5 Mark Question

Question 15 Marks

Ramanujan’s Magic square.

Add the four numbers in the rows, the columns and along the diagonals of this square.
What is the sum?
Is it the same every time?
What is the peculiarity?
Look at the numbers in the first row, 22 – 12 – 1887. Find out why this date is special.

Obtain and read a biography of the great Indian mathematician Srinivasa Ramanujan.

Answer

Sum of the numbers in each row:
i. 22 + 12 + 18 + 87 = 139
ii. 88 + 17 + 9 + 25 = 139
iii. 10 + 24 + 89 + 16 = 139
iv. 19 + 86 + 23 + 11 = 139

Sum of the numbers along the diagonals:
i. 22 + 17 + 89 + 11 = 139
ii. 87 + 9 + 24 + 19 = 139

Sum of the numbers in each column:
i. 22 + 88 + 10 + 19 = 139
ii. 12 + 17 + 24 + 86 = 139
iii. 18 + 9 + 89 + 23 = 139
iv. 87 + 25 + 16 + 11 = 139

∴ We observe that the sum of the numbers in each of the rows, the columns and along each diagonal remains the same every time. The numbers in the first row 22 – 12 – 1887 is the birth date of Srinivasa Ramanujan.

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Question 25 Marks

If we want to show the fractions $\frac{3}{10}, \frac{9}{20}, \frac{19}{40}$ on the number line, how big should the unit be?

Answer

The denominators of the given fractions are not equal.
The numbers in the denominators 10, 20 and 40 have common multiple 40.
∴ Making the denominators equal, we get

$\frac{3}{10}=\frac{3 \times 4}{10 \times 4}=\frac{12}{40} ; \quad \frac{9}{20}=\frac{9 \times 2}{20 \times 2}=\frac{18}{40} ; \quad \frac{19}{40}$

∴ To represent these fractions on the numbers line, each main unit should be divided into 40 equal sub-units.

Therefore,
$\frac{3}{10}=\frac{12}{40}$is represented on 12th mark from 0.
$\frac{9}{20}=\frac{18}{40}$is represented on 18th mark from 0 and $\frac{19}{40}$ is represented on 19 mark from 0.

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Question 35 Marks

i. $3 \frac{1}{3}-1 \frac{1}{4}$
ii. $5 \frac{1}{2}-3 \frac{1}{3}$
iii. $7 \frac{1}{8}-6 \frac{1}{10}$
iv. $7 \frac{1}{2}-3 \frac{1}{5}$

Answer

$\begin{aligned} & \text {} 3 \frac{1}{3}-1 \frac{1}{4} \\ & 3 \frac{1}{3}-1 \frac{1}{4}=\frac{3 \times 3+1}{3}-\frac{1 \times 4+1}{4} \\ & =\frac{9+1}{3}-\frac{4+1}{4} \\ & =\frac{10}{3}-\frac{5}{4} \\ & =\frac{10 \times 4}{3 \times 4}-\frac{5 \times 3}{4 \times 3} \\ & =\frac{40}{12}-\frac{15}{12} \\ & =\frac{40-15}{12}=\frac{25}{12} \\ & \therefore \quad 3 \frac{1}{3}-1 \frac{1}{4}=2 \frac{1}{12} \\ & \end{aligned}$

$\begin{aligned} & 5 \frac{1}{2}-3 \frac{1}{3} \\ & 5 \frac{1}{2}-3 \frac{1}{3}=\frac{5 \times 2+1}{2}-\frac{3 \times 3+1}{3} \\ & =\frac{10+1}{2}-\frac{9+1}{3} \\ & =\frac{11}{2}-\frac{10}{3} \\ & =\frac{11 \times 3}{2 \times 3}-\frac{10 \times 2}{3 \times 2} \\ & =\frac{33}{6}-\frac{20}{6} \\ & =\frac{33-20}{6}=\frac{13}{6} \\ & 5 \frac{1}{2}-3 \frac{1}{3}=2 \frac{1}{6} \\ & \end{aligned}$

$\begin{aligned} & 7 \frac{1}{8}-6 \frac{1}{10} \\ & 7 \frac{1}{8}-6 \frac{1}{10}=\frac{7 \times 8+1}{8}-\frac{6 \times 10+1}{10} \\ & =\frac{56+1}{8}-\frac{60+1}{10} \\ & =\frac{57}{8}-\frac{61}{10} \\ & =\frac{57 \times 5}{8 \times 5}-\frac{61 \times 4}{10 \times 4} \\ & =\frac{285}{40}-\frac{244}{40} \\ & =\frac{285-244}{40}=\frac{41}{40} \\ & 7 \frac{1}{8}-6 \frac{1}{10}=1 \frac{1}{40} \\ & \end{aligned}$

$\begin{aligned} 7 \frac{1}{2}-3 \frac{1}{5} & \\ 7 \frac{1}{2}-3 \frac{1}{5} & =\frac{7 \times 2+1}{2}-\frac{3 \times 5+1}{5} \\ & =\frac{14+1}{2}-\frac{15+1}{5} \\ & =\frac{15}{2}-\frac{16}{5} \\ & =\frac{15 \times 5}{2 \times 5}-\frac{16 \times 2}{5 \times 2} \\ & =\frac{75}{10}-\frac{32}{10} \\ & =\frac{75-32}{10}=\frac{43}{10} \\ 7 \frac{1}{2}-3 \frac{1}{5} & =4 \frac{3}{10}\end{aligned}$

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Question 45 Marks

$7 \frac{1}{2}-3 \frac{1}{5}$

Answer

$\begin{aligned} 7 \frac{1}{2}-3 \frac{1}{5} & \\ 7 \frac{1}{2}-3 \frac{1}{5} & =\frac{7 \times 2+1}{2}-\frac{3 \times 5+1}{5} \\ & =\frac{14+1}{2}-\frac{15+1}{5} \\ & =\frac{15}{2}-\frac{16}{5} \\ & =\frac{15 \times 5}{2 \times 5}-\frac{16 \times 2}{5 \times 2} \\ & =\frac{75}{10}-\frac{32}{10} \\ & =\frac{75-32}{10}=\frac{43}{10} \\ 7 \frac{1}{2}-3 \frac{1}{5} & =4 \frac{3}{10}\end{aligned}$

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Question 55 Marks

$7 \frac{1}{8}-6 \frac{1}{10}$

Answer

$\begin{aligned} & 7 \frac{1}{8}-6 \frac{1}{10} \\ & 7 \frac{1}{8}-6 \frac{1}{10}=\frac{7 \times 8+1}{8}-\frac{6 \times 10+1}{10} \\ & =\frac{56+1}{8}-\frac{60+1}{10} \\ & =\frac{57}{8}-\frac{61}{10} \\ & =\frac{57 \times 5}{8 \times 5}-\frac{61 \times 4}{10 \times 4} \\ & =\frac{285}{40}-\frac{244}{40} \\ & =\frac{285-244}{40}=\frac{41}{40} \\ & 7 \frac{1}{8}-6 \frac{1}{10}=1 \frac{1}{40} \\ & \end{aligned}$

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Question 65 Marks

$5 \frac{1}{2}-3 \frac{1}{3}$

Answer

$\begin{aligned} & 5 \frac{1}{2}-3 \frac{1}{3} \\ & 5 \frac{1}{2}-3 \frac{1}{3}=\frac{5 \times 2+1}{2}-\frac{3 \times 3+1}{3} \\ & =\frac{10+1}{2}-\frac{9+1}{3} \\ & =\frac{11}{2}-\frac{10}{3} \\ & =\frac{11 \times 3}{2 \times 3}-\frac{10 \times 2}{3 \times 2} \\ & =\frac{33}{6}-\frac{20}{6} \\ & =\frac{33-20}{6}=\frac{13}{6} \\ & 5 \frac{1}{2}-3 \frac{1}{3}=2 \frac{1}{6} \\ & \end{aligned}$

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Question 75 Marks

$3 \frac{1}{3}-1 \frac{1}{4}$

Answer

$\begin{aligned} & \text {} 3 \frac{1}{3}-1 \frac{1}{4} \\ & 3 \frac{1}{3}-1 \frac{1}{4}=\frac{3 \times 3+1}{3}-\frac{1 \times 4+1}{4} \\ & =\frac{9+1}{3}-\frac{4+1}{4} \\ & =\frac{10}{3}-\frac{5}{4} \\ & =\frac{10 \times 4}{3 \times 4}-\frac{5 \times 3}{4 \times 3} \\ & =\frac{40}{12}-\frac{15}{12} \\ & =\frac{40-15}{12}=\frac{25}{12} \\ & \therefore \quad 3 \frac{1}{3}-1 \frac{1}{4}=2 \frac{1}{12} \\ & \end{aligned}$

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Question 85 Marks

(1) $6 \frac{1}{3}+2 \frac{1}{3}$
(2) $1 \frac{1}{4}+3 \frac{1}{2}$
(3) $5 \frac{1}{5}+2 \frac{1}{7}$
(4) $3 \frac{1}{5}+2 \frac{1}{3}$

Answer

(1) $\begin{aligned} & \text { i. } 6 \frac{1}{3}+2 \frac{1}{3} \\ & 6 \frac{1}{3}+2 \frac{1}{3}=\frac{6 \times 3+1}{3}+\frac{2 \times 3+1}{3} \\ & =\frac{18+1}{3}+\frac{6+1}{3} \\ & =\frac{19}{3}+\frac{7}{3} \\ & =\frac{19+7}{3} \\ & =\frac{26}{3} \\ & \therefore \quad 6 \frac{1}{3}+2 \frac{1}{3}=8 \frac{2}{3} \\ & \end{aligned}$

(2) $\begin{aligned} & 1 \frac{1}{4}+3 \frac{1}{2} \\ & 1 \frac{1}{4}+3 \frac{1}{2}=\frac{1 \times 4+1}{4}+\frac{3 \times 2+1}{2} \\ & =\frac{4+1}{4}+\frac{6+1}{2} \\ & =\frac{5}{4}+\frac{7}{2} \\ & =\frac{5}{4}+\frac{7 \times 2}{2 \times 2} \\ & =\frac{5}{4}+\frac{14}{4}=\frac{5+14}{4}=\frac{19}{4} \\ & 1 \frac{1}{4}+3 \frac{1}{2}=4 \frac{3}{4} \\ & \end{aligned}$

(3) $\begin{aligned} & 5 \frac{1}{5}+2 \frac{1}{7} \\ & 5 \frac{1}{5}+2 \frac{1}{7}=\frac{5 \times 5+1}{5}+\frac{2 \times 7+1}{7} \\ & =\frac{25+1}{5}+\frac{14+1}{7} \\ & =\frac{26}{5}+\frac{15}{7} \\ & =\frac{26 \times 7}{5 \times 7}+\frac{15 \times 5}{7 \times 5} \\ & =\frac{182}{35}+\frac{75}{35} \\ & =\frac{182+75}{35} \\ & =\frac{257}{35} \\ & 5 \frac{1}{5}+2 \frac{1}{7}=7 \frac{12}{35} \\ & \end{aligned}$

(4) $\begin{aligned} & 3 \frac{1}{5}+2 \frac{1}{3} \\ & 3 \frac{1}{5}+2 \frac{1}{3}=\frac{3 \times 5+1}{5}+\frac{2 \times 3+1}{3} \\ &=\frac{15+1}{5}+\frac{6+1}{3} \\ &=\frac{16}{5}+\frac{7}{3} \\ &=\frac{16 \times 3}{5 \times 3}+\frac{7 \times 5}{3 \times 5} \\ &=\frac{48}{15}+\frac{35}{15} \\ &=\frac{48+35}{15} \\ &=\frac{83}{15} \\ & 3 \frac{1}{5}+2 \frac{1}{3}=5 \frac{8}{15}\end{aligned}$

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Question 95 Marks

$3 \frac{1}{5}+2 \frac{1}{3}$

Answer

$\begin{aligned} & 3 \frac{1}{5}+2 \frac{1}{3} \\ & 3 \frac{1}{5}+2 \frac{1}{3}=\frac{3 \times 5+1}{5}+\frac{2 \times 3+1}{3} \\ &=\frac{15+1}{5}+\frac{6+1}{3} \\ &=\frac{16}{5}+\frac{7}{3} \\ &=\frac{16 \times 3}{5 \times 3}+\frac{7 \times 5}{3 \times 5} \\ &=\frac{48}{15}+\frac{35}{15} \\ &=\frac{48+35}{15} \\ &=\frac{83}{15} \\ & 3 \frac{1}{5}+2 \frac{1}{3}=5 \frac{8}{15}\end{aligned}$

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Question 105 Marks

$5 \frac{1}{5}+2 \frac{1}{7}$

Answer

$\begin{aligned} & 5 \frac{1}{5}+2 \frac{1}{7} \\ & 5 \frac{1}{5}+2 \frac{1}{7}=\frac{5 \times 5+1}{5}+\frac{2 \times 7+1}{7} \\ & =\frac{25+1}{5}+\frac{14+1}{7} \\ & =\frac{26}{5}+\frac{15}{7} \\ & =\frac{26 \times 7}{5 \times 7}+\frac{15 \times 5}{7 \times 5} \\ & =\frac{182}{35}+\frac{75}{35} \\ & =\frac{182+75}{35} \\ & =\frac{257}{35} \\ & 5 \frac{1}{5}+2 \frac{1}{7}=7 \frac{12}{35} \\ & \end{aligned}$

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Question 115 Marks

$1 \frac{1}{4}+3 \frac{1}{2}$

Answer

$\begin{aligned} & 1 \frac{1}{4}+3 \frac{1}{2} \\ & 1 \frac{1}{4}+3 \frac{1}{2}=\frac{1 \times 4+1}{4}+\frac{3 \times 2+1}{2} \\ & =\frac{4+1}{4}+\frac{6+1}{2} \\ & =\frac{5}{4}+\frac{7}{2} \\ & =\frac{5}{4}+\frac{7 \times 2}{2 \times 2} \\ & =\frac{5}{4}+\frac{14}{4}=\frac{5+14}{4}=\frac{19}{4} \\ & 1 \frac{1}{4}+3 \frac{1}{2}=4 \frac{3}{4} \\ & \end{aligned}$

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Question 125 Marks

$6 \frac{1}{3}+2 \frac{1}{3}$

Answer

$\begin{aligned} & \text { i. } 6 \frac{1}{3}+2 \frac{1}{3} \\ & 6 \frac{1}{3}+2 \frac{1}{3}=\frac{6 \times 3+1}{3}+\frac{2 \times 3+1}{3} \\ & =\frac{18+1}{3}+\frac{6+1}{3} \\ & =\frac{19}{3}+\frac{7}{3} \\ & =\frac{19+7}{3} \\ & =\frac{26}{3} \\ & \therefore \quad 6 \frac{1}{3}+2 \frac{1}{3}=8 \frac{2}{3} \\ & \end{aligned}$

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Question 135 Marks

How to do this subtraction:$4 \frac{1}{4}-2 \frac{1}{2} ?$Is it same as$\left[4-2+\frac{1}{4}-\frac{1}{2}\right] ?$

Answer

$\begin{aligned} & 4 \frac{1}{4}-2 \frac{1}{2} \\ & 4 \frac{1}{4}-2 \frac{1}{2}=\frac{4 \times 4+1}{4}-\frac{2 \times 2+1}{2} \\ & =\frac{17}{4}-\frac{5}{2} \\ & =\frac{17}{4}-\frac{5 \times 2}{2 \times 2} \\ & =\frac{17}{4}-\frac{10}{4} \\ & =\frac{7}{4} \\ & =1 \frac{3}{4} \\ & \end{aligned}$

$\left[4-2+\frac{1}{4}-\frac{1}{2}\right]$
$\begin{aligned} 4-2+\frac{1}{4}-\frac{1}{2} & =2+\frac{1}{4}-\frac{1 \times 2}{2 \times 2} \\ & =2+\frac{1}{4}-\frac{2}{4} \\ & =2+\frac{-1}{4} \\ & =\frac{2 \times 4}{1 \times 4}+\frac{-1}{4} \\ & =\frac{8}{4}-\frac{1}{4} \\ & =\frac{8-1}{4}=\frac{7}{4}=1 \frac{3}{4}\end{aligned}$

The subtraction$4 \frac{1}{4}-2 \frac{1}{2}$ is the same as $\left[4-2+\frac{1}{4}-\frac{1}{2}\right]$

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Question 145 Marks

Convert into mixed numbers:
i. $\frac{30}{7}$
ii. $\frac{7}{4}$
iii. $\frac{15}{12}$
iv. $\frac{11}{8}$
V. $\frac{21}{4}$
V. $\frac{20}{7}$

Answer

i. $\frac{30}{7}$
$\frac{30}{7}=30 \div 7$
$\begin{array}{r}4 \\ 7 \longdiv { 3 0 } \\ -\frac{28}{02}\end{array}$
$\frac{30}{7}=4 \frac{2}{7}$

ii. $\frac{7}{4}$
$\frac{7}{4}=7 \div 4$
$\begin{array}{r}1 \\ 4 \longdiv { 7 } \\ -\frac{4}{03}\end{array}$
$\frac{7}{4}=1 \frac{3}{4}$

iii. $\frac{15}{12}$
$\frac{15}{12}=15 \div 12$
$\begin{array}{r}1 \\ 1 2 \longdiv { 1 5 } \\ -\frac{12}{03}\end{array}$
$\frac{15}{12}=1 \frac{3}{12}$ or $1 \frac{1}{4}$

iv. $\frac{11}{8}$
$\frac{11}{8}=11 \div 8$
$\begin{array}{r}5 \\ 4 \longdiv { 2 1 } \\ -\frac{20}{01}\end{array}$
$\frac{21}{4}=5 \frac{1}{4}$

v. $\frac{20}{7}$
$\frac{20}{7}=20 \div 7$
$\begin{array}{r}2 \\ 7 \longdiv { 2 0 } \\ -\frac{14}{06}\end{array}$
$\frac{20}{7}=2 \frac{6}{7}$

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Question 155 Marks

Convert into improper fractions:
i. $7 \frac{2}{5}$
ii. $5 \frac{1}{6}$
iii. $4 \frac{3}{4}$
iv. $2 \frac{5}{9}$
v. $1 \frac{5}{7}$

Answer

$\begin{aligned} & \text { i. } 7 \frac{2}{5} \\ & 7 \frac{2}{5}=7+\frac{2}{5}=\frac{7}{1}+\frac{2}{5} \\ & =\frac{7 \times 5}{1 \times 5}+\frac{2}{5}=\frac{35}{5}+\frac{2}{5}=\frac{35+2}{5} \\ & 7 \frac{2}{5}=\frac{37}{5} \\ & \end{aligned}$

$\begin{aligned} & \text { ii. } 5 \frac{1}{6} \\ & \begin{aligned} 5 \frac{1}{6} & =5+\frac{1}{6}=\frac{5}{1}+\frac{1}{6} \\ & =\frac{5 \times 6}{1 \times 6}+\frac{1}{6}=\frac{30}{6}+\frac{1}{6}=\frac{30+1}{6} \\ 5 \frac{1}{6} & =\frac{31}{6}\end{aligned}\end{aligned}$

$\begin{aligned} & \text { iii. } 4 \frac{3}{4} \\ & 4 \frac{3}{4}=4+\frac{3}{4}=\frac{4}{1}+\frac{3}{4} \\ & =\frac{4 \times 4}{1 \times 4}+\frac{3}{4}=\frac{16}{4}+\frac{3}{4}=\frac{16+3}{4} \\ & 4 \frac{3}{4}=\frac{19}{4} \\ & \end{aligned}$

$\begin{aligned} & \qquad \begin{aligned} 2 \frac{5}{9} & =2+\frac{5}{9}=\frac{2}{1}+\frac{5}{9} \\ & =\frac{2 \times 9}{1 \times 9}+\frac{5}{9}=\frac{18}{9}+\frac{5}{9}=\frac{18+5}{9} \\ 2 \frac{5}{9} & =\frac{23}{9}\end{aligned}\end{aligned}$

$\begin{aligned} & \text { v. } 1 \frac{5}{7} \\ & 1 \frac{5}{7}=1+\frac{5}{7}=\frac{1}{1}+\frac{5}{7} \\ & =\frac{1 \times 7}{1 \times 7}+\frac{5}{7}=\frac{7}{7}+\frac{5}{7}=\frac{7+5}{7} \\ & 1 \frac{5}{7}=\frac{12}{7} \\ & \end{aligned}$

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