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Observe the following sequence. What is the 100th term? $$7, 8, 1, 0, 0, 1, 0, 1, 1, 0, 2, 1, 0, 3, \ldots$$
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Observe the following sequence. What is the 100th term? $$7, 8, 1, 0, 0, 1, 0, 1, 1, 0, 2, 1, 0, 3, \ldots$$
Grouping the terms from the third term onward in blocks of three gives $$1, 0, 0$$ then $$1, 0, 1$$ then $$1, 0, 2$$ then $$1, 0, 3$$, so each block repeats the pair $$1, 0$$ followed by a number that increases by 1 in every block. Continuing this repeating structure out to the 100th position gives a value of 3.
A number is multiplied by 2 then by $$\frac{1}{3}$$, then by 4, then by $$\frac{1}{5}$$ then by 6 and finally by $$\frac{1}{7}$$. The answer is 16. Then the number is
Let the number be $$x$$. Then $$x \times 2 \times \frac{1}{3} \times 4 \times \frac{1}{5} \times 6 \times \frac{1}{7} = 16$$, which simplifies to $$x \times \frac{48}{105} = 16$$, so $$x = 35$$. Since 35 is odd, the number is odd.
Samrud bought a t-shirt for Rs. 250. His friend Shlok wanted to buy it. Samrud wants to have a $$10\%$$ profit on that. The selling price is (in rupees)
The cost price is Rs. 250 and the profit is $$10\%$$ of the cost price, which is $$250 \times \frac{10}{100} = 25$$. So the selling price is $$250 + 25 = 275$$ rupees.
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The value of $$1 + 21 + 4161 + 81 - 11 - 31 - 51 - 71 - 91$$ is
Pairing each subtracted term with the addition ten more than it gives $$1 - 11 = -10$$, $$21 - 31 = -10$$, $$41 - 51 = -10$$, $$61 - 71 = -10$$ and $$81 - 91 = -10$$. Adding these five results gives $$-50$$.
In the adjoining figure what portion of the figure is shaded?
Counting the shaded and unshaded unit triangles in the grid shown, the shaded portion of the whole figure works out to $$\frac{3}{10}$$.
The sum of the numbers in the three brackets ( ) is $$\frac{( )}{24} = \frac{20}{( )} = \frac{24}{18} = \frac{4}{( )}$$
Simplifying $$\frac{24}{18} = \frac{4}{3}$$ shows every fraction equals $$\frac{4}{3}$$. The first bracket is $$24 \times \frac{4}{3} = 32$$, the second is $$20 \div \frac{4}{3} = 15$$, and the third is $$4 \div \frac{4}{3} = 3$$. Their sum is $$32 + 15 + 3 = 50$$.
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A is the smallest three digit number which leaves a remainder 2 when divided by 17. B is the smallest three digit number which leaves a remainder 7 when divided by 12. Then $$A + B$$ is
The smallest three digit number leaving remainder 2 on division by 17 is $$A = 104$$, since $$17 \times 6 + 2 = 104$$. The smallest three digit number leaving remainder 7 on division by 12 is $$B = 103$$, since $$12 \times 8 + 7 = 103$$. So $$A + B = 104 + 103 = 207$$.
A square of side 3 cm is cut into 9 equal squares. Another square of side 4 cm is cut into 16 equal squares. Saket made a bigger square using all the smaller square bits. The length of the side of the bigger square is (in cm)
The 9 unit squares from the first square and the 16 unit squares from the second square together give $$9 + 16 = 25$$ unit squares of area 1 sq cm each. Since $$25 = 5^2$$, these can be arranged into one bigger square of side $$5$$ cm.
A contractor constructed a big hall, rectangular in shape, with length 32 meters and breadth 18 meters. He wanted to buy 1 meter by 1 meter tiles. But in the shop 3 meter by 2 meter tiles only were available. How many tiles he has to buy for tiling the floor?
Each 3 m by 2 m tile covers the same area as $$3 \times 2 = 6$$ of the 1 m by 1 m tiles. The number of 3 m by 2 m tiles needed is $$\frac{32}{2} \times \frac{18}{3} = 16 \times 6 = 96$$.
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The fraction to be added to $$2\frac{1}{3}$$ to get the fraction $$4\frac{4}{7}$$ is
Converting to improper fractions, $$2\frac{1}{3} = \frac{7}{3}$$ and $$4\frac{4}{7} = \frac{32}{7}$$. The difference is $$\frac{32}{7} - \frac{7}{3} = \frac{96 - 49}{21} = \frac{47}{21} = 2\frac{5}{21}$$.
In the adjoining figure $$\angle BAD = \angle DAF = \angle FAC$$. GE is parallel to DF, and $$\angle EGA = 90^\circ$$. If $$\angle ACE = 70^\circ$$, the measure of $$\angle FDE$$ is
TO BE FILLED - figure required. This problem was officially marked as a bonus question in the answer key, since the exact positions of points D, E, F and G on the figure are needed to pin down a unique value and cannot be recovered from the text alone.
$$ABC$$ is a triangle in which the angles are in the ratio $$3 \colon 4 \colon 5$$. PQR is a triangle in which the angles are in the ratio $$5 \colon 6 \colon 7$$. The difference between the least angle of $$ABC$$ and the least angle of PQR is $$a^\circ$$. Then $$a$$ =
For $$ABC$$, dividing $$180^\circ$$ in the ratio $$3 \colon 4 \colon 5$$ gives parts of $$15^\circ$$ each, so the least angle is $$3 \times 15 = 45^\circ$$. For $$PQR$$, dividing $$180^\circ$$ in the ratio $$5 \colon 6 \colon 7$$ gives parts of $$10^\circ$$ each, so the least angle is $$5 \times 10 = 50^\circ$$. The difference is $$50 - 45 = 5$$, so $$a = 5$$.
Samrud had to multiply a number by 35. By mistake he multiplied by 53 and got a result 720 more. The new product is
Let the number be $$x$$. The mistake gives $$53x - 35x = 720$$, so $$18x = 720$$ and $$x = 40$$. The new (mistaken) product is $$53 \times 40 = 2120$$.
Vishva plays football every 4th day. He played on a Tuesday. He plays football on a Tuesday again in is _____ days.
Since he plays every 4 days, the day of the week shifts each time, and it returns to Tuesday only after a number of days that is a multiple of both 4 and 7. The smallest such number is the LCM of 4 and 7, which is $$4 \times 7 = 28$$ days.
In an elementary school $$26\%$$ of the students are girls. If there are 240 less girls than boys, then the strength of the school is
Girls make up $$26\%$$ of the school, so boys make up $$74\%$$. The difference $$74\% - 26\% = 48\%$$ corresponds to 240 students, so $$1\%$$ is $$240 \div 48 = 5$$ students, and the total strength is $$5 \times 100 = 500$$.
There are three concentric circles as shown in the figure. The radii of them are $$2\text{ cm}$$, $$4\text{ cm}$$ and $$6\text{ cm}$$. The ratio of the area of the shaded region to the area of the dotted region is $$\frac{a}{b}$$ where $$a, b$$ are integers and have no common factor other than 1. Then $$a + b$$ =
The shaded ring lies between the circles of radius 2 and 4, with area proportional to $$4^2 - 2^2 = 12$$. The dotted ring lies between the circles of radius 4 and 6, with area proportional to $$6^2 - 4^2 = 20$$. The ratio is $$\frac{12}{20} = \frac{3}{5}$$, so $$a = 3$$, $$b = 5$$, and $$a + b = 8$$.
The value of $$\left(1+\frac{1}{9}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{7}\right)\left(1+\frac{1}{6}\right)\left(1+\frac{1}{5}\right)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{2}\right)$$ is
Each factor can be written as $$\frac{10}{9} \times \frac{9}{8} \times \frac{8}{7} \times \frac{7}{6} \times \frac{6}{5} \times \frac{5}{4} \times \frac{4}{3} \times \frac{3}{2}$$, and every term in the middle cancels with the next, leaving $$\frac{10}{2} = 5$$.
When a two digit number divides 265, the remainder is 5. The number of such two digit numbers is
Since the remainder is 5, the divisor must divide $$265 - 5 = 260$$ exactly and be greater than 5. Since $$260 = 2^2 \times 5 \times 13$$, its two digit divisors are 10, 13, 20, 26, 52 and 65, giving 6 such numbers.
If $$A \# B = \frac{A \times B}{A + B}$$, the value of $$\frac{12 \# 8}{8 \# 4} + \frac{10 \# 6}{6 \# 2}$$ is
Computing each part, $$12 \# 8 = \frac{96}{20} = \frac{24}{5}$$ and $$8 \# 4 = \frac{32}{12} = \frac{8}{3}$$, so $$\frac{12 \# 8}{8 \# 4} = \frac{24}{5} \times \frac{3}{8} = \frac{9}{5}$$. Similarly $$10 \# 6 = \frac{60}{16} = \frac{15}{4}$$ and $$6 \# 2 = \frac{12}{8} = \frac{3}{2}$$, so $$\frac{10 \# 6}{6 \# 2} = \frac{15}{4} \times \frac{2}{3} = \frac{5}{2}$$. Adding gives $$\frac{9}{5} + \frac{5}{2} = \frac{43}{10} = 4.3$$.
When water becomes ice, its volume increases by $$10\%$$. When ice melts into water its volume decreases by $$a\%$$. Then $$a$$ =
If the volume of water is $$x$$, the volume of ice is $$x + \frac{10}{100}x = \frac{11}{10}x$$. Melting back, the decrease from ice volume to water volume is $$\frac{11}{10}x - x = \frac{1}{10}x$$, which as a percentage of the ice volume is $$\frac{\frac{1}{10}x}{\frac{11}{10}x} \times 100 = \frac{100}{11} \approx 9.09$$. So $$a \approx 9.09$$.
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