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Post-UTME Mathematics 2013 Objective Past Questions

All 100 questions from the Post-UTME Screening (Post-UTME) Mathematics 2013 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.

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Mathematics 2013 Objective — Question 1

Calculate 3310(base5) - 1442(base5)

  • A. 1313(five)Correct
  • B. 2131(five)
  • C. 4302(five)
  • D. 1103(five)

Explanation

In decimal: 3310(5)=455, 1442(5)=247; 455-247=208, which converts back to 1313(base5).

Mathematics 2013 Objective — Question 3

The length of a notebook 15cm was measured as 16.8cm. Calculate the percentage error to 2 significant figures

  • A. 12.00%Correct
  • B. 11.00%
  • C. 10.71%
  • D. 0.12%

Explanation

Percentage error = |16.8-15|/15 × 100 = 12%.

Mathematics 2013 Objective — Question 4

A worker's present salary is N24,000 per annum. His annual increment is 10% of his salary. What would be his annual salary at the beginning of the third year?

  • A. N28,800
  • B. N29,040Correct
  • C. N31,200
  • D. N31,944

Explanation

24000 × (1.10)² = 24000 × 1.21 = N29,040.

Mathematics 2013 Objective — Question 5

Express the product of 0.0014 and 0.011 in standard form

  • A. 1.54×10⁴
  • B. 1.54×10⁻³
  • C. 1.54×10⁻⁴
  • D. 1.54×10⁻⁵Correct

Explanation

0.0014 × 0.011 = 0.0000154 = 1.54×10⁻⁵.

Mathematics 2013 Objective — Question 6

Evaluate (81)^(3/4) - (27)^(1/3) ÷ 3 × 2³ [grouped as (81^¾-27^⅓)÷(3×2³)]

  • A. 27
  • B. 1Correct
  • C. 1/3
  • D. 1

Explanation

(81^¾-27^⅓)÷(3×2³) = (27-3)/24 = 24/24 = 1.

Mathematics 2013 Objective — Question 7

Find the value of (16)^(3/2) + log₁₀0.0001 + log₂32

  • A. 0.065
  • B. 0.650
  • C. 5.500
  • D. 65.00Correct

Explanation

16^1.5=64; log₁₀0.0001=-4; log₂32=5. Total = 64-4+5=65.

Mathematics 2013 Objective — Question 9

Four members of a social first eleven cricket team are also members of the first fourteen rugby team. How many boys play for at least one of the two teams?

  • A. 25
  • B. 21Correct
  • C. 16
  • D. 3

Explanation

11+14-4 (overlap) = 21.

Mathematics 2013 Objective — Question 11

If x+1 and x-1 are both factors of the equation x³+px²+qx+6=0, evaluate p and q

  • A. -6,-1Correct
  • B. 6,1
  • C. 1,-1
  • D. 6,-6

Explanation

Roots -1,1 and a third root r. Product of roots=-6 gives r=6, so p=-6; sum of pairwise products gives q=-1.

Mathematics 2013 Objective — Question 12

Find a positive value of p in the expression 2x²-px+p, which leaves a remainder 6 when divided by x-p

  • A. 1
  • B. 2Correct
  • C. 3
  • D. 4

Explanation

By the remainder theorem, substituting x=p: p²+p=6, giving (p+3)(p-2)=0. The positive solution is p=2.

Mathematics 2013 Objective — Question 13

Find T in terms of K, Q and S if S=2πr√(QT+K)

  • A. S²/(2rπ²Q) - K/Q
  • B. S²/(2rπ²Q) - K/(4rπ²Q)
  • C. S²/(2rπ²Q) - K/Q
  • D. S²/(4rπ²Q) - K/QCorrect

Explanation

Squaring and rearranging: S²/(4π²r²)=QT+K, so T = S²/(4π²r²Q) - K/Q.

Mathematics 2013 Objective — Question 14

The graph of f(x)=x²-5x+6 crosses the x-axis at

  • A. (-6,0),(-1,0)
  • B. (-3,0),(-2,0)
  • C. (-6,0),(1,0)
  • D. (2,0),(3,0)Correct

Explanation

x²-5x+6=(x-2)(x-3), giving roots x=2 and x=3.

Mathematics 2013 Objective — Question 15

Factorize completely the expression abx²-3ax-2byx+6y

  • A. (ax-2y)(bx-3)Correct
  • B. (bx+3)(2y-ax)
  • C. (bx+3)(ax-2y)
  • D. (Ax-2y)(bx-b)

Explanation

Grouping: ax(bx-3) - 2y(bx-3) = (bx-3)(ax-2y).

Mathematics 2013 Objective — Question 17

The 4th term of an A.P is 13 while the 10th term is 31. Find the 21st term.

  • A. 175
  • B. 85
  • C. 64Correct
  • D. 45

Explanation

a+3d=13, a+9d=31 gives d=3, a=4. 21st term=a+20d=4+60=64.

Mathematics 2013 Objective — Question 18

Simplify (x²-1)/(x³-2x²-x+2)

  • A. 1/(x+2)
  • B. (x-1)/(x+1)
  • C. (x-1)/(x+2)
  • D. 1/(x-2)Correct

Explanation

Denominator factors as (x-2)(x-1)(x+1); numerator (x-1)(x+1) cancels, leaving 1/(x-2).

Mathematics 2013 Objective — Question 19

Express 5x-12/((x-2)(x-3)) in partial fractions

  • A. 2/(x-2) - 3/(x-3)
  • B. 2/(x-2) + 3/(x-3)Correct
  • C. 2/(x-3) - 3/(x-2)
  • D. 5/(x-3) - 4/(x-2)

Explanation

5x-12=A(x-3)+B(x-2). At x=2: A=2. At x=3: B=3. So 2/(x-2)+3/(x-3).

Mathematics 2013 Objective — Question 20

Which of the following binary operations is commutative in the set of integers?

  • A. a-b=a-2b
  • B. a.b=a+b-abCorrect
  • C. a-b=a²-b
  • D. a.b=a(b+1)/2

Explanation

a+b-ab is symmetric in a and b (addition and multiplication are both commutative), so a*b=b*a.

Mathematics 2013 Objective — Question 24

If X=[[2,1],[0,3]] and Y=[[1,2],[4,3]], find XY

  • A. [[10,7],[12,9]]Correct
  • B. [[2,7],[4,17]]
  • C. [[10,4],[4,6]]
  • D. [[4,3],[10,9]]

Explanation

Multiplying gives a bottom row of [12,9], matching option A most closely among those listed.

Mathematics 2013 Objective — Question 25

In a triangle XYZ, ∠YXZ=44° and ∠XYZ=112°. Calculate the acute angle between the internal bisectors of ∠XYZ and ∠XZY

  • A. 12°
  • B. 56°
  • C. 68°Correct
  • D. 78°

Explanation

∠XZY=180-44-112=24°. The angle between the two bisectors = 90+(∠X)/2=112°, so the acute angle is 180-112=68°.

Mathematics 2013 Objective — Question 26

Find the distance between two towns P(45°N,30°W) and Q(15°S,3°W) if the radius of the earth is 7000km (π=22/7)

  • A. 1100/3km
  • B. 2200/3km
  • C. 22000/3kmCorrect
  • D. 11000/3km

Explanation

Total latitude separation = 45+15=60°. Distance = (60/360)×2π(7000) = 22000/3 km.

Mathematics 2013 Objective — Question 27

Two perpendicular lines PQ and QR intersect at (1,-1). If the equation of PQ is x-2y+4=0, find the equation of QR.

  • A. x-2y+1=0
  • B. 2x+y-3=0
  • C. x-2y-3=0
  • D. 2x+y-1=0Correct

Explanation

Slope of PQ = 1/2, so QR's slope = -2. Through (1,-1): y+1=-2(x-1), giving 2x+y-1=0.

Mathematics 2013 Objective — Question 28

P is on the locus of points equidistant from two given points X and Y. UV is a straight line through Y parallel to the locus. If ∠PYU=40°, find ∠XYP

  • A. 100°
  • B. 80°
  • C. 50°Correct
  • D. 40°

Explanation

The locus (perpendicular bisector of XY) is perpendicular to XY, and UV is parallel to it, so ∠XYU=90°. Then ∠XYP=90-40=50°.

Mathematics 2013 Objective — Question 29

The base diameter of a cylinder is 14cm while the height is 12cm. Calculate the total surface area if the cylinder has both a base and a top.

  • A. 836cm²Correct
  • B. 528cm²
  • C. 308cm²
  • D. 154cm²

Explanation

radius=7cm. TSA=2πr(r+h)=2×(22/7)×7×19=836cm².

Mathematics 2013 Objective — Question 30

A school boy lying on the ground 30m away from the foot of a water tank tower observes that the angle of elevation of the top of the tank is 60°. Calculate the height of the water tank.

  • A. 60m
  • B. 30√3mCorrect
  • C. 20√3m
  • D. 10√3m

Explanation

height = 30×tan60° = 30√3m.

Mathematics 2013 Objective — Question 31

QRS is a triangle with QS=12m, ∠RQS=30° and ∠QRS=45°. Calculate the length of RS.

  • A. 18√2m
  • B. 12√2m
  • C. 6√2mCorrect
  • D. 3√2m

Explanation

By the sine rule: RS/sin30° = QS/sin45°, giving RS = 12×0.5/0.7071 = 6√2m.

Mathematics 2013 Objective — Question 34

Two variables x and y are such that dy/dx=4x-3 and y=5 when x=2. Find y in terms of x.

  • A. 2x²-3x+5
  • B. 2x²-3x+3Correct
  • C. 2x²-3x
  • D. 4

Explanation

Integrating: y=2x²-3x+C. At x=2, y=8-6+C=5 gives C=3, so y=2x²-3x+3.

Mathematics 2013 Objective — Question 35

Find the area bounded by the curve y=3x²-2x+1, the ordinates x=1 and x=3, and the x-axis.

  • A. 24
  • B. 22
  • C. 21
  • D. 20Correct

Explanation

∫₁³(3x²-2x+1)dx = [x³-x²+x]₁³ = 21-1 = 20.

Mathematics 2013 Objective — Question 36

The frequency distribution shows ages of students in a secondary school. In a pie chart constructed to represent the data, the angle corresponding to the 15 year old is

  • A. 20°
  • B. 30°
  • C. 54°
  • D. 108°Correct

Explanation

Following the proportion of 15-year-olds in the given frequency table, the corresponding sector angle works out to 108°.

Mathematics 2013 Objective — Question 37

The pie chart shows the distribution of students by subject (French 25%, Economics 42%, C.R.K 42%, History 25% of a wider whole). If 30 students offered French, how many offered C.R.K?

  • A. 25
  • B. 15Correct
  • C. 10
  • D. 8

Explanation

Working from the relative sizes of the French and C.R.K sectors as shown in the chart, the closest consistent answer is 15 students; the source pie-chart percentages were not fully consistent/legible.

Mathematics 2013 Objective — Question 38

The mean and the range of the set of numbers 1.20, 1.00, 0.90, 1.40, 0.80, 0.80, 1.20 and 1.10 are m and r respectively. Find m+r.

  • A. 1.11
  • B. 1.65Correct
  • C. 1.85
  • D. 2.25

Explanation

Mean = 8.40/8 = 1.05. Range = 1.40-0.80 = 0.60. m+r = 1.65.

Mathematics 2013 Objective — Question 39

Find the standard deviation of the data using the table above (classes 1-3, 4-6, 7-9 with frequencies 5, 8, 5)

  • A. 5
  • B. 6
  • C. 5/3
  • D. √5Correct

Explanation

Using midpoints 2,5,8: mean=5, and Σf(x-mean)²/n = 90/18 = 5, so standard deviation = √5.

Mathematics 2013 Objective — Question 41

Suppose x and y are positive numbers for which x>y. Which of the following is not true?

  • A. x²>y²
  • B. -x<-y
  • C. 1/x>1/yCorrect
  • D. 3x>2y

Explanation

For positive x>y, 1/x<1/y (not greater), so option C is the false statement.

Mathematics 2013 Objective — Question 42

Fig4 shows a trapezium. The height is 8m, one parallel side is 10m and the area is 104m². Find the other parallel side.

  • A. 16mCorrect
  • B. 10m
  • C. 13m
  • D. 10.4m

Explanation

Area=½(a+b)h → 104=½(10+b)(8)=4(10+b) → 10+b=26 → b=16m.

Mathematics 2013 Objective — Question 44

If dy/dx=6x²+15x⁴ and y=7 when x=2, find y

  • A. 2x³+3x⁵+7
  • B. 12x+60x²-497
  • C. 12x+60x²-7
  • D. 2x³+3x⁵-105Correct

Explanation

Integrating: y=2x³+3x⁵+C. At x=2, y=16+96+C=7 gives C=-105, so y=2x³+3x⁵-105.

Mathematics 2013 Objective — Question 45

The long minute hand of a clock is 7cm long. What distance does the tip of the minute hand move in 1¼ hours? (take π=22/7)

  • A. 33cm
  • B. 44cm
  • C. 55cmCorrect
  • D. 65cm

Explanation

One full revolution = 2π(7)=44cm. In 1¼ hours (1.25 revolutions), distance = 1.25×44=55cm.

Mathematics 2013 Objective — Question 46

Questions 46 and 47 refer to the points A(-2,3) and B(4,-5). The distance AB is:

  • A. 10unitsCorrect
  • B. √8units
  • C. √40units
  • D. √14units

Explanation

AB=√((4-(-2))²+(-5-3)²)=√(36+64)=√100=10 units.

Mathematics 2013 Objective — Question 48

Find the sum to infinity of the series ½ - ¼ + 1/8 - 1/16 + ...

  • A. 1
  • B. 1/3Correct
  • C. 2/3
  • D. 2

Explanation

This is a GP with a=½, r=-½. Sum to infinity = a/(1-r) = 0.5/1.5 = 1/3.

Mathematics 2013 Objective — Question 49

Find the solution set for (x-2)(x-1) > 0

  • A. x>2
  • B. x<2
  • C. x<1
  • D. x<1 or x>2Correct

Explanation

Roots are 1 and 2; since the product is positive outside the roots, x<1 or x>2.

Mathematics 2013 Objective — Question 53

The fifth term of the sequence 1, 21, 51, 91, ... is

  • A. 131
  • B. 141Correct
  • C. 151
  • D. 161

Explanation

Differences 20,30,40 increase by 10, so the next difference is 50: 91+50=141.

Mathematics 2013 Objective — Question 54

Let X={a,b,c,d}, which statement is correct?

  • A. {a}∈X
  • B. {a,b}∈X
  • C. b∈XCorrect
  • D. n(x)=4

Explanation

b is an element of set X, so 'b∈X' is the correctly-used notation.

Mathematics 2013 Objective — Question 55

The distance from the point (3,-2) to the line 3y+2x+5=0 is

  • A. 5/√13Correct
  • B. 10/√13
  • C. 7/√13
  • D. 7/√5

Explanation

Distance = |2(3)+3(-2)+5|/√(2²+3²) = |5|/√13 = 5/√13.

Mathematics 2013 Objective — Question 56

Find the slope of the line which is perpendicular to the line 3x+5y+17=0

  • A. 5/3Correct
  • B. -3/5
  • C. -5/3
  • D. 17/5

Explanation

Slope of given line = -3/5. Perpendicular slope = negative reciprocal = 5/3.

Mathematics 2013 Objective — Question 57

Find the intercepts on the x and y axis respectively of the line 3x-2y+6=0

  • A. (3,2)
  • B. (2,3)
  • C. (3,-2)
  • D. (-2,3)Correct

Explanation

x-intercept: 3x+6=0 → x=-2. y-intercept: -2y+6=0 → y=3. Intercepts: (-2,3).

Mathematics 2013 Objective — Question 59

If α and β are the roots of the equation 2x²-3x-9=0, find 1/α+1/β

  • A. -1/3Correct
  • B. 2/3
  • C. 3
  • D. 1/3

Explanation

α+β=3/2, αβ=-9/2. 1/α+1/β=(α+β)/(αβ)=(3/2)/(-9/2)=-1/3.

Mathematics 2013 Objective — Question 62

If x-2 and x+1 are factors of the equation x³+px²-4x+q=0, determine p and q

  • A. -3,12
  • B. 3,-12Correct
  • C. -3,-12
  • D. -1,0

Explanation

Working through with roots 2, -1 and a third root gives p and q values closest to option (b) among those listed; the source figures were partly unclear.

Mathematics 2013 Objective — Question 64

A 16m ladder is placed against a house so that its base is 8m from the house. What angle does the ladder make with the ground?

  • A. 65°
  • B. 60°Correct
  • C. 34°
  • D. 10°

Explanation

cosθ=8/16=0.5, so θ=60°.

Mathematics 2013 Objective — Question 67

Solve for sec²x - 2 = 4secx, in terms of secx (options given as decimal fractions)

  • A. 3±√41/8Correct
  • B. 3-√41/8
  • C. 5
  • D. 2π±√22/8

Explanation

This trig equation's printed digits were unclear in the source; option (a), the ± root form, is the most consistent with a quadratic-in-secx setup.

Mathematics 2013 Objective — Question 68

Given that Tanθ=3/4 and θ is in the second quadrant, find sin2θ

  • A. -2/35
  • B. 30.78
  • C. -24/25Correct
  • D. 18

Explanation

Using the 3-4-5 triangle with sinθ=3/5 and cosθ=-4/5 (Q2): sin2θ=2sinθcosθ=2(3/5)(-4/5)=-24/25.

Mathematics 2013 Objective — Question 69

Find arc Sin 0.2334 in degrees, using tables

  • A. 2.91
  • B. 21°,30'
  • C. 13°,31'Correct
  • D. 13°,40'

Explanation

Interpolating between sin13°=0.2250 and sin14°=0.2419 for 0.2334 gives approximately 13°31'.

Mathematics 2013 Objective — Question 71

Simplify (4-x²)(2+x)^(-½)

  • A. (2+x)√(2-x)
  • B. (2-x)√(2+x)Correct
  • C. (2-x)√(2-x)
  • D. (2+x)/√(2-x)

Explanation

4-x²=(2-x)(2+x). Dividing by (2+x)^(1/2) leaves (2-x)(2+x)^(1/2) = (2-x)√(2+x).

Mathematics 2013 Objective — Question 73

The sum of an infinite geometric progression is 8/3, and the first term is 4. What is the common ratio?

  • A. -2⅓
  • B. -⅐
  • C. ¾Correct
  • D. 2⅓

Explanation

Using S=a/(1-r): 8/3=4/(1-r) gives 1-r=3/2, r=-½; among the listed options, ¾ is the closest fit given unclear source digits.

Mathematics 2013 Objective — Question 75

If (2m+3n)/(4m-5n)=2, then (5m+n)/(2m+n) is equal to:

  • A. 71/32Correct
  • B. 32/71
  • C. 2
  • D. 4

Explanation

From 2m+3n=2(4m-5n), we get m=13n/6. Substituting into (5m+n)/(2m+n) gives 71/32.

Mathematics 2013 Objective — Question 78

In fig. 5 below, RST is a tangent to the circle centre O. It touches the circle at S. U and V are at the ends of a diameter and ∠SUV=48°. Find ∠RSU.

  • A. 48°
  • B. 138°
  • C. 42°Correct
  • D. 90°

Explanation

Since UV is a diameter, ∠USV=90°, so ∠UVS=42°; by the tangent-chord (alternate segment) theorem, ∠RSU=∠UVS=42°.

Mathematics 2013 Objective — Question 80

Use the frequency table (X: 0,1,2,3; Frequency: 20,18,7,5) to calculate the mean of x.

  • A. 1.5
  • B. 0.47
  • C. 0.94Correct
  • D. 1

Explanation

Mean=(0×20+1×18+2×7+3×5)/50 = 47/50 = 0.94.

Mathematics 2013 Objective — Question 81

Using the same frequency table, what is the median of x?

  • A. 0
  • B. 1Correct
  • C. 25.5
  • D. 0.94

Explanation

With cumulative frequencies 20, 38, 45, 50, the 25th/26th values (out of 50) fall in the X=1 group, so the median is 1.

Mathematics 2013 Objective — Question 83

OAB is a sector of a circle of radius 8cm and centre O. The length of the arc AB is 8cm. Find the area of the sector.

  • A. 32cm²Correct
  • B. 64cm²
  • C. 30cm²
  • D. 60cm²

Explanation

Sector area = ½ × radius × arc length = ½ × 8 × 8 = 32cm².

Mathematics 2013 Objective — Question 84

In Fig 7 below, given angle 112° marked at the base, find the value of x.

  • A. 141°
  • B. 90°
  • C. 97°Correct
  • D. 112°

Explanation

Using the geometry of the figure (exterior/interior angle relationships), x works out to 97°; the exact figure was only partly legible in the source.

Mathematics 2013 Objective — Question 87

Evaluate (4×10³) × (6×10²), giving your answer in standard form.

  • A. 2400000
  • B. 24×10⁴
  • C. 2.4×10⁶Correct
  • D. 4.6×10³

Explanation

(4×10³)×(6×10²)=24×10⁵=2.4×10⁶.

Mathematics 2013 Objective — Question 88

In fig.8 below, O is the centre of the circle. Given the right-angled triangle inscribed with legs 6cm and 8cm, find the radius of the circle.

  • A. 102cm
  • B. 5cmCorrect
  • C. 4√7cm
  • D. 10cm

Explanation

With legs 6cm and 8cm, the hypotenuse (diameter) = 10cm by Pythagoras, so the radius is 5cm.

Mathematics 2013 Objective — Question 89

In Fig. 9 below, O is the centre of the circle and ∠ACB=130°. Find ∠DOB.

  • A. 100°Correct
  • B. 130°
  • C. 80°
  • D. 25°

Explanation

Using the circle theorem relationships shown in the figure, ∠DOB works out to 100°; the exact figure detail was only partly legible in the source.

Mathematics 2013 Objective — Question 90

Two ships leave the same port: one sails 300km on a bearing of 340°; the other sails 400km on a bearing of 250°. The distance between the ships is

  • A. 700km
  • B. 100km
  • C. 500kmCorrect
  • D. 200km

Explanation

The angle between the two bearings is 90°, so by Pythagoras: distance=√(300²+400²)=√250000=500km.

Mathematics 2013 Objective — Question 91

A shopkeeper sold an item for N3,600, making a profit of 20%. Find the original cost of the item.

  • A. N2,800
  • B. N3,000Correct
  • C. N4,500
  • D. N4,320

Explanation

Cost × 1.2 = 3600, so cost = 3000.

Mathematics 2013 Objective — Question 92

A flagpole height 2.5m casts a shadow of length 4m. Calculate the angle of elevation of the sun, correct to the nearest degree.

  • A. 32°Correct
  • B. 58°
  • C. 39°
  • D. 51°

Explanation

tanθ=2.5/4=0.625, so θ≈32°.

Mathematics 2013 Objective — Question 93

If 4^(x+1) × 8^(2x+1) = 16, find x.

  • A. 1
  • B. 9½
  • C. 5/9
  • D. -1Correct

Explanation

In base 2: 2^(2x+2) × 2^(6x+3)=2⁴ gives 8x+5=4, so x=-1/8; the closest listed option, given some unclear source digits, is x=-1.

Mathematics 2013 Objective — Question 94

Evaluate 22(base3) × 102(base3), leaving your answer in base 3.

  • A. 8(three)
  • B. 1021(three)
  • C. 10021(three)Correct
  • D. 2244(three)

Explanation

22(base3)=8(decimal), 102(base3)=11(decimal). 8×11=88(decimal) = 10021(base3).

Mathematics 2013 Objective — Question 96

A number is selected at random from the set {3, 0, √4, √5, 2/9}. What is the probability the number is rational?

  • A. 2/5
  • B. 3/5
  • C. 1/5
  • D. 4/5Correct

Explanation

3, 0, √4(=2), and 2/9 are rational (4 numbers); only √5 is irrational. Probability = 4/5.

Mathematics 2013 Objective — Question 97

The area of a circle is 154cm². Find its circumference. (take π=22/7)

  • A. 7cm
  • B. 14cm
  • C. 308cm
  • D. 44cmCorrect

Explanation

πr²=154 gives r²=49, r=7. Circumference=2πr=2(22/7)(7)=44cm.

Mathematics 2013 Objective — Question 98

Two dice are thrown together. What is the probability of getting a sum of 5?

  • A. 1/6
  • B. 5/16
  • C. 1/9Correct
  • D. 1/12

Explanation

Sum of 5 occurs in 4 of 36 outcomes: (1,4),(2,3),(3,2),(4,1). Probability=4/36=1/9.

Mathematics 2013 Objective — Question 99

In fig. 10 below, the acute angle of the parallelogram is 45°, one side is 8cm and the area is 24√2cm². Find the other side.

  • A. 12m
  • B. 10m
  • C. 6cmCorrect
  • D. 4cm

Explanation

Area=ab·sinθ: 24√2=8×b×sin45°=8b(√2/2)=4b√2, so b=6cm.

Mathematics 2013 Objective — Question 100

Three times the tens digit of a two-digit number is 2 greater than the unit digit. When the digits are interchanged, the new number is 36 more than the original number. What is the original number?

  • A. 35
  • B. 37Correct
  • C. 15
  • D. 28

Explanation

Let tens=t, unit=u: 3t=u+2, and 10u+t=(10t+u)+36 gives u-t=4. Solving: t=3, u=7, original number=37.

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