{"id":2887,"date":"2023-10-03T15:58:37","date_gmt":"2023-10-03T15:58:37","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=2887"},"modified":"2023-10-03T16:03:34","modified_gmt":"2023-10-03T16:03:34","slug":"week-2-ss2-first-term-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-2-ss2-first-term-mathematics-notes\/","title":{"rendered":"Week 2 &#8211; SS2 First Term Mathematics Notes"},"content":{"rendered":"<p><strong>WEEK TWO<br \/>\n<\/strong><strong>TOPIC:  PERCENTAGE ERROR<br \/>\n<\/strong><strong>CONTENT<br \/>\n<\/strong><\/p>\n<ul>\n<li>Definition of percentage error\n<\/li>\n<li>Calculation of percentage error\n<\/li>\n<li>Percentage error (range of values via approximations)\n<\/li>\n<li>Calculations on percentage error in relation to approximation\n<\/li>\n<\/ul>\n<p><strong>Definition of Percentage Error<br \/>\n<\/strong>No measurement, however, carefully made is exact (accurate) i.e if the length of a classroom is measured as 2.8m to 2 s.f the actual length may be between 2.75 and 2.85, the error of this measurement is 2.75 \u2013 2.8 or 2.85 \u2013 2.8 = + 0.05.<br \/>\n\u00a0\u00a0\u00a0\u00a0      2.75\u00a0\u00a0\u00a0\u00a0              2.85\u00a0\u00a0\u00a0\u00a0 2.8           2.9\u00a0\u00a0\u00a0\u00a0<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1558_Week2SS2Fi1.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1558_Week2SS2Fi2.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1558_Week2SS2Fi3.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1558_Week2SS2Fi4.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1558_Week2SS2Fi5.png\" alt=\"\"\/><\/p>\n<p>\u00a0Percentage error  =\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0     error\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0         X   100<br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1558_Week2SS2Fi6.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0    Actual measurement\u00a0\u00a0\u00a0\u00a0                 1\u00a0\u00a0\u00a0\u00a0<br \/>\nerror  =+  0.05<br \/>\nactual measurement  2.8<br \/>\n% error  =0.05   x  100<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0        2.8\u00a0\u00a0\u00a0\u00a0         1<br \/>\n                =1.785% = 1.79%<br \/>\n<strong>Example 2<br \/>\n<\/strong>Suppose the length of the same room is measured to the nearest cm ,280cm i.e. (280cm) calculate the percentage error.<br \/>\nMeasurement = 280cm.<br \/>\nThe range of measurement will be between 279.5cm or 280cm<br \/>\nError  =  280 \u2013 279.5   =  0.5cm<br \/>\n% error  =       error          x  100<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0        Measurement        1<br \/>\n% error  = 0.5    x   100   =  0.178%<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0280         1<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0  =   0.18% (2sf)<br \/>\n<strong>Example 3<br \/>\n<\/strong>The length of a field is measured as 500m; find the percentage error of the length if the room is measured to<br \/>\ni.\u00a0\u00a0\u00a0\u00a0nearest metre      ii.\u00a0\u00a0\u00a0\u00a0nearest 10m         iii.\u00a0\u00a0\u00a0\u00a0one significant figure.<br \/>\n<strong>Solutions<br \/>\n<\/strong>i.\u00a0\u00a0\u00a0\u00a0To the  nearest  metre<br \/>\n\u00a0\u00a0\u00a0\u00a0Measurement = 500m<br \/>\n\u00a0\u00a0\u00a0\u00a0Actual measurement = between 499.5 &#8211; 500.5<br \/>\n\u00a0\u00a0\u00a0\u00a0Error =  + 0.5m<br \/>\n\u00a0\u00a0\u00a0\u00a0% error =            error        x 100<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0measurement<br \/>\n\u00a0\u00a0\u00a0\u00a00.5  x  100         =   0.10%<br \/>\n\u00a0\u00a0\u00a0\u00a0500\u00a0\u00a0\u00a0\u00a0 1<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=    <strong>0.10%<\/strong><br \/>\n\t\tii.\u00a0\u00a0\u00a0\u00a0Nearest 10m<br \/>\n\u00a0\u00a0\u00a0\u00a0Measurement = 500m,<br \/>\n\u00a0\u00a0\u00a0\u00a0range= 495m \u2013 505m<br \/>\n\u00a0\u00a0\u00a0\u00a0error  =+ 5m<br \/>\n\u00a0\u00a0\u00a0\u00a0error  =   5    x   100    =   1%<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 500\u00a0\u00a0\u00a0\u00a0   1<br \/>\niii.\u00a0\u00a0\u00a0\u00a0To 1 s.f.<br \/>\n\u00a0\u00a0\u00a0\u00a0measurement   =  500m<br \/>\n\u00a0\u00a0\u00a0\u00a0range   =   450 \u2013 550<br \/>\n\u00a0\u00a0\u00a0\u00a0error   =   +  50<br \/>\n\u00a0\u00a0\u00a0\u00a0% error  =  50   x    100    =  10%<br \/>\n                                500          1<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<br \/>\n\u00a0<strong>Evaluation<br \/>\n<\/strong>1.         The  length  of  each  side  of  a  square  is  3.6 cm to  2s.f. (a) Write down the  smallest  and  the  largest  of  each  side.  (b) Calculate the smallest and the largest values for the perimeter.<br \/>\n             (c)  Find the possible values of the area.<\/p>\n<p>\u00a0<strong>Percentage Error (range of values via approximations)<br \/>\n<\/strong>1.\u00a0\u00a0\u00a0\u00a0Range of values measured to the nearest whole number i.e. nearest tens, hundreds etc. e.g.<br \/>\nFind the range of values of N6000 to:<br \/>\ni.\u00a0\u00a0\u00a0\u00a0nearest naira\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0N5999.50\u00a0\u00a0\u00a0\u00a0&#8211;\u00a0\u00a0\u00a0\u00a06000.50<br \/>\nii.\u00a0\u00a0\u00a0\u00a0nearest N10\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0N5995\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0&#8211;\u00a0\u00a0\u00a0\u00a06005<br \/>\niii.\u00a0\u00a0\u00a0\u00a0nearest N100\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0N5950\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0&#8211;\u00a0\u00a0\u00a0\u00a06050<br \/>\niv.\u00a0\u00a0\u00a0\u00a0nearest N1000\u00a0\u00a0\u00a0\u00a0             =\u00a0\u00a0\u00a0\u00a0N5500\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0&#8211;\u00a0\u00a0\u00a0\u00a06,500<\/p>\n<p>\u00a02.\u00a0\u00a0\u00a0\u00a0Range of values measured to a given significant figure. E.g. find the range of value of 6000 to<br \/>\n\u00a0\u00a0\u00a0\u00a01 sf\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a05500  &#8211;  6500<br \/>\n\u00a0\u00a0\u00a0\u00a02 sf\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a05950  &#8211;  6050<br \/>\n\u00a0\u00a0\u00a0\u00a03 sf\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a05995  &#8211;  6005<br \/>\n\u00a0\u00a0\u00a0\u00a05 sf\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a05999.95- 6000.05<\/p>\n<p>\u00a03.\u00a0\u00a0\u00a0\u00a0Range of values measured to a given decimal places e.g. 39.8 to a 1d.p = 39.75 \u2013 39.85.<br \/>\n\u00a0\u00a0\u00a0\u00a0Note:   if it is 1 d.p, the range of values will be in 2 d.p, if 2 d.p, the range will be in 3 d.p etc. (i.e the range = d.p + 1). The same rule is also applicable to range of values to given significant figure. <\/p>\n<p>\u00a0<strong>Evaluation<br \/>\n<\/strong>Orally: From New General Mathematics SS 2 by J. B. Channon and Co 3<sup>rd<\/sup> edition exercise 46 no. 1a \u2013 f.<\/p>\n<p>\u00a0<strong>Calculations on percentage error:<br \/>\n<\/strong><strong>Example:<br \/>\n\t\t\t\t<\/strong>Calculate the percentage error if<br \/>\n1.\u00a0\u00a0\u00a0\u00a0The capacity of a bucket is 7.5 litres to 1 d.p.<br \/>\n2.\u00a0\u00a0\u00a0\u00a0The mass of a student is 62kg to 2 s.f.<\/p>\n<p>\u00a0<strong>Solutions<br \/>\n<\/strong>1.\u00a0\u00a0\u00a0\u00a0Measurement     =   7.5litres  ( 1d.p)<br \/>\n\u00a0\u00a0\u00a0\u00a0Range of values = 7.45  &#8211; 7.55<br \/>\n\u00a0\u00a0\u00a0\u00a0Error                   =   7.5 \u2013 7.45 = 0.05<br \/>\n\u00a0\u00a0\u00a0\u00a0% error =      error            x    100<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0measurement         1<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a00.05   x   100<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 7.51<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0    =   0.67%<br \/>\n2.\u00a0\u00a0\u00a0\u00a0Measurement = 62kg (2 s.f)<br \/>\n\u00a0\u00a0\u00a0\u00a0Range of values\u00a0\u00a0\u00a0\u00a0= 61.5kg to 62.5kg<br \/>\n\u00a0\u00a0\u00a0\u00a0error                  =  6.2  &#8211;  61.5   =  0.5kg<br \/>\n\u00a0\u00a0\u00a0\u00a0% error   =          error               x    100<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0measurement\u00a0\u00a0\u00a0\u00a0      1<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a00.5    x    100    =   <strong>0.81%<\/strong><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 62            1<\/p>\n<p>\u00a0<strong>EVALUATION<br \/>\n<\/strong>1. Calculate correct to 2 s.f. the percentage error in approximately 0.375 to 0.4.<\/p>\n<p>\u00a0<strong>GENERAL EVALUATION \/ REVISION QUESTION<br \/>\n<\/strong>1. A metal rod   was   measured as 9.20 m. If the real length is 9.43 m, calculate the percentage error to 3 s.f<br \/>\n2.A  student  measures  the  radius  of  a  circle  as  1.46 cm  instead  of   1.38 cm. Calculate  the  percentage  error.<br \/>\n3.The  weight  of  sugar  was  recorded  as  8.0 g  instead  of  8.2 g.  What is the percentage error?<br \/>\n4.A  student  mistakenly  approximated  0.03671  to  2 d.p  instead  of   2 s.f. What is the percentage error correct to 2  s.f<br \/>\n5.A  man&#8217;s  weight  was  measured  as  81.5 kg  instead  of  80 kg. Find the percentage error in the measurement.<\/p>\n<p>\u00a0<strong>WEEKEND ASSIGNMENT<br \/>\n<\/strong>What is the error in the following measurement<br \/>\n1. The distance between two towns is 60km to the nearest km.  (a) 5km     (b) 0.5km     (c) 8.3km     (d) 0.83km<br \/>\n2. The area of a classroom is 400m2 to 2 s.f.          (a) 50m<sup>2<\/sup>   (b) 1.25m<sup>2<\/sup>(c) 2.5m<sup>2<\/sup>     (d)  5m<sup>2<\/sup><br \/>\n\t\t3. A sales girl gave a girl a balance of N1.15 to a customer instead of N1.25, calculate the % error.<br \/>\n4. A student measured the length of a room and obtained the measurement of 3.99m, if the percentage error of his measurement was 5% and his own measurement was smaller than the length, what is the length of the room?(a) 3.78m   (b) 3.80m   (c) 4.18m   (d) 4.20m<br \/>\n5. A man is 1.5m tall to the nearest cm, calculate his percentage error.<br \/>\n(a)  0.05cm    (b) 0.33%   (c) 0.033%   (d)  0.05cm<\/p>\n<p>\u00a0<strong>THEORY<br \/>\n<\/strong>1.\u00a0\u00a0\u00a0\u00a0A classroom is 10m by 10m; a student measured a side as 9.5m and the other side as 10m and uses his measurement to calculate the area of the classroom. Find the percentage error in a. the length of one of the sidesb. the area of the room<br \/>\n2.\u00a0\u00a0\u00a0\u00a0Instead of recording the number 1.23cm for the radius of a tube, a student recorded 1.32cm, find the percentage error correct to 1 d.p.<\/p>\n<p>\u00a0<strong>Reading Assignment<br \/>\n<\/strong>Essential Mathematics for SSS2, pages 13-22, Exercise 2.4<\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<strong><br \/>\n\t\t\t<\/strong>\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"<p>WEEK TWO TOPIC: PERCENTAGE ERROR CONTENT Definition of percentage error Calculation of percentage error Percentage&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,233],"tags":[],"class_list":["post-2887","post","type-post","status-publish","format-standard","hentry","category-posts","category-first-term-ss2-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2887","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=2887"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2887\/revisions"}],"predecessor-version":[{"id":2888,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2887\/revisions\/2888"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=2887"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=2887"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=2887"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}