1. How many of the numbers x (x being integer) with 10<= x<= 99 are 18 more than the sum of their digits

a. 9

b. 12

c. 18

d. 10

Answer: d

Explanation:

Let the number be ab. So given that

⇒10a + b = 18 + a + b

⇒9a = 18

⇒a = 2

So 20, 21, … upto 29 there are total 10 numbers possible.

b.5

c.2

d.1

Answer: c

Explaination : The divisibility rule for 9 is sum of the digits is to be divisible by 9. So

We calculate separately, sum of the digits in hundreds place, tenths place, and units place.

Sum of the digits in hundreds place: 1 x 50 = 50

Sum of the digits in tenths place : 0 x 9 + 1 x 10 + 2 x 10 + 3 x 10 + 4 x 10 + 5 x 1 = 105

Sum of the digits in units place : (1 + 2 + 3 + …+ 9) x 5 = 225

So total = 380

So remainder = 380 / 9 = 2

3.7^1+7^2+7^3+…….+7^205. Find out how many numbers present which unit place contain 3?

a.44

b.51

c.25

d.65

Answer:b

Explaination: Units digits of first 4 terms are 7, 9, 3, 1. and this pattern repeats. So for every 4 terms we get one term with 3 in its unit digit. So there are total of 205/4 = 51 sets and each set contains one terms with 3 in its unit digit.

4. Given that 0 < a < b < c < d, which of the following the largest ?

a.(c+d) / (a+b)

b.(a+d) / (b+c)

c.(b+c) / (a+d)

d.(b+d) / (a+c)

Answer: A

Explanation: Take a = 1, b = 2, c = 3, d = 4. option A is clearly true.

5.The sum of the digits of a three digit number is 17, and the sum of the squares of its digits is 109. If we subtract 495 from the number, we shall get a number consisting of the same digits written in the reverse order. Find the number.

a. 773

b. 683

c. 944

d. 863

Answer: D

Explanation: Check options. Sum of the squares should be equal to 109. Only Options B and D satisfying. When we subtract 495, only 863 becomes 368.

6.2ab5 is a four digit number divisible by 25. If a number formed from the two digits ab is a multiple of 13, then ab is

a.52

b.45

c.10

d.25

Answer:a

Explaination: For a number to be divisible by 25, last two digits of that number should be divisible by 25. So b must be either 2 or 7

it is given that ab must be divisible by 13 and in the options only 52 is divisible by 13.

7.A school has 120, 192 and 144 students enrolled for its science, arts and commerce courses. All students have to be seated in rooms for an exam such that each room has students of only the same course and also all rooms have equal number of students. What is the least number of rooms needed?

a.18

b.12

c.19

d.20

Answer:c

Explaination: We have to find the maximum number which divides all the given numbers so that number of roots get minimized. HCF of 120,192 & 144 is 24. Each room have 24 students of the same course.

Then rooms needed

120/24 + 192/24 + 144/24 = 5 +8 + 6 = 19

8.A two digit number is 18 less than the square of the sum of its digits. How many such numbers are there?

a.1

b.2

c.3

d.4

Answer: Option 2

Explaination:Take N = 10a+b.

Given that, (10a+b)+18 = K2 = (a+b)2

Given number = K2 – 18 = (10a+b)

That means, when we add 18 to the given number it should be a perfect square. So K2 takes the following values. 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, ….

1 to 16 are ruled out as if we subtract 18 from them, the resulting number is a single digit number.

Now 25 – 18 = 7

36 – 18 = 18

49 – 18 = 31

64 – 18 = 46

81 – 18 = 63

100 – 18 = 82

121 – 18 = 103

Now 63, 82 satisfies.

9.Find the greatest number that will divide 148 246 and 623 leaving remainders 4 6 and 11 respectively?

a.20

b.12

c.6

d.48

Answer:b

Explanation:

Hcf ( (148-4), (246-6), (623-11))=12

10.Find L.C.M. of 1.05 and 2.1

a. 1.3

b. 1.25

c. 2.1

d. 4.30

Answer: c

Explaination: If numbers are in decimal form, convert them without decimal places. Therefore, the numbers are 105 and 210.