model 4 finding unit place of a number Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 6 EXERCISES

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The following question based on number system topic of quantitative aptitude

Questions : In a two–digit number, the digit at the unit’s place is 1 less than twice the digit at the ten’s place. If the digits at unit’s and ten’s place are interchanged, the difference between the new and the original number is less than the original number by 20. The original number is

(a) 47

(b) 59

(c) 35

(d) 23

The correct answers to the above question in:

Answer: (a)

Ten’s digit = x

Unit’s digit = 2x – 1

Original number = 10x + (2x – 1) = 12x – 1

New number = 10 (2x – 1) + x

= 20x – 10 + x = 21x – 10

(21x – 10) – (12x + 1) = 12x – 1 – 20

9x – 9 = 12x – 21

3x = 12 ⇒ x = 4

Original number = 12x – 1 = 12 × 4 – 1 = 47

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Read more unit place Based Quantitative Aptitude Questions and Answers

Question : 1

The unit digit in the expansion of $(2137)^754$ is

a) 9

b) 1

c) 7

d) 3

Answer: (a)

Expression = $(2137)^754$

Unit’s digit in 2137 = 7

Now, $7^1 = 7, 7^2 = 49, 7^3 = 343, 7^4 = 2401, 7^5$ = 16807, ...

Clearly, after index 4, the unit’s digit follow the same order.

Dividing index 754 by 4 we get remainder = 2

∴ Unit’s digit in the expansion of $(2137)^754$ = Unit’s digit in the expansion of $(2137)^2$ = 9

Question : 2

One’s digit of the number $(22)^23$ is

a) 2

b) 4

c) 8

d) 6

Answer: (c)

Unit’s digit in the expansion of $(22)^23$

= Unit’s digit in the expansion of $(2)^23$

Now, $2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 16, 2^5 = 32$

i.e. 2 repeats itself after the index 4.

On dividing 23 by 4, remainder = 3

∴ Unit’s digit in $(2)^23$ = Unit’s digit in $(2)^3$ = 8

Question : 3

The digit in the unit’s place of the product $(2464)^1793×(615)^317×(131)^491$ is

a) 5

b) 0

c) 3

d) 2

Answer: (b)

$(4)^{2m}$ gives 6 at unit digit.

$(4)^{2m +1}$ gives 4 at unit digit.

$(5)^n$ gives 5.

The same is the case with 1.

Required digit = Unit’s digit in the product of 4×5 × 1 = 0

Question : 4

The last digit of $(1001)^2008$ + 1002 is

a) 6

b) 0

c) 4

d) 3

Answer: (d)

Last digit of $(1001)^2008$ + 1002 = 1 + 2 = 3

Question : 5

By interchanging the digits of a two digit number we get a number which is four times the original number minus 24. If the unit’s digit of the original number exceeds its ten’s digit by 7, then original number is

a) 18

b) 29

c) 58

d) 36

Answer: (b)

Let the two–digit number be 10x + y where x < y.

Number obtained on reversing the digits =10y + x

According to the question,

10y + x = 4 (10x + y) – 24

40x + 4y – 10y – x = 24

39x – 6y = 24

13x – 2y = 8 ....(i)

Again, y – x = 7

y = x + 7 ....(ii)

13x – 2 (x + 7) = 8

13x – 2x – 14 = 8

11 x = 14 + 8 = 22

x = $22/11$ = 2

From equation (ii),

y – 2 = 7 ⇒ y = 2 + 7 = 9

Number = 10x + y =10×2+9= 29

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