{"id":43163,"date":"2020-10-15T12:36:36","date_gmt":"2020-10-15T07:06:36","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=43163"},"modified":"2020-10-15T12:36:36","modified_gmt":"2020-10-15T07:06:36","slug":"arithmetic-questions-for-ssc-stenographer-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/arithmetic-questions-for-ssc-stenographer-pdf\/","title":{"rendered":"Arithmetic questions for SSC Stenographer PDF"},"content":{"rendered":"

Arithmetic questions for SSC Stenographer PDF<\/strong><\/span><\/h2>\n

SSC Stenographer Arithmetic questions and answers pdf based on previous year question papers of SSC exam. Top 15 Very important questions \u00a0for Stenographer.<\/p>\n

Download Arithmetic questions for SSC Stenographer PDF<\/a><\/p>\n

15 Stenographer Mock Tests – Just Rs. 149<\/a><\/p>\n

Download All Important SSC Stenographer Questions PDF<\/a> (Topic-Wise)<\/p>\n

Question 1:\u00a0<\/b>Find the value of p if 3x + p, x – 10 and -x + 16 are in arithmetic progression.<\/p>\n

a)\u00a016<\/p>\n

b)\u00a036<\/p>\n

c)\u00a0-16<\/p>\n

d)\u00a0-36<\/p>\n

Question 2:\u00a0<\/b>If 9\/4th of 7\/2 of a number is 126, then 7\/2th of that number is …………..<\/p>\n

a)\u00a056<\/p>\n

b)\u00a0284<\/p>\n

c)\u00a072<\/p>\n

d)\u00a026<\/p>\n

Question 3:\u00a0<\/b>The 4th term of an arithmetic progression is 15, 15th term is -29, \ufb01nd the 10th term?<\/p>\n

a)\u00a0-5<\/p>\n

b)\u00a0-13<\/p>\n

c)\u00a0-17<\/p>\n

d)\u00a0-9<\/p>\n

SSC STENOGRAPHER PREVIOUS PAPERS<\/a><\/p>\n

SSC Stenographer Free Mock Test<\/a><\/p>\n

Question 4:\u00a0<\/b>(91 + 92 + 93 + \u2026\u2026\u2026 +110) is equal to<\/p>\n

a)\u00a04020<\/p>\n

b)\u00a02010<\/p>\n

c)\u00a06030<\/p>\n

d)\u00a08040<\/p>\n

Question 5:\u00a0<\/b>If 6\/7th of 8\/5th of a number is 192, then 3\/4th of that number is .———–<\/p>\n

a)\u00a0105<\/p>\n

b)\u00a077<\/p>\n

c)\u00a036<\/p>\n

d)\u00a080<\/p>\n

Question 6:\u00a0<\/b>40.36 – (9.347 – x ) – 29.02 = 3.68. Find x.<\/p>\n

a)\u00a0-56.353<\/p>\n

b)\u00a01.687<\/p>\n

c)\u00a0-17.007<\/p>\n

d)\u00a082.407<\/p>\n

Question 7:\u00a0<\/b>What is the value of (81 + 82 + 83 + \u2026\u2026\u2026 +130)?<\/p>\n

a)\u00a05275<\/p>\n

b)\u00a010550<\/p>\n

c)\u00a015825<\/p>\n

d)\u00a021100<\/p>\n

SSC STENOGRAPHER PREVIOUS PAPERS<\/a><\/p>\n

SSC Stenographer Free Mock Test<\/a><\/p>\n

Question 8:\u00a0<\/b>In an arithmetic progression if 13 is the 3rd term, \u00ad47 is the 13th term, then \u00ad30 is which term?<\/p>\n

a)\u00a09<\/p>\n

b)\u00a010<\/p>\n

c)\u00a07<\/p>\n

d)\u00a08<\/p>\n

Question 9:\u00a0<\/b>If 4\/5th of 6\/7th of a number is 216, then 8\/9th of that number will be<\/p>\n

a)\u00a0179<\/p>\n

b)\u00a0280<\/p>\n

c)\u00a0160<\/p>\n

d)\u00a0269<\/p>\n

Question 10:\u00a0<\/b>199994 x 200006 = ?<\/p>\n

a)\u00a039999799964<\/p>\n

b)\u00a039999999864<\/p>\n

c)\u00a039999999954<\/p>\n

d)\u00a039999999964<\/p>\n

Question 11:\u00a0<\/b>In an arithmetic progression, if 17 is the 3rd term, -25 is the 17th term, then -1 is which term?<\/p>\n

a)\u00a010<\/p>\n

b)\u00a011<\/p>\n

c)\u00a09<\/p>\n

d)\u00a012<\/p>\n

Question 12:\u00a0<\/b>In an arithmetic progression, if 9 is the 5th term, -26 is the 12th term, then -6 is which term?<\/p>\n

a)\u00a011<\/p>\n

b)\u00a08<\/p>\n

c)\u00a010<\/p>\n

d)\u00a07<\/p>\n

Question 13:\u00a0<\/b>2*[\u00ad0.3(1.3 + 3.7)] of 0.8 = ?<\/p>\n

a)\u00a0\u00ad1.92<\/p>\n

b)\u00a0\u00ad0.72<\/p>\n

c)\u00a0\u00ad2.16<\/p>\n

d)\u00a0\u00ad2.4<\/p>\n

Question 14:\u00a0<\/b>Find the value of p, if 2x – 4, 4x + p and 6x – 12 are in arithmetic progression.<\/p>\n

a)\u00a0-9<\/p>\n

b)\u00a0-10<\/p>\n

c)\u00a0-11<\/p>\n

d)\u00a0-8<\/p>\n

Question 15:\u00a0<\/b>If 7\/8th of 5\/4th of a number is 315, then 5\/9th of that number is _____ .<\/p>\n

a)\u00a0123<\/p>\n

b)\u00a081<\/p>\n

c)\u00a0140<\/p>\n

d)\u00a0160<\/p>\n

SSC Stenographer Previous Papers<\/a><\/p>\n

Daily Free SSC Practice Sets<\/a><\/p>\n

SSC STENOGRAPHER STUDY MATERIAL TOPIC-WISE<\/a><\/p>\n

Answers & Solutions:<\/strong><\/span><\/p>\n

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

Terms in arithmetic progression\u00a0: $(3x + p) , (x – 10) , (-x + 16)$<\/p>\n

=> Difference between first two terms is equal to the difference between last two terms<\/p>\n

=> $(x – 10) – (3x + p) = (-x + 16) – (x – 10)$<\/p>\n

=> $-2x -10 – p = -2x + 16 + 10$<\/p>\n

=> $-p = 26 + 10 = 36$<\/p>\n

=> $p = -36$<\/p>\n

=> Ans – (D)<\/p>\n

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

Let the number be $x$<\/p>\n

According to ques,<\/p>\n

=> $\\frac{9}{4} \\times \\frac{7}{2} \\times x = 126$<\/p>\n

=> $\\frac{63}{8} x = 126$<\/p>\n

=> $x = \\frac{126}{63} \\times 8$<\/p>\n

=> $x = 2 \\times 8 = 16$<\/p>\n

$\\therefore (\\frac{7}{2})^{th}$ of the number = $\\frac{7}{2} \\times 16$<\/p>\n

= $7 \\times 8 = 56$<\/p>\n

=> Ans – (A)<\/p>\n

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

The $n^{th}$ term of an A.P. = $a + (n – 1) d$, where ‘a’ is the first term , ‘n’ is the number of terms and ‘d’ is the common difference.<\/p>\n

4th term, $A_4 = a + (4 – 1) d = 15$<\/p>\n

=> $a + 3d = 15$ —————–(i)<\/p>\n

Similarly, 15th term, $A_{15} = a + 14d = -29$ ——————(ii)<\/p>\n

Subtracting equation (i) from (ii), we get\u00a0:<\/p>\n

=> $(14d – 3d) = -29 – 15$<\/p>\n

=> $d = \\frac{-44}{11} = -4$<\/p>\n

Substituting it in equation (i), => $a – 12 = 15$<\/p>\n

=> $a = 15 + 12 = 27$<\/p>\n

$\\therefore$ 10th term, $A_{10} = a + (10 – 1)d$<\/p>\n

= $27 + (9 \\times -4) = 27 – 36 = -9$<\/p>\n

=> Ans – (D)<\/p>\n

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

Expression\u00a0:\u00a0(91 + 92 + 93 + \u2026\u2026\u2026 +110)<\/p>\n

This is an arithmetic progression with first term, $a = 91$ , last term, $l = 110$ and common difference, $d = 1$<\/p>\n

Let number of terms = $n$<\/p>\n

Last term in an A.P. = $a + (n – 1)d = 110$<\/p>\n

=> $91 + (n – 1)(1) = 110$<\/p>\n

=> $n – 1 = 110 – 91 = 19$<\/p>\n

=> $n = 19 + 1 = 20$<\/p>\n

$\\therefore$ Sum of A.P. = $\\frac{n}{2} (a + l)$<\/p>\n

= $\\frac{20}{2} (91 + 110)$<\/p>\n

= $10 \\times 201 = 2010$<\/p>\n

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

Let the number be $x$<\/p>\n

According to ques,<\/p>\n

=> $\\frac{6}{7} \\times \\frac{8}{5} \\times x = 192$<\/p>\n

=> $\\frac{48}{35} x = 192$<\/p>\n

=> $x = \\frac{192}{48} \\times 35$<\/p>\n

=> $x = 4 \\times 35 = 140$<\/p>\n

$\\therefore (\\frac{3}{4})^{th}$ of the number = $\\frac{3}{4} \\times 140$<\/p>\n

= $3 \\times 35 = 105$<\/p>\n

=> Ans – (A)<\/p>\n

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

Expression\u00a0:\u00a040.36 – (9.347 – x ) – 29.02 = 3.68<\/p>\n

=> 40.36 – 9.347 + x = 3.68 + 29.02<\/p>\n

=>\u00a031.013 + x = 32.7<\/p>\n

=> x = 32.7 – 31.013<\/p>\n

=> x = 1.687<\/p>\n

=> Ans – (B)<\/p>\n

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

Expression\u00a0:\u00a0(81 + 82 + 83 + \u2026\u2026\u2026 +130)<\/p>\n

This is an arithmetic progression with first term, $a = 81$ , last term, $l = 130$ and common difference, $d = 1$<\/p>\n

Let number of terms = $n$<\/p>\n

Last term in an A.P. = $a + (n – 1)d = 130$<\/p>\n

=> $81 + (n – 1)(1) = 130$<\/p>\n

=> $n – 1 = 130 – 81 = 49$<\/p>\n

=> $n = 49 + 1 = 50$<\/p>\n

$\\therefore$ Sum of A.P. = $\\frac{n}{2} (a + l)$<\/p>\n

= $\\frac{50}{2} (81 + 130)$<\/p>\n

= $25 \\times 211 = 5275$<\/p>\n

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

The $n^{th}$ term of an A.P. = $a + (n – 1) d$, where ‘a’ is the first term , ‘n’ is the number of terms and ‘d’ is the common difference.<\/p>\n

3rd term, $A_3 = a + (3 – 1) d = 13$<\/p>\n

=> $a + 2d = 13$ —————–(i)<\/p>\n

Similarly, 13th term, $A_{13} = a + 12d = 47$ ——————(ii)<\/p>\n

Subtracting equation (i) from (ii), we get\u00a0:<\/p>\n

=> $(12d – 2d) = 47 – 13 = 34$<\/p>\n

=> $d = \\frac{34}{10} = 3.4$<\/p>\n

Substituting it in equation (i), => $a + 2 \\times 3.4 = 13$<\/p>\n

=> $a = 13 – 6.8 = 6.2$<\/p>\n

Let $n^{th}$ term = 30<\/p>\n

=> $a + (n – 1) d = 30$<\/p>\n

=> $6.2 + (n – 1) (3.4) = 30$<\/p>\n

=> $(n – 1) (3.4) = 30 – 6.2 = 23.8$<\/p>\n

=> $(n – 1) = \\frac{23.8}{3.4} = 7$<\/p>\n

=> $n = 7 + 1 = 8$<\/p>\n

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

Let the number be $x$<\/p>\n

According to ques,<\/p>\n

=> $\\frac{4}{5} \\times \\frac{6}{7} \\times x = 216$<\/p>\n

=> $x = 216 \\times \\frac{35}{24} = 9 \\times 35$<\/p>\n

$\\therefore$\u00a08\/9th of the number = $\\frac{8}{9} \\times (35 \\times 9)$<\/p>\n

= $8 \\times 35 = 280$<\/p>\n

=> Ans – (B)<\/p>\n

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

Expression\u00a0:\u00a0\u00a0199994 x 200006<\/p>\n

= (200000 – 6)\u00a0x (200000 + 6)<\/p>\n

= $(200000)^2 – (6)^2$<\/p>\n

= 40000000000 – 36 = 39999999964<\/p>\n

=> Ans – (D)<\/p>\n

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

The $n^{th}$ term of an A.P. = $a + (n – 1) d$, where ‘a’ is the first term , ‘n’ is the number of terms and ‘d’ is the common difference.<\/p>\n

3rd term, $A_3 = a + (3 – 1) d = 17$<\/p>\n

=> $a + 2d = 17$ —————–(i)<\/p>\n

Similarly, 17th term, $A_{17} = a + 16d = -25$ ——————(ii)<\/p>\n

Subtracting equation (i) from (ii), we get\u00a0:<\/p>\n

=> $(16d – 2d) = -25 – 17$<\/p>\n

=> $d = \\frac{-42}{14} = -3$<\/p>\n

Substituting it in equation (i), => $a – 6 = 17$<\/p>\n

=> $a = 17 + 6 = 23$<\/p>\n

Let $n^{th}$ term = -1<\/p>\n

=> $a + (n – 1) d = -1$<\/p>\n

=> $23 + (n – 1) (-3) = -1$<\/p>\n

=> $(n – 1) (-3) = -1 – 23 = -24$<\/p>\n

=> $(n – 1) = \\frac{-24}{-3} = 8$<\/p>\n

=> $n = 8 + 1 = 9$<\/p>\n

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

The $n^{th}$ term of an A.P. = $a + (n – 1) d$, where ‘a’ is the first term , ‘n’ is the number of terms and ‘d’ is the common difference.<\/p>\n

5th term, $A_5 = a + (5 – 1) d = 9$<\/p>\n

=> $a + 4d = 9$ —————–(i)<\/p>\n

Similarly, 12th term, $A_{12} = a + 11d = -26$ ——————(ii)<\/p>\n

Subtracting equation (i) from (ii), we get\u00a0:<\/p>\n

=> $(11d – 4d) = -26 – 9$<\/p>\n

=> $d = \\frac{-35}{7} = -5$<\/p>\n

Substituting it in equation (i), => $a – 20 = 9$<\/p>\n

=> $a = 9 + 20 = 29$<\/p>\n

Let $n^{th}$ term = -6<\/p>\n

=> $a + (n – 1) d = -6$<\/p>\n

=> $29 + (n – 1) (-5) = -6$<\/p>\n

=> $(n – 1) (-5) = -6 – 29 = -35$<\/p>\n

=> $(n – 1) = \\frac{-35}{-5} = 7$<\/p>\n

=> $n = 7 + 1 = 8$<\/p>\n

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

Expression\u00a0:\u00a02*[\u00ad0.3(1.3 + 3.7)] of 0.8 = ?<\/p>\n

= $2 \\times (0.3 \\times 5) \\times 0.8$<\/p>\n

= $10 \\times 0.24 = 2.4$<\/p>\n

=> Ans – (D)<\/p>\n

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

Terms in arithmetic progression\u00a0: $(2x – 4) , (4x + p) , (6x – 12)$<\/p>\n

=> Difference between first two terms is equal to the difference between last two terms<\/p>\n

=> $(4x + p) – (2x – 4) = (6x – 12) – (4x + p)$<\/p>\n

=> $2x + p + 4 = 2x – 12 – p$<\/p>\n

=> $2p = -12 – 4 = -16$<\/p>\n

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

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

Let the number be $x$<\/p>\n

=> $\\frac{7}{8} \\times \\frac{5}{4} \\times x = 315$<\/p>\n

=> $x = 9 \\times 8 \\times 4 = 32 \\times 9$<\/p>\n

$\\therefore$\u00a05\/9th of the number = $\\frac{5}{9} \\times 32 \\times 9$<\/p>\n

= 160<\/p>\n

SSC Stenographer Free Mock Test<\/a><\/p>\n

HIGHLY RATED PREPARATION APP<\/a><\/p>\n

We hope this Arithmetic questions pdf for SSC Stenographer exam will be highly useful for your Preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Arithmetic questions for SSC Stenographer PDF SSC Stenographer Arithmetic questions and answers pdf based on previous year question papers of SSC exam. Top 15 Very important questions \u00a0for Stenographer. Download All Important SSC Stenographer Questions PDF (Topic-Wise) Question 1:\u00a0Find the value of p if 3x + p, x – 10 and -x + 16 are […]<\/p>\n","protected":false},"author":49,"featured_media":43173,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3167,169,125,9,504,378,1493,3794,1459,1611,1741,1268,1441],"tags":[4257,461,2000,3093,1242],"class_list":{"0":"post-43163","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads-en","8":"category-downloads","9":"category-featured","10":"category-ssc","11":"category-ssc-cgl","12":"category-ssc-chsl","13":"category-ssc-cpo","14":"category-ssc-cpo-en","15":"category-ssc-gd","16":"category-ssc-je","17":"category-ssc-mts","18":"category-ssc-stenographer","19":"category-stenographer","20":"tag-arithmetic-questions","21":"tag-ssc-exams","22":"tag-ssc-mocks","23":"tag-ssc-previous-year-questions","24":"tag-ssc-stenographer"},"better_featured_image":{"id":43173,"alt_text":"Arithmetic questions for SSC Stenographer PDF","caption":"Arithmetic questions for SSC Stenographer PDF","description":"Arithmetic questions for SSC Stenographer PDF","media_type":"image","media_details":{"width":800,"height":420,"file":"2020\/10\/fig-14-10-2020_14-54-38.jpg","sizes":{"medium":{"file":"fig-14-10-2020_14-54-38-300x158.jpg","width":300,"height":158,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/10\/fig-14-10-2020_14-54-38-300x158.jpg"},"thumbnail":{"file":"fig-14-10-2020_14-54-38-150x150.jpg","width":150,"height":150,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/10\/fig-14-10-2020_14-54-38-150x150.jpg"},"medium_large":{"file":"fig-14-10-2020_14-54-38-768x403.jpg","width":768,"height":403,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/10\/fig-14-10-2020_14-54-38-768x403.jpg"},"tiny-lazy":{"file":"fig-14-10-2020_14-54-38-30x16.jpg","width":30,"height":16,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/10\/fig-14-10-2020_14-54-38-30x16.jpg"},"td_218x150":{"file":"fig-14-10-2020_14-54-38-218x150.jpg","width":218,"height":150,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/10\/fig-14-10-2020_14-54-38-218x150.jpg"},"td_324x400":{"file":"fig-14-10-2020_14-54-38-324x400.jpg","width":324,"height":400,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/10\/fig-14-10-2020_14-54-38-324x400.jpg"},"td_696x0":{"file":"fig-14-10-2020_14-54-38-696x365.jpg","width":696,"height":365,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/10\/fig-14-10-2020_14-54-38-696x365.jpg"},"td_80x60":{"file":"fig-14-10-2020_14-54-38-80x60.jpg","width":80,"height":60,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/10\/fig-14-10-2020_14-54-38-80x60.jpg"},"td_100x70":{"file":"fig-14-10-2020_14-54-38-100x70.jpg","width":100,"height":70,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/10\/fig-14-10-2020_14-54-38-100x70.jpg"},"td_265x198":{"file":"fig-14-10-2020_14-54-38-265x198.jpg","width":265,"height":198,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/10\/fig-14-10-2020_14-54-38-265x198.jpg"},"td_324x160":{"file":"fig-14-10-2020_14-54-38-324x160.jpg","width":324,"height":160,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/10\/fig-14-10-2020_14-54-38-324x160.jpg"},"td_324x235":{"file":"fig-14-10-2020_14-54-38-324x235.jpg","width":324,"height":235,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/10\/fig-14-10-2020_14-54-38-324x235.jpg"},"td_356x220":{"file":"fig-14-10-2020_14-54-38-356x220.jpg","width":356,"height":220,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/10\/fig-14-10-2020_14-54-38-356x220.jpg"},"td_356x364":{"file":"fig-14-10-2020_14-54-38-356x364.jpg","width":356,"height":364,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/10\/fig-14-10-2020_14-54-38-356x364.jpg"},"td_533x261":{"file":"fig-14-10-2020_14-54-38-533x261.jpg","width":533,"height":261,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/10\/fig-14-10-2020_14-54-38-533x261.jpg"},"td_534x462":{"file":"fig-14-10-2020_14-54-38-534x420.jpg","width":534,"height":420,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/10\/fig-14-10-2020_14-54-38-534x420.jpg"},"td_696x385":{"file":"fig-14-10-2020_14-54-38-696x385.jpg","width":696,"height":385,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/10\/fig-14-10-2020_14-54-38-696x385.jpg"},"td_741x486":{"file":"fig-14-10-2020_14-54-38-741x420.jpg","width":741,"height":420,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/10\/fig-14-10-2020_14-54-38-741x420.jpg"}},"image_meta":{"aperture":"0","credit":"","camera":"","caption":"","created_timestamp":"0","copyright":"","focal_length":"0","iso":"0","shutter_speed":"0","title":"","orientation":"0"}},"post":null,"source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2020\/10\/fig-14-10-2020_14-54-38.jpg"},"yoast_head":"\n\n\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\n\n\n