1. A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?

a) 45%

b) $$45\frac{5}{{11}}\% $$

c) $$54\frac{6}{{11}}\% $$

d) 55%

Explanation: Number of runs made by running

= 110 - (3 x 4 + 8 x 6)

= 110 - (60)

= 50

Required percentage

$$\eqalign{ & = \left( {\frac{{50}}{{110}} \times 100} \right)\% \cr & = 45\frac{5}{{11}}\% \cr} $$

2. Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:

a) 39, 30

b) 41, 32

c) 42, 33

d) 43, 34

Explanain : Let their marks be (x + 9) and x

$$\eqalign{ & x + 9 = \frac{{56}}{{100}}\left( {x + 9 + x} \right) \cr & 25\left( {x + 9} \right) = 14\left( {2x + 9} \right) \cr & 3x = 99 \cr & x = 33 \cr} $$

So, their marks are 42 and 33

3. A fruit seller had some apples. He sells 40% apples and still has 420 apples. Originally, he had:

a) 588 apples

b) 600 apples

c) 672 apples

d) 700 apples

Explanation: Suppose originally he had x apples

$$\eqalign{ & \left( {100 - 40} \right)\% \,{\text{of}}\,x = 420 \cr & \frac{{60}}{{100}} \times x = 420 \cr & x = {\frac{{420 \times 100}}{{60}}} = 700 \cr} $$

4. What percentage of numbers from 1 to 70 have 1 or 9 in the unit's digit?

a) 1%

b) 14%

c) 20%

d) 21%

Explanation: The numbers which have 1 or 9 in the unit's digit, have squares that end in the digit 1.

Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.

Number of such number = 14

$$ = \left( {\frac{{14}}{{70}} \times 100} \right)\% = 20\% $$

5. If A = *x*% of *y* and B = *y*% of *x*, then which of the following is true?

a) A is smaller than B

b) A is greater than B

c) Relationship between A and B cannot be determined

d) None of these

Explanation:

$$\eqalign{ & x\% \,{\text{of}}\,y = {\frac{x}{{100}} \times y} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\frac{y}{{100}} \times x} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = y\% \,{\text{of}}\,x \cr & So, A = B \cr} $$

6. Due to fall of 10% in the rate of sugar, 500 gm more sugar can be purchased for Rs. 140. Find the original rate?

a) Rs. 31.11

b) Rs. 29.22

c) Rs. 33.11

d) Rs. 32.22

Explanation: Money spent originally = Rs. 140

Less Money to be spent now

= 10% of 140

= Rs. 14

Rs. 14 now yield 500 gm sugar

So, Present rate of sugar = Rs. 28 per kg.

If the present value is Rs. 90, the original value = Rs. 100

If the present value is Rs. 28 the original value

$$ = {\text{Rs}}{\text{. }}\frac{{100}}{{90}} \times 28 = {\text{Rs}}{\text{. }}31.11$$

7. Two numbers are respectively 20% and 50% of a third number. What percent is the first number of second?

a) 10%

b) 20%

c) 30%

d) 40%

Explanation:

$$\eqalign{ & {\text{Let third number is x}}. \cr & {\text{Then}}\,{\text{first}}\,{\text{no}}{\text{.}} \cr & 20\% \,{\text{of}}\,x = \frac{{20x}}{{100}} \cr & {\text{Second}}\,{\text{number}} \cr & = 50\% \,{\text{of}}\,x = \frac{{50x}}{{100}} \cr & {\text{Percent of first no of second no,}} \cr & = {\frac{{ {\frac{{20x}}{{100}}} }}{{ {\frac{{50x}}{{100}}} }}} \times 100 \cr & = \frac{{ {2 \times 100} }}{{20}} \cr & = 40\% \cr} $$

8. An empty fuel tank of a car was filled with A type petrol. When the tank was half-empty, it was filled with B type petrol. Again when the tank was half-empty, it was filled with A type petrol. When the tank was half-empty again, it was filled with B type petrol. What is the percentage of A type petrol at present in the tank?

a) 33.5%

b) 37.5%

c) 40%

d) 50%

Explanation: Let the capacity of the tank be 100 litres

Initially, A type petrol = 100 litres

After first operation:

A type petrol = $$\frac{{100}}{2}$$ = 50 litres

B type petrol = 50 litres

After second operation:

A type petrol = $$\frac{{50}}{2}$$ + 50 = 75 litres

B type petrol = $$\frac{{50}}{2}$$ = 25 litres

After third operation:

A type petrol = $$\frac{{75}}{2}$$ = 37.5 litres

B type petrol = $$\frac{{25}}{2}$$ + 50 = 62.5 litres

Required percentage = 37.5%

9. For an examination it is required to get 36% of maximum marks to pass. A student got 113 marks and failed by 85 marks. The maximum marks for the examination are:

a) 500

b) 550

c) 565

d) 620

Explanation: 36% marks = 113 + 85

36% marks = 198

So, 1% marks = $$\frac{{198}}{{36}}$$ = 5.5

100% marks = 5.5 × 100 = 550

10. 1% of 1% of 25% 1000 is

a) 0.025

b) 0.0025

c) 0.25

d) 0.000025

Explanation:

$$\eqalign{ & 1\% \,{\text{of}}\,1\% \,{\text{of}}\,25\% \,1000 \cr & = 1\% \,{\text{of}}\,1\% \,{\text{of}}\,\, {\frac{{ {25 \times 1000} }}{{100}}} \cr & = 1\% \,{\text{of}}\,1\% \,{\text{of}}\,250 \cr & = 1\% \,{\text{of}}\,\, {\frac{{ {1 \times 200} }}{{100}}} \cr & = 1\% \,{\text{of}}\,\,2.5 \cr & = \frac{{2.5}}{{100}} \cr & = 0.025 \cr} $$