Aptitude Test-Problems of Numbers Questions(Mcq)

Numbers Aptitude Test Questions(Mcq)

All Questions Answers are Given Below Check it

Question 1.The Unit Digit in the Produt (784×618×917×463)

A)2
B)3
C)5
D)4

Question 2. 9587-X=7429-4358 then X=?

A)5011
B)6516
C)6511
D)6173

Question 3.(796×796-204×204)=?

A)592000
B)592001
C)582762
D)783623

Question 4.(87×87+61×61-2×87×61)=?

A)295
B)562
C)195
D)196

Question 5. What is the Unit Digit of  7¹⁰⁵   ?

A)1
B)5
C)7
D)9

Question 6.Which one of the following is not a prime number?

A)31
B)61
C)91
D)87 

Question 7.(112 x 54) = ?

A)77000
B)70000
C)82444
D)77733

Question 8.What least number must be added to 1056, so that the sum is completely divisible by 23 ?

A)3
B)5
C)6
D)2

Question 9. 1397 x 1397 = ?

A)1951609.
B)1951611
C)1826351
D)3682251

Question 10. How many of the following numbers are divisible by 132 ?
264, 396, 462, 792, 968, 2178, 5184, 6336

A)8
B)4
C)6
D)5

Question 11. If 6 + 12 + 18 + 24 + --- = 1800, then find the number of terms in the series.

A)24
B)25
C)30
D)21

Question 12. 1 + 2 + 3 + ---- 50 = ?

A)1276
B)1271
C)1275
D)1280

Question 13.Find the solution of(450 + 280)² – (450 – 280) ²/(450 × 280)

 A)1
B)4
 C)3
D)5 

Question 14. The remainder is 3, when a number is divided 5. If the square of this number is divided by 5, then what is  remainder?

 A)1
 B)2
 C)3
D)4

Question 15. Find the number of terms in geometric progression 3, 6, 12, 24, ---- , 384.

A)8
B)7
C)5
D)6

Show me the answers!

Question 1: The correct answer is the
Answer is A)
Unit Digit in the given product=Unit Digit in(4×8×7×3)=2

Question 2: The correct answer is 
Answer  is B)
9587-X=7429-4358 then X=?
then 9887-X=3071
X=9587-3071=6516

Question 3: The correct answer is 
Answer is C)
By Using this formula-(a²-b²)=(a+b)(a-b)
796×796-204×204=(796)²-(204)²
=(796+204)(796-204)=1000×593=592000

Question 4: The correct answer is 
Answer is D)
By Using the fomula-(a²+b²-2ab)=(a-b)² 
where a=87,b=61
(87²+61²-2×81×61)=(87-61)²=(26)²=(20+6)²
=20²+6²-2x20x6=(400+36-240)
=(436-240)=196.

Question 5: The correct answer is 
Answer is C)
 the Unit Digit of  7¹⁰⁵ =Unit Digit in [(7⁴)²⁶×7]
The Unit Digit in  (7⁴)²⁶=1  
^{then Unit Digit in}  7^{105 =(1x7)=7} 

Question 6: The correct answer is 
Answer is C)
91 is divisible by 7. So, it is not a prime number

Question 7: The correct answer is 
Answer is B)
(112 x 5⁴) = 112 x10/24=112 x 10⁴/16=1120000/6= 70000

Question 8: The correct answer is 
Answer is D)

 23) 1056 (45
      92
      ---
      136
      115
      ---
       21
      ---
     
 Required number = (23 - 21)    
                 = 2.

Question 9: The correct answer is 
Answer is A)

1397 x 1397	= (1397)2
	= (1400 - 3)2
	= (1400)2 + (3)2 - (2 x 1400 x 3)
	= 1960000 + 9 - 8400
	= 1960009 - 8400
	= 1951609.

Question 10: The correct answer is 

Answer is B)
132 = 4 x 3 x 11

So, if the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also.
264  11,3,4 (/)

396  11,3,4 (/)

462  11,3 (X)

792  11,3,4 (/)

968  11,4 (X)

2178  11,3 (X)

5184  3,4 (X)

6336  11,3,4 (/)

Therefore the following numbers are divisible by 132 : 264, 396, 792 and 6336.

Required number of number = 4.

Question 11: The correct answer is 

Answer is A)

This is an Arithmetic Progression, in which x = 6, y = 6, sum of terms = 1800

Sum of n terms =	n	[2x + (n – 1)y]
	2

Substituting the given values, we get

1800 =	n	[2 × 6 + (n – 1)6]
	2

Solving we get,
1800 = 3n (n + 1)
n(n +1) = 600
n² + n = 600
25 × 24 = 600
Therefore,
n² + 25n – 24n – 600 = 0
(n + 25) (n – 24) = 0
24 and -25 are the two solutions obtained. Only positive value can be considered. Hence, n = 24
The number of terms in the series = 24.

Question 12: The correct answer is 

Answer is C)

This is an Arithmetic Progression, here a = 1, last term = 50 and n = 50

Sum of n terms =	n	(a + l)
	2

Substituting the given values, we get

Sum of n terms =	50	(1 + 50)
	2

Sum of n terms = 25 × 51 = 1275

Question 13: The correct answer is 

Answer is B)

By Using Formula

(a + b)² = a² + 2ab + b²
(a – b)² = a² – 2ab + b²
If we substitute these formulae, we get

(a + b)2 – (a – b)2	=	(a2 + 2ab + b2 – a2 + 2ab – b2)	=	4ab	= 4
(a × b)		(a × b)		ab

Question 14: The correct answer is 

Answer is D)
Dividend = [(Divisor × Quotient)] + Remainder
It is given that, the remainder is 3, when a number (dividend) is divided 5(divisor).
Dividend and quotient are unknown, hence assume dividend as X and quotient as Y.
X = (5Y) + 3
The square of this number is divided by 5, therefore
X² = (5Y + 3)²
Simplifying we get,
X² = (25Y² + 30Y + 9)
On dividing this equation, we get
5(5Y² + 6Y + 1) + 4
The remainder is 4, because 9 is not exactly divisible by 5 and we get 4 as remainder

Question 15: The correct answer is 

Answer is A)

Here, x = 3, r =	6	= 2
	3

We have to find the number of terms
N^th term = 384
n^th term = xr ^{(n -1)}
384 = 3 x 2 ^{(n -1)}

2(n – 1) =	384	= 128
	3

2 ^{(n -1)} = 128
2⁷ = 128

Therefore,
2 ^{(n -1)} = 2⁷
(n – 1) = 7
n = 7 + 1 = 8

Hence, number of terms = 8

You answered them all right!