{"id":34942,"date":"2019-09-17T18:36:31","date_gmt":"2019-09-17T13:06:31","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=34942"},"modified":"2019-09-17T18:36:31","modified_gmt":"2019-09-17T13:06:31","slug":"arithmetic-questions-for-cat-set-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/arithmetic-questions-for-cat-set-2-pdf\/","title":{"rendered":"Arithmetic Questions for CAT Set-2 PDF"},"content":{"rendered":"

Arithmetic Questions for CAT Set-2 PDF<\/strong><\/span><\/h2>\n

Download important CAT Arithmetic\u00a0 Questions Set-2 with Solutions PDF based on previously asked questions in CAT exam. Practice Arithmetic Questions Set-2 with Solutions for CAT exam.<\/p>\n

Download Arithmetic Questions for CAT Set-2 PDF<\/a><\/p>\n

Crash course to Cracku CAT<\/a><\/p>\n

Download CAT Quant Questions PDF<\/a><\/p>\n

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

Question 1:\u00a0<\/b>The remainder, when $(15^{23} + 23^{23})$ is divided by 19, is<\/p>\n

a)\u00a04<\/p>\n

b)\u00a015<\/p>\n

c)\u00a00<\/p>\n

d)\u00a018<\/p>\n

Question 2:\u00a0<\/b>Let T be the set of integers {3,11,19,27,\u2026451,459,467} and S be a subset of T such that the sum of no two elements of S is 470. The maximum possible number of elements in S is<\/p>\n

a)\u00a032<\/p>\n

b)\u00a028<\/p>\n

c)\u00a029<\/p>\n

d)\u00a030<\/p>\n

Question 3:\u00a0<\/b>If x = $(16^3 + 17^3+ 18^3+ 19^3 )$, then x divided by 70 leaves a remainder of<\/p>\n

a)\u00a00<\/p>\n

b)\u00a01<\/p>\n

c)\u00a069<\/p>\n

d)\u00a035<\/p>\n

Question 4:\u00a0<\/b>Let N = 1421 * 1423 * 1425. What is the remainder when N is divided by 12?<\/p>\n

a)\u00a00<\/p>\n

b)\u00a09<\/p>\n

c)\u00a03<\/p>\n

d)\u00a06<\/p>\n

Question 5:\u00a0<\/b><\/del>A CAT aspirant appears for a certain number of tests. His average score increases by 1 if the first 10 tests are not considered, and decreases by 1 if the last 10 tests are not considered. If his average scores for the first 10 and the last 10 tests are 20 and 30, respectively, then the total number of tests taken by him is<\/p>\n

Take free mock tests for CAT<\/a><\/p>\n

Question 6:\u00a0<\/b>In an apartment complex, the number of people aged 51 years and above is 30 and there are at most 39 people whose ages are below 51 years. The average age of all the people in the apartment complex is 38 years. What is the largest possible average age, in years, of the people whose ages are below 51 years?<\/p>\n

a)\u00a027<\/p>\n

b)\u00a025<\/p>\n

c)\u00a026<\/p>\n

d)\u00a028<\/p>\n

Question 7:\u00a0<\/b>Ravindra and Rekha got married 10 years ago, their ages were in the ratio of 5 : 4. Today Ravindra\u2019s age is one sixth more than Rekha\u2019s age. After marriage, they had 6 children including a triplet and twins. The age of the triplets, twins and the sixth child is in the ratio of 3 : 2 : 1. What is the largest possible value of the present total age of the family?<\/p>\n

a)\u00a079<\/p>\n

b)\u00a093<\/p>\n

c)\u00a0101<\/p>\n

d)\u00a0107<\/p>\n

Question 8:\u00a0<\/b>The ratio of number of male and female journalists in a newspaper office is 5:4. The newspaper has two sections, political and sports. If 30 percent of the male journalists and 40 percent of the female journalists are covering political news, what percentage of the journalists (approx.) in the newspaper is currently involved in sports reporting?<\/p>\n

a)\u00a065 percent<\/p>\n

b)\u00a060 percent<\/p>\n

c)\u00a070 percent<\/p>\n

d)\u00a0None of the above<\/p>\n

Question 9:\u00a0<\/b>A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he had sold. What is the current proportion of water to milk?
\n[CAT 2004]<\/p>\n

a)\u00a02 : 3<\/p>\n

b)\u00a01 : 2<\/p>\n

c)\u00a01 : 3<\/p>\n

d)\u00a03 : 4<\/p>\n

Question 10:\u00a0<\/b>A student took five papers in an examination, where the full marks were the same for each paper. His marks in these papers were in the proportion of 6 : 7 : 8 : 9 : 10. In all papers together, the candidate obtained 60% of the total marks. Then the number of papers in which he got more than 50% marks is<\/p>\n

a)\u00a02<\/p>\n

b)\u00a03<\/p>\n

c)\u00a04<\/p>\n

d)\u00a05<\/p>\n

Download CAT Quant Questions PDF<\/a><\/p>\n

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

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

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

The remainder when $15^{23}$ is divided by 19 equals $(-4)^{23}$
\nThe remainder when $23^{23}$ is divided by 19 equals $4^{23}$
\nSo, the sum of the two equals$(-4)^{23}+(4)^{23}=0$<\/p>\n

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

No. of terms in series T , 3+(n-1)*8 = 467 i.e. n=59.<\/p>\n

Now S will have atleast have of 59 terms i.e 29 .<\/p>\n

Also the sum of 29th term and 30th term is less than 470.<\/p>\n

Hence, maximum possible elements in S is 30.<\/p>\n

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

We know that x = $16^3 + 17^3 + 18^3 + 19^3 = (16^3 + 19^3) + (17^3 + 18^3)$<\/p>\n

= $(16 + 19)(16^2 – 16 * 19 + 19^2) + (17 + 18)(17^2 – 17 * 18 + 18^2)$ = 35 \u00d7 odd + 35 \u00d7 odd = 35 \u00d7 even = 35 \u00d7 (2k)<\/p>\n

=>\u00a0x = 70k<\/p>\n

=> Remainder when divided by 70 is 0.<\/p>\n

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

The numbers 1421, 1423 and 1425 when divided by 12 give remainder 5, 7 and 9 respectively.<\/p>\n

5*7*9 mod 12 = 11 * 9 mod 12 = 99 mod 12 = 3<\/p>\n

5)\u00a0Answer:\u00a060<\/b><\/p>\n

Let the total number of tests be ‘n’ and the average by ‘A’
\nTotal score = n*A
\nWhen 1st 10 tests are excluded, decrease in total value of scores = (nA – 20 * 10) = (nA – 200)
\nAlso, (n – 10)(A + 1) = (nA – 200)
\nOn solving, we get 10A – n = 190……….(i)
\nWhen last 10 tests are excluded, decrease in total value of scores = (nA – 30 * 10) = (nA – 300)
\nAlso, (n – 10)(A – 1) = (nA – 300)
\nOn solving, we get 10A + n = 310……….(ii)
\nFrom (i) and (ii), we get n = 60
\nHence, 60 is the correct answer.<\/p>\n

Take free mock tests for CAT<\/a><\/p>\n

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

In an apartment complex, the number of people aged 51 years and above is 30 and there are at most 39 people whose ages are below 51 years. The average age of all the people in the apartment complex is 38 years. What is the largest possible average age, in years, of the people whose ages are below 51 years?<\/p>\n

The possible average age of people\u00a0whose ages are below 51 years will be maximum if the average age of the number of people aged\u00a051 years and above is minimum. Hence, we can say that that there are 30 people having same age 51 years.<\/p>\n

Let ‘x’ be the maximum average age of people whose ages are below 51.<\/p>\n

Then we can say that,<\/p>\n

$\\dfrac{51*30+39*x}{30+39} = 38$<\/p>\n

$\\Rightarrow$ $1530+39x = 2622$<\/p>\n

$\\Rightarrow$ $x = 1092\/39 = 28$<\/p>\n

Hence, we can say that option D is the correct answer.<\/p>\n

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

10 years ago, Let age of Ravindra be 5x and Rekha be 4x<\/p>\n

At present, Ravindra is 7\/6 times of Rekha’s age.<\/p>\n

5x + 10 = $\\frac{7}{6}$ (4x + 10)<\/p>\n

Solving, x =5<\/p>\n

Ravindra was 25 years (10 years ago) and Rekha was 20 years (10 years ago)<\/p>\n

Now, ages of their children is 3:2:1<\/p>\n

Maximum possible ages of children is 9,6,3 years.<\/p>\n

Total age of family is: 35 + 30 + 9*3 + 6*2 + 3 = 107 years.<\/p>\n

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

The ratio of number of male and female journalists in a newspaper office is 5:4. The newspaper has two sections, political and sports. If 30 percent of the male journalists and 40 percent of the female journalists are covering political news, what percentage of the journalists (approx.) in the newspaper is currently involved in sports reporting?<\/p>\n

Let ‘9x’ be the number of total journalists in the office. Then, we can say that the number of male and female journalists are ‘5x’ and ‘4x’ respectively.<\/p>\n

It is given that 30 percent of the male journalists and 40 percent of the female journalists are covering political news. Hence, total number of journalists who are covering political news = 0.3*5x + 0.4*4x = 3.1x<\/p>\n

Therefore, the total number\u00a0journalists who are covering sports news = 9x – 3.1x = 5.9x.<\/p>\n

Hence, the\u00a0percentage of the journalists in the newspaper is currently involved in sports reporting = $\\dfrac{5.9x}{9x}\\times 100$ $\\approx$ 65 percent. Therefore, option A is the correct answer.<\/p>\n

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

After selling 1\/4th of the mixture, the remaining quantity of water is 15 liters and milk is 60 liters. So the milkman would add 25 liters of water to the mixture. The total amount of water now is 40 liters and milk is 60 liters. Therefore, the required ratio is 2:3.<\/p>\n

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

Let the marks in the five papers be 6k, 7k, 8k, 9k and 10k respectively.
\nSo, the total marks in all the 5 papers put together is 40k. This is equal to 60% of the total maximum marks. So, the total maximum marks is 5\/3 * 40k
\nSo, the maximum marks in each paper is 5\/3 * 40k \/ 5 = 40k\/3 = 13.33k
\n50% of the maximum marks is 6.67k
\nSo, the number of papers in which the student scored more than 50% is 4<\/p>\n

Download CAT Previous Papers PDF<\/a><\/p>\n

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We hope this Arithmetic Questions for CAT Set-2 with Solutions PDF will be helpful to you.<\/p>\n","protected":false},"excerpt":{"rendered":"

Arithmetic Questions for CAT Set-2 PDF Download important CAT Arithmetic\u00a0 Questions Set-2 with Solutions PDF based on previously asked questions in CAT exam. Practice Arithmetic Questions Set-2 with Solutions for CAT exam. 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