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

Answer: (a)

We have to find numbers between 100 and 200 which are even and are neither divisible by 7 nor by 9.

∴ No. that are even and are divisible by 7 are 7 and no. which are even and divisible by 9 are 6.

Nos. even and divisible by 7 and 9 both are (e.g., 63) is only 126 :

∴ Required answer = 7 + 6 – 1 = 12

∴ 51 – 12 = 39.

Answer: (a)

Suppose the number is x.

$x × 5/6 = 720$

$x = {720 × 6}/5 = 864$

$864 × 45/100 = 388.8$

Answer: (b)

Decimal equivalent of given fractions:

$2/5 = 0.4; 3/4 = 0.75; 4/5 = 0.8;$

$5/7 = 0.714; 6/11 = 0.545$

Clearly, 0.4 < 0.545 < 0.714 < 0.75 < 0.8

∴ $2/5 < 6/11 < 5/7 < 3/4 < 4/5$

Answer: (d)

Unit digit in $7^{95}$

= [Unit digit in $(7^4)^23 × 7^3$]

= [1 × 343] = 343

Unit digit in $3^{58}$

= [Unit digit in $(3^4)^14 × 3^2$]

= [1 × 9] = 9

So unit digit in $7^{95} – 3^{58}$

= Unit digit in [343 – 9]

= Unit digit in 334 = 4

So the answer is 4.

Answer: (b)

I. The product of any three consecutive integers is divisible by 3! i.e., 6.

II. Here, 3k = {..., 6, 3,0,3,6,... - - }

3k + 1 = {..., 5, 2,1,4,7,... - - }

and 3k + 2 = {..., 4, 1, 2,5,8,... - - }

∴ ++ {3 ,3 1,3 2 kk k }

= ------ {..., 6, 5, 4, 3, 2, 1,0,1, 2,3, 4,5,6,...}

Hence, it is true.