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The fraction greater than $$8\frac{4}{9}$$ is
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The fraction greater than $$8\frac{4}{9}$$ is
$$8\frac{4}{9} = \frac{76}{9} \approx 8.44$$. Converting the options: $$8\frac{1}{3} = \frac{25}{3} \approx 8.33$$, $$\frac{150}{18} = \frac{25}{3} \approx 8.33$$, $$8\frac{2}{3} = \frac{26}{3} \approx 8.67$$, and $$\frac{216}{27} = 8$$. Only $$8\frac{2}{3}$$ exceeds $$8\frac{4}{9}$$, so option (c) is correct.
A car is slowly driven in a road full of fog. The car passes a man who was walking at the rate of 3 km an hour in the same direction. He could see the car for 4 minutes and was visible for up to a distance of 100 meters. The speed of the car is (in km per hours)
The car remains visible over a total relative distance of 100 m for 4 minutes, so the relative speed between car and man is $$\frac{0.1}{4/60} = 1.5$$ km/hr. Since they move in the same direction, the car's actual speed is the man's speed plus this relative speed: $$3 + 1.5 = 4.5$$ km/hr.
Kiran sells pens at a profit of 20% for Rs. 60. But due to lack of demand he reduced its price to Rs. 55. Then
Since selling at Rs. 60 gives a 20% profit, the cost price is $$\frac{60}{1.2} = 50$$. Selling at Rs. 55 gives a profit of $$55 - 50 = 5$$, which as a percentage of the cost price is $$\frac{5}{50} \times 100 = 10\%$$. So he still makes a profit of 10%.
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If 40% of a number is added to another number then it becomes 125% of itself. The ratio of the second to the first number is
Let the first number be $$x$$ and the second be $$y$$. Adding 40% of the first to the second gives 125% of the second: $$y + 0.4x = 1.25y$$, so $$0.4x = 0.25y$$, giving $$\frac{y}{x} = \frac{0.4}{0.25} = \frac{8}{5}$$. So the ratio of the second to the first number is $$8 \colon 5$$.
The length of a rectangular sheet of paper is 33 cm. It is rolled along its length into a cylinder so that width becomes height of the cylinder. The volume is 1386 cubic cms. The width of the rectangular sheet (in cm) is
Rolling along the length turns the length into the base circumference, so $$2\pi r = 33$$, giving $$r = \frac{33 \times 7}{2 \times 22} = 5.25$$ cm. From $$\pi r^2 h = 1386$$, we get $$h = \frac{1386}{86.625} = 16$$ cm, which is the width of the sheet.
If $$\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \ldots + \frac{1}{n \times (n+1)} = \frac{19}{20}$$ then $$n =$$
The sum telescopes to $$1 - \frac{1}{n+1} = \frac{n}{n+1}$$. Setting $$\frac{n}{n+1} = \frac{19}{20}$$ gives $$n = 19$$.
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$$a, b$$ are natural numbers. If $$9a^2 = 12a + 96$$ and $$b^2 = 2b + 3$$, the value of $$2018(a+b)$$ is
From $$3a^2 - 4a - 32 = 0$$ the natural number solution is $$a = 4$$. From $$b^2 - 2b - 3 = 0$$, i.e. $$(b-3)(b+1) = 0$$, the natural number solution is $$b = 3$$. So $$a+b=7$$ and $$2018 \times 7 = 14126$$.
Shanti has three daughters. The average age of them is 15 years. Their ages are in the ratio $$3 \colon 5 \colon 7$$. The age of the youngest daughter is (in years)
The total age of the three daughters is $$3 \times 15 = 45$$ years. Dividing 45 in the ratio $$3 \colon 5 \colon 7$$ (15 parts total), each part is $$\frac{45}{15} = 3$$, so the youngest is $$3 \times 3 = 9$$ years old.
In the adjoining figure, $$ABCD$$ is a quadrilateral. The bisectors of $$\angle B$$ and the exterior angle at $$D$$ meet at $$P$$. Given $$\angle C = 80^\circ$$, $$\angle ADC = \frac{1}{2}\angle A$$ and $$\angle A = \angle C + 40^\circ$$. Then $$\angle DPB$$ is
Here $$\angle A = 80^\circ+40^\circ=120^\circ$$ and $$\angle ADC = \frac{1}{2}\angle A = 60^\circ$$, so by the angle sum of the quadrilateral $$\angle B = 360^\circ-120^\circ-80^\circ-60^\circ=100^\circ$$. Chasing angles along the bisector of $$\angle B$$ and the bisector of the $$120^\circ$$ exterior angle at $$D$$ down to their meeting point $$P$$ gives $$\angle DPB = 70^\circ$$.
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The number of 3-digit number which contain 6 and 7 is
Of the 900 three digit numbers, those missing digit 6 number $$8\times9\times9=648$$, those missing digit 7 also number 648, and those missing both number $$7\times8\times8=448$$. By inclusion-exclusion, numbers missing at least one of 6 or 7 total $$648+648-448=848$$, so those containing both digits total $$900-848=52$$.
The difference between the biggest and the smallest three digit number each of which has different digits is
The biggest 3-digit number with all different digits is 987, and the smallest is 102 (the leading digit cannot be 0). Their difference is $$987-102=885$$.
If $$3x+1=2y-1=5z+3=7w+1=15$$, the value of $$6x-3y+5z-8w$$ is
Solving each part gives $$x=\frac{14}{3}$$, $$y=8$$, $$z=\frac{12}{5}$$ and $$w=2$$. Then $$6x-3y+5z-8w = 28-24+12-16=0$$, which is not among options a, b or c, so the answer is none of these.
Five years ago the average age of Aruna, Roy, David and salman is 45 years. Sita joins them now. The average age of all the five now is 49 years. The present age of sita is (in years)
Five years ago the four had total age $$4\times45=180$$, so now their total age is $$180+4\times5=200$$. With Sita, the total age of all five now is $$5\times49=245$$, so Sita's present age is $$245-200=45$$.
The fraction $$\frac{B}{3x-1}$$ is subtracted from the fraction $$\frac{A}{2x+3}$$. The resulting fraction is $$\frac{-11}{(2x+3)(3x-1)}$$. Then $$A+B=$$
Subtracting gives $$\frac{A(3x-1)-B(2x+3)}{(2x+3)(3x-1)}=\frac{-11}{(2x+3)(3x-1)}$$, so $$A(3x-1)-B(2x+3)=-11$$ for all $$x$$. Matching the $$x$$ coefficient gives $$3A=2B$$, and matching constants gives $$-A-3B=-11$$, which together give $$B=3$$, $$A=2$$, so $$A+B=5$$.
There are some cows and ducks. The total number of legs is equal to 14 more than twice the number of heads. The number of cows is
With $$c$$ cows and $$d$$ ducks, legs $$=4c+2d$$ and heads $$=c+d$$, so $$4c+2d=2(c+d)+14$$, which simplifies to $$2c=14$$, giving $$c=7$$.
The sum of 5% of a number and 9% another number is equal to sum of the 8% first number and 7% of the second number. The ratio between the numbers is
With numbers $$x$$ and $$y$$, the condition gives $$0.05x+0.09y=0.08x+0.07y$$, so $$0.02y=0.03x$$, giving $$\frac{x}{y}=\frac{2}{3}$$. So the ratio between the numbers is $$2 \colon 3$$.
The length of two sides of an isosceles triangle are 8 cm and 14 cm. The perimeter of the triangle (in cm) is
The third side may equal either given side. Sides 8, 8, 14 give perimeter $$8+8+14=30$$, valid since $$8+8>14$$. Sides 8, 14, 14 give perimeter $$8+14+14=36$$, valid since $$8+14>14$$. So the perimeter can be 30 or 36.
There are three cell phones A, B, C. A is 50% costlier than C and B is 25% costlier than C. A is a % costlier than B. Then $$a=$$
Taking C's price as 100, A costs 150 and B costs 125. The percentage by which A is costlier than B is $$\frac{150-125}{125}\times100=20\%$$.
Sushant wrote a two digit number. He added 5 to the tens digit and subtracted 3 from the unit digit of the number and got a number equal to twice the original number. The original number is
With tens digit $$t$$ and units digit $$u$$, the number is $$10t+u$$ and the new number is $$10(t+5)+(u-3)=10t+u+47$$. Setting this equal to twice the original gives $$10t+u+47=2(10t+u)$$, so $$10t+u=47$$. The original number is 47.
The units digit of $$5^{2018} - 3^{2018}$$ is
Powers of 5 always end in 5. Powers of 3 cycle with units digits 3, 9, 7, 1 in period 4, and since $$2018=4\times504+2$$, $$3^{2018}$$ ends in the same digit as $$3^2$$, which is 9. Subtracting requires a borrow, giving units digit $$15-9=6$$.
The smallest natural number that has to be added to 803642 to get a number which is divisible by 9 is
The digit sum of 803642 is $$8+0+3+6+4+2=23$$, and the next multiple of 9 above 23 is 27, so the number to be added is $$27-23=4$$.
The greatest two digit number that will divided 398, 436, and 542 leaving respectively 7, 11 and 15 as remainders is
Subtracting the remainders gives $$398-7=391$$, $$436-11=425$$ and $$542-15=527$$, which must each be exactly divisible by the required number. Since $$391=17\times23$$, $$425=17\times25$$ and $$527=17\times31$$, their HCF is 17.
$$\frac{2}{3}$$ is ___ of $$\frac{1}{3}$$.
Dividing $$\frac{2}{3}$$ by $$\frac{1}{3}$$ gives $$\frac{2}{3}\times\frac{3}{1}=2$$, so $$\frac{2}{3}$$ is 2 times $$\frac{1}{3}$$.
The sum of 5 positive integers is 280. The average of the first 2 number is 40. The average of the third and fourth number is 60. The fifth number is
The first two numbers sum to $$2\times40=80$$, and the third and fourth sum to $$2\times60=120$$. So the fifth number is $$280-80-120=80$$.
If $$a \colon b = 3 \colon 4$$ and $$\frac{p}{q}=\frac{a^2+b^2+ab}{a^2+b^2-ab}$$, where $$p, q$$ have no common divisors other than 1, $$p+q$$ is
Taking $$a=3k$$, $$b=4k$$ gives $$a^2+b^2=25k^2$$ and $$ab=12k^2$$, so $$\frac{p}{q}=\frac{25k^2+12k^2}{25k^2-12k^2}=\frac{37}{13}$$. Since 37 and 13 share no common factor, $$p+q=37+13=50$$.
$$a$$ is a natural number such that a has exactly two divisors and $$(a+1)$$ has exactly three divisors. The number of divisors of $$a+2$$ is
Since $$a$$ has exactly two divisors, $$a$$ is prime. Since $$a+1$$ has exactly three divisors, it must be the square of a prime, say $$p^2$$, so $$a=(p-1)(p+1)$$; for this to be prime, $$p-1=1$$, giving $$p=2$$ and $$a=3$$. Then $$a+2=5$$, which is prime and has exactly 2 divisors.
The first term of a series is $$\frac{2}{5}$$. If $$x$$ is a term of this series, the next term is $$\frac{1-x}{1+x}$$. If $$t_n$$ denotes the $$n$$ th term and $$t_{2018}-t_{2017}=\frac{p}{q}$$, where $$p, q$$ are integers having no common factors other than 1, $$p+q$$ is
From $$t_1=\frac{2}{5}$$ we get $$t_2=\frac{3}{7}$$ and $$t_3=\frac{2}{5}$$ again, so the series repeats with period 2: odd terms equal $$\frac{2}{5}$$ and even terms equal $$\frac{3}{7}$$. So $$t_{2018}-t_{2017}=\frac{3}{7}-\frac{2}{5}=\frac{1}{35}$$, giving $$p+q=1+35=36$$.
In the adjoining figure, the side of the square is $$\sqrt{\frac{2018}{\pi}}\text{cm}$$. The area of the unshaded region is $$\left(\frac{\pi-2}{\pi}\right)A$$ sq. cms. The value of $$A$$ is
The square's area is $$\frac{2018}{\pi}$$, and since its diagonal equals the circle's diameter, the circle's area works out to $$1009$$. The unshaded region is $$1009-\frac{2018}{\pi}=\left(1-\frac{2}{\pi}\right)\times1009=\left(\frac{\pi-2}{\pi}\right)\times1009$$, so $$A=1009$$.
$$n$$ is a natural number. The square root of the sum of the square of $$n$$ and 19 is equal to the next natural number to $$n$$. The value of $$n$$ is
The condition gives $$\sqrt{n^2+19}=n+1$$, so $$n^2+19=n^2+2n+1$$, which simplifies to $$2n=18$$, giving $$n=9$$.
Using only the digits 1, 2, 4, 5, two-digit numbers are formed. The digits of the two digit number may be the same or different. The number of such two-digit number is
Each of the two digit positions can independently be any of the 4 given digits, so the total number of two-digit numbers is $$4\times4=16$$.
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