Series Completion Questions for MAH – CET 2022 – Download PDF
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Question 1:Â What is the least number to be added to 2530 to make it a perfect square ?
a)Â 50
b)Â 65
c)Â 75
d)Â 80
e)Â None of these
Question 2:Â What should come in place of the question mark (?) in the following number series?
1 4 14 45 139 422 ?
a)Â 1268
b)Â 1234
c)Â 1272
d)Â 1216
e)Â None of these
Question 3:Â What should come in place of the question mark (?) in the following number series?
2 5 11 23 47 95 ?
a)Â 168
b)Â 154
c)Â 191
d)Â 172
e)Â None of these
Question 4:Â What approximate value should come in place of the question mark (?) in the following question?
$54.786 \div 10.121 \times 4.454 = ?$
a)Â 84
b)Â 48
c)Â 118
d)Â 58
e)Â 24
Question 5:Â What should come in place of the question mark (?) in the following number series ?
1, 5, 17, 53, 161, 485, ?
a)Â 1168
b)Â 1254
c)Â 1457
d)Â 1372
e)Â None of these
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Question 6:Â What will come in place of the question mark (?) in the following number series?
20Â 23Â 30Â 43Â 64Â ?
a)Â 95
b)Â 90
c)Â 100
d)Â 105
e)Â 96
Question 7:Â What will come in place of thequestion mark (?) in the following number series?
33 16.5 ? 24.75  49.5 123.75
a)Â 18.5
b)Â 16.5
c)Â 8.5
d)Â 8.25
e)Â None of these
Question 8:Â What will come in place of thequestion mark (?) in the following number series?
44 ? 99  148.5  222.75 334.125
a)Â 44
b)Â 55
c)Â 66
d)Â 33
e)Â 35
Question 9:Â What will come in place of thequestion mark (?) in the following number series?
121 238 472 ? 1876 3748
a)Â 1008
b)Â 948
c)Â 944
d)Â 940
e)Â 1005
Question 10:Â What will come in place of the question mark (?) in the following number series?
9 10 39 220 ? 14382
a)Â 1589
b)Â 1598
c)Â 1958
d)Â 1985
e)Â 1835
Question 11:Â What will come in the place of question mark (?) in the following series ?
2 Â 9 Â 28 Â 65 ?
a)Â 96
b)Â 106
c)Â 126
d)Â 130
e)Â None of these
Question 12:Â What would be the compound interest accrued on an amount of Rs. 9,000 at the rate of 11 p.c.p.a. in two years ?
a)Â Rs. 2089.90
b)Â Rs. 2140.90
c)Â Rs. 2068.50
d)Â Rs. 2085.50
e)Â None of these
Question 13:Â 16 8 12 30 ? 472.5
a)Â 104
b)Â 103
c)Â 106
d)Â 105
e)Â None of these
Question 14:Â 2, 5, 12, 27, 58, ?
a)Â 122
b)Â 121
c)Â 123
d)Â 120
e)Â None of these
Question 15:Â 18 19.7 16.3 23.1 9.5 ?
a)Â 36.5
b)Â 36.8
c)Â 36.7
d)Â 36.9
e)Â None of these
Question 16:Â 68, ?, 77, 104, 168, 293
a)Â 69
b)Â 70
c)Â 68
d)Â 74
e)Â None of these
Question 17: In how many different ways can the numbers ‘256974’ be arranged, using each digit only once in each arrangement, such that the digits 6 and 5 are at the extreme ends in each arrangement ?
a)Â 48
b)Â 720
c)Â 36
d)Â 360
e)Â None of these
Question 18:Â What will come in place of both the question marks (?) in the following question ?$\frac{(?)^{0.6}}{104}=\frac{26}{(?)^{1.4}}$
a)Â 58
b)Â -48
c)Â -56
d)Â 42
e)Â -52
Question 19:Â Out of the fractions $\frac{1}{2}, \frac{7}{8}, \frac{3}{4}, \frac{5}{6}$, and $\frac{6}{7}$ what is the difference between the largest and smallest fractions ?
a)Â $\frac{7}{13}$
b)Â $\frac{3}{8}$
c)Â $\frac{4}{7}$
d)Â $\frac{1}{6}$
e)Â None of these
Question 20:Â If $(11)^{3}$ is subtracted from $(46)^{2}$ . what will be the remainder ?
a)Â 787
b)Â 785
c)Â 781
d)Â 783
e)Â None of these
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Answers & Solutions:
1) Answer (E)
We know that $50^2 = 2500$ and $51^2 = 2601$
$\because$ 2500Â < 2530Â < 2601
$\therefore$ Required number = 2601 – 2530 = 71
2) Answer (C)
The pattern here followed is :
1Â * 3 +Â 1 = 4
4Â * 3 +Â 2 = 14
14Â * 3 + 3 = 45
45Â * 3 +Â 4 = 139
139Â * 3 + 5 = 422
422 * 3 + 6 =Â 1272
3) Answer (C)
The pattern here followed is :
2 * 2 + 1 = 5
5 *Â 2 + 1 =Â Â 11
11 *Â 2 + 1 =Â 23
23 *Â 2 + 1 =Â 47
47 *Â 2 + 1 =Â 95
95 *Â 2 + 1 =Â 191
4) Answer (E)
Expression : $54.786 \div 10.121 \times 4.454 = ?$
= $\frac{55}{10} \times 4.5$
= $24.75 \approx 24$
5) Answer (C)
The pattern here followed is :
1 * 3 + 2 = 5
5Â * 3 + 2 =Â 17
17Â * 3 + 2 =Â 53
53Â * 3 + 2 =Â 161
161Â * 3 + 2 =Â 485
485Â * 3 + 2 =Â 1457
6) Answer (A)
Numbers of the form $n^2 – (n-1)$ are added, where $n$ is an integer starting from 2
23 $+ 2^2 – 1$ = 23
23Â $+ 3^2 – 2$ =Â 30
30Â $+ 4^2 – 3$ =Â 43
43Â $+ 5^2 – 4$ =Â 64
64Â $+ 6^2 – 5$ =Â 95
7) Answer (B)
The pattern followed is :
33 $\times \frac{1}{2}$ = 16.5
16.5Â $\times \frac{2}{2}$ =Â 16.5
16.5Â $\times \frac{3}{2}$ =Â 24.75
24.75Â $\times \frac{4}{2}$ =Â 49.5
49.5Â $\times \frac{5}{2}$ =Â 123.75
8) Answer (C)
Each number is multiplied by $\frac{3}{2}$
44 $\times \frac{3}{2}$ =Â 66
66Â $\times \frac{3}{2}$ =Â 99
99Â $\times \frac{3}{2}$ =Â 148.5
148.5Â $\times \frac{3}{2}$ =Â 222.75
222.75Â $\times \frac{3}{2}$ =Â 334.125
9) Answer (D)
Each number is multiplied by 2 and then 4 is subtracted from it.
121 $\times 2 – 4$ = 238
238Â $\times 2 – 4$ =Â 472
472Â $\times 2 – 4$ =Â 940
940Â $\times 2 – 4$ =Â 1876
1876Â $\times 2 – 4$ =Â 3748
10) Answer (A)
The pattern followed is :
9 $\times 1 + 1^2$ = 10
10Â $\times 3 + 3^2$ =Â 39
39Â $\times 5 + 5^2$ =Â 220
220Â $\times 7 + 7^2$ =Â 1589
1589Â $\times 9 + 9^2$ =Â 14382
11) Answer (C)
Each number is of the form $(n^3+1)$ where $n$ is a natural number
$1^3+1$ = 2
$2^3+1$ =Â 9
$3^3+1$ =Â 28
$4^3+1$ =Â 65
$5^3+1$ =Â 126
=> Ans – (C)
12) Answer (E)
$C.I. = P [(1 + \frac{R}{100})^T – 1]$
= $9000 [(1 + \frac{11}{100})^2 – 1]$
= $9000 [(1.11)^2 – 1]$
= $9000 \times (1.2321 – 1)$
= $9000 \times 0.2321$ = Rs. $2,088.90$
13) Answer (D)
Odd multiples of $\frac{1}{2}$ are multiplied
16 $\times \frac{1}{2}$ = 8
8Â $\times \frac{3}{2}$ =Â 12
12Â $\times \frac{5}{2}$ =Â 30
30Â $\times \frac{7}{2}$ =Â 105
105Â $\times \frac{9}{2}$ =Â 472.5
14) Answer (B)
Each number is multiplied by 2 and then consecutive natural numbers are added
2 $\times 2 + 1$ = 5
5Â $\times 2 + 2$ =Â 12
12Â $\times 2 + 3$ =Â 27
27Â $\times 2 + 4$ =Â 58
58Â $\times 2 + 5$ =Â 121
15) Answer (C)
The pattern is :
18 $+ 1.7 \times 2^0$ = 19.7
19.7Â $- 1.7 \times 2^1$ =Â 16.3
16.3Â $+ 1.7 \times 2^2$ =Â 23.1
23.1Â $- 1.7 \times 2^3$ =Â 9.5
9.5Â $+ 1.7 \times 2^4$ =Â 36.7
16) Answer (A)
Cubes of consecutive natural numbers are added
68 $+ 1^3$ =Â 69
69Â $+ 2^3$ =Â 77
77Â $+ 3^3$ =Â 104
104Â $+ 4^3$ =Â 168
168Â $+ 5^3$ =Â 293
17) Answer (A)
Case 1 : 6 at left end and 5 is at right end : 6 _ _ _ _ 5
Now, four empty places can be filled by 2,9,7 and 4 in = $4!$ ways
= $4 \times 3 \times 2 \times 1 = 24$
Case 2 : 6 at right end and 5 at left end : 5 _ _ _ _ 6
Similarly, no. of ways = $4!$
= $4 \times 3 \times 2 \times 1 = 24$
$\therefore$ Total no. of ways = $24 + 24 = 48$
18) Answer (E)
$\frac{(x)^{0.6}}{104}=\frac{26}{(x)^{1.4}}$
${(x)^{0.6}} * {(x)^{01.4}}$ = 104*26
${(x)^{2}}$ = 104*26
x = ±52
19) Answer (B)
Given values are ,
$\frac{1}{2}$ = 0.5
$\frac{7}{8}$ = 0.87
$\frac{3}{4}$ = 0.75
$\frac{5}{6}$ = 0.83
$\frac{6}{7}$ = 0.86
∴ Required difference = $\frac{7}{8}$ – $\frac{1}{2}$ = (7-4)/8 = 3/8
20) Answer (B)
Here
$(46)^2$ = 2116
$(11)^3$ = 1331
So, 2116 – 1331 = 785