{"id":142805,"date":"2021-08-11T17:19:50","date_gmt":"2021-08-11T11:49:50","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=142805"},"modified":"2021-08-11T17:20:49","modified_gmt":"2021-08-11T11:50:49","slug":"number-system-questions-for-nmat","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/number-system-questions-for-nmat\/","title":{"rendered":"Number System Questions for NMAT"},"content":{"rendered":"

Number System Questions for NMAT:<\/strong><\/span><\/h1>\n

Download Number System Questions for NMAT PDF. Top 10 very important Number System Questions for NMAT based on asked questions in previous exam papers.<\/p>\n

Download Number System Questions for NMAT<\/a><\/p>\n

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Question 1:\u00a0<\/b>How many 4-digit numbers, each greater than 1000 and each having all four digits\u00a0distinct, are there with 7 coming before 3?<\/p>\n

Question 2:\u00a0<\/b>How many of the integers 1, 2, \u2026 , 120, are divisible by none of 2, 5 and 7?<\/p>\n

a)\u00a042<\/p>\n

b)\u00a041<\/p>\n

c)\u00a040<\/p>\n

d)\u00a043<\/p>\n

Question 3:\u00a0<\/b>If a, b and c are positive integers such that ab = 432, bc = 96 and c < 9, then the smallest possible value of a + b + c is<\/p>\n

a)\u00a049<\/p>\n

b)\u00a056<\/p>\n

c)\u00a059<\/p>\n

d)\u00a046<\/p>\n

Question 4:\u00a0<\/b>In a row at a bus stop, A is 7th from the left and B is 9th from the right. They both interchange their
\npositions. A becomes 11th from the left. How many people are there in the row?<\/p>\n

a)\u00a018<\/p>\n

b)\u00a019<\/p>\n

c)\u00a020<\/p>\n

d)\u00a021<\/p>\n

Question 5:\u00a0<\/b>The average of 4 distinct prime numbers a, b, c, d is 35, where a < b < c < d. a and d are equidistant from 36 and b and c are equidistant from 34 and a, b are equidistant from 30 and c and d are equidistant from 40. The difference between a and d is:<\/p>\n

a)\u00a030<\/p>\n

b)\u00a014<\/p>\n

c)\u00a021<\/p>\n

d)\u00a0Cannot be determined<\/p>\n

<\/a><\/div>\n

Question 6:\u00a0<\/b>In a six-digit number, the sixth, that is, the rightmost, digit is the sum of the first three digits, the fifth digit is the sum of first two digits, the third digit is equal to the first digit, the second digit is twice the first digit and the fourth digit is the sum of\u00a0fifth and sixth digits. Then, the largest possible value of the fourth digit is<\/p>\n

Question 7:\u00a0<\/b>If $\\sqrt[3]{7^a\\times 35^{b+1} \\times 20^{c+2}}$ is a whole number then which one of the statements below is consistent with it?<\/p>\n

a)\u00a0a = 2, b = 1, c = 1<\/p>\n

b)\u00a0a = 1, b = 2, c = 2<\/p>\n

c)\u00a0a = 2, b = 1, c = 2<\/p>\n

d)\u00a0a = 3, b = 1, c = 1<\/p>\n

e)\u00a0a = 3, b = 2, c = 1<\/p>\n

Question 8:\u00a0<\/b>While multiplying three real numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the product went up by 720. Then the minimum possible value of the sum of squares of the other two numbers is<\/p>\n

Question 9:\u00a0<\/b>An antique store has a collection of eight clocks. At a particular moment, the displayed times on seven of the eight clocks were as follows: 1:55 pm, 2:03 pm, 2:11 pm, 2:24 pm, 2:45 pm, 3:19 pm and 4:14 pm. If the displayed times of all eight clocks form a mathematical series, then what was the displayed time on the remaining clock?<\/p>\n

a)\u00a01:53 pm<\/p>\n

b)\u00a01:58 pm<\/p>\n

c)\u00a02:18 pm<\/p>\n

d)\u00a03:08 pm<\/p>\n

e)\u00a05:08 pm<\/p>\n

Question 10:\u00a0<\/b>A number is interesting if on adding the sum of the digits of the number and the product of the digits of the number, the result is equal to the number. What fraction of numbers between 10 and 100 (both 10 and 100 included) is interesting?<\/p>\n

a)\u00a00.1<\/p>\n

b)\u00a00.11<\/p>\n

c)\u00a00.16<\/p>\n

d)\u00a00.22<\/p>\n

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

Join 7K MBA Aspirants Telegram Group<\/a><\/p>\n

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

1)\u00a0Answer:\u00a0315<\/b><\/p>\n

Here there are two cases possible<\/p>\n

Case 1: When 7 is at the left extreme<\/p>\n

In that case 3 can occupy any of the three remaining places and the remaining two places can be taken by (0,1,2,4,5,6,8,9)<\/p>\n

So total ways 3(8)(7)= 168<\/p>\n

Case 2: When 7 is not at the extremes<\/p>\n

Here there are 3 cases possible. And the remaining two places can be filled in 7(7) ways.(Remember 0 can’t come on the extreme left)<\/p>\n

Hence in total 3(7)(7)=147 ways<\/p>\n

Total ways 168+147=315 ways<\/p>\n

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

The number of multiples of 2 between 1 and 120 = 60<\/p>\n

The number of multiples of 5 between 1 and 120 which are not multiples of 2 = 12<\/p>\n

The number of multiples of 7 between 1 and 120 which are not multiples of 2 and 5 = 7<\/p>\n

Hence,\u00a0number\u00a0of the integers 1, 2, \u2026 , 120, are divisible by none of 2, 5 and 7 = 120 – 60 – 12 – 7 = 41<\/p>\n

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

Since $c<9$, we can have the following viable combinations for\u00a0$b\\times\\ c\\ =96$ (given our objective is to minimize the sum):<\/p>\n

$48\\times\\ 2$ ;\u00a0$32\\times3$ ;\u00a0$24\\times\\ 4$ ;\u00a0$16\\times6$ ;\u00a0$12\\times8$<\/p>\n

Similarly, we can factorize\u00a0$a\\times\\ b\\ = 432$ into its factors. On close observation, we notice that\u00a0$18\\times24\\ and\\ 24\\ \\times\\ 4\\ $ corresponding to\u00a0$a\\times b\\ and\\ b\\times\\ c\\ $ respectively\u00a0together render us with the least value of the sum of\u00a0$a+b\\ +\\ c\\ \\ =\\ 18+24+4\\ =46$<\/p>\n

Hence, Option D is the correct answer.<\/p>\n

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

After interchanging the positions, A becomes 11th from the left. So B must have been 11th from the left and 9th from the right before the interchange. Hence the total number of people = 11+9-1=19<\/p>\n

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

Given,<\/p>\n

The average of the four prime numbers = 35.<\/p>\n

a + b + c + d = 35 * 4 = 140.<\/p>\n

Since a and d are equidistant from 36.<\/p>\n

a + d = 72 — Eq (1)<\/p>\n

b + c = 68 — Eq (2)<\/p>\n

a + b = 60 — Eq (3) and c + d = 80 — Eq (4)<\/p>\n

Using the equation (3) let us look for the prime values of a and b and the corresponding values of c and d using Eq 2 and 1.<\/p>\n

Also given that a < b < c < d.<\/p>\n

<\/figure>\n

(a, b, c, d) = 29, 31, 37, 43<\/p>\n

d – a = 43 – 29 = 14<\/p>\n

B is the correct answer.<\/p>\n\n

6)\u00a0Answer:\u00a07<\/b><\/p>\n

Let the six-digit number be ABCDEF<\/p>\n

F = A+B+C, E= A+B, C=A, B= 2A, D= E+F.<\/p>\n

Therefore D = 2A+2B+C = 2A + 4A + A= 7A.<\/p>\n

A cannot be 0 as the number is a 6 digit number.<\/p>\n

A cannot be 2 as D would become 2 digit number.<\/p>\n

Therefore A is 1 and D is 7.<\/p>\n

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

In the given statement, the expression becomes a whole number only when the powers of all the prime numbers are also whole numbers.<\/p>\n

Let us first simplify the expression a bit by expressing all terms in terms of prime numbers.<\/p>\n

$\\sqrt[3]{7^a\\times 35^{b+1} \\times 20^{c+2}}$<\/p>\n

$\\Rightarrow\u00a0\\sqrt[3]{7^a\\times 5^{b+1} \\times 7^{b+1} \\times 2^{2(c+2)} \\times 5^{c+2}}$<\/p>\n

$\\Rightarrow\u00a0\\sqrt[3]{2^{2c+4} \\times 5^{b+c+3}\u00a0\\times 7^{a+b+1}}$<\/p>\n

$\\Rightarrow 2^{\\frac{2c+4}{3}} 5^{\\frac{b+c+3}{3}} 7^{\\frac{a+b+1}{3}} $<\/p>\n

Now, from the given options, we can put in values of the variables and check the exponents of all the numbers.<\/p>\n

Option A :\u00a0a = 2, b = 1, c = 1 :<\/p>\n

In this case, we can see that exponent of 5 ie $\\frac{b+c+3}{3} = \\frac{5}{3} $ is not a whole number.<\/p>\n

Option B :\u00a0a = 1, b = 2, c = 2<\/p>\n

In this case, we can see that exponent of 2 ie $\\frac{2c+4}{3} = \\frac{8}{3}\u00a0$ is not a whole number.<\/p>\n

Option C :\u00a0a = 2, b = 1, c = 2<\/p>\n

In this case, we can see that exponent of 2 ie $\\frac{2c+4}{3} = \\frac{8}{3}\u00a0$ is not a whole number.<\/p>\n

Option D :\u00a0a = 3, b = 1, c = 1<\/p>\n

In this case, we can see that exponent of 5 ie $\\frac{b+c+3}{3} = \\frac{5}{3}\u00a0$ is not a whole number.<\/p>\n

Option E :\u00a0a = 3, b = 2, c = 1<\/p>\n

In this case, we can see that all exponents are whole numbers.<\/p>\n

Thus, option E is the correct option.<\/p>\n

8)\u00a0Answer:\u00a040<\/b><\/p>\n

We know that one of the 3 numbers is 37.
\nLet the product of the other 2 numbers be x.
\nIt has been given that 73x-37x = 720
\n36x = 720
\nx = 20<\/p>\n

Product of 2 real numbers is 20.
\nWe have to find the minimum possible value of the sum of the squares of the 2 numbers.
\nLet x=a*b
\nIt has been given that a*b=20<\/p>\n

The least possible sum for a given product is obtained when the numbers are as close to each other as possible.
\nTherefore, when a=b, the value of a and b will be $\\sqrt{20}$.<\/p>\n

Sum of the squares of the 2 numbers = 20 + 20 = 40.<\/p>\n

Therefore, 40 is the correct answer.<\/p>\n

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

Let us find out the difference between the times given to figure out the pattern.
\nThe times given are\u00a01:55 pm, 2:03 pm, 2:11 pm, 2:24 pm, 2:45 pm, 3:19 pm and 4:14 pm.
\nThe difference between 2 consecutive times given are\u00a0 8 minutes, 8 minutes, 13 minutes, 21 minutes, 34 minutes, and 55 minutes.
\nWe can observe that the difference between the times are in the Fibonacci series.
\n8 +\u00a013 = 21
\n21 +\u00a013 = 34
\n34 +\u00a021 = 55<\/p>\n

The Fibonacci series is as follows:
\n1,1,2,3,5,8,13,21,34,55.
\nBut the first difference in the times given is 8.
\nTherefore, the missing time must be such that it divides the interval of 8 minutes into 3 minutes and 5 minutes.
\nThe missing time should be 1:58 pm and hence, option B is the right answer.<\/p>\n

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

As the number is between 10 and 100 and 100 cannot be the number we are looking for, we can assume the number to be of two-digits.
\nLet the number be xy.
\nAccording to the question, for the number to be interesting
\nx + y + xy = 10x + y
\nOn solving, we get
\nxy = 9x
\nor, x (9 – y) = 0
\nx cannot be 0, because we need a number greater than or equal to 10.
\nSo, 9 – y = 0
\n=> y = 9
\nFor all the numbers whose unit digit is 9 will be an interesting number.
\nSo, the numbers are 19, 29, 39, 49, …..99
\nThere are 9 such numbers out of 91 total numbers between 10 and 100 including both.
\nRequired fraction = $\\dfrac{9}{91}$ = 0.0989
\nAs this is not given in any of the options, the answer will be “none of the above”.
\nHence, option E is the correct answer.<\/p>\n

<\/a><\/div>\n

We hope this Number System Questions for NMAT pdf for NMAT exam will be highly useful for your Preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Number System Questions for NMAT: Download Number System Questions for NMAT PDF. Top 10 very important Number System Questions for NMAT based on asked questions in previous exam papers. Take NMAT mock test Question 1:\u00a0How many 4-digit numbers, each greater than 1000 and each having all four digits\u00a0distinct, are there with 7 coming before 3? 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Get 5 NMAT Mocks for Rs. 499<\/a><\/p>\n