1) A train covers a distance in 50 min, if it runs at a speed of 48kmph on an average. The speed at which the train must run to reduce the time of journey to 40min will be.

1. Solution::

Time=50/60 hr=5/6hr

Speed=48mph

distance=S*T=48*5/6=40km

time=40/60hr=2/3hr

New speed = 40* 3/2 kmph= 60kmph

2) Vikas can cover a distance in 1hr 24min by covering 2/3 of the distance at 4 kmph and the rest at 5kmph. The total distance is?

2. Solution::

Let total distance be S

total time=1hr24min

A to T :: speed=4kmph

distance=2/3S

T to S :: speed=5km

distance=1-2/3S=1/3S

21/15 hr=2/3 S/4 + 1/3s /5

84=14/3S*3

S=84*3/14*3

= 6km

3) The average ages of three persons is 27 years. Their ages are in the proportion of 1:3:5.What is the age in years of the youngest one among them.

Sol: Let the age of three persons be x, 3x and 5x

-- > 9x/3 = 27 -- > x = 9

4) The average of 11 numbers is 50. If the average of first 6 numbers is 49 and that of last 6 is 52.Find

the 6th number.

Sol: The total sum of 11 results = 11 * 50 = 550

The total sum of first 6 results = 6 * 49 = 294

The total sum of last 6 results = 6 * 52 = 312

Sixth result = 294 + 312 – 550 = 56

5) The smallest number which when divided by 20, 25, 35, 40 leaves the remainder 6 When divided by 14, 19, 23 and 34 respectively is the difference between divisor and The corresponding remainder is 6.

: . Required number = (L.C.M of 20, 25, 35, 40) – 6

= 1400-6 = 1394

6) Sum of three even consecutive numbers is 48, and then least number is

1) 16

2) 18

3) 20

4) 14

Sol: 4) Let the numbers be 2n, 2n+2 and 2n+4

2n + (2n+2) + (2n+4) = 48

6n = 48-6 = 42, n = 7

Hence the numbers are -- > 14, 16 and 18

The least number is 14.

7) It being given that √ 15 = 3.88, the best approximation to √5/3 is

1) 0.43

2) 1.89

3) 1.29

4) 1.63

Sol: 3) x = √5/3 = √5*3/3*3 = √15 /√ 9 = √15/3 = 3.88/3 = 1.29

8) Of the two-digit numbers (those from 11 to 95, both inclusive) how many have a Second digit

greater than the first digit?

1) 37

2) 38

3) 36

4) 35

Sol: 3) 12 to 19 -- > 8

23 to 29 -- > 7

34 to 39 -- > 6

45 to 49 -- > 5

56 to 59 -- > 4

67 to 69 -- > 3

78 to 79 -- > 2

89 -- > 1

9) The Value of √24 + 3√64 + 4√28 is

Sol: 24*1/2 + 43*1/3 + 28*1/4

-- > 4 + 4 + 4 -- > 12

10) 3 . - 4/5 of 5/6 / 4 1/3 / 1/5 – ( 3/10 + 21 1/5 ) is equal to

Sol: 13/4 – 4/5 * 5/6 / 13/3 / 1/5 – ( 3/10 + 106/5 ) (use BODMASRULE)

-- > 13/4 – 4/6 / 13/3 / 1/5 – 215/10 -- > 31/12 / 13/3 * 5 – 215/10

-- > 31/12 / 65/3 – 43/2 -- > 31/12 / 130 – 129/6 -- > 31/12/1/6 = 31/12 * 6/1

-- > 31/2 = 15 1/2

10) 13 sheeps and 9 pigs were bought for Rs. 1291.85.If the average price of a sheep be Rs. 74. What

is the average price of a pig.

Sol: Average price of a sheep = Rs. 74

: . Total price of 13 sheeps

= (74*13) = Rs. 962

But, total price of 13 sheeps and 9 pigs

= Rs. 1291.85

Total price of 9 pigs

= Rs. (1291.85-962) = Rs. 329.85

Hence, average price of a pig

= (329.85/9) = Rs. 36.65

11) A batsman in his 18th innings makes a score of 150 runs and there by increasing his Average by 6.

Find his average after 18th innings.

Sol: Let the average for 17 innings is x runs

Total runs in 17 innings = 17x

Total runs in 18 innings = 17x + 150

Average of 18 innings = 17x + 150/18

: . 17x + 150/18 = x + 6 -- > x = 42

Thus, average after 18 innings = 42

12) The L.C.M of two numbers is 2310 and their H.C.F is 30. If one number is 210 the Other is

The other number

= L.C.M * H.C.F/given number

= 2310*30/210 = 330

13) The average of 50 numbers is 38. If two numbers namely 45 and 55 are discarded, The average of

remaining numbers is?

Total of 50 numbers = 50 * 38 = 1900

Average of 48 numbers = 1900-(45+55) / 48

= 1800 / 48 = 37.5

14) Divide 50 in two parts so that the sum of reciprocals is (1/12), the numbers are

1) 20,30

2) 24,36

3) 28,22

4) 36,14

Sol: 1) Let the numbers be x and y then

X+y = 50. ………(i)

1/x + 1/y = 12

1/x + 1/50-x = 1/12..From (i) y = 50 – x

-- > 50-x+x/x(50-x) = 1/12

-- > x2-50x+600 = 0

-- > (x-30) (x-20) = 0

&am

15) Five years ago the average age of a family of 3 members was 27 years. A child has Been born, due to which the average age of the family is 25 years today. What is the Present age of the child?

Sol: Average age of the family of 3 members

5 years ago = 27 years

Sum of the ages of the 3 members now

= (27 + 5) * 3 = 96 years

Average age of the family of 4 members now

= 25 years

Sum of the ages of the 4 numbers now

= 25*4 = 100 years

Age of child = 100 – 96 = 4 years

16) In a class of 20 students in an examination in Mathematics 2 students scored 100 Marks each, 3 get zero each and the average of the rest was 40. What is the average Of the whole class?

Sol: Total marks obtained by a class of 20 students

= 2 * 100 + 3 * 0 + 15 * 40

= 200 + 600 = 800

: Average marks of whole class = 800/20 = 40

17) The greatest number, which can divide 432, 534 and 398 leaving the same remainder 7 in each, is

Required number is the H.C.F of

(432-7), (534-7) and (398-7)

i.e., H.C.F. of 425, 527, 391

Required number = 17

18) The sum of two numbers is 216 and their H.C.F is 27. The numbers are?

Let the numbers be 27a and 27b

Then 27a + 27b = 216

a+b = 8

Value of co-primes a and b are (1,7) (3,5)

: . Numbers are (27*1, 27*7) = (27,189)

19) The greatest number of 4 digits which is divisible by each one of the number 12,18,21 and 28 is

Greatest number of 4 digits is 9999

L.C.M of 12, 18, 21, 28 = 252

On dividing 9999 by 252, the remainder is 171

: . Required number is (9999-171) = 9828

20) Four prime numbers are arranged in ascending order according to their magnitude. Product of first

three is 385 and the product of last three is 1001. The greatest number is.

1) 11

2) 13

3) 17

4) 19

Sol: 2) 385) 1001(2

770

231) 385(1

231 (1

154)231(1

154

77) 154(2

154

0

Hence the product of the middle terms = 77

Greatest prime number = 1001 / 77 = 13.

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