basic numeracy Model Questions & Answers, Practice Test for ssc cgl tier 1
ssc cgl tier 1 SYLLABUS WISE SUBJECTS MCQs
Computation of whole numbers, decimals, fractions and relationships between numbers
Percentage
Averages
How many even integers n, where 100 ≤ n ≤ 200, are divisible neither by seven nor by nine ?
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.
Five-sixth of a number is 720. What will 45% of that number be?
Answer: (a)
Suppose the number is x.
$x × 5/6 = 720$
$x = {720 × 6}/5 = 864$
$864 × 45/100 = 388.8$
If the fractions $2/5, 3/4, 4/5, 5/7$ and $6/11$ are arranged in ascending order of their values, which one will be the fourth ?
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
∴ $2/5 < 6/11 < 5/7 < 3/4 < 4/5$
The unit digit of $(7^95 – 3^58)$ is
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.
Consider the following statements:
- The product of any three consecutive integers is divisible by 6.
- Any integer can be expressed in one of the three forms 3k, 3k + 1, 3k + 2, where k is an integer.
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.
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