OBMEP Level 2 First Phase Mock Paper 1 with Solutions

C — R$29.

The larger purchase contains exactly two extra notebooks; the pencils are unchanged. Thus two notebooks cost $31-17=14$ reais, so each notebook costs $14\div2=7$ reais.

In the smaller purchase, the two pencils cost $17-7=10$ reais. Each pencil therefore costs $10\div2=5$ reais.

Two notebooks and three pencils cost $2\times7+3\times5=14+15=29$ reais. This is option C.

Useful beforehand: Subtraction; Equal unit prices. Find a concept to review.

A — 8 boys.

Initially there are $40\times3/5=24$ girls. There are still 24 girls after the boys join.

These 24 girls now form half the club, so the new total is $2\times24=48$. The club started with 40 members. Therefore $48-40=8$ boys joined. As a check, the original 16 boys become 24, matching the 24 girls.

Useful beforehand: Fractions of a quantity. Find a concept to review.

E — 86.

Let the digits be $a$ and $b$, where $1\le a\le9$ and $0\le b\le9$. The condition is $(10a+b)+(a+b)=100$, so $11a+2b=100$.

Since $0\le2b\le18$, we have $82\le11a\le100$. The only possible tens digits are 8 and 9. Also, $11a=100-2b$ is even; because 11 is odd, $a$ must be even. Hence $a=8$.

Then $2b=100-88=12$, giving $b=6$. The integer is 86, and $86+8+6=100$, so it satisfies the condition.

Useful beforehand: Decimal place value; Even and odd integers; Simple equations. Find a concept to review.

D — 60 cm.

Walk around the solid outline. Its side lengths are 18, 7, 6, 5, 12 and 12 centimetres: the right-hand vertical side is $12-5=7$, and the top horizontal side is $18-6=12$.

The perimeter is therefore $18+7+6+5+12+12=60$ cm.

Another check is to compare with the original rectangle. Removing the corner removes outer lengths 6 and 5, but exposes new lengths 6 and 5. The perimeter stays $2(18+12)=60$ cm.

Useful beforehand: Rectangle perimeter; Subtraction of lengths. Find a concept to review.

B — 09:06.

The multiples of 12 up to 60 are 12, 24, 36, 48 and 60. The multiples of 15 up to 60 are 15, 30, 45 and 60. Their first positive common multiple is 60.

Thus joint flashes are 60 minutes apart. The first after 08:06 is at 09:06, which is after 09:00. No earlier time between 09:00 and 09:06 is a joint flash, since it would be less than 60 minutes after the previous one.

Useful beforehand: Multiples; Reading a clock. Find a concept to review.

E — 5/12.

The dice are distinguishable. Each of the 6 red numbers can occur with each of the 6 blue numbers, giving $6\times6=36$ equally likely outcomes.

There are $1+2+3+2+1+6=15$ favourable outcomes. The probability is $15/36=5/12$. Each outcome appears in exactly one row of the table, so nothing is counted twice.

Useful beforehand: Probability of equally likely outcomes; Divisibility. Find a concept to review.

B — 36 cm².

The whole triangle has area $\tfrac12\times12\times8=48$ cm².

Use BC as the base line. Triangles ABD and ADC share the same perpendicular height from A to this line. In the formula $\text{area}=\tfrac12\times\text{base}\times\text{height}$, the height is therefore the same for both. Their areas have the same ratio as their bases, namely $1:3$.

The whole area is divided into four equal shares, with three shares in ADC. Its area is $\tfrac34\times48=36$ cm².

Useful beforehand: Triangle area; Ratios. Find a concept to review.

A — 6 orders.

Ana and Bruno can occupy any two of the four positions. The six possible pairs of positions are $(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)$.

For each pair, Ana must take the earlier position and Bruno the later one. The remaining two positions must contain Caio followed by Dina. Thus each pair of positions determines exactly one valid order.

There are six valid orders: ABCD, ACBD, ACDB, CABD, CADB and CDAB, using each child’s initial. This also checks that all six choices work.

Useful beforehand: Systematic counting; Ordered positions. Find a concept to review.

C — 48 beads.

Let the original total be $N$. There were $N/4$ red beads. After adding twelve red beads, there are $N/4+12$ red beads among $N+12$ beads in all.

The new fraction gives $N/4+12=2(N+12)/5$. Multiplying both sides by 20 gives $5N+240=8N+96$. Subtracting $5N+96$ from both sides gives $144=3N$, so $N=48$.

Check: originally 12 of 48 beads were red. Afterwards 24 of 60 are red, and $24/60=2/5$.

Useful beforehand: Fractions; One-variable linear equations. Find a concept to review.

D — 54 rectangles.

To contain S, the top boundary can be either the line above row 1 or the line above row 2: 2 choices. The bottom boundary can be below row 2, row 3 or row 4: 3 choices.

The left boundary can be to the left of column 1, 2 or 3: 3 choices. The right boundary can be to the right of column 3, 4 or 5: 3 choices.

Any combination of these choices makes a rectangle containing S. Different boundary choices make different rectangles. Therefore the count is $2\times3\times3\times3=54$.

Useful beforehand: Multiplication principle; Grid rows and columns. Find a concept to review.

A — 156.

The positive integers leaving remainder 2 on division by 5 begin $2,7,12,17,22,\ldots$. The first also leaving remainder 3 on division by 7 is 17, since $17=2\times7+3$.

If two integers satisfy both conditions, their difference is divisible by 5 and by 7. Since 5 and 7 have no common prime factor, their difference is divisible by $5\times7=35$. Conversely, adding or subtracting 35 preserves both remainders.

All solutions are therefore $17+35k$ for integers k. In the range 1 to 100 these are 17, 52 and 87. The previous value is −18 and the next is 122, outside the range. Their sum is $17+52+87=156$.

Useful beforehand: Remainders; Common multiples. Find a concept to review.

C — 38 cm².

The square has area $10\times10=100$ cm². The three unshaded corner regions are right triangles.

Triangle APQ has perpendicular legs 4 and 6, so its area is $4\times6/2=12$. Since $PB=10-4=6$, triangle PBC has area $6\times10/2=30$. Since $QD=10-6=4$, triangle QDC has area $4\times10/2=20$.

These three triangles and CPQ exactly fill the square, without overlapping interiors. Thus the shaded area is $100-12-30-20=38$ cm².

Useful beforehand: Area of a square; Area of a right triangle. Find a concept to review.

D — 15.

We need $1000=2^3\times5^3$, so the product must contain at least three factors of 5 and three factors of 2.

Among the positive integers less than 15, only 5 and 10 are multiples of 5. Each contributes one factor of 5. There is no multiple of 25 in that range, so no factor contributes two 5s. Hence $n!$ cannot contain $5^3$ when $n\lt15$.

For $n=15$, the factors 5, 10 and 15 supply three 5s. The factors 2, 4 and 6 alone already supply more than three 2s. Thus $15!$ is divisible by 1000, and 15 is the least possible n.

Useful beforehand: Prime factorisation; Multiplication. Find a concept to review.

B — 10 draws.

Nine draws need not give four of one colour: it is possible to draw three red, three blue and three green, because the bag contains at least three of each colour.

If ten draws still gave at most three of each colour, the total drawn would be at most $3+3+3=9$, a contradiction. Thus ten draws always give at least four of one colour.

Nine can fail and ten always works, so the smallest guarantee is 10. The counts of 5, 7 and 9 ensure that the nine-draw example is actually possible.

Useful beforehand: Counting; Worst-case reasoning. Find a concept to review.

E — 17 routes.

Every route has seven steps, three of them upward. Choose their three positions among seven. There are $(7\times6\times5)/(3\times2\times1)=35$ choices: the numerator chooses the three positions in order, and division by 6 removes the six orders of the same three positions.

To reach X requires two right steps and one up step. The up step can occupy any of the three positions, so there are 3 first parts. From X to B requires two right steps and two up steps. The positions of the two up steps can be chosen in $(4\times3)/2=6$ ways.

Each route through X has exactly one such first and second part. There are $3\times6=18$ of these routes. Therefore $35-18=17$ routes avoid X.

Useful beforehand: Multiplication principle; Choosing positions. Find a concept to review.

B — 6.

The units digits of $7^1,7^2,7^3,7^4$ are 7, 9, 3, 1. Multiplying by 7 then begins the same list again. Similarly, the units digits of powers of 3 repeat as 3, 9, 7, 1.

Since $2026=4\times506+2$, the units digit of $7^{2026}$ is the second in its list: 9. Since $2027=4\times506+3$, the units digit of $3^{2027}$ is the third in its list: 7.

The sum ends as $9+7=16$, so its units digit is 6. Higher place values add multiples of 10 and cannot change this units digit.

Useful beforehand: Powers; Remainders. Find a concept to review.

E — 4 sets.

Put the lengths in order as $a\le b\le10$. The perimeter condition gives $a+b=14$. Since $b\le10$, we have $a\ge4$. Since $a\le b$, we have $a\le7$.

The only possibilities are $(4,10,10),(5,9,10),(6,8,10),(7,7,10)$. Each works: the two shorter sides add to 14, greater than the longest side 10. This is the only triangle inequality that can be tight when the sides are ordered; the other two are automatic for positive sides.

All four are valid and every possible value of a has been considered. The answer is 4.

Useful beforehand: Triangle inequality; Positive integers. Find a concept to review.

D — 27 counters.

A pile of 0 is losing for the player whose turn would come next, because the other player has just taken the last counter. We claim that every multiple of 3 is losing and every other pile is winning.

From a positive multiple of 3, removing 1, 2 or 4 always leaves a non-multiple of 3. Conversely, from a pile leaving remainder 1, remove 1; from a pile leaving remainder 2, remove 2. These moves are legal and leave a multiple of 3, possibly zero.

Starting from the losing pile 0, this proves the claim successively for larger piles: every move from a multiple reaches a smaller winning pile, while a non-multiple has a move to a smaller losing pile. Therefore a player can keep returning the opponent to multiples of 3 and eventually take the last counter.

Of the five choices, only 27 is divisible by 3. Hence 27 is the losing starting position.

Useful beforehand: Remainders on division by 3; Working from smaller cases. Find a concept to review.

A — 36 cubes.

A small cube on an outer edge but away from the corners has two outside faces, so exactly two painted faces. A corner cube has three painted faces and must not be counted. A cube away from every edge has at most one painted face.

Each of the four edges of length 4 contributes $4-2=2$ cubes after its corners are excluded. The four edges of length 5 contribute $5-2=3$ each, and the four edges of length 6 contribute $6-2=4$ each.

Different edges meet only at corners, which have already been excluded. There is therefore no double counting. The total is $4\times2+4\times3+4\times4=8+12+16=36$.

Useful beforehand: Faces and edges of a cuboid; Counting without overlap. Find a concept to review.

C — 150.

There are only finitely many positive integer triples with sum 16, so one of them has the largest product. Put its entries in order $a\le b\le c$.

Suppose $c-a\ge2$. Replace a by $a+1$ and c by $c-1$. Both remain positive and the sum stays 16. The change in the product is

$$b(a+1)(c-1)-abc=b(c-a-1).$$

Because $b\gt0$ and $c-a-1\ge1$, this change is positive. That contradicts the assumption that the old product was largest.

Consequently the largest and smallest entries at a maximum differ by at most 1. If the smallest entry a were at most 4, both other entries would be at most 5, and their sum would be at most $4+5+5=14$. If a were at least 6, the sum would be at least $6+6+6=18$. Both contradict the sum 16. Thus a is 5. All entries must be 5 or 6; since $5+5+5=15$, precisely one entry must be 6. The triple is therefore $(5,5,6)$. Its product is $5\times5\times6=150$. This triple exists, so the bound is attained.

Useful beforehand: Expanding a product; Positive integers; Comparing quantities. Find a concept to review.