MCQ: Number System - 3 - SSC CGL MCQ

By trial, we find that the smallest number consisting entirely of fives and exactly divisible by 13 is 555555. On dividing 555555 by 13, we get 42735 as quotient.
∴ Req. smallest number =42735.

The only one even prime number is 2 and this is not in the form of 3k + 1.
Thus any prime number of the form 3k + 1 is an odd number.
This means 3k must be even number which implies that k must be even number also.
Let k = 2m.
Then a prime of the form 3k + 1 is of the form 3(2m) + 1 = 6m+1.
Every prime number of form 3k + 1 can be represented in the form 6m + 1 only, when k is even.

Let the fraction be x / y.
According to the question,
( x+2)/(y+1) = 1/2
⇒ 2x + 4 = y+1
⇒ 2x - y = -3 .........................(i)
Again according to the question,
(x+1)/(y-2) = 3/5
⇒ 5x +5 = 3y - 6
∴ 5x - 3y = -11..........................(ii)
on multiplying Eq. (i) by 3 and then subtracting from Eq. (ii) we get,
multiplying Eq. (i) by 3
6x - 3y = -9
subtracting from Eq. (ii)
5x- 3y - ( 6x - 3y ) = -11 - (-9)
5x- 3y - 6x + 3y = -11 + 9
- x = - 2
x= 2
On putting the value of x in Eq. (i) we get
2x- y = -3
2 x 2 - y = -3
⇒ 2 x 2 -y = -3
⇒ -y = -3 - 4
∴ y = 7
∴ Original fraction = x/y =2/7

Let the third numbers = p
Then, the first number = 3p
and second number = 3p/2
According to question,
( p + 3p + 3p/2 )/3 = 154
⇒ ( 2p + 6p + 3p )/6 = 154
∴ p = 154 x 6/11 = 84
∴ Required difference = 3p - p = 2p
= 2 x 84 = 168

• Let each student get 'y' sweets
• As there are 300 students, total sweets = 300 x y
• Now, 50 students are absent
• Present students = 300 - 50 = 250
• each students is getting 1 extra sweet because of the absentees
• total sweets = 250 x (x +1)
• Now, 300 x = 250 (x + 1)
• 300 x = 250 x + 250
• 300 x - 250 x = 250
• 50 x = 250
• x = 5
• number of sweets distributed = 250(5 + 1) = 250 x 6 = 1500
• Hence, option C is correct. 

Let number of students in examination halls P and Q is x and y, respectively
Then as per the first condition,
x - 10 = y + 10
⇒ x - y = 20 ..............(i)
As per the second condition,
⇒ x + 20 = 2( y - 20)
⇒ x - 2y = - 60 ...............(ii)
On subtraction Eq. (ii) from Eq. (i), we get
- y + 2y = 20 + 60
⇒ y = 80
Putting the value of y Eq. (i) we get,
x - 80= 20 ⇒ x = 100
Hence, number of students in examination halls P and Q is 100 and 80, respectively.

Let Ram spends p minute on each Mathematics question.
According to the question.
50 x p + 100 x p/2 + 50 x p/2 = 3 x 60
p(50+ 50 + 25) = 180
⇒ p = 180/125
∴ Required time = 50 x 180/125
∴ Required time = 2 x 180/5
∴ Required time = 2 x 36 = 72 min

Let total number of candidates in Exam = a
Number of candidates answered 5 questions = a x 5% = a x 5/100 = 5a/100 = a/20
Number of candidates answered not any questions = a x 5% = a x 5/100 = 5a/100 = a/20
∴ Remaining students = a - ( a/20 + a/20) = a - ( 2a/20 ) = a - ( a/10 ) = (10a - a)/10 = 9a/10
Number of candidates answered only 1 question = ( 9a/10 ) x 25% = ( 9a/10 ) x 25/100 = 9a/40
Number of candidates answered 4 questions = ( 9a/10 ) x 20% = ( 9a/10 ) x 20/100 = 9a/50
Given number of candidates awarded either 2 questions or 3 questions = 396
⇒ a - ( a/20+ a/20 + 9a/40 + 9a/50 ) = 396
⇒ a - ( a/10 + 9a/40 + 9a/50 ) = 396
⇒ a - ( ( a x 20 + 9a x 5 + 9a x 4 )/200) = 396
⇒ a - ( ( 20a + 45a + 36a )/200) = 396
⇒ a - ( 101a/200) = 396
⇒ ( 200a - 101a)/200 = 396
⇒ ( 99a)/200 = 396
∴ a = 396 x 200/99
∴ a = 4 x 200 = 800
⇒ a = 800
Hence, number of candidates = 800

Let ten's place digit be a and unit's place digit be a²
Original number = 10 x a + 1 x a² = 10a + a²
The number formed by interchanging the digits,
New number = 10 x a ² + 1 x a = 10a ² + a
According to the question
(10a² + a) - ( 10a + a²) = 54
⇒ 10a² + a - 10a - a ² = 54
⇒ 9a² - 9a = 54
⇒ 9( a² - a) = 54
⇒ ( a² - a) = 54/9
⇒ ( a² - a) = 6
⇒ a² - a - 6 = 0
⇒ a² - 3a + 2a - 6 = 0
⇒ a (a - 3) + 2 (a - 3) = 0
∴ (a - 3) (a + 2) = 0
∴ a = 3, - 2
∴ Ten,s digit = a = 3
Unit's digit = a² = 3² = 9
Original number = 39
∴ Required number = 39 x 40/100 = 15.6

Let the number be P.
According to the question,
P - P/7 = 180
⇒ 6P/7 = 180
∴ P = 180 x 7/6 = 210

Let two consecutive odd number be (A + 1) and (A + 3).
According to the question.
(A + 1) (A + 3) = 6723
⇒ A2 + 3A + A + 3 = 6723
⇒ A2 + 4A + 3 - 6723 = 0
⇒ A2 + 4A - 6720 = 0
⇒ A2 + 84A - 80A - 6720 = 0
⇒ A(A + 84) - 80 (A + 84) = 0
⇒ (A - 80) (A + 84) = 0
∴ A = 80, (A = ≠ - 84)
Hence, the greater number = 80 + 3 = 83

Let be the ten's digit be P and unit's digit be Q.
The two-digit number = 10P + Q
(where, P > Q)
According to the question,
         P + Q = 14 ......(i)
and   P - Q = 2 .......(ii)
solving Eqs. (i) and (ii), we get
P = 8 and Q = 6
∴ Required product = 8 x 6 = 48

LCM of 3, 2 and 6 = 6
Now we can write as below
∴ (3)^1/3 = (3²)^1/6 = (9)^1/6
2^1/2 = (2³)^1/6 = (8)^1/6
1 = (1)^1/6
(6)^1/6 = (6)^1/6

Clearly we can see that power 1/6 is same for all the base numbers. So the number which has greatest base with be greatest among these all number.
∴ (9)^1/6 = (3)^1/3 is the greatest number.

Numbers divisible by 3 and lying between 100 and 200 are : 102, 105, 108, 111,...................... 198
Let number of terms = n
As we know the formula
a_n = a + ( n – 1 )d
where a_n = Nth Term a = First term , n = Number of terms and d = Common Difference
∴  198 = 102 + ( n – 1 ) 3

Let us assume Ten’s digit = P
Unit’s digit = 2P – 1
∴  Original number = 10P + (2P – 1) = 12P – 1
New number = 10 (2P – 1) + P
New number = 20P – 10 + P = 21P – 10
According to given question,
∴  (21P – 10) – (12P – 1) = 12P – 1 – 20
⇒  9P – 9 = 12P – 21
⇒  3P = 12
⇒ P = 4
⇒  Original number = 12P – 1 = 12 × 4 – 1 = 47