[4 marks sum] - MATHS STD 6 Questions - Vidyadip

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[4 marks sum]

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17 questions · timed · auto-graded

Question 14 Marks

In buying a ready-made shirt-two-fifths of my pocket money is spent If Rs. 540 is still left with me, find:
(i) The money I had before I bought the shirt.
(ii) The emit of the shirt

Answer

Let total money in the pocket $=1$
Amount spent on shirt $=\frac{2}{5}$
Balance amount $=1-\frac{2}{5}=\frac{5-2}{5}=\frac{3}{5}$
Now $\frac{3}{5}$ of total money $=$ Rs. 540
(i) $\therefore$ Total money $=$ Rs. $540 \times \frac{5}{3}=180 \times 5$
$=$ Rs. 900
(ii) Cost of shirt $=\frac{2}{5}$ of Rs. $900=$ Rs. $2 \times 180$
$=$ Rs. 360

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Question 24 Marks

Evaluate:
$\frac{5}{6} \text { of } \frac{3}{4} \div \frac{7}{8} \times 1 \frac{1}{2}$

Answer

$\frac{5}{6} \text { of } \frac{3}{4} \div \frac{7}{8} \times 1 \frac{1}{2} $
$=\frac{5}{6} \text { of } \frac{3}{4} \div \frac{7}{8} \times \frac{3}{2} $
$=\frac{5}{8} \div \frac{7}{8} \times \frac{3}{2} \ldots \ldots . . \text { (first remove 'of') } $
$=\frac{5}{8} \times \frac{8}{7} \times \frac{3}{2} \ldots . . \text { (then remove } \div \text { ) } $
$=\frac{15}{14} $
$=1 \frac{1}{14}$

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Question 34 Marks

Simplify:
$\frac{3}{4} \times 1 \frac{1}{3} \div \frac{3}{7} \text { of } 2 \frac{5}{8}$

Answer

$\frac{3}{4} \times 1 \frac{1}{3} \div \frac{3}{7} \text { of } 2 \frac{5}{8}=\frac{3}{4} \times \frac{4}{3} \div \frac{9}{8} \ldots \ldots . .\left[\frac{3}{7} \text { of } 2 \frac{5}{8}=\frac{3}{7} \times \frac{21}{8}=\frac{9}{8}\right] $
$=\frac{3}{4} \times \frac{4}{3} \times \frac{8}{9}=\frac{8}{9}$

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Question 44 Marks

After finishing $\frac{3}{4}$ of my journey, I find that $12 km$ of my journey is covered. How much distance is still left to be covered?

Answer

Let the total journey $=x$,
$\text { distance covered }=\frac{3}{4}=12\ km$
Then, according to question $\frac{3}{4}$ of $x=12 km$
$x=12 \times \frac{4}{3} $
$\Rightarrow x=16 km$
Distance left $=$ total distance - distance cover
$=16-12 $
$=4\ km .$

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Question 54 Marks

In a school, $\frac{4}{5}$ of the students are boys, and the number of girls is $100.$ Find the number of boys.

Answer

Let the total number of boys and girls $=x$
Total number of boys $=\frac{4}{5}$ of $x=\frac{4 x}{5}$
According to question, the total strength of the school,
$x-\frac{4 x}{5}=100 $
$\frac{5 x-4 x}{5}=100 $
$\frac{x}{5}=100 $
$\Rightarrow x=500$
$\therefore$ Number of boys $=$ total strength - girls
$=500-100 $
$=400 .$

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Question 64 Marks

A man spends $\frac{3}{8}$ of his money and 8 still has Rs. 720 left with him. How much money did he have at first?

Answer

Let a man has money $=$ Rs. 1
Amount spent $=\frac{3}{8}$ of Rs. $1=$ Rs. $\frac{3}{8}$
Amount left $=1-\frac{3}{8}=\frac{8-3}{8}=$ Rs. $\frac{5}{8}$
$\therefore \frac{5}{8}$ of his total money $=$ Rs. 720
$\therefore$ Total money $=$ Rs. $\frac{720 \times 8}{5}$
$=$ Rs. $144 \times 8=$ Rs. 1152

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Question 74 Marks

A piece of cloth is $5$ metre long. After washing, it shrinks by $\frac{1}{25}$ of its length. What is the length of the cloth after washing?

Answer

Length of a piece of cloth $=5 m$
After washing, it is shrinked
$=\frac{1}{25} \text { of } 5 m =\frac{1}{5} m$
Length of cloth after washing
$=\left(5-\frac{1}{5}\right) m $
$=\frac{25-1}{5}$
$=\frac{24}{5} m$
$=4 \frac{4}{5} m$

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Question 84 Marks

Simplify:
$\left(1 \div 2 \frac{1}{5}\right) \div 2 \frac{1}{5} \text { of } 2 \frac{1}{2}-2$

Answer

$\left(1 \div 2 \frac{1}{5}\right) \div 2 \frac{1}{5} \text { of } 2 \frac{1}{2}-2 $
$=\left(1 \div \frac{11}{5}\right) \div \frac{11}{5} \text { of } \frac{5}{2}-2 $
$=\left(1 \times \frac{5}{11}\right) \div \frac{11}{5} \text { of } \frac{5}{2}-2 $
$=\frac{5}{11} \div \frac{11}{5} \times \frac{5}{2}-2=\frac{5}{11} \div \frac{11}{2}-2 $
$=\frac{5}{11} \times \frac{2}{11}-2=\frac{10}{121}-2 $
$=\frac{10-242}{121}=-\frac{232}{121}=-1 \frac{111}{121}$

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Question 94 Marks

Simplify:
$4 \frac{2}{3} \div\left(3-\frac{1}{2}\right)+\left(\frac{2}{5} \div 1 \frac{1}{5}\right)$

Answer

$4 \frac{2}{3} \div\left(3-\frac{1}{2}\right)+\left(\frac{2}{5} \div 1 \frac{1}{5}\right) $
$=\frac{14}{3} \div\left(3-\frac{1}{2}\right)+\left(\frac{2}{5} \div \frac{6}{5}\right) $
$=\frac{14}{3} \div\left(\frac{6-1}{2}\right)+\left(\frac{2}{5} \times \frac{5}{6}\right) $
$=\frac{14}{3} \div\left(\frac{5}{2}\right)+\frac{1}{3} $
$=\frac{14}{3} \times \frac{2}{5}+\frac{1}{3} $
$=\frac{28}{15}+\frac{1}{3}=\frac{28+5}{15} $
$=\frac{33}{15}=\frac{11}{55}=2 \frac{1}{5}$

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Question 104 Marks

Simplify:
$\left(\frac{1}{4}-\frac{1}{6}\right) \text { of }\left(\frac{2}{3}-\frac{5}{12}\right) \times\left(\frac{5}{8}-\frac{7}{12}\right)$

Answer

$\left(\frac{1}{4}-\frac{1}{6}\right) \text { of }\left(\frac{2}{3}-\frac{5}{12}\right) \times\left(\frac{5}{8}-\frac{7}{12}\right) $
$=\left(\frac{3-2}{12}\right) \text { of }\left(\frac{8-5}{12}\right) \times\left(\frac{15-14}{24}\right) $
$=\left(\frac{1}{12}\right) \text { of }\left(\frac{3}{12}\right) \times\left(\frac{1}{24}\right) $
$=\frac{1}{12} \text { of } \frac{3}{12} \times \frac{1}{24} $
$=\frac{1}{12} \times \frac{1}{96} $
$=\frac{1}{1152}$

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Question 114 Marks

Simplify:
$4 \frac{3}{5}-2 \frac{7}{9}-1 \frac{2}{15}-\frac{2}{45}$

Answer

$4 \frac{3}{5}-2 \frac{7}{9}-1 \frac{2}{15}-\frac{2}{45} $
$=\frac{23}{5}-\frac{25}{9}-\frac{17}{15}-\frac{2}{45} $
$=\frac{23 \times 9}{5 \times 9}-\frac{25 \times 5}{9 \times 5}-\frac{17 \times 3}{15 \times 3}-\frac{2 \times 1}{45 \times 1} $
$=\frac{207}{45}-\frac{125}{45}-\frac{51}{45}-\frac{2}{45} $
$=\frac{207-125-51-2}{45} $
$=\frac{207-178}{45}=\frac{29}{45}$

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Question 124 Marks

Simplify:
$6+\frac{3}{10}-1 \frac{8}{15}$

Answer

$6+\frac{3}{10}-1 \frac{8}{15} $
$=\frac{6}{1}+\frac{3}{10}-\frac{23}{15} $
$=\frac{6 \times 30}{1 \times 3}+\frac{3 \times 3}{10 \times 3}-\frac{23 \times 2}{15 \times 2} \ldots \ldots . .(\text { LCM of } 1,10,15=30) $
$=\frac{180}{30}+\frac{9}{30}-\frac{46}{30} $
$=\frac{180+9-46}{30}=\frac{189-46}{30} $
$=\frac{143}{30}=4 \frac{23}{30}$

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Question 134 Marks

Add the following fractions:
$1 \frac{7}{8}, 1 \frac{1}{2} \text { and } 1 \frac{3}{4}$

Answer

$1 \frac{7}{8}+1 \frac{1}{2}+1 \frac{3}{4} $
$=\frac{1 \times 8+7}{8}+\frac{1 \times 2+1}{2}+\frac{1 \times 4+3}{4} $
$=\frac{15}{8}+\frac{3}{2}+\frac{7}{4}=\frac{15 \times 1}{8 \times 1}+\frac{3 \times 4}{2 \times 4}+\frac{7 \times 2}{4 \times 2}$
$\left.=\frac{15}{8}+\frac{12}{8}+\frac{14}{8}=\frac{15+12+14}{8} \ldots . . \text { (L.C.M. of } 8,2 \text { and } 4 \text { is } 8\right) $
$=\frac{41}{8}=5 \frac{1}{8}$

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Question 144 Marks

Add the following fractions:
$\frac{2}{5}, 2 \frac{3}{15} \text { and } \frac{7}{10}$

Answer

$\frac{2}{5}, 2 \frac{3}{15} \text { and } \frac{7}{10} $
$=\frac{2}{5}+\frac{33}{15}+\frac{7}{10} $
$=\frac{2 \times 6}{5 \times 6}+\frac{33 \times 2}{15 \times 2}+\frac{7 \times 3}{10 \times 3} \ldots \ldots . .(\text { L.C.M. of 5, } 15 \text { and } 10=30) $
$=\frac{12}{30}+\frac{66}{30}+\frac{21}{30} $
$=\frac{12+66+21}{30}=\frac{99}{30}=\frac{99 \div 3}{30 \div 3} $
$=\frac{33}{10}=3 \frac{3}{10}$

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Question 154 Marks

Out of 50 identical articles, 36 are broken. Find the fraction of:
(i) The total number of articles and the articles broken.
(ii) The remaining articles and the total number of articles.

Answer

Total number of articles $=50$
Number of articles broken $=36$
Remaining articles $=50-36=14$
Now,
(i) The fraction of the total number of articles and articles broken $=\frac{50}{36}$
$
=\frac{50 \div 2}{36 \div 2}=\frac{25}{18}
$
(ii) The fraction between the remaining articles and total numbers of articles
$
=\frac{14}{50}=\frac{14 \div 2}{50 \div 2}=\frac{7}{25}
$

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Question 164 Marks

Arrange the given fractions in ascending order of magnitude:
$\frac{5}{6}, \frac{2}{7}, \frac{8}{9}, \frac{1}{3}$

Answer

$\frac{5}{6}, \frac{2}{7}, \frac{8}{9}, \frac{1}{3}$
L.C.M. of denominator $6,7,9,3=126$

$3$	$6,$	$7,$	$9,$	$3$
	$2,$	$7,$	$3,$	$1$

$=3 \times 2 \times 7 \times 3=126 $
$\therefore \frac{5}{6}=\frac{5 \times 21}{6 \times 21}=\frac{105}{126} ; \frac{2}{7}=\frac{2 \times 18}{7 \times 18}=\frac{36}{126} $
$\frac{8}{9}=\frac{8 \times 14}{9 \times 14}=\frac{112}{126} ; \frac{1}{3}=\frac{1 \times 42}{3 \times 42}=\frac{42}{126}$
Arranging in ascending order,
$\frac{36}{126}, \frac{42}{126}, \frac{105}{126}, \frac{112}{126}$
i.e. $\frac{2}{7}, \frac{1}{3}, \frac{5}{6}, \frac{8}{9}$

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Question 174 Marks

Arrange the given fractions in descending order of magnitude:
$\frac{4}{5}, \frac{7}{15}, \frac{11}{20}, \frac{3}{4}$

Answer

$\frac{4}{5}, \frac{7}{15}, \frac{11}{20}, \frac{3}{4}$
L.C.M. of denominator 5, 15, 20, $4=60$
Converting into like fractions,
$\frac{4}{5}=\frac{4 \times 12}{5 \times 12}=\frac{48}{60} ; \frac{7}{15}=\frac{7 \times 4}{15 \times 4}=\frac{28}{60} $
$\frac{11}{20}=\frac{11 \times 3}{20 \times 3}=\frac{33}{60} ; \frac{3}{4}=\frac{3 \times 15}{4 \times 15}=\frac{45}{60}$
Now, arranging in descending order
$\frac{4}{5}, \frac{3}{4}, \frac{11}{20}, \frac{7}{15}$

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[4 marks sum] - MATHS STD 6 Questions - Vidyadip