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Tambaya 1 Rahoto
Simplify x+2x+1 - x−2x+2
Bayanin Amsa
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)
Tambaya 2 Rahoto
If (IPO3)4 = 11510 find P
Bayanin Amsa
1 x 43 + P x 42 + 0 x 4 + 3 = 11510
16p + 67 = 115 p = 4816
= 3
Tambaya 3 Rahoto
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
Bayanin Amsa
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
Tambaya 6 Rahoto
Find correct to one decimal place, 0.24633 ÷ 0.0306
Bayanin Amsa
0.246330.03060
multiplying throughout by 100,000
= 246333060
= 8.05
= 8.1
Tambaya 7 Rahoto
Solve the following equation equation for x2 + 2xr2 + 1r4 = 0
Bayanin Amsa
x2 + 2xr2
+ 1r4
= 0
(x + 1r2
) = 0
x + 1r2
= 0
x = −1r2
Tambaya 8 Rahoto
If cos2θ + 18 = sin2θ , find tanθ
Bayanin Amsa
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
Tambaya 9 Rahoto
In the figure, PS = RS = QS and QRS = 50o. Find QPR
Bayanin Amsa
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
Tambaya 10 Rahoto
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
Bayanin Amsa
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
Tambaya 13 Rahoto
Evaluate 813×5231023 = 813×5231023
Bayanin Amsa
813×5322103
= (23)13×532(2×5)23
= 2×5223×532
= 21 - 23
= 213
= 3√2
Tambaya 14 Rahoto
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.
Bayanin Amsa
Tambaya 17 Rahoto
Given that 3x - 5y - 3 = 0, 2y - 6x + 5 = 0 the value of (x, y) is
Bayanin Amsa
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
)
Tambaya 19 Rahoto
A 5.0g of salt was weighted by Tunde as 5.1g. What is the percentage error?
Bayanin Amsa
% 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
Tambaya 20 Rahoto
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
Bayanin Amsa
Tambaya 21 Rahoto
Simplify 4a2−49b22a2−5ab−7b2
Bayanin Amsa
4a2−49b22a2−5ab−7b2
= (2a)2−(7b)2(a−b)(2a+7b)
= (2a+7b)(2a−7b)(a−b)(2a+7b)
= 2a−7ba−b
Tambaya 23 Rahoto
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?
Bayanin Amsa
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
Tambaya 24 Rahoto
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.
Bayanin Amsa
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
Tambaya 25 Rahoto
If cos 60o = 1/2, which of the following angle has cosine of -1/2?
Bayanin Amsa
cos60o = 1/2, cos(180o/60o) = -1/2
cos120o = -1/2
Tambaya 27 Rahoto
If cos ? = xy , find cosec?
Bayanin Amsa
Cos θ
= xy
= adjopp
(hyp2) = opp2 + adj2
(hyp2) = x2 + y2
hyp = √x2+y2
Cosecθ
= hyp
= x2 + y2
= 1y
√x2+y2
Tambaya 28 Rahoto
In the figure, PQ is a parallel to ST and QRS = 40∘
. Find the value of x.
Bayanin Amsa
From the figure, 3x + x - 40∘ = 180∘
4x = 180∘ + 40∘
4x = 220∘
x = 2204
= 55∘
Tambaya 30 Rahoto
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
Bayanin Amsa
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
Tambaya 31 Rahoto
If x varies inversely as the cube root of y and x = 1 when y = 8, find y when x = 3
Bayanin Amsa
Tambaya 32 Rahoto
The solution of the quadratic equation px2 + qx + b = 0 is
Bayanin Amsa
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
Tambaya 33 Rahoto
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
Bayanin Amsa
Mean ¯x
= 15610
= 15.6
Median = ¯y
= 15+162
312
= 15.5
xy
= 15.615.5
= 1.0065
1.0(1 d.p)
Tambaya 34 Rahoto
If two dice are thrown together, what is the probability of obtaining at least a score of 10?
Bayanin Amsa
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
Tambaya 35 Rahoto
Four interior angles of a pentagon are 90o - xo, 90o + xo, 110o - 2xo, 110o + 2xo. Find the fifth interior angle
Bayanin Amsa
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
Tambaya 36 Rahoto
In the figure, PS = 7cm and RY = 9cm. IF the area of parallelogram PQRS is 56cm2. Find the area of trapezium PQTS
Bayanin Amsa
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
Tambaya 37 Rahoto
In the figure, PQRS is a circle. If chords QR and RS are equal, calculate the value of x.
Bayanin Amsa
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∘
Tambaya 38 Rahoto
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
Bayanin Amsa
Tambaya 39 Rahoto
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
Bayanin Amsa
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
Tambaya 40 Rahoto
Scores(x)01234567Frequency(f)71167753
In the distribution above, the mode and median respectively are
Bayanin Amsa
From the distribution, Mode = 1 and
Median = 2+22
= 2
= 1, 2
Tambaya 42 Rahoto
For which of the following exterior angles is a regular polygon possible? i. 36o ii. 18o iii. 15o
Bayanin Amsa
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
Tambaya 43 Rahoto
If log102 = 0.3010 and log103 = 0.4771, evaluate; without using logarithm tables, log104.5
Bayanin Amsa
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
Tambaya 44 Rahoto
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
Bayanin Amsa
△
PUS is right angled
US5
= sin60o
US = 5 x √32
= 2.5√3
= 4.33cm
Tambaya 45 Rahoto
Find m such that (m + √3 )(1 - √3 )2 = 6 - 2√2
Bayanin Amsa
(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
Tambaya 46 Rahoto
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
Bayanin Amsa
Tambaya 47 Rahoto
In the figure, XR and YQ are tangents to the circle YZXP if ZXR = 45o and YZX = 55o, Find ZYQ
Bayanin Amsa
< RXZ = < ZYX = 45O(Alternate segment) < ZYQ = 90 + 45 = 135O
Tambaya 48 Rahoto
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
Bayanin Amsa
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
Tambaya 49 Rahoto
Simplify x−7x2−9 x x2−3xx2−49
Bayanin Amsa
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)
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