basic numeracy Model Questions & Answers, Practice Test for ssc cgl tier 1

Answer: (a)

$(n – 1)^2 + n^2 + (n + 1)^2 + (n + 2)^2 = 294$

$n^2 + 1 – 2n + n^2 + n^2 + 1 + 2n + n^2 + 4n + 4 = 294$

$4n^2 + 4n + 6 = 294$

$4n^2 + 4n – 288$ = 0

$n^2 + n – 77$ = 0

$n^2 + 9n – 8n – 72$ = 0

⇒ (n – 8) (n + 9) = 0

⇒ ∴ n = 8, –9

∴ number are 7, 8, 9, 10

Sum of number = 7 + 8 + 9 + 10 = 34

Answer: (d)

? = 1008 × $7/8 - 568 × 3/4$

? = 882 – 426 = 456

Answer: (c)

Four integers next lower than 81 are 80, 79, 78, 77

Four integers next higher then 81 are 82, 83, 84, 85

Sum = (80 + 82) + (79 + 83) + (78 + 84) + (77 + 85) = 81 + 81 + 81 + 81 = 4 × 81

Sum is divisible by 81, hence it is divisible by 9.

Answer: (c)

Largest 4-digit number = 9999

Required number = (9999 – 55) = 9944

Answer: (b)

Given $S_n = {n (n + 1)}/2$

Also $S_n$ = 861

⇒ 861 = ${n (n + 1)}/2$

$n^2$ + n – 1722 = 0

n = 41, – 42

(1) is true (i.e., there are exactly two values of n for which Sn = 861)

Given $S_n = S_n + 1$

⇒ ${n(n + 1)}/2 = {n(n + 1)(n + 2)}/2$

n = n + 2

2 ≠ 0 (this is not possible)

⇒ 2 is true.

∴ Option (c) is correct.