Math Problem Solutions

Below are the step‐by‐step solutions and final answers for each question.

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1.
We are given a three–digit number written as 52 (with “” as a missing digit) that is divisible by 6. For divisibility by 6 the number must be divisible by both 2 and 3.
– Divisibility by 2 is automatic since the last digit is 2 (an even number).
– For divisibility by 3 the sum of its digits must be divisible by 3. Here the sum is
  5 + * + 2 = 7 + *
Testing the options:
• If * = 2, then 7 + 2 = 9, which is divisible by 3.
• If * = 3, then 7 + 3 = 10 (not divisible by 3).
• If * = 6, then 7 + 6 = 13 (not divisible by 3).
• If * = 7, then 7 + 7 = 14 (not divisible by 3).

Answer 1: 2

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2.
Let the two numbers be $a$ and $b$ with
  $a + b = 12$ and $ab = 35$.
The sum of the reciprocals is

$$\frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab} = \frac{12}{35}.$$

Answer 2: $\frac{12}{35}$

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3.
We wish to evaluate

$$(963+476)^2 + (963-476)^2.$$

Recall the identity:

$$(a+b)^2 + (a-b)^2 = 2(a^2+b^2).$$

Thus the given expression equals

$$2\bigl(963^2 + 476^2\bigr).$$

Now note that one of the answer choices is “$963 \times 963 + 476 \times 476$”. Since our expression is exactly twice that quantity, the value of the ratio

$$\frac{(963+476)^2+(963-476)^2}{963^2+476^2} = 2.$$

Among the provided options the correct answer is 2.

Answer 3: 2

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4.
We need the greatest number $d$ that, when dividing 43, 91 and 183, leaves the same remainder. This happens when $d$ divides all the differences between the numbers. Compute the differences:

• $91-43 = 48$
• $183-91 = 92$
• $183-43 = 140$

Now find $\gcd(48,92,140)$.
First, $\gcd(48,92)$:
 48 = $2^4\times3$ and 92 = $2^2\times23$ so $\gcd(48,92)=2^2=4$.
Next, $\gcd(4,140)=4$ (since 140 is divisible by 4).

Answer 4: 4

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5.
We are to find the greatest four–digit number divisible by 15, 25, 40, and 75. First, determine the LCM of these numbers. Write each in prime–factorization form:

• $15 = 3 \times 5$
• $25 = 5^2$
• $40 = 2^3 \times 5$
• $75 = 3 \times 5^2$

Thus,

$$\text{LCM} = 2^3 \times 3 \times 5^2 = 8 \times 3 \times 25 = 600.$$

Now, the largest four–digit number is 9999. Dividing by 600,

$$\lfloor9999/600\rfloor = 16,\quad \text{and } 16 \times 600 = 9600.$$

Answer 5: 9600

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6.
The family consists of:

• 2 grandparents (avg = 67 years)
• 2 parents (avg = 35 years)
• 3 grandchildren (avg = 6 years)

Total age:

• Grandparents: $2 \times 67 = 134$
• Parents: $2 \times 35 = 70$
• Grandchildren: $3 \times 6 = 18$

Total = $134 + 70 + 18 = 222$
Number of persons = $2+2+3 = 7$

Average age = $222/7 \approx 31.714$ years, i.e. $31 \frac{5}{7}$ years.

Answer 6: $31\frac{5}{7}$ years

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7.
There are three batches:

• 55 students with an average of 50 → Total marks = $55 \times 50 = 2750$
• 60 students with an average of 55 → Total = $60 \times 55 = 3300$
• 45 students with an average of 60 → Total = $45 \times 60 = 2700$

Total students = $55 + 60 + 45 = 160$
Total marks = $2750 + 3300 + 2700 = 8750$

Overall average = $8750/160 \approx 54.6875$.

Answer 7: Approximately 54.68

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8.
100 oranges are bought for Rs.350, so cost price per orange = $350/100 = 3.5$ Rs.
They are sold at Rs.48 per dozen, so selling price per orange = $48/12 = 4$ Rs.

Profit per orange = $4 - 3.5 = 0.5$ Rs.
Profit percentage = $\frac{0.5}{3.5}\times100 \approx 14.29\%$.

Answer 8: Approximately 14.3% gain

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9.
Let the daily work rates of a man and a woman be $m$ and $w$ respectively.
We are told:

• 4 men + 6 women finish the work in 8 days:
   $(4m+6w)\times8 = 1$ → $4m+6w = \frac{1}{8}$. (1)
• 3 men + 7 women finish the work in 10 days:
   $(3m+7w)\times10 = 1$ → $3m+7w = \frac{1}{10}$. (2)

Subtract (2) from (1) suitably or solve by elimination.
Solve (1) for one variable:  $4m = \frac{1}{8} - 6w$ → $m = \frac{1}{32} - \frac{3}{2}w$.

Plug into (2):  $3\Bigl(\frac{1}{32} - \frac{3}{2}w\Bigr) + 7w = \frac{1}{10}$.
Compute:  $\frac{3}{32} - \frac{9}{2}w + 7w = \frac{1}{10}$
 $\frac{3}{32} + \frac{5}{2}w = \frac{1}{10}$.

Now solve for $w$:  $\frac{5}{2}w = \frac{1}{10} - \frac{3}{32}$.
Find a common denominator:
 $\frac{1}{10} = \frac{16}{160},\; \frac{3}{32} = \frac{15}{160}$
 So, $\frac{5}{2}w = \frac{1}{160}$ → $w = \frac{1}{160} \times \frac{2}{5} = \frac{1}{400}$ (work units per day).

Ten women work at a rate $10w = 10/400 = 1/40$ per day.
Thus, the time required = 1 ÷ (1/40) = 40 days.

Answer 9: 40 days

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10.
Let the speed of the boat in still water be 10 mph and the speed of the stream be $x$ mph. Then:

• Downstream speed = $10+x$
• Upstream speed = $10-x$

The boat covers 36 miles downstream and upstream with a time difference of 90 minutes (1.5 hours):

$$\frac{36}{10-x} - \frac{36}{10+x} = 1.5.$$

Combine the fractions:

$$36\left(\frac{(10+x) - (10-x)}{(10-x)(10+x)}\right) = 36\left(\frac{2x}{100-x^2}\right) = \frac{72x}{100-x^2} = 1.5.$$

Solving:

$$72x = 1.5(100-x^2) \quad\Longrightarrow\quad 72x = 150 - 1.5x^2.$$

Rearrange:

$$1.5x^2 + 72x -150 = 0.$$

Divide through by 1.5:

$$x^2 + 48x - 100 = 0.$$

The discriminant is:

$$\Delta = 48^2 + 4\times100 = 2304 + 400 = 2704,$$

and $\sqrt{2704}=52$. Thus:

$$x = \frac{-48 \pm 52}{2}.$$

Taking the positive root:

$$x = \frac{4}{2} = 2 \text{ mph}.$$

Answer 10: 2 mph

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11.
Let the boat’s speed in still water be $b$ and the stream’s speed be $s$.

• Going 2 km against the current in 1 hour: $b-s = 2$ km/h.
• Going 1 km with the current in 10 minutes ($1/6$ hour): $b+s = \frac{1}{(1/6)} = 6$ km/h.

Solve:

$$b = \frac{(b+s) + (b-s)}{2} = \frac{6+2}{2} = 4 \text{ km/h}.$$

In still water the speed is 4 km/h, so time to cover 5 km is:

$$\frac{5}{4} \text{ hours} = 1.25 \text{ hours} = 1 \text{ hr } 15 \text{ min}.$$

Answer 11: 1 hour 15 minutes

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12.
Question: What simple interest rate per annum did Anand pay to Deepak?

Statement I: Anand borrowed Rs.8000 for four years.
Statement II: Anand returned Rs.8800 to Deepak at the end of two years and settled the loan.

To calculate simple interest rate per annum we need the principal, the time period over which interest is charged, and the interest amount.
– Statement I gives the principal (Rs.8000) but not the time for which interest was actually charged (since the loan was repaid early, as per Statement II).
– Statement II tells us that he repaid Rs.8800 at the end of 2 years. Thus the interest for 2 years = Rs.8800 – Rs.8000 = Rs.800.

Using both statements:
Interest per annum = $800/2 = 400$ Rs.
Rate = $\frac{400}{8000} \times 100 = 5\%$ per annum.

Each statement by itself is insufficient; both are needed.

Answer 12: (E) Both I and II are necessary to answer

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13.
Question: What is the compound amount?
Statement I: Rs.200 was borrowed for 192 months at 6% compounded annually.
Statement II: Rs.200 was borrowed for 16 years at 6%.

Note that 192 months = 16 years. Thus the two statements are equivalent. Either one alone gives enough information to compute the compounded amount.

Answer 13: (C) Either I or II alone is sufficient to answer

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14.
Question: What is the area of the given rectangle?
Statement I: The perimeter of the rectangle is 60 cm.
Statement II: The breadth of the rectangle is 12 cm.
Statement III: The sum of two adjacent sides is 30 cm.

For a rectangle, if the perimeter is 60 cm then

$$2(l+b) = 60 \quad\Longrightarrow\quad l+b = 30.$$

Statement III gives the same information directly. Thus only one of I or III is needed along with II. With breadth = 12 cm, the length is:

$$l = 30 - 12 = 18 \text{ cm, and area } = 18 \times 12 = 216 \text{ cm}^2.$$

The redundant information is the extra statement giving the same sum of adjacent sides.

Answer 14: (E) Statement II and either I or III are needed

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15.
A and B run a 100 m race.
– A’s speed = 5 kmph. Convert to m/s:
 $5 \text{ kmph} = \frac{5\times1000}{3600} = \frac{25}{18}$ m/s.
A gives B an 8 m head start, so B runs only 92 m in the race.
Let $t_A$ be A’s time and $t_B$ be B’s time. We have:

$$t_A = \frac{100}{25/18} = 100\times\frac{18}{25} = 72 \text{ seconds.}$$

It is given that A beats B by 8 seconds, so:

$$t_B = 72 + 8 = 80 \text{ seconds.}$$

Then B’s speed (in m/s) is:

$$v_B = \frac{92}{80} = 1.15 \text{ m/s.}$$

Convert to kmph:

$$1.15 \times 3.6 = 4.14 \text{ kmph.}$$

Answer 15: 4.14 kmph

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16.
A clock started at noon; by 10 minutes past 5 (i.e. 5:10), the elapsed time is 5 hours 10 minutes = $5 + \frac{10}{60} = 5+\frac{1}{6} = \frac{31}{6}$ hours.
Since the hour hand moves 360° in 12 hours, its rate is 30° per hour.
Thus, the angle turned = $\frac{31}{6} \times 30 = 155^\circ.$

Answer 16: 155°

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17.
We want to form a committee of 5 men (from 8) and 6 women (from 10).
The number of ways is:

$$\binom{8}{5} \times \binom{10}{6}.$$

Note that $\binom{8}{5} = \binom{8}{3} = 56$ and $\binom{10}{6} = \binom{10}{4} = 210$.
Thus, total ways = $56 \times 210 = 11\,760.$

Answer 17: 11,760

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18.
A “true discount” is the difference between the amount due and its present value under simple interest. Let the rate (per period) be $r\%$ and time $T$. For a bill of Rs.110, the present value is:

$$PV = \frac{110}{1 + \frac{rT}{100}}.$$

Thus, the true discount (TD) is:

$$TD = 110 - \frac{110}{1+\frac{rT}{100}} = \frac{110\left(\frac{rT}{100}\right)}{1+\frac{rT}{100}}.$$

We are told that when $T$ is a certain time the true discount is Rs.10. Let $x = \frac{rT}{100}$. Then:

$$\frac{110x}{1+x} = 10 \quad\Longrightarrow\quad 110x = 10(1+x) \quad\Longrightarrow\quad 110x = 10 + 10x,$$

which gives:

$$100x = 10 \quad\Longrightarrow\quad x = 0.1.$$

Now, if the time is doubled (i.e. $2T$), then the new value is $2x = 0.2$. The new true discount is:

$$TD' = \frac{110 \times 0.2}{1+0.2} = \frac{22}{1.2} \approx 18.33.$$

Answer 18: Rs.18.33

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19.
The sequence is: 2, 5, 10, 17, 26, 37, 50, 64.
Notice the pattern:
 $1^2+1=2$
 $2^2+1=5$
 $3^2+1=10$
 $4^2+1=17$
 $5^2+1=26$
 $6^2+1=37$
 $7^2+1=50$
 $8^2+1=65$ —but the last term given is 64, not 65.

Thus, 64 is the odd one out.

Answer 19: 64

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20.
The sequence is: 10, 25, 45, 54, 60, 75, 80.
Notice that all numbers except 54 are divisible by 5. (In fact, 10, 25, 45, 60, 75, and 80 are multiples of 5, while 54 is not.)

Answer 20: 54

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Summary of Answers

1. 2
2. $\frac{12}{35}$
3. 2
4. 4
5. 9600
6. $31\frac{5}{7}$ years
7. Approximately 54.68
8. Approximately 14.3% gain
9. 40 days
10. 2 mph
11. 1 hour 15 minutes
12. (E) Both I and II are necessary to answer
13. (C) Either I or II alone is sufficient to answer
14. (E) Statement II and either I or III are needed
15. 4.14 kmph
16. 155°
17. 11,760
18. Rs.18.33
19. 64
20. 54