1000+ TNPSC Aptitude Study Materials Pdf - 1

Ans: C

Let the integer be x.

Then, x² - 20x = 96

⇔ x² - 20x – 96 = 0

⇔ (x + 4) (x – 24) = 0

⇔ x = 24.

Ans: C

Let the numbers be x and y.

Let a - b = k ----(i)

a + b = 7k ------(ii)

ab = 24k

Adding (i) and (ii), we get : 2a = 8k or a = 4k.

Putting a = 4k in (i), we get : b = 3k.

Putting a = 4k and b = 3k in (iii), we get : 4k × 3k = 24 k ⇔ 12² = 24k ⇔ k = 2.

Hence, product of numbers = 24k = 24 × 2 = 48.

Ans: C

120 = 2 × 2 × 2 × 3 × 5

= (2 × 2) × 5 × (2 × 3) = 4 × 5 × 6.

Clearly, the three consecutive integers whose product is 120 are 4, 5 and 6.

Required sum = 4 + 5 + 6 = 15.

Ans: B

Let the numbers be 3x, 2x and 5x.

Then, (3x)² + (2x)² + (5x)² = 1862 ⇒ 9x² + 4x² + 25x² = 1862

⇒ 38² = 1862 ⇒ x² = $${1862}/ {38}$$ = 49 ⇒ x = $$√49$$ = 7

Hence, the numbers are 21, 14 and 35.

Ans: B

Then, x² + (x + 2)² + (x + 4)² = 2531

⇒ 3x² + 12x - 2511 = 0 ⇒ x² + 4.

Hence, the required numbers are 27, 29 and 31.