Number System - Aptitude MCQ Questions and Solutions with Explanations

12345679 × 72
Shortcut method :
Since 72 is factor of 9, So the answer should be divisible by 9.
Now check with option by adding the digits of option and check which one is divisible by 9.
Here option (c) 888888888 is divisible by 9, so this is the multiple of 12345679 × 72

Alternate
12345679 x 72
= 12345679 x (70 +2)
= 12345679 x 70 + 12345679 x 2
= 864197530 + 24691358
= 888888888

$$\eqalign{ & \frac{3}{4} = 75\% ,\frac{5}{6} = 83.33\% , \cr & (1)\frac{2}{3} = 66.66\% , \cr & (2)\frac{1}{2} = 50\% , \cr & (3)\frac{4}{5} = 80\% , \cr & (4)\frac{9}{{10}} = 90\% \cr & {\text{Hence , }}\frac{4}{5}{\text{ is between the fraction}}{\text{.}} \cr} $$

81 × 82 × 83 ×..... × 89 take unit digit multiply
1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 = 0
Since there is 5 and 2 are present
So we get 5 × 2 = 10 and the last digit of 10 is 0, so the last digit of the above number is also 0

$$\eqalign{ & = \left( {999\frac{{999}}{{1000}} \times 7} \right) \cr & = \left( {999\frac{{999}}{{1000}}} \right) \times 7 \cr & = 6993 + \frac{{6993}}{{1000}} \cr & = 6993 + 6\frac{{993}}{{1000}} \cr & = 6993 + 6 + \frac{{993}}{{1000}} \cr & = 6999\frac{{993}}{{1000}} \cr} $$

$$\eqalign{ & {\text{2, 4, 6, 8, 10}}......{\text{198, 200 }} \cr & {2^{100}}(1 \times 2 \times 3 \times ..... \times 99 \times 100) \cr} $$
$$\eqalign{ & \underline {5\,\,\,|\,\,100} \cr & \underline {5\,\,\,|\,\,\,20} \,\,\,\,\, = 24{\text{ zero's (20 + 4)}} \cr & \,\,\,\,\,|\,\,\,4 \cr} $$
When we multiply the series of 1 × 2 × 3 ×..... × 100 then we find 24 zero at the end of the product

Let first number is x and second number is 2x.
Then, according to the question
x × 2x = 2048
x² = 1024
x = 32
Smaller number = 32

Given :
$$\eqalign{ & \Leftrightarrow \frac{{51.84}}{{4.32}} = 12 \cr & \Leftrightarrow \frac{{5184}}{{432}} = 12 \cr & \therefore \frac{{0.005184}}{{0.432}} \cr & = \frac{{5.184}}{{432}} \cr & = \frac{{5184}}{{432}} \times \frac{1}{{1000}} \cr & = 12 \times \frac{1}{{1000}} \cr & = 0.012 \cr} $$

$$\frac{{41\frac{2}{3}}}{{\frac{1}{6}}} = \frac{{125}}{3} \times \frac{6}{1} = 250{\text{ times}}$$

$$\eqalign{ & {\text{Let the number be x}} \cr & {\text{According to question}} \cr & \frac{x}{{7/8}} - \frac{7}{8}x = 15 \cr & \frac{8}{7}x - \frac{7}{8}x = 15 \cr & \frac{{64x - 49x}}{{56}} = 15 \cr & 15x = 15 \times 56 \cr & x = 56 \cr & {\text{Sum of digits }} \cr & = 5{\text{ }} + {\text{ }}6 \cr & = 11 \cr} $$

$$\eqalign{ & {\text{Let number are }}x,{\text{ }}x + 5,{\text{ }}x + 10 \cr & {\text{So, }}x{\text{ + }}x + 5 + x + 10 = 225 \cr & 3x + 15 = 225 \cr & 3x = 210 \cr & x = 70 \cr & {\text{largest number is }}70 + 10 = 80 \cr} $$