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## AP Grama Sachivalayam Aptitude Quiz Questions & Answers

Many of the candidates feel difficult to solve and answer the AP Grama Sachivalayam Aptitude Quiz. For all such candidates, we suggest to practice the below provided** AP Grama Sachivalayam Aptitude Free Mock Test Question & Answers**. Aptitude will be one of the section in AP Grama Sachivalayam Online Exam 2019. So, all the candidates need to practice the given important AP Grama Sachivalayam **Aptitude Mock Test** Q&A.

**1)** A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?

A. 120 metres

B. 180 metres

C. 324 metres

D. 150 metres

**Answer:** 150 metres

**Explanation:** Speed = (60 x 5/18)m/sec = 50/3 m/sec.

Length of the train = (Speed x Time).

Length of the train = (50/3 * 9)m = 150 m

**2)** A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:

A. 100 kmph

B. 110 kmph

C. 120 kmph

D. 130 kmph

**Answer:** 120 kmph

**Explanation:** Let speed of the car be x kmph.

Then, speed of the train = (150/100)x = (3/2 x) kmph

Therefore, (75/x) – (75/ (3/2)x) = (125/ 10×60)

Hence, (75/x) – (50/x) = 5/ 24

The value of x = (25*24/5) = 120 kmph

**3)** Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:

A. 39, 30

B. 41, 32

C. 42, 33

D. 43, 34

**Answer:** 42, 33

**Explanation:** Let their marks be (x + 9) and x.

Then, x+9 = (56/ 100 (x+9+x))

Implies, 25(x + 9) = 14(2x + 9)

3x = 99

x = 33

So, their marks are 42 and 33.

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

A. Rs. 1900

B. Rs. 2660

C. Rs. 2800

D. Rs. 2840

**Answer:** Rs. 2660

**Explanation:** For managing, A received = 5% of Rs. 7400 = Rs. 370.

Balance = Rs. (7400 – 370) = Rs. 7030.

Ratio of their investments = (6500*6):(8400*5):(10000*3) = 39000:42000:30000 = 13 : 14 : 10

Therefore, B’s share = Rs. (7030 * (14/37)) = Rs. 2660

**5)** The present ages of three persons in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years).

A. 8, 20, 28

B. 16, 28, 36

C. 20, 35, 45

D. None of these

**Answer:** 16, 28, 36

**Explanation:** Let their present ages be 4x, 7x and 9x years respectively.

Then, (4x-8) + (7x-8) + (9x-8) = 56

Implies, 20x = 80

x = 4.

Therefore, Their present ages are 4x = 16 years, 7x = 28 years and 9x = 36 years respectively.

**6)** What was the day of the week on 17th June, 1998?

A. Monday

B. Tuesday

C. Wednesday

D. Thursday

**Answer:** Wednesday

**Explanation:** 17th June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)

Odd days in 1600 years = 0

Odd days in 300 years = (5 x 3) 1

97 years has 24 leap years + 73 ordinary years.

Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.

Jan. Feb. March April May June

(31 + 28 + 31 + 30 + 31 + 17) = 168 days

Implies, 168 days = 24 weeks = 0 odd day.

Total number of odd days = (0 + 1 + 2 + 0) = 3.

As a result, Given day is Wednesday.

**7)** A man’s speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. The man’s speed against the current is:

A. 8.5 km/hr

B. 9 km/hr

C. 10 km/hr

D. 12.5 km/hr

**Answer:** 10 km/hr

**Explanation:** Man’s rate in still water = (15 – 2.5) km/hr = 12.5 km/hr.

Man’s rate against the current = (12.5 – 2.5) km/hr = 10 km/hr.

**8)** How many 4-letter words with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’, if repetition of letters is not allowed?

A. 40

B. 400

C. 5040

D. 2520

**Answer:** 5040

**Explanation:** ‘LOGARITHMS’ contains 10 different letters.

Required number of words = Number of arrangements of 10 letters, taking 4 at a time.

= 10P4

= (10 x 9 x 8 x 7)

= 5040.

**9)** A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is B’s share?

A. Rs. 500

B. Rs. 1500

C. Rs. 2000

D. None of these

**Answer:** Rs. 2000

**Explanation:** Let the shares of A, B, C and D be Rs.5x, Rs.2x, Rs.4x and Rs.3x respectively.

Then, 4x – 3x = 1000

Therefore, x = 1000.

B’s share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.

**10)** The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

A. 276

B. 299

C. 322

D. 345

**Answer:** 322

**Explanation:** Clearly, the numbers are (23 x 13) and (23 x 14).

Larger number = (23 x 14) = 322

**11)** A mixture of wheat is sold at Rs.3 per Kg. This mixture is formed by mixing the Wheat of Rs.2.10 per kg and Rs.2.52 per kg. What is the ratio of price of cheaper to the costlier quality in the mixture if the profit of 25% is being earned

A. 2:5

B. 2:1

C. 2:3

D. 2:7

E. 2:9

**Answer:** 2:5

**Explanation:** Selling Price = x + 25*x/100 = 3; x = 2.4

210………………….252

…………240…………..

12…………………….30

= 2:5

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**12)** The ratio of Solution “A” and Solution “B” in the container is 3:2 when 10 liters of the mixture is taken out and is replaced by the Solution “B”, the ratio become 2:3. The total quantity of the mixture in the container is:

A. 25L

B. 20L

C. 30L

D. 45L

E. None of these

**Answer:** 30L

**Explanation:**

Initial = 3:2 ; After replacement = 2:3

2/3 = (1 – 10/n)

n = 30L

**13)** 10 machines can wash 20 cars in 6 hours. In how many hours can 15 machines wash 40 cars?

A) 8

B) 7

C) 15

D) 12

E) 6

**Answer:** 8

**Explanation:** Washing cars is a work, so

M1*H1*W2 = M2*H2*W1

10*6*40 = 15*H2*20

Solve, H2 = 8 hrs

**14)** The present worth of a sum due sometime hence is Rs. 576 and the banker’s gain is Rs. 16. The true discount is:

A. Rs. 36

B. Rs. 72

C. Rs. 48

D. Rs. 96

**Answer:** 96

**Explanation:** T.D. = P.W. x B.G. = 576 x 16 = 96.

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### AP Grama Sachivalayam Aptitude Online Test Q&A

By taking more and more AP Grama Sachivalayam Aptitude Mock Tests, you can easily gain knowledge over the type of AP Grama Sachivalayam Aptitude Online Test. By this, you can gain more marks in the AP Grama Sachivalayam Exam 2019. So, keep checking the current article for more preparation.

**15)** Find the H.C.F, if the numbers are in the ratio of 4 : 5 : 6 and their L.C.M. is 2400.

A: 35

B: 20

C: 40

D: 67

**Answer:** 20

**Explanation:** Let the numbers be 3x, 4x and 5x.

Then, their L.C.M. = 120x.

So, 120x = 2400 or x = 20.

=> The numbers are (4 x 20), (5 x 20) and (6 x 20).

Hence, required H.C.F = 20.

**16)** The cost of setting up a magazine is Rs. 2800. The cost of paper and ink etc is Rs. 80 per 100 copies and printing cost is Rs. 160 per 100 copies. In last month 2000 copies were printed but only 1500 copies could be sold at Rs. 5 each. Total 25% profit on the sale price was realized. There is one more resource of income from magazine which is advertising. What sum of money obtained from the advertising in magazine?

A. Rs.1750

B. Rs.1150

C. Rs.2350

D. Rs.1975

**Answer:** Rs.1975

**Explanation:** Set up cost = Rs. 2800

Paper etc = Rs. 1600

Printing cost = Rs. 3200

Total cost = Rs. 7600

Total sale price = 1500*5 = 7500

Let amount obtained from advertising be x then,

(7500+x)-7600 = 25 % of 7500

x = 1975.

**17)** Find the odd man out? 396, 462, 572, 427, 671, 264.

A. 671

B. 462

C. 427

D. 264

**Answer:** 427

**Explanation:** In each number except 427, the middle digit is the sum of other two.

**18)** When 0.36 is written in simplest form, the sum of the numerator and the denominator is

A: 15

B: 34

C: 64

D: 13

**Answer:** 34

**Explanation:**

0.36 = 36/100 = 9/25.

Sum of the numerator and denominator is 9 + 25 = 34

**19)** Functions f and g are defined by:

f(x) = 1/x + 3x and g(x) = -1/x + 6x – 4

The domain of (f+g)(x) is:

A. (-infinity , 4) U (0 , + infinity)

B. (-infinity , 0) U (0 , + infinity)

C. (-infinity , 0) U (0 , 9)

D. None of these

**Answer:** (-infinity , 0) U (0 , + infinity)

**Explanation: **(f + g)(x) = f(x) + g(x) = (1/x + 3x) + (-1/x + 6x – 4)

(f + g)(x) = 9 x – 4

Domain of f + g is given by the interval (-infinity , 0) U (0 , + infinity)

**20)** It was Thursday on 2nd Jan 1993. What day of the week will be on 15th March 1993?

A. Monday

B. Wednesday

C. Friday

D. Saturday

**Answer:** Saturday

**Explanation:**

Today number of days = 29(Jan) + 28 (Feb) + (15) Mar = 72 days 10 Weeks + 2 days. Thus the given date will fall on two days beyond Thursday, that is Saturday.

**21)** Can you find the value for the following expression:

(29)^2 * 91 ÷ 7 + (8)^3 + 61 =(?) ^ 3 + 858

A. 22

B. 19

C. 31

D. 27

**22)** Find the center of the circle whose equation is x^2 + y^2 -10x + 12y -10 = 0

A. (5, -6)

B. (5,6)

C. (-5, -6)

D. (10, 12)

**Answer:** (5, -6)

**Explanation:**

For general format ofcircle: ax2+by2+cx+dy+e= 0

The center-radius form of the circle equation is in the format(x–h)2+ (y–k)2=r2, with the center being at the point(h, k) and the radius being “r”.

Given eqn is: x^2 + y^2 -10x + 12y -10 = 0

=> x^2 + y^2 -10x + 12y = 10

Group thex-stuff together. And theny-stuff together.

=>(x^2 – 10x) + (y^2 + 12y) = 10

Take thex-term coefficient, multiply it by one-half, square it, and then add this to both sides of the equation, as shown. Do the same with they-term coefficient.

=> (x^2 – 10x + 25) + (y^2 + 12y + 36) = 10 + 25 + 36

=> (x – 5)^2 + (y + 6)^2 = 71

=> (x – 5)^2 + [y – (-6)]^2 = 71

On comparing it with center-radius form of the circle equation: (x–h)^2+ (y–k)^2=r^2

=> The center is at (h, k) = (5, -6)

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**23)** A seller gains the cost of 40 dozen apples by selling 25 dozen of apples. Find out the gain percent

A. 60 %

B. 40 %

C. 58 %

D. 62 %

**Answer:** A

**Explanation: **Cost price (C.P) of 40 dozen of apples is equal to selling (S.P) of 25 dozen of apples.

Let the C.P of 1 dozen of apple = Rs.1

Therefore C.P of 25 dozen apples = Rs. 25

and C.P of 40 dozen apples = Rs.40

=> C.P of 40 dozen of apples = S.P of 25 dozen apples = Rs.40

Profit % = (S.P of 25 dozen apples – C.P of 25 dozen apples) / C.P of 25 dozen apples * 100%

= {(40 – 25) / 25} * 100%

= {15 / 25} * 100%

= 60%

Therefore, required Profit % = 60 %

**24)** Two people run around circular track and take 42 sec and 30 sec to make one complete round. If they start together after how much amount of time will they meet again in the same place ?

A. 3 min 30 sec

B. 4 min 40 sec

C. 3 min 20 sec

D. 2 min 50 sec

**Answer:** 3 min 30 sec

**Explanation:**

Two people run around circular track and take 42 sec and 30 sec.

LCM of ( 42, 30) = 210

They will meet in 210 seconds, which is 3 min 30 sec.

**25)** A and B start a business, with A investing the total capital of Rs.50000, on the condition that B pays A interest @ 10% per annum on his half of the capital. A is a working partner and receives Rs.1500 per month from the total profit and any profit remaining is equally shared by both of them. At the end of the year, it was found that the income of A is twice that of B. Find the total profit for the year?

A. Rs. 39000

B. Rs. 49000

C. Rs. 59000

D. Rs. 69000

**Answer:** Rs. 59000

**Explanation:**

A invested amount = Rs. 50,000

Interest received by A from B = 10% of half of Rs.50,000

= 10% of Rs. 25000

= Rs.2500.

Amount received by A per annum for being a working partner = 1500 x 12

= Rs.18000

Let ‘P’ be the part of the remaining profit amount that A receives as his share.

So,total income of A = (Rs.2500 + Rs.18000 + Rs. P ) = (Rs. 20500 + Rs. P)

Total income of B = only his share from the remaining profit amount = ‘P’, (Given as A and B share the remaining profit amount equally.)

Given, income of A = Twice the income of B

So, (20,500 + P ) = 2(P)

=> P = 20,500

Thus, the Total profit = 2P + Rs.18000

= 2 * (20500) + 18000

= Rs. 59000

Therefore, Total Profit = Rs. 59000

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