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Some applications of unitary methods:<\/strong><\/span><\/p>\n <\/span><\/p>\n The questions can be solved by two methods:<\/strong><\/span><\/p>\n <\/p>\n <\/p>\n 1. Priya went to the market to buy notebooks. The cost of 5 notebooks is Rs 250. Find the price of 3 such notebooks bought by Priya?<\/strong><\/p>\n Cost of 5 notebooks = Rs 250 (Given)<\/span><\/p>\n \u21d2 Cost of 1 notebook = Rs \\frac{250}{5}<\/span>= Rs 50<\/span><\/p>\n \u2234 Cost of 3 notebooks = Rs (50 \u00d7 3) = Rs 150<\/span><\/p>\n Hence, the cost of three notebooks bought by Priya is equal to Rs 150.<\/span><\/span><\/p>\n <\/p>\n 2. Neha\u2019s father Suresh bought 3 dozen pencils for Rs 360. What is the cost of 7 pencils?<\/strong><\/p>\n 1 dozen = 12<\/span><\/p>\n 3 dozens = (12 \u00d7 3) = 36<\/span><\/p>\n Cost of 36 pencils = Rs 360 (Given)<\/span><\/p>\n \u21d2 Cost of 1 pencil =\\operatorname{Rs} \\frac{360}{36}=\\operatorname{Rs} 10<\/span><\/span><\/p>\n \u2234 Cost of 7 pencil = Rs (10 \u00d7 7) = Rs 70<\/span><\/p>\n Hence, the cost of 7 pencils bought by Suresh is equal to Rs 70.<\/span> <\/span><\/p>\n 3. Vickey bought 6 apples for Rs 42 and Mohit bought 9 apples for Rs 81. Who bought the apples cheaper?<\/strong><\/p>\n Cost of 6 apples bought by Vickey = Rs 42<\/span><\/p>\n \u21d2 Cost of 1 apple bought by Vickey =\\operatorname{Rs} \\frac{42}{6}<\/span><\/span>= Rs 7<\/span><\/p>\n Cost of 9 apples bought by Mohit = Rs 81<\/span><\/p>\n \u21d2 Cost of 1 apple bought by Mohit =\\operatorname{Rs} \\frac{81}{9}<\/span><\/span>= Rs 9<\/span><\/p>\n Rs 7 < Rs 9<\/span><\/p>\n \u2234 Cost of 1 apple bought by Vickey < Cost of 1 apple bought by Mohit.<\/span><\/p>\n Hence, the apples bought by Vickey were cheaper.<\/span><\/p>\n <\/span><\/p>\n 4. The cost of 10 rings is equal to Rs 100. Find the maximum number of rings Sonia can buy if she has Rs 70.<\/strong><\/p>\n Cost of 10 rings = Rs 100<\/span><\/p>\n \u21d2 Cost of 1 ring =\\operatorname{Rs} \\frac{100}{10}<\/span><\/span>= Rs 10<\/span><\/p>\n Number of rings that can be bought with Rs 10 = 1<\/span><\/p>\n \u21d2 Number of rings that can be bought with Rs 70\u00a0 =\\frac{1}{10} \\times 70=7<\/span><\/span><\/p>\n Hence, Sonia can buy a maximum of 7 rings.<\/span><\/span><\/p>\n <\/p>\n 5. 5 packets of Pencil A costs Rs 600 having 10 pencils in each packet. 4 packets of Pencil B costs Rs 500 having 5 pencils in each packet. Which pencil is costlier?<\/strong><\/p>\n Pencil A<\/span> Number of Pencils in 5 packets = 10 \u00d7 5 = 50<\/span><\/p>\n Cost of 50 pencils = Rs 600<\/span><\/p>\n Hence, Cost of 1 pencil =\\mathrm{Rs} \\frac{600}{50}<\/span><\/span>= Rs 12<\/span><\/p>\n Pencil B\u00a0<\/span><\/p>\n Number of pencils in 1 packet = 5<\/span><\/p>\n Number of pencils in 4 packets = 5 \u00d7 4 = 20<\/span><\/p>\n Cost of 20 pencils = Rs 500<\/span><\/p>\n Cost of 1 pencil =\\operatorname{Rs} \\frac{200}{20}<\/span><\/span>= Rs 25<\/span><\/p>\n Since, Rs 25 > Rs 12.<\/span><\/p>\n \u2234 Cost of 1 Pencil B > Cost of 1 Pencil A.<\/span><\/p>\n Hence, Pencil B is costlier than Pencil A.<\/span><\/span><\/p>\n <\/p>\n 6. Pooja can complete building the dollhouse in 6 days and Vickey can complete building the same in 12 days. In how many days the construction of the dollhouse can be completed if they work together?<\/strong><\/p>\n Fraction of work done by Pooja in 6 days = 1<\/span><\/p>\n Fraction of work by Pooja in 1 day =\\frac{1}{6}<\/span><\/span><\/p>\n Fraction of work by Vickey in 12 days = 1<\/span><\/p>\n Fraction of work by Vickey in 1 day =\\frac{1}{12}<\/span><\/span><\/p>\n \u2234\u00a0 Fraction of work done by Vickey and Pooja together in 1 day =\\frac{1}{6}+\\frac{1}{8}=\\frac{1}{4}<\/span><\/span>\u00a0<\/span><\/p>\n No. of days taken by both of them to complete \\frac{1}{4}<\/span><\/span>of the total work = 1<\/span><\/p>\n No. of days taken by both of them to complete the total work =1 \\times \\frac{4}{1}=4<\/span> <\/span>days<\/span><\/p>\n Hence Vickey and Pooja can complete the dollhouse in 4 days if they work together.<\/span><\/p>\n <\/p>\n 7. A motorbike can travel from point A to point B in 2 hours if the speed of the motorbike is 60kmph. Find the time taken by the motorbike to travel the same distance if the speed is 40kmph?<\/strong><\/p>\n With the decrease in speed, the time taken by the motorbike will increase to travel from point A to point B.<\/span><\/p>\n Time taken by the motorbike\u00a0 when the speed is 60 kmph = 2 hours<\/span><\/p>\n Hence, time taken by the motorbike when the speed is 1 kmph = 2 \\times 60<\/span><\/span>hours = 120 hours<\/span><\/p>\n \u2234\u00a0 Time taken by the motorbike when the speed is 40 kmph =120 \\div 40<\/span><\/span>= 3 hours<\/span><\/p>\n Hence, the time taken by the motorbike when the speed is 40 kmph is equal to 3 hours.<\/span><\/p>\n <\/span><\/p>\n 8. Work can be completed by 4 men in 6 days. How many days will be required to complete the work by 3 men?<\/strong><\/p>\n With the increase in the number of men, the number of days to complete the work will be less.<\/span><\/p>\n No. of days taken to complete work by 4 men = 6 days<\/span><\/p>\n \u21d2 No. of days taken to complete the work by 1 man = 6 \u00d7 4 days = 24 days\u00a0<\/span><\/p>\n \u2234 No. of days taken to complete the work by 3 men =\\frac{24}{3}<\/span><\/span>days = 8 days<\/span><\/p>\n Hence, 3 men will take 8 days to complete the work.<\/span><\/p>\n <\/p>\n [\/et_pb_text][\/et_pb_column][et_pb_column type=”2_5″ module_id=”stickysideR” admin_label=”Column R” _builder_version=”4.10.4″ _module_preset=”default” background_color=”#fdefe0″ custom_padding=”25px|25px|25px|25px|true|true” sticky_position=”top” sticky_offset_top=”-280px” sticky_limit_top=”row” sticky_limit_bottom=”row” sticky_position_tablet=”none” sticky_position_phone=”none” sticky_position_last_edited=”on|desktop” sticky_limit_bottom_tablet=”” sticky_limit_bottom_phone=”” sticky_limit_bottom_last_edited=”on|phone” border_radii=”on|15px|15px|15px|15px” box_shadow_style=”preset3″ global_colors_info=”{}”][et_pb_image src=”https:\/\/eistudymaterial.s3.amazonaws.com\/1080×1080.png” alt=”Free Trial banner” title_text=”Mindspark Free Trial Banner” url=”https:\/\/mindspark.in\/free-trial” align=”center” module_class=”adsimg” _builder_version=”4.11.1″ _module_preset=”default” custom_padding=”||||false|false” border_radii=”on|10px|10px|10px|10px” global_colors_info=”{}” transform_styles__hover_enabled=”on|hover” transform_scale__hover_enabled=”on|hover” transform_translate__hover_enabled=”on|desktop” transform_rotate__hover_enabled=”on|desktop” transform_skew__hover_enabled=”on|desktop” transform_origin__hover_enabled=”on|desktop” transform_scale__hover=”102%|102%”][\/et_pb_image][et_pb_text admin_label=”Explore Other Topics [\/et_pb_text][et_pb_text _builder_version=”4.10.7″ _module_preset=”default” text_line_height=”2.2em” link_font_size=”16px” custom_margin=”||0px||false|false” custom_padding=”10px|15px|10px|28px|true|false” locked=”off” global_colors_info=”{}”]<\/p>\n [\/et_pb_text][et_pb_text admin_label=”Related Concepts [\/et_pb_text][et_pb_text _builder_version=”4.13.1″ _module_preset=”default” text_line_height=”2.2em” link_font_size=”16px” custom_margin=”||0px||false|false” custom_padding=”10px|15px|10px|28px|true|false” locked=”off” global_colors_info=”{}”]<\/p>\n\n
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Solved Examples\u00a0<\/b><\/h2>\n
<\/span><\/span><\/p>\n
<\/span>Number of Pencils in 1 packet = 10<\/span><\/p>\n
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\n” _builder_version=”4.9.11″ _module_preset=”default” header_font=”|700|||||||” header_font_size=”25px” text_orientation=”center” custom_margin=”0px||0px||true|false” custom_padding=”8px|15px|0px|15px|false|true” locked=”off” global_colors_info=”{}”]<\/p>\nRelated Concepts<\/h1>\n