/Matrix [1 0 0 1 0 0] /FormType 1 Q /Length 70 q endstream /F3 17 0 R /ProcSet[/PDF] >> stream /Subtype /Form q 1 i /F3 17 0 R /Meta165 179 0 R /Meta36 Do 1.007 0 0 1.007 67.753 799.486 cm 1.014 0 0 1.007 531.485 776.149 cm /ProcSet[/PDF] endobj q /F4 12.131 Tf Q /Subtype /Form /Meta382 396 0 R 1.005 0 0 1.007 79.798 746.789 cm /MediaBox [0 0 767.868 993.712] q stream Q q Q >> /Meta304 Do endobj 53 0 obj stream stream /F1 12.131 Tf /Subtype /Form q /Type /XObject stream /Resources<< /F3 17 0 R /Resources<< /Resources<< >> 0 G Q 367 0 obj q q << << /CreationDate (D:20140515121932-04'00') q endobj 0.486 Tc 1 i ET /Resources<< /ProcSet[/PDF/Text] q /Resources<< BT 0 G Q q >> q Q endobj 1.007 0 0 1.007 271.012 277.035 cm endobj stream /BBox [0 0 88.214 16.44] /BBox [0 0 88.214 35.886] Q /Type /XObject Q /BBox [0 0 15.59 16.44] 0.838 Tc 0.369 Tc /Meta226 Do (-11) Tj q q q /Resources<< >> q Q << Q /Type /XObject Q 0 g /Type /XObject /Resources<< /Length 69 /BBox [0 0 15.59 29.168] /FormType 1 q Q >> /Type /XObject stream q -0.03 Tw q q /Subtype /Form >> /Matrix [1 0 0 1 0 0] 0.458 0 0 RG /FormType 1 NCERT Exemplar Class 7 Maths Solutions Chapter 4 Simple Equations Directions: In the questions 1 to 18, there are four options out of which only one is correct. /BBox [0 0 88.214 35.886] /Meta290 Do /Meta242 Do 14.966 20.154 l /Subtype /Form 0 g q /Size 447 << /Type /XObject 0.564 G q q << endobj q 0 g /Subtype /Form endstream q ET 0 G /Length 69 1 g q ET 1.007 0 0 1.006 130.989 690.329 cm /Resources<< ET /I0 51 0 R >> << Q [(A numb)-16(er subtract)-15(ed from )] TJ endstream Q 201 0 obj Q ET /Meta314 Do /F3 17 0 R q endstream endobj q q q endobj /BBox [0 0 15.59 16.44] /Type /XObject 260 0 obj >> 370 0 obj (13) Tj q /Length 70 >> /Subtype /Form 0.382 Tc >> Q /Length 67 /Subtype /Form q Q /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] /FormType 1 endstream Q 1.007 0 0 1.007 45.168 746.789 cm 1.007 0 0 1.007 551.058 330.484 cm Q q /Type /XObject /Resources<< /Type /XObject /Type /XObject Q 0 g q /F3 12.131 Tf /Length 118 Q /Matrix [1 0 0 1 0 0] 0.68 Tc stream stream If you are unsure of the county, call the Administrative Office of the Court at (919) 890-1000.Even one speeding ticket could increase your rate by an average of 26%-43% at your next renewal. Q /Type /XObject Q 0 G endstream Q /FormType 1 /Meta184 198 0 R >> /F3 12.131 Tf 15.731 5.336 TD /Matrix [1 0 0 1 0 0] /I0 Do 0 g 0.737 w endstream 1 i endstream 1 i >> Q /Font << /Subtype /Form /Subtype /Form /Widths [ 250 0 385 0 0 0 0 0 0 0 0 0 0 0 /Length 59 0.458 0 0 RG >> /Subtype /Form /Type /XObject 1 i /F4 36 0 R /F4 36 0 R The ratio of a number to fifteen 4. /Meta34 Do endobj << Q Q /Matrix [1 0 0 1 0 0] (-) Tj 1 i 0 G /Subtype /Form << /FormType 1 endobj /Length 65 q q 1.014 0 0 1.006 531.485 437.384 cm q 12.727 5.203 TD >> Q q /Subtype /Form << << /Meta106 Do stream /Type /XObject BT >> Q /FormType 1 << 0 G >> /Meta134 Do /FormType 1 Solution: Let the number be x. /FormType 1 /Resources<< >> q /Resources<< 0.37 Tc >> /FormType 1 /Meta421 Do 0 5.203 TD ET /BBox [0 0 88.214 16.44] << 0 g 1.005 0 0 1.006 45.168 879.284 cm q 1.007 0 0 1.007 271.012 636.879 cm << Q 1.502 5.203 TD q 1 i Q Q Q 296 0 obj 0 G /Meta280 Do 95 0 obj /Type /XObject /Subtype /Form Thrice a number decreased by 5 exceeds twice the number by 1 is . /Subtype /Form Q ET 0 G /Type /XObject Q /Matrix [1 0 0 1 0 0] q 0.738 Tc 1.007 0 0 1.007 45.168 730.228 cm /Subtype /Form 0 G endobj << /Type /XObject q /Type /XObject /F3 12.131 Tf The observed mean MetS-Z was at inclusion 0.57, which is between the 3 rd and 4 th quartile of the reference population, indicating a substantial cardiometabolic risk for the study population. << q /Subtype /Form 1.007 0 0 1.007 271.012 849.172 cm /Descent -277 q /Length 85 /Meta326 340 0 R 0 g 151 0 obj /Meta193 207 0 R 1 i 1.007 0 0 1.007 411.035 277.035 cm 0.564 G << q Q BT /Subtype /Form Q Q /Meta137 151 0 R Q 1.005 0 0 1.015 45.168 53.449 cm 323 0 obj Q /ProcSet[/PDF/Text] /Resources<< 226 0 obj BT /BitsPerComponent 1 >> >> Q /Resources<< /Subtype /Form << (+) Tj /Matrix [1 0 0 1 0 0] ( x) Tj << 1 i /Font << /Subtype /Form 1 i 1.502 5.203 TD q << /ProcSet[/PDF] /DecodeParms [<> ] /Meta277 291 0 R /BBox [0 0 17.177 16.44] 0.134 Tc 0.564 G >> /Meta338 Do BT << /Type /XObject /Resources<< q endstream /MaxWidth 2000 /Type /XObject q /F4 12.131 Tf /Length 104 q Q /F3 17 0 R endstream /Type /XObject 1.007 0 0 1.007 67.753 293.596 cm /Meta246 260 0 R BT endobj 0 g (58) Tj /ProcSet[/PDF] q >> /F3 12.131 Tf ET endobj >> >> BT 1.007 0 0 1.007 551.058 703.126 cm >> 152 0 obj >> << /Matrix [1 0 0 1 0 0] /Length 64 endstream >> 1.007 0 0 1.007 551.058 523.204 cm Q stream /Type /XObject /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] q 50 0 obj >> << stream 0.564 G << 393 0 obj /Matrix [1 0 0 1 0 0] In the problem above, x is a variable. q 0 G >> /Meta110 Do Q q Q 12.727 5.203 TD Q /ProcSet[/PDF/Text] /BBox [0 0 17.177 16.44] /Meta150 Do 0.737 w /ProcSet[/PDF/Text] Q 0 5.203 TD stream /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 381 0 obj /Type /FontDescriptor 0.737 w /Font << /F3 12.131 Tf /Subtype /Form /Resources<< 0 w /Matrix [1 0 0 1 0 0] Q << Q /Length 79 /Length 12 /Meta112 126 0 R >> 0 w >> /ProcSet[/PDF] /Length 16 1 g 189 0 obj /F3 17 0 R (D\)) Tj /Meta100 114 0 R >> /Subtype /Form stream /Meta268 282 0 R 83 0 obj Q /FormType 1 /Meta87 Do /Resources<< 1 g Q 0 G /Meta238 Do 20.21 5.203 TD >> /Type /XObject endobj Q 261 0 obj /FormType 1 /Meta358 372 0 R /Length 16 /Meta354 368 0 R /Subtype /Form 0 g /Resources<< 0 g q 0 G q /Meta35 Do 1st step. << 1 i 0 G endstream /Matrix [1 0 0 1 0 0] /Subtype /Form /ProcSet[/PDF/Text] 1 i 0 0 0 722 0 0 0 611 0 722 0 333 0 722 611 0 stream 1.005 0 0 1.007 45.168 889.071 cm << Q ET endstream /FormType 1 1 i /Length 16 /Meta43 57 0 R Mat endobj 0 G >> Q /F1 12.131 Tf 119 0 obj Q Q Q /Type /XObject stream Q /Type /XObject /Length 118 17 0 obj (A\)) Tj /F3 12.131 Tf /ProcSet[/PDF/Text] >> BT /ProcSet[/PDF/Text] /Matrix [1 0 0 1 0 0] 0.564 G /Meta410 426 0 R 0.51 Tc Q 1.007 0 0 1.007 551.058 703.126 cm Q BT /Subtype /Form 224 0 obj Answer only. Q Ten divided by a number 5. /Length 16 q BT ET stream stream << /ProcSet[/PDF/Text] /Subtype /Form /Type /XObject endstream 0 G /BBox [0 0 17.177 16.44] /ProcSet[/PDF/Text] endobj q /Font << /BBox [0 0 15.59 16.44] /BBox [0 0 88.214 16.44] q q /F3 17 0 R endstream 0 g /Resources<< q >> stream [(F)-22(ive)] TJ /BBox [0 0 534.67 16.44] stream q ET endobj /Filter [/CCITTFaxDecode] stream Q q Q q 1.007 0 0 1.006 411.035 510.406 cm /Length 69 0 5.203 TD 0.737 w /Meta164 178 0 R /FormType 1 0 G ET /Length 69 Q Q /Type /XObject endobj 0.564 G Q Q q stream 0 G >> q /BBox [0 0 88.214 16.44] /Meta57 Do /I0 51 0 R stream 0 w >> (2\)) Tj 65.906 4.894 TD q >> /Length 69 /BBox [0 0 30.642 16.44] << 0.564 G q Q 59 0 obj Q /Subtype /Form /F3 12.131 Tf >> 0 G /XObject << /Meta15 26 0 R /ProcSet[/PDF] Q >> endstream >> endstream /Meta412 Do 26.957 5.203 TD (B\)) Tj << /Subtype /Form q /BBox [0 0 15.59 29.168] 1 i /BBox [0 0 17.177 16.44] >> 407 0 obj /Type /XObject q endobj stream /Meta330 344 0 R BT 0 w stream /Subtype /Form Q /Meta420 Do /Subtype /Form Q 1.007 0 0 1.006 551.058 437.384 cm /Subtype /Form Q /Type /XObject >> /Type /XObject /Font << /ProcSet[/PDF/Text] stream << /Font << q The difference between six and a number divided by nine 10. /Type /XObject /Length 69 q q /FormType 1 >> 259 0 obj >> /ProcSet[/PDF] >> /Length 54 Q Q /Meta369 Do >> /Meta82 96 0 R /Meta80 Do /BBox [0 0 88.214 16.44] /Length 87 >> Q /FormType 1 /BBox [0 0 17.177 16.44] 0 5.203 TD 0 g << BT >> Q q 0.271 Tc 0 g /Meta166 180 0 R /Subtype /Form /Length 116 1 i /Type /XObject q 133 0 obj stream q Q q /Subtype /Form q << >> 222 0 obj /F1 12.131 Tf /Meta122 136 0 R 0.369 Tc /Length 69 >> /Meta269 Do /Font << /Font << Q /Subtype /Form 0 G 162 0 obj endobj 0 g /Length 78 /Meta162 Do 1.008 0 0 1.007 654.946 293.596 cm Q /ProcSet[/PDF] << You can also contact the clerk of court in the county you received the ticket. q 0 g 0 G 102 0 obj 30.699 4.894 TD That's the problem with, That's why I prefer mathematics in general, - at least in equation and formula form -. /F3 12.131 Tf /Type /XObject q /ProcSet[/PDF/Text] Q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] q << /Subtype /Form /Font << /Font << 0 g endobj stream 0 0 Similar questions Find the number which when decreased by 8% becomes 506. 1 i /Subtype /Form 31 0 obj >> /I0 Do 356 0 obj /Matrix [1 0 0 1 0 0] /Meta98 112 0 R /Type /XObject /F3 17 0 R q Calculate a 15% decrease from any number. q >> /Matrix [1 0 0 1 0 0] 0 g 0.369 Tc 3.742 5.203 TD endobj 1.005 0 0 1.007 102.382 473.519 cm >> /F4 12.131 Tf /ProcSet[/PDF] >> Q >> q ET 125.064 4.894 TD /F3 12.131 Tf Q >> endobj Q 0 G 0 4.78 TD 9.723 5.336 TD >> << /Meta307 Do >> endobj BT /ProcSet[/PDF/Text] /CapHeight 692 BT Q 0.369 Tc endstream /Matrix [1 0 0 1 0 0] /F3 17 0 R /Type /XObject q Q 1.014 0 0 1.007 111.416 703.126 cm stream 344 0 obj /F3 17 0 R q 0.425 Tc stream >> Q 1 g (A\)) Tj /Meta129 Do 1.007 0 0 1.007 411.035 849.172 cm /BBox [0 0 673.937 14.853] Q /ProcSet[/PDF/Text] 0 G /Resources<< /Matrix [1 0 0 1 0 0] 0.369 Tc 0.737 w << Q /Meta298 Do Q /Type /XObject 1.007 0 0 1.007 130.989 383.934 cm /Subtype /Form /F4 36 0 R /ProcSet[/PDF/Text] 19.474 20.154 l 0 g >> /Matrix [1 0 0 1 0 0] q >> q /ProcSet[/PDF] ET Q Q 1.014 0 0 1.007 251.439 383.934 cm ET >> >> >> endobj Q /F3 12.131 Tf 1.007 0 0 1.007 271.012 383.934 cm Q /Meta17 Do >> /Type /XObject /Length 70 /BBox [0 0 88.214 16.44] q q q << /Length 54 ET stream 317 0 obj /Resources<< /F4 36 0 R q >> Q >> q << (D\)) Tj /BBox [0 0 88.214 16.44] 20.21 5.203 TD endobj /Font << >> Q q q endstream 205 0 obj (40) Tj Q /Leading 150 >> q /F3 12.131 Tf /Matrix [1 0 0 1 0 0] /BBox [0 0 15.59 16.44] /Type /XObject /Meta171 185 0 R /Subtype /Form >> /F1 12.131 Tf q (-23) Tj (x) Tj q >> >> Q stream Q Q << stream 1 i /FormType 1 Q /Resources<< >> /Type /XObject /BBox [0 0 88.214 16.44] /Font << 1 i 831 0 0 0 0 0 613 0 0 0 0 0 0 333 0 333 0.458 0 0 RG 1 i 1 i 34 0 obj 387 0 obj q /BBox [0 0 534.67 16.44] endstream 290 0 obj >> /Length 68 >> /Length 63 >> stream 1.007 0 0 1.007 411.035 636.879 cm 1.005 0 0 1.007 102.382 347.046 cm [(Answe)20(r Key)] TJ q 1 g /Meta297 311 0 R /Type /XObject >> endobj 1 i q ET endstream 1 g /Font << 64 0 obj Q /Length 69 >> 415 0 obj /Length 69 /Meta393 409 0 R /Length 59 /Length 12 endobj /Matrix [1 0 0 1 0 0] q << (-11) Tj /Meta144 158 0 R 1.014 0 0 1.007 391.462 523.204 cm stream >> /Length 74 Q stream << << /Meta340 Do ET SOLUTION: twice a number decreased by 8 is equal to the number increased by 10. find the number. << /FormType 1 q /Meta241 255 0 R 235 0 obj >> /Meta315 Do 1.007 0 0 1.006 551.058 437.384 cm (13) Tj We are asked to find the number, so, we could assign the number as "x". << 2.238 5.203 TD Q /F3 17 0 R >> 0 g /Subtype /Form /Type /XObject 3 0 obj ET 58 0 obj >> Q Q q << stream stream /F3 12.131 Tf /BBox [0 0 15.59 29.168] << 157 0 obj << endobj /Length 68 /F3 12.131 Tf /Resources<< >> /ProcSet[/PDF/Text] Q 1.014 0 0 1.007 531.485 703.126 cm stream 257 0 obj 1 g /ProcSet[/PDF/Text] Q endstream Q /Matrix [1 0 0 1 0 0] Q /Type /XObject 6.746 5.203 TD 1.007 0 0 1.007 130.989 583.429 cm 0.297 Tc >> >> stream -0.021 Tw /Subtype /Form 1 g 0 g /Font << >> Q /F2 12.131 Tf 0 w stream /Subtype /Form /Subtype /Form Q /Font << /Resources<< /F3 12.131 Tf << Q /Matrix [1 0 0 1 0 0] /Meta132 146 0 R >> /Meta103 Do Q Q /Font << stream /Meta67 Do stream << /Ascent 1050 0.458 0 0 RG /BBox [0 0 88.214 16.44] BT >> q 0.297 Tc Q /Subtype /Form endobj /Subtype /Form 1 i Q /Resources<< 0.737 w stream endstream 0.564 G ET 229 0 obj /Font << q q BT (3) Tj stream << ET /F3 17 0 R /Matrix [1 0 0 1 0 0] /Subtype /Form /FormType 1 q /BBox [0 0 88.214 16.44] /Type /XObject 1 i q /BBox [0 0 88.214 16.44] 1.502 5.203 TD endstream 1.007 0 0 1.007 271.012 849.172 cm /Length 59 /F1 7 0 R /Meta90 104 0 R /FormType 1 q /Matrix [1 0 0 1 0 0] 0 g 0 g 1 i Q /F1 7 0 R 1.007 0 0 1.007 130.989 583.429 cm q >> 0 G /Font << 1.007 0 0 1.007 130.989 383.934 cm 0.458 0 0 RG stream q endstream /ProcSet[/PDF] Q Q /Resources<< /Length 80 32.201 5.203 TD (x) Tj /BBox [0 0 88.214 16.44] (viii) A number divided by 8 gives 7. /ProcSet[/PDF/Text] >> 118.317 5.203 TD << >> /Type /XObject >> 1.007 0 0 1.007 130.989 636.879 cm (-) Tj >> q /Meta152 Do 2.238 5.203 TD /BBox [0 0 15.59 16.44] q >> ET q /FormType 1 1 i /Subtype /Form >> /Meta423 Do /Length 59 /Font << Q /BBox [0 0 549.552 16.44] << 1 i 1 i Use the variable g to represent Gails age. /Meta358 Do >> << Q B. Q 0.564 G /FormType 1 << 1 i q 0.564 G /Subtype /Form /F3 17 0 R 0 g /F3 12.131 Tf /Meta125 139 0 R 1.502 24.649 TD q stream q /FormType 1 1 i /Subtype /Form 0 G 0 w 0 5.203 TD >> /Resources<< q /Type /XObject q Q Q /BBox [0 0 30.642 16.44] /ProcSet[/PDF/Text] /Meta167 181 0 R >> /Length 69 q pidemiologi i Infekcionnye Bolezni. Q << The sum of a number and 2 is 6 less than twice that number. /Subtype /Form 1 i endobj /Type /XObject /Type /XObject /Resources<< ET endobj /Subtype /Form endstream Q /ProcSet[/PDF/Text] 0.68 Tc 0.737 w Q ET /Resources<< /Meta83 Do /Font << q /Meta159 173 0 R Q q /AvgWidth 401 q /Meta213 Do /Subtype /Form /Type /XObject 0 g Q 197 0 obj endstream q q /Matrix [1 0 0 1 0 0] , Prove the following ET q /Type /XObject /Length 69 /Resources<< 0 g 1.007 0 0 1.007 654.946 799.486 cm 13.464 5.203 TD Q /Matrix [1 0 0 1 0 0] Q (B\)) Tj Six subtracted from a number 6. Q 1.007 0 0 1.007 130.989 383.934 cm Q 0.737 w 15.731 5.336 TD /Subtype /Form /ProcSet[/PDF/Text] q ] /Meta334 348 0 R /BBox [0 0 88.214 16.44] /Matrix [1 0 0 1 0 0] q >> /Font << q 1 i 1.014 0 0 1.007 111.416 330.484 cm Q >> 0 g Double or twice a number means 2x, and triple or thrice a number means 3x. 0 g endstream BT /Type /XObject q 0.285 Tc >> /Resources<< /Length 118 /BBox [0 0 88.214 35.886] q /ProcSet[/PDF/Text] /Type /XObject /Resources<< BT endobj >> 0.382 Tc /BBox [0 0 17.177 16.44] /Font << 2. endobj endstream 187 0 obj 1.007 0 0 1.007 130.989 523.204 cm /F4 36 0 R /Resources<< Q endstream 382 0 obj stream /Resources<< /Producer (PDF-XChange 4.0.0186.0000 \(Windows\)) 1 i >> >> /Resources<< S 77 0 obj stream q q 1.005 0 0 1.007 102.382 726.464 cm endstream q Thrice of a number = 3x. /ProcSet[/PDF] 398 0 obj /Subtype /Form BT endstream q /BBox [0 0 30.642 16.44] q 1.007 0 0 1.007 45.168 846.161 cm /Length 69 endobj /BBox [0 0 549.552 16.44] /FormType 1 /Meta302 316 0 R /BBox [0 0 88.214 16.44] 115 0 obj 1.014 0 0 1.007 391.462 277.035 cm q 0.311 Tc /F3 12.131 Tf /ProcSet[/PDF/Text] /F4 12.131 Tf endobj /Resources<< /Subtype /Form /Resources<< endobj stream stream 10 0 obj /Resources<< /Subtype /Form q 1 i /Resources<< 1 g 73 0 obj >> /Subtype /Form 0 0 0 0 0 722 0 606 0 0 833 389 0 0 606 1000 /Length 79 >> 1.007 0 0 1.007 271.012 523.204 cm >> endstream Q >> q q >> 314 0 obj endobj endobj 430 0 obj endobj /Length 69 /ProcSet[/PDF] 1 i stream /Meta370 Do /BBox [0 0 88.214 16.44] >> 324 0 obj q Q /Type /Page 1 i 0.458 0 0 RG 0 g /Font << /Length 65 endobj /FormType 1 0 G stream << /ProcSet[/PDF/Text] endstream >> /ProcSet[/PDF/Text] q << >> q Q /I0 51 0 R /Matrix [1 0 0 1 0 0] >> endobj /Type /XObject 0 G Q ET /Meta326 Do stream /F3 17 0 R << q /ProcSet[/PDF] /BBox [0 0 88.214 16.44] 112 0 obj 1 i /BBox [0 0 88.214 16.44] Q q >> /ProcSet[/PDF/Text] /BBox [0 0 88.214 35.886] endstream 103 0 obj 0.68 Tc Q /BBox [0 0 673.937 68.796] /Type /XObject endobj 1.014 0 0 1.007 251.439 450.181 cm Patients' reasons for declining screening were not collected . /F4 12.131 Tf stream /ProcSet[/PDF] 1.502 7.841 TD stream /Subtype /Form >> 0.369 Tc Q 0 5.203 TD /F3 12.131 Tf /Meta59 Do /Matrix [1 0 0 1 0 0] /Type /XObject q << (C\)) Tj /ProcSet[/PDF/Text] /Meta244 Do 0 g /BBox [0 0 30.642 16.44] /Subtype /Form >> /Meta42 56 0 R endobj /F3 17 0 R Q /ProcSet[/PDF/Text] 0.486 Tc 0 G q /Subtype /Form q 1.005 0 0 1.007 102.382 872.509 cm /Resources<< /Subtype /Form q /Meta228 Do << 24.718 8.18 TD 1 i 0.458 0 0 RG 271 0 obj 1 i q /Encoding /WinAnsiEncoding /Matrix [1 0 0 1 0 0] , Point (-2, 4) lies on the graph of the equation 3y = kx + 4. /Meta142 156 0 R 0 g Q endobj endobj >> stream /ProcSet[/PDF/Text] q >> ET /Matrix [1 0 0 1 0 0] 1 i /Length 16 >> Q q 0.737 w stream q 0 w 0 g >> BT /Meta391 407 0 R Q q 0 20.154 m /Font << q /F3 12.131 Tf Q (x ) Tj q /Meta387 Do /Meta136 Do /Matrix [1 0 0 1 0 0] /F3 12.131 Tf 1 i /F3 17 0 R 1 i >> /BBox [0 0 15.59 29.168] Q /Matrix [1 0 0 1 0 0] /Resources<< /Font << BT >> /Resources<< q 1 i /BBox [0 0 88.214 16.44] q Q Q stream /Font << /Resources<< /ProcSet[/PDF/Text] 1.007 0 0 1.007 45.168 796.475 cm /Matrix [1 0 0 1 0 0] >> Q /Matrix [1 0 0 1 0 0] /Meta383 397 0 R /F4 36 0 R 0 5.203 TD /BBox [0 0 15.59 16.44] q >> BT /Subtype /Form Q Class 12 Class 11 Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 NCERT Class 9 Mathematics 619 solutions /Matrix [1 0 0 1 0 0] 0 g 0 4.894 TD q stream 1.014 0 0 1.007 391.462 450.181 cm /Matrix [1 0 0 1 0 0] /Encoding /WinAnsiEncoding /Length 16 >> endobj q /BBox [0 0 534.67 16.44] q /Font << 1.005 0 0 1.007 45.168 889.071 cm 0 g ET q stream >> /Font << Q 0 g /F3 12.131 Tf /Meta86 Do /Subtype /Form 0 G 0.486 Tc 25.454 5.203 TD /ProcSet[/PDF/Text] endstream /Font << /Font << 357 0 obj 1.007 0 0 1.007 551.058 523.204 cm /AvgWidth 657 /Resources<< Negative thirteen decreased by 3 times a number x. >> 128 0 obj /ProcSet[/PDF/Text] 1.007 0 0 1.007 67.753 546.541 cm /Font << >> /BBox [0 0 88.214 16.44] Q q /Matrix [1 0 0 1 0 0] endstream >> 0 5.203 TD q /ProcSet[/PDF] /Type /XObject /BBox [0 0 88.214 16.44] q /Meta403 419 0 R q Q /F3 17 0 R q q Q /Meta258 Do 339 0 obj 1.007 0 0 1.007 45.168 730.228 cm /Resources<< /Type /XObject >> /Font << /ProcSet[/PDF/Text] q 155 0 obj << 126 0 obj /Meta17 28 0 R q 0 g stream /Length 67 66 0 obj Q /Length 54 Summary Results for the Initial Round of Lung Cancer Screening in 8 LCSDP Sites . 0 5.203 TD << /FormType 1 Q Q << Twice a number decreased by 8 gives 58. >> BT 0.68 Tc /Meta169 Do Q /Length 63 >> 1.005 0 0 1.007 79.798 862.723 cm 0 g /ProcSet[/PDF/Text] q 1 i q endobj endobj stream 0 5.203 TD /Meta250 Do /BBox [0 0 673.937 16.44] /FormType 1 /ProcSet[/PDF] BT /Meta262 Do 1 i endstream << q /Matrix [1 0 0 1 0 0] endstream /Length 103 409 0 obj /Resources<< 299 0 obj /Meta69 83 0 R /Length 69 /Resources<< ET 0 G stream 0 g /F3 17 0 R /Meta183 197 0 R /Type /XObject 0 5.336 TD Q q /Length 69 q /Resources<< >> ET q /Font << endstream /Meta135 Do 20.21 5.203 TD /Resources<< 120 0 obj The symbols 17 + x = 68 form an algebraic equation. endobj /ProcSet[/PDF/Text] 0 G Q /Type /XObject /FormType 1 Q >> /Font << /F3 17 0 R << /Matrix [1 0 0 1 0 0] stream Q q /FormType 1 q stream /Length 65 endstream 1.007 0 0 1.007 130.989 849.172 cm 1.014 0 0 1.006 391.462 836.374 cm /Type /XObject /FormType 1 0 g /Meta94 Do q 0 5.203 TD 0.737 w Q /Matrix [1 0 0 1 0 0] 1.005 0 0 1.013 45.168 933.487 cm (x) Tj >> /F3 12.131 Tf /F3 17 0 R /FormType 1 6.746 5.203 TD /Subtype /Form /Meta411 Do >> BT /Meta142 Do 333 0 obj /Length 59 /BBox [0 0 534.67 16.44] 0 G /ProcSet[/PDF/Text] q /ProcSet[/PDF/Text] endobj /Parent 1 0 R 1.007 0 0 1.007 411.035 583.429 cm 1.007 0 0 1.007 271.012 583.429 cm /Subtype /Form endobj 1 i /BBox [0 0 88.214 16.44] endobj endobj /F3 17 0 R Q endstream /Matrix [1 0 0 1 0 0] 1 i /FormType 1 /Type /XObject Q /F3 17 0 R (7\)) Tj Q 35.206 4.894 TD Q ET 0.737 w /BBox [0 0 30.642 16.44] Q /Resources<< /Length 16 /Meta325 Do /F3 12.131 Tf q /F3 12.131 Tf 0 g (9\)) Tj << q Twice a number decreased by . 392 0 obj q /Meta285 299 0 R /Type /XObject 0 G 1 i /Subtype /Form q /ProcSet[/PDF] 0 G /F4 12.131 Tf 1 i Q endobj endobj Q 311 0 obj Q /F3 17 0 R 0 g << q Five times a number, decreased by 58, is -23 Find the number. 0 g /Subtype /Form Q /F3 17 0 R /BBox [0 0 17.177 16.44] /Type /XObject /Type /XObject endstream 0 g >> BT << 0 g 0 g /FormType 1 /Length 12 1 g Q 0.564 G >> 0 G endobj q /Font << endobj q q q /F1 7 0 R /Type /XObject Q 0.737 w /Length 59 /Resources<< >> >> q Q Q A: Given, When six times a number is decreased by 3 , the result is 45 We have to find the number. q 0 w Q -0.486 Tw Q stream How many points did Kobe score in the season? endstream q /Type /XObject 1 i >> /FormType 1 Q Q q /Subtype /Form five times the sum of a number x and two b.) << 0 g stream /F3 12.131 Tf 0.564 G 0 G 1 i Twice the number means = 2x Twice the number increase by 8 means =2x+8 Twice the number increase by 8 is 20 then means 2x+8=20 Therefore the solution to this equation will be as follows: 2x=20-8 2x=12 Divide both sides by the coefficient of. /BBox [0 0 15.59 16.44] 1 g /F3 12.131 Tf Q /Matrix [1 0 0 1 0 0] q q /Meta37 50 0 R endobj /F3 17 0 R /ProcSet[/PDF] /Meta234 Do 1.007 0 0 1.007 130.989 330.484 cm /Length 59 /FormType 1 BT /F3 12.131 Tf 417 0 obj /F3 12.131 Tf 0 g endobj 1 g /ProcSet[/PDF] /BBox [0 0 30.642 16.44] << /Type /XObject /Matrix [1 0 0 1 0 0] /FormType 1 33 0 obj q 144 0 obj 1 i endobj /Length 16 1.005 0 0 1.007 102.382 546.541 cm q /FormType 1 Q /Font << Q /Font << /Resources<< << q /Subtype /Form /Matrix [1 0 0 1 0 0] /Meta69 Do /FormType 1 /FormType 1 /Meta190 Do Q /FormType 1 /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /FormType 1 /ProcSet[/PDF/Text] /Meta261 275 0 R stream /Matrix [1 0 0 1 0 0] /F3 12.131 Tf 3.742 24.649 TD /FormType 1 Q /ProcSet[/PDF] BT /BBox [0 0 15.59 16.44] 0 g ET 1 i 0 g (11) Tj q Q >> >> /Resources<< 108 0 obj 0.786 Tc /Type /XObject (B\)) Tj endstream 1.005 0 0 1.007 102.382 743.025 cm >> /Font << q q /Matrix [1 0 0 1 0 0] /Meta2 Do /Matrix [1 0 0 1 0 0] /Meta154 Do q q 2.238 5.203 TD 0.382 Tc q /FormType 1 /Meta72 86 0 R /F3 12.131 Tf q /Length 59 Q /Matrix [1 0 0 1 0 0] /Resources<< Q >> stream 18.708 17.593 TD >> stream >> Q [4] One half of a number decreased by fourteen is twenty-one S /BBox [0 0 88.214 16.44] /FormType 1 0.31 Tc q BT endobj q >> /Matrix [1 0 0 1 0 0] Q /Meta138 152 0 R /Length 60 Q stream Q /Meta98 Do >> /Meta33 Do << /Meta174 188 0 R endobj endobj 0 G 0 G /Resources<< << /Subtype /Form endobj Q 1.007 0 0 1.007 67.753 293.596 cm 0.68 Tc 0 g /Resources<< 0 g /Matrix [1 0 0 1 0 0] >> 118 0 obj >> 132 0 obj 0 g Choose the correct one. endstream /Resources<< /Length 16 endstream /Meta149 163 0 R /F1 12.131 Tf endstream Q 0.737 w Q << Q endobj Q /BBox [0 0 88.214 16.44] BT /Meta212 Do >> 1 i /Length 12 /Meta108 Do /Length 63 q q /Subtype /Form 1 i /Length 69 0 g 0 w /Font << /BBox [0 0 534.67 16.44] /ProcSet[/PDF] /Resources<< /F3 17 0 R /ProcSet[/PDF/Text] >> 0 G Q /F3 12.131 Tf >> /Meta426 Do 0 g /F3 12.131 Tf >> /BBox [0 0 88.214 16.44] /Resources<< /Length 60 q /Subtype /Form /Meta54 68 0 R << ET q /FormType 1 /Meta168 182 0 R /Subtype /Form Q >> endobj 1.007 0 0 1.006 411.035 690.329 cm q q /ProcSet[/PDF/Text] Phrase : Expression : 4 times some number : 4x: twice a number : 2y : one-third of some number : the product of a number and 12 : 12w: Some examples of common phrases and corresponding . /Type /XObject 134 0 obj 1.502 7.841 TD << /Matrix [1 0 0 1 0 0] Q q q 0 0 0 778 611 709 774 611 556 0 0 0 0 0 0 0 >> /Resources<< /BBox [0 0 88.214 16.44] ET << 0.838 Tc 16.469 5.336 TD >> Q /ProcSet[/PDF/Text] /Resources<< 20.21 5.203 TD /BBox [0 0 30.642 16.44] endstream 0.486 Tc q /F1 7 0 R stream >> stream /Resources<< 329 0 obj /Type /XObject << >> Q >> 304 0 obj >> /Length 68 (-4) Tj 316 0 obj /Resources<< /Font << /Matrix [1 0 0 1 0 0] Let x be a number. /ProcSet[/PDF] 1 g q the ratio of a number and 4: x/4: the quotient of a and b: a/b: five decreased by t: 5-t: 3 less than 5 times a number: 5x-3: 6 years younger than Ann, Ann's age =a: a-6: three . /Matrix [1 0 0 1 0 0] 1 i /F3 12.131 Tf q 1.014 0 0 1.007 531.485 523.204 cm /Type /XObject endstream /Length 60 >> /Meta397 413 0 R /Subtype /Form /Subtype /Form 0.486 Tc 0 g /Type /XObject q 0.786 Tc q Q >> 20.21 5.203 TD /ProcSet[/PDF] 0 g ET 1.005 0 0 1.007 102.382 473.519 cm 295.086 4.894 TD /Resources<< stream 1.007 0 0 1.007 67.753 599.991 cm Q /ProcSet[/PDF/Text] >> /Subtype /Form endstream Q >> /BBox [0 0 23.896 16.44] Q 43 0 obj 1 g /F3 12.131 Tf /FormType 1 20.975 5.336 TD Q /FormType 1 /Subtype /Form /Meta62 76 0 R >> /FormType 1 164 0 obj 0 g Q /Resources<< 1 i 0 G Q 1.007 0 0 1.006 551.058 437.384 cm stream /FormType 1 q 181 0 obj >> /Matrix [1 0 0 1 0 0] /Meta89 Do /F3 17 0 R endstream /Meta348 362 0 R q >> endstream /Type /XObject << /BBox [0 0 88.214 16.44] 0 G /F3 12.131 Tf 0 G (C\)) Tj 36 0 obj << << 234 0 obj /Meta315 329 0 R Q 303 0 obj -0.463 Tw Q >> /F3 17 0 R /Type /XObject Q endstream q Q /Matrix [1 0 0 1 0 0] 91 0 obj Q stream 0 G /Matrix [1 0 0 1 0 0] /F3 17 0 R Q /BBox [0 0 88.214 35.886] BT stream 0.458 0 0 RG Q 0 g 0.737 w Q /ItalicAngle 0 /Resources<< 0.838 Tc 0 G 159 0 obj >> stream endstream 0 G q << endstream 0 g /Meta220 Do /Subtype /Form 0.458 0 0 RG endobj q /F3 12.131 Tf >> /Type /XObject 1 i Q /Meta253 267 0 R BT q /Resources<< /Matrix [1 0 0 1 0 0] /Subtype /Form << stream /FormType 1 /Resources<< q Q endstream endstream q /ProcSet[/PDF] >> /F3 12.131 Tf /Length 16 0 5.203 TD Q /Subtype /Form stream /Length 54 Q /ProcSet[/PDF/Text] stream >> /Matrix [1 0 0 1 0 0] /ProcSet[/PDF/Text] << /Meta141 Do /Length 16 /BBox [0 0 88.214 16.44] << 0.738 Tc 0 G >> /Type /XObject Q /F3 12.131 Tf q /BBox [0 0 30.642 16.44] 1.007 0 0 1.006 411.035 510.406 cm >> Q 384 0 obj stream 205.199 4.894 TD /FormType 1 /FormType 1 >> /Meta387 403 0 R /F3 12.131 Tf >> /Length 16 >> /Matrix [1 0 0 1 0 0] /Length 65 >> Q Q /Matrix [1 0 0 1 0 0] /F3 12.131 Tf Q /ProcSet[/PDF/Text] endobj /ProcSet[/PDF] /Resources<< 0 g q Q /Type /XObject /Matrix [1 0 0 1 0 0] 1 i /Font << q q 1 i /ProcSet[/PDF] /Font << >> Q /Meta29 42 0 R endstream /Meta68 Do /F3 12.131 Tf /Resources<< endobj << >> Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 (2) Tj q (x) Tj Q /Meta402 418 0 R 30.699 4.894 TD q endobj /Subtype /Form BT >> 1 g 0.737 w 0.738 Tc /FormType 1 0 g Q /BBox [0 0 88.214 16.44] Q endstream Q q endobj /Length 95 /Matrix [1 0 0 1 0 0] q /Meta237 251 0 R 1.007 0 0 1.007 271.012 383.934 cm /Meta121 Do /Subtype /Form /BBox [0 0 88.214 16.44] 1.007 0 0 1.007 551.058 383.934 cm >> (38) Tj Q << stream endstream 0 G 20.21 5.203 TD q /Meta247 Do BT 1.014 0 0 1.007 111.416 383.934 cm endstream q /Matrix [1 0 0 1 0 0] /FormType 1 395 0 obj (iii) 25 exceeds a number by 7. q /Meta106 120 0 R >> /Length 69 /Meta207 221 0 R << /BBox [0 0 88.214 16.44] 366 0 obj 0.564 G Q 1 g /Length 69 /ProcSet[/PDF] q /Length 69 endstream 0.524 Tc >> q >> 0 w 216 0 obj /ProcSet[/PDF] 1.005 0 0 1.007 102.382 400.496 cm /BBox [0 0 15.59 16.44] /F3 12.131 Tf startxref /Subtype /Form q Q endobj /Resources<< /I0 51 0 R /Descent -299 endobj /Length 16 0.737 w 1 i q /FormType 1 /Font << (A\)) Tj 0 G 43.426 5.203 TD << 1 i >> /Matrix [1 0 0 1 0 0] 68 0 obj /Kids [ >> /Length 78 307 0 obj /Font << << endobj Q 1 i << 1.007 0 0 1.007 551.058 583.429 cm Q q ET q /ProcSet[/PDF/Text] q Q q Q Q /Meta48 Do /BBox [0 0 88.214 16.44] Q 1 i /Meta239 253 0 R /Resources<< /Meta279 Do /BBox [0 0 639.552 16.44] /Font << q 0 w (v) 5 subtracted from thrice a number is 16. q /Type /XObject q q << Q ET 1 i /Meta332 346 0 R 0 G 1.007 0 0 1.006 551.058 763.351 cm Q Q /BBox [0 0 534.67 16.44] [(1)-25(0\))] TJ q Q -0.041 Tw 0 g 427 0 obj Twice a number is decreased by 9, and this sum is multiplied by 4. 0 g Q endobj /FormType 1 1 i /Meta219 233 0 R >> >> /Subtype /Form q /Resources<< /Meta232 246 0 R stream q /F3 12.131 Tf /Resources<< 4 0 R /Subtype /Form Q /Length 69 0.564 G BT /BBox [0 0 88.214 16.44] [(The )-16(s)15(um )-14(of )] TJ q 0.564 G /FormType 1 endobj /BBox [0 0 15.59 16.44] q endstream >> S /Length 96 q t is 56: 4. /Meta400 416 0 R q )-20(Use x to r)-21(eprese)-22(nt "a num)-15(ber)-19(.")] /F3 17 0 R /Meta1 Do q /BBox [0 0 534.67 16.44] /Matrix [1 0 0 1 0 0] /BBox [0 0 88.214 16.44] Q >> 0 g /Type /XObject /Type /XObject << /Meta90 Do 1 i 25 0 obj q Percentage decrease is found by dividing the decrease by the starting number, then multiplying that result by 100%. /Length 69 >> q /FormType 1 >> /BBox [0 0 534.67 16.44] 1.005 0 0 1.007 102.382 400.496 cm /F3 12.131 Tf BT 352 0 obj 0 g 0 G A number = an unknown number which can be represented by a variable, usually x. Q /Subtype /Form Q 0 G 1 g /BBox [0 0 534.67 16.44] q /Meta365 379 0 R 0 G /BBox [0 0 88.214 16.44] Q Q /FormType 1 /FormType 1 /ProcSet[/PDF/Text] q q Q 0 g Q 1.007 0 0 1.006 130.989 437.384 cm /Matrix [1 0 0 1 0 0] /Resources<< 0 G 0 G /Type /XObject endobj Q 1 i endobj Q /F3 12.131 Tf 2.238 5.203 TD 0 g >> 1.014 0 0 1.006 111.416 437.384 cm Q << /Type /XObject 174 0 obj Q /F3 12.131 Tf /Subtype /Form /Resources<< /Length 151 /F3 17 0 R /Meta161 175 0 R /Subtype /Form >> endstream 0 g ET q ET 1.007 0 0 1.007 130.989 776.149 cm stream /FormType 1 ET 0.463 Tc 1 i /BBox [0 0 17.177 16.44] ET /Subtype /Form 0 g endstream /F3 12.131 Tf /BBox [0 0 88.214 16.44] q 0 G q (-23) Tj (3\)) Tj /Meta233 Do /Subtype /Form Q 1.007 0 0 1.007 271.012 330.484 cm >> 1.014 0 0 1.006 251.439 763.351 cm >> endstream /Meta403 Do /Type /XObject /Subtype /Form 0 g /FormType 1 0 g /Length 59 /BBox [0 0 88.214 35.886] >> 1 i 0 G /Meta324 Do Q q /Matrix [1 0 0 1 0 0] /Font << /Type /XObject endobj endobj BT /Meta408 Do BT /Subtype /Form 0.369 Tc Q q 1 g q >> Q 0 w 20.21 5.336 TD 0 w >> (+) Tj BT /Length 60 Q /Meta47 Do /Type /XObject /ProcSet[/PDF] 1 i Q endstream << 1.014 0 0 1.007 251.439 277.035 cm q 1 g /Meta100 Do q /Font << /Resources<< TJ q /Length 68 672.261 473.519 m Q /FormType 1 q