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

Answer: (c)

Let the two-digits numbers less than 50 which when divided by 4 yield unity as remainder be 13, 17, ..., 49.

Here, first term, a = 13, common difference, d = 4 and n = 10.

∴ Required sum = $n/2[2a + (n - 1)d]$

= $10/2[2 × 13 + (10 - 1)4]$

= $10/2[26 + 36] = {10 × 62}/2 = 310$

Answer: (a)

Unit digit of $(3^98 – 3^89)$

= Unit digit of $(3^96 – 3^2 – 3^88 – 3)$

= Unit digit of (1 × 9 – 1 × 3)

= 6

Answer: (c)

Let the two-digit number be = 10x + y, where x > y

According to the question,

10x + y – 10y – x = 18

or, 9x – 9y = 18

or, 9(x – y) = 18

or, x – y = $18/9$ = 2 ...(i)

and, x + y = 12 ...(ii)

From equations (i) and (ii)

2x = 14 ⇒ x = $14/2$ = 7

From equation (ii)

y = 12 – 7 = 5

∴ Required product = xy = 7 × 5 = 35

Answer: (d)

999 × abc = def132

We can write the above equation as

(1000 – 1) × abc = def132

abc000 – abc = def 000 + 132 = (def + 1) × 1000 – 868

on comparing the LHS and RHS, we get

a = 8, b = 6, and c = 8 and d = a = 8

Now, 999 × 868 = 867132

∴ d = 8, e = 6, f = 7

Answer: (d)

$(3^25 + 3^26 + 3^27 + 3^28) = 3^25 × (1 + 3 + 3^2 + 3^3) = 3^25 × 40$

= $3^24 × 3 × 4 × 10 = (3^24 × 4 × 30)$, which is divisible by 30.