# Top 1499+ Number System Aptitude MCQ Test SSC CGL Tier-2

## Top 1499+ Number System Aptitude MCQ Test

Hello Students,

Dear students, you know that QUANT is a part of getting points and every chapter is important. Top 1499+ Number System Aptitude MCQ Test SSC CGL Tier-2 Therefore, we are providing 15 questions of quant. Solve all these quizzes every day so that you can improve your accuracy and speed. We also provide lots of quant questions. So you can practice that chapter which takes more time to solve the questions.
प्रिय पाठकों, आप सभी जानते हैं कि संख्याताम्क अभियोग्यता का भाग बहुत ही महत्वपूर्ण है. इसलिए हम आपको संख्यात्मक अभियोग्यता कि 15 प्रश्नों कि प्रश्नोत्तरी प्रदान कर रहे हैं. इन सभी प्रश्नोत्तरी को दैनिक रूप से हल कीजिये ताकि आप अपनी गति और सटीकता में वृद्धि कर सकें. हम आपको अन्य कई संख्यात्मक अभियोग्यता के प्रश्न प्रदान करेंगे. ताकि आप पाठ्यक्रम अनुसार उन्हें हल कर पायें.

Today we are sharing an important pdf in hindi Top 1499+ Number System Aptitude MCQ Test SSC CGL Tier-2  एक ऐसा number होता है जिसे की express किया जाता है एक ratio के तोर पर दो integers (इसलिए इसका नाम है “rational“) का. इसे लिखा जाता है एक fraction के तोर पर जिसमें top number (numerator) को divide किया गया होता है bottom number (denominator) के द्वारा.

Top 1499+ Number System Aptitude MCQ Test SSC CGL Tier-2 सभी integers rational numbers होते हैं क्यूंकि उन्हें 1 के द्वारा divide किया जा सकता है, जो की प्रदान करता है एक ratio दो integers का. बहुत से floating point numbers भी rational numbers होते हैं चूँकि उन्हें fractions के तोर पर express किया जा सकता है. उदाहरण के तोर पर, 1.5 rational होता है चूँकि इसे लिखा जा सकता है 3/2, 6/4, 9/6 या दुसरे fraction या दो integers. Pi (π) irrational होता है क्यूंकि इसे fraction में लिखा नहीं जा सकता है.

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Question :21 The greatest common divisor of 33333 + 1 and 33334 + 1 is ?

A. 2
B. 1
C. 33333 + 1
D. 20

Question :22
A number when divided by 91 gives a remainder 17. When the same number is divided by 13, the remainder will be ?

A. 0
B. 4
C. 6
D. 3

Here, the first divisor (91) is a multiple of the second divisor (13).

∴ Required remainder = Remainder obtained on dividing 17 by 13

⇒ 17 = ( 13 × 1 ) + 4

Hence Required remainder = 4

Question :23
Each member of a picnic party contributed twice as many rupees as the total number of members and the total collection was 3042. The number of members present in the party was ?

A. 2
B. 32
C. 40
D. 39

Let the total number of members be X.

Then, each member’s contribution =Rs. 2X

Given, X × 2X = 3042

⇒ 2X2 = 3042

⇒ X2 = 1521

⇒ X = 39

Question :24
The expression 26n – 42n, where n is a natural number is always divisible by

A. 15
B. 18
C. 36
D. 48

The expression 26n − 42n = (26)n − (42)n

= 64n − 16n

Which is divisible by 64 –16= 48

Therefore the required answer is 48.

Question :25
A student was asked to divide a number by 6 and add 12 to the quotient. He, however, first added 12 to the number and then divided it by 6, getting 112 as the answer. The correct answer should have been ?

A. 124
B. 122
C. 118
D. 114

Let the number be p.

According to the question,

p + 12
6
=112

⇒ p + 12 = 672

⇒ p = 672 – 12 = 660

600
6
=112

Correct answer = 110 + 12 = 122

Question :1
All natural numbers and 0 are called as _______ numbers.

A. Rational
B. Integers
C. Whole
D. Prime

The counting numbers {1, 2, 3, …} are commonly called natural numbers; however, include 0, so that the non-negative integers {0, 1, 2, 3, …} are also called natural numbers.

Natural numbers including 0 are also called whole numbers.

Also, Refer to Types of Numbers.

Question :2
Consider the following statements about natural numbers: (1) There exists a smallest natural number. (2) There exists a largest natural number. (3) Between two natural numbers, there is always a natural number. Which of the above statements is/are correct?

A. None
B. Only 1
C. 1 and 2
D. 2 and 3

1) There exists the smallest natural number – which is 1. So, true.

2) There exists a largest natural number – there is no upper limit to numbers. So, false.

3) Between two natural numbers there is always a natural number – for example, between 1 and 2 there is no natural number. So, false.

Hence, only option 1 is true.

Question :3
Every rational number is also

A. an integer
B. a real number
C. a natural number
D. a whole number

Both rational and irrational numbers taken together are called Real numbers.

Therefore, every real number is either a rational number or an irrational number. Hence, every rational number is a real number.

Hence, (B) is the correct option.

Question :4
The number π is

A. a fraction
B. a recurring decimal
C. a rational number
D. an irrational number

Pi (π) is an irrational number

π is an irrational number that has a value of 22/7 or 3.142…and is a never-ending and non-repeating number.

Irrational numbers are non-terminating

Question :5
√2 is a/an

A. rational number
B. natural number
C. irrational number
D. integer

1. इसमें limited number की digits होती है decimal point के बाद (e.g., 5.467)

2. इसमें एक infinitely repeating number मेह्ज्दु होता है decimal point के बाद (e.g., 2.333333…)

3. इसमें infinitely repeating pattern होती है numbers की वो भी decimal point के बाद (e.g. 3.151515…)
अगर numbers वो भी decimal point के बाद repeat होती है infinitely बिना किसी pattern के, तब number न ही rational होता है या “irrational.” यहाँ निचे में कुछ examples दिए गए हैं rational और irrational numbers के.

1 – rational
0.5 – rational
2.0 – rational
√2 – irrational
3.14 – rational
π (3.14159265359…) – irrational
√4 – rational
√5 – irrational
16/9 – rational
1,000,000.0000001 – rational

Question :6
The number √3 is

A. a finite decimal
B. an infinite recurring decimal
C. equal to 1.732
D. an infinite non-recurring decimal

Question :7
There are just two ways in which 5 may be expressed as the sum of two different positive (non-zero) integers, namely 5 = 4 + 1 = 3 + 2. In how many ways, 9 can be expressed as the sum of two different positive (non-zero) integers?

A. 3
B. 4
C. 5
D. 6

9 = 1+8 = 2+7 = 3+6 = 4+5

So it can be expressed in 4 ways

Question :8
P and Q are two positive integers such that PQ = 64. Which of the following cannot be the value of P + Q?

A. 16
B. 20
C. 35
D. 65

The possible combinations of (P, Q)

So that the product is 64 are

(1, 64), (2, 32), (4, 16) and (8, 8)

∴ P+Q cannot be 35

Question :9
If n is an integer between 20 and 80, then any of the following could be n+7 except

A. 47
B. 58
C. 84
D. 88

Given, 20 < n < 80

Now n + 7 = ?

20 + 7 < n + 7 < 80 + 7

i.e., 27 < n + 7 < 87

∴ 88 does not lies between 27 to 87.

Question :10
If x, y, z be the digits of a number beginning from the left, the number is

A. xyz
B. x + 10y + 100z
C. 10x + y + 100z
D. 100x + 10y + z

In this problem, the digits of the number from the left are x, y and z respectively.

So, x is at hundreds places, y is at tens place and z is at ones place.

Thus, the required number be

= (100×x) + (10×y) + (1×z)

= (100x + 10y + z)

Question :16
How many numbers between 1000 and 5000 are exactly divisible by 225 ?

A. 16
B. 18
C. 19
D. 12

First number (a) = 1125, Last number (an) = 4950,

Divisible by (d) = 225, No. of terms (n) = ?

No. of term formula, an= a+(n-1)d

Then, 4950= 1125 + (n – 1) 225

3825= 225n – 225

4050= 225n

n = 4050/225 = 18

Therefore, 18 numbers between 1000 and 5000 are exactly divisible by 225.

Question :17
If m and n are positive integers and (m – n) is an even number, then (m2 – n2) will be always divisible by

A. 4
B. 6
C. 8
D. 12

In this problem, put any even positive value for both m & n,

For example m = 4 & n = 2

∴ (m2 – n2) = (42 – 22) = 16 – 4

= 12 is always divisible by 4.

Question :18
When a number is divided by 56, the remainder obtained is 29. What will be the remainder when the number is divided by 8 ?

A. 4
B. 5
C. 3
D. 7

When the second divisor is a factor of the first divisor, the second remainder is obtained by dividing the first remainder by the second divisor.

Hence, on dividing 29 by 8, the remainder is 5.

Question :19
(49)15 – 1 is exactly divisible by :

A. 50
B. 51
C. 29
D. 8

As we know, xn – an is exactly divisible by (x – a) if n is odd.

∴ (49)15 – (1 )15 is exactly divisible by 49 – 1 = 48, that is a multiple of 8.

Question :20
If a and b are two odd positive integers, by which of the following integers is (a4 – b4) always divisible ?

A. 3
B. 6
C. 8
D. 12

a4−b4=(a−b)(a+b)(a2+b2),

Where a and b are odd positive integers.

If two positive integers are odd, then their sum, difference and sum of their squares are always even.

∴ (a – b) (a + b) and (a2+b2) are divisible by 2.

Hence (a – b) (a + b) x (a2+b2)=a4−b4 is always divisible by 23=8