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# Basic Numbers - Quantitative Aptitude Questions And Answers (MCQs)

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Basic Numbers - Aptitude Questions And Answers ( MCQs )

Quantitative Aptitude Questions and Answers Quantitative Aptitude Questions and Answers contains concepts and test papers on many topics such as average, numbers, compound interests, partnership, problem on ages, calendar, boats and streams, clock, height and distance, percentage, pipes and cisterns, profit and loss, speed, time and distance, simple interest, problem on trains, time and work, etc. The Aptitude Questions and Answers are very helpful for competitive exams like SSC, CAT (Common Aptitude Test), MAT, GMAT, GRE, UGC, UPSC Exams, ICET, Bank PO, Defence Exams, Bank Exams, Railway Exams and placements like TCS, Wipro, Capgemini, Accenture, Infosys, IBM, Cognizant, Adobe, Cisco, Amazon, Samsung, Google, Yahoo, Facebook, Ola, Paypal, Microsoft, Oracle, etc.

Q1. DIRECTIONS for the question:
In the following questions, select a suitable replacement for the word in bold/underlined.

After the division of a number successively by 3, 4, and 7, the remainders obtained are 2, 1 and 4 respectively. What will be the remainder if 84 divide the same number ?

a) 80

b) 76

c) 41

d) 53

Answer is d) 53
Explanation : Since Dividend = Divisor × Quotient + Remainder, we have three divisors D1, D2, D3, as 3,4,7 and their respective remainders as R1, R2, R3, as 2,1,4. Taking quotient Q3 = 1, we have D3 × Q3 + R3 = 7 × 1 + 4 = 11. Now 11 becomes Q2 for D2 and so on. Now solving 4 × 11 + 1 = 45, 3 × 45 + 2 = 137. Now when 84 divides 137 remainder is 53.

Q2. Ghosh Babu thinks of a natural number, which is less than 100. This number has a property of being a perfect square as well as a perfect cube. What is the number of possible values that Ghosh Babu could have thought of?

a) No value

b) 2 values

c) 1 value

d) > 2 values

Answer is b) 2 values
Explanation : Number is a perfect square as well as perfect cube So number should be in form N6 and less than 100. So it should be 16 and 26 There are only Two no’s i.e. 1 and 64. So, the correct answer is option B.

Q3. In a weight lifting competition each candidate has to lift similar slabs, the candidate who lifts maximum total weight wins. 3 candidates lifted total 399, 494 and 513 kgs weight. What could the 4th candidate lift if it is known that he did not win the game?

a) 418 kgs

b) 400 kg

c) 512 kgs

d) 495 kgs

Answer is a) 418 kgs
Explanation : The HCF of the weights is 19. As the fourth candidate was not a winner, so the 4th candidate will lift less than 513. So basically we have to check a multiple of 19 less than 513. Only 418 is the multiple of 19 among the options and hence is the answer.

Q4. A natural number when divided by 6, 7, 8, 9 and 10 leaves remainders of 4, 5, 6, 7 and 8 respectively, but when divided by 54 leaves remainder 34. If x is the smallest such 5-digit natural number and y is the largest such 6-digit natural number, what is the value of |x – y|?

a) 982,800

b) 504,000

c) 987,056

d) 987,842

Answer is a) 982,800
Explanation : The number is of the form n × LCM (6, 7, 8, 9, 10) – 2 = 2520n – 2. Since (2520n - 2) = (54×46n + 36n - 2) ...... (1)
And, on dividing the number by 54, we can get a remainder of 34, 54×1+34, 54×2+34, 54×3+34 i.e. 34, 88, 142, 196 and so on ..... (2)
Hence, remainders from (1) and (2) should be equal or, 36n – 2 = 34, 88, 142, 196, and so on.
In other words, 36n = 36, 90, 144, 198, …
Thus, n = 1, 2.5, 4, 5.5, 7, and so on. (The values of n forms an A.P. with a common difference of 1.5)
The smallest 5-digit natural number can be written as 10000 = 2520 × 3 + 2440.
So, x = 2520 × 4 – 2 = 10080 – 2 = 10078. (4 appears in the above A.P.)
The largest 6-digit natural number can be written as 999999 = 2520 × 396 + 2079.
So, y = 2520 × 394 – 2 = 992878. (394 appears in the above A.P.)
Thus, |x - y| = 992,878 – 10,078 = 982,800.
OR, else use answer options.
Only first two options are divisible by 2520, further only first option is divisible by 54 as well.

Q5. What is the remainder when (4! + 5! + 6! +…….+999!) × 169×170×171 is divided by 84?

a) 0

b) 83

c) 60

d) 1

Answer is c) 60
Explanation : As 84=2x2x3x7, so all numbers 7! onwards are exactly divisible by 84. Now4!+5!+6!=24+120+720=864.
So now we have to find the remainder when 864 x169 x170 x171 is divided by 84.each number in given multilication on further divided by 84 gives remainders 24,1,2,3 res. so we have to find remainder when 24x1x2x3 i.e. 144 is divided by 84. so 144divided by 84 gives us remainder 60 . hence option 3 is the answer.

Q6. DIRECTIONS for the question: Solve the following question and mark the best possible option.
At 9:00 a.m., A0 people enter room A. After 5 minutes, half of these people move to room B. After 5 minutes, half of these move to room C, while another A0 people enter room B. If the number of people entering the rooms are integers, which of the following represents the minimum number of people A0?

a) 2

b) 4

c) 6

d) 8

Answer is b) 4
Explanation : Let us use the table to judge
Time A B C 9:00 A0 - - 9:05 A0/2 A0/2 - 9:10 A0/4 3A0/2 A0/4
So min value of A0 is 4. The question is not clear about whether the half that move at 9:10 are from room A or room B. But it does not matter.

Q7. When x3 + kx2 + x + 1 is divided by x – 1, the remainder is 6. The value of k is

a) 0

b) 1

c) 2

d) 3

Explanation : Using remainder theorem x3 + kx2 + x + 1=6 when x=1.
Therefore by putting x = 1
1+k+1+1=6 or k=3

Q8. What is the value of a if x3 + 3x2 + ax + b leaves the same remainder when divided by (x – 2) and (x + 1)?

a) 18

b) 3

c) -6

d) cannot be determined

Answer is c) -6
Explanation : Suppose the remainder is R.
Substituting x = 2 and x = –1, we get R = 8 + 12 + 2a + b = –1 + 3 – a + b => 3a = –18 => a = –6.

Q9. Find the smallest 6-digit natural number which when divided by 7, 11, 14 and 18 leaves a remainder of 6 in each case, but when divided by 12 and 15 leaves no remainder.

a) 101,190

b) 95,640

c) 109,500

d) 999,300

Answer is c) 109,500
Explanation : When divided by 7, 11, 14 and 18, the number is of the form n × LCM(7, 11, 14, 18) + 6 = 1386n + 6. Since this number leaves no remainder when divided by 12 and 15, we know that the number is divisible by 60.
When we divide by 60, we get a remainder of 6n + 6 = 0, 60, 120, 180 and so on.
In other words, 6n = –6, 54, 114, 174, ….
So n = –1, 9, 19, 29, 39, and so on.
Consider the smallest 6-digit natural number as 100000 = 1386 x 72 + 208.
The number we are interested in must have a quotient > 72.
Referring to the above series, the quotient for the smallest number must be 79.
Thus the required number is 1386 × 79 + 6 = 109494 + 6 = 109,500.

Q10. If 5n + 9n + 2 is always divisible by m for all positive integer values of n, which of the following represents m?

a) 6

b) 4

c) 8

d) None of these

Answer is b) 4
Explanation : 5n and 9n when divided by 4 will always leave remainders of 1 each.
So, the effective remainder will be 1 + 1 + 2 = 4.
Thus the given expression is always divisible by 4.

Q11. Find the last two digits of 2324 ?

a) 26

b) 32

c) 16

d) 52

Answer is c) 16
Explanation : We have 2324 = (210)32 x 24 = (24)32 x 16 = 76 x 16 = 1216. So, the last two digits of 2324 is 16.(24even=76(last two digits)

Q12. a, b, c are 3 digits of a number such that 49a + 7b + c = 286. Find a + b + c.

a) 15

b) 26

c) 12

d) 16

Answer is d) 16
Explanation : Best worked with trial and error:
Now since a, b and c are all single digit numbers, most of the contribution to 286 will come from 49a.
So we look at a = 5.
So 49a = 245
Now we need to look at a value of b which gets us as close to 286 as possible (but lower than it). Again try b = 5, So 7b = 35.
Total is 49a + 7b = 280
So c = 6
Hence a + b + c = 5 + 5 + 6 = 16

Q13. If j and k are integers divisible by 5, which of the following is not necessarily true?

a) j – k is divisible by 5

b) j3 + k3 is divisible by 5

c) j + k is divisible by 10

d) j + k is divisible by 5

Answer is c) j + k is divisible by 10
Explanation : Let us assume j and k to be 20 and 15 respectively as the multiples of 5.
Thus, j - k = 20 - 15 = 5 is divisible by 5.
j3 + k3 = 203 + 153 = 8000 + 3375 = 11375 is also divisible by 5.
j + k = 20 + 15 = 35 is divisible by 5 but not divisible by 10.
Thus, statement given in option 3 is not necessarily true.

Q14. Let S = 254 + 364 + 494. Then S divided by 11 leaves a remainder of

a) 0

b) 3

c) 6

d) 9

Answer is c) 6
Explanation : S = 254 + 364 + 494 = (22 + 3)4 + (33 + 3)4 + (44 + 5)4. The remainder would be the same as when 34 + 34 + 54 is divided by 11. The answer is 6.

Q15. What least digit should come in place of # in the 9 digit number 15549#325, for which the number is divisible by 3?

a) 0

b) 1

c) 2

d) Any of these

Answer is c) 2
Explanation : A number if divisible by 3 if sum of digits is multiple of 3. In 15549#325, Sum of the digits= 1 + 5 + 5 + 4 + 9 + # + 3 + 2 + 5 = 34 + # = 34 + 2 = 36 which is divisible by 3

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