 120, P=d and Q=120/d give a sum of d+120/d. This produces the possible sums 22, 23, 26, 29, 34, 43, 62, 121. The value 128 is not among these possible sums, so it can NOT be P+Q.

Since PQ = 72, P and Q must be a factor pair of 72. Listing the factor pairs of 72 and their sums: for each divisor d of 72, P=d and Q=72/d give a sum of d+72/d. This produces the possible sums 17, 18, 22, 27, 38, 73. The value 76 is not among these possible sums, so it can NOT be P+Q.

Since PQ = 84, P and Q must be a factor pair of 84. Listing the factor pairs of 84 and their sums: for each divisor d of 84, P=d and Q=84/d give a sum of d+84/d. This produces the possible sums 19, 20, 25, 31, 44, 85. The value 90 is not among these possible sums, so it can NOT be P+Q.

We need pairs (a, b) with a+b=19, a≠b, both positive, counting each pair once. Pairing 1 with 18, 2 with 17, and so on: each pair (a, 19-a) is valid as long as a < 19-a, i.e. a < 19/2 = 9.5. So a ranges from 1 to 9, giving 9 distinct unordered pairs. Check: the last pair is (9, 10), and 9 ≠ 10 confirms they're different numbers.

We need pairs (a, b) with a+b=13, a≠b, both positive, counting each pair once. Pairing 1 with 12, 2 with 11, and so on: each pair (a, 13-a) is valid as long as a < 13-a, i.e. a < 13/2 = 6.5. So a ranges from 1 to 6, giving 6 distinct unordered pairs. Check: the last pair is (6, 7), and 6 ≠ 7 confirms they're different numbers.

We need pairs (a, b) with a+b=16, a≠b, both positive, counting each pair once. Pairing 1 with 15, 2 with 14, and so on: each pair (a, 16-a) is valid as long as a < 16-a, i.e. a < 16/2 = 8.0. So a ranges from 1 to 7, giving 7 distinct unordered pairs. Check: the last pair is (7, 9), and 7 ≠ 9 confirms they're different numbers.

We need pairs (a, b) with a+b=18, a≠b, both positive, counting each pair once. Pairing 1 with 17, 2 with 16, and so on: each pair (a, 18-a) is valid as long as a < 18-a, i.e. a < 18/2 = 9.0. So a ranges from 1 to 8, giving 8 distinct unordered pairs. Check: the last pair is (8, 10), and 8 ≠ 10 confirms they're different numbers.

We need pairs (a, b) with a+b=22, a≠b, both positive, counting each pair once. Pairing 1 with 21, 2 with 20, and so on: each pair (a, 22-a) is valid as long as a < 22-a, i.e. a < 22/2 = 11.0. So a ranges from 1 to 10, giving 10 distinct unordered pairs. Check: the last pair is (10, 12), and 10 ≠ 12 confirms they're different numbers.

We need pairs (a, b) with a+b=11, a≠b, both positive, counting each pair once. Pairing 1 with 10, 2 with 9, and so on: each pair (a, 11-a) is valid as long as a < 11-a, i.e. a < 11/2 = 5.5. So a ranges from 1 to 5, giving 5 distinct unordered pairs. Check: the last pair is (5, 6), and 5 ≠ 6 confirms they're different numbers.

Let the required number be y. We need y x 16 = 120 x 252. First compute the right side: 120 x 252 = 30240. Then solve for y: y = 30240/16 = 1890. Check: 1890 x 16 = 30240, which equals 120 x 252 = 30240.

Let the required number be y. We need y x 16 = 83 x 304. First compute the right side: 83 x 304 = 25232. Then solve for y: y = 25232/16 = 1577. Check: 1577 x 16 = 25232, which equals 83 x 304 = 25232.

Let the required number be y. We need y x 18 = 294 x 339. First compute the right side: 294 x 339 = 99666. Then solve for y: y = 99666/18 = 5537. Check: 5537 x 18 = 99666, which equals 294 x 339 = 99666.

Let the required number be y. We need y x 24 = 144 x 279. First compute the right side: 144 x 279 = 40176. Then solve for y: y = 40176/24 = 1674. Check: 1674 x 24 = 40176, which equals 144 x 279 = 40176.

Let the required number be y. We need y x 48 = 92 x 276. First compute the right side: 92 x 276 = 25392. Then solve for y: y = 25392/48 = 529. Check: 529 x 48 = 25392, which equals 92 x 276 = 25392.

Let the required number be y. We need y x 36 = 154 x 252. First compute the right side: 154 x 252 = 38808. Then solve for y: y = 38808/36 = 1078. Check: 1078 x 36 = 38808, which equals 154 x 252 = 38808.

Let the required number be y. We need y x 24 = 356 x 234. First compute the right side: 356 x 234 = 83304. Then solve for y: y = 83304/24 = 3471. Check: 3471 x 24 = 83304, which equals 356 x 234 = 83304.

Let the required number be y. We need y x 32 = 205 x 352. First compute the right side: 205 x 352 = 72160. Then solve for y: y = 72160/32 = 2255. Check: 2255 x 32 = 72160, which equals 205 x 352 = 72160.

Let S1 = a+b+c = 6 and S2 = ab+bc+ca = 5. We need a³+b³+c³-3abc, which factors as (a+b+c)(a²+b²+c²-ab-bc-ca). First find a²+b²+c² using (a+b+c)² = a²+b²+c²+2(ab+bc+ca): a²+b²+c² = S1² – 2xS2 = 6² – 2×5 = 36 – 10 = 26. Then a²+b²+c²-ab-bc-ca = 26 – 5 = 21. Finally, a³+b³+c³-3abc = S1 x 21 = 6 x 21 = 126.

Let S1 = a+b+c = 8 and S2 = ab+bc+ca = 20. We need a³+b³+c³-3abc, which factors as (a+b+c)(a²+b²+c²-ab-bc-ca). First find a²+b²+c² using (a+b+c)² = a²+b²+c²+2(ab+bc+ca): a²+b²+c² = S1² – 2xS2 = 8² – 2×20 = 64 – 40 = 24. Then a²+b²+c²-ab-bc-ca = 24 – 20 = 4. Finally, a³+b³+c³-3abc = S1 x 4 = 8 x 4 = 32.

Let S1 = a+b+c = 15 and S2 = ab+bc+ca = 21. We need a³+b³+c³-3abc, which factors as (a+b+c)(a²+b²+c²-ab-bc-ca). First find a²+b²+c² using (a+b+c)² = a²+b²+c²+2(ab+bc+ca): a²+b²+c² = S1² – 2xS2 = 15² – 2×21 = 225 – 42 = 183. Then a²+b²+c²-ab-bc-ca = 183 – 21 = 162. Finally, a³+b³+c³-3abc = S1 x 162 = 15 x 162 = 2430.

Let S1 = a+b+c = 7 and S2 = ab+bc+ca = 8. We need a³+b³+c³-3abc, which factors as (a+b+c)(a²+b²+c²-ab-bc-ca). First find a²+b²+c² using (a+b+c)² = a²+b²+c²+2(ab+bc+ca): a²+b²+c² = S1² – 2xS2 = 7² – 2×8 = 49 – 16 = 33. Then a²+b²+c²-ab-bc-ca = 33 – 8 = 25. Finally, a³+b³+c³-3abc = S1 x 25 = 7 x 25 = 175.