Saket wanted to add two two-digit numbers. Instead, he multiplied them and obtained $$629$$. The sum of the two two-digit numbers is
NMTC gauss ( primary level) 5th and 6th Aug 31 2024
For the following questions answer them individually
Solution
Factorising gives $$629=17 \times 37$$. Both factors are two-digit numbers, and their sum is $$17+37=54$$.
The sum of three integers is $$1$$ and their product is $$36$$. The greatest of these three integers is
Solution
The integers $$6,-3,-2$$ have product $$6 \times (-3) \times (-2)=36$$. Their sum is $$6-3-2=1$$, so the greatest integer is $$6$$.
The sum of five consecutive even numbers is $$150$$. When written in ascending order, the fourth number is
Solution
The middle number equals the average, which is $$\frac{150}{5}=30$$. The five numbers are $$26,28,30,32,34$$, so the fourth number is $$32$$.
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The price of a cell phone is decreased by $$25\%$$. What percentage increase in the decreased price is required to restore the original price?
Solution
Assume the original price is $$100$$. After a $$25\%$$ decrease, the price is $$75$$, and the required increase is $$25$$. Therefore, the percentage increase is $$\frac{25}{75}\times100=33\frac{1}{3}\%$$.
A rectangular carpet is placed in an $$8\text{ m} \times 8\text{ m}$$ room as shown. What fraction of the floor is not covered?
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Solution
From the grid, the carpet has perpendicular side lengths $$6\sqrt2$$ and $$2\sqrt2$$. Its area is therefore $$6\sqrt2 \times 2\sqrt2=24$$, while the room area is $$8 \times 8=64$$. The uncovered fraction is $$\frac{64-24}{64}=\frac{5}{8}$$.
$$p$$ and $$p+1$$ are both prime numbers. Then the value of $$\frac{p(p+1)}{2p+1}$$ lies between
Solution
Two consecutive integers can both be prime only when they are $$2$$ and $$3$$, so $$p=2$$. The expression becomes $$\frac{2 \times 3}{5}=\frac{6}{5}$$. This lies between $$1$$ and $$\frac{7}{5}$$.
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Real numbers $$a,b,c,d$$ satisfy $$a-2023=b+2024=c-2025=d+2026$$. The greatest among $$a,b,c,d$$ is
Solution
Let the common value be $$k$$. Then $$a=k+2023$$, $$b=k-2024$$, $$c=k+2025$$ and $$d=k-2026$$. Therefore, $$c$$ is the greatest.
In the adjoining figure, $$AD=AE$$. The measure of $$\angle EAD$$ is
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Solution
In right triangle $$ABC$$, the other angles are $$2x$$ and $$\frac{x}{2}$$. Thus $$2x+\frac{x}{2}=90^\circ$$, giving $$x=36^\circ$$. Since $$AD=AE$$, the base angles of triangle $$ADE$$ are both $$36^\circ$$, so $$\angle EAD=180^\circ-72^\circ=108^\circ$$.
The largest three-digit number exactly divisible by the HCF of $$24$$ and $$36$$ is $$n$$. Then $$n+4$$ equals
Solution
The HCF of $$24$$ and $$36$$ is $$12$$. The largest three-digit multiple of $$12$$ is $$996$$. Hence $$n+4=996+4=1000$$.
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In the given figure, $$AB$$, $$HG$$, $$CD$$ and $$FE$$ are parallel. Also, $$AB=6$$, $$GH=4$$, $$CD=5$$, $$FE=9$$ and $$BC=8$$. The distances between each indicated pair of parallel lines are $$3$$. The area of the complete figure is
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Solution
Divide the figure into two trapezia and the central rectangle shown in the diagram. Their areas are $$\frac{1}{2}(3+4)\times3$$, $$3 \times 11$$ and $$\frac{1}{2}(5+6)\times3$$. The total is $$\frac{21}{2}+33+\frac{33}{2}=60$$.
The number of pairs of two-digit square numbers whose sum or difference is also a square is
Solution
The two-digit squares are $$16,25,36,49,64,81$$. We have $$25-16=9$$ and $$36+64=100$$, both of which are squares. Checking the remaining pairs gives no other square sum or difference, so there are $$2$$ pairs.
There are $$20$$ people around a table. Each person shakes hands with the person immediately to the left and the person immediately to the right. The total number of handshakes is
Solution
Each person has two neighbouring handshakes, giving $$20 \times 2=40$$ counts. Every handshake is counted twice, once by each participant. Therefore, the number of distinct handshakes is $$\frac{40}{2}=20$$.
In the adjoining figure, $$A,B,C,D$$ are the vertices of a square of side $$3$$ units. All the semicircles are equal. The area of the shaded region in square units is
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Solution
Each small semicircle has radius $$\frac{1}{2}$$ and area $$\frac{\pi}{8}$$. The outer boundary adds twelve such semicircles to the square, while the central white region removes a unit square and four such semicircles. Hence the shaded area is $$9+12\left(\frac{\pi}{8}\right)-\left(1+4\left(\frac{\pi}{8}\right)\right)=8+\pi$$.
Two box types $$A$$ and $$B$$ hold $$38$$ and $$20$$ candles respectively. Exactly $$288$$ candles fill all the boxes. If $$m$$ boxes are of type $$A$$ and $$n$$ boxes are of type $$B$$, then $$\frac{m}{n}$$ is
Solution
The equation is $$38m+20n=288$$, or $$19m+10n=144$$. The positive integer solution is $$m=6$$ and $$n=3$$. Therefore, $$\frac{m}{n}=\frac{6}{3}=2$$.
For a positive integer $$r$$, let $$[r]$$ denote the sum of its divisors other than $$1$$ and $$r$$ itself. For example, $$[6]=2+3=5$$. The value of $$[[[12]]]$$ is
Solution
The proper divisors of $$12$$ other than $$1$$ are $$2,3,4,6$$, so $$[12]=15$$. Next, $$[15]=3+5=8$$. Finally, $$[8]=2+4=6$$, so $$[[[12]]]=6$$.
The fraction $$\frac{(2 \times 3 \times 4)+(4 \times 6 \times 8)+(6 \times 9 \times 12)+\cdots+(20 \times 30 \times 40)}{(1 \times 2 \times 3)+(2 \times 4 \times 6)+(3 \times 6 \times 9)+\cdots+(10 \times 20 \times 30)}$$ reduces to
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Solution
The $$k$$th term in the numerator is $$(2k)(3k)(4k)=24k^3$$. The corresponding denominator term is $$k(2k)(3k)=6k^3$$. Every numerator term is therefore four times its matching denominator term, so the complete fraction equals $$4$$.
In the adjoining figure, $$ABCD$$ is a rectangle, and $$AE$$ and $$CF$$ are quadrants. The length of the rectangle is twice its breadth. Taking $$\pi=\frac{22}{7}$$, the shaded area is $$21\text{ cm}^2$$. The area of the rectangle is
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Solution
Let the breadth be $$x$$, so the length is $$2x$$ and the rectangle area is $$2x^2$$. The two quadrants together have area $$\frac{\pi x^2}{2}=\frac{11x^2}{7}$$. Thus $$2x^2-\frac{11x^2}{7}=21$$, giving $$x^2=49$$ and rectangle area $$2x^2=98$$.
The number $$m$$ has $$2,5,6$$ as factors, and the number $$n$$ has $$4,8$$ as factors. The smallest possible value of $$m+n$$ is
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Solution
The smallest number divisible by $$2,5,6$$ is their LCM, which is $$30$$. The smallest number divisible by $$4$$ and $$8$$ is $$8$$. Hence the smallest value is $$30+8=38$$.
Bus route $$X$$ arrives every $$15$$ minutes and bus route $$Y$$ arrives every $$40$$ minutes. Both buses have arrived now. They will next arrive simultaneously after how many hours?
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Solution
The next simultaneous arrival occurs after the LCM of $$15$$ and $$40$$ minutes. This is $$120$$ minutes, which equals $$2$$ hours.
In the adjoining figure, $$\angle ABC=60^\circ$$ and $$\angle ACB=80^\circ$$. $$AD$$ bisects $$\angle A$$. Through $$C$$, a line making an angle of $$\frac{\angle A}{2}$$ with $$BC$$ is drawn. It intersects the bisector and the perpendicular from $$B$$ to $$AD$$ as shown. The value of $$x$$ in degrees is
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Solution
Since the angles at $$B$$ and $$C$$ are $$60^\circ$$ and $$80^\circ$$, we have $$\angle A=40^\circ$$, so each half is $$20^\circ$$. The bisector line makes $$80^\circ$$ with $$BC$$, and its perpendicular through $$B$$ makes $$10^\circ$$ with $$BC$$ on the other side. The required exterior angle is therefore $$10^\circ+20^\circ=30^\circ$$.
For real numbers $$a$$ and $$b$$, define $$a \ast b=\left(a+\frac{b}{2}\right)\left(b+\frac{a}{2}\right)$$. The value of $$(2 \ast 8)\ast2$$ is
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Solution
First, $$2\ast8=(2+4)(8+1)=6 \times 9=54$$. Then $$54\ast2=(54+1)(2+27)=55 \times 29=1595$$.
A two-digit number has repeated digits. The number of such numbers having exactly $$4$$ positive divisors is
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Solution
The repeated-digit numbers are $$11,22,33,44,55,66,77,88,99$$. Exactly four divisors occur for $$22=2 \times 11$$, $$33=3 \times 11$$, $$55=5 \times 11$$ and $$77=7 \times 11$$. Therefore, the count is $$4$$.
The salaries of Peter and Ali are in the ratio $$3 \colon 2$$. Their expenditures are in the ratio $$5 \colon 3$$, and each saves Rs. $$5000$$. Peter's income in rupees is
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Solution
Let their salaries be $$3x$$ and $$2x$$, and their expenditures be $$5y$$ and $$3y$$. Then $$3x-5y=5000$$ and $$2x-3y=5000$$. Solving gives $$x=10000$$, so Peter's income is $$3x=30000$$.
In the given figure, $$\angle A \colon \angle B \colon \angle C=14 \colon 3 \colon 1$$. The two additional lines are drawn as described in the figure. The measure of $$\angle EGF$$ in degrees is
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Solution
The angle ratio gives $$\angle A=140^\circ$$, $$\angle B=30^\circ$$ and $$\angle C=10^\circ$$. The exterior angle $$\angle CAD$$ is $$40^\circ$$, so the line through $$A$$ makes $$10^\circ$$ with $$AC$$ and is parallel to $$BC$$. The line through $$B$$ also makes $$10^\circ$$ with $$BC$$, so the vertically opposite angle $$\angle EGF$$ is $$10^\circ$$.
Five people have different odd amounts of money, each less than Rs. $$100$$. The largest possible total amount in rupees is
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Solution
To maximise the total, choose the five largest distinct odd numbers below $$100$$. They are $$99,97,95,93,91$$. Their sum is $$475$$.
The cost price of $$10$$ articles equals the selling price of $$9$$ articles. The profit percentage is $$11\frac{1}{a}\%$$. The value of $$a$$ is
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Solution
If the cost price of one article is $$x$$, the selling price of one article is $$\frac{10x}{9}$$. The profit fraction is therefore $$\frac{1}{9}$$, giving a profit percentage of $$\frac{100}{9}\%=11\frac{1}{9}\%$$. Hence $$a=9$$.
When $$2\frac{6}{11}$$ of $$1\frac{2}{7}$$ is divided by $$3\frac{3}{11}$$, the result is
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Solution
Convert the mixed numbers to improper fractions. The expression is $$\left(\frac{28}{11}\times\frac{9}{7}\right)\div\frac{36}{11}$$. The product in brackets is $$\frac{36}{11}$$, so the final value is $$1$$.
Two cell phones are sold at the same selling price. One gives a $$10\%$$ gain and the other gives a $$10\%$$ loss. The overall percentage loss is
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Solution
Let the selling price of each phone be $$100$$. Their cost prices are $$\frac{1000}{11}$$ and $$\frac{1000}{9}$$, so the total cost price is $$\frac{20000}{99}$$ while the total selling price is $$200$$. The loss percentage is $$1\%$$.
An office employee works for $$4$$ consecutive days and then has one day off, repeating this pattern. Today is an off-day and a Sunday. The minimum number of days the employee must work before another off-day falls on Sunday is
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Solution
Consecutive off-days are $$5$$ calendar days apart. The first return to Sunday occurs after $$7$$ such cycles because $$5 \times 7=35$$ is divisible by $$7$$. Each cycle contains $$4$$ working days, so the employee works $$7 \times 4=28$$ days.
In the adjoining figure, triangle $$BCD$$ is equilateral. If $$\angle AFB=90^\circ$$ and $$AH$$ bisects $$\angle FAE$$, then the measure of $$\angle HGE$$ in degrees is
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Solution
Since triangle $$BCD$$ is equilateral, the corresponding angle at $$B$$ is $$60^\circ$$. The exterior angle at $$A$$ gives $$\angle BAE=80^\circ$$, while right triangle $$ABF$$ gives $$\angle BAF=30^\circ$$. Hence $$\angle FAE=50^\circ$$, and its bisector makes $$25^\circ$$ with $$AF$$. Therefore, the angle at $$G$$ is $$180^\circ-90^\circ-25^\circ=65^\circ$$, and the vertically opposite angle $$\angle HGE$$ is also $$65^\circ$$.
