1. Ram left $$\frac{1}{3}$$ of his property to his widow and $$\frac{3}{5}$$ of the remainder to his daughter. He gave the rest to his son who received Rs. 6,400. How much was his original property worth?

a) Rs. 16,000

b) Rs. 24,000

c) Rs. 30,000

d) Rs. 32,000

Explanation:

$$\eqalign{ & {\text{Let}}\,{\text{total}}\,{\text{property}}\,{\text{of}}\,{\text{Ram}} = x \cr & {\text{His}}\,{\text{widow}}\,{\text{gets}} \cr & \frac{1}{3}\,of\,x = \frac{x}{3} \cr & {\text{His}}\,{\text{daughter}}\,{\text{gets}}\, \cr & = \frac{3}{5}\,of\,\left[ {x - {\frac{x}{3}} } \right] \cr & = \frac{3}{5} \times {\frac{{2x}}{3}} \cr & = \frac{{2x}}{5} \cr & {\text{His}}\,{\text{son}}\,{\text{gets}}\,{\text{his}}\,{\text{rest}}\,{\text{property}} \cr & {\text{His}}\,{\text{son's}}\,{\text{property}} \cr & = x - \left[ { {\frac{x}{3}} + {\frac{{2x}}{5}} } \right] \cr & 6400 = x - {\frac{{11x}}{{15}}} \cr & 6400 = \frac{{ {15x - 11x} }}{{15}} \cr & 4x = 6400 \times 15 \cr & x = 1600 \times 15 \cr & x = 24000 \cr & {\text{Total}}\,{\text{Property}} = 24000 \cr} $$

2. If x : y be the ratio of two whole numbers and z be their HCF, then the LCM of those two numbers is:

a) $${\text{yz}}$$

b) $$\frac{{{\text{xz}}}}{{\text{y}}}$$

c) $$\frac{{{\text{xy}}}}{{\text{z}}}$$

d) $${\text{xyz}}$$

Explanation: Ratio of the numbers = x : y

HCF of the numbers = z

So, z is the common factor of the numbers

Then, First number = xz

Second Number = yz

First Number × Second Number = HCF and LCM of the numbers

xzyz = z × LCM

LCM = xyz

3. The sum of the digits of a two-digits numbers is 12 and when the digits of the to digit number are interchanged, the new number is 36 more than the original. What is the Original number?

a) 93

b) 48

c) 39

d) 84

Explanation: Let the Original number is (X + 10Y)

Sum of the digit of the number = 12

X + Y = 12 ----------- (1) When digit interchanged, number become 36 more than original number,

10X + Y = 10Y + X + 36

X - Y = 4 ------------------- (2)

On solving equation (1) and (2),

X = 8

Y = 4

Original number,

= X + 10Y = 8 + 10 × 4 = 48

4. A certain number of the capsule were purchased for Rs. 176. Six more capsule could have been purchased in the same amount if each capsule were cheaper by Rs. 3. What was the number of capsules purchased?

a) 13

b) 16

c) 17

d) 8

Explanation: Let Price of each capsule were Rs. X, originally.

So, Number of capsule purchased originally,

= $$ \frac{{176}}{{\text{X}}}$$

If price of capsule is Rs. 3 less then 6 more capsule were purchased. In this case, the number of capsules,

= $$ \frac{{176}}{{{\text{X}} - 3}}$$

$$\eqalign{ & \frac{{176}}{{{\text{X}} - 3}} - \frac{{176}}{{\text{X}}} = 6 \cr & \frac{{176{\text{X}} - 176{\text{X}} + 528}}{{{\text{X}} \times \left( {{\text{X}} - 3} \right)}} = 6 \cr & 6{{\text{X}}^2} - 18{\text{X}} - 528 = 0 \cr & {{\text{X}}^2} - 3{\text{X}} - 88 = 0 \cr & {{\text{X}}^2} - 11{\text{X}} + 8{\text{X}} - 88 = 0 \cr & \left( {{\text{X}} - 11} \right)\left( {{\text{X}} + 8} \right) = 0 \cr} $$

Either, X = 11 Or, X = -8 (Price cannot be negative)

Thus, X = Rs. 11

Number of the capsule purchased = $$\frac{{176}}{{11}}$$ = 16

5. An even number can be expressed as the square of an integer as well as cube of another integer. Then the number has to be necessarily divisible by?

a) 32

b) 64

c) 128

d) Both (a) and (b)

Explanation: Check the option. 32 and 128 are neither of a square and nor cube root. 64 is the square of 8 and a cube of 4

6. The remainder when (12^{13} + 23^{13}) is divided by 11.

a) 0

b) 1

c) 2

d) 3

Explanation:

$$\eqalign{ & \frac{{{{12}^{13}}}}{{11}}\,{\text{gives}}\,{\text{Remainder}}\,1 \cr & {\text{It}}\,{\text{can}}\,{\text{be}}\,{\text{written}}\,{\text{as}}\, \cr & \frac{{\left( {12 \times 12 \times 12 \times 12\,........\,13\,\text{times}} \right)}}{{11}} \cr & {\text{On}}\,{\text{dividing}}\,{\text{it}}\,{\text{gives}}\,{\text{remainder}}\,{\text{1,}}\,{\text{each}}\,{\text{time}}. \cr & 1 \times 1 \times 1 \times 1\,.........\,13\,{\text{times}} \cr & {\text{So,}}\,{\text{final}}\,{\text{remainder}}\,{\text{will}}\,{\text{be}}\,1 \cr & \frac{{{{23}^{13}}}}{{11}} \Rightarrow \,{\text{Remainder}}\,1 \cr & {\text{The}}\,{\text{remainder}}\,{\text{of}}\,\frac{{\left( {{{12}^{13}} + {{23}^{13}}} \right)}}{{11}} \cr & = \left( {1 + 1} \right) \cr & = 2 \cr} $$

7. The remainder when 75^{7575} is divide by 37.

a) 0

b) 1

c) 5

d) 7

Explanation: $$\frac{{{{75}^{{{75}^{75}}}}}}{{37}}$$

When 75 is divided by 37, it leaves remainder of 1.

The expression will become,

$$\frac{{{1^{{{75}^{75}}}}}}{{37}}$$

Now, for any power of 1, we always get 1 and expression becomes $$\frac{1}{{37}}$$ that leaves remainder 1

8. The product of the digits of a three digit number is a perfect square and perfect cube both is :

a) 126

b) 256

c) 18

d) None of these

Explanation: The only possible number which is both perfect square and perfect cube is 729 which is cube of 9 and square of 27

7 × 2 × 9 = 126

9. The LCM of two numbers is 1020 and their HCF is 34 the possible pair of number is:

a) 255, 34

b) 102, 204

c) 204, 170

d) None of these

Explanation: Check options one by one: LCM of 204 and 170 = 1020.

10. Sunny gets $$\frac{7}{9}$$ times as many marks in QA as in ENGLISH. If his total combined marks in both the papers is 90. His marks in QA is:

a) 50

b) 60

c) 70

d) 80

Explanation:

$$\eqalign{ & {\text{Total}}\,{\text{Marks}} = 90 \cr & {\text{Marks}}\,{\text{in}}\,{\text{QA}} = \frac{7}{9}\,\text{of}\,\,90 \cr & = \frac{{\left( {7 \times 90} \right)}}{9} \cr & = 7 \times 10 \cr & = 70 \cr} $$