graduate level exam (cgle)] maths practice set 15 ...€¦a tank is fitted with two taps. the first...

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MATHS PRACTICE SET 15 SOLVED [CombinedGraduate Level Exam (CGLE)]

MAXIMUM TIME: 20 minutes

1. The number 2564 x 6425 is square of a natural number n. Find the sum of thedigits of the number n.

A) 10B) 14B) 11B) 25

2. An electronic device makes a beep after every 60 sec. Another device makes abeep after every 62 sec. They beeped together at 10 am. The time when they willnext make a beep together at the earliest, is?

A) 10:28 amB) 10:30 amC) 10:31 amD) 10:15 am

3. 325 + 326 + 327 + 328 is divisible by?

A) 16B) 11C) 25D) 30

4. If (x + 1/x)2 = 3; then value of x206 + x200 + x90 + x84 + x12 + x6 +1 is?

A) 84B) 206C) 0D) 1

5. In 2 type of brass, ratio of copper and zinc are 8 : 3 and 15 : 7. The ratio in whichthese two type of brass should be mixed so that the ratio of copper and zinc in thisnew type of brass becomes 5 : 2 is?

A) 5 : 2

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B) 1 : 2 C) 4 : 3 D) 7 : 4

6. The average of marks scored by 35 students is 72. If the marks secured by 1student were written as 36 instead of 86, find the current average.

A) 73B) 73.43C) 75D) 74.25

7. A, B and C jointly thought of engaging themselves in a business venture. It wasagreed that A would invest Rs 6500 for 6 months, B Rs 8400 for 5 months and CRs 10,000 for 3 months. A wants to be the working member for which, he was toreceive 5% of the profits. The profit earned was Rs. 7400. Calculate the share of Bin the profit.

A) Rs. 1900B) Rs. 2660C) Rs. 2800D) Rs. 2840

8. A man spends 75% of his income. If his income is increased by 20% and heincreases his expenditure by 10%, his savings are increased by?

A) 45%B) 50%C) 55%D) 57%

9. A fruitseller buys x guavas for Rs y and sells y guavas for Rs x. If x > y, then findhis profit or loss?

A) (x² y²)/xy % lossB) (x² y²)/xy % gainC) (x² y²)/y2 % lossD) (x² y²)/y2 % gain

10. A sum of Rs 10 is lent to be returned in 11 monthly installments of Re 1 each,interest being simple. The rate of interest is?

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A) 100/11%B) 10%C) 11%D) 240/11%

11. A tank is fitted with two taps. The first tap can fill the tank completely in 45minutes and the second tap can empty the full tank in one hour. If both the tapsare opened alternatively for one minute, then in how many hours the empty tankwill be filled completely?

A) 2 hours 55 minutesB) 3 hours 40 minutesC) 4 hours 48 minutesD) 5 hours 53 minutes

12. A man can row upstream 10 km/h and downstream 20 km/h. Find the man'srate in still water and rate of the stream.

A) 0 km/h, 5 km/hB) 5 km/h, 15 km/hC) 15 km/h, 5 km/hD) 10 km/h, 5 km/h

13. If the radii of the circular ends of a conical bucket are 28 cm and 7 cm,whose height is 45 cm. Find the capacity of the bucket.

A) 48510 cm³B) 84510 cm³C) 45005 cm³D) 48000 cm³

14. Two circles C1 and C2 of radii 2 cm and 3 cm respectively touch each other asshown in the figure. If AD and BD are tangents then the length of BD is?

A) 15√10 cm

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B) 1.5√10 cmC) 10√10 cmD) √10 cm

15. From the top of a mountain 90 m high, the angle of depression of top andbottom point of a tower are 30° & 60° respectively. Find the height of the tower?

A) 45 mB) 60 mC) 75 mD) 30 m

QUESTIONS 16 to 20. The following piechart shows the contents of insects and rodents in anaverage Indian Household. Examine the chart and answer the questions.

16. If the percentage of spider is x% of the cockroach, then x is equal to?

A) 734 ∕7%

B) 752∕7%

C) 784 ∕7%

D) 754 ∕7%

17. The total percentage of spider, rat and cockroach is greater than thepercentage of ant by?

A) 34B) 32C) 28D) 24

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18. If the percentage of rat is y% of the total percentage of ant and cockroach, theny is equal to?

A) 30B) 35C) 25D) 20

19. The total percentage of common lizard, spider and cockroach is greater thanthe percentage of rat by?

A) 37B) 47C) 35D) 25

20. If the difference of the percentage of rat and miscellaneous is z% of thepercentage of cockroach, then z is equal to?

A) 25%B) 35%C) 20%D) 15%

SOLUTION

1. n² = 2564 x 6425

=> n² = (5²)64 x (26)25

=> n2 = 5128 x 2150

=> n2 = 5128 x 2(128+22)

=> n2 =(564)2 x (264)2 x (211)2

=> n = 564 x 264 x 211

=> n =1064 x 2048

If 10 raised to power any natural number multiplied with a number gives only zeros at the right,so the sum of the digit of the number n = 2 + 0 + 4+ 8 = 14 (option 'B')

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2. Obviously both the bells will make a beep together in the time equal to the LCM of the timethey make a beep.

Now the LCM of the times they make a beep = LCM of 60 and 62 seconds i.e. 1860 seconds= 1860/60 minutes = 31 minutes

Thus the time at which they will now make a beep together = 10 am + 31 minutes = 10:31 am(option 'C')

3. 325 + 326 + 327 + 328 = 324 x (3 + 32 + 33 + 34)

= 324 x (3 + 9 + 27 + 81) = 324 x 120

It's clear from the above that 120 is a factor of the given expression. We know that if a factor ofan expression is divisible by any number, that whole expression too is divisible by that number.

We can see 120 is divisible only by 30, so the given expression too is divisible by 30

Hence 30 (option 'D') is the answer.

4. Notice that there is a difference of 6 between the exponents of every set of two terms of thegiven expression

So the given expression can be rewritten: x203 (x3 + 1/x3) + x87(x3 + 1/x3) + x9 (x3 + 1/x3) +1

Now (x + 1/x)² = 3 (given)=> x + 1/x = √3 (squaring root of both the above sides)

Now cubing both side we get(x³ + 1/x³) + 3 (x + 1/x) = 3√3=> (x³ + 1/x³) + 3 √3 = 3√3 (putting x + 1/x = root 3 as shown above)=> (x³ + 1/x³) = 0

By putting it in the asked expression, you'll see the answer is 1 (option 'D')

5. First of all see QUERY 2 here to know how to do such types

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Here p/(p+q) = 8/(8+3) = 8/11r/(r+s) = 15/(15+7) = 15/22m/(m+n) = 5/(5+2) = 5/7

Here we see that 5/7 is greater than 15/22 and smaller than 8/11Now, 5/7 15/22 = 5/154 & 8/11 5/7 = 1/77

=> 5/154 : 1/77 = 5 : 2

So as explained above 5 : 2 is the answer (option 'A')

6. Total of marks to be increased due to error = 86 36 = 50

But this error will effect the average of all 35 studentsTherefore, average to be increased = 50/35 = 1.43 (appox)

Hence the current average = 72 + 1.43 = 73.43 (option 'B')

7. Money received by A for managing the business = 5% of Rs 7400 = Rs 370

Thus, the remaining profit = 7400 370 = Rs 7030

You should remember that when the profit sharing ratio is not given, the profit between thepartners is distributed in the ratio of their investments/capitals.

But here partners'investments are not comparable as they are given for different periods. Insuch a case it's better to find one month equivalent capital so that they become comparable.

Now one month equivalent investment of A = 6500*6 = 39000One month equivalent investment of B = 8400*5 = 42000One month equivalent investment of C = 10000*3 = 30000

So, the ratio of investments = 39000 : 42000 : 30000 = 13 : 14 : 10

Therefore, B's share = 7030 x 14/37 = Rs 2660 (option 'B')

8. Let his initial income = 100

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Then his expenditure = 75So, his savings = 100 75 = 25

His income after = 100 + 20% of 100 = 120And expenditure after = 75 + 10% of 75 = 82.50So his savings now = 120 82.50 = 37.50

Therefore increase in his savings = 37.50 25 = 12.50

Now increase on 25 = 12.50Hence increase on 100 = (12.50/25)*100 = 50% (option 'B')

9. Let the seller buys xy guavas (take LCM of x & y to make the calculation easier)

Now we have to find the profit earned or loss suffered, so we need to know the cost price as wellas the selling price.

Thus the CP of xy guavas = y/x*xy = y²and the SP of xy guavas = x/y*xy = x²

As x is greater than y, so the selling price is more than the cost price. Obviously it will be a gain.So gain = x² y²and gain in percentage = [(x² y²)/y²]*100 = (x² y²)/y² % (option 'D')

10. When it's a monthly installment it's better to find the one month equivalent principal.

As the sum of Rs 10 has be returned in 11 monthly equal installments, the payment of theprincipal to be made will end on the completion of 10 months, Re 1 being for every month.

And the principal for the first month will be Rs 10, for the second month Rs 9, for the thirdmonth Rs 8, and so on.

Thus the one month equivalent sum = 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55

As this sum of Rs 10 has to be paid in 11 monthly installments of Re 1 each, the amount(principal + interest) will be Rs 11, and the interest to be paid = 11 10 = 1

Now the principal (equivalent to one month) = 55;interest = 1

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time = 1 month = 1/12 years (as the principal has been converted equivalent to one month)

So rate = 100*12/55*1 (rate = 100*interest/principal*time)=> 1200/55=> 240/11 So option 'D' is correct.

11. Both the taps are kept open alternatively for 1 minute each, we'll find the part of the tankfilled in 2 minutes i.e. one interval of time; so it is 1/45 1/60 = 1/180

It's obvious that the second tap which is emptying the tank will not have to be opened in the lastinterval, otherwise the tank will never fill; so we need to know the work of the first tap in thelast 1 minute.

So, work of tap first in 1 minute = 1/45It does mean both the taps work simultaneously to fill = 1 1/45 = 44/45 of the tank.

Time taken to fill 1/180 of the tank by both the taps working together = 2 minutesHence, time taken in all to fill 44/45 of the tank = 352 minutes

So, total time to fill the full tank = time taken for 44/45 of the part + time taken for 1/45 of thepart= 352 + 1 = 353 minutes = 5 hours 53 minutes (option 'D')

12. Remember,Speed of a body in still water = (speed downstream + speed upstream)/2Speed of the current/stream = (speed downstream speed upstream)/2

Therefore, man's rate in still water = (20 + 10)/2 = 15 km/h (option 'C')And rate of the stream = (20 10)2 = 5 km/h (option 'C')

13. You see, the bucket forms a frustum of a cone such that the radii of its circular ends are R =28 cm, r = 7 cm and height is = 45 cm.

Therefore, capacity of the bucket (volume of the frustum)

= 1∕3πh (R² + Rr + r²)

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= 1∕3 π*45(28² + 7² + 28 *7)

= 22*15*(28*4 + 7 + 28)= 330*147 cm³= 48510 cm³ (option 'A')

14. Let the point of intersection of the tangent AD to the circle C2 be T and the center of C2 be Oand join O and T to form OT. Thus ΔOTA is a right angle triangle, where ∠OTA = 90°, AO =diameter of C1+radius of C2 = 4+3 = 7 and OT = radius of C2 = 3, and AO being the hypotenuse.

Therefore AT = √(7² 3²) = √40 = 2√10 ...............by Pythagoras

Now, as BD is another tangent to C2, B being the point of tangent, and it's joined with thecenter of the circle, so ΔABD too is a right triangle, where angle ABD = 90°, AB = sum of thediameters of the two circles given = 4+6 = 10 cm and AD being the hypotenuse.

Note that BT and TD are equal as they both are tangents to the same circle joined together (herejoined at D)Now let BD = x cm, therefore AD = (2√10 + x) cm

Hence by Pythagoras BD² = (2√10 +x)² 10²

=> x² = 40 + x² + 4x√10 100

=> 4x√10 60 = 0

=> x√10 = 15

=> x = 15/√10 = (15/10)√10 ............by rationalization of √10

=> x = 1.5√10 (option 'B')

15.

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In ΔABD = tan 60 = DB/AB=> √3 = 90/AB=> AB = 30√3

But EC = AB, therefore EC = 30√3

Now, in ΔECD, tan 30 = CD/EC=> 1/√3 = CD/30√3=> CD = 30

Hence, height of the tower i.e. AB = BD CD = 90 30 = 60 m (option 'B')

16. According to the questionx% = 26/28 × 100 = 550/7

=> x = 784 ∕7% (option 'C')

17. Total percentage of spider, rat and cockroach = 22 + 11 + 28 = 61%Therefore by how much this is greater than the percentage of ant = 61 27 = 34% (option 'A')

18. Total percentage of ant and cockroach = 27 + 28 = 55%

Now according to the questiony% = 11/55 × 100 = 20

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=> y = 20%

19. Total percentage of common lizard, spider and cockroach = 8 + 22 + 28 = 58%Therefore by how much this is greater than the percentage of rat = 58 11 = 47% (option 'B')

20. Difference of the percentage of rat and miscellaneous = 11 4 = 7%

Now, according to the questionz% = 7/28 × 100 = 25=> z = 25% (option 'A')

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graduate level exam (cgle)] maths practice set 15 ...€¦a tank is fitted with two taps. the first...

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