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

Answer: (b)

$(1- 1/5)$ of the number = 84

∴ number = $({84 × 5}/4) = 105$

Answer: (d)

We know that first 45 even numbers are

2, 4, 6, 8, 10, 12, ..........., 90 .

Product of these number is

2 . 4 . 6 . 8 . 10 . 12 ........... 90

= $2^{45}$[1 . 2 . 3 . 4 . 5 . 6 . 7 . 8 . 9 . 10 . 11 ....... 45]

= $2^{45}$[(5 . 20) . 1 . 2 . 3 . 4 . 6 . 7 . 8 . 9 . 10 . 11 .......... 18 . 19 . 21 . 22 ......... 45]

$2^{45}$[(100) . 1 . 2 . 3 . 4 . 6 . 7 ....... 18 . 19 . 21 . 22 ....... 45]

Now the product will consist 0 at hundred place.

Answer: (d)

Sum of digits = 35 and so it is not divisible by 3.

Now, (Sum of digits at odd places)

– (Sum of digits at even places)

= (19 – 16) = 3, not divisible by 11.

So, the given number is neither divisible by 3 nor by 11.

Answer: (d)

Amount received by each person

= Rs.$50176/32 = 1568$

Answer: (d)

= ${4^{1012}}/7 = {4^{3 × 337 + 1}}/7 = {4^{3 × 337} × 4}/7$

= ${(64)^337 × 4}/7 = {(9 × 7 + 1)^337 × 4}/7$

We know that $={(ax + 1)^n}/a$ then its Remainder is 1

= 1 × 4 = 4