XAT Number System Questions PDF [Important]
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Question 1: What is the difference of mean and median of the given data : 4, 13, 8, 15, 9, 21, 18, 23, 35, 1?
a) 0.7
b) 1.7
c) 1.2
d) 2.1
1) Answer (A)
Solution:
Mean:
No. of samples (n) = 10
Mean = $\frac{\sum x}{n} = \frac{4+13+8+15+9+21+18+23+35+1}{10}= \frac{147}{10} = 14.7$
Median:
Arranging the data in ascending order, we get:
1, 4, 8, 9, 13, 15, 18, 21, 23, 35
n = 10 (even)
Therefore, median is the average of 5th and 6th term.
Median = $ \frac{13+15}{2} = 14$
Mean – Median = 14.7 – 14 = 0.7
Therefore, Option A is correct.
Question 2: 60% of a number is 168, then what is the number?
a) 280
b) 320
c) 240
d) 200
2) Answer (A)
Solution:
60% of the number is 168.
Let’s assume the number is ‘y’.
60% of y = 168
0.6y = 168
y = 280
Question 3: What is the value of: $5 of 5 of 5 \div 5 + 5 – 6 \div 3 \times 4 + 2 + (3 \div 6 \times 2)?$
a) 21
b) 25
c) 28
d) 19
3) Answer (B)
Solution:
$5\times5\times\frac{5}{5}+5-\frac{6}{3}\times4+2+\frac{3}{6}\times2$
$25+5-8+2+1$
25
Question 4: What is the median of the given data?
41, 43, 46, 50, 85, 61, 76, 55, 68, 95
a) 61
b) 58
c) 57
d) 55
4) Answer (B)
Solution:
Arrange the given data in ascending order.
41, 43, 46, 50, 55, 61, 68, 76, 85, 95
n = number of given data
n = 10
median = $\frac{\left(\frac{n}{2}\right)^{th\ term}\ +\ \left(\frac{n}{2}+1\right)^{th\ term}}{2}$
= $\frac{\left(\frac{10}{2}\right)^{th\ term}\ +\ \left(\frac{10}{2}+1\right)^{th\ term}}{2}$
= $\frac{5^{th\ term}\ +\ \left(5+1\right)^{th\ term}}{2}$
= $\frac{5^{th\ term}\ +\ 6^{th\ term}}{2}$
= $\frac{55+61}{2}$
= $\frac{116}{2}$
= 58
Question 5: If A is the smallest three-digit number divisible by both 6 and 7 and B is the largest four-digit number divisible by both 6 and 7, then what is the value of (B – A)?
a) 9912
b) 9870
c) 9996
d) 9954
5) Answer (B)
Solution:
LCM of (6, 7) = 42
6 = $2\times3$
7 = 7
As per the information given in the question, the value of both A and B will be divisible by both 6 and 7. The LCM of these is 42. It means that the value of both A and B will be the multiple of 42.
If A is the smallest three-digit number divisible by both 6 and 7.
The smallest three-digit number is 100. When we divide 100 by 42, then the remainder will be 16. So we need to add (42-16) in 100.
So A = 100+(42-16)
A = 100+26
A = 126
B is the largest four-digit number divisible by both 6 and 7.
The largest four-digit number is 9999. When we 9999 by 42, then the remainder will be 3. So we need to subtract 3 from 9999.
So B = 9999-3
B = 9996
Value of (B – A) = 9996-126
= 9870
Question 6: A set of data is as under: 4, 2, 3, 2, 7, 4, 8, 5, 2, 4, 5, 6, 2, 5, 6, 6, 5, 4, 6, 5, 3, 5, 4, 3 What is the mode of the set?
a) 2
b) 5
c) 6
d) 4
6) Answer (B)
Solution:
4, 2, 3, 2, 7, 4, 8, 5, 2, 4, 5, 6, 2, 5, 6, 6, 5, 4, 6, 5, 3, 5, 4, 3
Arrange the given data in ascending order.
2, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 8
Mode = maximum number of times a number is available in the given data set
= 5 (It is available six times in the given data set.)
Question 7: The median of the first seven prime numbers is:
a) 7
b) 5
c) 13
d) 11
7) Answer (A)
Solution:
first, seven prime numbers are 2, 3, 5, 7, 11, 13, 17.
Here the number of terms are 7. So n = 7.
median = $\left(\frac{n+1}{2}\right)^{th}\ term$
= $\left(\frac{7+1}{2}\right)^{th}\ term$
= $\left(\frac{8}{2}\right)^{th}\ term$
= $4^{th}\ term$
= 7
Question 8: What is the sum of the digits of the largest six-digit number divisible by 3, 4, 5 and 6?
a) 45
b) 42
c) 39
d) 48
8) Answer (B)
Solution:
As we know that the largest six-digit number is 999999.
The required number should be divisible by 3, 4, 5 and 6.
So the LCM of 3, 4, 5 and 6 is 60.
When 999999 is divided by 60, then the remainder will be 39.
$999999=60\times16666+39$
Required number = 999999-39 = 999960
the sum of the digits of the required number = 9+9+9+9+6+0 = 42
Question 9: What is the difference between the mean and median of the given data:
4, 6, 3, 7, 10, 13, 16 and 5?
a) 5
b) 1.5
c) 3
d) 4.5
9) Answer (B)
Solution:
$Mean=\frac{sum\ of\ data}{number\ of\ data}$
$=\frac{4+6+3+7+10+13+16+5}{8}$
$=\frac{64}{8}$
= 8
For median, first we need to arrange the given data in ascending order from left to right.
3, 4, 5, 6, 7, 10, 13, 16
n = number of data = 8
median = $\frac{\left(\frac{n}{2}\right)^{th}\ term\ +\ \left(\frac{n}{2}+1\right)^{th}\ term}{2}$
= $\frac{\left(\frac{8}{2}\right)^{th}\ term\ +\ \left(\frac{8}{2}+1\right)^{th}\ term}{2}$
= $\frac{4^{th}\ term\ +\ \left(4+1\right)^{th}\ term}{2}$
= $\frac{4^{th}\ term\ +\ 5^{th}\ term}{2}$
= $\frac{6+7}{2}$
= $\frac{13}{2}$
= 6.5
difference between the mean and median = 8-6.5
= 1.5
Question 10: What is the difference between the mean and the median of the following data:
5, 7, 8, 13, 12, 14, 9, 2, 26, 10 ?
a) 2.3
b) 0.4
c) 1.8
d) 1.1
10) Answer (D)
Solution:
First, arrange the given data in ascending order from left to right.
2, 5, 7, 8, 9, 10, 12, 13, 14, 26
There are ten numbers. So n = 10.
median = $\frac{\left(\frac{n}{2}\right)^{th}\ term\ +\ \left(\frac{n}{2}+1\right)^{th}\ term}{2}$
= $\frac{\left(\frac{10}{2}\right)^{th}\ term\ +\ \left(\frac{10}{2}+1\right)^{th}\ term}{2}$
= $\frac{5^{th}\ term\ +\ \left(5+1\right)^{th}\ term}{2}$
= $\frac{5^{th}\ term\ +\ 6^{th}\ term}{2}$
= $\frac{9+10}{2}$
= $\frac{19}{2}$
= 9.5
mean = $\frac{sum\ of\ data}{number\ of\ data}$
= $\frac{2+5+7+8+9+10+12+13+14+26}{10}$
= $\frac{106}{10}$
= 10.6
difference between the mean and the median = 10.6-9.5
= 1.1
Question 11: Let x be the median of data: 33, 42, 28, 49, 32, 37, 52, 57, 35, 41.
If 32 is replaced by 36 and 41 by 63, then the median of the data, so obtained, is y. What is the value of (x + y)?
a) 78
b) 78.5
c) 79.5
d) 79
11) Answer (B)
Solution:
33, 42, 28, 49, 32, 37, 52, 57, 35, 41
Arrange the given data in ascending order from left to right.
28, 32, 33, 35, 37, 41, 42, 49, 52, 57
Median = $\frac{\left(\frac{n}{2}\right)^{th}\ term\ +\left(\frac{n}{2}+1\right)^{th}\ term}{2}$
Here n = the number of data
Median = x = $\frac{\left(\frac{10}{2}\right)^{th}\ term\ +\left(\frac{10}{2}+1\right)^{th}\ term}{2}$
= $\frac{5^{th}\ term\ +\left(5+1\right)^{th}\ term}{2}$
= $\frac{5^{th}\ term\ +6^{th}\ term}{2}$
= $\frac{37+41}{2}$
= $\frac{78}{2}$
x = 39
If 32 is replaced by 36 and 41 by 63, then the median of the data, so obtained, is y.
28, 36, 33, 35, 37, 63, 42, 49, 52, 57
Arrange the given data in ascending order from left to right.
28, 33, 35, 36, 37, 42, 49, 52, 57, 63
Median = y = $\frac{\left(\frac{10}{2}\right)^{th}\ term\ +\left(\frac{10}{2}+1\right)^{th}\ term}{2}$
= $\frac{5^{th}\ term\ +\left(5+1\right)^{th}\ term}{2}$
= $\frac{5^{th}\ term\ +6^{th}\ term}{2}$
= $\frac{37 + 42}{2}$
= $\frac{79}{2}$
y = 39.5
value of (x + y) = (39+39.5)
= 78.5
Question 12: The following table shows the weight of 20 students:
The mode and median of the above data are, respectively:
a) 51 and 48
b) 53 and 56
c) 60 and 53
d) 48 and 53
12) Answer (D)
Solution:
Mode is the maximum number of students have the same weight. So here 6 students have 48 kg weight. Hence 48 will be the mode.
For Median, first, we need to arrange the given weight in the ascending order from left to right.
Total number of students = n = 20
Median = $\frac{\left(\frac{n}{2}\right)th+\left(\frac{n}{2}+1\right)th}{2}$
= $\frac{\left(\frac{20}{2}\right)th+\left(\frac{20}{2}+1\right)th}{2}$
= $\frac{\left(10\right)th+\left(10+1\right)th}{2}$
= $\frac{\left(10\right)th+\left(11\right)th}{2}$
= $\frac{53+53}{2}$
= 53
Question 13: The mode of the given data 2, 5, 5, 7, 2, 6, 8, 6, 9, 6 is:
a) 7
b) 6
c) 5
d) 2
13) Answer (B)
Solution:
Arrange the given data in ascending order which is given below.
2, 2, 5, 5, 6, 6, 6, 7, 8, 9
The maximum number of times a number is occurring is called mode. Here ‘6’ is occurring three times. So ‘6’ will be the mode.
Question 14: x is the greatest number by which, when 2460, 2633 and 2806 are divided, the remainder in each case is the same. What is the sum of digits of x?
a) 11
b) 10
c) 13
d) 9
14) Answer (A)
Solution:
As given in the question, the remainder is the same in each. So (2633-2460), (2806-2460) and (2806-2633) should be completely divisible by ‘x’.
(2633-2460) = 173
(2806-2460) = 346
(2806-2633) = 173
As we know that 173 is a prime number. So the HCF of (173, 346 and 173) will be 173.
Hence x = 173
Sum of digits of x = 1+7+3
= 11
Question 15: What is the sum of the median and mode of the data
8, 1, 5, 4, 9, 6, 3, 6, 1, 3, 6, 9, 1, 7, 2, 6, 5 ?
a) 13
b) 11
c) 12
d) 14
15) Answer (B)
Solution:
Arrange the given data in ascending order which is given below.
1, 1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 6, 6, 7, 8, 9, 9
Median = middle term of the given data after arranging them in ascending order
Here 17 numbers are given. So the median will be the 9th number from the left end. It’s 5.
Mode = maximum number of times a number is occurring in the given data
= 6 (It is coming four times in the given data)
Sum of the median and mode of the data = 5+6
= 11
Question 16: The mode of 2, 2, 3, 3, 5, 5, 5, 7, 8, 8, 9, 10 is:
a) 5
b) 2
c) 3
d) 6
16) Answer (A)
Solution:
Mode : The value that appears most often in a set of given data values.
Given Data : 2, 2, 3, 3, 5, 5, 5, 7, 8, 8, 9, 10
- Most number repeated in above data is 5.
So, Mode of the given data is 5.
Hence, Option A is correct.
Question 17: The mode of the following data is 36. What is the value of x ?
a) 11
b) 15
c) 13
d) 12
17) Answer (D)
Solution:
As per given data,
Class interval of 30-40 has highest frequency, thatswhy it is modal class
As we know,
M = $l+\left\{\frac{\left(f_1-f_0\right)}{2f_1-f_0-f_2}\right\}\times\ h$
where, h= size of the class interval,
l = lower limit of the modal class,
$f_{1=}$ frequency of the modal class,
$f_{_0=}$ frequency of the class preceding the modal class
$f_{_2=}$ frequency of the class succeeding the modal class
putting the values from the given data :
$36=30+\frac{\left(16-10\right)}{2\times\ 16-10-x}\times\ 10$
$36-30=\frac{6}{22-x}\times\ \ 10$
$22-x=10$
$x=12$
Hence, Option D is correct.
Question 18: When 6892, 7105 and 7531 are divided by the greatest number x, then the remainder in each case is y. What is the value of (x – y)?
a) 123
b) 137
c) 147
d) 113
18) Answer (B)
Solution:
We have to find HCF of given numbers : 6892, 7105, 7531
7105 – 6892 = 213
7531 – 7105 = 426
426 – 213 = 213
So, Either the difference or the factor of difference is the HCF of those given number.
Here , 213 is the HCF.
When 6892, 7105, 7531 is divided by 213 we get 76 as an remainder
So, x = 213 and y = 76
According to Question :
x – y = 213 – 76 = 137
Hence, Option B is correct.
Question 19: The sum of the perfect square between 120 and 300 is:
a) 1400
b) 1024
c) 1296
d) 1204
19) Answer (A)
Solution:
Sum of the squares of n consecutive numbers =
The sum of the perfect square between 120 and 300 = $11^2+12^2+13^2+14^2+15^2+16^2+17^2$
$=\frac{17\left(17+1\right)\left(2\left(17+1\right)\right)}{6}-\frac{10\left(10+1\right)\left(2\left(10\right)+1\right)}{6}$
$=\frac{17\left(18\right)\left(35\right)}{6}-\frac{10\left(11\right)\left(21\right)}{6}$
$=51\times35-11\times35$
$=35\left(51-11\right)$
$=35\left(40\right)$
$=1400$
Hence, the correct answer is Option A
Question 20: The difference between the greatest and the least four digit numbers that begins with 3 and ends with 5 is:
a) 990
b) 900
c) 909
d) 999
20) Answer (A)
Solution:
The greatest four digit number that begins with 3 and ends with 5 = 3995
The least four digit number that begins with 3 and ends with 5 = 3005
$\therefore\ $The difference between the greatest and the least four digit numbers that begins with 3 and ends with 5 = 3995 – 3005 = 990
Question 21: If a nine-digit number 389 x 6378 y is divisible by 72, then the value of $\sqrt{6x + 7y}$ will be:
a) 6
b) $\sqrt{13}$
c) $\sqrt{46}$
d) 8
21) Answer (D)
Solution:
389x6378y is divisible by 72,
Factor of 72 = 8 $\times$ 9
So, number is divisible by 8 and 9 both.
Divisibility rule for 8,
78y (last three digits should be divisible by 8)
784 is divisible by 8 so,
Value of y = 4
Divisibility rule of 9,
3 + 8 + 9 + x + 6 + 3 + 7+ 8 + 4
= 48 + x
54 is divisible by 9
So, x = 54 – 48 = 6
Value of $\sqrt{6x + 7y}$
= $\sqrt{6 \times 6 + 7 \times 4}$
= $\sqrt{36 + 28}$ =$\sqrt{64}$
= 8
Question 22: When 7897, 8110 and 8536 are divided by the greatest number x, then the remainder in each case is the same. The sum of the digits of x is:
a) 14
b) 5
c) 9
d) 6
22) Answer (D)
Solution:
Let the remainder be k.
7897 – k = ax
8110 – k = bx
8536 – k = cx
Common factor is x.
So difference between the numbers,
8110 – 7897 = 213
8536 – 8110 = 426
8536 – 7897 = 639
HCF of 213, 426 and 639 is 213.
x = 213
Sum of the digits of x = 2 + 1 + 3 = 6
Question 23: One of the factors of $(8^{2k} + 5^{2k})$, where k is an odd number, is:
a) 86
b) 88
c) 84
d) 89
23) Answer (D)
Solution:
$(8^{2k} + 5^{2k})$,k is odd nuber so,
Let the k be 1.
= $(8^{2} + 5^{2})$
= 64 + 25 = 89
Question 24: The sum of the digits of a two-digit number is $\frac{1}{7}$ of the number. The units digit is 4 less than the tens digit. If the number obtained on reversing its digits is divided by 7, the remainder will be:
a) 4
b) 5
c) 1
d) 6
24) Answer (D)
Solution:
Let the number be (10a + b).
ATQ,
a + b = $\frac{10a + b}{7}$
7a + 7b = 10a + b
6b = 3a
2b = a —(1)
a – b = 4 —(2)
From eq (1) and (2),
2b – b = 4
b = 4
a = 4 $\times 2$ = 8
Number = 10a + b = 10 $\times 8 + 4 = 84
reverse of the number = 48
Remainder after divide by 7 = 48/7 = 6
Question 25: If the 11-digit number 5678x43267y is divisible by 72, then the value of $\sqrt{5x + 8y}$ is:
a) 6
b) 4
c) 7
d) 8
25) Answer (A)
Solution:
11-digit number 5678x43267y is divisible by 72.
It will be divisible by 9 and 8.
For the divisiblity by 8,
67y divisible by 8.
So value of y = 2
For the divisiblity by 9,
(5+6+7+8+x+4+3+2+6+7+y = 50 + x) divisible by 8.
So value of x = 54 – 50 = 4
$\sqrt{5x + 8y}$
= $\sqrt{5\times 4 + 8\times 2}$
= $\sqrt{36}$ = 6
