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Frage 1 Bericht
If (IPO3)4 = 11510 find P
Antwortdetails
1 x 43 + P x 42 + 0 x 4 + 3 = 11510
16p + 67 = 115 p = 4816
= 3
Frage 2 Bericht
Solve the following equation equation for x2 + 2xr2 + 1r4 = 0
Antwortdetails
x2 + 2xr2
+ 1r4
= 0
(x + 1r2
) = 0
x + 1r2
= 0
x = −1r2
Frage 3 Bericht
Scores(x)01234567Frequency(f)71167753
In the distribution above, the mode and median respectively are
Antwortdetails
From the distribution, Mode = 1 and
Median = 2+22
= 2
= 1, 2
Frage 5 Bericht
If x and y represent the mean and the median respectively of the following set of numbers 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, find the xy correct to one decimal place
Antwortdetails
Mean ¯x
= 15610
= 15.6
Median = ¯y
= 15+162
312
= 15.5
xy
= 15.615.5
= 1.0065
1.0(1 d.p)
Frage 6 Bericht
Simplify x+2x+1 - x−2x+2
Antwortdetails
x+2x+1
- x−2x+2
= (x+2)(x+2)−(x−2)−(x−2)(x+1)(x+1)(x+2)
= (x2+4x+4)−(x2−x−2)(x+1)(x+2)
= x2+4x+4−x2+x+2(x+1)(x+2)
= 5x+6(x+1)(x+2)
Frage 7 Bericht
In the figure, STQ = SRP, PT = TQ = 6cm and QS = 5cm. Find SR
Antwortdetails
From similar triangle, QSQP=TQQR=512=6QR
QT = 6×125=725=SR=QR−QS
= 725−5=72−255
= 475
Frage 8 Bericht
In the figure, a solid consists of a hemisphere surmounted by a right circular cone, with radius 3.0cm and height 6.0cm. Find the volume of the solid
Antwortdetails
The volume of the solid = vol. of cone + vol. of hemisphere
volume of cone = 12π2h
= 1π3×(3)2x6=18πcm2
vol. of hemisphere = 4πr36=2πr33
= 2π3×(3)3=18πcm3
vol. of solid = 18π + 18π
= 36π cm3
Frage 9 Bericht
A 5.0g of salt was weighted by Tunde as 5.1g. What is the percentage error?
Antwortdetails
% error = actual errortrue value
x 100
Where actual error = 5.1 - 5.0 = 0.1
true value = 5.0g
% error = 0.15.0
x 100
= 105
= 2
Frage 10 Bericht
If the sum of the 8th and 9th terms of an arithmetic progression is 72 and the 4th term is -6, find the common difference
Antwortdetails
Frage 11 Bericht
In a class of 150 students, the sector in a pie chart representing the students offering Physics has angle 12o. How many students are offering Physics?
Antwortdetails
No of students offering Physics are 12360
x 150
= 5
Frage 12 Bericht
if x is the addition of the prime numbers between 1 and 6; and y the H.C.F. of 6, 9, 15. Find the product of x and y
Antwortdetails
Prime numbers between 1 and 6 are 2, 3 and 5
x = 2 + 3
= 5 = 10
H.C.F. of 6, 9, 15 = 3
∴ y = 3
X x y = 10 x 3
= 30
Frage 13 Bericht
If cos 60o = 1/2, which of the following angle has cosine of -1/2?
Antwortdetails
cos60o = 1/2, cos(180o/60o) = -1/2
cos120o = -1/2
Frage 15 Bericht
Simplify x−7x2−9 x x2−3xx2−49
Antwortdetails
x−7x2−9
x x2−3xx2−49
= x−7(x−3)(x+3)
x x(x−3)(x−7)(x+7)
= x(x+3)(x+7)
Frage 16 Bericht
If 7 and 189 are the first and fourth terms of a geometric progression respectively find the sum for the first
three terms of the progression
Antwortdetails
Frage 17 Bericht
Two sisters, Taiwo and Keyinde, own a store. The ratio of Taiwo's share to Kehinde's is 11:9. Later, Keyinde sells 23 of her share to Taiwo for ₦720.00. Find the value of the store.
Antwortdetails
Let value of store = X
Ratio of Taiwo's share to kehine's is 11:9 Keyinde sells 23
of her share to Taiwo for ₦720
23
of 9 = 6
∴ Sum of the ratio = 11 + 9 = 20
620
of x = ₦720
6x20
= 720
∴ x = 720×206
x = ₦24,000
Frage 18 Bericht
PQR is a triangle in which PQ = 10cm and QPR = 60oS is a point equidistant from P and Q. Also S is a point equidistant from PQ and PR. If U is the foot of the perpendicular from S on PR, find the length SU in cm to one decimal place
Antwortdetails
△
PUS is right angled
US5
= sin60o
US = 5 x √32
= 2.5√3
= 4.33cm
Frage 19 Bericht
Find m such that (m + √3 )(1 - √3 )2 = 6 - 2√2
Antwortdetails
(m + √3
)(1 - √3
)2 = 6 - 2√2
(m + √3
)(4 - 2√3
) = 6 - 2√2
= 6 - 2√3
4m - 6 + 4 - 2m√3
= 6 - 2√3
comparing co-efficients,
4m - 6 = 6.......(i)
4 - 2m = -2.......(ii)
in both equations, m = 3
Frage 20 Bericht
In triangle PQR, PQ = 1cm, QR = 2cm and PQR = 120o Find the longest side of the triangle
Antwortdetails
PR2 = PQ2 + QR2 - 2(QR)(PQ) COS 120o
PR2 = 12 + 22 - 2(1)(2) x - cos 60o
= 5 - 2(1)(2) x -12
= 5 + 2 = 7
PR = √7
cm
Frage 22 Bericht
The solution of the quadratic equation px2 + qx + b = 0 is
Antwortdetails
px2 + qx + b = 0
Using almighty formula
−b±√b2−4ac2a
.........(i)
Where a = p, b = q and c = b
substitute for this value in equation (i)
= −q±√q2−4bp2p
Frage 25 Bericht
In the figure, XR and YQ are tangents to the circle YZXP if ZXR = 45o and YZX = 55o, Find ZYQ
Antwortdetails
< RXZ = < ZYX = 45O(Alternate segment) < ZYQ = 90 + 45 = 135O
Frage 26 Bericht
In the figure, PQRS is a circle. If chords QR and RS are equal, calculate the value of x.
Antwortdetails
SRT is a straight line, where QRT = 120
SRQ = 180∘ - 120∘ = 60∘ - (angle on a straight line)
also angle QRS = 180∘ - 100∘ (angle on a straight line) . In angles where QR = SR and angle SRQ = 60∘
x = 100 - 60 = 40∘
Frage 27 Bericht
In the figure, PS = RS = QS and QRS = 50o. Find QPR
Antwortdetails
In the figure PS = RS = QS, they will have equal base QR = RP
In angle SQR, angle S = 50O
In angle QRP, 65 + 65 = 130O
Since RQP = angle RPQ = 180−1302
= 502=25o
QPR = 25O
Frage 28 Bericht
If x varies inversely as the cube root of y and x = 1 when y = 8, find y when x = 3
Antwortdetails
Frage 29 Bericht
In the figure, PS = 7cm and RY = 9cm. IF the area of parallelogram PQRS is 56cm2. Find the area of trapezium PQTS
Antwortdetails
From the figure, PS = QR = YT = 7cm
Area of parallelogram PQRS = 56cm
56 = base x height, where base = 7
7 x h = 56cm,
h = 567
= 8cm
Area of trapezium 12 (sum of two sides)x height where two sides are QT and PS but QT = QR + RY + YT = 7 +9 + 7 = 23cm
Area of trapezium PQTS = 12 (23 + 7) x 8
12 x 30 x 8 = 120cmsq
Frage 30 Bericht
For which of the following exterior angles is a regular polygon possible? i. 36o ii. 18o iii. 15o
Antwortdetails
for a regular polygon to be possible, it must have all sides angles equal. 36018
= 20 sides and 36015
= 24 sides
(ii) and (iii) are right
Frage 32 Bericht
Tope bought X oranges at N5.00 each and some mangoes at N4.00 each. if she bought twice as many mangoes as oranges and spent at least N65.00 and at most N130.00, find the range of values of X.
Antwortdetails
Frage 34 Bericht
If a metal pipe 10cm long has an external diameter of 12cm and a thickness of 1cm find the volume of the metal used in making the pipe
Antwortdetails
The volume of the pipe is equal to the area of the cross section and length.
let outer and inner radii be R and r respectively.
Area of the cross section = (R2 - r2)
where R = 6 and r = 6 - 1
= 5cm
Area of the cross section = (62 - 52)π
= (36 - 25)π
cm sq
vol. of the pipe = π
(R2 - r2)L where length (L) = 10
volume = 11π
x 10
= 110π
cm3
Frage 35 Bericht
Find correct to one decimal place, 0.24633 ÷ 0.0306
Antwortdetails
0.246330.03060
multiplying throughout by 100,000
= 246333060
= 8.05
= 8.1
Frage 36 Bericht
Simplify 4a2−49b22a2−5ab−7b2
Antwortdetails
4a2−49b22a2−5ab−7b2
= (2a)2−(7b)2(a−b)(2a+7b)
= (2a+7b)(2a−7b)(a−b)(2a+7b)
= 2a−7ba−b
Frage 37 Bericht
The solutions of x2 - 2x - 1 = 0 are the points of intersection of two graphs. if one of the graphs is y = 2 + x - x2, find the second graph
Antwortdetails
Frage 38 Bericht
A tax player is allowed 18
th of his income tax-free, and pays 20% on the remainder. If he pays ₦490.00 tax, what is his income?
Antwortdetails
He pays tax on 1 - 78
= 17
th of his income
20% is 490, 100% is 100020
x 490, ₦2,450.00
= 78
of his income = ₦2,450.00
178
x 2450
= 8×24507
= 196007
= ₦2800.00
Frage 39 Bericht
Evaluate 813×5231023 = 813×5231023
Antwortdetails
813×5322103
= (23)13×532(2×5)23
= 2×5223×532
= 21 - 23
= 213
= 3√2
Frage 40 Bericht
Given that 3x - 5y - 3 = 0, 2y - 6x + 5 = 0 the value of (x, y) is
Antwortdetails
3x - 5y = 3, 2y - 6x = -5
-5y + 3x = 3........{i} x 2
2y - 6x = -5.........{ii} x 5
Substituting for x in equation (i)
-5y + 3(1924
) = 3
-5y + 3 x 1924
= 3
-5y = 3−198
-5 = 24−198
= 58
y = 58×5
y = −18
(x, y) = (1924,−18
)
Frage 41 Bericht
The thickness of an 800 pages of book is 18mm. Calculate the thickness of one leaf of the book giving your answer in meters and in standard form
Antwortdetails
Thickness of an 800 pages book = 18mm to meter
18 x 103m = 1.8 x 10-2m
One leaf = 1.8×10−2800
= 1.8×10−28×102
= −1.88
x 10-4
= 0.225 x 10-4
= 2.25 x 10-5m
Frage 42 Bericht
If cos ? = xy , find cosec?
Antwortdetails
Cos θ
= xy
= adjopp
(hyp2) = opp2 + adj2
(hyp2) = x2 + y2
hyp = √x2+y2
Cosecθ
= hyp
= x2 + y2
= 1y
√x2+y2
Frage 43 Bericht
If log102 = 0.3010 and log103 = 0.4771, evaluate; without using logarithm tables, log104.5
Antwortdetails
If log102 = 0.3010 and log103 = 0.4771,
log104.5 = log10 (3×3)2
log103 + log103 - log102 = 0.4771 + 0.4771 - 0.0310
= 0.6532
Frage 44 Bericht
If cos2θ + 18 = sin2θ , find tanθ
Antwortdetails
cos2θ
+ 18
= sin2θ
..........(i)
from trigometric ratios for an acute angle, where cosθ
+ sin2θ
= 1 - cosθ
........(ii)
Substitute for equation (i) in (i) = cos2θ
+ 18
= 1 - cos2θ
= cos2θ
+ cos2θ
= 1 - 18
2 cos2θ
= 78
cos2θ
= 72×3
716
= cosθ
√716
= √74
but cos θ
= adjhyp
opp2 = hyp2 - adj2
opp2 = 42 (√7
)2
= 16 - 7
opp = √9
= 3
than θ
= opphyp
= 3√7
3√7
x 7√7
= 3√77
Frage 45 Bericht
If two dice are thrown together, what is the probability of obtaining at least a score of 10?
Antwortdetails
The total sample space when two dice are thrown together is 6 x 6 = 36
1234561.1.11.21.31.41.51.622.12.22.32.42.52.633.13.23.33.43.53.644.14.24.34.44.54.655.15.25.35.45.55.666.16.26.36.46.56.6
At least 10 means 10 and above
P(at least 10) = 636
= 16
Frage 46 Bericht
A basket contain green, black and blue balls in the ratio 5 : 2 : 1. If there are 10 blue balls. Find the corresponding new ratio when 10 green and 10 black balls are removed from the basket
Antwortdetails
Let x represent total number of balls in the basket.
If there are 10 blue balls, 18
of x = 10
x = 10 x 8 = 80 balls
Green balls will be 58
x 80 = 50 and black balls = 28
x 80 = 20
Ratio = Green : black : blue
50 : 20 : 10
-10 : 10 : -
------------------
New Ratio 40 : 10 : 10
4 : 1 : 1
Frage 48 Bericht
In the figure, PQ is a parallel to ST and QRS = 40∘
. Find the value of x.
Antwortdetails
From the figure, 3x + x - 40∘ = 180∘
4x = 180∘ + 40∘
4x = 220∘
x = 2204
= 55∘
Frage 49 Bericht
Four interior angles of a pentagon are 90o - xo, 90o + xo, 110o - 2xo, 110o + 2xo. Find the fifth interior angle
Antwortdetails
Let the fifth interior angle be y: sum of interior angle of a pentagon
= (2 x 5 - 4) x 90o
= 6 x 90o
= 540o
(90 - x) + (90 + x) + (110 - 2x) + (110 + 2x) + y = 540o
400o + y = 540o
y = 540 - 400o
y = 140o
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