{"id":19566,"date":"2022-05-19T17:32:30","date_gmt":"2022-05-19T12:02:30","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=19566"},"modified":"2024-03-29T14:27:31","modified_gmt":"2024-03-29T08:57:31","slug":"cat-questions-on-ap-gp-hp","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/cat-questions-on-ap-gp-hp\/","title":{"rendered":"CAT Questions on AP GP HP"},"content":{"rendered":"

CAT Questions on AP GP HP:<\/strong><\/span><\/p>\n

Download CAT Questions and Answers PDF on AP, GP & HP. Progressions and series is an important topic in CAT exam. Practice solved questions on Arithmetic, Harmonic & Geometric progression.<\/p>\n

Download CAT Questions on AP GP HP<\/a><\/p>\n

CAT Test Series @ Just Rs. 399<\/a><\/p>\n

Download LR Puzzles Set-1 PDF<\/a><\/p>\n

Download All CAT LR Questions PDF<\/p>\n

Take Free Mock Test for CAT<\/a><\/p>\n

Question 1:<\/strong>\u00a0Find the sum of all the 3-digit integers that leave a remainder of 5 when divided by 7.<\/p>\n

a) 59489
\nb) 66879
\nc) 71079
\nd) 78659<\/p>\n

Question 2:\u00a0<\/strong>If 2x + y = 28, then find the maximum value of $x^4y^3$.<\/p>\n

a) $3^{18}*2^3$
\nb) $3^{15}*2^5$
\nc) $2^{15}*3^5$
\nd) $2^{18}*3^3$<\/p>\n

Question 3:\u00a0<\/strong>Find the sum of all the odd integers between 150 to 246 that do not end in 3.<\/p>\n

a) 9504
\nb) 7524
\nc) 8634
\nd) 8164<\/p>\n

Question 4:<\/strong> If 0.43232323232… = $\\frac{a}{b}$ such that both a and b are integers, then find the least value of a+b.<\/p>\n

a) 635
\nb) 684
\nc) 709
\nd) 736<\/p>\n

Question 5:<\/strong>\u00a0Every term of the series starting from the third term is the sum of two preceding terms. If the first term is odd, the second term is even and the total number of terms in the series in 150, then find the ratio of number of even terms to the number of odd terms.<\/p>\n

a) 1:3
\nb) 3:1
\nc) 2:1
\nd) 1:2<\/p>\n

Factors of a number – Formulas for CAT<\/a><\/p>\n

Formulas on Number system and factorials\u00a0<\/a><\/p>\n

Question 6:<\/strong>\u00a0If the first term of an infinite geometric progression is equal to twice the sum of terms that follow, then what is the ratio of the third term to the sixth term ?<\/p>\n

a) 27 : 1
\nb) 1 : 27
\nc) 9 : 1
\nd) cannot be determined<\/p>\n

Question 7:<\/strong>\u00a0Find the sum to infinite terms of $\\frac{1}{3^2-4}+\\frac{1}{4^2-4}+\\frac{1}{5^2-4}+\\frac{1}{6^2-4}+…$<\/p>\n

a) $\\frac{33}{64}$
\nb) $\\frac{35}{64}$
\nc) $\\frac{25}{48}$
\nd) $\\frac{23}{48}$<\/p>\n

Question 8:<\/strong>\u00a0Find the sum of the following series: 1 + 2 + 5 + 10 + 17 + 26….till 50 terms.
\na) 43375
\nb) 40475
\nc) 42250
\nd) None of these<\/p>\n

Question 9:<\/strong>\u00a0A sequence is such that in any set of four consecutive terms, the sum of first and third term is equal to the sum of second and fourth term. If the third term is equal to 5, 26th term is equal to 9 and the sum of first 18 terms is equal to 58, find the first term.
\na) 1
\nb) 2
\nc) 3
\nd) 4<\/p>\n

Question 10:<\/strong>\u00a0The internal angles of a convex polygon are in arithmetic progression with a common difference of 10 degrees. If the smallest angle is 100 degrees, what is the number of sides of the polygon?
\na) 10
\nb) 7
\nc) 8
\nd) 9<\/p>\n

Take a free CAT online mock test<\/a><\/p>\n

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

1) Answer (C)<\/strong><\/p>\n

The first three digit number that leaves a remainder of 5 when divided by 7 is 103
\n103, 110, 117, 124,….
\nThis forms an Arithmetic Progression.
\nThe last three digit number that leaves a remainder of 5 when divided by 7 is 999.
\n$T_n$ = first_term + (n-1)(common_difference)
\n999 = 103 + (n-1)7
\n=> 7(n-1) = 896
\n=> n-1 = 128
\n=> n = 129
\nSum = $\\frac{129}{2}*(103+999)$
\n= $\\frac{129}{2}*(1102)$
\n= 129 * 551
\n= 71079<\/p>\n

2) Answer (D)<\/strong><\/p>\n

2x + y = 28 => ($\\frac{x}{2}$) + ($\\frac{x}{2}$) + ($\\frac{x}{2}$) + ($\\frac{x}{2}$) + ($\\frac{y}{3}$) + ($\\frac{y}{3}$) + ($\\frac{y}{3}$) = 28<\/p>\n

As AM>=GM
\n=> $\\frac{(\\frac{x}{2}) + (\\frac{x}{2}) + (\\frac{x}{2}) + (\\frac{x}{2}) + (\\frac{y}{3}) + (\\frac{y}{3}) + (\\frac{y}{3})}{7} \\geq ((\\frac{x}{2})(\\frac{x}{2})(\\frac{x}{2})(\\frac{x}{2})(\\frac{y}{3})(\\frac{y}{3})(\\frac{y}{3})^{\\frac{1}{7})}$
\n=> $\\frac{28}{7} \\geq ((\\frac{x}{2})(\\frac{x}{2})(\\frac{x}{2})(\\frac{x}{2})(\\frac{y}{3})(\\frac{y}{3})(\\frac{y}{3})^{\\frac{1}{7})}$
\n=> $(\\frac{x^4y^3}{2^4*3^3})^{\\frac{1}{7}} \\leq 4$
\n=> $\\frac{x^4y^3}{2^4*3^3} \\leq 2^{14}$
\n=> $x^4y^3 \\leq 2^{18}*3^3$<\/p>\n

3) Answer (B)<\/strong><\/p>\n

Sum of all odd integers upto 245 = $123^2$
\nSum of all odd integers upto 149 = $75^2$
\n=> Sum of all odd integers between 150 and 246 = $123^2-75^2$ = 198*48 = 9504\\\\
\nNumbers between 150 and 246 that end in 3 are 153, 163, 173, 183, 193, 203, 213, 223, 233 and 243
\nThere are in Arithmetic Progression.
\n=> Sum = $\\frac{10}{2}(153+243)$ = 1980
\n=> Required sum = 9504 – 1980 = 7524<\/p>\n

4) Answer (C)<\/strong><\/p>\n

0.43232323232… = 0.4 + 0.032 + 0.00032 + 0.0000032 + …
\n= $\\frac{4}{10}$ + $\\frac{32}{10^3}$ + $\\frac{32}{10^5}$ + $\\frac{32}{10^7}$ + …
\n= $\\frac{4}{10}$ + $\\frac{32}{10^3} * (1 + \\frac{1}{10^2} + \\frac{1}{10^4} + \\frac{1}{10^6} + \\frac{1}{10^8} + …)$
\n= $\\frac{4}{10}$ + $\\frac{32}{10^3} * (\\frac{1}{1-\\frac{1}{10^2}})$
\n= $\\frac{4}{10}$ + $\\frac{32}{10^3} * (\\frac{100}{99})$
\n= $\\frac{4}{10}$ + $\\frac{32}{990}$
\n= $\\frac{396+32}{990}$
\n= $\\frac{428}{990}$
\n= $\\frac{214}{495}$ = $\\frac{a}{b}$
\n=> a + b = 214 + 495 = 709<\/p>\n

5) Answer (D)<\/strong><\/p>\n

The first term is odd and the second term is even.
\nO, E, O, O, E, O, O, E and so on.
\nIt means the second term, the fifth term, the eigth term and so on till 149th term are even terms.
\nNumber of even terms = x
\n149 = 2+(x-1)3
\nx = 50
\nNumber of odd terms = 100
\nRatio = 50:100 = 1:2<\/p>\n

6) Answer (A)<\/strong><\/p>\n

Let the first term of the infinite G.P be a and r the common ratio.
\nGiven, a = 2(ar + ar$^2$ + ar$^3$ …..)
\na = 2($\\frac{ar}{1-r}$)
\n1 – r = 2r , thus r = 1\/3
\nThe ratio of the third term to the ratio of the sixth term = $\\frac{ar^2}{ar^5} = \\frac{27}{1}$
\nThus, A is the right choice.<\/p>\n

7) Answer (C)<\/strong><\/p>\n

Let S = $\\frac{1}{3^2-4}+\\frac{1}{4^2-4}+\\frac{1}{5^2-4}+\\frac{1}{6^2-4}+…$
\n=> S = $\\frac{1}{(3+2)(3-2)}+\\frac{1}{(4+2)(4-2)}+\\frac{1}{(5+2)(5-2)}+\\frac{1}{(6+2)(6-2)}+…$
\n=> S = $\\frac{1}{(1)(5)}+\\frac{1}{(2)(6)}+\\frac{1}{(3)(7)}+\\frac{1}{(4)(8)}+…..$
\n=> S = $\\frac{1}{4}(1-\\frac{1}{5})+\\frac{1}{4}(\\frac{1}{2}-\\frac{1}{6})+\\frac{1}{4}(\\frac{1}{3}-\\frac{1}{7})+\\frac{1}{4}(\\frac{1}{4}-\\frac{1}{8})+\\frac{1}{4}(\\frac{1}{5}-\\frac{1}{9})+….$
\n=> S = $\\frac{1}{4}(1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4})$
\n=> s = $\\frac{25}{48}$<\/p>\n

8) Answer (B)<\/strong><\/p>\n

S = 1 + 2 + 5 + 10 + 17 + 26 +….+ tn
\nS-S = 1 + (2-1) + (5-2) + (10-5) + (17-10) +….. +[tn-t(n-1)]-tn
\n0 = 1 + (1 + 3 + 5 + 7 +……+ (n-1) terms) – tn<\/p>\n

=> tn = 1 + (n-1)\/2 (2 + (n-2)2) = 1 + (n-1)( 1 + n-2) = 1 + $(n-1)^2$
\n=>tn = $(n^2 – 2n + 2)$
\nSum = (n)(n+1)(2n+1)\/6 – n(n+1) + 2n = 50*51*101\/6 – 50*51 + 100 = 40475<\/p>\n

9) Answer (A)<\/strong><\/p>\n

Let the first term be a, second term be x and the third term be c.
\nFourth term = a+c-x
\nFifth term = a
\nSixth term = x
\nIn this way pattern will repeat.
\n26th term = 2nd term = x = 9
\nc = 5
\n2(a+c)*4+a+x = 58
\n9a+8c+x = 58
\n9a+40+9 = 58
\na = 1<\/p>\n

10) Answer (C)<\/strong><\/p>\n

Let the number of sides of the polygon be n.
\nHence, the sum of the internal angles of the polygon equals (n-2)*180.
\nThe sum of the arithmetic progression equals n\/2*(200 + (n-1)*10)
\nHence, (n-2)*180 = n(100 + 5n – 5) = n(5n+95)
\nTherefore, $n^2 + 19n = 36n – 72$
\nSo, $n^2 – 17n +72 = 0$
\nHence, $n=8$ or $n=9$
\nIf n=9, the biggest angle becomes 100+(9-1)*10 = 180 degrees.
\nAs this is not possible in a convex polygon, the correct answer is 8.<\/p>\n

CAT Online Most Trusted Courses<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

CAT Questions on AP GP HP: Download CAT Questions and Answers PDF on AP, GP & HP. Progressions and series is an important topic in CAT exam. Practice solved questions on Arithmetic, Harmonic & Geometric progression. Download LR Puzzles Set-1 PDF Download All CAT LR Questions PDF Take Free Mock Test for CAT Question 1:\u00a0Find […]<\/p>\n","protected":false},"author":34,"featured_media":19590,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3,169,125],"tags":[6,5119],"class_list":{"0":"post-19566","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-cat","8":"category-downloads","9":"category-featured","10":"tag-cat","11":"tag-cat-2022"},"better_featured_image":{"id":19590,"alt_text":"CAT Questions on AP GP HP","caption":"CAT Questions on AP GP HP","description":"CAT Questions on AP GP 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