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Mental Ability Questions & Answers (Compiled from UPSC, SSC ,PSC ,IBPS previous question papers)

4 years ago, the respective ratio between 1/2 of A's age at that time and four times of B's age at that time was 5:12. Eight years hence, 1/2 of A's age at time will be less than B's age at that time by 2 years. What is B's present age
A10 years

B14 years
C12 years
D5 years
E8 years
Explanation:

Let the present age of A = x years

Let the present age of B = y years

According to question,

x-4/2÷4(y-4) = 5/12

3x-12=10y-40

3x-10y=-28

y+8-x+8/2=2

2y-x= -4

On solving eq. (i) and (ii),

x=24, y=10

Present age of B = 10 years

A sum of money was invested for 14 years in Scheme A which offers simple interest at a rate of 8% p.a. The amount received from Scheme A after 14 years was then invested for two years in Scheme B which offers compound interest (compounded annually) at a rate of 10% p.a. If the interest received from Scheme B was INR 6678/- what was the sum invested in Scheme A
AINR 15500/-
BINR 14500/-
CINR 16000/-
DINR 12500/-
EINR 15000/-

Explanation:

Let the sum of money invested in scheme A be INR x.

Amount = x×8×14/100+x = 53/25x

Now, C.I. for 2 years = 53/25x[(1+10/100)²-1]

6678=53/25×[(11/10)²-1]

6678=53/25×[121-100/100]

6678=53/25x×21/100

6678×100×25/21×53=x

x=INR 15000/-

There are three positive numbers, 1/3rd of average of all the three numbers is 8 less than the value of the highest number. Average of the lowest and the second lowest number is 8. Which is the highest number
A11

B14
C10
D9
E13
Explanation:

Let the three numbers be x, y and z

It is given that, y+z/2 = 8

⇒ y+z=16

and x-1/3[x+y+z/3] = 8

⇒x-x+y+z/9=8

⇒x-x+16/9=8

⇒9x-x-16=72

⇒8x=88

∴x=11

Directions (next five questions) : Refer to the pie charts and answer the given questions.

Distribution of number of cellular phones (both Nokia

and Samsung) sold by 5 stores in 2012

Total number = 6400

Distribution of number of Nokia cellular

phones sold by 5 stores

Total number = 3000

Number of Samsung cellular phones sold by store E is what percent of the number of phones (both Nokia and Samsung) sold by store C
A55%

B45%
C60%
D65%
E50%
Explanation:

Number of Samsung cellular phones sold by store E

= 12/100×6400-8/100×3000=528

Number of phones (both) sold by store C

= 15/100×6400=960

∴ Required percentage = [528/960×100]%

=55%

What is the difference between number of cellular phones (both Nokia and Samsung) sold by store A and total number of Nokia cellular phones sold by stores B and E together
A442
B436

C432
D428
E426
Explanation:

Number of cellular phones (both) sold by store A

= 19/100×6400=1216

Total number of Nokia cellular phones sold by stores B and E = (18+8)/100×3000=780

Required difference = 436

What is the average number of Samsung cellular phones sold by stores B,C and D
A796
B792

C780
D812
E784
Explanation:

Samsun cellular phones sold by store B

21/100×6400-18/100×3000 = 804

Samsung cellular phones sold by store C

= 15/100×6400-20/100×3000 = 360

Samsung cellular phones sold by store D

= 33/100×6400-30/100×3000 = 1212

∴Required average = 804+360+1212/3 = 792

What is the central angle corresponding to number of cellular phones (both Nokia and Samsung) sold by store D
A124.2⁰
B112.4⁰
C115.2⁰
D120.8⁰
E118.8⁰

Explanation:

Central angle = 33/100×360⁰ =118.8⁰

Number of cellular phones (both Nokia and Samsung) sold by store B is what percent more than the number of Nokia cellular phones sold by store A
A82 1/3%
B86 2/3%

C84 2/3%
D92 2/3%
E78 1/3%
Explanation:

Number of cellular phones both sold by store B

=21/100×6400=1344

Number of Nokia phones sold by store A

=24/100×3000=720

∴ Required percentage = [1344-720/720×100]%

A, B and C started a business by investing INR 20000/-, INR 28000/- and INR 36000/- respectively. After 6 months, A and B withdrew an amount of INR 8000/- each and C invested an additional amount of INR 8000/-. All of them invested for equal period of time. If at the end of the year, C got INR 12550/- as his share of profit, what was the total profit earned
AINR 25100/-

BINR 26600/-
CINR 24300/-
DINR 22960/-
EINR 21440/-
Explanation:

A:B:C

=20000×6+12000×6:28000×6+20000×6

:36000×6+44000×6

=20000+12000:28000+20000 : 36000+44000

=20+12:28+20 : 36+44

=5+3:7+5:9+11

=8:12:20=2:3:5

Let the total profit earned be INR x.

5/10×x = 12550

x = INR 25100/-

A vessel contains a mixture of milk and water in the respective ratio of 14:3.25.5 litre of the mixture is taken out from the vessel and 2.5 litre of pure water and 5 litre of pure milk is added to the mixture. If the resultant mixture contains 20% water, what was the initial quantity of mixture in the vessel before the replacement ? (in litre)
A51
B102
C68

D85
E34
Explanation:

Let the quantity of milk in vessel = 14x

Let the quantity of water in vessel = 3x

According to question,

14x-25.5×14/17+5÷3x-25.5×3/17+2.5 =80/20

14x-21+5/3x-4.5+2.5 = 4/1

14x-16x=12x-8

2x=8

x=4

∴ quantity of mixture in vessel = 17x⇒17×4

= 68 litre

8 5.5 8.5 23 89.5 ?
A455
B420.5
C445

D415.5
E433
Explanation:

8(×1-2.5)5.5(×2-2.5)8.5(×3-2.5)23(×4-2.5)89.5(×5-2.5)

31 15 21 50 ? 767.25
A160.5
B171.5

C156.5
D122.5
E143.5

402 400 388 358 302 ?
A212

B236
C190
D182
E210

16 19 24 33 50 ?
A83

B66
C99
D74
E102

158 78 38 18 8 ?
A3

B5
C2
D7
E6
Explanation:

158(÷2-1)78(÷2-1)38(÷2-1)18(÷2-1)8(÷2-1)3

Directions (next three questions) : Study the following information to answer the given questions.

S is the sister of B. K is the brother of B.
K is the son of L. L is married to Y, Y is the daughter of J.
Y has only one daughter. J is married to C.
K is the brother D. D is married to T.

If Z is the daughter of T, then how is B related to Z
AUncle

BFather
CAunt
DCannot be determined
EMother
Explanation:

How is T related to S
ABrother-in-law
BCousin
CSister-in-law

DSister
EBrother
Explanation:

Four circle having equal radii are drawn with centres at the four corners of a square. Each circle touches the other two adjacent circles. If the remaining area of the square is 168 cm², what is the size of the radius of the circle?(in cm)
A14

B1.4
C35
D21
E3.5
Explanation:

Area of shaded region = (4-∏)r²

∴ (4-∏)r²= 168

r²= 168÷4-22/7

r²= 168/6×7

r²= 28×7

r=√196=14 cm

Directions (next five questions): What approximate value will come in place of the question mark (?) in the given questions

299÷12×13.95+?=24.02²
A285
B225

C325
D150
E185
Explanation:

299÷12×13.95+?=24.02²

300÷12×14+?=(24)²

?=576-350

?=225

820.01-21×32.99+?=240
A105
B173
C113

D234
E143
Explanation:

820-21×32+?=240

820-693+?=240

127+?=240

?=240-127=113

11.92²+16.01² = ?²×3.85²
A15
B2
C4
D5

E12
Explanation:

(11.92)²+(16.01)² = ?²×(3.85)²

(12)²+(16)²=?²×(4)2

?=√400/16

∴ ? = 5

Directions (next five questions) : Study the table to answer the given questions.

Data regarding number of candidates appearing for

Civil Services (CS) and Engineering Services (ES)

Examinations in the years 2007, 2008, 2009 and

2010 in the country XYZ

Year	Civil Services		Engineering Services
Year	Total Number of Candidates Appeared	Graduates Out of the Total Candidates Appeared (in Percentage)	Total Number of Candidates Appeared	Graduates Out of the Total Candidates Appeared (in Percentage)
2007	58	75	30	52
2008	60	60	36	50
2009	70	65	52	40
2010	76	50	40	60

(Note : Figures with regards to total number of candidates appeared are given in thousands)

What is the total number of graduates who appeared for both CS and ES together in the year 2009
A66300

B64200
C60800
D62800
E66800
Explanation:

In 2009, total number of graduates who appeared in both CS and ES together = 70×65/100+52×40/100

=45500+20800

=66300

What is the difference between the average number of candidates who appeared CS in the year 2007 and 2008 and average number of candidates who appeared for ES in the same years together
A38400
B24400
C26000

D26400
E24000
Explanation:

Required difference = [58+60/2-30+36/2] thousand

= (59-33) thousand = 26000

Total number of graduates who appeared for ES in 2008 is what percent of the total number of graduates who appeared for CS in the same year
A75
B40
C55
D60
E50

Explanation:

Required 36×50/100÷60×60/100×100 = 50%

What is the respective ratio between the number of graduates who had appeared for ES in 2010 and the number of graduates who appeared for CS in 2010
A13:21
B12:17
C12:19

D11:17
E11:19
Explanation:

Required ratio = 40×60/100÷76×50/100 = 12/19 = 12:19

Total number of candidates who appeared for CS and ES together in 2011 was 25% more than the total number of candidates who appeared for the same together in 2010. How many female candidates appeared for both the exams together in 2011, if they formed 2/5th of the total number of candidates appearing for both CS and ES that year
A52000
B58000

C60000
D62000
E64000
Explanation:

In 2011, total number of candidates who appeared for CS and ES together

=(76+40)×125/100 = 145 (in thousands)

Hence, total female candidates = 145×2/5

=58000

Abhay gave 30% of his money to Vijay, Vijay gave 2/3rd of what he received to his mother. Vijay's mother gave 5/8th of the money she received from Vijay to the grocer. Vijay's mother is now left with INR 600/-. How much money did Abhay have initially
AINR 6200/-
BINR 8000/-

CINR 6000/-
DINR 8200/-
EINR 10200/-
Explanation:

Let the total money of Abhay be INR x.

Vijay received = x×30/100

Vijay's mother received = x×30/100×2/3 = x/5

Grocer received = x/5×5/8 = x/8

According to question,

x/5×3/8 = 600

x = INR 8000/-

A rectangular plot 55 m long and 45 m broad, has two concrete crossroads (of equal width) running in the middle of it. One parallel to the length and the other parallel to the breadth. The rest of the plot is used as lawn. If the area of the lawn is 1911 m², what is the width of each of the crossroads? (in m)
A6

B8
C10
D2
E11
Explanation:

Let the width of path be x metre.

According to question,

55×45-[55×x+45×x-x²] = 1911

2475-100x+x²=1911

x²-100x+564=0

x²-94x-6x+564=0

x(x-94)-6(x-94)=0

x=6

x≠94

Hence, the width of path = 6m

A bag contains 4 red balls, 6 green balls and 5 blue balls. If three balls are picked at random, what is the probability that two of them are green and one of them is blue in colour
A20/91
B1091
C15/91

D5/91
E25/91
Explanation:

Required probability = ⁶C₂×⁵C₁/¹⁵C₃

6!/2!.4!×5!/1!.4!÷15!/3!.12!

=15/91

In a 90 litres mixture of milk and water, percentage of water is only 30%. The milkman gave 18 litres of this mixture to a customer and then added 18 litres of water to the remaining mixture. What is the percentage of milk in the final mixture
A64%
B48%
C52%
D68%
E56%

Explanation:

Remaining mixture = 90-18 = 72 litre

Quantity of milk = 72×70/100

=504 litre

Quantity of water = 72-50.4

=21.6 litre

On adding 18 litre of water,

Required percentage of milk

= 50.4/90×100

=56%