1. The sum of four consecutive two-digit odd numbers, when divided by 10, become a perfect square. Which of the following can possibly be one of these four numbers?

a) 21

b) 67

c) 25

d) 41

Explanation: Using options,

We find that four consecutive odd numbers are 37, 39, 41 and 43

The sum of these 4 numbers is 160, when divided by 10 we get 16 which is a perfect square.

Thus, 41 is one of the odd numbers

2. If 381A is divisible by 9, find the value of the smallest natural number A?

a) 5

b) 8

c) 6

d) 9

Explanation: A number is divisible by 9 when the sum of its digit is divisible by 9

So, (3 + 8 + 1 + A) = must be divisible by 9

Thus, smallest natural number be 6

(3 + 8 + 1 + 6) = 18, this is divisible by 9

3. The HCF of two numbers, each having three digits, is 17 and their LCM is 714. The sum of the numbers will be:

a) 289

b) 391

c) 221

d) 731

Explanation: Let the numbers be 17x and 17y where x and y are co-prime.

LCM = 17xy

17xy = 714

xy = 42 = 6 × 7

→x = 6 and y = 7

x = 7 and y = 6

1

^{st}number = 17 × 6 = 102

2

^{nd}number = 17 × 7 = 119

Sum = 102 + 119 = 221

4. Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?

a) 1

b) 2

c) 4

d) 0

Explanation: Any four digit number in which first two digits are equal and last two digits are also equal will be in the form 11 × (11a + b) i.e. it will be the multiple of 11 like 1122, 3366, 2244, . . . .

Now, let the required number be aabb.

Since aabb is a perfect square, the only pair of a and b that satisfy the above mentioned condition is a = 7 and b = 4

Hence, 7744 is a perfect square

5. A forester wants to plant 44 apples tree, 66 banana trees and 110 mango trees in equal rows (in terms of number of trees). Also, he wants to make distinct rows of tree (i.e. only one type of tree in one row). The number of rows (minimum) that required is:

a) 2

b) 3

c) 10

d) 11

Explanation:

$$\eqalign{ & {\text{we}}\,{\text{first}}\,{\text{need}}\,{\text{to}}\,{\text{find}}\,{\text{the}}\,{\text{HCF}}\,{\text{of}}\,{\text{44,66,110}}. \cr & 44 = 2 \times 2 \times 11 \cr & 66 = 2 \times 3 \times 11 \cr & 110 = 2 \times 5 \times 11 \cr & {\text{HCF}} = 2 \times 11 = 22 \cr & {\text{Then,}}\,{\text{the}}\,{\text{required}}\,{\text{numbers}}\,{\text{of}}\,{\text{rows}}, \cr & = { {\frac{{44}}{{22}}} + {\frac{{66}}{{22}}} + {\frac{{110}}{{22}}} } \cr & = 10 \cr} $$

6. Which among $${2^{\frac{1}{2}}}$$, $${3^{\frac{1}{3}}}$$, $${4^{\frac{1}{4}}}$$, $${6^{\frac{1}{6}}}$$ and $${12^{\frac{1}{{12}}}}$$ is largest?

a) $${3^{\frac{1}{3}}}$$

b) $${4^{\frac{1}{4}}}$$

c) $${12^{\frac{1}{{12}}}}$$

d) $${2^{\frac{1}{2}}}$$

Explanation: LCM in power 2, 3, 4, 6, 12 is 12

We multiplied the LCM to the power of the numbers.

$${2^{\frac{{1 \times 12}}{2}}},{3^{\frac{{1 \times 12}}{3}}},{4^{\frac{{1 \times 12}}{4}}},{6^{\frac{{1 \times 12}}{6}}}{\kern 1pt} {\text{and}}\,{12^{\frac{{1 \times 12}}{{12}}}}$$

We get,

= 2

^{6}, 3

^{4}, 4

^{3}, 6

^{2}, 12

= 64, 84, 64, 36, 12

Hence, greatest number would be $${3^{\frac{1}{3}}}$$

7. The last digit of the number obtained by multiplying the numbers 81 × 82 × 83 × 84 × 86 × 87 × 88 × 89 will be

a) 0

b) 6

c) 7

d) 2

Explanation: The last digit of multiplication depends on the unit digit of (81 × 82 × 83 × 84 × 86 × 87 × 88 × 89) which is given by the remainder obtained on dividing it by 10.

$$\eqalign{ & \frac{{\left( {81 \times 82 \times 83 \times 84 \times 86 \times 87 \times 88 \times 89} \right)}}{{10}} \cr & {\text{We}}\,{\text{take}}\,{\text{individual}}\,{\text{remainder}}\,{\text{of}}\,{\text{each}}\,{\text{digit,}} \cr & \frac{{\left( {1 \times 2 \times 3 \times 4 \times 6 \times 7 \times 8 \times 9} \right)}}{{10}} \cr & {\text{Numbers}}\,{\text{multiplied}}, \cr & \frac{{\left( {24 \times 42 \times 72} \right)}}{{10}} \cr & {\text{Individual}}\,{\text{Remainder}}\,{\text{has}}\,{\text{been}}\,{\text{taken}}, \cr & \frac{{\left( {4 \times 2 \times 2} \right)}}{{10}} \cr & =\,\frac{{\left( {16} \right)}}{{10}} \cr & =\,6 \cr & {\text{Remainder}} = 6 \cr & {\text{So,}}\,{\text{the}}\,{\text{last}}\,{\text{digit}}\,{\text{will}}\,{\text{be}}\,6 \cr} $$

8. In a 4-digit number, the sum of the first two digits is equal to that of last two digits. The sum of the first and last digits is equal to third digit. Finally, the sum of the second and fourth digits is twice the sum of the other two digits. What is the third digit of the number?

a) 8

b) 1

c) 5

d) 4

Explanation: Let the 1

^{st}, 2

^{nd}, 3

^{rd}and 4

^{th}digits be a, b, c and d respectively.

a + b = c + d ----------(i)

a + d = c ----------(ii )

b + d = 2(a + c) ----------(iii)

from eqn. (i) and (ii),

a + b = a + 2d

→b = 2d

and eqn (iii);

2d + d = 2(a + a + d)

→ 3d = 2(2a + d)

→ d = 4a

a = $$\frac{{\text{d}}}{4};$$

→ Now, from eqn. (ii),

a + d = $$\frac{{\text{d}}}{4}$$ + d = $$\frac{{{\text{5d}}}}{4}$$ = c

c = $$\frac{5}{{4{\text{d}}}}$$

The value of d can be either 4 or 8.

If d = 4, then c = 5

If d = 8, then c = 10

But the value of c should be less than 10

Hence, value of c would be 5

9. On a road three consecutive traffic lights change after 36, 42 and 72 seconds respectively. If the lights are first switched on at 9:00 AM sharp, at what time will they change simultaneously?

a) 9:08:04

b) 9:08:44

c) 9:08:24

d) None of these

Explanation: LCM of 36, 42 and 72,

36 = 2 × 2 × 3 × 3

42 = 2 × 3 × 7

72 = 2 × 2 × 2 × 3 × 3

LCM = 2 ×2 × 2 × 3 × 3 × 7 = 504 seconds.

LCM of 36, 42 and 72 is 504

Hence, the lights will change simultaneously after 8 minutes and 24 seconds.

10. The rightmost non-zero digit of the number 30^{2720} is

a) 1

b) 3

c) 7

d) 9

Explanation: (30)

^{2720}, we can write it as[(30)

^{4}]

^{680}

= [(10 × 3)

^{4}]

^{680}

The right most non-zero digit depends on the unit digit of [(3)

^{4}]

^{680}

Unit digit of [(3)

^{4}]

^{680},

= (81)

^{680}

The unit digit of 81 is 1 so any power of 81 will always give its unit digit as 1

Thus, required unit digit is 1