TNPSC previous year maths question paper with answers

 TNPSC previous year maths question paper with answers

1. The place value of 5 in the number 6,52,37,401 is

a) five crores

b) five lakhs

c) Fifty lakhs

d) five million.

 

Solution          (50 lakhs & 5 million are same

(10 lakh = 1 million)

 

Fifty lakhs

 

2. The difference between the place value and the face value of 6 in the numeral 856973 is

a) 973              b) 6973            C) 5994           d) None

Solution:

Place value of 6 is       = 6000 -

Face value of 6 is        =       6

--------------

= 5994

5994  

3. The difference between the place values of two sevens in the numeral is 69758472 is

(a) 0                 b) 6993            c) 699930        d) None

Solution:

Place value of 1^ST 7 is            = 700000         -

Face value of 2^ND 7 is             =         30

--------------

= 699930

699930

4. The unit digit in the product

274 x 318 × 577 x 313.

a) 2

b) 3

c) 4

d) 5.

 

Solution: 4 x 8 x 7 x 3

2

5. The unit digit in (264) ¹⁰²+ (264) ¹⁰³ is

A) 0

b) 2

C) 4

d) b

0

6. 2a x 1a = 41a,         2 bx 1b = 376.

Then a+b= ?

a) 10                b) 11                (c) 12               d) 13. 

Solution:

Use possibilities 1x1 = 1 or 5x5 = 5 or 6x6 = 6

 

11

7. 4 ⁶¹ +4 ⁶² +4 ⁶³+ 4 ⁶⁴ is divisible by

a) 9      b)10     c) 11    d)12

 

Using unit place method;

4 + 6 + 4 + 6 = 20

last digit for the addition of given number is 0

it is divisible by 10

10

8. If the options are given in another form,

a) 13

b)19

c) 17

d) 21

we us factor idea,

4⁶¹ [1+4 +16+64]

= 4 ⁶¹ X 85 = 4 ⁶⁷ X 5 X 17.

:. It must be divisble by 4, 5 & 17

So in the option we have 17.

 

17

 

9. Units place digit in 7 ¹⁰⁵ is

a) 5

b) 1

c) 9

d) 1

 Solution:

[Tip: 7 4 = 1, 7 8 = 1 .. .. … 7 20 = 1 …

The unit digit of 7 is 1 when power of 7 is multiple of 4.

 

ex :7 148 = 1

ex : 7 150 à This can be written as 7 148 x 7 2.

= 1 x 9 = 9 à unit digit of 7 150

7 97     = 7 96 x 7 1

= 1x7 = 7 à unit digit of 7 97].

So

  7 105 = 7 104 x 7 = 1x7=7

ex: 7 99 = 7 96 x 7 3 = 1x 3 = 3, à unit digit of 7 99

 

1

10. Find the units digit pa in the product (2467) ¹⁵³ × (341) ⁷²

a) 7      b) 3      c) 9      d)1

solution:

7 ¹⁵³ x 1 ⁷² = 7 ¹⁵² × 7 × 1 = 1 x 7 x 1 = 7

 

7

11.  What is the unit digit in the product (3⁶⁵ x 6 ⁵⁹ x 7")? 

A) 1     B) 2     C) 4     D) 6 

Solution:

[Tip: 3⁴ =1, 3⁸ = 1 similarly 7.

The unit digit of 3 is 1 when the power of 3 is the multiple of 4]

 

3 ¹⁵¹ x 7 ⁷¹ = 3 ⁶⁴ x 3 x 6 x 7 ⁶⁸ x 7 ³

= 1 x 3 x 6 x 1 x 3.

= 54

 

4

 

Tip: For 9

9 ^odd  = 9          9 ^even = 1 

ex)       9 ⁹⁹  = 9            9 ¹⁴² = 1           9 ¹³⁷= 9

 

12. If the number 653xy is divisible by 90 then (x+y)?

A) 2

B) 3

C) 4

D) 6

Solution:

90à9x10

Y = 0  [ 10 is a factor for 653 xy]

we got 653 x0

Now check, 653 x 0 is divisible by 9

6+5+3+x+0 = multiple of 9

14+x = multiple of 14+4=18

X = 4

Y= 0

X + y = 4

 

4

13. The units digit of (7⁹⁵-3⁵⁸) is

a) 0

b) 6

c) 3

d) 4

 Solution:

7⁹² x 7³ - (3⁵⁶ x 3²)

= 1× 3 – (1x9)

= 3 – 9 (here 3 & 9 areà unit digits only we have previous digits also)

= …. 4 unit digit 

4

14. The unit digit of 6324 ¹⁷⁹⁷ × 615 ³¹⁶ x 341 ⁴⁷⁶ is

a)1

b)5

c) 7

d) 0

 Solution:

 4 x 5 x x 1 à 20 unit digit 0

0

15.  If 'a' is a positive integer, and if the unit digit of a² is a and the units of (a+1)2 is 4, what is the unit digit of (a+2)?

a) 1

b) 3

c) 6

d) 5 

Solution:

a² = …..9         3² = …..9        7² = …..9

a = 3 or 1.

If a = 3            then (3+1)² = 16 = 4

If a =7             then (7+1)² = 64 = 4

: a = 7

:. (a + 2) 2 = (1 + 2) 2 = 8  à1 unit digit.

 

1

16. A number when divided by 342 gives a remainder 47. When the same number is divided by 19. What would be the remainder?

a)8       b)9       c)6       D)1.

 

9