2017 - JAMB Mathematics Past Questions and Answers - page 3

4(\sqrt{27}) + 5(\sqrt{12}) − 3(\sqrt{75})

= 4(\sqrt{3}) × 9 + 5(\sqrt{3}) × 4 − 3(\sqrt{3}) × 25

= 4 × 3(\sqrt{3}) + 5 × 2(\sqrt{3}) − 3 × 5(\sqrt{3})

= 12(\sqrt{3}) + 10(\sqrt{3}) − 15(\sqrt{3})

= (12 + 10 − 15)(\sqrt{3})

= 7(\sqrt{3})

Let the speed of the 1st trip be x miles/hr

and the speed of the 2nd trip be 3x miles/hr

Speed = distance/time

∴ Time taken to cover a distance of 50 miles on the 1st trip

= (\frac{50}{xhr})

time taken to cover a distance of 300 miles on the next trip

= (\frac{300}{3xhr})

= (\frac{100}{xhr})

∴the new time compared with the old time is twice as much

Sum of angle at a point = 360^o

2x + 3x + 4x = 360

9x = 360

x = (\frac{360}{9})

x = 40^o

Total Cost Price = N(250,000 + 70,000)

= N 32,000

Selling Price = N 400,000(Given)

Gain = Selling Price - Cost Price

= 400,000 - 300,000

= 80,000

% gain = (\frac{\text{Gain}}{\text{Cost Price}}) × 100

= (\frac{80,000}{320,000}) × 100

Gain % = 25%

EXCELLENCE

It is a ten letter word = 10!

Since we have repeating letters, we have to divide to remove the duplicates accordingly. There are 4 Es, 2 Cs, 2 Ls

∴ there are

(\frac{10!}{4!2!2!}) ways to arrange

(\frac{0.00000231}{0.007}) to standard form

= (\frac{231 \times 10^{-8}}{7 \times 10^{-3}})

= 33 × 10(^{-8 − (−3)})

= 33 × 10(^{− 8 + 3})

= 33 × 10^-5

Actual measurement = 10cm

approximated value of measurement = 10.5cm

% error = (\frac{\text{Actual measurement − Approximated}}{\text{Actual measure}}) × 100

= (\frac{10 − 10.5}{10}) × 100

= (\frac{-0.5}{10}) × 100

ignore -sign i.e take absolute value

= (\frac{0.5}{10}) × 100

= 5 %

Principal = P, Simple Interest = I, Amount = A

Amount = Principal + Simple Interest

I = (\frac{PRT}{100})

R = rate, T = time

I = (\frac{P \times 5 \times 2}{100})

I = (\frac{10P}{100})

I = (\frac{P}{10})

Amount A = P + I

5500 = P + (\frac{P}{10})

Multiply through by 100

5500 = 10P + P

5500 = 11P

p = (\frac{5500}{11})

p = ₦5000

Total angle at a point = 3600

∴ To get the angle occupied by fertilizer we have,

40 + 50 + 80 + 70 + 30 + fertilizer(x) = 360

270 + x = 360

x = 360 - 270

x = 90

Total amount allocated to the farm

= ₦ 80,000

∴Amount allocated to the fertilizer

= (\frac{\text{fertilizer (angle) × Total amount}}{\text{total angle}})

= (\frac{90}{360}) × 80,000

= ₦20,000