XAT Progression Questions PDF [Most Important]
Download Progression Questions for XAT PDF – XAT Progression questions PDF by Cracku. Practice XAT solved Progression Questions paper tests, and these are the practice question to have a firm grasp on the Progression topic in the XAT exam. Top 20 very Important Progression Questions for XAT based on asked questions in previous exam papers. The XAT question papers contain actual questions asked with answers and solutions.
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Question 1: Out of 6 numbers, the sum of the first 5 numbers is 7 times the $6^{th}$ number. If their average is 136, then the $6^{th}$ number is:
a) 116
b) 102
c) 96
d) 84
1) Answer (B)
Solution:
Average of 6 numbers = 136
(average = sum of terms/number of terms)
Sum of 6 numbers = 136 $\times$ 6 = 816 —(1)
Let the 6th number be x.
Sum of 5 numbers = 7x
From eq(1),
7x + x = 816
x = 816/8 = 102
$\therefore$ The $6^{th}$ number is 102.
Question 2: If x is subtracted from each of the numbers 20, 37, 54 and 105, then the numbers so obtained in this order are in proportion. What is the mean proportional between (7x – 5) and (x + 1)?
a) 8
b) 6
c) 12
d) 9
2) Answer (A)
Solution:
As per question $\dfrac {20-x}{37-x} = \dfrac{54-x}{105-x}$
$\Rightarrow 2100-20x-105x+x^2= 1998-37x-54x+x^2$
$\Rightarrow 125x-91x= 2100-1998$
$\Rightarrow 34x= 102 $
$\Rightarrow x =3 $
mean proportional between $(7x-5) and (x+1) = \sqrt{(7x-5) (x+1)}$
$\Rightarrow \sqrt {(7\times 3-5)(3+1)}$ put the value $x$
$\Rightarrow \sqrt {16\times 4}$
$\Rightarrow \sqrt{64}$
$\Rightarrow 8 $ Ans
Question 3: The most stable measure of Central tendency is
a) Range
b) Mean
c) Median
d) Mode
3) Answer (B)
Solution:
The Mean is widely preferred as the best measure of central tendency because it is the measure that includes all the values in the data set for its calculation, and any change in any of the scores will affect the value of the mean.
Hence, the correct answer is Option B
Question 4: What is the greatest number which when divides 460, 491 and 553, leave 26 as remainderin each case?
a) 33
b) 27
c) 35
d) 31
4) Answer (D)
Solution:
Subtract each number by 26
so we get
460 – 26 = 434
491 – 26 = 465
553 – 26 = 527
now taking HCF (434, 465 , 527) = 31 ( since 31 divides all the 3 numbers)
Question 5: The greatest number,that divides 43, 91 and 183 so as to leave the same remainder in each case, is
a) 9
b) 8
c) 4
d) 3
5) Answer (C)
Solution:
3 numbers are 43,91,183
largest number is 183
smallest number is 43
subtract smallest number from both the highest number
so 183 – 43 = 140
91 – 43 = 48
91 is smaller than 183, so subtract 91 from 183
183 – 91 = 92
now we have 3 numbers 140,48,92
so HCF of 140,48,92 = 4
thus the greatest number,that divides 43, 91 and 183 so as to leave the same remainder in each case, is 4
Question 6: The greatest number among $ 3^{50}, 4^{40}, 5^{30}, 6^{20}$ is
a) $ 4^{40} $
b) $ 5^{30} $
c) $ 6^{20} $
d) $ 3^{50} $
6) Answer (A)
Solution:
$ 3^{50} = 243^{10} $
$4^{40} = 1024^{10}$
$5^{30} = 125^{10}$
$6^{20} = 36^{10}$
So, the greatest would be $4^{40}$.
So , the answer would be option a)$4^{40}$.
Question 7: If the 10-digit number 897359y7x2 is divisible by 72, then what is the value of(3x-y), for the possible greatest value of y ?
a) 3
b) 8
c) 7
d) 5
7) Answer (C)
Solution:
Given, 897359y7x2 is divisible by 72 then the number is divisible by 8 and 9
If the number is divisible by 8, then the last three digits should be divisible by 8
$\Rightarrow$ 7×2 is divisible by 8
$\Rightarrow$ x = 1, 5, 9
If the number is divisible by 9, then the sum of the digits of the number should be divisible by 9
$\Rightarrow$ 8 + 9 + 7 + 3 + 5 + 9 + y + 7 + x + 2 = multiple of 9
$\Rightarrow$ 50 + y + x = multiple of 9
For x = 1
50 + y + 1 = multiple of 9
$\Rightarrow$ 51 + y = multiple of 9
The only possible value of y is 3
For x = 5
50 + y + 5 = multiple of 9
$\Rightarrow$ 55 + y = multiple of 9
The only possible value of y is 8
For x = 9
50 + y + 9 = multiple of 9
$\Rightarrow$ 59 + y = multiple of 9
The only possible value of y is 4
So when x = 5, y has the greatest value 8.
$\therefore\ $3x – y = 3(5) – 8 = 7
Hence, the correct answer is Option C
Question 8: The greatest number that will divide 146, 248 and 611 leaving remainders 2,8 and 11 respectively is:
a) 47
b) 144
c) 24
d) 612
8) Answer (C)
Solution:
The greatest number that will divide 146, 248 and 611 leaving remainders 2,8 and 11 respectively
= H.C.F. of (146-2), (248-8), (611-11)
= H.C.F. (144,240,600) = 24
=> Ans – (C)
Question 9: Which of the following is the greatest among all?
a) 8% discount on marked price ₹400
b) 12% discount on marked price ₹240
c) 7% discount on marked price ₹500
d) 10% discount on marked price ₹320
9) Answer (C)
Solution:
Discount $=Discount\% \times Marked Price$
8% discount on marked price ₹400 = $400\times \frac{8}{100} =$₹ 32
12% discount on marked price ₹240 = $240\times \frac{12}{100} =$₹ 28.8
7% discount on marked price ₹500 =$500\times \frac{7}{100} =$₹ 35
10% discount on marked price ₹320 =$320\times \frac{10}{100} =$₹ 32
Option C is correct.
Question 10: If $A = \frac{1}{1 \times 2} + \frac{1}{1 \times 4} + \frac{1}{2 \times 3} + \frac{1}{4 \times 7} + \frac{1}{3 \times 4} + \frac{1}{7 \times 10}…$ upto 20 terms, then what is the value of $A?$
a) $\frac{379}{308}$
b) $\frac{171}{140}$
c) $\frac{379}{310}$
d) $\frac{420}{341}$
10) Answer (D)
Solution:
A=$\frac{1}{1 \times 2} + \frac{1}{1 \times 4} + \frac{1}{2 \times 3} + \frac{1}{4 \times 7} + \frac{1}{3 \times 4} + \frac{1}{7 \times 10}…$ upto 20 terms
It has two series i.e $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + …\frac{1}{10 \times 11} $and
$\frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \frac{1}{7 \times 10}+….\frac{1}{28 \times 31}$
=$\frac{1}{1 } – \frac{1}{2}+ \frac{1}{2 } + \frac{1}{3 } + … \frac{1}{10}-\frac{1}{ 11} +\frac{1}{3} (\frac{1}{1} – \frac{1}{4} + \frac{1}{4}+….\frac{1}{28}-\frac{1}{31})$
=(1-(1/11))+(1/3)(1-(1/31))
=10/11 +10/31
=420/31
Question 11: What is the value of $S = \frac{1}{1 \times 3 \times 5} + \frac{1}{1 \times 4} + \frac{1}{3 \times 5 \times 7} + \frac{1}{4 \times 7} + \frac{1}{5 \times 7 \times 9 } + \frac{1}{7 \times 10}+…..$upto 20 terms, then what is the value of S?
a) $\frac{6179}{15275}$
b) $\frac{6070}{14973}$
c) $\frac{7191}{15174}$
d) $\frac{5183}{16423}$
11) Answer (B)
Solution:
This series consists of two different series one having 2 numbers as product in denominator and the other has three numbers as product in the denominator
lat terms in each series is $ \frac{1}{19 \times 21 \times 23}$ and $\frac{1}{28 \times 31}$
$\frac{1}{1 \times 3 \times 5} + \frac{1}{1 \times 4} + \frac{1}{3 \times 5 \times 7} + \frac{1}{4 \times 7} + \frac{1}{5 \times 7 \times 9 } + \frac{1}{7 \times 10}+…..$upto 20 terms
=$\frac{1}{1 \times 3 \times 5} +\frac{1}{3 \times 5 \times 7}+…\frac{1}{1 \times 4} +\frac{1}{4 \times 7}+..$
=$\frac{1}{4}( \frac{1}{1 \times 3 }- \frac{1}{3 \times 5}+ \frac{1}{5 \times 7}+\frac{1}{7 \times 9}…-\frac{1}{21 \times 23 }+\frac{1}{1 \times 3} (\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}….-\frac{1}{31})$
=$\frac{1}{4}( \frac{1}{1 \times 3 }-\frac{1}{21 \times 23})+\frac{1}{1 \times 3} (\frac{1}{1}-\frac{1}{31})$
=$\frac{40}{483}+\frac{10}{31}$
=$\frac{6070}{14973}$
Question 12: If the Arithmetic mean of 7, 5, 13, x and 9 is 10, then the value of x is
a) 10
b) 12
c) 14
d) 16
12) Answer (D)
Solution:
Arithmetic mean of 7, 5, 13, x and 9 = 10
=> $\frac{(7+5+13+x+9)}{5}=10$
=> $34+x=10 \times 5=50$
=> $x=50-34=16$
=> Ans – (D)
Question 13: If 7 times the seventh term of an Arithmetic Progression (AP) is equal to 11 times its eleventh term, then the 18th term of the AP will be
a) 1
b) 0
c) 2
d) -1
13) Answer (B)
Solution:
Let the first term of the AP be $a$ and the common difference = $d$
7th term = $A_7=a+6d$
11th term = $A_{11}=a+10d$
According to ques,
=> $7 \times (a+6d)=11 \times (a+10d)$
=> $7a+42d=11a+110d$
=> $11a-7a=42d-110d$
=> $4a=-68d$
=> $a=-17d$
=> $a+17d = 0 = A_{18}$
=> Ans – (B)
Question 14: What is the sum of the first 13 terms of an arithmetic progression if the first term is -10 and last term is 26?
a) 104
b) 140
c) 84
d) 98
14) Answer (A)
Solution:
$S_{n}=\frac{n}{2}[a+l]$
$S_{13}=\frac{13}{2}[-10+26]$
$S_{13}=\frac{13}{2}[16]=104$
So the answer is option A.
Question 15: The 3rd and 7th term of an arithmetic progression are -9 and 11 respectively. What is the 15th term?
a) 28
b) 87
c) 51
d) 17
15) Answer (C)
Solution:
$T_{3}$ = a + 2d = -9——-(1)
$T_{7}$ = a + 6d = 11——-(2)
on solving (1) AND (2)
d = 5 & a = -19
$T_{15}$ = a + 14d = -19 + 14(5) = 51
so the answer is option C.
Question 16: The 2nd and 6th term of an arithmetic progression are 8 and 20 respectively. What is the 20th term?
a) 56
b) 65
c) 62
d) 59
16) Answer (C)
Solution:
$T_{2}$ = a + d = 8——-(1)
$T_{6}$ = a + 5d = 20——-(2)
on solving (1) AND (2)
d = 3 & a = 5
$T_{20}$ = a + 19d = 5 + 19(3) = 62
So the answer is option C.
Question 17: The 3rd and 6th term of an arithmetic progression are 13 and -5 respectively. What is the 11th term?
a) -29
b) -41
c) -47
d) -35
17) Answer (D)
Solution:
$T_{3}$ = a + 2d = 13——-(1)
$T_{6}$ = a + 5d = -5——-(2)
on solving (1) AND (2)
d = -6 & a = 25
$T_{11}$ = a + 10d = 25 + 10(-6) = -35
so the answer is option D.
Question 18: If the 3rd and the 5th term of an arithmetic progression are 13 and 21, what is the 13th term?
a) 53
b) 49
c) 57
d) 61
18) Answer (A)
Solution:
$T_{3}$ = a + 2d = 13——-(1)
$T_{5}$ = a + 4d = 21——-(2)
on solving (1) AND (2)
d = 4 & a = 5
$T_{13}$ = a + 12d = 5 + 12(4) = 5 + 48 = 53
So the answer is option A.
Question 19: What is the sum of the first 11 terms of an arithmetic progression if the 3rd term is -1 and the 8th term is 19?
a) 204
b) 121
c) 225
d) 104
19) Answer (B)
Solution:
$T_{3}$ = a + 2d = -1——-(1)
$T_{8}$ = a + 7d = 19——-(2)
on solving (1) AND (2)
d = 4 & a = -9
$S_{n}=\frac{n}{2}[2a+(n-1)d]$
$S_{11}=\frac{11}{2}[2(-9)+(11-1)(4)]$
$S_{11}=\frac{11}{2}[(-18)+(40)]$
$S_{11}=\frac{11}{2}[22]$
$S_{11}=121$
So the answer is option B.
Question 20: The 2nd and 8th term of an arithmetic progression are 17 and -1 respectively. What is the 14th term?
a) -22
b) -25
c) -19
d) -28
20) Answer (C)
Solution:
$T_{2}$ = a + d = 17——-(1)
$T_{8}$ = a + 7d = -1——-(2)
on solving (1) AND (2)
d = -3 & a = 20
$T_{14}$ = a + 13d = 20 + 13(-3) = -19
so the answer is option C.
