Practice Number system - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) A number exceeds its one-fifth by 20. The number is
(a)
(b)
(c)
(d)
Let the number be x.
According to the question,
x = $\text"x"/5$ + 20⇒ x - $\text"x"/5$ = 20
$\text"4x"/5$ = 20
x = ${20×5}/4 = 25$
Q-2) The last digit of $(1001)^2008$ + 1002 is
(a)
(b)
(c)
(d)
Last digit of $(1001)^2008$ + 1002 = 1 + 2 = 3
Q-3) The unit digit in the expansion of $(2137)^754$ is
(a)
(b)
(c)
(d)
Expression = $(2137)^754$
Unit’s digit in 2137 = 7
Now, $7^1 = 7, 7^2 = 49, 7^3 = 343, 7^4 = 2401, 7^5$ = 16807, ...
Clearly, after index 4, the unit’s digit follow the same order.
Dividing index 754 by 4 we get remainder = 2
∴ Unit’s digit in the expansion of $(2137)^754$ = Unit’s digit in the expansion of $(2137)^2$ = 9
Q-4) One’s digit of the number $(22)^23$ is
(a)
(b)
(c)
(d)
Unit’s digit in the expansion of $(22)^23$
= Unit’s digit in the expansion of $(2)^23$
Now, $2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 16, 2^5 = 32$
i.e. 2 repeats itself after the index 4.
On dividing 23 by 4, remainder = 3
∴ Unit’s digit in $(2)^23$ = Unit’s digit in $(2)^3$ = 8
Q-5) The digit in the unit’s place of the product $(2464)^1793×(615)^317×(131)^491$ is
(a)
(b)
(c)
(d)
$(4)^{2m}$ gives 6 at unit digit.
$(4)^{2m +1}$ gives 4 at unit digit.
$(5)^n$ gives 5.
The same is the case with 1.
Required digit = Unit’s digit in the product of 4×5 × 1 = 0
Q-6) If 3 times a number exceeds its $3/5$ by 60, then what is the number ?
(a)
(b)
(c)
(d)
Suppose required number is x Then,
3x -$\text"3x"/5$ = 60 ⇒ $\text"12x"/5$ =60
= x = ${60 × 5}/12$ = 25
Q-7) If 1 is added to the denominator of a fraction it becomes $1/2$. If 1 is added to the numerator it becomes 1. The product of numerator and denominator of the fraction is
(a)
(b)
(c)
(d)
Let the numerator = x and denominator = y
Fraction = $\text"x"/ \text"y"$ and $\text"x"/ \text"y+1" = 1/2$
2x = y + 1 ⇒ x =$\text"y+1"/2$
$\text"x+1"/\text"y"$ =1⇒ x+1 = y
$\text"y+1"/2$+1 = y
$\text"y+1+2"/2$ = y
y +3= 2y ⇒ y = 3
x +1= 3 ⇒ x = 2
∴ xy = 2 × 3 = 6
Q-8) By interchanging the digits of a two digit number we get a number which is four times the original number minus 24. If the unit’s digit of the original number exceeds its ten’s digit by 7, then original number is
(a)
(b)
(c)
(d)
Let the two–digit number be 10x + y where x < y.
Number obtained on reversing the digits =10y + x
According to the question,
10y + x = 4 (10x + y) – 24
40x + 4y – 10y – x = 24
39x – 6y = 24
13x – 2y = 8 ....(i)
Again, y – x = 7
y = x + 7 ....(ii)
13x – 2 (x + 7) = 8
13x – 2x – 14 = 8
11 x = 14 + 8 = 22
x = $22/11$ = 2
From equation (ii),
y – 2 = 7 ⇒ y = 2 + 7 = 9
Number = 10x + y =10×2+9= 29
Q-9) In a two–digit number, the digit at the unit’s place is 1 less than twice the digit at the ten’s place. If the digits at unit’s and ten’s place are interchanged, the difference between the new and the original number is less than the original number by 20. The original number is
(a)
(b)
(c)
(d)
Ten’s digit = x
Unit’s digit = 2x – 1
Original number = 10x + (2x – 1) = 12x – 1
New number = 10 (2x – 1) + x
= 20x – 10 + x = 21x – 10
(21x – 10) – (12x + 1) = 12x – 1 – 20
9x – 9 = 12x – 21
3x = 12 ⇒ x = 4
Original number = 12x – 1 = 12 × 4 – 1 = 47
Q-10) There is a number consisting of two digits, the digit in the units’ place is twice that in the tens’ place and if 2 be subtracted from the sum of the digits, the difference is equal to $1/6$ th of the number. The number is
(a)
(b)
(c)
(d)
Ten’s digit of original number = x
Unit’s digit = 2x
Number = 10x + 2x = 12x
According to the question,
3x – 2 = $1/6$ × 12x
3x – 2 = 2x
3x – 2x = 2
x = 2
Number = 12x = 12 × 2 = 24
Q-11) 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)
(b)
(c)
(d)
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.
Q-12) The sum of all even numbers between 21 and 51 is
(a)
(b)
(c)
(d)
22 + 24 + 26 + ... + 50
= 2 (11 + 12 + 13 + .... + 25)
= 2 [(1 + 2 + 3 + ... + 25) – (1 + 2 + 3 ... + 10)]
= 2$({25×26}/2 - {10×11}/2)$
= 2 (325 – 55) = 2 × 270 = 540
Method 2 :
Sum of first n even numbers = n (n + 1)
Required sum = Sum of 25 even numbers from 1 to 50 – sum of 10 even numbers from 1 to 20
= 25×26 – 10 × 11 = 650 – 110 = 540
Q-13) If a and b are two odd positive integers, by which of the following integers is (a4 – b4) always divisible ?
(a)
(b)
(c)
(d)
$a^4 - b^4 = (a - b) (a + b) (a^2 + b^2)$,
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 $(a^2 + b^2)$ are divisible by 2.
Hence (a - b) (a + b) x $(a^2 + b^2) = a^4 - b^4$ is always divisible by $2^3 = 8$
Q-14) The sum of first 20 odd natural numbers is equal to :
(a)
(b)
(c)
(d)
Series of first 20 odd natural numbers is an arithmetic progression with 1 as the first term and the common difference 2.
Sum of n terms in arithmetic progression is given by.
$S_n =1/2n[2a + (n – 1)d]$
Where a : First term; d : common difference
$S_{20}= 1/2×20[(2×1) + (20 -1) × 2]$
= 10 [2 + 38]=10 × 40 = 400
Note :Sum of first n consecutive odd numbers = $n^2$
Q-15) The greatest common divisor of 33333 + 1 and 33334 + 1 is ?
(a)
(b)
(c)
(d)
Q-16) The greatest value among the fractions $2/7,1/3,5/6,3/4$ is :
(a)
(b)
(c)
(d)
$2/7 = 0.286 ; 1/3 = 0.33$
$5/6 = 0.833; 3/4 = 0. 75$
Q-17) Which of the following fraction is the smallest? $7/6, 7/9, 4/5, 5/7$
(a)
(b)
(c)
(d)
$7/6$ = 1.166 ; $7/9$ = 0.777; $4/5$ = 0.8 and $5/7$ = 0.714
Therefore, the smallest number is $5/7$
Q-18) Which of the following is the largest fraction ? $6/7, 5/6, 7/8, 4/5$
(a)
(b)
(c)
(d)
The decimal equivalents of :
$6/7 = 0.857, 5/6 = 0.833$
$∴7/8 = 0.875, 4/5 = 0.8$
Obviously, 0.875 is the greatest.
$7/8$ is the largest fraction.
Q-19) The least number of five digits which has 123 as a factor is
(a)
(b)
(c)
(d)
The smallest number of 5 digits = 10000
Remainder on dividing 10000 by 123 = 37
∴ Required number = 10000 + (123 – 37) = 10086
Q-20) The smallest number of five digits exactly divisible by 476 is
(a)
(b)
(c)
(d)
The smallest number of 5 digits = 10000
Now,
$10000/476$=21$4/476$
∴Required number = 10000 + (476 – 4)
= 10000 + 472 = 10472