| Column-I | Column-II |
| (i) Divide the sum of \[\frac{12}{5}\]and \[\frac{13}{7}\]by the product of \[\frac{-4}{7}\] and \[\frac{-1}{2}\] The result obtained is | (p) \[\frac{7}{10}\] |
| (ii) Niharika spent \[\frac{3}{4}\]of her pocket money. She spent \[\frac{1}{2}\]of it on a book, \[\frac{1}{6}\] on a movie and rest for a dress. ___ part of her pocket money she spend on the dress. | (q) \[3\frac{19}{28}\] |
| (iii) If 35 shirts of equal size can be stitched from \[\frac{49}{2}\] metres of cloth. The length (in m) of doth required for each shirt is | |
| (r) 14? | |
| (iv) Two packets of chocolates weigh \[\frac{9}{4}\] kg and \[\frac{10}{7}\] kg respectively. The total weight (in kg) of the chocolates is | (s) \[\frac{1}{4}\] |
A) (i)\[\to \](p); (ii)\[\to \](q); (iii)\[\to \](r); (iv)\[\to \](s)
B) (i)\[\to \](r); (ii)\[\to \](s); (iii)\[\to \](p); (iv)\[\to \](q)
C) (i)\[\to \](p); (ii)\[\to \](r); (iii)\[\to \](s); (iv)\[\to \](q)
D) (i)\[\to \](r); (ii)\[\to \](p); (iii)\[\to \](s); (iv)\[\to \](q)
Correct Answer: B
Solution :
(i) Sum of \[\frac{12}{5}\] and \[\frac{13}{7}=\frac{12}{5}+\frac{13}{7}=\frac{84+65}{35}=\frac{149}{35}\] Product of \[\frac{-4}{7}\] and \[\frac{-1}{2}=\frac{-4}{7}\times \left( \frac{-1}{2} \right)=\frac{2}{7}\] Division of \[\frac{149}{35}\]and \[\frac{2}{7}=\frac{149}{35}\div \frac{2}{7}=\frac{149}{35}\times \frac{7}{2}\] \[=\frac{149}{10}=14\frac{9}{10}\] (ii) Let x be the total pocket money of Niharika. Total amount of money spent by her \[=\frac{3}{4}x\] Money spent on a book \[=\frac{1}{2}\times \frac{3}{4}\times =\frac{3}{8}x\] Money spent on movie \[=\frac{1}{6}\times \frac{3}{4}\times =\frac{1}{8}x\] \[\therefore \] Amount of money spent on a dress \[=\frac{3}{4}x-\left( \frac{3}{8}+\frac{1}{8} \right)x=\frac{3}{4}x-\left( \frac{3+1}{8} \right)x=\frac{3x}{4}-\frac{4x}{8}\]\[=\frac{3x}{4}-\frac{x}{2}=\frac{3x-2x}{4}=\frac{x}{4}\] Hence, \[\frac{1}{4}\] of her pocket money was spent on the dress. (iii) Length of cloth required for each shirt \[=\frac{Total\text{ }length\text{ }of\text{ }cloth}{Number\text{ }of\text{ }shirts~}\] \[=\frac{\frac{49}{2}}{35}=\frac{49}{2}\times \frac{1}{35}=\frac{7}{10}\,m\] (iv) Weight of one packet of chocolates \[=\frac{9}{4}kg\] Weight of other packet of chocolates \[=\frac{10}{7}kg\] \[\therefore \] Total weight of the chocolates \[=\frac{9}{4}+\frac{10}{7}\] \[=\frac{63+40}{28}=\frac{103}{28}=3\frac{19}{28}kg\]
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