2020/2021 WAEC/NECO Answers: Neco Gce 2021. Mathematics answers

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31-40: BCBCABABBCThis is actually the Neco Gce 2018. Mathematics answers Answers as questions sent to our subscribers on WHATSAPP, if you are looking for 2020 WAEC/NECO Answers midnight before the exam, try contacting us.

41-50: DACCCCDBDB

**(10)**

**Draw the triangle **

**Hence we have A+B+C = 180**

**50+80+C = 180**

**130+C = 180**

**C = 180 - 130**

**C = 50**

**(i)**

**The bearing of B from C = 90 - 50 = 40°**

**(ii) **

**Bearing of A from B**

**90+50(alternate angle to A) **

**= 140°**

**(iii) **

**Distance between B and C**

**Using sine rule **

**C/sinC = a/SinA**

**40/sin50 = a/sin50**

**Cross multiply **

**asin50 = 40sin50**

**a = 40km**

**Hence distance between B and C = 40km**

**(iv) **

**Using cosine rule **

**b²= a²+c² - 2acCosB**

**where a = 40, c= 40 and B=80**

**b² = 40²+40² -2*40*40cos80**

**b² = 1600+1600 - 3200(0.1736)**

**b² = 3200 - 555.52**

**b² = 2644.48**

**b = âˆš2644.48**

**b = 51.42km**

**(v)**

**Height of Î”ABC**

**Draw the triangle **

**Area of the triangle **

**=1/2acSinB**

**=1/2*40*40sin80**

**=800(0.9848)**

**=787.84km²**

**Hence 1/2bh=787.84**

**bh=2*787.84**

**h=2*787.84/b**

**h=2*787.84/51.42**

**h=30.64**

**=31**

**(12)**

**TABULATE:**

**Marks | frq | c.f | boundary**

**11-20 | 8 | 8 | 10.5 â€“ 20.5**

**21-30 | 6 | 14 | 20.5 â€“ 30. 5**

**31-40 | 10 | 24 | 30.5 â€“ 40.5**

**41-50 | 12 | 36 | 40.5 â€“ 50.5**

**51-60 | 8 | 44 | 50.5-60.5**

**61-70 | 6 | 50 | 60.5 â€“ 70.5**

**(12b)**

**DRAW THE CUMULATIVE FREQUENCE CURVE**

**(12ci)**

**Medium mark = N + ½**

**= 50 + 1/2 = 51/2**

**= 25. 5**

**= 48**

**(12cii)**

**Lower quartile Q1**

**= ¼ x 50**

**= 12.5**

**= 28**

**(12ciii)**

**Upper quartile Q2**

**= ¾ x 50**

**= 37.5**

**=53**

**(12civ)**

**Both percentile **

**70/100 x 50 **

**= 35**

**= 50**

**====================================**

**(1) **

**TABULATE:**

**No| Log**

**3081| 3.4887**

**0.775 | 1.8893**

**0.456 | 1. 6589**

**| UNDER Log**

**| 3.4887**

**| 1. 5482**

**| 3.9405**

**| 0.9851**

**Square root of 4/3081/0.775.0456**

**Antilog = 9663**

**= 9.663**

**= 2.9**

**(2a)**

**1101â‚‚ = 2x + 1**

**1 x 2² + 0 x 2¹+ 1 x 2º = 2x + 1**

**8 + 4 + 1 = 2x +1**

**13 â€“ 2x**

**X = 12/2 = 6**

**(2b)**

**Sin x = 12/13 = 0.9231**

**Z = sin-¹ 0.9231 = 67.4º**

**Therefore 3 sin x + ½ cos x**

**= 3 sin 67.4 + ½ cos 67.4**

**= 3 x 0.9232 + ½ x 0.3843**

**= 7.7696 + 0.1921**

**(3a)**

**PQ x 90º [angle in a semicircle]**

**QPO = 90º - 62 [ angle in a triangle]**

**= 28º**

**Therefore POZ = 28º [alternative angles]**

**(ii)**

**PXZ = ½ x 28º = 14º [angle at center is twice angle at cirumfeence]**

**(3b)**

**6/5 + 3/x+3 â€“ 9/5(x+3)**

**= 6 (x+3) + 3 (5) â€“ 9/5(x+3)**

**= 6x + 18 + 15 â€“ 9/5 (x+3)**

**= 6x + 24/5(x+3) **

**= 6(x+4)/5(x+3)**

**(4a)**

**1/3(y-1)+2>1/2(2y-1)+1**

**2y-2+12>6y-3+1**

**2y-6y>-2-10**

**-4y>-12**

**y<3**

**(4bi)**

**M=y2-y1/X2-X1**

**=-1-2/2-3=-3/-1=3**

**(4bii)**

**2X+1=X+3**

**2X-X=3-1**

**X=2**

**(6)COMPLETED **

**(6)**

**126y = 86**

**1 x y² 2 x y¹ + 6 x yº = 86**

**y² + 2y + 6 = 86**

**y² + 2y + 6 â€“ 86 = 0**

**y² + 2y â€“ 80 = 0**

**y + 10y â€“ 8y â€“ 80 = 0**

**y (y+10)-8y (y+10) = 0 **

**y â€“ 8 = 0 or y + 10 = 0**

**the positive value of y**

**(6b) Area of triangle = ½ x b x h**

**Let the Acheal Area = x**

**Area = ½ x b x h**

**Base = x â€“ 9x/100 = 93/100**

**Height = 9x/100 + x = 107x/100**

**Area = ½ x 93x/100 x 107x/100**

**=963x/2000**

**% error = actual Area â€“ wrong/actual area x 100**

**= x â€“ 963x/2000 /x 100**

**= 20000x â€“ 963x/20000 x 100**

**= 19037/20000 x 100**

**= 95.185%**

**(6c)**

**p/100 + 2p + 7 = 11.02 x 100**

**p + 200p + 700 = 1102**

**201p = 1102 â€“ 700**

**201p = 402**

**P = 402/201**

**P = N2**

**P = 200K**

**(8)**

**S²â‚ (3x â€“ 1) (x+2) dx**

**By expansionâ€™**

**S²â‚ 3x² + 6x â€“ x â€“ 2 dx**

**S²â‚ 3x² + 5x â€“ 2 dx**

**By integrating using d formula**

**Xn+1/n+1**

**Therefore 3x ²+¹/2+1 + 5x¹+¹/1+1**

**= 2xº+¹/0+1 + c**

**Therefore 3x³/3 + 5x²/2 + 2x/1 + c**

**Therefore x³ + 5x²/2 + 2x**

**But x = 2 at higher and are at lower by substituting 2 in x than of value of 1**

**(2)³ + 5/2(2)² + 2(2) â€“ ((1)³ + 3/2(1) + 2 (1)**

**8 + 5/2 x 4 + 4 â€“ (1 + 5/2 + 2)**

**8 + 10 + 4 â€“ (3 + 5/2)**

**22 â€“ 11/2 = = 22-11/2**

**= 44- 11/2 **

**= 33/2**

**(8bi)**

**T = 2Ï€ square l/g**

**Dividing both side by 2Ï€**

**T/2Ï€ = square root of L/g**

**By squaring both side**

**(T/2 Ï€)² = (L/g)²**

**T²/4 Ï€² = l/g**

**Cross multiplication**

**gT²/T² = 4Ï€²L/T² **

**g = 4Ï€²L/T²**

**(8bii)**

**T = (0.4)1/2 = square root of 0.4**

**L = 0.04 T1 = 3.14**

**G = 4Ï€²L/T2 **

**= 4 x (3.14)² x 0.04/(0.4)²**

**= 4 x 9.8596 x 0.04/0.4 **

**= 1.578/0.4**

**g = 3.94**

**(9a)**

**2p-q=10.......(1)**

**3p+q^2=22........(2)**

**Eq(1) x3 and eq(2) x2**

**6p-3q=30**

**6p+2q=44**

**Subtracting eq(1) from eq(2)**

**-3q-2q= -14**

**2q^2+3q= 14**

**2q^2+3q-14=0**

**2q^2+7q-4q-14=0**

**q(2q+7)-2(2q+7)=0**

**(q-2)(2q+7)=0**

**q-2=0 or 2q+7=0**

**q=2 or q= -7/2**

**Substitute q into eq(1)**

**2p-q=10**

**2p-2=10**

**2p=10+2**

**2p=12**

**p=12/2=6**

**When q= -7/2**

**2p-q=10**

**2p-(-7/2)=10**

**2p+7/2=10**

**4p+7=10**

**4p=10-7**

**4p=3**

**p=3/4**

**(9b)**

**z^2 -25/z^2-9z+20**

**If Z is undefined **

**Z^2-9z+20=0**

**Using factorization method **

**-4z and -5z**

**Z^2 -4z-5z+20=0**

**Z(z-4) - 5(z-4)=0**

**(z-4)(z-5)=0**

**Z-4=0 or Z-5=0**

**Z=4 or Z=5**

**Z=4 or 5**

**Z is undefined when it is equal to 4 or 5**

**Answers loading.... **

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