Saturday, 7 March 2020

Arithmetic Ability Practice Mcq Question And Answer with Explanation Part 1

Arithmetic Ability Practice Mcq Question And Answer with Explanation 

For

Technical exam and other entrance exams of GMAT, CAT, SAT, MAT and also bank aspirants can use this Maths tips and tricks app for banking exams etc

Part-1


Based on - AGE - ALLIGATION OR MIXTURE - AREA  - AVERAGE



1 Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit's age. After further 8 years, how many times would he be of Ronit's age? 

2 times
2 1 times
2

2 3 times
4

3 times 

Answer: Option A
Explanation:

Let Ronit's present age be x years.
Then, father's present age
       =(x + 3x) years = 4x years.
(4x + 8) = 5 (x + 8)
2
 8x + 16 = 5x + 40
 3x = 24
 x = 8.
Hence, required ratio
= (4x + 16) = 48 = 2.
(x + 16)24

2 The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child? 

4 years 
8 years 
10 years 
None of these 

Answer: Option A
Explanation:

Let the ages of children be
x, (x + 3), (x + 6), (x + 9)
            and (x + 12) years.
Then, x + (x + 3) + (x + 6)
    + (x + 9) + (x + 12) = 50
 5x = 20
 x = 4.
 Age of the youngest child
            x = 4 years.

3 A father said to his son, "I was as old as you are at the present at the time of your birth". If the father's age is 38 years now, the son's age five years back was: 

14 years 
19 years 
33 years 
38 years 

Answer: Option A
Explanation:

Let the son's present age be x years.
Then, (38 - x) = x
 2x = 38.
 x = 19.
 Son's age 5 years back
    (19 - 5) = 14 years.

4 A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, the how old is B? 

7 
8 
6 
10 

Answer: Option D
Explanation:

Let C's age be x years.
Then, B's age = 2x years.
A's age = (2x + 2) years.
 (2x + 2) + 2x + x = 27
 5x = 25
 x = 5.
Hence,
   B's age = 2x = 10 years.

5 Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand's present age in years? 

24 
27 
40 
Cannot be determined 

Answer: Option A
Explanation:

Let the present ages of Sameer and Anand be 5x years and 4x years respectively.
Then, 5x + 3 = 11
4x + 39
 9(5x + 3) = 11(4x + 3)
 45x + 27 = 44x + 33
 45x - 44x = 33 - 27
 x = 6.
 Anand's present age
        4x = 24 years.

6 A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of his son is: 

14 years 
18 years 
20 years 
22 years 

Answer: Option D
Explanation:

Let the son's present age be x years.
Then, man's present age
       = (x + 24) years.
 (x + 24) + 2 = 2(x + 2)
 x + 26 = 2x + 4
 x = 22.

7 Six years ago, the ratio of the ages of Kunal and Sagar was 6 : 5. Four years hence, the ratio of their ages will be 11 : 10. What is Sagar's age at present? 

16 years 
18 years 
20 years 
Cannot be determined 

Answer: Option A
Explanation:

Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.
Then, (6x + 6) + 4 = 11
(5x + 6) + 410
 10(6x + 10) = 11(5x + 10)
 5x = 10
 x = 2.
 Sagar's present age
       = (5x + 6) = 16 years.

8 The sum of the present ages of a father and his son is 60 years. Six years ago, father's age was five times the age of the son. After 6 years, son's age will be: 

12 years 
14 years 
18 years 
20 years 

Answer: Option D
Explanation:

Let the present ages of son and father be x and (60 -x) years respectively.
Then, (60 - x) - 6 = 5 (x - 6)
 54 - x = 5x - 30
 6x = 84
 x = 14.
 Son's age after 6 years
      = (x+ 6) = 20 years.

9 At present, the ratio between the ages of Arun and Deepak is 4 : 3. After 6 years, Arun's age will be 26 years. What is the age of Deepak at present ? 

12 years 
15 years 
19 and half 
21 years 

Answer: Option B
Explanation:

Let the present ages of Arun and Deepak be 4x years and 3x years respectively. Then,
4x + 6 = 26        4x = 20
x = 5.
 Deepak's age
         3x = 15 years.

10 Sachin is younger than Rahul by 7 years. If the ages of Sachin and Rahul are in the respective ratio of 7 : 9, how old is Sachin? 

16 years 
18 years 
28 years 
24.5 years 

Answer: Option D
Explanation:

Let Rahul's age be x years.
Then, Sachin's age
      = (x - 7) years.
x - 7 = 7
x9
 9x - 63 = 7x
 2x = 63
 x = 31.5
Hence, Sachin's age
        = (x - 7) = 24.5 years.

11 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). 

8, 20, 28 
16, 28, 36 
20, 35, 45 
None of these 

Answer: Option B
Explanation:

Let their present ages be 4x, 7x and 9x years respectively.
Then, (4x - 8) + (7x - 8) + (9x - 8) = 56
 20x = 80
 x = 4.
 Their present ages are
4x = 16 years,
7x = 28 years and
9x = 36 years respectively.

12 A person's present age is two-fifth of the age of his mother. After 8 years, he will be one-half of the age of his mother. How old is the mother at present? 

32 years 
36 years 
40 years 
48 years 

Answer: Option C
Explanation:

Let the mother's present age be x years.
Then, the person's present age
= 2xyears.
5
2x + 8 = 1(x + 8)
52
 2(2x + 40) = 5(x + 8)
 x = 40.

13 Q is as much younger than R as he is older than T. If the sum of the ages of R and T is 50 years, what is definitely the difference between R and Q's age? 

1 year 
2 years 
25 years 
Data inadequate 

Answer: Option D
Explanation:
Given that:

1. The difference of age b/w R and Q = The difference of age b/w Q and T.

2. Sum of age of R and T is 50
     i.e. (R + T) = 50.

Question: R - Q = ?.

Explanation:

R - Q = Q - T

(R + T) = 2Q

Now given that, (R + T) = 50

So, 50 = 2Q and therefore Q = 25.

Question is (R - Q) = ?

Here we know the value(age) of Q (25), but we don't know the age of R.

Therefore, (R-Q) cannot be determined.


14 The age of father 10 years ago was thrice the age of his son. Ten years hence, father's age will be twice that of his son. The ratio of their present ages is: 

5 : 2 
7 : 3 
9 : 2 
13 : 4 

Answer: Option B
Explanation:

Let the ages of father and son 10 years ago be 3x and x years respectively.
Then, (3x + 10) + 10
          = 2[(x + 10) + 10]
 3x + 20 = 2x + 40
 x = 20.
 Required ratio
       = (3x + 10) : (x + 10)
       = 70 : 30 = 7 : 3.

15 A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

1
3

1
4

1
5

1
7


Answer: Option C
Explanation:

Suppose the vessel initially contains 8 litres of liquid.
Let x litres of this liquid be replaced with water.
Quantity of water in new mixture
=3 - 3x + xlitres
8
Quantity of syrup in new mixture
=5 -  5xlitres
8
3 - 3x + x=5 - 5x
88
 5x + 24 = 40 - 5x
 10x = 16
 x = 8.
5
So, part of the mixture replaced
=8 x 1 = 1.
585

16 Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, the price of the third variety per kg will be: 

Rs. 169.50 
Rs. 170 
Rs. 175.50 
Rs. 180 

Answer: Option C
Explanation:

Since first and second varieties are mixed in equal proportions.

So, their average price
= Rs.126 + 135= Rs. 130.50
2
So, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. x per kg in the ratio 2 : 2,
i.e., 1 : 1. We have to find x.
By the rule of alligation, we have:
Cost of 1 kg
of 1st kind
Rs. 130.50Cost of 1 kg
tea of 2nd kind
Rs. x
Mean Price
Rs. 153
(x-153)22.50
x - 153 = 1
22.50
 x - 153 = 22.50
 x = 175.50

17 A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially? 

10 
20 
21 
25

Answer: Option C
Explanation:

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.
Quantity of A in mixture left
=7x - 7x 9litres =7x - 21 litres.
124
Quantity of B in mixture left
=5x - 5x 9litres =5x - 15 litres.
124
7x - 21
4
 = 7
5x -15 + 9
4
9
28x - 21 = 7
20x + 219
 252x - 189 = 140x + 147
 112x = 336
 x = 3.
So, the can contained 21 litres of A.

18 A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5? 

4 litres, 8 litres 
6 litres, 6 litres 
5 litres, 7 litres 
7 litres, 5 litres 

Answer: Option B
Explanation:

Let the cost of 1 litre milk be Re. 1
Milk in 1 litre mix. in 1st can =3 litre,
4
C.P. of 1 litre mix. in 1st can Re.3
4
Milk in 1 litre mix. in 2nd can =1 litre,
2
C.P. of 1 litre mix. in 2nd can Re.1
2
Milk in 1 litre of final mix. =5 litre,
8
Mean price = Re.5
8
By the rule of alligation, we have:
C.P. of 1 litre
mixture
in 1st canC.P. of 1 litre
mixture
in 2nd can
3
4
Mean Price
5
8
1
2
1
8
1
8
 Ratio of two mixtures = 1 : 1 = 1 : 1.
88
So, quantity of mixture taken from
each can =1x 12= 6 litres.
2

19 In what ratio must a grocer mix two varieties of pulses costing Rs. 15 and Rs. 20 per kg respectively so as to get a mixture worth Rs. 16.50 kg? 

3 : 7 
5 : 7 
7 : 3 
7 : 5 

Answer: Option C
Explanation:

By the rule of alligation:
Cost of 1 kg
pulses of 1st kindCost of 1 kg
pulses of 2nd kind
Rs. 15Mean Price
Rs. 16.50		Rs. 20
3.501.50
 Required rate = 3.50 : 1.50 = 7 : 3.

20 A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is: 

4%
61%
4
20% 
25% 

Answer: Option C
Explanation:

Let C.P. of 1 litre milk be Re. 1
Then, S.P. of 1 litre of mixture = Re. 1, Gain = 25%.
C.P. of 1 litre mixture = Re.100x 1=4
1255
By the rule of alligation, we have:
C.P. of 1
litre of milkC.P. of 1
litre of water
Re. 1Mean Price
Re.4
5
0
4
5
1
5
 Ratio of milk to water = 4 : 1 = 4 : 1.
55
Hence, percentage of water in the
mixture =1x 100%= 20%.
5


21 How many kilogram of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per kg so that there may be a gain of 10% by selling the mixture at Rs. 9.24 per kg? 

36 kg 
42 kg 
54 kg 
63 kg 

Answer: Option D
Explanation:

S.P. of 1 kg of mixture = Rs. 9.24, Gain 10%.
 C.P. of 1 kg of mixture
= Rs.100x 9.24= Rs. 8.40
110
By the rule of allilation, we have:
C.P. of 1 kg sugar
of 1st kindCost of 1 kg sugar
of 2nd kind
Rs. 9Mean Price
Rs. 8.40		Rs. 7
1.400.60
 Ratio of quantities of 1st and 2nd kind
= 14 : 6 = 7 : 3.
Let x kg of sugar of 1st be mixed with 27 kg of 2nd kind.
Then, 7 : 3 = x : 27
 x =7 x 27= 63 kg.
3
22 A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container? 

26.34 litres 
27.36 litres 
28 litres 
29.16 litres 

Answer: Option D
Explanation:

Amount of milk left after 3 operations
=401 - 43 litres
40
   =40 x9x9x9= 29.16 litres.
101010
23 A jar full of whisky contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohol and now the percentage of alcohol was found to be 26%. The quantity of whisky replaced is:

1
3

2
3

2
5

3
5


Answer: Option B
Explanation:

By the rule of alligation, we have:
Strength of first jarStrength of 2nd jar
40%Mean
Strength
26%		19%
714
So, ratio of 1st and 2nd quantities 
            = 7 : 14 = 1 : 2
 Required quantity replaced =2
3

24 In what ratio must water be mixed with milk to gain 16(2/3)% on selling the mixture at cost price?


1 : 6 
6 : 1 
2 : 3 
4 : 3 

Answer: Option A
Explanation:

Let C.P. of 1 litre milk be Re. 1.
S.P. of 1 litre of mixture = Re.1,
Gain =50%.
3
 C.P. of 1 litre of mixture
=100 x3x 1=6
3507
By the rule of alligation, we have:
C.P. of 1 litre
of waterC.P. of 1 litre
of milk
0Mean Price
Re.6
7
Re. 1
1
7
6
7
 Ratio of water and milk =1:6= 1 : 6.
77
25 Find the ratio in which rice at Rs. 7.20 a kg be mixed with rice at Rs. 5.70 a kg to produce a mixture worth Rs. 6.30 a kg. 

1 : 3 
2 : 3 
3 : 4 
4 : 5 

Answer: Option B
Explanation:

By the rule of alligation:
Cost of 1 kg
of 1st kindCost of 1 kg
of 2nd kind
720 pMean Price
630 p		570 p
6090
 Required ratio = 60 : 90 = 2 : 3.
26 In what ratio must a grocer mix two varieties of tea worth Rs. 60 a kg and Rs. 65 a kg so that by selling the mixture at Rs. 68.20 a kg he may gain 10%? 

3 : 2 
3 : 4 
3 : 5 
4 : 5 

Answer: Option A
Explanation:

S.P. of 1 kg of the mixture = Rs. 68.20, Gain = 10%.
C.P. of 1 kg of the mixture
= Rs.100x 68.20= Rs. 62.
110
By the rule of alligation, we have:
Cost of 1 kg
tea of 1st kind.Cost of 1 kg
tea of 2nd kind.
Rs. 60Mean Price
Rs. 62		Rs. 65
32
 Required ratio = 3 : 2.
27 The cost of Type 1 rice is Rs. 15 per kg and Type 2 rice is Rs. 20 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2 : 3, then the price per kg of the mixed variety of rice is: 

Rs. 18 
Rs. 18.50 
Rs. 19 
Rs. 19.50 

Answer: Option A
Explanation:

Let the price of the mixed variety be Rs. x per kg.
By rule of alligation, we have:
Cost of 1 kg
of Type 1 riceCost of 1 kg
of Type 2 rice
Rs. 15Mean Price
Rs. x		Rs. 20
(20-x)(x - 15)
(20 - x)=2
(x - 15)3
 60 - 3x = 2x - 30
 5x = 90
 x = 18.
28 8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of water is 16 : 65. How much wine did the cask hold originally? 

18 litres 
24 litres 
32 litres 
42 litres

Answer: Option B
Explanation:

Let the quantity of the wine in the cask originally be x litres.
Then, quantity of wine left in cask after 4 operations
=x1 - 84 litres.
x
x(1 - (8/x))4=16
x81
1 -84=24
x3
x - 8=2
x3
 3x - 24 = 2x
 x = 24.
29 A merchant has 1000 kg of sugar, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The quantity sold at 18% profit is: 

400 kg 
560 kg 
600 kg 
640 kg

Answer: Option C
Explanation:

By the rule of alligation, we have:
Profit on 1st partProfit on 2nd part
8%Mean Profit
14%		18%
46
Ration of 1st and 2nd parts = 4 : 6 = 2 : 3
 Quantity of 2nd kind
=3x 1000kg= 600 kg.
5
30 The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is: 

15360 
153600 
30720 
307200

Answer: Option B
Explanation:

Perimeter = Distance covered in 8 min
=12000x 8m = 1600 m.
60
Let length = 3x metres
       and breadth = 2x metres.
Then, 2(3x + 2x) = 1600 
             or x = 160.
 Length = 480 m 
        and Breadth = 320 m.
 Area = (480 x 320) m2
              = 153600 m2.
31 An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is: 

2% 
2.02% 
4% 
4.04% 


Answer: Option D
Explanation:

100 cm is read as 102 cm.
 A1 = (100 x 100) cm2 and 
      A2 (102 x 102) cm2.
(A2 - A1) = [(102)2 - (100)2]
= (102 + 100) x (102 - 100)
= 404 cm2.
 Percentage error
=404x 100%= 4.04%
100 x 100
32 The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle? 

16 cm 
18 cm 
24 cm 
Data inadequate 

Answer: Option B
Explanation:

2(l + b)=5
b1
 2l + 2b = 5b
 3b = 2l
b =2l
3
Then, Area = 216 cm2
 l x b = 216
l x2l= 216
3
 l2 = 324
 l = 18 cm.
33 The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is: 

40% 
42% 
44% 
46% 

Answer: Option C
Explanation:

Let original length = x metres and 
      original breadth = y metres.
Original area = (xy) m2.
New length =120xm=6xm.
1005
New breadth =120ym=6ym.
1005
New Area =6x x6ym2=36xym2.
5525
The difference between the original area = xy and new area 36/25 xy is
= (36/25)xy - xy
= xy(36/25 - 1)
= xy(11/25) or (11/25)xy
 Increase %
=11xy x1x 100%= 44%.
25xy
34 A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road? 

2.91 m 
3 m 
5.82 m 
None of these

Answer: Option B
Explanation:

Area of the park 
    = (60 x 40) m2 = 2400 m2.
Area of the lawn = 2109 m2.
 Area of the crossroads 
    = (2400 - 2109) m2 = 291 m2.
Let the width of the road be x metres. Then,
60x + 40x - x2 = 291
 x2 - 100x + 291 = 0
 (x - 97)(x - 3) = 0
 x = 3.
35 The diagonal of the floor of a rectangular closet is 7  feet. The shorter side of the closet is 4  feet. What is the area of the closet in square feet? 

26 
25 
27 
37 

Answer: Option C
Explanation:

Other side=
152-92
22
ft
=
225-81
44
ft
=
144
4
ft
=6 ft.
 Area of closet 
      = (6 x 4.5) sq. ft = 27 sq. ft.
36 A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is: 

10% 
10.08% 
20% 
28% 

Answer: Option D
Explanation:

Let original length = x and 
      original breadth = y.
Decrease in area
= xy -80xx90y
100100
=xy -18xy
25
=7xy.
25
 Decrease %
=7xy x1x 100%= 28%.
25xy
37 A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges? 

20 
24 
30 
33 

Answer: Option C
Explanation:

Let the side of the square(ABCD) be x metres.



Then, AB + BC = 2x metres.
AC = 2x = (1.41x) m.
Saving on 2x metres = (0.59x) m.
Saving %
=0.59xx 100%
2x
        = 30% (approx.) 
38 The diagonal of a rectangle is 41 cm and its area is 20 sq. cm. The perimeter of the rectangle must be: 

9 cm 
18 cm 
20 cm 
41 cm 

Answer: Option B
Explanation:

l2 + b2 = 41.
Also, lb = 20.
(l + b)2 = (l2 + b2) + 2lb 

            = 41 + 40 = 81
 (l + b) = 9.
 Perimeter = 2(l + b) = 18 cm.
39 What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad? 

814 
820 
840 
844 

Answer: Option A
Explanation:

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.
Area of each tile 
       = (41 x 41) cm2.Required number of tiles
=1517 x 902= 814.
41 x 41
40 The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is: 

1520 m2 
2420 m2 
2480 m2 
2520 m2 

Answer: Option D
Explanation:

We have: (l - b) = 23 and 
2(l + b) = 206 or (l + b) = 103.
Solving the two equations, 
we get: l = 63 and b = 40.
 Area = (l x b) 
      = (63 x 40) m2 = 2520 m2.
41 The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area? 

25% increase 
50% increase 
50% decrease 
75% decrease 

Answer: Option B
Explanation:

Let original length = x and 
      original breadth = y.
Original area = xy.
New length =x.
2
New breadth = 3y.
New area
=xx 3y=3xy.
22
Increase %
=1xy x1x 100%
2xy
        = 50%. 
42 The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ 26.50 per metre is Rs. 5300, what is the length of the plot in metres? 

40 
50 
120 
None of these 

Answer: Option E
Explanation:

Let breadth = x metres.
Then, length = (x + 20) metres.
Perimeter =5300m
26.50
               = 200 m.
 2[(x + 20) + x] = 200
 2x + 20 = 100
 2x = 80
 x = 40.
Hence,
     length = x + 20 = 60 m.
43 A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required? 

34 
40 
68 
88 

Answer: Option D
Explanation:

We have: l = 20 ft and
           lb = 680 sq. ft.
So, b = 34 ft.
 Length of fencing 
     = (l + 2b) = (20 + 68) ft
     = 88 ft.
44 A tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paise per sq. m, is: 

Rs. 456 
Rs. 458 
Rs. 558 
Rs. 568 

Answer: Option C
Explanation:

Area to be plastered
= [2(l + b) x h] + (l x b)
= {[2(25 + 12) x 6] + (25 x 12)} m2
= (444 + 300) m2
= 744 m2.
Cost of plastering
= Rs.744 x75= Rs. 558.
100
45 In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs? 

6.25 
6.5 
6.75 
7 6.

Answer: Option A
Explanation:
Required run rate
=282 - (3.2 x 10)
40
=250   = 6.25
40


46 A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. What is the average age of the family?

28 4 years
7

31 5 years
7

32 1 years
7
None of these

Answer: Option B
Explanation:
Required average
=67 x 2 + 35 x 2 + 6 x 3
2 + 2 + 3
=134 + 70 + 18
7
=222
7
= 31 5years.
7

47 A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500? 

Rs. 4991 
Rs. 5991 
Rs. 6001 
Rs. 6991 

Answer: Option A
Explanation:

Total sale for 5 months 
= Rs. (6435 + 6927 + 6855 + 7230 + 6562) 
= Rs. 34009.
 Required sale 
  = Rs. [ (6500 x 6) - 34009 ]
   = Rs. (39000 - 34009)
   = Rs. 4991.

48 The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero? 

0 
1 
10 
19 

Answer: Option D
Explanation:

Average of 20 numbers = 0.
 Sum of 20 numbers (0 x 20) = 0.
It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a).
49 The average weight of 8 person's increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person? 

76 kg 
76.5 kg 
85 kg 
Data inadequate 

Answer: Option C
Explanation:

Total weight increased 
    = (8 x 2.5) kg = 20 kg.
Weight of new person 
    = (65 + 20) kg = 85 kg.
50 The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team? 

23 years 
24 years 
25 years 
None of these 


Answer: Option A
Explanation:

Let the average age of the whole team by x years.
 11x - (26 + 29) = 9(x -1)
 11x - 9x = 46
 2x = 46
 x = 23.
So, average age of the team is 23 years.



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