{"id":211299,"date":"2022-05-11T13:58:51","date_gmt":"2022-05-11T08:28:51","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=211299"},"modified":"2022-05-11T14:03:23","modified_gmt":"2022-05-11T08:33:23","slug":"number-series-questions-for-mba-cet-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/number-series-questions-for-mba-cet-pdf\/","title":{"rendered":"Number Series Questions for MBA-CET PDF"},"content":{"rendered":"

Here you can download the important Number series questions PDF with solutions for MAH MBA CET by Cracku. These questions will help you to make practice and solve the Number series aptitude section<\/strong> questions in the MAH MBA CET exam. Utilize this best PDF practice set<\/strong> which is included answers in detail. Click on the below link to download the Number series questions PDF for MBA-CET 2022.<\/p>\n

Download Number Series Questions for MBA-CET PDF<\/a><\/p>\n

Enroll to MAH-CET Crash Course<\/a><\/p>\n

Question 1:\u00a0<\/b>What will come in the place of question mark (?) in the following series?
\n2 \u00a09 \u00a028 \u00a065?<\/p>\n

a)\u00a096<\/p>\n

b)\u00a0106<\/p>\n

c)\u00a0126<\/p>\n

d)\u00a0130<\/p>\n

e)\u00a0None of these<\/p>\n

1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Solution:<\/b><\/p>\n

Each number is of the form $(n^3+1)$ where $n$ is a natural number<\/p>\n

$1^3+1$ = 2<\/p>\n

$2^3+1$ =\u00a09<\/p>\n

$3^3+1$ =\u00a028<\/p>\n

$4^3+1$ =\u00a065<\/p>\n

$5^3+1$ =\u00a0126<\/strong><\/p>\n

=> Ans – (C)<\/p>\n

Question 2:\u00a0<\/b>What will come in place of the question mark (?) in the following number series?
\n9\u00a0 10\u00a0 39\u00a0 220\u00a0 ?\u00a0 14382<\/p>\n

a)\u00a01589<\/p>\n

b)\u00a01598<\/p>\n

c)\u00a01958<\/p>\n

d)\u00a01985<\/p>\n

e)\u00a01835<\/p>\n

2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Solution:<\/b><\/p>\n

The pattern followed is\u00a0:<\/p>\n

9 $\\times 1 + 1^2$ = 10<\/p>\n

10\u00a0$\\times 3 + 3^2$ =\u00a039<\/p>\n

39\u00a0$\\times 5 + 5^2$ =\u00a0220<\/p>\n

220\u00a0$\\times 7 + 7^2$ =\u00a01589<\/strong><\/p>\n

1589\u00a0$\\times 9 + 9^2$ =\u00a014382<\/p>\n

Question 3:\u00a0<\/b>What will come in place of the question mark (?) in the following number series?
\n121\u00a0 238\u00a0 472\u00a0 ?\u00a0 1876\u00a0 3748<\/p>\n

a)\u00a01008<\/p>\n

b)\u00a0948<\/p>\n

c)\u00a0944<\/p>\n

d)\u00a0940<\/p>\n

e)\u00a01005<\/p>\n

3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Solution:<\/b><\/p>\n

Each number is multiplied by 2 and then 4 is subtracted from it.<\/p>\n

121 $\\times 2 – 4$ = 238<\/p>\n

238\u00a0$\\times 2 – 4$ =\u00a0472<\/p>\n

472\u00a0$\\times 2 – 4$ =\u00a0940<\/strong><\/p>\n

940\u00a0$\\times 2 – 4$ =\u00a01876<\/p>\n

1876\u00a0$\\times 2 – 4$ =\u00a03748<\/p>\n

Question 4:\u00a0<\/b>What will come in place of the question mark (?) in the following number series?
\n44\u00a0 ?\u00a0 99\u00a0 \u00a0148.5 \u00a0222.75\u00a0 334.125<\/p>\n

a)\u00a044<\/p>\n

b)\u00a055<\/p>\n

c)\u00a066<\/p>\n

d)\u00a033<\/p>\n

e)\u00a035<\/p>\n

4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Solution:<\/b><\/p>\n

Each number is multiplied by $\\frac{3}{2}$<\/p>\n

44 $\\times \\frac{3}{2}$ =\u00a066<\/strong><\/p>\n

66\u00a0$\\times \\frac{3}{2}$ =\u00a099<\/p>\n

99\u00a0$\\times \\frac{3}{2}$ =\u00a0148.5<\/p>\n

148.5\u00a0$\\times \\frac{3}{2}$ =\u00a0222.75<\/p>\n

222.75\u00a0$\\times \\frac{3}{2}$ =\u00a0334.125<\/p>\n

Question 5:\u00a0<\/b>What will come in place of the question mark (?) in the following number series?
\n33\u00a0 16.5?\u00a0 24.75\u00a0 \u00a0 49.5\u00a0 123.75<\/p>\n

a)\u00a018.5<\/p>\n

b)\u00a016.5<\/p>\n

c)\u00a08.5<\/p>\n

d)\u00a08.25<\/p>\n

e)\u00a0None of these<\/p>\n

5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Solution:<\/b><\/p>\n

The pattern followed is\u00a0:<\/p>\n

33 $\\times \\frac{1}{2}$ = 16.5<\/p>\n

16.5\u00a0$\\times \\frac{2}{2}$ =\u00a016.5<\/strong><\/p>\n

16.5\u00a0$\\times \\frac{3}{2}$ =\u00a024.75<\/p>\n

24.75\u00a0$\\times \\frac{4}{2}$ =\u00a049.5<\/p>\n

49.5\u00a0$\\times \\frac{5}{2}$ =\u00a0123.75<\/p>\n

Question 6:\u00a0<\/b>What will come in place of the question mark (?) in the following number series?
\n20\u00a0 23\u00a0 30\u00a0 43\u00a0 64\u00a0 ?<\/p>\n

a)\u00a095<\/p>\n

b)\u00a090<\/p>\n

c)\u00a0100<\/p>\n

d)\u00a0105<\/p>\n

e)\u00a096<\/p>\n

6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Solution:<\/b><\/p>\n

Numbers of the form $n^2 – (n-1)$ are added, where $n$ is an integer starting from 2<\/p>\n

23 $+ 2^2 – 1$ = 23<\/p>\n

23\u00a0$+ 3^2 – 2$ =\u00a030<\/p>\n

30\u00a0$+ 4^2 – 3$ =\u00a043<\/p>\n

43\u00a0$+ 5^2 – 4$ =\u00a064<\/p>\n

64\u00a0$+ 6^2 – 5$ =\u00a095<\/strong><\/p>\n

Question 7:\u00a0<\/b>What should come in place of the question mark (?) in the following number series?
\n1, 5, 17, 53, 161, 485, ?<\/p>\n

a)\u00a01168<\/p>\n

b)\u00a01254<\/p>\n

c)\u00a01457<\/p>\n

d)\u00a01372<\/p>\n

e)\u00a0None of these<\/p>\n

7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Solution:<\/b><\/p>\n

The pattern here followed is\u00a0:<\/p>\n

1 * 3 + 2 = 5<\/p>\n

5\u00a0* 3 + 2 =\u00a017<\/p>\n

17\u00a0* 3 + 2 =\u00a053<\/p>\n

53\u00a0* 3 + 2 =\u00a0161<\/p>\n

161\u00a0* 3 + 2 =\u00a0485<\/p>\n

485\u00a0* 3 + 2 =\u00a01457<\/strong><\/p>\n

Question 8:\u00a0<\/b>What approximate value should come in place of the question mark (?) in the following question?
\n$54.786 \\div 10.121 \\times 4.454 = ?$<\/p>\n

a)\u00a084<\/p>\n

b)\u00a048<\/p>\n

c)\u00a0118<\/p>\n

d)\u00a058<\/p>\n

e)\u00a024<\/p>\n

8)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

Solution:<\/b><\/p>\n

Expression\u00a0: $54.786 \\div 10.121 \\times 4.454 = ?$<\/p>\n

= $\\frac{55}{10} \\times 4.5$<\/p>\n

= $24.75 \\approx 24$<\/p>\n

Question 9:\u00a0<\/b>What should come in place of the question mark (?) in the following number series?
\n2 5 11 23 47 95 ?<\/p>\n

a)\u00a0168<\/p>\n

b)\u00a0154<\/p>\n

c)\u00a0191<\/p>\n

d)\u00a0172<\/p>\n

e)\u00a0None of these<\/p>\n

9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Solution:<\/b><\/p>\n

The pattern here followed is\u00a0:<\/p>\n

2 * 2 + 1 = 5<\/p>\n

5 *\u00a02 + 1 =\u00a0\u00a011<\/p>\n

11 *\u00a02 + 1 =\u00a023<\/p>\n

23 *\u00a02 + 1 =\u00a047<\/p>\n

47 *\u00a02 + 1 =\u00a095<\/p>\n

95 *\u00a02 + 1 =\u00a0191<\/strong><\/p>\n

Question 10:\u00a0<\/b>What should come in place of the question mark (?) in the following number series?
\n1 4 14 45 139 422 ?<\/p>\n

a)\u00a01268<\/p>\n

b)\u00a01234<\/p>\n

c)\u00a01272<\/p>\n

d)\u00a01216<\/p>\n

e)\u00a0None of these<\/p>\n

10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Solution:<\/b><\/p>\n

The pattern here followed is\u00a0:<\/p>\n

1\u00a0* 3 +\u00a01 = 4<\/p>\n

4\u00a0* 3 +\u00a02 = 14<\/p>\n

14\u00a0* 3 + 3 = 45<\/p>\n

45\u00a0* 3 +\u00a04 = 139<\/p>\n

139\u00a0* 3 + 5 = 422<\/p>\n

422 * 3 + 6 =\u00a01272<\/strong><\/p>\n

Take Free MAH-CET mock tests here<\/a><\/p>\n

Question 11:\u00a0<\/b>What is the least number to be added to 2530 to make it a perfect square?<\/p>\n

a)\u00a050<\/p>\n

b)\u00a065<\/p>\n

c)\u00a075<\/p>\n

d)\u00a080<\/p>\n

e)\u00a0None of these<\/p>\n

11)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

Solution:<\/b><\/p>\n

We know that $50^2 = 2500$ and $51^2 = 2601$<\/p>\n

$\\because$ 2500\u00a0< 2530\u00a0< 2601<\/p>\n

$\\therefore$ Required number = 2601 – 2530 = 71<\/p>\n

Question 12:\u00a0<\/b>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?<\/p>\n

a)\u00a0Rs. 2089.90<\/p>\n

b)\u00a0Rs. 2140.90<\/p>\n

c)\u00a0Rs. 2068.50<\/p>\n

d)\u00a0Rs. 2085.50<\/p>\n

e)\u00a0None of these<\/p>\n

12)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

Solution:<\/b><\/p>\n

$C.I. = P [(1 + \\frac{R}{100})^T – 1]$<\/p>\n

= $9000 [(1 + \\frac{11}{100})^2 – 1]$<\/p>\n

= $9000 [(1.11)^2 – 1]$<\/p>\n

= $9000 \\times (1.2321 – 1)$<\/p>\n

= $9000 \\times 0.2321$ = Rs. $2,088.90$<\/p>\n

Question 13:\u00a0<\/b>16 8 12 30 ? 472.5<\/p>\n

a)\u00a0104<\/p>\n

b)\u00a0103<\/p>\n

c)\u00a0106<\/p>\n

d)\u00a0105<\/p>\n

e)\u00a0None of these<\/p>\n

13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Solution:<\/b><\/p>\n

Odd multiples of $\\frac{1}{2}$ are multiplied<\/p>\n

16 $\\times \\frac{1}{2}$ = 8<\/p>\n

8\u00a0$\\times \\frac{3}{2}$ =\u00a012<\/p>\n

12\u00a0$\\times \\frac{5}{2}$ =\u00a030<\/p>\n

30\u00a0$\\times \\frac{7}{2}$ =\u00a0105<\/strong><\/p>\n

105\u00a0$\\times \\frac{9}{2}$ =\u00a0472.5<\/p>\n

Question 14:\u00a0<\/b>2, 5, 12, 27, 58, ?<\/p>\n

a)\u00a0122<\/p>\n

b)\u00a0121<\/p>\n

c)\u00a0123<\/p>\n

d)\u00a0120<\/p>\n

e)\u00a0None of these<\/p>\n

14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Solution:<\/b><\/p>\n

Each number is multiplied by 2 and then consecutive natural numbers are added<\/p>\n

2 $\\times 2 + 1$ = 5<\/p>\n

5\u00a0$\\times 2 + 2$ =\u00a012<\/p>\n

12\u00a0$\\times 2 + 3$ =\u00a027<\/p>\n

27\u00a0$\\times 2 + 4$ =\u00a058<\/p>\n

58\u00a0$\\times 2 + 5$ =\u00a0121<\/strong><\/p>\n

Question 15:\u00a0<\/b>18 19.7 16.3 23.1 9.5 ?<\/p>\n

a)\u00a036.5<\/p>\n

b)\u00a036.8<\/p>\n

c)\u00a036.7<\/p>\n

d)\u00a036.9<\/p>\n

e)\u00a0None of these<\/p>\n

15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Solution:<\/b><\/p>\n

The pattern is\u00a0:<\/p>\n

18 $+ 1.7 \\times 2^0$ = 19.7<\/p>\n

19.7\u00a0$- 1.7 \\times 2^1$ =\u00a016.3<\/p>\n

16.3\u00a0$+ 1.7 \\times 2^2$ =\u00a023.1<\/p>\n

23.1\u00a0$- 1.7 \\times 2^3$ =\u00a09.5<\/p>\n

9.5\u00a0$+ 1.7 \\times 2^4$ =\u00a036.7<\/strong><\/p>\n

Question 16:\u00a0<\/b>68, ?, 77, 104, 168, 293<\/p>\n

a)\u00a069<\/p>\n

b)\u00a070<\/p>\n

c)\u00a068<\/p>\n

d)\u00a074<\/p>\n

e)\u00a0None of these<\/p>\n

16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Solution:<\/b><\/p>\n

Cubes of consecutive natural numbers are added<\/p>\n

68 $+ 1^3$ =\u00a069<\/strong><\/p>\n

69\u00a0$+ 2^3$ =\u00a077<\/p>\n

77\u00a0$+ 3^3$ =\u00a0104<\/p>\n

104\u00a0$+ 4^3$ =\u00a0168<\/p>\n

168\u00a0$+ 5^3$ =\u00a0293<\/p>\n

Question 17:\u00a0<\/b>In how many different ways can the numbers \u2018256974\u2019 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?<\/p>\n

a)\u00a048<\/p>\n

b)\u00a0720<\/p>\n

c)\u00a036<\/p>\n

d)\u00a0360<\/p>\n

e)\u00a0None of these<\/p>\n

17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Solution:<\/b><\/p>\n

Case 1 : 6 at left end and 5 is at right end : 6 _ _ _ _ 5<\/p>\n

Now, four empty places can be filled by 2,9,7 and 4 in = $4!$ ways<\/p>\n

= $4 \\times 3 \\times 2 \\times 1 = 24$<\/p>\n

Case 2 : 6 at right end and 5 at left end : 5 _ _ _ _ 6<\/p>\n

Similarly, no. of ways = $4!$<\/p>\n

= $4 \\times 3 \\times 2 \\times 1 = 24$<\/p>\n

$\\therefore$ Total no. of ways = $24 + 24 = 48$<\/p>\n

Question 18:\u00a0<\/b>What will come in place of both the question marks (?) in the following question ?$\\frac{(?)^{0.6}}{104}=\\frac{26}{(?)^{1.4}}$<\/p>\n

a)\u00a058<\/p>\n

b)\u00a0-48<\/p>\n

c)\u00a0-56<\/p>\n

d)\u00a042<\/p>\n

e)\u00a0-52<\/p>\n

18)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

Solution:<\/b><\/p>\n

$\\frac{(x)^{0.6}}{104}=\\frac{26}{(x)^{1.4}}$<\/p>\n

${(x)^{0.6}} * {(x)^{01.4}}$ = 104*26<\/p>\n

${(x)^{2}}$ = 104*26<\/p>\n

x = \u00b1<\/span><\/span>52<\/span><\/p>\n

Question 19:\u00a0<\/b>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 ?<\/p>\n

a)\u00a0$\\frac{7}{13}$<\/p>\n

b)\u00a0$\\frac{3}{8}$<\/p>\n

c)\u00a0$\\frac{4}{7}$<\/p>\n

d)\u00a0$\\frac{1}{6}$<\/p>\n

e)\u00a0None of these<\/p>\n

19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Solution:<\/b><\/p>\n

Given values are ,
\n$\\frac{1}{2}$ = 0.5<\/p>\n

$\\frac{7}{8}$ = 0.87<\/p>\n

$\\frac{3}{4}$ = 0.75<\/p>\n

$\\frac{5}{6}$ = 0.83<\/p>\n

$\\frac{6}{7}$ = 0.86<\/p>\n

\u2234 Required difference = $\\frac{7}{8}$ – $\\frac{1}{2}$ = (7-4)\/8 = 3\/8<\/p>\n

Question 20:\u00a0<\/b>If $(11)^{3}$ is subtracted from $(46)^{2}$ . what will be the remainder?<\/p>\n

a)\u00a0787<\/p>\n

b)\u00a0785<\/p>\n

c)\u00a0781<\/p>\n

d)\u00a0783<\/p>\n

e)\u00a0None of these<\/p>\n

20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Solution:<\/b><\/p>\n

Here<\/p>\n

$(46)^2$ = 2116<\/p>\n

$(11)^3$ = 1331<\/p>\n

So, 2116 – 1331 = 785<\/p>\n

Take MAH-CET Mock Tests<\/a><\/p>\n

Enroll to CAT 2022 course<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Here you can download the important Number series questions PDF with solutions for MAH MBA CET by Cracku. These questions will help you to make practice and solve the Number series aptitude section questions in the MAH MBA CET exam. Utilize this best PDF practice set which is included answers in detail. Click on the […]<\/p>\n","protected":false},"author":32,"featured_media":211301,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,3167,125,4409],"tags":[5420,3260,5556],"class_list":{"0":"post-211299","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads","8":"category-downloads-en","9":"category-featured","10":"category-mah-mba-cet","11":"tag-mah-cet-2022","12":"tag-number-series-questions","13":"tag-numbers"},"better_featured_image":{"id":211301,"alt_text":"Number series Questions PDF","caption":"Number series Questions","description":"Number series 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